An Introduction to Algebra

T. Shaska

February 3, 2020

Contents

1 Fundamentals 31.1 Algebraic operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Congruences modulo n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Symmetries of a regular n-gon, dihedral groups . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.5 Linear groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.6 Complex numbers and groups associated to them . . . . . . . . . . . . . . . . . . . . . . . . . 351.7 The group of points in an algebraic curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2 Basic properties of groups 452.1 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.2 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.3 Cyclic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532.4 Cosets and Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3 Quotient Groups and Homomorphisms 653.1 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.2 Normal subgroups and factor groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.3 Isomorphism theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.4 Cauchy’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.5 Conjugacy classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 763.6 Cayley’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4 Groups acting on sets 834.1 Groups acting on sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.2 Some classical examples of group action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.3 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904.4 The modular group and the fundamental domain . . . . . . . . . . . . . . . . . . . . . . . . . 91

5 Sylow theorem 955.1 Groups acting on themselves by conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.2 p-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995.3 Automorphisms of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1015.4 Sylow theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1035.5 Simple groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

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6 Direct products and Abelian groups 1176.1 Direct products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1176.2 Finite Abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.3 Free groups and Finitely generated Abelian groups . . . . . . . . . . . . . . . . . . . . . . . . 1276.4 Canonical forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

7 Solvable Groups 1297.1 Normal series and the Schreier theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1297.2 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1317.3 Nilpotent Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

8 Extension and Cohomology 1398.1 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1398.2 More on automorphism groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1398.3 Semidirect Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1408.4 Cocycles and coboundaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1408.5 The second cohomology group and the Schreier theorem . . . . . . . . . . . . . . . . . . . . . 1408.6 Schur-Zassenhaus lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1408.7 Projective Representations and the Schur Multiplier . . . . . . . . . . . . . . . . . . . . . . . . 140

I Ring theory 143

9 Rings 1459.1 Introduction to rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1459.2 Ring homomorphisms and quotient rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1519.3 Ideals, nilradical, Jacobson’s radical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1549.4 Ring of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1579.5 Chinese remainder theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

10 Euclidean rings, PID’s, UFD’s 16110.1 Integral domains and fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16110.2 Euclidean domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16310.3 Principal ideal domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16610.4 Unique factorization domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

11 Polynomial rings 17311.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17311.2 Polynomials over UFD’s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18111.3 Irreducibility of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18311.4 Symmetric polynomials and discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18711.5 Formal power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

12 Local and Notherian rings 19712.1 Introduction to local rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19712.2 Introduction to Notherian rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19912.3 Hilbert’s basis theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20012.4 Hilbert’s basis theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

II Module theory 205

13 Introduction to modules 20713.1 Introduction to modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20713.2 Module homomorphisms and quotient modules . . . . . . . . . . . . . . . . . . . . . . . . . . 20913.3 Direct sums and free modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21313.4 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21713.5 Exact sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22113.6 Projective, injective, and flat modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22513.7 The Snake Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

14 Modules over a Principal Ideal Domains 23114.1 Notherian Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23114.2 Torsion modules over a PID . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23214.3 Finitely generated modules over a Principal Ideal Domain . . . . . . . . . . . . . . . . . . . . 23314.4 Endomorphisms of vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23514.5 The rational canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24114.6 The Jordan canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247

III The theory of fields 253

15 Field theory 25515.1 Introduction to fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25515.2 Field extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25815.3 Finitely generated and finite extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26115.4 Simple extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26615.5 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

16 Algebraic Closure 27516.1 Algebraic extensions revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27516.2 Splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28016.3 Normal extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28916.4 Algebraic closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29016.5 Some classical problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292

17 Galois theory 30117.1 Automorphisms of fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30117.2 Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30417.3 Galois extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30717.4 Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31117.5 Norm and trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31317.6 Cyclic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31617.7 Fundamental theorem of Galois theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31817.8 Solvable extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32317.9 Fundamental theorem of Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326

18 Computing Galois groups of polynomials 32918.1 The Galois group of a polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32918.2 Galois groups of quartics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33118.3 Galois groups of quintics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33418.4 Determining the Galois group of higher degree polynomials . . . . . . . . . . . . . . . . . . . 33818.5 Polynomials with non-real roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340

19 Abelian Extensions 34319.1 Abelian extensions and Abelian closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34319.2 Roots of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34319.3 Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34319.4 Cyclic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34619.5 Kumer extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34719.6 Artin-Schreier theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347

20 Finite Fields 34920.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34920.2 Separable extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35020.3 Constructing Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35020.4 Irreducibility of polynomials over finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35120.5 Artin-Schreier extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35120.6 The algebraic closure of a finite field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351

21 Transcendental Extensions 35321.1 Transcendental Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35321.2 Lüroth and Castelnuovo theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35321.3 Noether Normalization Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35421.4 Linearly disjoint extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35421.5 Separable and Inseparable extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354

22 Field Extensions 355

23 Norms and Traces 35723.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357

24 Solutions 361

Preface

There are many books in abstract algebra. However, I was never happy with any of them. Some of theones that I really liked were just too difficult for the student and some other ones, whose authors madea real effort to make the subject accessible to the student, spend way too much time on basic elementaryconcepts. Was there a middle road?

The majority of the undergraduate students who take algebra for the first time find the subject difficult.Why is it that this very concrete subject which started with solving equations, multiplying matrices, orstudying symmetries of a regular n-gon has been transformed in such a formal and non-intuitive approachthat students dread? For the most part we have nobody to blame but ourselves. Mathematicians likeelegant approaches, when a theory is developed with the minimum amount of definitions and theorems.But for the young minds who are trying to learn this subject, that comes at a cost. In the process algebrahas become dry and intimidating. In this book, we attempt to change that.

There is also another problem in the teaching of algebra that comes from a more pedagogical pointof view. In most of Eastern Europe, Russia, and many of the Western countries a first course in algebrawas taught to the students of mathematics at the freshmen level. This made it necessary to have someprerequisites of set theory, functions, logic, and number theory introduced at the beginning of the course.Almost all introductory books in algebra have such material.

On contrary, in US universities algebra is taught at the junior or senior level, after students usually havetaken three semesters of calculus and one semester of linear algebra, discrete mathematics, introductionto proofs, and elementary number theory. After such courses it becomes rather unnecessary to have suchbackground material introduced in an algebra course. However, most books still cover this material andinstructors spend weeks covering unnecessary topics.

It is rather shocking how little of linear algebra is used in introductory books in abstract algebra. Afterall, all students who take an algebra course have taken at least one semester of college level linear algebra.Incorporating more linear algebra will make the material more concrete, it will give many more examples,and it would relate the subject to topics that students are already familiar with.

The goal of this book is to provide an introduction of the subject for the students who have been exposedto calculus, basic logic and set theory, discrete mathematics, and one semester of linear algebra.

If all the above reasons weren’t enough motivation to undertake the task of writing a book on algebra,there something even deeper and more important. Most books of abstract algebra are disconnected to thehistory of mathematics, to how the concepts of algebra made its way into the mathematics. It is as theyhave no soul and they make you feel as the subject was developed in the last two decades. You see Abeliangroups and Abelian rings all over such books, but very few people know what the real contribution of Abelwas to the subject. There have been some efforts to bring some sparks from history into the subject, butthat has been limited to biographies of some mathematicians. Ideas are not there, and it is not easy evenfor an expert to connect the subject with its historical development.

While the history of mathematics per se is not the focus of this book, it my belief that ideas thatshaped development of mathematics should be part of the mathematics culture of a student majoringin mathematics. Such ideas not only give the student an overall view of mathematics, but also help tounderstand why things are the way they are and it avoids simply learning definitions and theorems just

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for the sake of it.

What is abstract algebra?

What is algebra and why a student should study algebra? There always been some discussion of that.There is some universal understanding among mathematicians that abstract algebra is the study of

groups, rings, fields, modules and their use in higher mathematics. This viewpoint has dominated thelast several decades and has influenced what we teach and how we teach in algebra courses. There are,of course, pros and cons to this approach. One definitely appreciates the formal approach of learning alltechnical details of groups, rings, modules, and fields before trying to understand how this is used in otherareas of mathematics such as algebraic geometry, number theory, computational algebra, etc.

However, one could argue that through problems in such areas one can motivate the study of algebra,after all that is how the subject was developed. But somehow our students learn about categories andhomological algebra, but they never learned how to standardize the equation of a quadratic surface, theresultants, or binary forms.

Is there a way to do both? To give the student the motivation and the problems that algebra is concernedwith while also accomplishing the technical expertise needed to continue further with the subject.

A short historical perspective

[?Weber] [?salmon], [?Jacobson-1], [?Jacobson-2], [?Jacobson-3], [?birkhoff], [?herstein] [?fraleigh][?DF] [?H] [?La]

[?bourbaki],[?DF],[?kleiner]

Organization of the course

It is assumed that students taking a course in abstract algebra have basic knowledge of sets and logic inthe level of [?mth-302]. Furthermore, familiarity with Linear Algebra concepts in the level of [?lin-alg] isrequired. The course is not intended to review the material that was covered in previous courses.

I find it quite often that students taking abstract algebra have no knowledge of set theory, Zorn’s lemma,complex numbers, equivalence relations, permutations, etc. The first advise I give to such students is to goback and take the classes where these concepts are covered. However, providing some minimal knowledgeof basics is necessary for the course even for the sake of establishing some notation. In order to avoidconfusion we provide two appendices which cover some very minimal mathematics which must be knownbefore taking this course.

In Appendix A we provide some basic concepts on sets, relations, mathematics induction, etc. Suchmaterial is usually covered in a mathematical proofs course which must be a prerequisite for this course.However, there is always some choice of the instructor what to cover in such a course.

An appendix is provided with some basic material on elementary number theory. We give a definitionof congruences, division algorithm, greatest common divisor, etc. Some of these facts are used throughoutthe book and sometimes students have a hard time finding the right references for them. [?L] [?lin-alg],

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Chapter 1

Fundamentals

Groups are fundamental objects in algebra. The first examples of groups appeared in the study of thesymmetries of roots of polynomial equations. The study of group theory became one of the major areas ofmathematics during the XX century, culminating with the classification of finite simple groups in 1970’s.Groups as structures appear in many branches of mathematics and applications including chemistry,physics, digital communications, cryptography, etc.

In this chapter we will introduce the basic definition of a group and describe some classical examplesof groups such as congruences, symmetries of a regular n-gon, linear groups, the group of points in theunit circle, permutations, and the group of points of algebraic curves. Each one of these examples leads tobeautiful branches of mathematics and will be used extensively throughout this book.

1.1 Algebraic operations

Let A be a set and An denote the n-th Cartesian product. An n-ary algebraic operation is called a map

ϕ : An−→ A

An n-ary operation is called associative if for every a,b,c ∈ An we have

ϕ(a,ϕ(b,c)

)= ϕ

(ϕ(a,b),c

).

If n = 2 then we say a binary operation. In this book we will focus on binary operations. Hence, a binaryoperation ? on a set G is called a function

? : G×G→ G.

For every a,b ∈ G we write a? b instead of ?(a,b). A binary operation ? is called associative if for everya,b,c ∈ G we have

a? (b? c) = (a?b)? c

Exercise 1.1. For any (x1,x2), (y1, y2) ∈R2 define

(x1,x2)(y1, y2) = (x1y2,x2y2) ∈R2.

Is this a binary operation? If so, is it associative?

Notice that not all operations are associative. Indeed, some very important operations in algebra arenot associative.

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Exercise 1.2. Let G = GL2(R) be the set of all invertible 2 by 2 matrices with entries inR; see [?lin-alg] for details.In G define the following binary operation [A,B] such that

[A,B] = ABA−1B−1 (1.1)

Is this a well defined binary operation? Is it associative.

The Jacobi identity is a relationship

[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0

between three elements A, B, and C, where [A,B] is the commutator as in Eq. (3.1). It can be generalized toany operation.

Exercise 1.3. Prove that Jacobi identity for the cross product of vectors in Rn.

1.1.1 Groups

A set G together with a binary operation, which satisfies certain properties is called a group. We make thisdefinition more precise below.

Definition 1.1. Let G be a nonempty set and ? a binary operation on it. Then, the ordered pair (G,?) is called agroup if the following hold:

1. There is an element e ∈ G such that for every g ∈ G, g? e = e? g = g.

2. For every a,b,c ∈ G, (a?b)? c = a? (b? c).

3. For every a ∈ G, there is an element b ∈ G such that a?b = b? a = e.

The element e, sometimes denoted by eG, is called identity of the group G. For every a ∈G, the element b of3) is called inverse of a and is denoted by a−1. A group G is called Abelian if When the defined operationin a group is the usual addition "+ ", the group is called an additive group. The identity of this group isdenoted by e = 0 and the inverse of an element a is denoted by −a. We usually call it the opposite of a.

Example 1.1. The following (R,+), (Q,+), where + and × is the usual operation of addition, are Abelian groups.

Example 1.2. The pair (Z,+) is an Abelian group, but (Z∗,×) is not a group because there is no multiplicativeinverse for an integer.

The cardinality of G is called the order of the group G and denoted by |G|. When the set G is finite wesay that the group has finite order, otherwise we say that the group has infinite order.

Lemma 1.1. Let G be a group. Then, the following are true:

a) the identity is unique,

b) for every element a ∈ G, the inverse a−1 is unique.

Proof. a) Assume that there exist e, e′ such that ae = ea = a and ae′ = e′a = a for every a ∈ G. Then, we haveee′ = e′ which implies that e′ = e.

b) Let a ∈ G. Assume that there exist b,b′ ∈ G such that ab = ba = e and ab′ = b′a = e. Then,

b′ = b′e = b′(ab) = (b′a)b = eb = b.

Thus, b′ = b.

From now on, we will refer to the element e as the identity element of G and a−1 as the inverse of a.

4

Shaska T. 1

Lemma 1.2. In a group G the cancellation property from the left and from the right holds true. Thus, ba = ca impliesb = c and ab = ac implies b = c.

Proof. Assume that ba = ca. Let a−1 be the inverse of a. Then, multiplying from the right with a−1 we have

(ba)a−1 = (ca)a−1,

and from the associativity we have b (aa−1) = c (aa−1). Then, be = ce and thus b = c. Similarly, multiplyingfrom the left with a−1, we prove that the equality ab = ac implies b = c.

Corollary 1.1. For every a ∈ G we have (a−1)−1 = a.

Proof. Assume that there is an element g ∈ G such that a−1g = ga−1 = e and this element is unique. Then,g = a and (a−1)−1 = a.

We use the exponential symbols for multiplicative groups. If G is group and g ∈ G, then define g0 = e.For n ∈N, define

gn = g · g · · · g︸ ︷︷ ︸n− times

andg−n = g−1

· g−1· · · g−1︸ ︷︷ ︸

n− times

.

In a group, the usual power laws are true.

Lemma 1.3. For every g,h ∈ G we have

i) gmgn = gm+n, for every m,n ∈Z

ii) (gm)n = gmn, for every m,n ∈Z

iii) (gh)n = (h−1g−1)−n, for every n ∈Z.

Moreover, if G is Abelian, then (gh)n = gn hn.

We leave the proof of the Lemma as an exercise. Notice that in general (gh)n , gn·hn, since the group is not

necessarily Abelian.

Lemma 1.4. Suppose that a finite set G is closed under an associative multiplication and both cancelation propertieshold. Prove that G must be a group.

Proof. Exercise.

Definition 1.2. Let G a group. The order of an element g ∈ G is called the smallest positive integer n (if it exists)such that gn = e. If such number does not exist we say that the element g has infinite order.

The order of an element is denoted by |g|. When the operation of the group is addition then the orderof an element g is the smallest positive integer such that n g = 0.

Thus, to find the order of an element g it is enough to find the sequence of powers g, g2, g3, · · · until wearrive at the identity. If we never arrive at the identity we say that element g has infinite order.

Example 1.3. Find the order of the elements of the group (Z,+).

Solution: Since for every m from Z (not zero) and for every n from N we have that m ·n is different from zero, we

get that the order of every element from (Z,+) is infinity.

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1 Shaska T.

The Klein 4-group; Viergrouppe.

Next, we will see a classical example of a group.

C

D

A

B

Figure 1.1: Symmetries of the rhom-bus

Consider the symmetries of a rhombus. Then we have the identity e,the reflection α around the vertical axis, and the reflection β aroundthe horizontal axis. Let us see what happens to the rhombus underthese symmetries:

e : (A), (B), (C), (D)α : B→D, D→ B, (A), (C)β : A→ C, C→ A, (B), (D)

We also have the symmetry which is obtained by rotation first aroundthe vertical axis and then around the horizontal axis,

αβ : A→ C, C→ A, B→D, D→ B

Verify that all the symmetries of the rhombus are obtained by any ofthese movements. Then,

α1 = e, β2 = e, (αβ)2 = e.

The set V = e,α,β,αβ forms a group under the composition of symmetries. The above group is called theKlein 4-group or Viergrouppe and denoted by V4. Every element has order 2 other than the identity e. Itwas first discovered by Felix Klein in 1884.

It is sometimes convenient to use tables to present the mul-tiplication of elements in a given group. Such tables were firstused by Cayley and are called Cayley’s tables.

In the case of the Klein 4-group the multiplication can berepresented by the following Cayley’s table in Table 1.1. In thistable, the product xiy j of any two elements gives the element inthe position (i, j). The reader can prove that this multiplicationin the above set forms a group. Cayley’s table can be constructedfor any finite group.

∗ e α β γ

e e α β γα α e γ ββ β γ e αγ γ β α e

Table 1.1: Cayley’s table for V4

The idea of the symmetries of an n-gon will be explored further in Section 1.3.

Exercises:

1.1. If x, y ∈ G, denote yxy−1 by xy and [x, y] = xyx−1y−1. If x, y,z ∈ G, prove that

[x, yz] = [x, y][x,z]y and [xy,z] = [y,z]x [x,z]

1.2 (Jacobi identity). If x, y,z ∈ G, denote [x, [y,z]] by [x, y,z]. Prove the Jacobi identity

[x, y−1,z]y [y,z−1,x]z [z,x−1, y]x = eG

1.1.2 Rings

One of the very first things that every human learns in his/her education is the set of integers Z andtwo basic operations; addition and multiplication. This triple (Z,+,?) is basically our playground fromkindergarten until very late in life. This is the first example of a ring even though nobody ever told us thatin elementary school.

The set R with two algebraic operations (R,+,?) (addition and multiplication) that satisfies the followingconditions is called a ring.

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1. (R,+) is Abelian group

2. the multiplication is associative with respect to addition. In other words,

(a? b)? c = a? (b? c), ∀a,b,c ∈ R

3. It is true the distributive property. In other words, for every a,b,c ∈ R,

(a + b)? c = a? c + b? ca? (b + c) = a?b + a? c

A ring R in which multiplication is commutative is called commutative ring or Abelian ring. The ringR has identity when there exists the element eR ∈ R such that

∀a ∈ R, a? eR = a.

The identity of the group (R,+) is called the zero of the ring and denoted by 0R or simply 0. The symbol R×

will always denote R× =R \ 0R.

Exercise 1.4. Let (Z,+,?) be the set of integers with the usual addition and multiplication. Prove that (Z,+,?) isa ring.

Exercise 1.5. Let Mat2(Z) be the set 2 by 2 matrices with coefficients in Z. Is Mat2(Z) together with addition andmultiplication of matrices a ring? Justify your answer.

Exercise 1.6. Consider the sets Z, Q, R, C together with the usual addition and scalar multiplication. Are theyrings? Justify your answer.

Let us consider another example from linear algebra. In [?lin-alg] we have shown that Mat2(R)together with addition of matrices and multiplication by scalars is a vector space.

Exercise 1.7. Is Mat2(Z) together with addition of matrices and multiplication by scalars a vector space?

1.1.3 Fields

A ring R with identity (eR , 0) in which every element a ∈ R\ 0 has inverse with multiplication is called adivision ring. An Abelian ring which is also a division ring is called a field .

A field is a triple (F,+,?) such that (F,+) is an Abelian group and (F×,?) is also an Abelian group (F× isis F\ 0).

Exercise 1.8. Let (Q,+,?) be the set of integers with the usual addition and multiplication. Prove that (Q,+,?) is afield.

Exercise 1.9. Let Mat2(Q) be the set 2 by 2 matrices with coefficients in Z. Is Mat2(Q) together with addition andmultiplication of matrices a field? Justify your answer. What about Mat2(R)?

Exercise 1.10. Consider the sets Z, Q, R, C together with the usual addition and scalar multiplication. Are theyfields? Justify your answer.

Exercise 1.11. Let k be one of the followingQ,R,C. Is Mat2(k) together with addition of matrices and multiplicationby scalars a vector space? Justify your answers.

We will denote a field of q elements by Fq. In ?? we will see that q must be a power of a prime. Fieldsare studied in detail in the last part of this book.

Exercises:

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1 Shaska T.

1.3. Prove that R× =R \ 0 and Q× =Q\ 0 form groups with multiplication of numbers.

1.4. Let (V,+, ·) be a vector space with scalars from R; see [?lin-alg, Chap. 2]. Prove that (V,+) is an Abeliangroup.

1.5. Let the groups (R×, ·) and (Z,+) be given and denote with G the set G =R××Z. Define the operation in Gsuch that

(a,m) (b,n) = (ab,m + n).

Prove that (G,) is a group.

1.6. Let G be given as follows,G = x ∈R |x > 0 and x , 1.

Define the binary operation ∗ on the set G as

a ∗b = alnb for every a,b ∈ G.

Prove that (G,∗) is an Abelian group.

1.7. Let be given the set S =R \ −1. Define in S the operation ∗ such that

a ∗b = a + b + ab.

Prove that (S,∗) is an Abelian group.

1.8. Let be given u = (u1,u2) ∈ R2. Denote with C(u,r) the circle in R2 with center u = (u1,u2) and radius r > 0.Denote with S the set of all circles in R2 with r > 0,

S = all circles C(u,r) ⊂R2 such that r > 0.

Define binary operation ? : S×S→S, such that

C(u,r)?C(v,s) = C(u + v,rs),

where u + v is the sum of vectors in R2. Prove that (S,?) is a group.

1.9. Prove that a group which has every element, different from identity, of order two is an Abelian group.

1.10. If G is a group in which (a ·b)i = ai·bi for three consecutive integers i ∈Z for all a,b ∈G, show that G is Abelian.

1.11. Show that the conclusion of the problem above is not true if we assume the relation (a · b)i = ai· bi for two

consecutive integers.

1.12. Prove that every group with 3, 4, 5 elements is Abelian.

1.13. Show that if every element of the group G is its own inverse, then G is Abelian.

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1.2 Congruences modulo n

Next we will see some very basic groups from the set of integers Z. For a fixed n ∈ Z, we define thefollowing relation

∀x, y ∈Z, x ∼ y ⇐⇒ n | (x− y).

This relation is called the congruence modulo n.

Exercise 1.12. The reader must verify that this is an equivalence relation.

Denote the equivalence class of x ∈Z, under the congruence modulo n, by the symbol [x].

Lemma 1.5. The following are true:

i) This equivalence relation has n distinct equivalence classes.

ii) If a ≡ b mod n and c ≡ d mod n then

a± c ≡ (b±d) mod nac ≡ bc mod n

(1.2)

iii) If ab ≡ ac mod n and (a,n) = 1, thenb ≡ c mod n.

iv) If a ≡ b mod m and d | m, then a ≡ b mod d.

v) If (m,n) = 1, a ≡ b mod m , and a ≡ b mod n, then

a ≡ b mod mn.

Proof. i) Let a ∈Z. Then, [a] contains all integers x ∈Z such that n |a−x. In other words, a−x = kn, for somek ∈Z. Hence, a = x + kn for some k ∈Z. Then, we have the following characterization of the equivalenceclass of a;

[a] = a + kn |k ∈Z = a,a±n,a±2n,a±3n, · · · , .

From the Euclidean Algorithm we can write

a = rn + y, for some r, y ∈Z, 0 ≤ y < a,

where y is the remainder of the division by n. Hence, y∼ a and 0≤ y< a. We usually pick y as a representativeof the equivalence class [a]. There are n−1 such remainders and therefore n−1 equivalence classes, namely

[0], [1], . . . , [n−1].

ii) Suppose a≡ b mod n, i.e., a−b is divisible by n. Then a = b+sn for some integer s. Similarly, c≡ d mod nmeans c = d + tn for some integer t. Then a + c = (b + d) + (s + t)n so that a + c ≡ b + d mod n, which shows thatthe sum of residue classes is independent of the representatives chosen.

Similarly, ac = (b + sn)(d + tn) = bd + (bt + ds + stn)n shows that ac ≡ bd mod n and so the product of theresidue classes is also independent of the representatives chosen.

iii) Suppose ab ≡ ac mod n, i.e., ab− ac is divisible by n. Then, ab = ac+ rn for some integer r. So, we haveab− ac = rn⇒ a(b− c) = rn. Since a is relatively prime to n we get that b− c is divisible by n, which showsthat b ≡ c mod n.

We leave parts iv) and v) as exercises. As described above, the equivalence classes for the congruence modulo n are:

[0], [1], . . . , [n−1].

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1 Shaska T.

Denote by Zn or Z/nZ the set of equivalence classes

Zn := [0], [1], . . . , [n−1] .

In this set, we define the addition modulo n, denoted by mod n, as follows

[a] + [b] = [a + b]

and the multiplication modulo n, denoted by mod n, as

[a] · [b] = [a ·b]

Exercise 1.13. Prove that the addition modulo n and the multiplication modulo n are well defined binary operations.

Lemma 1.6. The following properties hold:

1. [a] + [b] = [b] + [a]

2. [a][b] = [b][a]

3. ([a] + [b]) + [c] = [a] + ([b] + [c])

4. ([a][b]) [c] = [a] ([b][c])

5. [a] ([b] + [c]) = [a][b] + [a][c]

6. [0] + [a] = [a]

7. [1][a] = [a]

Proof. Exercise.

Exercise 1.14 (The group of integers modulo n). Verify that the set Zn, for n ≥ 1, forms a group under additionmod n.

This group is called the group of integers mod n and denoted by (Zn,+) or simply Zn. Usually thesymbol Z/nZ is also used for this group.

1.2.1 The group of units of Z/nZ

Let’s denote by U (n) the set of all nonzero elements of Zn which have a multiplicative inverse. UsingLemma 1.4, prove the following:

Lemma 1.7. U (n) together with multiplication modulo n, forms a group.

From now on we will refer to U (n) as the the group of units of Z/nZ. Next are given a couple ofelementary examples of this group.

Example 1.4. Consider U (8). Then this is a group with multiplication modulo 8. Its elements are

U (8) = [1], [3], [5], [7]

Its multiplication is given by Table 1.2, which is Cayley’s table for U (8).

So the following is a natural question.

Question 1.1. for what n the set Zn together with multiplication modulo n is a group?

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Shaska T. 1

· [1] [3] [5] [7][1] [1] [3] [5] [7][3] [3] [1] [7] [5][5] [5] [7] [1] [3][7] [7] [5] [3] [1]

Table 1.2: Multiplication table for U (8)

For example, [2] ∈Z6, but [2] does not have a multiplicative inverse. We check the product of [2] withall other elements

[0] · [2] = [0], [1] · [2] = [2],[2] · [2] = [4], [3] · [2] = [0],[4] · [2] = [2], [5] · [2] = [4].

Remark 1.1. It is very important, at this point, that we warn the reader about the use of notation for the elementsof Zn or U (n). An element of Zn or U (n) is an equivalence class. Therefore, it must be denoted by the symbol [x].However, many times in literature the brackets are dropped and simply the symbol x is used.

Let n be a positive integer and [x] ∈Zn. What are the necessary and sufficient conditions that [x] has amultiplicative inverse?

Lemma 1.8. Every nonzero element [k] has inverse in Zn if and only if k is relatively prime with n.

Proof. Suppose [k] ∈ Zn has a multiplicative inverse. This means that there exists [x] ∈ Zn such that[k] · [x] = [1]. Hence, kx ≡ 1 mod n which implies that n | kx−1. In other words, there exists a ∈Z such thatna = kx−1. Thus, kx−na = 1. Since −a ∈Z, then we have kx + n(−a) = 1. Therefore, (k,n) = 1.

Conversely, suppose that k and n are relatively prime. This implies that exist a,b ∈Z such that 1 = ak+bn.Then, ka−1 = n(−b). Hence, n | ka−1 which implies that ak ≡ 1 mod n. Therefore, [a] is the multiplicativeinverse of [k].

Exercise 1.15. a) Using the result of Lemma 1.4, prove that the nonzero integers modulo p, p a prime number, forma group under multiplication mod p.

b) Do part a) for the nonzero integers relatively prime to n under multiplication mod n.

Example 1.5. The set Z/nZ together with addition and multiplication modulo n is a ring. The reader should checkthat all the properties in the definition are satisfied. Is it a field?

Example 1.6. Find the orders of elements of the group U (15).

Solution: The set U (15) has elements

U (15) = 1,2,4,7,8,11,13,14.

First we find the order of 7. Hence,

72 = 4 mod 1573 = 13 mod 1574 = 1 mod 15.

Thus, the order of 7 is 4. Similarly for 11 we have

111 = 11 mod 15112 = 1 mod 15.

Thus, |11| = 2. Similarly we prove that |1| = 1, |2| = 4, |4| = 2, |8| = 4, |13| = 4 and |14| = 2.

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1 Shaska T.

Exercise 1.16. Let p be a prime. Prove that (Z/pZ, + mod n, ? mod n) is a field.

The field above will be denoted by the symbol Fp. Every field of p elements is algebraically the same asFp; we will make this precise when we talk about homomorphisms. Hence, from now on we will say thefinite field of p elements.

A natural question is; are there finite fields with the number of elements not a prime number? In ?? wewill show that every finite field has pn elements, where p is a prime.

Exercise 1.17. Let Zn = 0,1, . . . ,n−1 as above, for n ≥ 2. Denote by

Zrp =Zn× · · ·Zn

the following setZr

n = (a1,a2, . . . ,an) | ai ∈Zn .

Define in Zrn the binary operation such that

(a1,a2, . . . ,ar) + (b1,b2, . . . ,br) = (a1 + b1,a2 + b2, . . . ,ar + br),

where ai + bi denotes the addition in Zn. Prove that Zrn forms a group with this operation.

Next we see a classical result known in the math folklore as the Chinese Remainder Theorem.

1.2.2 The Chinese remainder theorem

Suppose that we want to solve the a system of congruences of different moduli:x ≡ a1 mod n1

x ≡ a2 mod n2

· · · · · · · · ·

x ≡ ar mod nr

(1.3)

We assume that al ni’s are pairwise coprime (i.e., (ni,n j) = 1 for each i , j).

Theorem 1.1 (Chinese Remainder Theorem). There is a solution x to the above system of equations, and any twosolutions are congruent modulo N = n1 ·n2 · · ·nr.

Proof. Let N := n1 ·n2 · · ·nr and Ni := Nni

, for all i = 1, . . . ,r. For each i we have gcd (ni,Ni) = 1, since (ni,n j) = 1for each j , i. Hence, there exists Pi such that

NiPi ≡ 1 mod ni.

Let

x =

r∑i=1

aiNiPi,

where ai’s are given as in Eq. (1.3). For each i , j, ni | a jN jP j. Hence,

x ≡ aiNiPi mod ni

andx ≡ ai mod ni,

since NiPi ≡ 1 mod ni.

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Let y and z be two solutions of the system and x = y− z. By Lemma 1.5 we have

x ≡ ai mod ni, for each i = 1, . . .r.

Hence, x ≡ 0 mod N. This completes the proof.

Next, we see an example how the above theorem can be used in exercises.

Example 1.7. Solve the following system of congruencesx ≡ 1 mod 2x ≡ 2 mod 3x ≡ 1 mod 5x ≡ 4 mod 7

(1.4)

Solution: Let N := 2 ·3 ·5 ·7. Then we have the following

N1 = 105 N2 = 70 N3 = 42 N4 = 30

P1 = 1 P2 = 1 P3 = 3 P4 = 4

Hence,x = 105 + 2 ·70 + 42 ·3 + 4 ·30 ·4 = 851

Then, x ≡ 11 mod 210.

Exercises:

1.14. Find the smallest non-negative solution for the following system of congruencesx ≡ 1 mod 3x ≡ 3 mod 4x ≡ 4 mod 5x ≡ 5 mod 7

1.2.3 Fermat and Euler theorems

Definition 1.3. Let n be a positive integer. The Euler phi-function ϕ(n) is defined to be the number of non-negativeintegers x less then n which are relatively prime to n:

ϕ(n) := | 0 ≤ x ≤ n | (x,n) = 1 |

The following properties are easy to prove:

Lemma 1.9. The Euler function satisfies the following:

i) ϕ(1) = 1

ii) ϕ(p) = p−1

iii) ϕ(pα) = pα(1− 1

p

)13

1 Shaska T.

Proof. We will comment only on iii). Notice that the numbers between 0 and pα−1 which are not relativelyprime to pα are exactly the multiples of p. There are exactly pα−1 of such numbers. Hence,

ϕ(pα) = pα−pα−1 = pα(1−

1p

).

Corollary 1.2. The Euler φ-function is multiplicative. In other words, if (m,n) = 1 then

ϕ(mn) = ϕ(m) ·ϕ(n)

Proof. We want to compute ϕ(mn) which is the number of integers between 0 and mn− 1 which have nocommon factors with mn. For each 0 ≤ j ≤ mn− 1 let j1, j2 denote the residues of j mod m and mod nrespectively. So we have

j ≡ j1 mod mj ≡ j2 mod n

(1.5)

From the Chinese Remainder Theorem we have that for each pair ( j1, j2) there is only one 0 ≤ j ≤ mn− 1such that system is satisfied. j has no common factors with mn iff has no common factors with m and withn. j has no common factors with m (resp., n) iff j1 has no common factors with m (resp., n). Thus we haveϕ(m) (resp., ϕ(n)) choices for j1 (resp., j2). Hence, there are ϕ(m) ·ϕ(n) choices for ( j1, j2). This completesthe proof.

Let n be any integer which is factored as a product of powers of primes as follows:

n = pα11 · · ·p

αrr .

Then,

ϕ(n) = ϕ(pα11 ) · · ·ϕ(pαr

r ) = pα11

(1−

1p 1

)· · ·pαr

r

(1−

1p r

)= n

∏p|n

(1−

1p

)(1.6)

Theorem 1.2 (Euler’s theorem). Let a and n be positive integers, such that gcd (a,n) = 1. Then, aϕ(n)≡ 1 (mod n).

Proof. From Corollary 2.4 the order of U (n) is ϕ(n). Hence, aϕ(n) = 1 for all a ∈U (n) or aϕ(n)−1 is divisible

by n. Thus, aϕ(n)≡ 1 (mod n).

Consider the special case for the Euler’s theorem, when n = p is a prime number and recall that

ϕ(p) = p−1, we get the following result discovered by Pierre de Fermat.

Theorem 1.3 (Fermat’s Little Theorem). Let p be a prime number, a ∈Z an integer, and assume that p6 | a. Then,

ap−1≡ 1 (mod p).

Moreover, for any nonzero b ∈Z we havebp≡ b (mod p).

Proof. Since p is a prime number, then the group U (p) has order ϕ(p) = p−1. Since p6 | a, then a ∈ [x] mod p,where 1 ≤ x ≤ p− 1. Then, aϕ(p) = ap−1 = 1 mod p. This implies that ap

≡ p mod p, for all nonzero a ∈ Z.This completes the proof.

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Corollary 1.3. If a is not divisible by p and if m ≡ n mod (p−1), then

an≡ am mod p

We can actually do better then the above.

Lemma 1.10. Let a, n be integers such that (a,n) = 1 and n is factored as a product of powers of primes as follows:

n = pα11 · · ·p

αrr

and let l = lcm(ϕ(pα1

1 ), . . . ,ϕ(pαrr )

). Then, al

≡ 1 mod n.

Proof. The proof is immediate from the proof of Euler’s theorem.

Example 1.8. Let n = 105 and (a,n) = 1. Then n = 3 ·5 ·7 and

ϕ(3) = 2, ϕ(5) = 4, ϕ(7) = 6.

Hence, l = lcm(2,4,6) = 12 anda12≡ 1 mod 105.

Lemma 1.11. If (a,m) = 1 and if n′ is the smallest positive integer such that

n′ ≡ n mod ϕ(m),

thenan≡ an′ mod m

Proof. Similarly to the proof of Corollary 1.3.

Example 1.9. Compute 21000000 mod 77.

Solution: We have ϕ(77) = ϕ(7) ·ϕ(11) and lcm(10,6) = 30.

Then 230≡ 1 mod 77. Notice that

1000000 ≡ 30 ·33333 + 10

and from the previous lemma we have;21000000

≡ 210≡ 23 mod 77

We have one last formula that we will need:

Lemma 1.12. Let n be a positive integer. Then,∑

d|nϕ(d) = n

Proof. Let

f (n) :=∑d|n

ϕ(d)

Claim: f (n) is multiplicative (i.e., if (m,n) = 1 then f (mn) = f (m) f (n)).

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1 Shaska T.

Let d | mn. Then d = d1d2 such that d1 | m and d2 | n. We have gcd (d1,d2) = 1 since gcd (m,n) = 1. Hence,ϕ(d1 ·d2) = ϕ(d1) ·ϕ(d2). Thus,

f (mn) =∑d1 |m

∑d2 |n

ϕ(d1)ϕ(d2) =

∑d1 |m

ϕ(d1)

∑

d2 |n

ϕ(d2)

= f (m) f (n)

This completes the proof of the claim.

Let first prove the Lemma for n = pα.

f (pα) =∑

d | pαϕ(d) =

α∑j=0

ϕ(p j

)= 1 +

α∑j=1

p j(1−

1p

)

= 1 + (p−1)α∑

j=1

p j−1 = 1 + (p−1)(1 + p + p2 + · · ·+ pα−1

)= 1 + (pα−1) = pα

(1.7)

Now let n = pα11 · · ·p

αrr . Then

f (n) = f (pα11 ) · · · f (pαr

r ) = pα11 · · ·p

αrr = n.

This completes the proof.

Exercises:

1.15. Use Fermat’s Little Theorem to prove that if p = 4n + 3 is prime, then the equation x2≡ −1 (mod p) does not

have a solution.

1.16. Find ϕ(n) for the following n = 12,13,15,23,34,36,16,18.

1.17. Find all ϕ(n) for all n between 100 and 110.

1.18. Give an example of a group G with elements g,h ∈ G such that (gh)n , gn hn.

1.19. Let a and b be elements of the group G. Prove that abna−1 = (aba−1)n.

1.20. Prove that if G is a finite group with even order, then there is an element a ∈G different from identity and a2 = e.

1.21. Prove that if n > 2 then there is an element k ∈U (n) such that k2 = 1 and k , 1.

1.22. Let G be a group and assume that (ab)2 = a2b2 for every a and b in G. Prove that G is Abelian.

1.23. If we have that xy = x−1y−1 for every x and y in G prove that G is Abelian.

1.24. Let a and b any two elements of G. Prove that ab = ba if and only if a−1b−1 = b−1a−1.

1.25. Prove that (Zp,+) does not have proper subgroups if p is prime.

1.26. If g and h have orders respectively 15 and 16 in a group G what is the order of 〈g〉∩ 〈h〉?

1.27. Let a be an element of a group G. What is the generator of the subgroup 〈am〉∩ 〈an

〉?

1.28. Prove that Zn for n > 2 has an even number of generators.

1.29. Let G be a group and a,b ∈ G. Prove that if |a| = m and |b| = n and gcd (m,n) = 1, then we have 〈a〉∩ 〈b〉 = e.

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Pierre de Fermat (1601-1665)

Pierre de Fermat (August 17, 1601 – 12 January 1665) was a Frenchlawyer at the Parlement of Toulouse, France, and a mathematicianwho is given credit for early developments that led to infinitesimalcalculus, including his technique of adequality. In particular, he isrecognized for his discovery of an original method of finding thegreatest and the smallest ordinates of curved lines, which is analogousto that of differential calculus, then unknown, and his research intonumber theory. He made notable contributions to analytic geometry,probability, and optics. He is best known for his Fermat’s principle forlight propagation and his Fermat’s Last Theorem in number theory,which he described in a note at the margin of a copy of Diophantus’Arithmetica.

Fermat’s pioneering work in analytic geometry (Methodus addisquirendam maximam et minimam et de tangentibus linearum cur-varum) was circulated in manuscript form in 1636 (based on resultsachieved in 1629), predating the publication of Descartes’ famous Lagéométrie. This manuscript was published posthumously in 1679in Varia opera mathematica, as Ad Locos Planos et Solidos Isagoge(Introduction to Plane and Solid Loci).

In Methodus ad disquirendam maximam et minimam and in De tangentibus linearum curvarum, Fermatdeveloped a method for determining maxima, minima, and tangents to various curves that was equivalentto differential calculus. In these works, Fermat obtained a technique for finding the centers of gravity ofvarious plane and solid figures, which led to his further work in quadrature.

In number theory, Fermat studied Pell’s equation, perfect numbers, amicable numbers and what wouldlater become Fermat numbers. It was while researching perfect numbers that he discovered Fermat’s littletheorem. He invented a factorization method?Fermat’s factorization method?as well as the proof techniqueof infinite descent, which he used to prove Fermat’s right triangle theorem which includes as a corollaryFermat’s Last Theorem for the case n = 4. Fermat developed the two-square theorem, and the polygonalnumber theorem, which states that each number is a sum of three triangular numbers, four square numbers,five pentagonal numbers, and so on.

Although Fermat claimed to have proved all his arithmetic theorems, few records of his proofs havesurvived. Many mathematicians, including Gauss, doubted several of his claims, especially given thedifficulty of some of the problems and the limited mathematical methods available to Fermat. His famousLast Theorem was first discovered by his son in the margin in his father’s copy of an edition of Diophantus,and included the statement that the margin was too small to include the proof. It seems that he had notwritten to Marin Mersenne about it.

Although he carefully studied and drew inspiration from Diophantus, Fermat began a different tradition.Diophantus was content to find a single solution to his equations, even if it were an undesired fractionalone. Fermat was interested only in integer solutions to his Diophantine equations, and he looked for allpossible general solutions. He often proved that certain equations had no solution, which usually baffledhis contemporaries.

Through their correspondence in 1654, Fermat and Blaise Pascal helped lay the foundation for thetheory of probability. From this brief but productive collaboration on the problem of points, they are nowregarded as joint founders of probability theory. Fermat is credited with carrying out the first ever rigorousprobability calculation. In it, he was asked by a professional gambler why if he bet on rolling at least onesix in four throws of a die he won in the long term, whereas betting on throwing at least one double-six in24 throws of two dice resulted in his losing. Fermat showed mathematically why this was the case.

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Leonard Euler (1707-1783)

Leonhard Euler (15 April 1707 ? 18 September 1783) was aSwiss mathematician, physicist, astronomer, logician and engineerwho made important and influential discoveries in many branches ofmathematics like infinitesimal calculus and graph theory while alsomaking pioneering contributions to several branches such as topologyand analytic number theory. He also introduced much of the modernmathematical terminology and notation, particularly for mathemati-cal analysis, such as the notion of a mathematical function. He is alsoknown for his work in mechanics, fluid dynamics, optics, astronomy,and music theory.

Euler was one of the most eminent mathematicians of the 18thcentury, and is held to be one of the greatest in history. He is alsowidely considered to be the most prolific mathematician of all time.His collected works fill 60 to 80 quarto volumes, more than anybodyin the field. He spent most of his adult life in Saint Petersburg, Russia,and in Berlin, then the capital of Prussia.

A statement attributed to Pierre-Simon Laplace expresses Euler’sinfluence on mathematics: "Read Euler, read Euler, he is the master ofus all.

Euler worked in almost all areas of mathematics, such as geometry, infinitesimal calculus, trigonometry,algebra, and number theory, as well as continuum physics, lunar theory and other areas of physics. He isa seminal figure in the history of mathematics; if printed, his works, many of which are of fundamentalinterest, would occupy between 60 and 80 quarto volumes. Euler’s name is associated with a large numberof topics.

Euler is the only mathematician to have two numbers named after him: the important Euler’s numberin calculus, e, approximately equal to 2.71828, and the Euler - Mascheroni constant γ sometimes referred toas just "Euler’s constant", approximately equal to 0.57721. It is not known whether γ is rational or irrational.

Euler proved Newton’s identities, Fermat’s little theorem, Fermat’s theorem on sums of two squares, andhe made distinct contributions to Lagrange’s four-square theorem. He also invented the totient functionϕ(n), the number of positive integers less than or equal to the integer n that are coprime to n. Usingproperties of this function, he generalized Fermat’s little theorem to what is now known as Euler’s theorem.He contributed significantly to the theory of perfect numbers, which had fascinated mathematicians sinceEuclid. He proved that the relationship shown between perfect numbers and Mersenne primes earlierproved by Euclid was one-to-one, a result otherwise known as the Euclid?Euler theorem. Euler alsoconjectured the law of quadratic reciprocity. The concept is regarded as a fundamental theorem of numbertheory, and his ideas paved the way for the work of Gauss. By 1772 Euler had proved that 231

− 1 =2,147,483,647 is a Mersenne prime. It may have remained the largest known prime until 1867.

In 1735, Euler presented a solution to the problem known as the Seven Bridges of Königsberg. The cityof Königsberg, Prussia was set on the Pregel River, and included two large islands that were connected toeach other and the mainland by seven bridges. The problem is to decide whether it is possible to follow apath that crosses each bridge exactly once and returns to the starting point. It is not possible: there is noEulerian circuit. This solution is considered to be the first theorem of graph theory, specifically of planargraph theory.

Euler also discovered the formula V?E + F = 2 relating the number of vertices, edges and faces of aconvex polyhedron, and hence of a planar graph. The constant in this formula is now known as the Eulercharacteristic for the graph (or other mathematical object), and is related to the genus of the object. Thestudy and generalization of this formula, specifically by Cauchy and L’Huillier, is at the origin of topology.

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1.3 Symmetries of a regular n-gon, dihedral groups

In this part we will investigate the symmetries of a regular n-gon. We will see that a regular n-gon has2n symmetries, n of which are rotations and the other n are reflections. This set of symmetries togetherwith their composition forms a group, which is called the dihedral group. Dihedral groups are amongthe simplest examples of finite groups, and they play an important role in group theory, geometry, andchemistry.

1.3.1 Symmetries of regular polygons

Symmetries of a regular triangle

There are six motions that can bring an equilateral triangle back into its original position. They are

1. Do nothing

2. Rotate 120 degrees counterclockwise

3. Rotate 240 degrees counterclockwise

4. Flip about the symmetry axis through the upper vertex

5. Flip about the symmetry axis through the lower left-hand vertex

6. Flip about the symmetry axis through the lower right-hand vertex

We label all these six motions respectively as e,σ1,σ2,τ1,τ2,τ3.

Figure 1.2: Symmetries of the regular triangle

There are other motions but they are "equivalent" to those listed above. For example rotating the triangle360 degrees is "equivalent" to doing nothing since the basic orientation of the triangle is unchanged.

We’ve labelled the vertices A, B and C and have shown the 6 symmetry motions in Fig. 1.2. Thesymmetries of a regular triangle form a non Abelian group. This group will be denoted by S3 or D3, seeFig. 1.2. Its Cayley’s table is given in Table 1.3.

Notice that σ2 = σ21, and σ1

1 = e. Also, τ21 = e and

σ1τ1 = τ2, σ21τ1 = τ3.

Hence,D3 = e,σ1,σ

21,τ1,σ1τ1,σ

21τ1.

Another way to write this fact is that

D3 = 〈σ1,τ1 |σ31 = τ2 = e, σ1τ1σ1 = τ−1

1 〉

Before we attempt to generalize this for every n ≥ 3, let us check out the case when n = 4.

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∗ e σ1 σ2 τ1 τ2 τ3

e e σ1 σ2 τ1 τ2 τ3σ1 σ1 σ2 e τ2 τ3 τ1σ2 σ2 e σ1 τ3 τ1 τ2τ1 τ1 τ3 τ2 e σ2 σ1τ2 τ2 τ1 τ3 σ1 e σ2τ3 τ3 τ2 τ1 σ2 σ1 e

Table 1.3: Cayley’s table for the dihedral group D3

Symmetries of a square

Next we try to figure out the symmetries of the square. Let us have a square ABCD on a plane and denoteits vertices as below. We perform the following movements:

A

B C

D

a) σ : rotation with +900 around the center, clockwise, which gives

A→ B→ C→D

b) σ2 : rotation with +1800 which gives

A→ C→ A and B→D→ B

c) σ3 : rotation with +2700 which gives

A→D→ C→ B→ A

d) e : rotation with +3600 which fixes every point.

e) τ : rotation around the vertical axis which gives

A→D→ A and B→ C→ B

f) σ2τ : rotation around the horizontal axis:

A→ B→ A and D→ C→D

g) στ : rotation around the diagonal BD:

A→ C→ A,B→ B and D→D

h) σ3τ : rotation around the diagonal AC:

B→D→ B,A→ A and C→ C

Exercise 1.18. Prove that the set of these symmetries with the composition forms a group.

The above group is called the groups of symmetries of the square and denoted by D4 or sometimeswith D8.

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Figure 1.3: Symmetries of the square

Remark 1.2. Notice that operations are performed from left to right. In other words, σ2τ means that we rotate twiceby 90 degrees and then flip around the vertical axis. The orientation of the rotation does not matter as long as we areconsistent.

Then, the elements of the dihedral group D4 are

D4 = e,σ,σ2,σ3,τ,στ,σ2τ,σ3τ

Thus, all symmetries of the square are generated by the elements σ and τ. Below are all elements of D4presented visually where the vertices are labeled by numbers 1 through 4.

Exercise 1.19. Construct the Cayley’s table for the group D4.

1.3.2 Dihedral groups

We generalize the previous results for n = 3,4 in the following theorem.

Theorem 1.4. The group Dn, for n ≥ 3, contains all products of elements r and s which satisfy

rn = id, s2 = id, srs = r−1.

Proof. Possible symmetries of a regular n-gon are rotations and reflections. We have exactly n rotationswith angles

id,2πn,2 ·

2πn, . . . , (n−1) ·

2πn.

The rotation with angle 2πn we denote it with r. This rotation generates all the other ones. In other words,

rk = k ·2πn.

We label n reflections with s1,s2, . . . ,sn, where sk is the reflection that fixes the vertex k. There are twocases, depending if n is even or odd.

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If n is even there are two vertices fixed by the reflection, namely k and k + n2 . Hence, if n = 2m for some

integer m, then si = si+m for 1 ≤ i <m. Then, |sk| = 2. Let s = s1. Then s2 = id and rn = id.Since every symmetry t of a regular n-gon replaces the k-th vertex with vertex k+1 or k−1 then t = rk or

t = rks. Hence, r and s generate Dn. In other words, Dn contains all products of r and s. It is easy to checkthat srs = r−1.

If n is odd, then there is only the vertex k fixed by the reflection sk. This reflection can be obtained bysk = rk−1

· s1. Let s = s1. It can be easily checked that srs = r−1. The following is an immediate consequence of the above theorem.

Corollary 1.4. The dihedral group Dn, is a subgroup of Sn with order 2n. Moreover,

Dn = 〈r,s |rn = s2 = 1,srs = r−1〉

The elements r and s are called the generators of the group Dn and relations rn = s2 = 1 and srs = r−1 arecalled relations.

Exercises:

1.30. How many lines of symmetries have a regular n-gon?

1.31. Let σ and τ be elements of order 2 in any group G. Show that if α = στ, then ασ = σα−1.

1.32. If n is odd and n ≥ 3, prove that the only element in D2n which commutes with all elements of D2n is theidentity.

1.33. Show that the group of rigid motions of a tetrahedron in R3 has order 12.

1.34. Show that the group of rigid motions of a cube in R3 has order 24.

1.35. Show that the group of rigid motions of a octahedron in R3 has order 24.

1.36. Show that the group of rigid motions of a dodecahedron in R3 has order 60.

1.37. Show that the group of rigid motions of a icosahedron in R3 has order 60.

1.38. Can you write a computer program that would list all symmetries of a regular n-gon if you label the vertices1, . . . ,n?

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1.4 Permutations

The reader must have seen an introduction to permutations in an introductory course on discrete mathe-matics. However, no primary knowledge is assumed here. We will define permutations and permutationgroups. As the reader will hopefully realize later is that the concept of a permutation group is a mainconcept of abstract algebra. This is due to a theorem of Cayley that every group can be represented as apermutation group.

Recall that a permutation of a set X is called a bijective function f : X→ X. The set of permutationsof X, which is denoted by SX, together with composition of functions forms a group, which is called thesymmetric group of X.

Exercise 1.20. Let X be a set and SX the set of all bijections f : X→ X. Show that SX is a group together withcomposition of functions.

We will focus mainly on permutations of finite sets. We label the elements of X as 1,2,3, · · · ,n, wheren = card (X). Permutations of finite sets are given in an explicit way by listing the value of the function forevery element x ∈ X.

For example, take a permutation α over the set 1,2,3,4, which is given as follows

α(1) = 2, α(2) = 3, α(3) = 1, α(4) = 4.

Another way to express this correspondence is

α =

(1 2 3 42 3 1 4

).

Similarly the permutation β of the set 1,2,3,4,5,6 is given from

β(1) = 5, β(2) = 3, β(3) = 1, β(4) = 6, β(5) = 2, β(6) = 4

or by

β =

(1 2 3 4 5 65 3 1 6 2 4

).

Composition of permutations is found as the composition of any functions. For example, if we have:

σ =

(1 2 3 4 52 4 3 5 1

)and γ =

(1 2 3 4 55 4 1 2 3

)then

1

σ

2

3

4

5

γσ

~~

2

γ

4

3

5

1

4 2 1 3 5

Thus,

γσ = γσ =

(1 2 3 4 55 4 1 2 3

)(1 2 3 4 55 4 1 2 3

)=

(1 2 3 4 54 2 1 3 5

).

The following theorem says that Sn is a group, it is called the symmetric group in n letters.

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Theorem 1.5. The symmetric group in n letters, Sn, is a group with n! elements, where binary operation is thecomposition of functions.

Proof. The identity of Sn is the identity function id : x 7→ x. If f : Sn→ Sn is a permutation, then f−1 exists,because f is bijective. Composition of functions is associative, hence even binary operation is associative.The elements of Sn have the form:

α =

(1 2 · · · nα(1) α(2) · · · α(n)

).

It is easy to determine the order of symmetric group Sn. The element α(1) can take values from 1 to n.Once α(1) has been determined then we have n− 1 options for α(2) (since α is bijective function we haveα(1), α(2) ) and so on α(n−1) we have 2 options and to determine α(n) we have only one possibility. Hence,Sn has n(n−1)(n−2) · · ·3 ·2 ·1 = n! elements.

Symmetric groups have many subgroups. For example the group S4 has 30 subgroups and S5 has over100 subgroups. A subgroup of Sn is called a permutation group.

.

Remark 1.3. Notice that in the literature the composition above is called multiplication from the right, but bothmultiplication from the left and right are used. It is very important that we are aware if we are using multiplicationfrom the left or from the right since the results are different. However, the fact that Sn is a group and all its algebraicproperties are independent from the way we multiply.

In this book we will always use the multiplication from the right. Hence the symbol αβ always means

αβ = αβ = α(β)

Example 1.10. Multiplication of permutations usually is not commutative. Let

σ =

(1 2 3 44 1 2 3

)and τ =

(1 2 3 42 1 4 3

).

Then,

στ =

(1 2 3 41 4 3 2

), however, τσ =

(1 2 3 43 2 1 4

).

1.4.1 Presentation of permutations in cyclic notation

Another way to represent permutations is in the cycle notation. For example,

α =

(1 2 3 4 5 62 1 4 6 5 3

).

is written with cycles asα = (12) (346)(5).

We look at another example. Let a permutation β be presented as follows:

β =

(1 2 3 4 5 65 3 1 6 2 4

).

then writing it with cycles we have

β = (1523)(46) or β = (46) (3152)

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A permutation σ ∈ SX is a cycle with length k if there exist elements a1,a2, . . . ,ak ∈ X such that

σ(a1) = a2

σ(a2) = a3

...

σ(ak) = a1

and σ(x) = x for all other elements x ∈ X. We will use (a1,a2, . . . ,ak) to present σ.

Example 1.11. The permutation

σ =

(1 2 3 4 5 6 76 3 5 1 4 2 7

)= (162354)

is a cycle with length 6, however

τ =

(1 2 3 4 5 61 4 2 3 5 6

)= (243)

is a cycle with length 3.Not every permutation is a cycle. Consider the permutation(

1 2 3 4 5 62 4 1 3 6 5

)= (1243)(56).

This permutation contains a cycle with length 2 and a cycle with length 4.

Example 1.12. Let σ and τ be given as follows

σ =

(1 2 3 4 5 66 4 3 1 5 2

), τ =

(1 2 3 4 5 63 2 1 5 6 4

)In cycle notation we have

σ = (1624), τ = (13) (456), στ = (136) (245), τσ = (143) (256).

1.4.2 Dihedral groups as permutation groups

Another kind of permutation group which is important in algebra is the group of symmetries of a regularn -gon. We have already seen the group of symmetries of a regular triangle. In this section we will studythe group of symmetries of a regular n -gon.

For n = 3,4, . . ., define the n -th dihedral group as the group of solid motions of a regular n -gon. Wedenote such group with Dn or sometimes D2n.

Label the vertices of a regular n -gon by 1,2, . . . ,n. From Chapter 1 we have the following:

Lemma 1.13. i) The dihedral group, Dn, is a subgroup of Sn with order 2n.ii) The group Dn, n ≥ 3, contains all products of two elements r and s which satisfy the relations

rn = id,s2 = id,= srs = r−1.

Example 1.13. The group D4 of symmetries of a square contains 8 elements. We label vertices with 1, 2, 3, 4. Then,rotations are

r = (1234), r2 = (13)(24), r3 = (1432), r4 = id

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and reflections ares1 = (24), s2 = (13).

The order of D4 is 8. Two elements which are left are

rs1 = (12)(34), r3s1 = (14)(23).

Example 1.14. Find the generators of Dn in Sn.

Proof. Since Dn must have an element of order n then we take σ = (123 . . .n). Fix the vertex labeled as 1 andtake τ as the following permutation

τ =

(1 2 3 . . . i . . . n−1 n1 n n−1 . . . n + 2− i . . . 3 2

).

Hence, τ is given as product of transpositions by

τ =∏

2≤i<n+2−i

(i, n + 2− i).

Obviously, σn = τ2 = e. The reader can verify that Dn is generated by σ and τ.

Exercise 1.21. Can you find in D3, D4 an element that commutes with all elements of the group? Can you generalizethis for any Dn, n > 4? The geometry of the regular n-gon should provide some clues for the general case.

Exercises:

1.39. Let be given r and s the elements of Dn as in the Theorem 1.4.

1. Prove that srs = r−1.

2. Prove that rks = sr−k in Dn.

3. Prove that the order of rk∈Dn is n/gcd (k,n).

1.40. Prove that the dihedral group Dn is the group generated by the complex matrices

x =

[1 00 −1

], y =

[ξn 00 ξn

]where ξn = e

2πn is the n-th primitive root of unity.

1.4.3 Properties of permutations

In this section we study some properties of permutations.

Definition 1.4. Let α = (a1a2 · · ·ak) and β = (b1b2 · · ·bm) two cyclic permutations for a set X. We say that α and β,are disjoint if for every i, j we have that ai , b j.

If αβ = βα we say that they commute. Not every two permutations commute.

Proposition 1.1. If α = (a1a2 · · ·ak) and β = (b1b2 · · ·bm) are disjoint cycles in Sn, then αβ = βα.

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Proof. Let’s say that α and β are permutations of the set A :

A = a1,a2, · · · ,ak,b1,b2, · · · ,bm,c1,c2, · · · ,cn

where c are elements of A that are fixed from two permutations α,β. To prove that αβ = βα we must provethat for every x ∈ A we have that (αβ)(x) = (βα)(x). First we assume that x = ai, then we have

(αβ)(ai) = α(β(ai)) = α(ai) = ai+1,

since β fixes all the elements ai we have that β(ai) = ai. Similarly we have:

(βα)(ai) = β(α(ai)) = β(ai+1) = ai+1.

Thus, for all the elements a ∈ A we have that αβ = βα. Similarly we can prove that for every b ∈ A we havethat αβ = βα. Then, since α and β fix all the elements c we have

(αβ)(ci) = α(β(ci)) = α(ci) = ci

and(βα)(ci) = β(α(ci)) = β(ci) = ci

This completes the proof. Let α be a permutation. Composing α several times with itself will give again a permutation for some

positive integer i. We can writeαi = αα · · ·α︸ ︷︷ ︸

i− times

and for every two positive numbers m,n we have:

αmαn = αm+n and (αm)n = αmn

Theorem 1.6. Every permutation in Sn can be written as a unique product (up to permutation of cycles) of disjointcycles.

Proof. Let α ∈ Sn. If α fixes all objects then α = id and we can write

α = (1)(2)(3) · · · (n),

hence as a product of n disjoint cycles. Next, assume that α moves k objects. Without loss of generality weassume that α moves the first k objects. Hence we have,(

1 2 3 · · · k k + 1 · · · nα(1) α(2) α(3) · · · α(k) k + 1 · · · n

).

Since α is a bijection on the set 1,2, · · · ,k, then for any i ∈ 1, . . . ,n the element α(i) will never appear twice inthe second row. For example when we start writing the cycles (1,α(1) . . . , we know that α(1) will not appearagain in another cycle since in that case α would not be a bijection. Hence, in the cycle decomposition of αno element will appear twice. That concludes the proof.

Definition 1.5. Let α ∈ Sn. The smallest positive integer m such that αm = 1 is called order of permutation α.

Theorem 1.7. Let α ∈ Sn, which is written as product of disjoint cycles. Then, the order of α is the least commonmultiple of the lengths of its cycles.

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Proof. If α = (a1a2 · · ·am) then α has order m. Moreover, we know that if αk = e then m |k.If

α = (a1a2 · · ·am)(b1b2 · · ·br)

is given as a product disjoint cycles, then we have that

α j = (a1a2 · · ·am) j(b1b2 · · ·br) j,

since (a1a2 · · ·am) commutes with (b1b2 · · ·br). If α j = e then we have that (a1a2 · · ·am) j = e and also and(b1b2 · · ·br) j = e since all (b1b2 · · ·br) j fix all the elements ai and (a1a2 · · ·am) j fix all the elements bi. Thishappens if and only if m | j and r | j and lcm (m,r) is a divisor of j. The smallest number which has thisproperty is lcm (m,r).

Example 1.15. The permutation (1537)(284) has order 12 in S8 however the permutation (153)(284697) has order6 in S9.

1.4.4 Transpositions and involutions

The simplest permutations are cycles of length 2. Such cycles are called transpositions. Since

(a1,a2, . . . ,an) = (a1an)(a1an−1) · · · (a1a3)(a1a2),

then every cycle can be written as a product of transpositions.

Theorem 1.8. Every permutation in Sn, for n > 1, can be written as a product of transpositions. This product is notnecessarily unique.

Proof. First notice that if a permutation is the identity then it can be written as (12)(12), hence is the productof transpositions. Otherwise, from Theorem 1.7 we know that every permutation can be written in theform:

(a1a2 · · ·ak)(b1b2 · · ·bt)(c1c2 · · ·cs)

By multiplying through we see that this is equal to

(a1ak)(a1ak−1) · · · (a1a2)(b1bt)(b1bt−1) · · · (b1b2)(c1cs)(c1cs−1) · · · (c1c2).

This completes the proof. As seen above, writing a permutation as a product of transpositions is not unique. However, there is

something unique in some sense about such products as we will see in the next Theorem. But first theauxiliary Lemma.

Lemma 1.14. If the identity permutation is written as product of r transpositions,

id = τ1τ2 · · ·τr,

then r is an even number.

Proof. We will use induction on r. A transposition can not be identity. Thus, r > 1. If r = 2, then the lemmais true. Assume that r > 2. In this case the product of last two transpositions τr−1τr, can be one of thefollowing

(ab)(ab) = id(bc)(ab) = (ab)(ac)(cd)(ab) = (ab)(cd)(bc)(ac) = (ab)(bc).

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The first equation shows that a transposition is equal to its inverse. In this case we delete τr−1τr fromthe product and we have

id = τ1τ2 · · ·τr−3τr−2.

From the induction hypothesis r−2 is even, hence r is even.In each of the other three cases, we can substitute τr−1τr with the right side of the equation to obtain a

new product of r transpositions for the identity. In this new product the position of a is in the former positionof the last transposition. We continue this process with τr−2τr−1 to obtain a product of r−2 transpositionsor a product r transpositions, where the last position of a is in τr−2. If identity is the product of r− 2transpositions, then from induction hypothesis the proof is complete. Otherwise we repeat the procedurewith τr−3τr−2.

In some points there might be some overlaying. The last case can not happen since the identity cannot fix a in this case. Thus, the identity will be the product of r−2 transpositions and from the inductionhypothesis this is true. This completes the proof.

Theorem 1.9 (Always even or always odd). If a permutation is written in two ways as a product of transpositionsthen the number of transpositions in both cases is even or odd.

Proof. Assume thatσ = σ1σ2 · · ·σm = τ1τ2 · · ·τn,

where m is even. We must show that n is also an even number. The inverse of σ−1 is σm · · ·σ1. Since

id = σσm · · ·σ1 = τ1 · · ·τnσm · · ·σ1,

from Lemma 1.14 n must be even. The proof for the case where σ can be written as a number of oddtranspositions is left as an exercise for the reader.

Based on the above theorem we have the following definition.

Definition 1.6. A permutation is called even if it can be written as a product of an even number of transpositionsand it is called odd if it can be written as a product of an odd number of transpositions.

Exercise 1.22. The number of r-cycles in Sn is

#r− cycles =n!

(n− r)! · r!

Exercises:

1.41. Write the permutations α = (12345) and β = (1632)(457) as a product of transpositions.

1.42. Check that the order of α = (145)(23) ∈ S5 is |α| = gcd (3,2) = 6. Hence, α6 = e.

1.43. Let be given permutations σ = (123)(352) and τ = (34)(25). Find which of the following permutations σ, τ,στ, στ−1 are even or odd.

1.44. Let be given permutations σ = (1423)(352) and τ = (342)(25). Find which of the following permutations σ,τ, στ, στ−1 are even or odd.

1.45. Let be given σ ∈ Sn. Prove that σ can be written as product of at most (n−1) transpositions.

1.46. Let be given σ ∈ Sn. If σ is not a cycle, prove that σ can be written as product of at most (n−2) transpositions.

1.47. If σ is a cycle with odd length, prove that σ2 is also a cycle of odd length.

1.48. Prove that a 3-cycle is an even permutation.

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1.49. Prove that An is the only subgroups of Sn of index 2.

1.50. The alternating group A4 has no subgroups of order 6.

1.51. Prove that Sn is non Abelian for n ≥ 3.

1.52. Prove that An is non Abelian for n ≥ 4.

1.53. Prove that Dn is non Abelian for n ≥ 3.

1.54. Prove that in An, for n ≥ 3, every permutation is a product of cycles with length 3.

1.55. Prove that:

a) Sn is generated from (12), (13), . . . (1n).

b) Sn is generated from (12), (23), . . . , (i i + 1), . . . , (n−1, n).

c) Sn is generated from (12) and (1 . . .n).

1.56. The Frobenius group F20 is given by F20 := 〈(2354), (12345)〉. Find the order of F20 and draw its lattice.

1.57. Let G be a group and λg : G→ G the function defined by λg(a) = ga. Prove that λg is a permutation of G.Let G be a group and λg : G→ G be the function defined by λg(a) = ga. WTS λg is bijective.Injectivity: Let a1,a2 ∈ G and let g ∈ G be fixed. If λg(a1) = λg(a2)⇐⇒ ga1 = ga2 ⇐⇒ a1 = a2 since we can

multiply on the left by g−1.Surjectivity: Consider some h ∈ G. Take a = g−1h, then λg(g−1h) = h.Therefore, λg is bijective.

1.58. Find the center of D8. What can you say about the center of D10? What is the center of Dn?

1.59. Let be given τ = (a1,a2, . . . ,ak) a cycle with length k.

1. Prove that if σ is a permutation, then

στσ−1 = (σ(a1),σ(a2), . . . ,σ(ak))

is a cycle with length k.

2. Let be given µ a cycle with length k. Prove that there exists a permutation σ such that στσ−1 = µ.

1.60. Let be given α ∈ Sn for n ≥ 3. If αβ = βα for every β ∈ Sn, prove that α is the identity permutation. Thus, thecenter of Sn is the trivial subgroup.

1.61. If α is even, prove that α−1 is also even. Is it true that the same holds if α is odd?

1.62. Prove that α−1β−1αβ is even for α,β ∈ Sn.

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1.5 Linear groups

In this section we will see that the concept of a group is not totally new to us. In fact we have seen it beforein linear algebra; see [?lin-alg] for details. Throughout this section we will assume that the reader hasknowledge of linear algebra basics in the level of [?lin-alg].

Let (V, +, ·) be a vector space over R. As an elementary exercise prove that

Example 1.16. (V,+) is an Abelian group.

Let Matn(R) be the vector space of n×n matrices with entries in R.

Example 1.17. Prove directly, using the properties of matrices, that Matn(R) together with matrix addition is agroup.

1.5.1 The general linear group

Denote byGLn(R) = all n×n invertible matrices from Matn(R)

Show that GLn(R) is a group with matrix multiplication. This is the general linear group. The speciallinear group is defined as

SLn(R) = all n×n invertible matrices with det = 1

The orthogonal linear group On(R) is the group of invertible symmetric matrices. Hence,

On(R) = M ∈ GLn(R) |MT M = I

The special orthogonal linear group SOn(R) is the group of all orthogonal matrices with determinant 1. Inother words,

SOn(R) = M ∈On(R) | detM = 1

The reader should prove that GLn(R), SLn(R), On(R), and SOn(R) are groups for any n ≥ 2.The reader must be familiar with all these linear groups from [?lin-alg]. Next we recall the following

elementary example.

Example 1.18. Let Mat2(R) denote the set of all 2× 2 matrices with entries from R. Let GL2(R) be the subset of

Mat2(R) of all invertible matrices. Thus, a matrix A =

(a bc d

)is in GL2(R) if there is a matrix A−1 such that

AA−1 = A−1A = I,

where I is the 2×2 identity matrix. In order for a matrix A to have an inverse the determinant det(A) = ad−bc , 0.The identity of this group is the identity matrix

I =

(1 00 1

).

The inverse of A ∈ GL2(R) is

A−1 =1

ad− bc

(d −b−c a

).

The product of two matrices which have inverses is an invertible matrix. We know from linear algebra that themultiplication of matrices is associative. However, the multiplication of matrices is not commutative. In generalAB , BA. Thus, GL2(R) is non Abelian group.

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A permutation matrix is a matrix obtained by permuting the columns (rows) of the identity matrix.Such matrices are also called elementary matrices.

Let the set of all n×n permutation matrices with entries in a field k be denoted by P(n,k).

Exercise 1.23. Show thati) P(n,k) is a subgroup of GLn(k).ii) P(n,k)Sn.

The following theorem is quite useful.

Theorem 1.10 (Dickson). Let G be a finite subgroup of PGL2(C). Then G is isomorphic to one of the following:Zn,Dn,A4,S4,A5.

Proof. We will see the proof at a later stage.

1.5.2 Matrices modulo n

Let p be a prime integer and Fp denote the set of congruence classes modulo p together with addition andmultiplication modulo p as defined in the previous section. Hence, as sets Fp =Z/pZ. The triple (Fp,+, ·)we have called it a finite field in [?lin-alg]. We will study finite fields in more detail the the last part of thebook.

Let Matn(Fp) denote the set of n by n matrices with entries in Fp. The reader should be able to do thefollowing exercises or recall the results from [?lin-alg].

Exercise 1.24. Prove that Matn(Fp) is a vector space over Fp. Prove that this vector space has

|Matn(Fp)| = pn2

elements.

Exercise 1.25. Find the order of GLn(Fp) and SLn(Fp) are given by the formulas

|GLn(Fp)| = (pn−1)(pn

−p)(pn−p2) . . . (pn

−pn−1)

and

|SLn(Fp)| = pn(n−1)

2

i=n∏i=2

(pi−1).

Exercise 1.26. Sometimes in cryptography we use matrices which are invertible. We pick randomly a matrix A anduse a transformation x 7→ Ax to encrypt the message. The receiver knows A−1 and therefore decrypt the message byA−1· (Ax) = x. This is an elementary private key crypto-system.

If the matrix A is picked from Matn(R) then the probability that A is not invertible is zero. However, incryptography we normally use matrices from Matn(Fp), where p is the size of the alphabet in use. What is theprobability that A has an inverse when picked from Matn(Fp)?

Next we consider the case when n is not necessarily a prime. Let us denote again by Matn(Z/nZ) theset of all n by n matrices with entries in Z/nZ. Is Matn(Z/nZ) a vector space in this case?

From linear algebra you are probably aware that the answer is "No". What are such spaces? We willdiscuss such question when we study the module theory.

Exercise 1.27. Prove that the following are groups under multiplication of matrices.

a) The set of matricesSLn(Z) = M ∈Matn(Z) | detM = 1

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b) The setSLn(Z/nZ) = M ∈Matn(Z/nZ) | detM = 1,

for any integer n ≥ 2.

Linear groups are some of the most important objects of mathematics. We will revisit them again andagain in the coming chapters. We suggest the reader do all the following problems.

Exercises:

1.63. Let us have

1 =

(1 00 1

), I =

(0 1−1 0

), J =

(0 ii 0

), K =

(i 00 −i

),

where i2 = −1. Prove that,

I2 = J2 = K2 = −1, IJ = K, JK = I, KI = J,JI = −K, KJ = −I and IK = −J.

Moreover, show that the set Q8 = ±1,±I,±J,±K is a group. This is called the quaternion group. Notice that Q8 isnot Abelian.

1.64. Prove that the set of matrices of the type 1 a b0 1 c0 0 1

forms a group with the multiplication of matrices. This group is known as the Heisenberg group, and it is importantin quantum physics.

1.65. Give an example of two elements A and B in GL2(R) such that AB , BA.

1.66. Let G be the set of matricesG =

M ∈ GL2(R) | detM ∈Q×

Prove that G forms a group under matrix multiplication.

1.67. Let G be the set of matrices

G =

[a b0 d

]∈Mat2(R) | ad , 0

Prove that G forms a group under matrix multiplication. Is G Abelian?

1.68. Let G be the set of matrices

G =

[a 00 a−1

]∈Mat2(R) | a , 0

Prove that G is an Abelian group under matrix multiplication.

1.69. Let G be the set of matrices

G =

[a bc d

]∈Mat2(F2) | ad−bc , 0

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1.70. Let GL2(3) be the set of matrices

GL2(3) =

[a bc d

]∈Mat2(F3) | ad− bc , 0

Prove that GL2(3) has order 48.

1.71. Let SL2(3) be the set of matrices

SL2(3) =

[a bc d

]∈Mat2(F3) | ad− bc = 1

What is the order of SL2(3)?

1.72. Let p be a prime and GL2(p) be the set of matrices

GL2(p) =

[a bc d

]∈Mat2(Fp) | ad− bc , 0

What is the order of GL2(p)?

1.73. Let p be a prime and SL2(p) be the set of matrices

SL2(p) =

[a bc d

]∈Mat2(Fp) | ad− bc = 1

What is the order of SL2(p)?

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1.6 Complex numbers and groups associated to them

In this section we see some applications of group theory on the set of complex numbers.

1.6.1 A brief introduction of complex numbers

We denote by i the symbol such that i2 = −1. The set ofcomplex numbers, C, is the set

C := a + bi |a,b ∈R.

Two complex numbers z1 = a + bi and z2 = c + di are equal if andonly if a = b and c = d. Further, we define the addition andmultiplication as follows:

i) (a + bi) + (c + di) = (a + c) + (b + d)iii) (a + bi) · (c + di) = (ac− bd) + (ad + bc)i

The scalar multiplication is defined as

r · (a + bi) = ra + (rb) i,

for any r ∈ R and a + b i ∈ C. Both addition and scalar multipli-cation are illustrated geometrically in Section 1.6.1.

If z = a + bi, then a is called the real part of z and b is calledthe imaginary part of z. It is obvious that R ⊂ C.

x

y

z1 = a + bi

z2

z1 = a− bi

z1 + z2

z1z2

Every nonzero complex number has a multiplicative inverse. Hence there is a z−1∈ C× such that

zz−1 = z−1z = 1. If z = a + bi, then

z−1 =a−bi

a2 + b2 .

Example 1.19. Let’s denote with C× the set of nonzero complex numbers. The set C× with multiplication forms agroup. The identity of the group is 1. If z = a + bi is a nonzero complex number, then

z−1 =a−bi

a2 + b2 ,

is its inverse. It is easy to check the rest of group axioms.

The conjugate of a complex number z = a + bi is another complex number z = a− bi.

Exercise 1.28. Prove the following:

i) z + w = z + w

ii) zw = z ·w

The modulus or magnitude of z = a+bi is called the distance from z to the origin, which is given by theformula

|z| =√

a2 + b2.

We can represent a complex number z = a + bi as an ordered pair points in the plane xy where a is thecoordinate x (or real part) and b is the coordinate y (imaginary part). This is called the Cartesian form orrectangular coordinates.

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Nonzero complex numbers can be given using polar coordinates. To present a point in the plane,different form the origin, it is enough to give an angle θ from the positive x -axis, counter clockwise andthe distance r of the point from the origin. We can see that

z = a + bi = r (cosθ+ isinθ) .

Hence,a = r cosθ and b = rsinθ.

The following result leads to a geometric interpretation of the multiplication of complex numbers.

Lemma 1.15. Let z = r (cosθ+ isinθ) and w = s(cosφ+ isinφ

)be two complex numbers different from zero. Then,

zw = rs(cos(θ+φ) + isin(θ+φ)

).

In the literature, the following elementary result is known as the De Moivre’s formula.

Lemma 1.16 (De Moivre). Let z = r (cosθ+ isinθ) be a nonzero complex number. Then,

zn = rn (cosnθ+ isinnθ)

for all n ∈Z+.

Proof. We will use induction over n. For n = 1 the theorem is true. Assume that theorem is true for all ksuch that 1 ≤ k ≤ n. Then,

zn+1 = zn· z

= rn (cosnθ+ isinnθ) · r (cosθ+ isinθ)

= rn+1[

(cosnθcosθ− sinnθsinθ) + i (sinnθcosθ+ cosnθsinθ)]

= rn+1[

cos(nθ+θ) + i sin(nθ+θ)]

= rn+1[

cos(n + 1)θ+ isin(n + 1)θ].

This completes the proof.

Exercise 1.29. Prove that ∀u,v ∈ C,|u ·v| = |u| · |v|.

1.6.2 The unit circle and roots of unity

The group C×, contains some interesting subgroups. First consider the group of the unit circle,

T = z ∈ C : |z| = 1.

Exercise 1.30. The group of the unit circle is a subgroup of C×.

Even though the unit circle group has infinite order, it has many interesting subgroups of finite order.Assume that H = 1,−1, i,−i. Then, H is a subgroup of the unit circle group. Also 1, −1, i, and −i are exactlythose complex numbers which satisfy the equation z4 = 1.

Complex numbers that satisfy the equation zn = 1 are called the n -th roots of unity. We will denote thecyclic group of the n-th roots of unity by H(n).

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Lemma 1.17. The solutions of the equation zn = 1 are given by

z = cos2kπ

n+ isin

2kπn

where k = 0,1, . . . ,n−1. Moreover, these n-th roots of unity form a cyclic group of order n.

Proof. From De Moivre’s Theorem

zn = cos(n

2kπn

)+ isin

(n

2kπn

)= cos(2kπ) + isin(2kπ) = 1.

Numbers z are different since numbers 2kπ/n are all different and are≥ 0 but ≤ 2π.The fact that these numbers are the only roots of zn = 1 comes fromFundamental Theorem of Algebra ?? which says that a polynomialwith degree n can have at most n roots. We leave as an exercise toprove that n-th roots of unity form a cyclic group.

A generator of the group of n -th roots of unity is called n -th primitive root of unity and usually

denoted by ξn. The group of n-th roots of unity will be denoted by µn.

Exercise 1.31. Find all the 8-th roots of unity and draw them in the complex plane. Find all the subgroups of µ8 andcolor them with different colors. Do you see any pattern?

1.6.3 Möbius transformations

The extended complex plane is the complex plane added the point at infinity. Hence,

C = C∪∞

Figure 1.4: example caption

The general form of a Möbius transformation is given by

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1 Shaska T.

f (z) =az + bcz + d

where a, b, c, d are any complex numbers satisfying ad− bc , 0. If ad = bc, the rational function definedabove is a constant since

f (z) =az + bcz + d

=a(cz + d)c(cz + d)

−ad− bc

c(cz + d)=

ac

and this is not a Möbius transformation.In case c , 0, this definition is extended to the whole Riemann sphere by defining

f(−dc

)=∞ and f (∞) =

ac.

If c = 0, we definef (∞) =∞.

Thus a Möbius transformation is always a bijective holomorphic function from the Riemann sphere to theRiemann sphere.

Lemma 1.18. The set of all Möbius transformations forms a group under composition.

1.6.4 Fixed points of Möbius transformations

How the Möbius transformations move around the points of P1 is interesting for many reasons. But firstlet us consider the following question:

Question 1.2. How many fixed points can a Mobius transformation have?

So we want to solve the equationf (α) = α

which is equivalent with solving the quadratic

cα2− (a−d)α− b = 0

Lemma 1.19. The fixed points of a Möbius transformation f (z) = az+bcz+d are

Exercises:

1.74. For what integers n we have that −1 is an n -th root of unity?

1.75. Prove that αm = 1 and αn = 1 if and only if αd = 1 for d = gcd (m,n).

1.76. Let z ∈ C×. Prove that if |z| , 1, then the order of the element z is infinite.

1.77. Solve the following equationzn−1 = 0

1.78. Factor completely the following polynomial p(z) = z7−1.

1.79. Factor over Q the polynomial p(z) = z5−1.

1.80. Does the equationz4 + z3 + z2 + z + 1 = 0

have any rational solutions?

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1.7 The group of points in an algebraic curve

Let us revisit the material in Section 1.6. The unit circle is an algebraic curve in R2 with equation

x2 + y2 = 1

or a planar curve. Loosely speaking an algebraic curve is the set of points on the graph

f (x, y) = 0,

for some polynomial f with coefficients from some field k.We saw that the unit circle was group (under the complex multiplication). Here is a natural question:

what other curves can be made into a group?It might sound as an innocent question, but it has this question and its implications have occupied

some of the greatest minds of science for the last 200 years. In this lecture we will give some examples ofalgebraic curves which can be made into groups.

1.7.1 Conics

A conic section is technically the curve obtained by cutting a double cone by a plane. A general conic hasequation

ax2 + bxy + cy2 + dx + ey + f = 0, (1.8)

with a,b,c not all zero. It can be written in matrix notation as(x y

)( a b/2b/2 c

)(xy

)+

(d e

)(xy

)+ f = 0.

What is the shape of the graph with Eq. (1.8)? From elementary algebra we know that it is one ofthe following: parabola, hyperbola, ellipse, or an intersection of lines (in this case the conic is calleddegenerate).

From [?lin-alg] we know how to write any conic in a standard form and determine its shape. Let’squickly review that procedure. The symmetric matrix

M =

[a b/2

b/2 c

]is called the corresponding matrix to this conic. The discriminant of the conic is defined as

∆ = b2−4ac.

Notice that∆ = −4detM.

The following lemma determines the shape of the graph.

Lemma 1.20. The shape of the graph is determined as follows:

1. If ∆ > 0, then the graph is an hyperbola

2. If ∆ < 0, then the graph is an ellipse

3. If ∆ = 0, then the graph is a parabola

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The above quadratic can be written (with the appropriate substitutions) as

λ1X2 +λ2Y2 = D,

where λ1,λ2 are the eigenvalues of M. Moreover, from [?lin-alg, Lem. 7.5] we have that the curve

ax2 + bxy + cy2 = D.

is an ellipse if both eigenvalues of M are positive and a hyperbola if one is positive and the other is negative.The above procedure can be generalized to ternary quadratics; see [?lin-alg, Chap. 7] for details. The

following to exercises are elementary if the material in [?lin-alg, Chap. 7] is understood.

Example 1.20. Consider the quadratic form

q(x1,x2,x3) = x21 + x2

2−x23−2x1x2 + 4x1x3−6x2x3

Write this quadratic in the diagonal form p(x1,x2,x3). Sketch the surface

p(x1,x2,x3) = 1

Exercise 1.32. Determine the definiteness of the quadratic form

q(x1,x2) = x21 + 4x1x2 + x2

2

Conics as groups

Can conics be made into groups? Let us just attempt the following procedure to define an operation on aconic C.

• Fix a point O on the conic C. This will be our identity of the group.

• For any two points P,Q ∈ C, draw the line parallel to PQ and going through O. This line will intersectthe conic on a second point R.

• Define P⊕Q := R.

We have to check if the above procedure defines a group operation on any conic.The reader should, at this point investigate if the above procedure would work for all conics. The next

two examples show that at least it works for some conics. The following is due to F. Lemmermeyer.

Example 1.21. Consider the conicC : Y2

−∆X2 = 4,

and put N = (2,0). Prove that the group law on C with neutral element N is given by

(r,s) + (t,u) =( rt +∆su

2,

ru + st2

).

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1.7.2 Elliptic curves

One of the most celebrated groups and part of the mathematics folklore is the example of elliptic curves.

1.81. Let E(R) be the set of points in R2 of the graph given by the equation

y2 = x3 + ax2 + bx + c

Define the following operation on E(R). Given any two points P,Q ∈ E, we find the third point of intersection of theline PQ with E. From this point of intersection we drop the perpendicular line to the x-axis. Since E(R) is symmetricwith respect to the x-axis then this vertical line will intersect this graph in a second point. This point is denoted byP⊕Q. Prove the following:

i) (E(R),⊕) is an Abelian group.ii) Let E(Q) denote the set of points with rational coordinates. Prove that E(Q) is a subgroup of E(R).

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Exercises:

1.82. Let P1 = C∪∞ be the Riemann sphere.a) Show that functions α : P1

→ P1 such that

α(x) = x,1x, 1−x,

11−x

,x

x−1,x−1

x

form a group under composition of functions.b) Write a multiplication table for this group.c) Show that this group is isomorphic to S3.

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Nils Abel (1802-1829)

Niels Henrik Abel, (born August 5, 1802, island of Finney, nearStavanger, Norway–died April 6, 1829, Froland), Norwegian mathe-matician, a pioneer in the development of several branches of modernmathematics.

Abel’s father was a poor Lutheran minister who moved his familyto the parish of Gjerstad, near the town of Risor in southeast Nor-way, soon after Niels Henrik was born. In 1815 Niels entered thecathedral school in Oslo, where his mathematical talent was recog-nized in 1817 with the arrival of a new mathematics teacher, BerntMichael Holmboe, who introduced him to the classics in mathemati-cal literature and proposed original problems for him to solve. Abelstudied the mathematical works of the 17th-century Englishman SirIsaac Newton, the 18th-century German Leonhard Euler, and his con-temporaries the Frenchman Joseph-Louis Lagrange and the GermanCarl Friedrich Gauss in preparation for his own research.

Abel’s father died in 1820, leaving the family in straitened circumstances, but Holmboe contributedand raised funds that enabled Abel to enter the University of Christiania (Oslo) in 1821. Abel obtaineda preliminary degree from the university in 1822 and continued his studies independently with furthersubsidies obtained by Holmboe.

Abel’s first papers, published in 1823, were on functional equations and integrals; he was the first personto formulate and solve an integral equation. His friends urged the Norwegian government to grant hima fellowship for study in Germany and France. In 1824, while waiting for a royal decree to be issued, hepublished at his own expense his proof of the impossibility of solving algebraically the general equationof the fifth degree, which he hoped would bring him recognition. He sent the pamphlet to Gauss, whodismissed it, failing to recognize that the famous problem had indeed been settled.

Abel spent the winter of 1825–26 with Norwegian friends in Berlin, where he met August LeopoldCrelle, civil engineer and self-taught enthusiast of mathematics, who became his close friend and mentor.With Abel’s warm encouragement, Crelle founded the Journal für die reine und angewandte Mathematik("Journal for Pure and Applied Mathematics"), commonly known as Crelle’s Journal. The first volume(1826) contains papers by Abel, including a more elaborate version of his work on the quintic equation.Other papers dealt with equation theory, calculus, and theoretical mechanics. Later volumes presentedAbel’s theory of elliptic functions, which are complex functions (see complex number) that generalize theusual trigonometric functions.

In 1826 Abel went to Paris, then the world centre for mathematics, where he called on the foremostmathematicians and completed a major paper on the theory of integrals of algebraic functions. His centralresult, known as Abel’s theorem, is the basis for the later theory of Abelian integrals and Abelian functions,a generalization of elliptic function theory to functions of several variables. However, Abel’s visit to Pariswas unsuccessful in securing him an appointment, and the memoir he submitted to the French Academyof Sciences was lost.

Abel returned to Norway heavily in debt and suffering from tuberculosis. He subsisted by tutoring,supplemented by a small grant from the University of Christiania and, beginning in 1828, by a temporaryteaching position. His poverty and ill health did not decrease his production; he wrote a great numberof papers during this period, principally on equation theory and elliptic functions. Among them arethe theory of polynomial equations with Abelian groups. He rapidly developed the theory of ellipticfunctions in competition with the German Carl Gustav Jacobi. By this time Abel’s fame had spread to allmathematical centres, and strong efforts were made to secure a suitable position for him by a group fromthe French Academy, who addressed King Bernadotte of Norway-Sweden; Crelle also worked to secure aprofessorship for him in Berlin.

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Carl Gustav Jacobi (1804-1851)

Carl Gustav Jacob Jacobi (10 December 1804 – 18 February 1851)was a German mathematician, who made fundamental contributionsto elliptic functions, algebraic geometry, dynamics, differential equa-tions, and number theory. His name is occasionally written as CarolusGustavus Iacobus Iacobi in his Latin books, and his first name is some-times given as Karl.

One of Jacobi’s greatest accomplishments was his theory of ellipticfunctions and their relation to the elliptic theta function. This wasdeveloped in his great treatise Fundamenta nova theoriae functionumellipticarum (1829), and in later papers in Crelle’s Journal. Thetafunctions are of great importance in mathematical physics becauseof their role in the inverse problem for periodic and quasi-periodicflows. The equations of motion are integrable in terms of Jacobi’selliptic functions in the well-known cases of the pendulum, the Eulertop, the symmetric Lagrange top in a gravitational field and the Keplerproblem (planetary motion in a central gravitational field).

He also made fundamental contributions in the study of differen-tial equations and to rational mechanics, notably the Hamilton–Jacobitheory.

It was in algebraic development that Jacobi’s peculiar power mainly lay, and he made importantcontributions of this kind to many areas of mathematics, as shown by his long list of papers in Crelle’sJournal and elsewhere from 1826 onwards. One of his maxims was: ’Invert, always invert’ (’man mussimmer umkehren’), expressing his belief that the solution of many hard problems can be clarified byre-expressing them in inverse form.

In his 1835 paper, Jacobi proved the following basic result classifying periodic (including elliptic)functions: If a univariate single-valued function is multiply periodic, then such a function cannot havemore than two periods, and the ratio of the periods cannot be a real number. He discovered many of thefundamental properties of theta functions, including the functional equation and the Jacobi triple productformula, as well as many other results on q-series and hypergeometric series.

The solution of the Jacobi inversion problem for the hyperelliptic Abel map by Weierstrass in 1854required the introduction of the hyperelliptic theta function and later the general Riemann theta functionfor algebraic curves of arbitrary genus. The complex torus associated to a genus g algebraic curve, obtainedby quotienting Cg by the lattice of periods is referred to as the Jacobian variety. This method of inversion,and its subsequent extension by Weierstrass and Riemann to arbitrary algebraic curves, may be seen as ahigher genus generalization of the relation between elliptic integrals and the Jacobi, or Weierstrass ellipticfunctions

Jacobi was the first to apply elliptic functions to number theory, for example proving of Fermat’s two-square theorem and Lagrange’s four-square theorem, and similar results for 6 and 8 squares. His otherwork in number theory continued the work of Gauss: new proofs of quadratic reciprocity and introductionof the Jacobi symbol; contributions to higher reciprocity laws, investigations of continued fractions, andthe invention of Jacobi sums.

He was also one of the early founders of the theory of determinants; in particular, he invented theJacobian determinant formed from the n2 differential coefficients of n given functions of n independentvariables, and which has played an important part in many analytical investigations. In 1841 he reintro-duced the partial derivative ∂ notation of Legendre, which was to become standard. Students of vectorfields and Lie theory often encounter the Jacobi identity, the analog of associativity for the Lie bracketoperation.

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Chapter 2

Basic properties of groups

2.1 Subgroups

In this section we will develop criteria when a subset of a group G is also a group. Such subsets of G whichare also groups under the operation of G we will call them subgroups.

Definition 2.1. If a subset H of the group G forms a group with the operation of G, then we say that H is a subgroupof the group G.

To show that H is subgroup of G we use the notation H ≤G. When H is a proper subset of G then we saythat H is a proper subgroup of G and denoted by H < G. The subgroup e is called the trivial subgroup ofG.

2.1.1 Subgroup tests

In this section we see different methods of proving that a subset is a subgroup.

Theorem 2.1 (First Subgroup test). Let G be a group and H a nonempty subset of G. Then the following areequivalent:

i) H is subgroup of Gii) For every a, b ∈H, ab−1

∈H.

Proof. Obviously, i) =⇒ ii). Next, we want to show that ii) =⇒ i).Since the operation of H is the same with that of G then the associativity property is true.Next, we want to show that the identity of H is eG. Since H is nonempty then there exists at least an

element x ∈H. If we take a = x and b = x−1 then we have that eG = xx−1 = ab−1 which is in H.To verify that x−1

∈H when x ∈H it is enough to substitute a = eG and b = x.The proof is completed if we show that H is closed under the multiplication of G. Let x, y be any two

elements in H. From above, we know that for any y ∈H then y−1∈H. Take a = x and b = y−1 and we have

that xy = x(y−1)−1 = ab−1∈H. This completes the proof.

Example 2.1. Let G be a group and A and B be subgroups of G. When is A∪B a subgroup of G?

Solution: Obviously if (A⊆ B or B⊆A), then A∪B is a subgroup. Hence, assume that (A\B, ∅ and B\A,

∅).Suppose that A∪B is a subgroup. Let a ∈ A\B and b ∈ B\A. Since A∪B is a subgroup then

ab−1∈ A∪B =⇒ ab−1

∈ A or ab−1∈ B.

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2 Shaska T.

Without loss of generality assume that ab−1∈ A which implies that b−1

∈ A. Hence, b ∈ A. Since b ∈ B \A, thenB ⊂ A. Thus, the only case when A∪B is a subgroup is A ⊆ B or B ⊆ A.

Theorem 2.2 (Second subgroup test). A subset H of G is a subgroup if and only if it satisfies the following:i) Identity e of G is in H.ii) If h1,h2 ∈H, then h1h2 ∈H.iii) If h ∈H, then h−1

∈H.

Proof. First assume that H is a subgroup of G. We must show that the three conditions are satisfied. Since His group it has identity of eH. We want to show that eH = e where e is identity of the group G. We know thateHeH = eH and eeH = eHe = eH. Hence, eeH = eHeH. From the left cancellation property we have that e = eH.The second condition is true because the subgroup H is a group. To prove the third condition take h ∈ H.Since H is a group then there is an element h′ ∈ H such that hh′ = h′h = e. From uniqueness of the inversein G we have that h′ = h−1.

Conversely, if the three conditions are satisfied then H is a group since these are the group axioms.

Example 2.2. Let G the group of nonzero real numbers with multiplication and let H

H = x ∈ G |x = 1 or x is irrational.

Prove that H is not a subgroup of G.

Solution: We have that√

2 ∈H but√

2 ·√

2 = 2 <H, hence the operation in H is not closed.

Theorem 2.3 (Third subgroup test). Let H a nonempty finite subset of G. If H is closed with the operation of Gthen H is subgroup of G.

Proof. Using the second test it is enough to show that for every a ∈H there exists a−1∈H. If a = e then this

is true because e−1 = e ∈H.Let a, e. Consider the sequence a,a2,a3, · · · . Since, H is closed, then all these elements are in H. However,

H is finite. Hence, not all of them are different. Thus, we have that ai = a j and i > j, for some i, j. Then,ai− j = e and since a , e we have that i− j−1 > 0. Thus, aai− j−1 = ai− j = e and ai− j−1 = a−1. However, i− j−1 ≥ 1which implies ai− j−1

∈H.

Example 2.3. The set of all matrices 2×2 with elements from R, denoted by Mat2(R), forms a group with addition.The set GL2(R) is a subset of Mat2(R). However, it is not a subgroup of Mat2(R) under addition. The sum of

two matrices does not necessary have an inverse. Notice that(1 00 1

)+

(−1 00 −1

)=

(0 00 0

),

however the zero matrix is not in GL2(R).

Exercises:

2.1. Let G be an Abelian group with identity element e and H a subset of G such that

H = x ∈H |x2 = e.

Prove that H is a subgroup of G.

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2.2. Let (G, ·) be an Abelian group with identity e and H a subset of G such that

H =x2|x ∈ G

.

Prove that H is a subgroup of G.

2.3. Consider the set of nonzero real numbers Rtimes together with multiplication. The identity of this group is 1and the inverse of every element a ∈Rtimes is 1/a. We will prove that

Q× = p/q : pandq are nonzero integers

is a subgroup of Rtimes.

2.4. Recall that with C× we have denoted the group of nonzero complex numbers with multiplication. Let be givenH = 1,−1, i,−i. Prove that H is a subgroup of C×.

2.5. Let SL2(R) be the subset of GL2(R) which contains all matrices with determinant 1. Prove that SL2(R)<GL2(R)

2.1.2 The center and centralizers

Next we see some example of subgroups. Let G be a group. Denoted by Z(G) the set

Z(G) = a ∈ G |xa = ax, ∀x ∈ G.

Hence, Z(G) is the set of elements of G which commute with all elements of G. We call Z(G) the center of G.

Proposition 2.1. The center Z(G) of a group G is a subgroup.

Proof. We use the second subgroup test. Clearly, e ∈ Z(G) because ∀x ∈ G we have ex = xe. Thus, Z(G) is anonempty set. Next, we show that Z(G) is closed under the multiplication of G. Let a,b ∈ Z(G), then

(ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab), ∀ x ∈ G.

Thus, ab ∈ Z(G).Assume that a ∈ Z(G) then we have that ax = xa. Multiply both sides with a−1 and we have:

a−1(ax)a−1 = a−1(xa)a−1

(a−1a)xa−1 = a−1x(aa−1)

exa−1 = a−1xe

xa−1 = a−1x

(2.1)

Thus, we proved that for every a ∈ Z(G) we have a−1∈ Z(G). Therefore, Z(G) is a subgroup of G.

Definition 2.2. Let A be any subset of G. We call the centralizer of A in G the set of elements of G which commutewith all elements a ∈ A. This is denoted by CentG(A)

CentG(A) = g ∈ G | ga = ag, ∀a ∈ A.

Lemma 2.1. Centralizer CentG(A) is subgroup of G.

Proof. The proof is similar as for Z(G) and it is left as an exercise.

Exercise 2.1. Prove thatZ(G) =

⋂g∈G

CentG(g),

where CentG(g) is the centralizer of g.

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2.1.3 Alternating groups

One of the most important subgroups of Sn is the set of even permutations, An. The group An is called thealternating group in n letters.

Theorem 2.4. The set An is a subgroup of Sn.

Proof. Since the product of two even permutations is also even, then An is closed. The identity permutationis an even permutation, hence it is in An. If σ is an even permutation, then

σ = σ1σ2 · · ·σr,

where σi is a transposition and r is an even number. Since the inverse of a transpositions is equal to itself,then

σ−1 = σrσr−1 · · ·σ1,

it is in An. Next we determine the number of even permutations in Sn.

Proposition 2.2. The number of even permutations in Sn, n ≥ 2, is equal with the number of odd permutations.Thus, the order of An is n!/2.

Proof. Let An the set of even permutations in Sn and Bn the set of odd permutations. If we prove that thereexists a bijection between these sets then they have the same number of elements. Fix a transposition σ inSn. Since n ≥ 2, there exists such a σ. Define the map

λσ : An→ Bn

such asλσ(τ) = στ.

The reader can check that this map is well defined.Assume that λσ(τ) = λσ(µ). Then, στ = σµ and

τ = σ−1στ = σ−1σµ = µ.

Hence, λσ is injective. The reader can show that λσ is surjective.

Lemma 2.2. The alternating group An is generated from 3-cycles, for n ≥ 3.

Proof. To prove that 3-cycles generate An, we must prove only that every even transpositions can be writtenas a product 3-cycles. Since (ab) = (ba), then every even transpositions can be one of the following types

(ab)(ab) = id(ab)(cd) = (acb)(acd)(ab)(ac) = (acb).

This completes the proof.

Exercises:

2.6. Let G a Abelian group andCube (G) = g3

| g ∈ G.

Is Cube (G) a subgroup of G? Justify your answer.

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2.7. Let be given G an Abelian group, where n > 1 is an integer and

Gn := gn| g ∈ G.

Prove that Gn is a subgroup of G.

2.8. Let H = 2k : k ∈Z. Prove that H is a subgroup of Q×.

2.9. Let the set G which contains all matrices 2×2 of the type(cosθ −sinθsinθ cosθ

)where θ ∈R. Prove that G is a subgroup of SL2(R).

2.10. Prove thatG = a + b

√

2 : a,b ∈Q and a,b are not both zero

is a subgroup of R× with multiplication.

2.11. Let G the group of 2×2 matrices together with the addition and

H =

(a bc d

): a + d = 0

.

Prove that H is a subgroup of G.

2.12. Prove or disprove that SL2(Z), the set of 2×2 matrices with integer entries and with determinant 1, is subgroupof SL2(R).

2.13. Find the subgroups of the quaternion group Q8.

2.14. Prove that the intersection of two subgroups of a group G is also a subgroup of G.

2.15. Prove or disprove the following: if H and K are subgroups of a group G, then HK = hk : h ∈H and k ∈ K issubgroup of G. What can you say if G is Abelian?

2.16. In GL2(R) find the centralizer of [1 10 1

]

2.2 Homomorphisms

One of the most important concept in algebra is that of a homomorphism. Homomorphisms are mapsbetween groups which preserve the algebraic structure.

2.2.1 Group homomorphisms

Let groups (G,∗) and (H,?). The function f : G −→H is called a group homomorphism if we the followingis satisfied,

f (a ∗b) = f (a)? f (b)

for every a,b ∈ G. The following is an easy exercise.

Lemma 2.3. Let φ : G1→ G2 be a group homomorphism. Then,

i) If e is the identity of G1, then φ(e) is the identity of G2 ;

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2 Shaska T.

ii) ∀g ∈ G1, φ(g−1) =(φ(g)

)−1;

Proof. i) Assume that e and e′ are respectively identities of the groups G1 and G2, then

e′φ(e) = φ(e) = φ(ee) = φ(e)φ(e).

Hence, we have φ(e) = e′.ii) This comes from the fact that

φ(g−1)φ(g) = φ(g−1g) = φ(e) = e.

If f is injective then f is called monomorphism and if is surjective is called epiomorphism.If the function f is bijective, then f is called isomorphism. The group G is called isomorphic with the

group H and denoted by GH.For any map f : G→H, the preimage of H is

f−1(H) := g ∈ G | f (g) ∈H

and the image of G isImg ( f ) := h ∈H | ∃g ∈ G such that f (g) = h

Let A ⊂ G and B ⊂H. Then, similarly

f−1(B) := g ∈ G | f (g) ∈ Bf (A) := h ∈H | ∃g ∈ A such that f (g) = h

Obviously, f (G) = Img ( f ).

Example 2.4. Let G be a group and g ∈ G. Define the function φ :Z→ G such that φ(n) = gn. Then, φ is a grouphomomorphism because

φ(m + n) = gm+n = gmgn = φ(m)φ(n).

This homomorphism maps the group Z in the cyclic subgroup of G generated by g.

Example 2.5. Let G = GL2(R). If

A =

(a bc d

)is in G, then the determinant is nonzero. Thus, det(A) = ad− bc , 0. Also, for every two elements A and B in G,det(AB) = det(A)det(B). Using the determinant we can define a homomorphism

φ :GL2(R)→Z∗,

A→ det(A).

Exercise 2.2. Let be given the sets Ω and ∆ such that |Ω| = |∆|. Prove that S∆SΩ.Hint: Since |Ω| = |∆| then there is a bijection θ : Ω −→ ∆. Define

φ : S∆ −→ SΩ

σ −→ θσθ−1

for every σ ∈ S∆. Prove thata) φ is a functionb) φ is a bijection (For example find a ϕ such that ϕφ = eSΩ

, φϕ = eS∆.)

c) φ is a homomorphism.

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2.2.2 Properties of homomorphisms

The following statement gives some of the main properties of homomorphisms

Proposition 2.3. Let φ : G1→ G2 be a group homomorphism. Then,

a) If H1 ≤ G1, then φ(H1) ≤ G2 ;

b) If H2 ≤ G2, then φ−1(H2) ≤ G1.

Proof. a) The set φ(H1) is nonempty because identity of H2 is in φ(H1). Assume that H1 is subgroup of G1and take x and y in φ(H1). There exist elements a,b ∈H1 such that φ(a) = x and φ(b) = y. Since

xy−1 = φ(a)[φ(b)]−1 = φ(ab−1) ∈ φ(H1),

then φ(H1) is subgroup of G2 from Theorem 2.1.b) Let H2 a subgroup of G2 and define H1 such thatφ−1(H2). Thus, H1 is the set of all elements g ∈G1 such

that φ(g) ∈H2. The identity element is in H1 because φ(e) = e. If a and b are in H1, then φ(ab−1) =φ(a)[φ(b)]−1

is in H2 since H2 is subgroup of G2. Thus, ab−1∈H1 and H1 is subgroup of G1.

Exercise 2.3. Let f : G1→ G2 be a homomorphism of groups and H ≤ G1. Is f (H) a subgroup of G2? Justify youranswer.

Let be given the homomorphism f : G −→H. The kernel of f is called the set

ker( f ) := g ∈ G : f (g) = eH.

The reader must be familiar with the concept of the kernel from linear algebra; see [?lin-alg].

Lemma 2.4. Let f : G −→H be a group homomorphism with K = ker( f ). Then, the following hold:

i) ker( f ) ≤ G.

ii) f is injective if and only if ker( f ) = eG

iii) f (G) ≤H

Proof. Part i) is an easy exercise.ii) Assume that f is injective. We know that f (eG) = eH. Hence eG ∈ ker( f ). Since f is injective, then

ker( f ) = eG.Assume that f is not injective. Then, there exist a,b ∈G such that a , b and f (a) = f (b) = c ∈H. Since a , b

then ab−1 , eG. We see thatf (ab−1) = f (a) f (b−1) = f (a) f (b)−1 = cc−1 = eH.

Then, ab−1∈ ker( f ) and ab−1 , eG, so we have a contradiction.

iii) We must show that f (G) is group. Since f (eG) = eH then eH ∈ f (G). Take x, y ∈ f (G). From thesubgroup tests it is enough to prove that xy−1

∈ f (G). Since x, y ∈ f (G) there exist a,b ∈ G that f (a) = x andf (b) = y. Then,

xy−1 = f (a) f (b)−1 = f (a) f (b−1) = f (ab−1).

Thus, ab−1∈ G. Thus , f (ab−1) = xy−1

∈ f (G). Finally f (G) is a subgroup of H.

Example 2.6. Consider the homomorphism

φ : GL2(R) −→R∗

A −→ det(A),

as in Example 2.5. Since 1 is the identity ofR∗, the kernel of this homomorphism contains from all 2×2 matrices thathave determinant 1. Thus,

kerφ = SL2(R).

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Exercises:

2.17. Let G be a finite group of even order, show that G has an odd number of elements of order 2.

2.18. Let G be a finite group of odd order. Show that

s : G→ G

x→ x2

is a surjective map. When is it a homomorphism?

2.19. Let G be e group and g ∈G. Show that there exists a unique homomorphism φ : (Z,+)→G, such that φ(1) = g.

2.20. Recall that the circle group,T, contains all numbers complex z such that |z|= 1. We can define a homomorphismφ, from the additive group of integers Z to T, such that

φ : θ 7→ cosθ+ isinθ.

Prove that this is a homomorphism.

2.21. If G is an Abelian group and n ∈N, prove that φ : G→ G such that g 7→ gn is a homomorphism.

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2.3 Cyclic groups

Let a be an element of a group, then by 〈a〉we denote the set of all the powers of a, namely the set an|n ∈Z.

If a has finite order then with 〈a〉we denote

〈a〉 := e = a0,a,a2,a3, . . . ,an−1.

Theorem 2.5. Let G a group and a ∈ G. Then, 〈a〉 is subgroup of G.

Proof. Since a belongs to 〈a〉, then 〈a〉 is not empty. Let am and an be any two powers of a. Then, an,am∈ 〈a〉

and an(am)−1 = an−m∈ 〈a〉. Thus, from the first subgroup test, 〈a〉 is subgroup of G.

The subgroup 〈a〉 is called the cyclic subgroup of G generated by the element a.

Definition 2.3. A group G is called cyclic if there is an element a ∈ G such that

G = an|n ∈Z

and such an element (if it exists) is called generator of G and we write G = 〈a〉.

A cyclic group can have many generators. Below we give a few examples of such groups. Notice that,

aia j = ai+ j = a j+i = a jai.

Hence, every cyclic group is Abelian.

Example 2.7. The group of integers with addition (Z,+) is a cyclic group with generators 1 or -1.

Example 2.8. The group of integer with addition modulo n is a cyclic group and 1 or n− 1 are generators of thisgroup.

Example 2.9. Prove that the group Z8 is generated by the element 3, hence Z8 = 〈3〉.

Solution: In the group Z8 with addition mod 8 the element 3 generates these elements:

〈3〉 = 3,3 + 3,3 + 3 + 3, · · · = 3,6,1,4,7,2,5,0 =Z8

Hence we say that 3 is a generator of Z8. Similarly, Z8 = 〈1〉 = 〈3〉 = 〈5〉 = 〈7〉.

Example 2.10. Prove that U (10) = 〈3〉 = 〈7〉.

Solution: First we have seen that U (10) = 1,3,7,9. Numbers 3 and 7 are generators of this group because

〈3〉 = 30,31,33,32 = 1,3,7,9

and also〈7〉 = 70,73,71,72

= 1,3,7,9

Theorem 2.6. Let G a group and a ∈ G. If a has infinite order then all powers of a are distinct elements. If a hasfinite order, say |a| = n, then

〈a〉 = e,a,a2,a3, . . . ,an−1

and ai = a j if and only if n | (i− j).

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2 Shaska T.

Proof. Assume that the element a has infinite order, so it does not exist any positive integer n such thatan = e. We have to prove that all powers of a form distinct group elements. Assume the contrary, there existai,a j where i , j such that ai = a j. Then, we have that ai− j = e which is true only for i− j = 0. Hence, for i = jwhich is a contradiction.

Now let’s prove that if |a| = n, then we have that:

〈a〉 = e,a,a2,a3, . . . ,an−1.

First let’s prove that the elements e,a, · · · ,an−1 are different between them. Assume the contrary, so fori, j such that 0 ≤ j < i ≤ n− 1 we have that ai = a j. Then we have that ai− j = e for i− j < n. However, thiscontradicts the fact that n is the smallest positive integer such that an is identity, which is a contradiction.

Assume that ak is any element of 〈a〉. From the division algorithm we know that there exist two integersq and r such that

k = qn + r where 0 ≤ r < n

Then, ak = aqn+r = aqnar = (an)qar = ear = ar, so we have that ak∈ e,a,a2, · · · ,an−1

. This proves that 〈a〉 =e,a,a2, · · · ,an−1

.Now we prove that if ai = a j, then n divides i− j. Assume the contrary, so n does not divide i− j. Then

from division algorithm we know that there exist two integers q and r such that

i− j = qn + r where 0 ≤ r < n.

Substituting in the above equality we get ai− j = aqn+r and therefore

e = ai− j = aqn+r = (an)qar = eqar = ear = ar.

Since n is the smallest positive integer such that an = e then we must have r = 0 and therefore n divides i− j.Conversely, if i− j = nq then ai− j = anq = eq = e so ai = a j.

Corollary 2.1. For any element a in a group we have that |a| = |〈a〉|.

Corollary 2.2. Let G a group and let a ∈ G such that |a| = n. If ak = e then n divides k.

Proof. Since ak = e = a0 from the above theorem we have that n divides k−0 = k

Theorem 2.7. Let G be a finite cyclic group such that |G| = n and G = 〈a〉. Then, G = 〈ak〉 if and only if k and n are

relatively prime.

Proof. First we prove that if (k,n) = 1, then the group G is generated from 〈ak〉, so G = 〈ak

〉.Since (k,n) = 1, then there exist two integers u and v that satisfy

1 = ku + nv.

Then, we can writea = aku+nv = aku

· anv = aku· (an)v = aku

· ev =(ak

)u,

which implies that a ∈ 〈ak〉. Hence, all powers of a belong to 〈ak

〉. For example, ap = (aku)p = (ak)up so ap∈ 〈ak〉.

Thus, G = 〈ak〉 and 〈ak

〉 is generator of G.Now let’s prove that if G = 〈ak

〉 then (k,n) = 1. Assume that k and n are not coprime, (k,n) = d > 1 thenwe have

k = td and n = sd

where s < n. Thus,(ak)s = (atd)s = (at)ds = (at)n = (an)t = et = e

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Since (ak)s = e we have that the order of ak is smaller or equal to s. Thus,

|ak| ≤ s < n.

This means that ak is not a generator for the group G because the order of the group G is n, but |〈ak〉| = s

where s < n. Above we explained that Zn has more then one generator. From the above theorem, generators of Zn

are all numbers k ∈Zn such that (k,n) = 1.

Corollary 2.3. The number k is a generator of the group Zn if and only if (k,n) = 1.

Example 2.11. Not every group is a cyclic group. Consider the group S3 of symmetries of the regular triangle. Allsymmetries are shown in Fig. 1.2.

Subgroups of S3 are shown in Figure Fig. 2.1. Notice that every subgroup is cyclic even though no single elementgenerates all the group.

S3

id

id,ρ1,ρ2 id,µ1 id,µ2 id,µ3

!!!!

!!!!

aaaaaaaa

SSS

SSS

aaaa

aaaa

!!!!!!!!

Figure 2.1: Subgroups of S3

The diagram above is called the lattice of subgroups of S3. We can construct such lattice of subgroups forany given group. As we will see in the coming lectures, knowing all the subgroups and their intersectionsis very important to understand the structure of the group.

Example 2.12. Let the group U (50) be given. Find the order of this the group and other generators when we knowthat 3 is already a generator.

Solution: The elements of U (50) are

U (50) = 1,3,7,9,11,13,17,19,21,23,27,29,31,33,37,39,41,43,47,49.

Hence |U (50)| = 20. From the above theorem, generators are all elements of the form 3k for k such that (k,20) = 1hence all elements

31 = 3, 33 = 9, 37 = 37, 39 = 49, 311 = 47, 313 = 23, 317 = 13, 319 = 17.

Keep in mind that all operations are mod 50.

Let n be a positive integer. How many integers k are that (n,k) = 1? Let ϕ(n) denote the number of suchintegers k.

The Euler’s functionϕ :N→N

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is defined as ϕ(n) = 1 for n = 1 and for n > 1, ϕ(n) is the number of integers m, where 1 ≤ m < n andgcd (m,n) = 1.

Then, we have the following,

Corollary 2.4. Let U (n) be the group of unit elements in Zn. Then,

|U (n)| = ϕ(n).

2.3.1 Subgroups of cyclic groups

Theorem 2.8. Every subgroup of a cyclic group is cyclic.

Proof. Let G be cyclic group, G = 〈a〉 and assume that H ≤ G. If H = e then H is cyclic. Assume that Hcontains another element g different from identity. Then, for some integer n, the element g can be writtenas an. Assume that n > 0. Let m be the smallest integer such that am

∈ H. Such number m exists from thewell ordering principle.

Assume that h = am is a generator for H. We must show that every h′ ∈H can be written as a power of h.Since h′ ∈H and H is a subgroup of G then for some integer k we have h′ = ak. Using the division algorithmwe find q and r such that k = mq + r where 0 ≤ r <m. Thus,

ak = amq+r = (am)qar = hqar.

Hence, ar = akh−q. Since ak and h−q are in H, then ar∈ H. However m was the smallest integer such that

am∈H and therefore r = 0. Hence, k = mq. Finally,

h′ = ak = amq = hq

and H is generated by h.

Corollary 2.5. Subgroups of the group Z are exactly nZ for n = 0,1,2, . . ..

Proposition 2.4. Let G be a cyclic group such that |G| = n and assume that G = 〈a〉. Then, ak = e if and only if n |k.

Proof. First assume that ak = e. From division algorithm we have k = nq + r where 0 ≤ r < n. Hence,

e = ak = anq+r = anqar = e · ar = ar.

Since n is the smallest nonzero integer such that an = e, then r = 0.Conversely, if n divides k then k = ns for some integer s. Hence, we have that

ak = ans = (an)s = es = e.

This completes the proof.

Theorem 2.9. Let G cyclic group such that G = 〈a〉 and |G| = n. If b = ak then |b| = nd , where d = gcd (k,n).

Proof. We want to find the smallest integer m such that bm = akm = e. However, this is the smallest integerm such that n |km or n/d divides m · k

d . Since, d is the greatest common divisor of n and k then n/d and k/dare relatively prime. Thus, for n/d to divide m · k

d it must divide m. The smallest such number m is n/d.

Theorem 2.10 (Fundamental Theorem of Cyclic Groups). Let G = 〈a〉 and |G| = n. Then for every divisor k ofn, G has exactly one subgroup of order k, namely 〈a

nk 〉.

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Proof. We want to show that⟨a

nk⟩

is the one and only subgroup of order k of 〈a〉. From the previous Theoremwe know that ∣∣∣〈a n

k 〉

∣∣∣ =n

gcd (n, nk )

=n

(n/k)= k

So, this is a subgroup of order k. Let us show now that this is the only subgroup of order k. Let H = 〈am〉 be

a subgroup of G with order k. We know that m is a divisor of n and m = gcd (n,m). Thus,

k =∣∣∣am

∣∣∣ =∣∣∣∣agcd (n,m)

∣∣∣∣ =n

gcd (n,m)= n/m

So, k = n/m, and m = n/k implying H =⟨a

nk⟩.

Exercise 2.4. If G is cyclic and has infinite order, then G (Z,+).

Proof. Indeed, Φ(n) = xn, where Φ :Z→ G is a homomorphism and bijection.

Example 2.13. Let G be a group possessing no proper subgroups. Show that G is cyclic, finite of prime order.

Solution: Let x ∈ G such that x , e. Then 〈x〉 is a subgroup of G. But 〈x〉 can not be proper. So 〈x〉 = G. Thus, G is

cyclic.If G is cyclic and has infinite order, then from the above Exercise we have that G (Z,+). But (Z,+) has many

proper subgroups. It implies that G is not isomorphic to (Z,+). Hence, G is finite.Assume that G is not of prime order, say |G| = d1d2 · · ·dn. Then G would have proper subgroups of order

d1,d2, · · · ,dn, see Theorem 2.10 which contradicts the assumption of the problem.

The following is an important result for cyclic groups. We leave it as an exercise for the reader.

Proposition 2.5. Let p be a prime. A group G of order |G| = pn is cyclic if and only if it is an Abelian group with aunique subgroup of order p.

Exercises:

2.22. Prove or disprove the following statement: if G is a group such that every proper subgroup is cyclic then G iscyclic.

2.23. Prove or disprove the followinga) The group U (8) is cyclic.b) All generators of Z60 are prime numbersc) A group with a finite number of subgroups is finite.

2.24. What are all cyclic subgroups of the quaternion group Q8?

2.25. Let G Abelian group with order pq where gcd (p,q) = 1. Show that if G contains the elements a and b withorders respectively p and q then G is cyclic.

2.26. How many generators does a cyclic group of order n have?

2.27. Let p and q be two distinct prime numbers. How many generators has Zpq?

2.28. Let p be a prime number and r a positive integer. How many generators has Zpr ?

2.29. Let G be a finite Abelian group in which the number of solutions in G of the equation xn = e is at most n forevery positive integer n. Prove that G must be a cyclic group.

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2.4 Cosets and Lagrange’s Theorem

In this section we will study cosets and Lagrange’s theorem which is the starting point of our excursion tothe theory of groups.

2.4.1 Cosets

Let G be a group and H a subgroup of G. We call a left coset of H with representative g ∈ G the set

gH := gh : h ∈H.

Similarly, we define the right coset asHg := hg : h ∈H.

If left and right cosets are the same then we simply use the term coset. For the rest of this section we willtalk mostly about right cosets, but all the properties hold in the case of left cosets as well.

Let H ≤ G. Define a relation on G as follows

a ∼ b if and only if a and b belong to the same right coset.

Lemma 2.5. i) Show that ∼ is an equivalence relation.ii) Prove that a ∼ b⇔ ab−1

∈H.

Proof. We leave i) as an exercise for the reader.ii) Assume that a ∼ b. Then, Ha = Hb. Since H ≤ G, then e ∈H. Thus, there exists s ∈H such that ea = sb.

Therefore, s = ab−1∈H.

Let s = ab−1∈H. Then, a = s−1b ∈ Sb. Thus, a ∼ b.

Lemma 2.6. Let H ≤ G. Then, Ha = Hb if and only if ab−1∈H.

Proof. If Ha = Hb, then there is h ∈H such that 1 · a = h ·b. Hence, ab−1∈H.

Now let’s prove that if ab−1 = h ∈H then Ha = Hb. We must prove that Ha ⊆Hb and Hb ⊆Ha.a) Take x ∈ Ha. Then, x = h1a for some h1 ∈ H. Since ab−1

∈ H, then denote by h2 := ab−1∈ H. Then,

a = h2b. Thus,x = h1a = h1h2b = (h1h2)b ∈Hb

since h1h2 ∈H. Hence, x ∈Hb.b) Similarly we prove that Hb ⊆Ha. Thus, Ha = Hb.

Lemma 2.7. Let H ≤ G and a,b ∈ G. Then, every two right (left) cosets of H in G are equal or their intersection isempty. In other words,

Ha = Hb or Ha∩Hb = ∅.

Proof. Suppose that Ha∩Hb , ∅. Let x ∈ Ha∩Hb. Then for some h1,h2 ∈ H we have that x = h1a = h2b.Hence, a = h−1

1 h2b and ab−1 = h−11 h2 ∈H. Thus, Ha = Hb.

Corollary 2.6. Let H a subgroup of the group G. Left (resp. right) cosets of H in G partition G.

The following facts are useful when working with cosets. We leave the proof as exercise.

Lemma 2.8. Let H be a subgroup of the group G and assume that g1, g2 ∈G. The following statements are equivalent.

1. g1H = g2H

2. Hg−11 = Hg−1

2

3. g1H ⊆ g2H

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4. g2 ∈ g1H

5. g−11 g2 ∈H.

Proof. Exercise.

Theorem 2.11. Let H be a subgroup of the group G. The number of left cosets of H in G is equal with the number ofright cosets of H in G.

Proof. Let’s denote with LH and RH respectively the set of left and right cosets. If we can define a bijectivefunction φ :LH→RH, then the theorem is proved.

If gH ∈ LH, let φ(gH) = Hg−1. From Lemma 2.8, the map φ is well defined. Thus, if g1H = g2H, thenHg−1

1 = Hg−12 . To prove that φ is injective assume that

Hg−11 = φ(g1H) = φ(g2H) = Hg−1

2 .

Again from Lemma 2.8, g1H = g2H. The function φ is surjective since φ(g−1H) = Hg. Let G be a group and H a subgroup of G. Define the index of H in G to be the number of left cosets of H

in G. The index of H in G we will denote with [G : H].

Example 2.14. Let G =Z6 and H = 0,3. Then, [G : H] = 3.

Example 2.15. Assume that G = S3, H = (1), (123), (132) and K = (1), (12). Then, [G : H] = 2 and [G : K] = 3.

If H is a subgroup of the group G, then left cosets are not always the same with right cosets. Thus, notalways gH = Hg for every g ∈ G. A subgroup H of the group G is normal in G, denoted by H C G, if

gH = Hg, for every g ∈ G.

Thus, a normal subgroup of the group G is the subgroup in which left cosets and right cosets are the same.

2.4.2 Lagrange’s Theorem

Next we will prove a theorem which even though elementary it is one of the most important theorem inthe theory of groups. First we prove the auxiliary lemma.

Lemma 2.9. Let H be a subgroup of G. Then, |H| = |gH| for every g ∈ G.

Proof. Let g ∈ G be fixed. Define a function

φ : H→ gH,

such that φ(h) = gh. If we show that the function φ is bijective then the number of elements in H is the sameas the number of elements in gH.

First we have to show that φ is a well-defined function. Clearly, for every h ∈H there is φ(h) = gh ∈ gH.If h1 = h2 then gh1 = gh2. Hence, φ(h1) = φ(h2).

To show injectivity we assume that φ(h1) = φ(h2) for the elements h1,h2 ∈ H. Then, gh1 = gh2, whichimplies that h1 = h2. Hence, φ is injective.

To prove that φ is surjective let gh ∈ gH for some h ∈H. Then, φ(h) = gh.

Theorem 2.12 (Lagrange). Let G be a finite group and let H ≤ G. Then, [G : H] = |G|/|H|. In particular, the orderof H must divide the order of G.

Proof. The group G is partitioned in [G : H] distinct cosets. Each coset has H elements, therefore |G| = [G :H]|H|.

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Corollary 2.7. Assume that G is finite group and g ∈ G. Then, the order of g divides the order of G.

Corollary 2.8. Let |G| = p, where p is a prime number. Then, the group G is cyclic and there is an element g ∈ G,such that g , e, is a generator.

Proof. Let g be from G such that g , e. Then, from Corollary 2.7, the order of g must divide the order of G.Since |〈g〉| > 1, the order of g is p. Thus, g generates G.

Corollary 2.8 says that groups of prime order p are cyclic and algebraically similar to Zp.

Corollary 2.9. Let H and K subgroups of a finite group G such that K ≤H ≤ G. Then,

[G : K] = [G : H][H : K].

Proof. Notice that

[G : K] =|G||K|

=|G||H|·|H||K|

= [G : H][H : K].

The converse of the Lagrange’s theorem is not true: namely, if G is a finite group and n divides |G|, then

G does not necessarily have a subgroup of order n. However, under certain conditions the converse of theLagrange’s theorem hold.

Proposition 2.6 (Cauchy). If G is finite Abelian group and p is prime such that p | |G|, then G contains an elementwith order p.

Proof. We will prove Cauchy’s theorem by induction on |G|. Let n = |G|. Since p | n then p < n. The basecase is n = p. When |G| = p , any non-identity element of G has order p because p is prime. Now supposen > p, p /n, and the theorem is true for all groups with size less than n and divisible by p. Let G be a groupof size n.

Since p /n and n > p, |G| is not prime. Therefore G has a proper non-trivial subgroup, say H. Since G isabelian, G/H is a group. Since |H||(G/H)| = |G| = n, the prime p divides either |H| or |(G/H)| (we don’t knowwhich). Therefore, by induction, H or G/H has an element with order p. If H does, then so does G. If G/Hhas an element with order p, say g, then what can we say about the order of g (in G)? Let m be the order ofg. Then gm = e ∈ G =⇒ gm = e ∈ G/H =⇒ p /m. Thus, g has order divisible by p, so gm/p is an element of Gwith order p.

In the next chapters we will prove the following results.

Theorem 2.13. If G is an Abelian group and n divides |G|, then there is a subgroup H ≤ G such that |H| = n.

Proof. See Remark 6.4. Next we study a useful result. Let H and K be subgroups of a group G. By HK we denote the set

HK := hk |h ∈H,k ∈ K

Notice that HK is not necessarily a group.

Lemma 2.10. Let H and K finite subgroups of the group G. Then,

|HK| =|H| · |K||H∩K|

.

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Proof. Recall that thatHK = hk : h ∈H,k ∈ K.

Obviously, |HK| ≤ |H| · |K| since an element in HK can be written as product of different elements in H andK it is possible that h1k1 = h2k2 for h1,h2 ∈H and k1,k2 ∈ K. In this case let

a = (h1)−1h2 = k1(k2)−1.

Notice that a ∈H∩K, since (h1)−1h2 is in H and k2(k1)−1 is in K. Thus,

h2 = h1a−1

k2 = ak1.

Thus, let h = h1b−1 and k = bk1 for b ∈ H∩K. Then, hk = h1k1, where h ∈ H and k ∈ K. Thus, an elementhk ∈ HK can be written in the form hiki for hi ∈ H and ki ∈ K, as many times as elements we have in H∩K.Thus, |H∩K| times. Thus, |HK| = (|H| · |K|)/|H∩K|.

Proposition 2.7. Let H and K be subgroups of a group G. HK is a subgroup of G if and only if HK = KH.

Proof. Assume first that HK = KH and let a,b ∈ HK. We prove that ab−1∈ HK so HK is a subgroup by the

subgroup criterion. Let a = h1k1 and b = h2k2, for some h1,h2 ∈ H and k1,k2 ∈ K. Then, b−1 = k−12 h−1

2 . Soab−1 = h1k1k−1

2 h−12 .

Let k3 = k1k−12 ∈ K and h3 = h−1

2 . Thus ab−1 = h1k3h3. Since HK = KH, then k3h3 = h4k4, for some h4 ∈ Hand k4 ∈ K. Thus, ab−1 = h1h4k4 and since h1h4 ∈H and k4 ∈ K, we obtain ab−1

∈HK, as desired.Conversely, assume that HK is a subgroup of G. Since H ≤ HK and K ≤ HK by the closure property of

subgroups, KH ⊆HK. To show the reverse containment let hk ∈HK. Since HK is assumed to be a subgroup,write hk = a−1, for some a ∈HK. If a = h1k1 then

hk = (h1k1)−1 = k−11 h−1

1 ∈ KH.

This completes the proof.

Exercises:

2.30. Assume that G is a finite group that has an element g with order 5 and an element h with order 7. Why do wehave that |G| ≥ 35?

2.31. Prove or disprove the following: Every subgroup of integers has finite index.

2.32. Describe left cosets of SL2(R) in GL2(R). What is the index of SL2(R) in GL2(R)?

2.33. Show that the group of integers has infinite index in the additive group of rational numbers.

2.34. Show that the additive group of real numbers has infinite index in the additive group of complex numbers.

2.35. If ghg−1∈H for every g ∈ G and h ∈H, prove that right cosets are identical with left cosets.

2.36. Let G be a group and g ∈ G such that gn = e. Show that the order of g divides n.

2.37. If |G| = 2n, prove that number of elements with order 2 is odd. Use this result to prove that G contains asubgroup with order 2.

2.38. Let H and K be subgroups of the group G. Prove that gH∩ gK is a coset of H∩K in G.

2.39. Let H and K be subgroups of the group G. Define a relation ∼ on G, where a ∼ b if there exist elements h ∈Hand k ∈ K such that hak = b. Show that this relation is an equivalence relation. Corresponding equivalence classesare called double cosets. Find double cosets of H = (1), (123), (132) in A4.

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2.40. Let G be a group and A,B subgroups of G. If x, y ∈ G define the relation ∼ as follows:

x ∼ y if y = axb, for some a ∈ A,b ∈ B.

Prove thata) The relation ∼ is an equivalence relation in G.b) The equivalence class of x is

[x] = AxB = axb | a ∈ A,b ∈ B.

The set AxB for x ∈ G is called a double coset of A and B in G.

2.41. Prove that if G is a finite group, then the number of elements in the double coset AxB is

|A| · |B||A∩xBx−1|

2.42. If G is a finite group and A is a subgroup of G such that all double cosets AxA have the same number of elements,show that

gAg−1 = A,

for all g ∈ G.

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Joseph-Louis Lagrange (1736-1813)

Joseph-Louis Lagrange, born Giuseppe Lodovico Lagrangia (25January 1736 – 10 April 1813), was an Italian and French Enlighten-ment Era mathematician and astronomer. He made significant con-tributions to the fields of analysis, number theory, and both classicaland celestial mechanics.In 1766, on the recommendation of Euler and d’Alembert, Lagrangesucceeded Euler as the director of mathematics at the PrussianAcademy of Sciences in Berlin, Prussia, where he stayed for overtwenty years, producing volumes of work and winning several prizesof the French Academy of Sciences. Lagrange’s treatise on analyticalmechanics, written in Berlin and first published in 1788, offered themost comprehensive treatment of classical mechanics since Newtonand formed a basis for the development of mathematical physics inthe nineteenth century.

In 1787, at age 51, he moved from Berlin to Paris and became a member of the French Academy.He remained in France until the end of his life. He was significantly involved in the decimalisation inRevolutionary France, became the first professor of analysis at the École Polytechnique upon its openingin 1794, was a founding member of the Bureau des Longitudes, and became Senator in 1799.

Lagrange was one of the creators of the calculus of variations, deriving the Euler?Lagrange equationsfor extrema of functionals. He also extended the method to take into account possible constraints, arrivingat the method of Lagrange multipliers. Lagrange invented the method of solving differential equationsknown as variation of parameters, applied differential calculus to the theory of probabilities and attainednotable work on the solution of equations. He proved that every natural number is a sum of four squares.His treatise Theorie des fonctions analytiques laid some of the foundations of group theory, anticipatingGalois. In calculus, Lagrange developed a novel approach to interpolation and Taylor series. He studiedthe three-body problem for the Earth, Sun and Moon (1764) and the movement of Jupiter?s satellites (1766),and in 1772 found the special-case solutions to this problem that yield what are now known as Lagrangianpoints. But above all, he is best known for his work on mechanics, where he has transformed Newtonianmechanics into a branch of analysis, Lagrangian mechanics as it is now called, and presented the so-calledmechanical "principles" as simple results of the variational calculus.

The greater number of his papers during this time were, however, contributed to the Prussian Academyof Sciences. Several of them deal with questions in algebra.

• His discussion of representations of integers by quadratic forms (1769) and by more general algebraicforms (1770).

• His tract on the Theory of Elimination, 1770.

• Lagrange’s theorem that the order of a subgroup H of a group G must divide the order of G.

• His papers of 1770 and 1771 on the general process for solving an algebraic equation of any degreevia the Lagrange resolvents. This method fails to give a general formula for solutions of an equationof degree five and higher, because the auxiliary equation involved has higher degree than the originalone. The significance of this method is that it exhibits the already known formulas for solvingequations of second, third, and fourth degrees as manifestations of a single principle, and wasfoundational in Galois theory. The complete solution of a binomial equation of any degree is alsotreated in these papers.

• In 1773, Lagrange considered a functional determinant of order 3, a special case of a Jacobian. He alsoproved the expression for the volume of a tetrahedron with one of the vertices at the origin as the onesixth of the absolute value of the determinant formed by the coordinates of the other three vertices.

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64

Chapter 3

Quotient Groups and Homomorphisms

The concept of the homomorphism is a fundamental concept in algebra. In this chapter we will study inmore detail homomorphisms, isomorphisms, and the isomorphism theorems.

3.1 Isomorphisms

Let f : G1→G2 be an homomorphism. We say that f is an isomorphism if f is bijective. Assume f : G1→G2is an isomorphism. Since f : G1→ G2 is bijective, then there exists its inverse f−1 : G2→ G1. The followinglemma shows that f−1 : G2→ G1 is a homomorphism as well.

Lemma 3.1. Let f := G→H be a bijective homomorphism of groups. Then, f−1 : H→ G is a homomorphism.

Proof. Let h1,h2 ∈H. We want to show that f−1(h1h2) = f−1· f−1(h2).

Since f is bijective, then exist g1, g2 ∈ G such that f (g1) = h1 and f (g2) = h2. Hence,

f−1 ( f (g1) · f (g2))

= f−1 ( f (g1g2))

= g1g2 = f−1(h1) · f−1(h2)

This completes the proof.

Theorem 3.1. Let φ : G→H a isomorphism between two groups. Then, the following statements are true.

1. φ−1 : H→ G is an isomorphism.

2. |G| = |H|.

3. If G is Abelian, then H is Abelian.

4. If G is cyclic, then H is cyclic.

5. If G has a subgroup with order n, then H has a subgroup with order n.

Proof. Statements (1) and (2) follow from the fact that φ is a bijection and from the above Lemma. We provestatement (3) and other parts are left as an exercise.

(3) Assume that h1 and h2 are elements of H. Since φ is surjective, there exist the elements g1, g2 ∈ Gsuch that φ(g1) = h1 and φ(g2) = h2. Thus,

h1h2 = φ(g1)φ(g2) = φ(g1g2) = φ(g2g1) = φ(g2)φ(g1) = h2h1.

The following characterizes infinite cyclic groups.

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Theorem 3.2. All cyclic groups with infinite order are isomorphic to Z.

Proof. Let G a cyclic group with infinite order. Assume that a is a generator of G. Define a map φ :Z→ Gsuch that φ : n 7→ an. Then,

φ(m + n) = am+n = aman = φ(m)φ(n).

To prove that φ is injective, assume that m and n are two elements in Z where m , n. Let m > n. We mustshow that am , an. Assume the contrary, so am = an. In this case am−n = e, where m−n > 0, which contradictsthe fact that a has infinite order. The map is surjective since every element in G can be written as an forsome integer n and φ(n) = an.

Theorem 3.3. If G is a cyclic group with order n, then G is isomorphic to Zn.

Proof. Let G be a cyclic group with order n generated from a. Define φ :Zn→ G such that φ : k 7→ ak where0 ≤ k < n. The proof that φ is an isomorphism is an exercise at the end of the chapter.

Corollary 3.1. If G is a group with order p, where p is a prime number, then G is isomorphic to Zp.

Proof. The proof follows directly from Corollary 2.8.

The main goal of the group theory is to classify all groups. As mentioned from the beginning, from thealgebraic point of view the set that holds the group is not interesting to us, instead the algebraic structureis important. As we will see next the isomorphism of groups plays this important role.

Let S be the class of all groups. Define a relation in S as follows

G1 ∼ G2 if and only if G1 is isomorphic to G2

Prove the following theorem.

Exercise 3.1. The isomorphism of groups defines an equivalence relation in the class of all groups.

An isomorphism f : G −→ G of the group G in itself is called an automorphism of G.

Exercise 3.2. Let be given the function f : (C,+) −→ (C,+) such that

f (a + bi) = a− bi.

This function is called the conjugation map. Prove that this function is an automorphism.

Let G be a group. Denote by Aut (G) the set of all automorphisms of G

Exercise 3.3. Prove that Aut (G) forms a group with composition of functions.

The group Aut (G) is called automorphism group of G. An inner automorphism of the group G,

ig : G→ G,

is given by the functionig(x) = gxg−1,

for every g ∈ G. The set of inner automorphisms is denoted by Inn(G). The automorphism ig is also calledthe conjugation by g.

Exercise 3.4. Prove that ig ∈ Aut (G). Prove that Inn(G) is a subgroup of Aut (G).

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We have seen examples of the conjugation in linear algebra; see [?lin-alg]. For example, let G = GL2(R)and C ∈ Gl2(R). The map,

iC(M) = CMC−1

is a conjugation map. Two matrices A and B such that iC(B) = A for some matrix C ∈GL2(R) we have calledthem similar matrices.

Exercises:

3.1. Prove that the exponential function f (x) = ex is an isomorphism of groups (R,+) and (R+, ·).

3.2. The group (Z,+) is isomorphic to the subgroup of (Q×, ·), which consists in all the elements of the form 2n.

3.3. Prove that Aut (G) is a subgroup of the group of permutations of G. Thus, Aut (G) ≤ SG.

3.4. Prove that A 7→ B−1AB is a automorphism of SL2(R) for every B in GL2(R).

3.2 Normal subgroups and factor groups

If H is a subgroup of the group G, then left cosets are not always the same with right cosets. Thus, notalways gH = Hg for every g ∈ G. Subgroups for which this property is true play an important role in thetheory of groups.

A subgroup H of the group G is normal in G, denoted by H C G, if

gH = Hg, for every g ∈ G.

Thus, a normal subgroup of the group G is the subgroup in which left cosets and right cosets are the same.Let G be an Abelian group. Every subgroup H of G is a normal subgroup. Since gh = hg for every g ∈ G

and h ∈H, we have that gH = Hg.The following theorem is fundamental to understand normal subgroups.

Theorem 3.4. Let G be a group and N a subgroup of G. Then, the following are equivalent.i) The subgroup N is normal in G.ii) For every g ∈ G, gNg−1

⊂N.iii) For every g ∈ G, gNg−1 = N.

Proof. (1)⇒ (2). Since N is normal in G we have that gN = Ng for every g ∈ G. Thus, for a given g ∈ G andn ∈N, there exists a n′ in N such that gn = n′g. Thus, gng−1 = n′ ∈N or gNg−1

⊂N.(2) ⇒ (3). Let g ∈ G. Since gNg−1

⊂ N, it is enough to prove that N ⊂ gNg−1. For n ∈ N we haveg−1ng = g−1n(g−1)−1

∈N. Thus, g−1ng = n′ for a n′ ∈N. Thus, n = gn′g−1 is in gNg−1.(3)⇒ (1). Assume that gNg−1 = N for every g ∈ G. Then, for every n ∈N there exists a n′ ∈N such that

gng−1 = n′. Thus, gn = n′g or gN ⊂Ng. Similarly it can be proven that Ng ⊂ gN.

A group G is called simple group if does not have proper normal subgroups.

Lemma 3.2. Let G be a finite group, p the smallest prime divisor of |G|, and N C G such that |N| = p. Show thatN ≤ Z(G).

Solution: Let |G| = n. For every x ∈ G, denote by

σx :N→N

g→ x−1gx

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be the conjugation by x map, i.e. x−1Nx→ N. We want to show that N ≤ Z(G) which is enough to showthat σx = id.

Let x ∈ G \N such that xm = 1. It implies that m ≥ p, where m doesn’t have any prime factor less than p.Thus (m,p−1) = 1, and a(p−1) + bm = 1, for some integers a,b ∈Z.

Then we have (σx)p−1 = id, and σxm = id which implies that (σx)m = id. Hence, σx = (σx)1 = σa(p−1)+bmx =

(σ(p−1)x )a

· (σmx )b = id. Thus, σx = id and we are done.

3.2.1 Factor groups

Let N be a normal subgroup in G. Denote by G/N the set of cosets of N in G. Define the following binaryoperation in G/N.

aN ·bN = (ab)N

Theorem 3.5. Let N be a normal subgroup of the group G. G/N with the operation above forms a group.

Proof. We must show that the operation is well-defined, hence independent from the choice of the repre-sentatives. Let aN = bN and cN = dN. We must prove that

(aN)(cN) = acN = bdN = (bN)(dN).

Let a = bn1 and c = dn2 for some n1,n2 ∈N. Thus,

acN = bn1dn2N = bn1dN = bn1Nd = bNd = bdN.

This operation is associative since

aN · (bN · cN) = aN · (bc)N = (a(bc))N = (ab)cN = (aN ·bN))cN

The other part of the theorem is simple, eN = N is the identity and g−1N is the inverse of gN. The group G/N is called the factor group of G and N.

Example 3.1. Consider the normal subgroup N = (1), (123), (132) in S3. Cosets of N in S3 are N and (12)N. Thefactor group S3/N has multiplication table as follows.

N (12)NN N (12)N

(12)N (12)N N

This group is isomorphic to Z2.

3.2.2 The natural projection map

We can use factor groups to study homomorphisms. We know that for every homomorphism the groupφ : G→H we get a normal subgroup of G, namely kerφ. The converse is also true. Every normal subgroupof the group G gives a group homomorphism.

Let H be a normal subgroup of G. The natural projection or canonical homomorphism is called thehomomorphism

φ : G→ G/H

such thatφ(g) = gH.

This is a homomorphism since:

φ(g1g2) = g1g2H = g1Hg2H = φ(g1)φ(g2).

The kernel of this homomorphism is H.

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Example 3.2. Prove or disprove: Is G is a finite group and H,K are normal subgroups of G with G/HG/K, thenHK.

Solution: The statement is not true. Consider G =Z4×Z2. Let H be the isomorphic copy of Z4 in G. Then, H C G

as a subgroup of index 2.Let K be the isomorphic copy of Z2 ×Z2 in G. Also, K C G. However, G/HZ2 and G/KZ2 since all the

groups of order 2 are isomorphic to Z2. Hence, G/KG/H even though H is not isomorphic to K.

Here is another example of a very famous group that we have seen before.

Example 3.3 (The modular group). We have shown in Chapter 1 that SLn(Z) is a group. Of course the identitymatrix I and −I commute with all the matrices of SLn(Z). Hence ±I is normal in SL2(Z).

The quotient group Γ = SL2(Z)/±I is called the modular group. It is probably one of the most important groupsin number theory and arithmetic geometry. We will briefly describe some of its properties and will revisit it again inSection 4.4.

Exercises:

3.5. Let be given H C G and K C G. Prove that H∩K C G.

3.6. Let be given H C G,K C G. Prove that HK C G.

3.7. If H is a subgroup of G and N is a normal subgroup of G, show that H∩N is normal subgroup of H.

3.8. Suppose H is the only subgroup of order |H| in the finite group G. Prove that H is a normal subgroup of G.

3.9. If H is a subgroup of G, let the normalizer of H be N(H) = g ∈ G|gHg−1 = H. Prove

a) N(H) is a subgroup of G.

b) H is normal in N(H).

c) If H is a normal subgroup of a subgroup K of G, then K ⊂N(H) (that is, N(H) is the largest subgroup of G inwhich H is normal).

d) H is normal in G if and only if N(H) = G

3.10. If p is a prime number, prove that any group G of order 2p must have a subgroup of order p, and that thissubgroup is normal in G.

3.11. Let G be a group in which for some integer n > 1, (ab)n = anbn for all a,b ∈ G. Show that

a) G(n) = xn| x ∈ G is a normal subgroup of G.

b) G(n−1) = xn−1| x ∈ G is a normal subgroup of G.

3.12. Let G be a se of all real 2×2 matrices(a b0 d

)where ad , 0, under matrix multiplication. Let N =

(1 b0 1

).

Prove that

a) N is a normal subgroup of G.

b) G/N is abelian.

3.13. Let be given the group G and Z(G) its center. Prove that Z(G) C G.

3.14. Prove that if G/Z(G) is cyclic, then G is Abelian.

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3.15. Denote with GL2(R) the group of 2×2 matrices with terms inR and nonzero determinant. Denote with SL2(R)the group of 2×2 matrices with terms in R and determinant 1. Prove that SL2(R) C GL2(R).

3.16. Let be given H and K normal subgroups of G such that H ∩ K = e. Prove that hk = kh for every h ∈ H andk ∈ K.

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3.3 Isomorphism theorems

The following theorems describe the relations between homomorphisms, normal subgroups, and factorgroups.

Theorem 3.6 (First Isomorphism Theorem). Let ϕ : G→H be a surjective homomorphism. Then,

G/ker(ϕ)H.

Proof. Let K := kerϕ. We know that K C G. As usual, let π : G 7→ G/K ne the natural projection. Define thefunction ψ as follows:

ψ : G/K −→HgK −→ ϕ(g)

First we prove that ψ is a function. Assume that aK = bK. Then,b−1a ∈ K. Since K is the kernel, then ϕ(b−1a) = eH. We know that ϕ is ahomomorphism so

Gϕ

//

π##

HOOψ

G/kerϕ

ϕ(b−1b) = ϕ(b)−1ϕ(a) = eH.

Hence, ϕ(a) = ϕ(b) and we have ψ(aK) = ψ(bK). Thus, ψ is a well defined map. Next we will prove that ψ isa homomorphism. We have that

ψ(aK ·bK) = ψ(abK) = ϕ(ab) = ϕ(a) ·ϕ(b) = ψ(aK) ·ψ(bK).

Hence, ψ is a homomorphism.Also we must prove that ψ is a injective function. Let aK ∈ ker(ψ). Then, ψ(aK) = ϕ(a) = eH. Thus, a ∈ K

and this implies that aK = eG/K. Therefore, we have that ker(ψ) = eG/K and the function ψ is injective.Finally we must prove that this function is surjective. For every ϕ(g) ∈H there is a g ∈ G and for every

g ∈ G there is a gK ∈ G/K. This completes the proof.

Remark 3.1. Notice that in the diagram above we have that

ϕ = ψπ.

In such case we say that the diagram is commutative. We will see many cases when such diagrams are very helpfulin proofs later on in this book.

The following is a useful corollary of the above theorem.

Corollary 3.2. Any cyclic group G is isomorphic to Z or Z/nZ.

Proof. Let G = 〈g〉. Define the function

φ :Z→ Gn 7→ gn.

This function is a surjective and clearly a homomorphism since

φ(m + n) = gm+n = gmgn = φ(m)φ(n).

If the order of G is infinite, then so is the order of g. Hence, kerφ = 0 and φ is injective. Thus, GZ. If|G| = n then for every m ∈ kerφ we have

φ(m) = gm = e.

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Hence, n |m. Thus, kerφ = mZ. From the First Isomorphism Theorem we have that

GZ/kerφ =Z/mZ

This completes the proof. From now on when we will talk of the cyclic group of order n we will mean a group isomorphic to Z/nZ.

Example 3.4. Are the groups Z and Q isomorphic under addition? The answer is obviously "no" since Z is cyclicand Q is not.

The next result is usually called the Second Isomorphism Theorem.

Theorem 3.7 (Second Isomorphism Theorem). Let H ≤ G and N C G. Then, HN is a subgroup of G, H∩N is anormal subgroup of H, and

H/H∩NHN/N.

Proof. First we prove that HN = hn : h ∈H,n ∈N is a subgroup of the group G. Assume that h1n1,h2n2 ∈HN.Since N is normal, (h2)−1n1h2 ∈N.

Thus,(h1n1)(h2n2) = h1h2((h2)−1n1h2)n2

is in HN. The inverse of hn ∈HN is in HN since

(hn)−1 = n−1h−1 = h−1(hn−1h−1).

Now let’s prove that H∩N is normal in H. We take h ∈ H andn ∈H∩N. Then, h−1nh ∈H since for every element is in H. Also,h−1nh ∈N since N is normal in G. Thus, h−1nh ∈H∩N.

Now let define the function φ from H in HN/N suchthat h 7→ hN. The function φ is injective since for every cosethnN = hN is the image of h in H.

G

HN

H N

H∩N

1The function φ is a homomorphism because

φ(hh′) = hh′N = hNh′N = φ(h)φ(h′).

From the First Isomorphism Theorem, the image of φ is isomorphic me H/kerφ. Thus,

HN/N = φ(H)H/kerφ.

Since kerφ = h ∈H : h ∈N = H∩N, we have HN/N = φ(H)H/H∩N. This completes the proof.

As we have seen above, determining lattices of subgroups of a given group is an important tool forstudying groups (this will become more evident in the coming sections). The following theorem, sometimescalled the Lattice Theorem, is an important tool in determining such lattices.

Theorem 3.8 (Third Isomorphism Theorem). Let K ≤H C G such that K C G. Then,

H/K C G/K

and(G/K)/(H/K)G/H.

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Proof. First we must check that H/K C G/K. We leave this as an exercise for the reader.

Define the map

f : (G/K)→ G/HgK→ gH

First we prove that f is a function. If g1K = g2K, then g1g−12 ∈ K and

g1g−12 ∈H. Hence, g1H = g2H.Now let’s prove that f is a homomorphism, so let’s show that

f (g1K · g2K) = f (g1K) · f (g2K).

We see that

f (g1K · g2K) = f (g1g2 K)= g1g2H = (g1H) (g2K) = f (g1K) f (g2K).

G G/K

H H/K

K eG/K

1

Now we prove that the function f is surjective. We have that for every Hg ∈ G/H there is a Kg ∈ G/Ksuch that f (Kg) = Hg. It can easily be shown that the function is injective.

Example 3.5. From Third Isomorphism Theorem,

Z/mZ (Z/mnZ)/(mZ/mnZ).

Since |Z/mnZ| = mn and |Z/mZ| = m, we have |mZ/mnZ| = n.

Next we see another important theorem which is called Correspondence Theorem or sometimes theFourth Isomorphism Theorem.

Theorem 3.9 (Correspondence Theorem). Let N be a normal subgroup of the group G. Then,

H 7→H/N =: H

is a one to one correspondence between the set of subgroups H of G which contain N and the set of subgroups of G/N.Moreover, normal subgroups of G correspond to normal subgroups of G/N. This correspondence has the followingproperties:

i) H1 ≤H2 if and only if H1 ≤H2

ii) if H1 ≤H2 then [H2 : H1] = [H2 : H1]

iii) 〈H1,H2〉 = 〈H1,H2〉

iv) H C G if and only if H C G

Proof. Let H a subgroup of G which contains N. Since N is normal in G, then it is normal in H. Hence, thereexists the factor group H/N. Let aN and bN be elements of H/N for a,b ∈H. Then,

(aN)(b−1N) = ab−1N ∈H/N.

Thus, H/N is subgroup of G/N by the second subgroup test.

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Let S be a subgroup of G/N. This subgroup is a set of cosetsof N. If H = g ∈ G : gN ∈ S, then for h1,h2 ∈H, we have that

(h1N)(h2N) = hh′N ∈ S

and h−11 N ∈ S. Thus, H must be a subgroup of G. It is clear that

H contains N. Hence, S = H/N. Thus, the function H 7→ H/H issurjective.

Assume that H1 and H2 are subgroups of the group G thatcontain N such that H1/N = H2/N. If h1 ∈ H1, then h1N ∈ H1/N.Thus, h1N = h2N ⊂ H2 for a h2 in H2. However, since N is con-tained in H2, we know that h1 ∈H2 or H1 ⊂H2. Similarly, H2 ⊂H1.Since H1 = H2, the function H 7→H/H is bijective.

G G = G/N

H H = H/N

N eG/N

1

Assume that H is normal in G and N is a subgroup of H. Then, it is easy to prove that the functionG/N→ G/H defined by gN 7→ gH is a homomorphism. The kernel of this homomorphism is H/N, whichproves that H/N is normal in G/N.

Conversely, assume that H/N is normal in G/N. The homomorphism given by

G→ G/N→G/NH/N

has kernel H. Thus, H must be normal in G.

Next we see an illustration of the Correspondence Theorem in construction the lattice of the dihedralgroup of 8 elements.

Example 3.6. Construct the lattice of the group D4.

We know that D4 is given by

D4 =σ,τ |σ4 = 1,τ2 = 1,τστ = σ−1

,

see Theorem 1.4 for details.So the two obvious subgroups are the cyclic ones

〈σ〉 and 〈τ〉with order 4 and 2 respectively. Since τ hasoder 2, then 〈τ〉 has no proper subgroups. However, σhas order 4 and therefore 〈σ〉 has a subgroup of ordertwo, namely 〈σ2

〉.The elements σ2τ, στ, σ3τ all have order 2 and

generate subgroups of order 2. 〈e〉

〈τ〉 〈σ2τ〉 〈σ2〉 〈στ〉 〈σ3τ〉

〈σ〉〈σ2,τ〉 〈σ2,στ〉

D4

"""

bbbXXX

XXXXX

"""

bbb

ZZZ

HHHH

H

The elements σ2 and τ, both have order 2 and together they generate a Klein 4-group. The same can besaid for σ2 and στ. This completes the lattice on the right.

Exercises:

3.17. Prove that for every n, Sn is isomorphic to a subgroup of An+2.

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3.4 Cauchy’s theorem

Theorem 3.10. If p is a prime and p divides the order of a group G, then G contains an element of order p.

Proof.

Remark 3.2. The proof above is due to J. McKay; see [?mckay] for details.

Corollary 3.3. Let G be a group of order |G| = pq, where p and q are primes and p > q. If a ∈G is an element of orderp then 〈a〉 C G.

Exercises:

3.18. Prove that a group of order 35 is cyclic.

3.19. Let G be a group of order |G| = pn m, such that p is prime and -m. Prove that if G is a subgroup P of order pn,then P C G.

3.20. Prove that a group of order 99 has no proper normal subgroups.

3.21. Prove that a group of order 42 has a normal subgroup of order 21.

3.22. Prove that any two non Abelian groups of order 21 are isomorphic to each other.

3.23. Prove that any group of order 99 is Abelian

3.24. Let p and q be primes such that q | p−1. Prove that there exists a non Abelian group of order pq.

3.25. Let p and q be primes such that q | p−1. Prove that any two non Abelian groups of order pq are isomorphic.

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3.5 Conjugacy classes

Let G be a group. Define the relation in G as follows:

y is conjugate of x in G, if and only if y = gxg−1 for a g ∈ G,

Exercise 3.5. Show that this is an equivalence relation.

The equivalence class of a ∈G with the above equivalence is called conjugacy class of a and denoted byaG.

The conjugacy class of aG is the set of all elements of G conjugated to a. A subgroup is normal if and onlyif is a union of conjugacy classes. If a and b are conjugated in G, so b = gag−1, then there is a isomorphismρg : G→ G, which is called the conjugation by g where

a→ gag−1.

Exercise 3.6. Let G be a group and g a fixed element in G. Define ρg : G→ G, such that a→ gag−1. Prove that ρgis an isomorphism.

Since isomorphisms preserve orders of elements then all elements in the same conjugacy class have thesame order. For two elements x, y ∈ G, the elements xy and yx have the same order, since xy is conjugatedto y · (xy) · y−1 = yx.

Corollary 3.4. If a ∈ G then the number of conjugates of a is equal to the index of its centralizer,

|aG| = [G : CentG(a)]

and this number is a divisor of |G| when G is finite.

Proof. Let H = CG(a) and denote with λ the set of all left cosets of H in G. We must show that the numberof elements of aG is equal to the number of elements of λ. Define the function f such that

f : aG→ λ

gag−1→ gH

First let’s prove that f is function. Let’s assume that gag−1 = hah−1 then we have h−1gag−1h = a. Hence wehave (h−1g)a(g−1h) = a and from this equality we can write (h−1g)a(h−1g)−1 = a so h−1g ∈H and hH = gH.

Prove that f is injective. If we have g1H = g2H then g−11 g2 ∈H and we have that g−1

1 g2a = ag−11 g2 because

a is in the centralizer. Hence we have g2a = g1ag−11 g2 and therefore g2ag−1

2 = g1ag−11 . Also the function f is

surjective because for every gH ∈ λ we have f (gHg−1) = gH.

Definition 3.1. Let H ≤ G then the conjugate of H in G from g is

gHg−1 = ghg−1|h ∈H

and denoted by Hg.

Lemma 3.3. Let H ≤ G. Then, we have that gHg−1≤ G for every g ∈ G.

Proof. We prove that H ≤ G, so for every a,b ∈ gHg−1 we have ab−1∈ gHg−1. Let

a = gh1g−1∈ gHg−1

b = gh2g−1∈ gHg−1

then ab−1 = gh1g−1gh2g−1 and therefore gh1h2g−1∈ gHg−1.

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Definition 3.2. If H ≤ G then the normalizer of H in G, which denoted by NG(H), is the set

NG(H) := g ∈ G | gHg−1 = H.

A subgroup H < G is called normal in G if gH = Hg, for all g ∈ G.

Lemma 3.4. Let G be a group and H ≤ G. Then, we have that:a) NG(H) ≤ Gb) H CNG(H)c) NG(H) is the largest subgroup of G in which H is normal.

Proof. a)Prove that NG(H) ≤ G, using the first subgroup test. The identity e ∈NG(H) because eHe−1 = H, soNG(H) is a nonempty set. Second, if g1, g2 ∈NG(H), then g2Hg−1

2 = H and g1Hg−11 = H. Thus,

g1g−12 Hg2g−1

1 = g1Hg−11 = H.

b) and c) are consequences of the definition of NG(H).

Theorem 3.11. Let H ≤ G. Then, the number of conjugates of H in G is the index of the normalizer [G : NG(H)].Moreover,

aHa−1 = bHb−1⇔ b−1a ∈NG(H)

Proof. Let’s denoteH the set of all conjugates of H in G, N = NG(H), and withLN the set of all left cosets ofN in G. Define the map f such that

ϕ : H −→LN

aHa−1−→ aN

First we prove that the map ϕ is a function. Let aHa−1 = bHb−1 then b−1aHa−1b = H implies b−1a ∈ N andtherefore aN = bN.

Let’s prove first that the function ϕ is injective. Let aN = bN then b−1a ∈N implies b−1aHa−1b = H. Thus,aHa−1 = bHb−1.

Finally, we prove that the function ϕ is surjective. Let aN ∈ λN then there is a aHa−1∈ H such that

f (aHa−1) = aN. Hence, the function ϕ is bijective.

For a given group G, we define the commutator of G as

G′ = 〈xyx−1y−1| x, y ∈ G. (3.1)

It is easy to check that G′ is a subgroup of G which we will refer as the commutator subgroup of G.

Exercise 3.7. Let G be a group. Find an example that the setaba−1b−1

|a,b ∈ G

is not necessarily a group.

We will use commutators extensively in later chapters when we study solvable groups.

Exercises:

3.26. Prove that

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a) Prove that G′ is normal in G.

b) Prove that G/G′ is abelian.

c) If G/N is abelian, prove that N ⊃ G′.

d) Prove that if H is a subgroup of G and H ⊃ G′, then H is normal in G.

3.5.1 Conjugacy in Sn

In this section we see what it means for two permutations to be conjugate in Sn.

Theorem 3.12. If two cycles τ and µ in Sn have the same length then there exists σ ∈ Sn such that µ = στσ−1.

Proof. Assume thatτ = (a1,a2, . . . ,ak) and µ = (b1,b2, . . . ,bk).

Define σ to be the permutation

σ(a1) = b1

σ(a2) = b2

...

σ(ak) = bk.

Then, µ = στσ−1.Conversely, assume that τ = (a1,a2, . . . ,ak) is a k -cycle and σ ∈ Sn. If σ(ai) = b and σ(a(i mod k)+1) = b′, then

µ(b) = b′. Thus,µ = (σ(a1),σ(a2), . . . ,σ(ak)).

Since σ is injective and surjective, µ is a cycle that has the same length with τ.

Let α ∈ Sn such that α is a disjoint product of si, n1 -cycles for i = 1, . . . ,r. We say that α is of type

ns11 · · ·n

srr .

Exercise 3.8. Let be given C, the conjugacy class in Sn of elements α. Find a formula for the number of elements inC.

Theorem 3.13. Any two permutations in Sn are conjugate if and only if they are of the same type.

Proof. Assume that σ has cycle type k1,k2 . . .kl. Then, σ can be written as a product of disjoint cycles:σ = α1α2 . . .αl where αi is a ki-cycle. Let τ ∈ Sn. Then,

τστ−1 = τα1α2 . . .αlτ−1 = (τα1τ

−1)(τα2τ−1) . . . (ταlτ

−1).

For each i such that 1 ≤ i ≤ l, ταiτ−1 is also a ki-cycle; see Them. Theorem 3.12. For any i, j ∈ 1,2, . . . l suchthat i , j we know that αi and α j are disjoint and so ταiτ−1 and τα jτ−1 must be disjoint since τ is one to onefunction. Thus, τστ−1 written as a product of disjoint cycles and is of type k1,k2 . . .kl.

Conversely, let σ,µ ∈ Sn both be cycle type k1,k2 . . . ,kl. Let σ and µ be written as products of disjointcycles as

σ = α1α2 . . .αl and µ = β1β2 . . .βl,

where αi and βi are ki-cycle for 1 ≤ i ≤ s. For each i we write

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αi = (ai1ai2 . . .aiki ) and βi = (bi1bi2 . . .biki )

Now define τ by τ(ai j) = bi j for every i, j such that 1 ≤ i ≤ l and 1 ≤ j ≤ ki. Now we know that ταiτ−1 = βiand so we have

τστ−1 = (τα1τ−1)(τα2τ−1) . . . (ταlτ−1) = β1β2 . . .βl = µ.

Any two elements of Sn with the same cycle type are in the same conjugacy class.

Exercise 3.9. Let σ ∈ Sn such that m1, . . . ,ms are the distinct integers which appear in the cycle type of σ (including1-cycles). For each i = 1, . . . ,s assume that σ has ki cycles of length mi. Prove that the number of conjugates of σ is

n!

(k1! ·mk11 ) · · · (k1! ·mks

s )

Exercise 3.10. Let α ∈ Sn. Show that α−1 is conjugate to α.

Exercises:

3.27. Let H be a subgroup of the group G. Prove or disprove that normalizer of H is normal in G.

3.28. Let H be a subgroup of a finite group G. Prove that gN(H)g−1 = N(gHg−1) for every g ∈ G.

3.29. Let be given p a prime number and let C a cyclic subgroup with order p in Sp. Determine the order of NSp (C).

3.6 Cayley’s theorem

Theorem 3.14. Every group G can be embedded to subgroup of SG. In particular if | G |= n then G → Sn

Proof. Let a ∈ G. Define the function

λa : G→ Gx→ ax

We prove that this function λa is a bijection and therefore λa ∈ SG. Let’s define the function Φ : G→ SG suchthat

Φ(a) = λa

and prove that this function is injective and homomorphism. Let’s prove first thatλa is an injective function.IfLa =Lb then ax = bx so we have that a = b for every x ∈G. It is easy to prove that the function is surjective.Finally, to prove that this function is homomorphism we must prove that Φ(ab) = Φ(a)Φ(b) soLab(x) =λaλb.This equality is true because lambdaab(x) = (ab)x and also

(λa λb)(x) = λa(Lb(x)) = λa(bx) = a(bx) = (ab)x.

This completes the proof. The embedding φ : G → SG is called the (left) regular representation of G.

Corollary 3.5. Let k be a field and G a finite group of order n. Then G can be embedded in GLn(k).

Proof. From Cayley’s theorem Theorem 3.14 G → SnP(n,k) → GLn(k).

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Theorem 3.15 (Core Theorem). If H ≤ G and [G : H] = n then there is a homomorphism φ : G→ Sn such thatkerφ ≤H.

Proof. Let a ∈ G and denote with λ the set of all cosets left of H in G. Then, define the function

ρa : λ→ λ

gH→ agH

We prove that ρa is a bijective function and therefore ρa ∈ Sn. Define the function φ such that

φ : G→ SαSn

g→ ρg

The function φ is a homomorphism because ∀g ∈ G we have:

φ(g1g2) = ρg1 g2 (gH) = g1g2gH = g1(ρg2 (gH)) = ρg1 (ρg2 (gH)) = ρg1 ρg2 = φ(g1)φ(g2)

Now let’s prove that kerφ ≤ H. Let a ∈ kerφ then φ(a) = 1 and as a consequence ρa = e. Thus, for everyg ∈ G we have that

ρa(gH) = agH = gH so aH = H

from which is clear that a ∈H.

Corollary 3.6. Let G be a simple group which contains a subgroup H such that [G : H] = n. Then,

G → Sn

Proof. From the above theorem, we have φ : G→ Sn such that kerφ ≤ G < G. But kernels are normal and Gis simple, which implies that kerphi = 1G. Thus, φ is an embedding.

Theorem 3.16 (Index Theorem). Let G be a finite group and H ≤G such that [G : H] = n. If |G| - n!, then G is notsimple group.

Proof. From Theorem 3.15 we know that there is a function θ : G→ SX such that kerθ ≤H. Since |G| - n! thefunction θ is not injective so kerθ , e. Hence there is a K = kerθ ≤ H ≤ G and also K C G which impliesthat the group G is not simple.

Theorem 3.17. Let H ≤G and X be the set of conjugates of H in G. There exists a homomorphism ψ : G→ SX suchthat kerψ ≤NG(H).

Proof. For every element a ∈ G defineψa : X→ X

ghg−1→ agHg−1a−1

It can be easily shown that this function is bijective. Let’s define the function ψ such that

ψ : G→ SX

a→ ψa

This function is a homomorphism. Now let’s prove that kerψ ≤ NG(H). Let an element a ∈ kerψ then forevery g ∈ G we have agHg−1a−1 = gHg−1 and for g = e we have aHa−1 = H so a ∈NG(H).

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Example 3.7. Let G be a finite group, where p is the smallest prime divisor of |G| and N a subgroup of G such that[G : N] = p. Show that N C G.

Solution: Let |G| = pα1pα22 · · ·p

αnn , where p < p2 < p3 < · · ·pn. We want to show that N C G, where N ≤ G and

[G : N] = p. There exists a homomorphism σ such that σ : G→ Sp. Then, σ(G) ≤ Sp. Therefore |σ(G)| | p!. But|σ(G)| =

∣∣∣G/kerσ∣∣∣ implies that |σ(G)| | |G|

Thus, |σ(G)| | p!, and |σ(G)| | |G| = pα1pα22 · · ·p

αnn which means that |σ(G)| = 1 or p. So | G/kerσ |= 1 or p.

If∣∣∣G/kerσ

∣∣∣ = 1 then kerσ = G which can’t happen because kerσ ⊆N and N is a proper subgroup of G. Hence,∣∣∣G/kerσ∣∣∣ = p which implies that |kerσ| = |N|. Thus, kerσ = N. Since kernels are normal, then N C G.

Exercises:

3.30. Let be given G a finite group with order |G| = 2n·m, where m is odd. Prove that if G contains an element with

order 2n, then the set of elements with odd order is a normal subgroup of G.

3.31. Find all subgroups of D4. Which ones are normal? Which are factor groups?

3.32. Find all subgroups of the quaternion group Q8. Which are normal? Which are factor groups of Q8?

3.33. Let T the group of 2×2 upper triangular matrices with elements from Z, i.e., matrices of the form(a b0 c

),

where a, b, c ∈Z and ac , 0. Let U be the set which contains matrices of the form:(1 x0 1

),

where x ∈Z.a) Prove that U is a subgroup of T.b) Prove that U is Abelian.c) Prove that U C T.d) Prove that T/U is Abelian.e) Is T normal in GL2(Z)?

3.34. If φ : G→H is a homomorphism and G is Abelian, prove that φ(G) is also Abelian.

3.35. If φ : G→H is a homomorphism and G is cyclic, prove that φ(G) is also cyclic.

3.36. If G is Abelian, prove that G/H must be also Abelian.

3.37. Let G be a finite group, N a normal subgroup of G and

φ : G→ G/N

the natural projection. If H is a subgroup of G/N, prove that φ−1(H) is subgroup in G with order |H| · |N|.

3.38. Let G1 and G2 groups, H1 C G1, H2 C G2, and

φ : G1→ G2

a group homomorphism. Then, φ induces a natural homomorphism

φ : (G1/H1)→ (G2/H2)

if φ(H1) ⊆H2. Define φ and prove that it is a well defined map and a homomorphism.

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3.39. Find an automorphism of a group G that is not an inner automorphism.

3.40. Let G be group and ig an inner automorphism of G. Define the function

G→ Aut (G)

such thatg 7→ ig.

Prove that this function is a homomorphism with image Inn(G) and kernel Z(G). Use this result to generalize that

G/Z(G) Inn(G).

3.41. Determine Aut (S3) and Inn(S3).

3.42. Determine Aut (D4) and Inn(D4).

3.43. Find all automorphisms of Z8. Prove that Aut (Z8)U (8).

3.44. For k ∈Zn, define the function φk :Zn→Zn such that a 7→ ka. Prove that φk is a homomorphism.

3.45. Prove that the group of n -th roots of unity is isomorphic to Zn.

3.46. Prove that the set of all matrices of the form:

B =

(±1 n0 1

),

where n ∈Zn is a group isomorphic to Dn.

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Chapter 4

Groups acting on sets

Groups acting on sets is a fundamental concept in mathematics which is used in many areas such asgeometry, topology, etc. In this chapter, we will give the basics of group actions and some applications.

In the last two sections of the chapter we will give two exciting applications of the group action that ofthe modular group acting on the complex upper-half plane and the action of the general linear group onthe space of binary forms. Both these actions have played a fundamental role in the history of mathematicsand are important in higher mathematics.

4.1 Groups acting on sets

Let X be a set and G a group. We say that the group G acts over X if there is a function

f : C×X→ X(g,x)→ gx

which satisfies the following properties:

a) ex = x for every x ∈ X

b) g(hx) = (gh)x, for every g,h ∈ G.

The set X is called a G-set. Notice that gx is just a symbol and has nothing to do with multiplication in G.After all, x < G.

If G acts on the set X and x, y ∈ X, then we say that x is G-equivalent with y if there exists a g ∈ G suchthat gx = y. If two elements are G-equivalent, we write x ∼G y or x ∼ y.

Proposition 4.1. Let X be a G -set. Then, G-equivalent is an equivalence relation in X.

Proof. The relation ∼ is symmetric because ex = x. To show that is well defined suppose that x ∼ y forx, y ∈X, then there is a g such that gx = y. In this case g−1y = x. Thus, y ∼ x. To prove the transitive propertyassume that x ∼ y and y ∼ z. Then, must exist the elements g and h such that gx = y and hy = z. Thus,z = hy = (hg)x and x is equivalent with z.

The kernel of the action is the set of elements

ker( f ) = g ∈ G | gx = x, for all x ∈ X

For x ∈ X, the stabilizer of x ∈ G isStabG(x) = g ∈ G | gx = x

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sometimes denoted by Gx.

Exercise 4.1. The stabilizer StabG(x) is a subgroup of G.

Lemma 4.1. Let X be a G -set and assume that x ∼ y. Then, the stabilizer StabG(x) is isomorphic to the stabilizerStabG(y).

Proof. Since x ∼ y, there is a g ∈ G such that y = gx. Let a ∈ StabG(x). Since

gag−1· y = ga · g−1y = ga ·x = g ·x = y,

we define the function φ : StabG(x)→ StabG(y) such that φ(a) = gag−1. The function φ is a homomorphismbecause

φ(ab) = gabg−1 = gag−1gbg−1 = φ(a)φ(a).

Assume that φ(a) = φ(b). Then, gag−1 = gbg−1 or a = b. Thus, the function is injective. To prove that φ issurjective, let b in StabG(y). Then, g−1bg is in StabG(x) since

g−1bg ·x = g−1b · gx = g−1b · y = g−1· y = x

and φ(g−1bg) = b. The action of G on X is called faithful if its kernel is the identity. The orbit of x ∈X (or G-orbit) is the set

Orb(x) = gx ∈ X | g ∈ G

An action is called transitive if for every x1,x2 ∈ X, there is g ∈ G such that x2 = gx1.

Lemma 4.2. Let G act on a set X and x ∈ X. Then, the cardinality of the orbit Orb(x) is the index of the stabilizer

|Orb(x)| = [G : StabG(x)]

Proof. Fix x ∈ X. Let L be the family of all left cosets of H := StabG(x) in G. Define

φ : Orb(x)→Lgx→ gH

Show that φ is a bijection.

A G-set is transitive if it has only one G-orbit. Prove that this definition is equivalent with the abovedefinition of the transitive. Let X be a finite G-set and XG the set of fixed points in X. Thus,

XG = x ∈ X : gx = x for every g ∈ G .

The set XG is also called the set of invariants of the G-action.Since the orbits partition X we have

|X| = |XG|+

n∑i=k

|Orb(xi)|,

where xk, . . . ,xn are representative of orbits of X.Let g be a fixed element in G. The set of fixed points of g in X, which we denote with Xg, is the set of

all points x ∈ X such that gx = x. Thus,Xg = x ∈ X | gx = x

The term invariant by g is also used for the set Xg.

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Theorem 4.1 (Orbit counting theorem). Let G be a finite group acting on X. If N is number of orbits, then

N =1|G|

∑g∈G

|Xg|.

Proof. Consider the sum ∑g∈G

|Xg|

Notice that x ∈ Xg if and only if g ∈ StabG(x). Hence, every element x ∈ X contributes |StabG(x)| to the sum.Thus, we have ∑

g∈G

|Xg| =∑x∈X

|StabG(x)| .

Let us now compute the right hand sum by grouping elements in each orbit. Since there are N orbitsO1, . . . ,ON we have ∑

g∈G

|Xg| =∑x∈X

|StabG(x)| =∑x∈O1

|StabG(x)|+ · · ·+∑

x∈ON

|StabG(x)|

Now, two elements x and y in the same orbit have isomorphic stabilizers; see Lem. Lemma 4.1. So eachorbit Oi contributes

|Oi| · |StabG(xi)|

to the sum, for some representative xi ∈ Oi. So we have

∑g∈G

|Xg| =∑x∈X

|StabG(x)| =N∑

i=1

|Oi| · |StabG(xi)| =N∑

i=1

|G| = N · |G|

This completes the proof.

Remark 4.1. The above theorem says that the number of orbits is equal to the average number of points fixed by anelement of G.

Corollary 4.1. Let G be a finite group and X a finite set such that |X| > 1. If G acts on X transitively then thereexists τ ∈ G with no fixed points.

Proof. Let |G| = n. Since the action is transitive then there is only one G-orbit. Form the above theorem wehave that

|G| = F(1G) + F(g1) + · · ·F(gn) = |X|+ · · ·

If F(σ) ≥ 1 for all σ ∈ G then|G| = |X|+

∑σ∈G

F(σ) ≥ |X|+ (n−1)

Thus, |G| > n which is a contradiction. Hence, there must be some σ ∈ G such that F(σ) = 0. Let a group G act on the sets X and Y. A function f : X→ Y is called an equivariant function if and

only if∀σ ∈ G,∀x ∈ X : σ f (x) = f (σx).

The concept of a group acting on a set is one of the most important concepts of algebra. Below we willsee some classical examples of group actions which were studied and understood well before the conceptof a group was well established.

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4.2 Some classical examples of group action

Here we give some classical examples of group actions which have played a major role in algebra, geometry,topology and other areas of mathematics.

4.2.1 Transformations of R2

LetR2 be the Euclidean space that we are familiar from linear algebra and G = GL2(R). For any point P ∈R2

with coordinates P(x, y), consider the vector

~v =−−→OP =

[xy

]Then we define the following action

GL2(R)×R2→R2([

a bc d

],

[xy

])→

[ax + bycx + dy

]

Exercise 4.2. Prove that the above is a group action. What is the kernel of this action? Is it a transitive action?

Let P(a,b) and G = GL2(R).

Exercise 4.3. Determine the stabilizer StabG(P).

In the last section of this chapter, we will revisit again this action and study it in more detail.

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4.2.2 Change of coordinates

Consider the following problem, which looks very innocent and elementary.

Exercise 4.4. Let be given a curve with equation

13x2−18xy + 9y2

−40x = −64 (4.1)

Graph the set of solutions of this equation.

We don’t really know a methodological approach how tograph the set of solutions of this equation. We are not evensure what shape the graph might have. If we were to use anycomputational algebra packages then we can get the graph inFig. 4.1.

The graph looks like an ellipse. Is it really an ellipse? Can youfind algebraic substitutions which do not change the shape of thegraph and make the equation easier to graph? After all, if thisis really an ellipse, shouldn’t we be able to move the coordinatesystem such that it is right in the center of this ellipse? In otherwords, can we find algebraic substitutions for x and y such thatthis equation becomes

X2

A2 +Y2

B2 = 1

for some real numbers A and B?It can be easily verified that Eq. (4.1) can be written as

9(x− y)2 + 4(x−5)2 = 36Figure 4.1: The graph of Eq. (4.1)

Thus, by lettingX = x−5 and Y = x− y

we getX2

32 +Y2

22 = 1 (4.2)

which definitely seems nicer than Eq. (4.1). This "new" ellipse has axes of length 6 and 4 and they seem, atleast visually, to be close to the original ellipse.

Question 4.1. Can we find a methodological approach to solve this problem or similar problems like this one?

Stated a bit differently, given a degree 2 equation, can we determine a methodological approach so wecan make the right substitutions and that the equation is transformed into a nicer one?

Question 4.2. Can this be done for quadratic surfaces (degree 2 equations in space)? What about higher degreeequations?

In the process, we will have to understand and answer the following three questions:

Question 4.3. i) Which substitutions preserve the shape of graphs?

ii) Which substitutions preserve the size of graphs?

ii) How do we determine such substitutions?

A student must have learnt to answer all of three parts of the question above in a linear algebra course.Can a problem like this be stated in terms of group actions?

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4.2.3 The space of binary forms

A binary form is a homogenous polynomial f (x, y) of fixed combined degree. Let k be a field and Vd(k) thespace of all degree d ≥ 2 binary forms f (x, y) with coefficients in k.

Exercise 4.5. Prove that Vd(k) is a vector space.

There is an action of GL2(k) on Vd(k) as follows:

GL2(k)×Vd→ Vd([a bc d

], f (x, y)

)→ f (ax + by,cx + dy)

We will denote f (ax + by,cx + dy) by f M.

Example 4.1. Show that the function defined above is a group action. What are the orbits of this action?

Two binary forms f and g are called equivalent when they are in the same orbit of the GL2(k)-action.Sometimes the term GL2(k)-equivalent is used to avoid confusion with other group actions.

Let us now consider the space of quadratic forms V2.

Example 4.2. Let M =

[λ1 λ2λ3 λ4

]∈ GL2(k) and

f (x, y) = ax2 + bxy + cy2

Find f M. What is the stabilizer Stab( f )? What is the orbit Orb( f )?

Example 4.3. Let M =

[λ1 λ2λ3 λ4

]∈ SL2(k) and

f (x, y) = ax2 + bxy + cy2

Can you find an invariant of f ? For example a quantity in terms of the coefficients with is fixed by the SL2(k) action.

Example 4.4. Can you generalize the results from the above exercise to degree d > 2 binary forms? Justify youranswers.

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Paul Gordan

Paul Albert Gordan (27 April 1837 – 21 December 1912) wasa German mathematician, a student of Carl Gustav Jacobi atthe University of Königsberg before obtaining his Ph.D. at theUniversity of Breslau (1862), and a professor at the University ofErlangen-Nuremberg.

He was born in Breslau, Germany (now Wroc?aw, Poland),and died in Erlangen, Germany.

He was known as "the king of invariant theory". His mostfamous result is that the ring of invariants of binary forms offixed degree is finitely generated. He and Alfred Clebsch gavetheir name to Clebsch–Gordan coefficients. Gordan computed,by hand, all 70 invariants of binary sextics. Together with F.Klein, M. Noether, et al. they made Erlangen one of the mostimportant mathematical centers of the world durinf XIX century.

Gordan also served as the thesis advisor for Emmy Noether. Figure 4.2: Paul Albert Gordan

A famous quote attributed to Gordan about David Hilbert’s proof of Hilbert’s basis theorem, a resultwhich vastly generalized his result on invariants, is "This is not mathematics; this is theology." The proofin question was the (non-constructive) existence of a finite basis for invariants. It is not clear if Gordanreally said this since the earliest reference to it is 25 years after the events and after his death, and nor is itclear whether the quote was intended as criticism, or praise, or a subtle joke. Gordan himself encouragedHilbert and used Hilbert’s results and methods, and the widespread story that he opposed Hilbert’s workon invariant theory is a myth (though he did correctly point out in a referee’s report that some of thereasoning in Hilbert’s paper was incomplete).

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4.2.4 Algebraic curves

Let us consider now the geometric aspects of such actions. Take k =R. Then this action becomes the actionin Section 4.2.1

In ?? we saw how we could identify the group S3 with the set of six transformations of the Riemannsphere. Consider now the set

X = 0, 1,∞, t

where t is a parameter. Then S3 acts on X via the transformations

α(x) = x,1x, 1−x,

11−x

,x

x−1,x−1

x

An invariant of this action is an expression in t which is fixed under all the transformations above.

Example 4.5. Show that the above is a well-defined action. Find an invariant of this action.

Exercises:

4.1. A group acts faithfully on a G -set X if identity is the only element of G which fixes every element of X. Provethat G acts faithfully on X when there are no two elements of G acting the same way on an element of X.

4.3 Symmetries

This section is based on [?Indra]

4.3.1 Translations and reflections of the plane

4.3.2 Preserving distances

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4.4 The modular group and the fundamental domain

Let P1 be the Riemann sphere and GL2(C) the group of 2×2 matrices with entries in C. The group GL2(C)acts on P1 by linear fractional transformations as follows(

α βγ δ

)z =

αz +β

γz +δ(4.3)

where(α βγ δ

)∈ GL2(C) and z ∈ P1. It is easy to check that this is a group action. If a group G acts on a set

S, we say that G acts transitively if for each x, y ∈ S there exists some g ∈ G such that g(x) = y.

Lemma 4.3. The GL2(C) action on P1 is a transitive action, i.e. has only one orbit. Moreover, the action of SL2(C)on P1 is also transitive.

Proof. For every z ∈ C, (z z−11 1

)∞ = z

and∣∣∣∣∣z z−11 1

∣∣∣∣∣ = 1. So the orbit of infinity passes through all points.

For the rest of this section we will consider the action of SL2(R) on the Riemann sphere. Notice that this

action is not transitive, because as we will see below for M =

(α βγ δ

)∈GL2(R) we have

Img(Mz) =(αδ−βγ)|γz +δ|2

Imgz.

This action has three orbits, as we will prove below. Therefore we restrict this action to the upper half-plane.LetH2 be the complex upper half plane, i.e.

H2 =z = x + iy ∈ C

∣∣∣∣ y > 0⊂ C.

The group SL2(R) acts onH2 via linear fractional transformations. In the following lemma we prove thatthis action is transitive.

Lemma 4.4. i) The group SL2(R) preserves H2 and acts transitively on it, further for g ∈ SL2(R) and z ∈ H2 wehave

Img(gz) =Imgz|γz +δ|2

ii) The action of SL2(R) on P1 has three orbits, namely R∪∞, the upper half plane, and the lower-half plane.

Proof. Let us first prove thatH2 is preserved under an SL2(R) action. Consider(α βγ δ

)· z =

αz +β

γz +δ

But γz +δ = γx + iγy +δ = γx +δ+γiy, therefore its conjugate is (γx +δ)− iγy = γz +δ and

(γz +δ)(γz +δ) = |γz +δ|2 = (γx +δ)2 + (γy)2.

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Hence,

αz +β

γz +δ=αz +β

γz +δ·γz +δ

γz +δ=

(αz +β)(γz +δ)|γz +δ|2

=αγzz +αδz +βγz +βδ

|γz +δ|2

=αγ|z|2 +βδ+αδx +αδiy +βγx−βγiy

|γz +δ|2

=αγ|z|2 +βδ+αδx +βγx

|γz +δ|2+

i(αδ−βγ)y|γz +δ|2

Therefore we see that

Img(gz

)=

(αδ−βδ) Imgz|γz +δ|2

=Imgz|γz +δ|2

> 0.

To show that SL2(R) action onH2 is transitive, pick any a + ib ∈H2. Then if g ∈ SL2(R) such that

g =

(a b0 1

): z→ a + bz

we have g(i) = a + ib. Thus the orbit of i passes through all points inH2 and so SL2(R) is transitive inH2.ii) The result is obvious from above.

Recall that a group action G×X→ X is called faithful if there are no group elements g, except theidentity element, such that gx = x for all x ∈ X. The group SL2(R) does not act faithfully on H2 since theelements ±I act trivially on H2. Hence, consider the above action as PSL2(R) = SL2(R)/±I action. Thisgroup acts faithfully onH2.

Let S be a set and G a group acting on it. Two points s1,s2 are said to be G-equivalent if s2 = gs1 forsome g ∈ G. For any group G acting on a set S to itself we call a fundamental domain F, if one exists, asubset of S such that any point in S is G-equivalent to some point in F, and no two points in the interior ofF are G-equivalent.

The group Γ = SL2(Z)/±I is called the modular group. It is easy to prove that the Γ action onH2 vialinear fractional transformations is a group action. This action has a fundamental domain F

F =

z ∈H2

∣∣∣∣ |z|2 ≥ 1 and |Re(z)| ≤ 1/2

as proven in the following theorem, as well as [?serre], and displayed in Fig. 4.3.

Figure 4.3: The action of the modular group on the upper half plane.

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Theorem 4.2. i) Every z ∈H2 is Γ-equivalent to a point in F.ii) No two points in the interior of F are equivalent under Γ. If two distinct points z1,z2 of F are equivalent under

Γ then Re(z1) = ±1/2 and z1 = z2±1 or |z1| = 1 and z2 = −1/z1.iii) Let z ∈ F and I(z) = g | g ∈ Γ, gz = z the stabilizer of z ∈ Γ. One has I(z) = 1 except in the following cases:z = i, in which case I(z) is the group of order 2 generated by S;z = ρ = e2πi/3, in which case I(z) is the group of order 3 generated by ST;z = −ρ = eπi/3, in which case I(z) is the group of order 3 generated by TS.

Proof. i) We want to show that for every z ∈ H2, there exists g ∈ Γ such that gz ∈ F. Let Γ′ be a subgroup ofΓ generated by

S =

(0 −11 0

): z→−

1z

and T =

(1 10 1

): z→ z + 1.

Note that when we apply an appropriate T j to z then we can get a point equivalent to z inside the strip−

12 ≤ Re(z) ≤ 1

2 . If the point lands outside the unit circle then we are done, otherwise we can apply S to getit outside the unit circle and then apply again an appropriate Tn to get it inside the strip − 1

2 ≤ Re(z) ≤ 12 .

Let g ∈ Γ′. We have seen that Img(gz) =Imgz|cz+d|2 . Since, c and d are integers, the number of pairs (c,d) such

that |cz + d| is less then a given number is finite. Hence, there is some g =

(a bc d

)∈ Γ′ such that Img(gz) is

maximal (|cz + d| is minimal).Without loss of generality, replacing g by Tng for some n we can assume that gz is inside the strip

−12 ≤ Re(z) ≤ 1

2 . If |gz| ≥ 1 we are done, otherwise we can apply S. Then

Img(Sgz) =Img(gz)|gz + 0|2

=Img(gz)|gz|2

> Img(gz).

But this contradicts our choice of g ∈ Γ′ so that Img(gz) is maximal.ii), and iii) Suppose z1,z2 ∈ F are Γ-equivalent. Without loss of generality assume Img(z1) ≥ Img(z2). Let

g =

(a bc d

)∈ Γ be such that z2 = gz1. Since

Img(gz1) =Img(z1)|cz + d|2

,

we get |cz + d| ≤ 1. But z1 ∈ F, d ∈ Z, and Img(z1) ≥√

32 hence the inequality does not hold for |c| ≥ 2, i.e.

c = 0,±1.

Case 1: c = 0. Since ad− bc = 1 and c = 0, we have a,d = ±1 and g = ±

(1 b0 1

). Since Re(z1) and Re(z2) are

both between − 12 and 1

2 , this implies either b = 0 and g = ±

(1 00 1

)or b = ±1 and g =

(1 ±10 1

)in which case

either Re(z1) = 12 and Re(z2) = − 1

2 , or the other way around.Case 2: c = 1. Since |1z1 +d| < 1, then d = 0 except when z1 = ρ, or −ρ in which cases d = 0,1 and d = 0,−1.Let us first consider the case c = 1, d = 0. In this case z1 is in the unit circle since otherwise |1z + 0| ≤ 1

is not fulfilled, and since ad− bc = 1, we have b = −1 and g = ±

(a −11 0

): z1→ a− 1

z1. The case |a| > 1 is not

possible, since z1 and gz1 are both in F.If a = 0 the z1,z2 are symmetrically located on the unit circle with respect to the imaginary axis. And

for a = ±1, g = ±

(±1 −11 0

)= ±T±1S from case 1 we have that Re(z1) = 1

2 and Re(z2) = − 12 , or the other way

around i.e. z1,z2 = ρ,−ρ.

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The case z = ρ, d = 1 gives a− b = 1 and gρ = a− 11+ρ = a +ρ, hence a = 0,1; we can argue similarly when

z = −ρ, d = −1.Finally to prove the case when c = −1, we just need to change the signs of a,b,c,d.

The following corollary is obvious.

Corollary 4.2. The canonical map F→H2/Γ is surjective and its restriction to the interior of F is injective.

The following theorem determines the generator of the modular group and their relations.

Theorem 4.3. The modular group Γ is generated by S =

(0 −11 0

)and T =

(1 10 1

), where S2 = 1 and (ST)3 = 1.

Proof. Let Γ′ be a subgroup of Γ generated by

S =

(0 −11 0

): z→−

1z

and T =

(1 10 1

): z→ z + 1.

We want to show that Γ is a subgroup of Γ′. Assume g ∈ Γ. Choose a point z1 in the interior of F, andlet z2 = gz1 ∈ H2. From the definition of the fundamental domain we have that there exists a g′ ∈ Γ′ suchthat g′z2 ∈ F. But z1 and g′z2 of F are Γ-equivalent, and one of them is in the interior of F, hence fromTheorem 4.2 these points coincide and g′g = 1. Thus, g ∈ Γ′.

Note that S2 = 1, so S has order 2, while Tk =

(1 k0 1

)for any k ∈ Z, so T has infinite order. For more

details on the modular group and related arithmetic questions the reader can see [?serre] among others.

Exercises:

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Chapter 5

Sylow theorem

Sylow’s theorem is one of the most important theorems of elementary group theory. It tells us for whatdivisors of the group order we are guaranteed a subgroup. As you will see in the applications of thistheorem, we will be able to prove many important results of group theory.

5.1 Groups acting on themselves by conjugation

Some of the most interesting group actions are when groups act on themselves. Let G be a group and takeX = G. The G acts on itself, when we have a function

ϕ : G×G −→ G

which is a group action.

5.1.1 Groups acting on themselves by left multiplication

Let G be a group and take X = G. One of the most common actions of G onto itself is the left multiplication

G×G −→ G(g,x) −→ gx

(5.1)

It is obvious that this is a group action. We have seen this action before in the Cayley’s theorem.If H is a subgroup of the group G, then G is a H -set under the action of multiplication from the left from

the elements of H. Let LH be the set of left cosets of H in G. Then, |LH | = [G : H] = |G||H| , 1. Then H act on

L−H as follows

H×LH −→LH

(h, gH) −→ (hg)H(5.2)

Denote byL0 the fixed set of this action. We have the following observation which will be used in the proofof the following lemma.

Lemma 5.1. gH ∈ L0 if and only if g ∈NG(H).

Proof. If gH ∈L0, then (hg)H = gH, for all h ∈H. Hence, g−1hgH = H, for all h ∈H. In other words, g−1Hg⊂Hwhich implies that g ∈NG(H).

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Conversely, if g ∈NG(H) then gH = Hg. Hence, we have

ϕ(h, gH) = h(gH) = h(Hg) = Hg = gH.

This completes the proof.

5.1.2 Groups acting on themselves by conjugation

Consider next another group action onto itself. Let G be a group and

ϕ : G×G −→ G

(g,x) −→ gxg−1 (5.3)

This is called G acts on itself by conjugation.

Exercise 5.1. Prove that the above is a group action.

Remark 5.1. Notice that when the group action is by conjugation, then for every x ∈ G,

StabG(x) = CentG(x)

The same can be said for any subset A ⊂ G.

The conjugation action turns out to be very useful in establishing properties of finite groups. Thefollowing example is very constructive.

Groups acting on their power sets

For a given G the power setP(G) is the set of all subsets of G. Let S ⊂G. Then, the conjugate of S is definedas

gSg−1 = gsg−1| s ∈ S

Then, we have the following action

ϕ : G×P(G) −→P(G)

(g,S) −→ gSg−1 (5.4)

Exercise 5.2. Prove that this is a group action.

Lemma 5.2. The number of conjugates nS of S is the index of the normalizer of S, namely

nS = [G : NG(S)].

Proof. The proof is just a simple application of Lemma 4.2, which says that

|Orb(S)| = [G : StabG(S)]

whereStabG(S) = g ∈ G | gSg−1 = g = NG(S)

This completes the proof.

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The Class Equation

The center of the group G,Z(G) = x : xg = gx for every g ∈ G ,

is the set of points which are fixed by conjugation. Nontrivial orbits of this action are called conjugacyclasses of G. If x1, . . . ,xk are representatives for every conjugacy class of G and |Ox1 | = n1, . . . , |Oxk | = nk, then

|G| = |Z(G)|+ n1 + · · ·+ nk.

The stabilizer subgroup of each xi, C(xi) = g ∈ G : gxi = xig, is called centralizer subgroup of xi. Then weobtain the class equation:

|G| = |Z(G)|+ [G : C(x1)] + · · ·+ [G : C(xk)].

It follows that the order of any conjugation class must divide the order of the group G. WE summarize inthe following theorem.

Theorem 5.1 (Class equation). Let G be a finite group and g1, . . . , gr the representatives of distinct conjugacyclasses of G not contained in Z(G). Then,

|G| = |Z(G)|+r∑1

[G : CG(gi)] (5.5)

Proof. Let Cg denote the conjugacy class of g ∈ G. Then, |Cg| = 1 if and only if g ∈ Z(G). Let

Z(G) = 1G,z2, . . . ,zm

and C1, . . . ,Cr be the conjugacy classes of G not contained in Z(G). We denote by g1, . . . , gr their representa-tives respectively. Then,

|G| = |Z(G)|+r∑1

|Ci| (5.6)

This completes the proof.

Definition 5.1. Conjugation by g is called an inner automorphism of G. The set of all inner automorphismsdenoted by Inn(G) forms a group with composition functions.

Lemma 5.3. Inn(G) C Aut (G).

Proof. First we need to show that Inn(G) is a subgroup of Aut (G). For every g ∈G we have the conjugationisomorphism σg:

σg :G→ G

h→ ghg−1

For every a,b,x ∈ G we have that:

σaσb(x) = abxb−1a−1 = (ab)x(ab)−1 = σabx,

so we have that σaσb(x) = σabx. It can easily be shown that σe is the identity and also (σa)−1 = σa−1 soInn(G) ≤Aut (G).

To prove that Inn(G) CAut (G) we take a α ∈Aut (G) and σα ∈ Inn(G). For a x ∈ G we have:

ασαα−1(x) = α(α(α−1(x))α−1) = (α(α))(αα−1(x))(α(α−1) = (α(α))(x)(α(α))−1 = bxb−1 = σb(x)

for an element b = α(a). Thus, ασαα−1∈ Inn(G). Thus , Inn(G) is a normal subgroup of Aut (G).

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5 Shaska T.

Example 5.1. It is easy to prove that conjugacy classes of S3 are:

(1), (123), (132), (12), (13), (23).

The class equation is 6 = 1 + 2 + 3.

Example 5.2. Conjugacy classes for D4 are

(1), (13), (24), (1432), (1234), (12)(34), (14)(23), (13)(24).

The class equation is 8 = 1 + 2 + 2 + 3.

Example 5.3. There are 2 conjugacy classes of 5-cycles in A5, each of which has 12 elements.

Solution: We know that all elements of the same type are conjugate in Sn. There are 24 elements which are 5-cycles

in S5 and they are all conjugate. Let α be a 5-cycle in S5. Then,

CS5 (α) = 1,α,α2,α3,α4

Since all of them are 5-cycles then CS5 (α) = CA5 (α). Then,

|OrbA5 (α)| =|A5|

|CA5 |=

605

= 12

Exercise 5.3. Find all conjugacy classes of S5. What about the conjugacy classes of A5?

Exercises:

5.1. Write the class equation for S5 and for A5.

5.2. Let a ∈ G. Prove that for some g ∈ G, we have gCentG(a)g−1 = CentG(gag−1).

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5.2 p-groups

Here we will study some special groups which play an important role in the study of the structure ofgroups. They are called p-groups. We we have seen before some cases of p-groups. Throughout this sectionp denotes a prime integer.

A p-group is a group such that of the group has order a power of p. We will focus on finite p-groups.

Exercise 5.4. Prove that a finite p-group has order pn for some integer n ≥ 1.

Lemma 5.4. If G is a finite p-group, then G has nontrivial center.

Proof. Let |G| = pn. From class equation we have

|G| = |Z(G)|+∑

[G : Cx]

Since every summand [G : Cx] is a power pi of p, for i ≥ 1, then we know that the centralizer of x is asubgroup and therefore we have that

|Z(G)| = |G| −∑

[G : Cx] = pn−

∑pi

is divisible by p. This completes the proof.

Corollary 5.1. A group with order p2 where p is a prime number is Abelian.

Proof. Since G = p2, then Z(G) has order p or p2. If |Z(G)| = p2, then we are done. Otherwise, G/Z(G) hasorder p and is therefore cyclic. From 3.14, G is Abelian.

Lemma 5.5. Let G be a finite p-group. Show that if H is a nontrivial normal subgroup of G then H∩Z(G) , 1.

Proof. Recall that if H C G and C a conjugacy class of G then C ⊂H or C∩H = ∅. Pick representatives fromthe conjugacy classes of G

g1, g2, . . . , gr,

such thatg1, . . . , gs ∈H and gs+1, . . . , gr <H

Then,

|H| = |H∩Z(G)|+r∑

i=s+1

[G : CG(gi)]

Since p | |H| and p | [G : CG(gi)] for all i = s + 1, . . . ,r, then p | |H∩Z(G)|. This completes the proof.

Now we are ready to prove the following lemma which will be used in the proof of the Sylow’s theorem.

Lemma 5.6. Let G be a finite p-group and H a proper subgroup of G. Then,

i) H is proper in NG(H),

ii) Every maximal subgroup M is normal in G and [G : M] = p.

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5 Shaska T.

Proof. i) LetLH be the set of left cosets of H in G. Then, |LH |= [G : H] = |G||H| , 1. Let H act onLH as described

in Eq. (5.4).Denote by S0 the fixed set of this action. From the above xH ∈ S0 if and only if x ∈NG(H). Thus, |S0| is

the number of cosets xH with x ∈NG(H). Hence, |S0| = [NG(H) : H].Let Si for i = 1, . . .r denote the orbits of length > 1. Then,

|S| = |S0|+

r∑i=1

|Si| = |S0|+

r∑i=1

[G : StabG(si)]

Since, p | |S| = [G : H] and p divides the sum then p | |S0|. Hence, p | [NG(H) : H]. Therefore H is proper inNG(H).

ii) Let M be a maximal subgroup in G. From part i) we know that M is proper in NG(M). Therefore,M = G. This implies that M C G.

By the Theorem 3.9 we have that [G : M] = p.

Exercises:

5.3. If G is non Abelian group of order p3, then Z(G) = G′.

5.4. Let G be the additive group of real numbers. Let θ ∈G and consider the action on the real planeZ2 which rotatesthe plane counterclockwise, around the origin, with angle θ. Let P be a point different from the origin

a) Prove that Z2 is a G -set.b) Describe geometrically the orbit which contains P.c) Find the group StabG(P).

5.5. Let |G| = pn and assume that |Z(G)| = pn−1 for a prime number p. Prove that G is Abelian.

5.6. Let G be a group with order pr, where p is a prime number. Prove that G contains a normal subgroup with orderpr−1.

5.7. Assume that G is a finite group with order pnk, where k < p. Prove that G must have a normal subgroup.

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5.3 Automorphisms of groups

Let G be a group. An isomorphismσ : G→ G

is called an automorphism of G. The set of all automorphisms of G is denoted by Aut (G).

Exercise 5.5. Show that Aut (G) is a group under the composition of functions.

Exercise 5.6. Let g ∈ G be fixed and define the following

σ :G→ G

h→ ghg−1 (5.7)

Show that σ ∈ Aut (G).

The automorphisms defined as above is called conjugation by g. Conjugations by g for some g ∈ G arecalled inner automorphisms of G. The set of all inner automorphisms of G is denoted by Inn(G).

Lemma 5.7. In any group G the following are true:i) Inn(G) C Aut(G)ii) G/Z(G) Inn(G).

Proof. Let σ ∈Aut (G). For the first part it is enough to show that σ Inn(G)σ−1⊂ Inn(G). Let fg ∈ Inn(G) such

thatfg(h) = ghg−1, for all h ∈ G.

Then, σ fgσ−1 : G→ G where for every x ∈ G we have(σ fgσ

−1)(x) = σ

(fg

(σ−1(x)

))= σ

(gσ−1(x)g−1

)=

= σ(g)σ(σ−1(x)

)σ(g−1

)= σ(g) ·xσ

(g−1

) (5.8)

Hence, σ fgσ−1∈ Inn(G).

For the second part define a homomorphismsφG/Z(G)→ Inn(g) such thatφ(gZ(G)) = fg as above. Showthat this is an isomorphism.

The group Aut (G)/ Inn(G) is called the group of outer automorphisms of G and denoted by Out (G).

A subgroup H ≤ G is called characteristic in G, denoted by HcC G, if

∀σ ∈Aut (G),σ(H) = H.

Lemma 5.8. Show that

i) If HcC G then H C G

ii) If HcC K

cC G then H

cC G

iii) HcC K C G then H C G.

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Proof. i) Let HcC G. Then, for all g ∈ G the conjugation by g fixes H. Hence, gHg−1 = H which means that

H C G.ii) Let σ ∈Aut(G). Then σ(K) = K since K

cCG. Therefore the restriction σ |K∈Aut (K). Hence, σ |K (H) = H

since HcC K. Thus, σ(H) = H. This shows that H

cC G.

iii) We want to show that gHg−1 = H for all g ∈G. Fix g ∈G. Take σg : G→G such that σg(h) = ghg−1, forall h ∈ G. Then, σg(K) = K since K C G. Therefore, σg |K∈ Aut (K) which implies that σg |K (H) = H. Hence,gHg−1 = H for all G ∈ G.

Let G be a group and K < G. Then H ≤ G is called a complement of K in G if H∩K = 1G and KH = G.Hence, G is a direct product of K and H.

Example 5.4. Prove that every subgroup of a cyclic group is a characteristic subgroup.

Solution: Let G = 〈a〉 and H ≤ G. Show that HcC G which is equivalent to say σ(H) = H, for all s ∈ Aut(G).

Case 1: |G| =∞.Then G (Z,+) and so Aut(G)Aut(Z). But Aut(Z) = id, because

σ(n) = σ(1 + 1 + 1 + · · ·+ 1) = σ(1) +σ(1) + · · ·+σ(1) = nσ(1)

andσ(−n) = σ(−1) ·σ(n) = −σ(n) = −nσ(1).

So ∀n ∈Z,σ(n) = nσ(1). So σ(1) and σ(−1) have to generateZ. Thus, σ(1) = 1 or σ(1) =−1 which implies σ(n) =−n.But σ(m ·n) = −mn , (−m)(−n) and so σ is not a homomorphism.

Case 2: |G| = 〈a〉 = n.Let |H| = m. Then σ(H) ≤ G. Also |σ(H)| = |H| = m because σ is a bijection. But from the Fundamental Theorem

of Cyclic Groups we have that since G is cyclic, then G has a unique subgroup of order m.

Thus, σH = H and so HcC G.

Example 5.5. If G is an abelian group and k a positive integer, then

Gk = ak| a ∈ G

is a characteristic subgroup of G.

Solution: Let σ be an automorphism of G. We want to show that σ(Gk) = Gk.

Let a ∈Gk. Then there exists b ∈G such that σ(b) = a. It implies that σ(bk) = ak. But bk∈Gk and so σ(bk) ∈ σ(Gk)

and ak∈ σ(Gk).

Let b ∈ σ(Gk). Then there exists a ∈ G such that b = σ(ak) = σ(a)k. But σ(a) ∈ G implies σ(a)k∈ Gk and so

b ∈ Gk

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5.4 Sylow theorems

Let G be a group and p a prime number such that |G| = pαm where (p,m) = 1. A subgroup of G with order pα

is called a Sylow p -subgroup of G. The set of all Sylow p -subgroups is denoted by Sylp(G) and the numberof Sylow p -subgroups of G by np.

Lemma 5.9. Let P ∈ Sylp(G). If Q is a p -subgroup of G then Q∩NG(P) = Q∩P.

Proof. Let’s denote with H = Q∩NG(P). Since P ⊆NG(P) then P∩Q ≤H. We want to show that H ≤ P∩Q.Since H ≤Q, it is enough to show that H ≤ P.

Consider PH. It is a subgroup of G, hence it is a p -subgroup and |PH| = p j, for some j < n. Since P is amaximal subgroup, p -subgroup of G we have that PH = P. Then H ≤ PH and H ≤ P. Hence, H ≤Q∩P.

Theorem 5.2 (Sylow). Let |G| = pαm where p is a prime number such that (p,m) = 1. The following are true.

a) Sylp(G) , ∅

b) If P is a Sylow p -subgroup of G and Q is a p -subgroup of G then Q is contained in a conjugate of P. Inparticular, any two Sylow p -subgroups are isomorphic.

c) The number np of Sylow p -subgroups is

np1 mod p and np = [G : NG(P)].

Moreover, np |m.

Proof. We will prove a) by induction on |G|. If the order of |G| = 1 then the theorem is clear. Assume that itis true for all groups with order < |G|.

If p | |Z(G)|, then from Proposition 2.6, Z(G) has a subgroup with order p, say N ≤ Z(G), which of courseis normal in G. Let G := G/N. Then we have

|G| = pα−1·m

and so G has a Sylow p -subgroup P such that |P| = pα−1. By the Theorem 3.9 there is a subgroup P of Gsuch that N ≤ P. Then, [P : N] = |P| and

|P| = |N| · |P| = p ·pα−1 = pα.

If p - |Z(G)|, then from the Class Equation Eq. (5.5) we have

|G| = |Z(G)|+r∑1

[G : CG(gi)]

Then, there is an i ≤ r, such that p - [G : CG(gi)]. Let H = CG(gi). Thus,

p - [G : H] =pα ·m|H|

Hence, |H| = pα · s where (p,s) = 1. Since H is a proper subgroup of G then by induction hypothesis thereexists a Sylow p-subgroup P such that |P| = pα. Hence, P < G and |P| = pα. This completes the proof of parta).

To prove the other two parts we proceed as follows. From a) we have that there is a P ∈ Sylp(G). Let’sdenote by

X = P1,P2, . . . ,Pr

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5 Shaska T.

the set of all conjugated subgroups of P. From the action of the group on the set X we have

G×X −→ X

(g,Pi) −→ Pgi := gPig−1

Let Q be a p-subgroup of G. Then, Q acts on X by restricting the action of G on Q. Thus,

X = Orb1∪· · ·∪Orbs

and r = |X| = |Orb1 |+ · · · |Orbs |, where |Orbi | = [Q : NQ(Pi)]. Since from Lemma 5.9 we have

NQ(Pi) = Q∩NG(Pi) = Q∩Pi,

then |Orbi | = [Q : Pi∩Q].Now take Q = P1. Hence, |Orb1 | = [Q : Q] = 1 and for all i > 1 we have |Orbi | = [Pi : P1∩Pi] > 1. Hence,

p | |Orbi | for al l i > 1 andr ≡ 1 mod p

b) Assume there is a p-subgroup Q of G which is not contained in any of the conjugates of P. In otherwords, there is a Q such that Q Pi, for all i = 1, . . .r. Then, Q∩Pi is proper in Q for all i. Thus,

|Orbi | = [Q : Q∩Pi] > 1.

Hence, p | |Orbi | for all i = 1, . . . ,r. Then, p | r since r =∑s

i1|Orbi |. This contradicts the fact that r1 mod p.

Hence, Q ≤ gPg−1, for some g ∈ G.It is left to show that any two Sylow p-subgroups are conjugate. Since every Sylow p-subgroup is

contained in some Pg, for some g ∈ G and the sizes of Pg and Q are the same then Q = Pg.From Lemma 5.2 we know that the number of conjugates of a subgroup P is exactly the index of its

normalizer. Hence,np = [G : NG(P)]

This completes the proof.

Remark 5.2. The original proof of Sylow can be found at [?Sylow]. As G. Frobenius said:"as every educated person knows the Pythagorean theorem so does every mathematician speak of Abel’s theorem

and Sylow’s theorem".

Corollary 5.2. A Sylow p -subgroup of a finite order group G is a normal subgroup of G if and only if it is the onlySylow p -subgroup of G.

Exercise 5.7. Let P ∈ Sylp(G). Prove that the following are equivalent.i) np = 1ii) P C Giii) P

cC G

Lemma 5.10. Suppose that G is a finite group.a) If H C G and P a Sylow p -subgroup of H then G = HNG(P) (Frattini’s argument)b) Let P a Sylow p subgroup of the group G then we have that NG(NG(P)) = NG(P).

Proof. a) Since H C G then H ·NG(P) is a subgroup of the group G and

H ·NG(P) = NG(P) ·H ≤ G.

Let a g ∈ G. We want to show that g ∈H ·NG(P). If P ≤H then g−1Pg < g−1Hg = H (Hnormal).

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Thus, g−1Pg is a Sylow p -subgroup of H. From Sylow Theorem g−1Pg is conjugate with P. Hence, thereis a h ∈H such that g−1Pg = h−1Ph then hg−1Pgh−1 = P. From the definition of the normalizer we have thatif (hg−1)−1Pgh−1 = P. Then hg−1

∈NG(P). If g−1∈ h−1NG(P) then g−1

∈H ·NG(P). Hence, g ∈H ·NG(P).

b) 1) The fact that NG(P) <NG(NG(P)) is true for every group.

2)If P is a Sylow p -subgroup of the group G then P is a Sylow p -subgroup for every H ≤ G such thatP < H. Hence P is a Sylow p -subgroup of NG(P) but P C NG(P) therefore it is the only Sylow p -subgroupof NG(P). Let a x ∈NG(NG(P)) Then,

x−1NG(P)x = NG(P) then x−1Px <NG(P).

Hence x−1Px = P and finally we say that x ∈NG(P).

Lemma 5.11. Let P be a Sylow p -subgroup of a finite group G and let x be an element with order a power of p. Ifx−1Px = P, then x ∈ P.

Proof. Obviously x ∈N(P) and the cyclic subgroup, 〈xP〉 ⊂N(P)/P, has order a power of p. From Theorem 3.9there is a subgroup H of N(P) such that H/P = 〈xP〉. Since |H| = |P| · |〈xP〉|, the order of H must be a powerof p. However, P is a Sylow p -subgroup which is contained in H. Since the order of P is the largest powerof p that divides |G|, then H = P. Thus, H/P is subgroup trivial and xP = P, or x ∈ P.

Next we see some applications of the Sylow’s Theorem.

Example 5.6. Prove that there is no simple group with order 30.

Proof. Every group G with order 30 = 2 ·3 ·5 can have

n2 = 1,3,5,15n3 = 1,10n5 = 1,6

If n2 = 1 or n3 = 1 or n5 = 1 then one of Sylow p -subgroups is normal and G is not simple.Suppose that

n2 > 1 and n3 > 1 and n5 > 1

Counting the elements of G we have at least 3 subgroups P2 and each one has an element different fromidentity, 10 subgroups P3 and each has 2 elements different from identity, and 6 subgroups P5 and each has4 elements different from identity. Thus,

1 + 3 ·1 + 10 ·2 + 6 ·4 = 48 > 30

Thus, one of n2,n3,n5 is 1 and G has a normal subgroup.

Example 5.7. Let a group G with order 40. What does the Sylow’s theorem tell us about this subgroup.

Proof. The group G has order:|G| = 40 = 5 ·8 = 23

·5

Then, n2 = 1,5 and n5 = 1. Hence P5 is normal in G. If n2 = 1 then P2 CG and GP2×P5. If n2 = 5 then thereexist 5 subgroups each with order 8. These subgroups are conjugated to each other (hence isomorphic) andcontribute exactly 35 elements to the group. Hence each contributes 7 elements, since their intersection iseG.

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Example 5.8. Find all groups which have precisely 2 subgroups, 3 subgroups.

Solution: Every group |G| > 1 has at least two subgroups, namely and e and G. Hence, we want to determine all

groups which have no proper subgroups. So we want groups whose order n has no divisors other then 1 and itself.Otherwise the group would have p-groups (Sylow p-groups). So |G| = p, where p is prime number. Thus, the groupsare isomorphic to Cp for prime p.

In the second case, We are looking for groups with exactly one proper subgroup. So G| = pα for some prime p andα ∈Z+. If α > 2, then from properties of p-groups we know that G would have subgroups of order p,p2.

So |G| = p2. In this case G is abelian. Since p is a divisor of |G| then there is an element of order p in G. Thus, Ghas a subgroup of order p.

Example 5.9. Prove that a group of order G = 5 ·7 ·17 is cyclic.

Example 5.10. Let G be given such that |G| = 495 = 32·5 ·11. Prove that G has a normal subgroup of order 5 or 11.

Proof. From the Sylow’s Theorem we get

n3 = 1,55, n5 = 1,11, n11 = 1,45

We see that if n5 = 11 and n11 = 45, then by counting the elements of the group we have at least

1 + 1 ·8 + 11 ·4 + 45 ·10 > 495 = |G|

Hence,n5 = 1 or n11 = 1.

Hence, P5 C G or P11 C G.

5.8. Prove that if |G| = 462 then G is not simple.

Proof. From the Sylow’s Theorem we get

n3 =

n5 =

n11 =

5.9. Prove that if |G| = 132 then G is not simple.

Proof. From the Sylow’s Theorem we get

n3 =

n5 =

n11 =

Example 5.11. Prove that a group of order 105 has an element of order 15.

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Proof. From the Sylow’s Theorem we get

n3 = 1,7,n5 = 1,21,

n11 = 1,15

If n3 = 7 then [G : NG(P3)] = 7. Hence, |NG(P3)| = 15. We know that every group of order 15 is cyclic, sayNG(P3) = 〈a〉. Then, |a| = 15.

If n3 = 1, then P3 C G. Hence, P3P5 is a subgroup of G and has order 15. Then, there exists a ∈ G suchthat P3P5 = 〈a〉. Thus, |a| = 15.

Exercises:

5.10. If G is a group with order pn, where p is a prime number, prove that G has a proper subgroup with order p. Ifn ≥ 3, is it true that G must have a proper subgroup with order p2?

5.11. If |G| is pq where p and q are distinct prime numbers and if G has a normal subgroup of order p and a normalsubgroup of order q, prove that G is cyclic.

5.12. Let |G| be pq,p > q are primes, prove

a) G has a subgroup of order p and subgroup of order q.

b) If q - p−1, then G is cyclic.

c) Given two primes p, q, such that q | p−1, there exists a non-abelian group of order pq.

d) Any two non-abelian groups of order pq are isomorphic.

5.13. Show that the group of upper triangular matrices with 1’s in the main diagonal is Sylow subgroup in GLn(Fp).

5.14. Let p be prime, n > 1 an integer, and G = GLn(Fp). Show that there are p-Sylow subgroups H1, H2 of G suchthat H1∩H2 = e.

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Ludwig Sylow

A high school teacher who became a university professorwhen he was 65 years of age, Sylow for sure left his mark inmathematics. He studied the theory of equations and grouptheory. His main contribution is the celebrated Sylow’s theorem.In 1902 Sylow gave the welcoming address at a conference tomark the centenary of Niels Abel’s birth. He said (see [2]):- Inthe early nineteenth century, applied mathematics had alreadyachieved great triumphs, especially in the fields of astronomyand physics. But just at the same time mathematics ... started toturn its gaze back to the pure and abstract theories. [Gauss andCauchy] initiated that great movement, which has run throughthe whole of the previous [19th] century, and which has reformedmathematics from its foundations at the same time it enriched itwith new theories. ... It was in this movement that Niels Abeltook such a significant part that he will forever he counted asone of the greatest mathematicians ever.

In 1902 Sylow, in collaboration with Elling Holst, published Abel’s correspondence. Further Abeldocuments had been discovered after the Sylow/Lie book came out in 1881 and, at the ’Third ScandinavianCongress of Mathematicians’ which was held in Kristiania in 1913, Sylow discussed this new material.

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5.5 Simple groups

In this lecture we will use the Sylow’s theorem to study the structure of groups of given order. Let usremind ourselves of the following important definition.

Definition 5.2. A group G is called simple group if does not have proper normal subgroups.

Now you posses the knowledge to complete the following project.

Exercise 5.8. Prove that if G is a group of order |G| < 60, then G is not simple.

5.5.1 Alternating groups An

In this section we will prove that the alternating group An is simple for n ≥ 5. First, as a simple exerciselet’s consider only A5.

Lemma 5.12. If |G| = 60 and it has more than one Sylow 5-subgroup, then G is simple.

Proof. From the Sylow theorem, since |G| = 22·3 ·5, we have that

n2 = 1,3,5,15 n3 = 1,4,10, n5 = 1,6.

Suppose that G is not simple. Then, n5 = 6

Corollary 5.3. A5 is simple.

Proof. Since H = 〈(12345)〉 and K = 〈(13245)〉 are two distinct Sylow 5-subgroups of A5, then A5 is simple.

Lemma 5.13. Any simple group of order 60 is isomorphic to A5.

Proof. Let G be a simple group of order 60. Since |G| = 22·3 ·5 then from the Sylow theorem we have that

n1 = 1,3,5, or 15, n3 = 1,4, or 10, and n5 = 1, or 6. Since, G is simple that n2,n3, and n5 can not be equal to 1.By the same argument we ca show that n2 , 2.

Assume that n3 = 4. Since |G| - 4! then G is not simple Theorem 3.16. Hence, n3 , 4. Thus, n3 = 10, n5 = 6,and n2 = 5 or 15.Case I: Assume n2 = 5. Then, [G : NG(P2) = 5. Hence, there is a subgroup K = NG(P2) of G which has index5. From Theorem 3.15 there is σ : G→ S5 such that kerσ ⊂ K. Since kernels are normal and G is simple then|kerσ| = 1. Thus, σ is an embedding and we can identify G with σ(G). Hence, take G < S5.

If G ⊂ A5 then we are done since both have the same order. Assume G 1 A5. Then, S5 = GA5 because|GA5| > 60 and GA5 is a subgroup since both G and A5 are normal in S5.

Then,

|GA5| =60 ·60|G∩A5|

= 120

Hence, |G∩A5| = 30. This means that G∩A5 CG as a subgroup of order 2. This is a contradiction since G issimple.Case II: Let n2 = 15 which implies that we have 15 Sylow 2-subgroups (each or order 4). They can’t all bedisjoint since then we would h ave at least 1 + 15 ·3 + n2 ·4 = 69 elements. Hence, assume that two of them,say P and Q intersect nontrivially. Then, |P∩Q| = 2. Let H = NG(P∩Q). Since P and Q are Abelian thenboth are contained in H. So |H| > 4 and a multiple of 4, then |H| = 12 or 20.

If |H| = 20 then its index is 3 and Theorem 3.16 implies that G is not simple. If |H| = 12 then its index is 5and this case is reduced to Case 1.

Example 5.12. a) Show that A5 does not contain subgroups of order 15, 20, or 30.b) Show that A5 contains a subgroup of order 10 and a subgroup of order 6.

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Proof. The proof of part a) is an immediate consequence of the Theorem 3.16 and the fact that A5 is simple.b) Since 60 = 22

·3 ·5 then from the Sylow’s Theorem we get

n2 = 1,3,5,15n3 = 1,4,10n5 = 1,6

Since A5 is simple than

n2 = 3,5,15n3 = 4,10n5 = 6

Let’s assume that n3 = 4. Since n3 is the index of the normalizer of P3 then exists a subgroup H := NA5 (P3)such that [A5 : H] = 4. But 606 | 4!, hence A5 is not simple by the Theorem 3.16. Hence, n3 = 10. But thenagain n3 is the index of the normalizer of P3. Thus, H := NA5 (P3) has order 6.

We know that n5 = 6. Since n5 is the index of the normalizer of P5 then this normalizer has order 10.This completes the proof.

Let us now try to generalize some of the results for An, n ≥ 5.

Lemma 5.14. Let N be a normal subgroup of An, where n ≥ 3. If N contains a 3 -cycle, then N = An.

Proof. First we prove that An is generated from 3-cycles of type (i jk), where i and j are fixed in 1,2, . . . ,nand let k vary. Every 3-cycle is the product of 3-cycles of this type, because

(ia j) = (i ja)2

(iab) = (i jb)(i ja)2

( jab) = (i jb)2(i ja)(abc) = (i ja)2(i jc)(i jb)2(i ja).

Assume that N is a normal nontrivial subgroup of An for n ≥ 3 such that N contains a 3-cycle of the form(i ja). Using the fact that N is normal we have that

[(i j)(ak)](i ja)2[(i j)(ak)]−1 = (i jk)

is in N. Thus, N must contain all 3-cycles (i jk), for 1 ≤ k ≤ n. From Lemma 2.2, these 3-cycles generate An.Thus, N = An.

Lemma 5.15. For n ≥ 5, for every normal subgroup N of An contains a 3 -cycle.

Proof. Let σ be any element in a normal subgroup N. Then, the possible cyclic types for σ are:

• σ is a 3-cycle.

• σ is the product of disjoint cycles, σ = τ(a1a2 · · ·ar) ∈N, where r > 3.

• σ is the product of disjoint cycles, σ = τ(a1a2a3)(a4a5a6).

• σ = τ(a1a2a3), where τ is the product of disjoint 2-cycles.

• σ = τ(a1a2)(a3a4), where τ is the product of an even number of disjoint 2-cycles.

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If σ is a 3 -cycle, then this completes the proof. If N contains a product of disjoint cycles, σ and at leastone of these cycles has length bigger than 3, so σ = τ(a1a2 · · ·ar). Then

(a1a2a3)σ(a1a2a3)−1

is in N because N is normal. Thus,σ−1(a1a2a3)σ(a1a2a3)−1

is also in N. Since

σ−1(a1a2a3)σ(a1a2a3)−1

= σ−1(a1a2a3)σ(a1a3a2)= (a1a2 · · ·ar)−1τ−1(a1a2a3)τ(a1a2 · · ·ar)(a1a3a2)= (a1arar−1 · · ·a2)(a1a2a3)(a1a2 · · ·ar)(a1a3a2)= (a1a3ar),

N contains a 3-cycle. Thus, N = An.Assume that N contains a disjoint product of the form

σ = τ(a1a2a3)(a4a5a6).

Then,σ−1(a1a2a4)σ(a1a2a4)−1

∈N

since(a1a2a4)σ(a1a2a4)−1

∈N.

Thus,

σ−1(a1a2a4)σ(a1a2a4)−1

= [τ(a1a2a3)(a4a5a6)]−1(a1a2a4)τ(a1a2a3)(a4a5a6)(a1a2a4)−1

= (a4a6a5)(a1a3a2)τ−1(a1a2a4)τ(a1a2a3)(a4a5a6)(a1a4a2)= (a4a6a5)(a1a3a2)(a1a2a4)(a1a2a3)(a4a5a6)(a1a4a2)= (a1a4a2a6a3).

Hence N contains a cycle with length bigger than 3 and we proceed as above.Assume that N contains a disjoint product of the type σ = τ(a1a2a3), where τ is the product of disjoint

2-cycles. Since σ ∈N, σ2∈N, and

σ2 = τ(a1a2a3)τ(a1a2a3)= (a1a3a2).

Hence, N contains a 3-cycle.It is left only the case when we have a disjoint product of the form:

σ = τ(a1a2)(a3a4),

where τ is the product of an even number of disjoint 2-cycles. However

σ−1(a1a2a3)σ(a1a2a3)−1

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is in N since (a1a2a3)σ(a1a2a3)−1 is in N. Thus, :

σ−1(a1a2a3)σ(a1a2a3)−1

= τ−1(a1a2)(a3a4)(a1a2a3)τ(a1a2)(a3a4)(a1a2a3)−1

= (a1a3)(a2a4).

Since n ≥ 5, we can find b ∈ 1,2, . . . ,n such that b , a1,a2,a3,a4. Let µ = (a1a3b). Then,

µ−1(a1a3)(a2a4)µ(a1a3)(a2a4) ∈N

and

µ−1(a1a3)(a2a4)µ(a1a3)(a2a4)= (a1ba3)(a1a3)(a2a4)(a1a3b)(a1a3)(a2a4)= (a1a3b).

Thus, N contains a 3-cycle. The proof is completed.

Theorem 5.3. The alternating group, An, is simple for n ≥ 5.

Proof. Let N a normal subgroup of An. From Lemma 5.15, N contains a 3-cycle. From Lemma 5.14, N = An.Thus, An does not contain proper normal subgroups for n ≥ 5.

5.5.2 Other simple groups

Let’s first see the following elementary exercise.

Lemma 5.16. Let p and q be prime numbers such that p < q and G a group with order |G| = pq. Let P ∈ Sylp(G) andQ ∈ Sylq(G). Then the following are true.

i) Q C G and G is not simple.

ii) If q . 1 (mod p), then G is cyclic.

iii) If P C G, then G is cyclic.

Proof. i) From Sylow’s theorem we know that G contains a subgroup Q with order q. Also, nq ≡ 1 mod q.Since q > p, then nq - p and so nq = 1. Thus, Q C G. Thus, G is not simple.

ii) There exists P ∈ Sylp(G) with order p. Then, np = 1 or np = q and q ≡ 1 mod p. Since q . 1 (mod p),then np = 1 the P C G. Since P∩Q = eG then GP×Q. Thus, GZp×Zq. However Zp×ZqZpq when(p,q) = 1.

iii) complete ...

Exercise 5.9. Prove that every group of order 30 has a subgroup of order 15.

Remark 5.3. Notice that the above Exercise could be stated that "Every subgroup of order 30 has a cyclic, normalsubgroup of order 15."

Exercise 5.10. Prove that every group of order 12 either has a normal Sylow 3-subgroup or is isomorphic to A4.

Lemma 5.17. Let p and qbe prime numbers. Any group G of order |G| = p2 q, is not simple.

Proof.

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5.5.3 Classification of simple groups

The following result is one of the most celebrated and most important results of the XX-century mathematics.

Theorem 5.4. Every finite simple group is isomorphic to one of the following groups:

• a member of one of three infinite classes of such, namely:

– the cyclic groups of prime order,

– the alternating groups of degree at least 5,

– the groups of Lie type

• one of 26 groups called the "sporadic groups"

• the Tits group (which is sometimes considered a 27th sporadic group).

Below we give a brief history of the problem and the people involved.

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Exercises:

5.15. Let G be a finite group and let H and K be subgroups of G such that HCp×Cp and KCp2 . Prove that p3| |G|.

5.16. Let G be a finite group which has a unique maximal subgroup. Show that G is cyclic if and only if it has primepower order.

5.17. Let p and q be primes. Show that there is no simple group of order p2q.

5.18. Let G be a finite group, P ∈ Sylp(G), and H C G. Prove that H∩P is a Sylow subgroup of H and HP/H is aSylow subgroup of G/H.

5.19. Let G be a finite group and P ∈ Sylp(G). Give an example of a subgroup H of G with H∩P not a Sylowsubgroup of H.

5.20. Let G be a finite group, P ∈ Sylp(G) and H C G. Prove that H∩P is a Sylow subgroup of H and HP/H is aSylow subgroup of G/H.

5.21. Let be given G a finite group and P ∈ Sylp(G). Give an example of a subgroup H of G where H∩P is not aSylow subgroup of H.

5.22. Let be given G a finite group in which every Sylow subgroup is normal. Prove that G is isomorphic to the directproduct of its Sylow subgroups.

5.23. Let be given p a prime number, n> 1 an integer and G = GLn(Fp). Prove that there exist two Sylow p -subgroupsH1 and H2 in G such that H1∩H2 = e.

5.24. Let be given |G| = pqr where p,q,r are primes and p < q < r. Prove that G has a Sylow normal subgroup for p,q or r.

5.25. Let be given the group G such that |G| = 495 = 32·5 ·11. Prove that G has a normal subgroup with order 5 or

11.

5.26. Prove that a group with order 105 has a element with order 15.

5.27. What is the order all Sylow p - subgroups when the group G of has the order 18, 24, 54, 72 and 80?

5.28. Find all Sylow 3-subgroups of S4 and prove that they are all conjugated.

5.29. Prove that every group with order 45 has a normal subgroup with order 9.

5.30. Let H be a Sylow p -subgroup of the group G. Prove that H is the only Sylow p -subgroup of G that is containedin N(H).

5.31. Prove that no group with order 96 is simple.

5.32. Prove that no group with order 160 is simple.

5.33. If H is a normal subgroup of a finite group G and |H| = pk for a prime number p, prove that H is contained insome Sylow p -subgroup of G.

5.34. Let G be a group with order p2q2, where p and q are two distinct prime numbers such that q 6 | p2−1 and p 6 | q2

−1.Prove that G must be Abelian. Find three pair of numbers p and q which satisfy these conditions.

5.35. Show that a group with order 33 has only a Sylow 3-subgroup.

5.36. Prove that a group with order 108 must have a normal subgroup.

5.37. Classify all groups with order 175 up to isomorphisms.

5.38. Prove that every group with order 255 is cyclic.

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5.39. Prove that a Sylow 2-subgroup of S5 is isomorphic to D4.

5.40. Prove that any group G of order 20 is not simple.

5.41. Prove that groups with order 4, 8, 9, 16, 25, 27, 32, 49, 64 and 81 are not simple and groups with order 4, 9,25 and 49 are Abelian.

5.42. A group of order 56 = 23·7 is not simple.

5.43. Prove that a group G with order 48 is not simple.

5.44. Prove that for every simple group with order 60 is isomorphic to A5.

5.45. Prove that there are no simple groups of order 264.

5.46. Prove that there are no simple groups of order 3159.

5.47. Let be given G a simple group with order 168. Prove thata) n2 = 21, n3 = 7, n7 = 8.b) Sylow 2-subgroups of G are dihedral, Sylow 3-subgroups and Sylow 7-subgroups are cyclic.

5.48. Prove that GL3(2) is a simple group with order 168.

5.49. Prove that if |G| = 462, then G is not simple.

5.50. Prove that if |G| = 132, then G is not simple.

5.51. Let G be a simple group of order 168. Show that i) n2 = 21, n3 = 7, n7 = 8.ii) Sylow 2-subgroups of G are dihedral, Sylow 3-subgroups and Sylow 7-subgroups are cyclic.

5.52. Prove that SL2(F4)A5.

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John Griggs Thompson

John Griggs Thompson, (born October 13, 1932, Ottawa, Kansas,U.S.), American mathematician who was awarded the Fields Medalin 1970 for his work in group theory. In 2008 the Norwegian Academyof Science and Letters awarded Thompson and Jacques Tits of Francethe Abel Prize for their ?profound achievements in algebra and inparticular for shaping modern group theory.?

Thompson earned a B.A. from Yale University in 1955 and a Ph.D.from the University of Chicago in 1959. After a year at HarvardUniversity (1961–62), he returned to the University of Chicago (1962–68), and he subsequently moved to Churchill College, Cambridge,England.

Thompson was awarded the Fields Medal at the InternationalCongress of Mathematicians in Nice, France, in 1970. His work waslargely in group theory. In 1963 he and Walter Feit published theirfamous theorem that every finite simple group that is not cyclic hasan even number of elements, a proof requiring more than 250 pages.Because every finite group is made up of composition factors, buildingblocks that are finite simple groups theorems about simple groupshave ramifications for all finite groups.

The subsequent work that resulted in Thompson’s receiving the Fields Medal was the determinationof all the minimal simple finite groups?that is, those groups all of whose proper subgroups are builtonly of cyclic composition factors. Thompson’s revolutionary ideas inspired and permeated an effort,hitherto considered hopeless, to determine all the finite simple groups. The solution of this problem, theso-called "Enormous Theorem", was announced in 1981 and represents the combined efforts of hundredsof mathematicians in separate journal articles consuming well over 10,000 pages. Thompson made furthercontributions to Galois theory, representation theory, coding theory, and, working on the proof of thenonexistence of a plane of order 10, the theory of finite projective planes.

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Chapter 6

Direct products and Abelian groups

One of the main problems in group theory is that of classifying all groups, up to isomorphism. For example,we showed in the previous section that all cyclic groups are isomorphic to Z or Cn. Can we accomplishthis in general? This was one of the main mathematical problems of the XX century. For more historicalcomments see section three in this chapter.

6.1 Direct products

The direct products of groups is a concept that we are familiar from linear algebra, even though we havenot called it by this name there.

Consider two groups (G1,?) and (G2,). Is it possible to construct a new group on the Cartesian productG1×G2?

6.1.1 The outer direct product

Let (G1,?) and (G2,) be two groups, then taking their Cartesian product we can form a new group on theset G×H. We can define a binary operation in G1×G2 such that

(G1×G2)× (G1×G2) 7→ (G1×G2) (6.1)((x1,x2), (y1, y2)

)= (x1?x2, y1 y2). (6.2)

Exercise 6.1. Prove that G1×G2 together with the operation defined above forms a group.

The group G1 ×G2 is called the outer direct product of G1 and G2. We have seen examples of outerdirect products before. For example, the Euclidean space R×R together with addition is an outer directproduct; see [?lin-alg] for details.

We can also construct the outer direct product of more then two groups. Thus, the n direct product

n∏i=1

Gi = G1×G2× · · ·×Gn

of groups G1,G2, . . . ,Gn is defined in the same way. Hence, G is merely the Cartesian product of groupsG1, . . . ,Gn with a new binary operation defined component-wise. We will denote by ei the identity elementfor each Gi, i = 1, . . . ,n.

Lemma 6.1. Let (g1, g2) ∈ G1×G2. If g1 and g2 have finite orders, respectively m and n, then

|(g1, g2)| = lcm (m,n).

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Proof. Assume that l = lcm (m,n) and s = |(g1, g2)|. Then, obviously

(g1, g2)l = (gl1, g

l2) = (e1,e2)

(gs1, g

s2) = (g1, g2)s = (e1,e2).

Thus, s must divide l and s ≤ l. However, m and n also must divide s. Thus, s = lcm (m,n). The reader should prove the following:

Exercise 6.2. Let G =∏n

i=1 Gi and g ∈ G such that g = (g1, . . . , gn). If gi has finite order ri in Gi, then

|(g1, . . . , gn)| = lcm (r1, . . . ,rn).

The following result will be quite useful later on. We leave its proof as an exercise for the reader.

Lemma 6.2. Cm×CnCmn if and only if gcd (m,n) = 1.

The i-th projection πi is defined as follows:

πi : G1×G2× · · ·×Gn 7→ Gi (6.3)(x1, . . . ,xi, . . .xn) 7→ xi (6.4)

(6.5)

Exercise 6.3. Is πi a homomorphism?

Consider now the other maps φi for i = 1, . . . ,n defined as follows:

φi : G1×G2× · · ·×Gn 7→ G1×G2× · · ·×Gn (6.6)(x1, . . . ,xi−1,xi,xi+1, . . .xn) 7→ (x1, . . . ,xi−1,ei,xi+1, . . . ,xn) (6.7)

(6.8)

So φi is the identity map on each G j, j , i and it is the constant map xi→ ei on Gi.

Exercise 6.4. Prove that φi a homomorphism for each i = 1, . . . ,n.

Exercise 6.5. Interpret geometrically the maps φi, when G =R2 or G =R3 with addition of vectors.

Let Gi = ker(φi). Then Gi is isomorphic to Gi. Moreover, Gi is a subgroup of G and Gi C G since it is thekernel of a homomorphism.

Given an element (x1, . . . ,xn) ∈ G we have

(x1, . . . ,xn) = (x1,e2, . . . ,en)(e1,x2,e3, . . . ,en) . . . (e1, . . . ,en−1,xn) = x1 · · ·xn,

where xi ∈ Gi. Moreover, this expression of an element (x1, . . . ,xn) ∈ G as a product of elements in G1, . . . ,Gnis unique. So, summarizing we have:

Remark 6.1. Thus, G = G1×G2×· · ·×Gn is constructed by certain normal subgroups G1, . . . ,Gn and every elementg ∈ G can be written in a unique way as a product g = g1 · · · gn, for gi ∈ Gi.

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Before we go on, let’s just recall that this is not totally new to us. Consider x ∈Rn, such x =

x1x2...

xn

. Then,

x =

x1x2...

xn

=

x10...0

+

0x2...0

+ · · ·+

00...

xn

since R is an additive group and ei = 0 for i = 1, . . . ,n. The fact that x can be written uniquely this way inlinear algebra means that x it is written uniquely as a linear combination in terms of the standard basis; see[?lin-alg, Chap. 2] for details.

Let us now get back to Remark 6.1. So the group G is given as G = G1G2 · · ·Gn, where every element ofg ∈ G is written uniquely as g = g1 · · · gn, for gi ∈ Gi. Moreover, from construction we have that Gi C G andG∩G j = e for all i , j.

This motivates the following definition.

Definition 6.1. A group G is said to be the inner direct product of its normal subgroups N1, . . . , Nn, if every g ∈Gis written uniquely as g = g1 · · · gn, for gi ∈Ni.

6.1.2 The inner direct product

We will see some additional ways of characterizing the inner direct products. First the following lemma.

Lemma 6.3. Let G be a group and H and K normal subgroups of G such that H∩K = e. Then, for every h ∈H andk ∈ K we have that hk = kh.

Proof. start here

Then we have the following.

Lemma 6.4. If G is the inner direct product of its normal subgroups N1, . . . , Nn, then for i , j, Ni ∩N j = e.Moreover, for i , j and gi ∈Ni and g j ∈N j we have that gig j = g jgi.

Proof. start here

Hence we have proved that the definition of the inner product in G to a collection of subgroups

H1,H2, . . . ,Hn of G, which must have the following properties

• G = H1H2 · · ·Hn = h1h2 · · ·hn : hi ∈Hi

• Hi∩〈∪ j,iH j〉 = e

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• hih j = h jhi for every hi ∈Hi and h j ∈H j.

Sometimes the inner product is defined via the properties above. Now we have the following theorem.

Theorem 6.1. If G be the inner direct product of its normal subgroups

G = H1 · · ·Hn,

then G is isomorphic to the outer direct product

GH1× · · ·×Hn.

Proof. From now on we will speak of simply direct product and drop the adjectives inner or outer.

Exercises:

6.1. Let G be an inner direct product of subgroups H and K. Prove that the function φ : G→ H×K such thatφ(g) = (h,k) for g = hk, where h ∈H and k ∈ K, is injective and surjective.

6.2. Prove or disprove: If every element of G has finite order, then G is finite.

6.3. Let n1, . . . ,nk be positive integers. Prove that

k∏i=1

Cni Cn1···nk

if gcd (ni,n j) = 1 for i , j.

6.4. If m = pe11 · · ·p

ekk , where pi are distinct prime numbers, then

CmCpe11× · · ·×Cp

ekk.

6.5. If G1 and G2 are groups, prove that G1×G2G2×G1.

6.6. Let G be a group. The diagonal group D in G×G is the group

D = (g, g) ∈ G×G | g ∈ G

Prove that

1. Prove that DG.

2. Prove that D C G×G if and only if G is Abelian.

6.7. Let G be a finite group, H1, . . . ,Hn normal subgroups of G such that G = H1 · · ·Hn and |G| = |H1| · · · |Hn|. Provethat G is the direct product of H1, . . . ,Hn.

6.8. Let G be a group and H1, . . . ,Hn normal subgroups of G such that

1. G = H1 · · ·Hn

2. For each i, Hi∩ (H1 . . .Hi−1Hi+1 · · ·Hn = e

Prove that G is the direct product of H1, . . . ,Hn.

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6.9. If G1, G2, G3 are groups, prove that

(G1×G2)×G3G1×G2×G3.

Generalize to G1, . . . ,Gn.

6.10. Use Lemma 6.2 to prove the Chinese Remained Theorem: if m and n are relatively prime integers and u,v ∈Z,there there exists x ∈Z such that x ≡ u mod m and x ≡ v mod n.

6.11. Give an example of a group G and normal subgroups N1, . . . ,Nn such that G = N1 . . .Nn and Ni∩N j = e forall i , j, but G is not the inner direct product of N1, . . .Nn.

6.12. Let G be a group and N1, . . . ,Nn subgroups of G such that N1∩N2∩· · ·∩Nn = e. Let Vi = G/Ni. Prove that

GV1× · · ·×Vn.

6.13. Let G be a finite Abelian group which contains a proper subgroup H0 which in contained in every propersubgroup of G. Prove that G is cyclic. What is the order of G?

6.14. Let G be a finite Abelian group. Use ?? to prove that G is isomorphic to a subgroup of a direct product of a finitenumber of cyclic groups.

6.15. Let G be a group of order p2, where p is a prime. Prove that

GCp2 or GCp×Cp.

6.16. Find Aut (Cp×Cp).

6.17. Let G = H1× · · ·×Hn. What is the center Z(G)?

6.18. if G = H1× · · ·×Hn and g ∈ G, can you determine

N(g) = x ∈ G | xg = gx

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6.2 Finite Abelian groups

We proved that every group with prime order is isomorphic to Cp, where p is a prime number. We alsoknow that CmnCm ×Cn, if and only if gcd (m,n) = 1, see Lemma 6.2. In this section we will classify allfinitely generated Abelian groups.

Let G be a group and let A be a subset of G, not necessarily finite. The smallest subgroup of G whichcontains all elements of A is called the subgroup generated by A and denoted by 〈A〉. If this subgroup of Gcontains completely G, then we say that the group G is generated by the set A and denoted by G = 〈A〉. Inthis case elements of A are called generators of G. If there is a finite set A which generates G, then we saythat the group G is finitely generated.

Let us first see some elementary examples of finitely generated groups.

Example 6.1. All finite groups are finitely generated. For example, the group S3 is generated from permutations(12) and (123). The group Z×Zn is a infinite group but is finitely generated by (1,0), (0,1).

Example 6.2. Let G be a group and H a proper subgroup of G. Show that 〈G\H〉 = G.

Proof. Notice that the group 〈G \H〉, generated by G \H, is a subgroup of G. Also, 〈G \H〉∪H = G. FromExample 2.1 we have that either 〈G\H〉 ⊆H or H ⊆ 〈G\H〉. But 〈G\H〉 ⊆H is impossible. Thus, H ⊆ 〈G\H〉and so G = 〈G\H〉.

Lemma 6.5. Let G be a group and H ≤ G, such that

H = 〈gi ∈ G : i ∈ I〉.

Then, h ∈H if and only if h is a product of the form

h = gα1i1· · · gαn

in,

where gik are not necessarily distinct.

Proof. Let K be the set of all products of the form gα1i1· · · gαn

in, where gik are not necessarily distinct. Hence K

is a subset of H. We must prove that K is a subgroup of the group G. If this holds then K = H because H isthe smallest subgroup which contains all gi.

The set K is closed under the operation of the group. Since g0i = 1, identity is in K. The inverse of

g = gk11 · · · g

knin

in K is also in K, since

g−1 = (gk11 · · · g

knin

)−1 = (g−kn1 · · · g−k1

in) ∈ K.

This completes the proof.

Remark 6.2. Notice that we allow that gi to repeat because we are not assuming that G is an Abelian group. In thecase of Abelian group each gi appears only once.

The following is an immediate consequence of the Sylow’s theorem.

Theorem 6.2. Every finite Abelian group G is the direct product of its Sylow subgroups.

Proof. Assume that|G| = pα1

1 · · ·pαnn ,

where p1, . . . ,pn are primes. Let Pi ∈ Sylpi(G). Since G is an Abelian group, then all Pi are normal in G for

i = 1, . . . ,n. Hence, G = P1 · · ·Pn. Hence,G = P1× · · ·×Pn,

since their intersection is just the identity.

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Corollary 6.1. Let G be an Abelian group such that its Sylow subgroups are cyclic, then G is cyclic.

Proof. From the above theorem, G is the direct product of its cyclic Sylow subgroups. Since their orders arecoprime, then this direct product is a cyclic group.

In our quest of classifying all finite Abelian groups the main task now becomes classifying all isomor-phism types of finite Abelian p-groups.

Let G be an Abelian p-group, say |G| = pα. Can we write down all the isomorphism classes for G?

Lemma 6.6. If G be an Abelian p-group such that |G| = pα, then G is isomorphic to one of the following

GCpβ1 ×Cpβ2 × · · ·×Cpβt

with β1 ≥ β2 ≥ · · · ≥ βt ≥ 1 and β1 +β2 + · · ·βt = α.

Proof. Start here ....

Notice that the above result, gives many groups as possibilities to be isomorphic to G. More precisely,

for each choice of integers(β1,β2, . . . ,βt), such that β1 +β2 + · · ·βt = α,

gives us a possible group. Hence, to list all possible groups we have to list all the partitions of the positiveinteger α.

There is a huge amount of literature in elementary number theory on the number of partitions and howto represent partition through Ferrers diagrams, Young diagrams and other diagrams. To avoid doublecounting we will follow the usual conventions that β1 ≥ β2 ≥ · · · ≥ βt ≥ 1. The integers β1, . . . ,βt are calledthe invariant factors and integers pβi above are called the elementary divisors of G.

Example 6.3. Classify all Abelian groups of order 35.

Solution: Hence, we are looking first for the partitions of 5. The seven partitions of 5 and the corresponding

groups are:

5 C35 C4054 + 1 C34 ×C3 C81×C33 + 2 C33 ×C32 C27×C93 + 1 + 1 C33 ×C3×C3 C27×C3×C32 + 2 + 1 C32 ×C32 ×C3 C9×C9×C32 + 1 + 1 + 1 C32 ×C3×C3×C3 C9×C3×C3×C31 + 1 + 1 + 1 + 1 C3×C3×C3×C3×C3 C3×C3×C3×C3×C3

which consists of all Abelian groups of order 405.

Example 6.4. Write out, up to isomorphism, all Abelian groups G of order 540 = 22·33·5.

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Solution: From the theorem we know that G is isomorphic to the direct product of its Sylow subgroups,

GP2×P3×P5

Since |P2| = 4 then P2 is isomorphic toP2C4, P2C2×C2.

The Sylow 3-subgroup P3 has order 33. The partitions of the exponent 3 are

3, 3 = 2 + 1, 3 = 1 + 1 + 1.

Hence, P3 is isomorphic toC27, C9×C3, C3×C3×C3.

The Sylow 5-subgroup has order 5 and therefore it is isomorphic to C5. Putting all cases together we have:

• C4×C27×C5

• C4×C9×C3×C5

• C4×C3×C3×C3×C5

• C2×C2×C27×C5

• C2×C2×C9×C3×C5

• C2×C2×C3×C3×C3×C5

Remark 6.3. Every group G with order |G| = pα has a subgroup of order pβ, for every β such that β ≤ α.

Solution: Indeed

GCβ1p ×Cβ2

p × · · ·×Cβnp ,

each one has a subgroup of order p since they are cyclic.

Remark 6.4. Let m and n be positive integers such that m | n. Every Abelian group of order n has a subgroup oforder m.

Proof. Let |G| = pα11 ·p

α22 · · ·p

αnn and m = pβ1

1 ·pβ22 · · ·p

βnn . Then,

GA1×A2× · · ·×An,

where |Ai| = pαii . Thus, Cp×Cp× · · ·×Cp ≤ Ai.

We turn our attention to our initial problem; that of classifying all finite Abelian groups of a given order.From the previous two lemmas we know that such groups are a direct product of its Sylow subgroups andeach such Sylow subgroup is a direct product of cyclic groups. We combine these two result is a singletheorem which is usually called the Fundamental Theorem of Finite Abelian Groups or as sometimes calledthe Primary Decomposition Theorem.

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Theorem 6.3 (Primary Decomposition Theorem). Let G be a finite Abelian group of order

|G| = n = pα11 ·p

α22 · · ·p

αkk

Then, the following hold:

1) GG1×G2× · · ·×Gk, where |Gi| = pαii .

2) For each Gi with |Gi| = pα we haveGiCpβ1 ×Cpβ2 × · · ·×Cpβt

with β1 ≥ β2 ≥ · · · ≥ βt ≥ 1 and β1 +β2 + · · ·βt = α.

Moreover the decomposition above is unique.

Proof. The theorem is a consequence of the previous two Lemmas. The decomposition of G as above is called the invariant factor decomposition and integers n1 =

pα11 , . . . ,nk = pαk

k are called the invariant factors of G. The group G with invariant factors as above iscalled a group of type (n1, . . . ,nk).

Example 6.5. List all possible invariant factors and the corresponding Abelian groups G of order

|G| = 252 = 22·32·7

Solution: The choices for n1 are

2 ·3 ·7, 22·3 ·7, 22

·32·7, 2 ·32

·7.

We consider each case below:Case n1 = 2 ·3 ·7: Then, n2 = 2 ·3. The decomposition is

C42×C6

Case n1 = 22·3 ·7: Then, n2 = 3. The decomposition is

C84×C3

Case n1 = 22·32·7: Then, n2 = 1. The decomposition is

C252

Case n1 = 2 ·32·7: Then, n2 = 2. The decomposition is

C126×C2

Example 6.6. Let G be a finite abelian group of order pn. Prove that if G has exactly one subgroup of order p, then Gis cyclic. Is "abelian" condition necessary?

Solution: Let |G| = pn. From the Fundamental Theorem of Finitely Generated Abelian Groups we have

G = Cpi1 ×Cpi2 × · · ·×Cpin

where i1 + i2 + · · ·+ ik = n.Recall that all Cpin are cyclic. So

G = 〈g1〉× 〈g2〉× · · · 〈gn〉

of order p,pi, · · ·pi precisely. But from Fundamental Theorem of Cyclic Groups, we know that every cyclic group oforder pi has a subgroup of order p. So we have many other subgroups of order p, which can’t be true. So G = 〈g1〉.

Yes, the condition that G is abelian is necessary. Consider Q8.

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Exercises:

6.19. Prove that the infinite direct product G =Z2×Z2× · · · is not finitely generated.

6.20. A group G is called a torsion group if every element of G has finite order. Prove that a torsion group is finitelygenerated.

6.21. Let G, H and K be finitely generated Abelian groups. Prove that if G×HG×K, then HK. Give acounterexample to prove that this result is not in general true.

6.22. (Q,+) is not finitely generated.

6.23. Prove that a group with order G = 5 ·7 ·17 is cyclic.

6.24. Find all Abelian groups with smallest order or equal with 40 up to isomorphisms.

6.25. Find all Abelian groups with order 200 up to isomorphisms.

6.26. Find all Abelian groups with order 720 up to isomorphisms.

6.27. Let G be a finite abelian group of order n and p a prime divisor of n. Show that G contains an element of order p.

6.28. Let G be a finite Abelian group with |G| square free. Show that G is cyclic.

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6.3 Free groups and Finitely generated Abelian groups

Now that we know how to classify the isomorphism classes of finite Abelian groups we turn our attentionto a larger class of groups, namely finitely generated Abelian groups. Recall that a group G is finitelygenerated if there is a finite set A such that G = 〈A〉.

Let G be any Abelian group. An element g ∈G is called a torsion element if it has a finite order. The setof all torsion elements of G forms a subgroup of G which is denoted by Tor(G) or Gtor and called the torsionsubgroup of G.

Exercise 6.6. Prove that Gtor is a subgroup of G.

An Abelian group G is called torsion group if G = Gtor. Next, we will see how we can decompose thefinitely generated Abelian groups into a direct product of the torsion part and the free part of G. The freepart of G is a group isomorphic to G/Gtor which we will discuss in detail next.

6.3.1 Free groups

Lemma 6.7. Let X be a set of cardinality n. Then the free group F(X) generated by X is isomorphic to

F(X)Z× · · ·Z =Zn

6.3.2 Finitely generated Abelian groups

Theorem 6.4. Let G be a finitely generated Abelian group. Then, Gtor is finite. Moreover,

GGtor×Zr

for some r ≥ 0.

Proof. See ?? for details. The integer r is called the rank of the group G (or Betti number) and is the same as the rank of the free

group G/Gtor.

6.3.3 An application: elliptic curves

In 1.81 we showed that the set of rational points (E(Q) of an elliptic curve is an Abelian group. Indeed, itcan be shown that (E(Q) is finitely generated.

Theorem 6.5 (Mordel-Weil). Let E be an elliptic curve defined over Q. Then, (E(Q) is finitely generated.

From Theorem 6.4 we have thatEEtor×Z

r

The integer r ≥ 0 is called the rank of the elliptic curve.

Conjecture There are elliptic curves with arbitrary large rank.

Determining the torsion part of n elliptic curve is interesting on its own right. There is a theorem ofMazur which determines all possibilities

Exercises:

6.29. List all the Abelian groups of order 420.

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6.30. Prove that S3×C2 is isomorphic to D6. What can you guess for D2n? Prove your guess.

6.31. Prove or disprove: Every Abelian group with order divisible by 3 contains a subgroup with order 3.

6.32. Prove or disprove: Every non Abelian group with order divisible by 6 contains a subgroup with order 6.

6.33. Prove or disprove: Let G, H, and K be groups. If G×KH×K, then GH.

6.34. Let p be a prime number. Prove that the number of the distinct Abelian groups with order pn (up to isomorphism)is the same as the number of conjugacy classes in Sn.

6.4 Canonical forms

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Chapter 7

Solvable Groups

7.1 Normal series and the Schreier theorem

We will start first by defining normal series of groups. The main result here is Schreier’s theorem that anytwo normal series are equivalent when refined appropriately.

Let G be a group. A normal series of a group G is a chain of subgroups

1 C G1 C . . . C Gn = G.

The groups Gi+1/Gi are called factor groups of the normal series and n is the length of the series. Twonormal series of G are equivalent if they have the same length and isomorphic factor groups.

Let1 C G1 C . . . C Gn = G

be a normal series of G. Then1 C H1 C . . . C Hm = G

is a refinement of the first normal series if

G1, . . . ,Gn ⊂ H1, . . . ,Hm.

The following useful technical result will be needed later on.

Lemma 7.1 (Butterfly Lemma (Zassenhaus)). Let A?,B? ≤ G and A C A?, B C B?. Then,

A(A?∩B) C A(A?

∩B?),

B(B?∩A) C B(B?∩A?).

Moreover,A(A?

∩B?)/A(A?∩B)B(B?∩A?)/B(B?∩A)

Proof. to be completed ...

Theorem 7.1 (Schreier). Any two normal series of an arbitrary group G have refinements that are equivalent.

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7 Shaska T.

A? B?

A(A?∩B?)

==

(A?∩B?)B

==A?∩B?

==A(A?∩B) (A∩B?)B

A B

A∩B? A?∩B

Figure 7.1: Butterfly Lemma

Proof. Homework

Definition 7.1. A normal series1 C G1 C . . . C Gn = G

is a composition series if each Gi is maximal normal in Gi+1.

Theorem 7.2 (Jordan - Hölder). Any two composition series of a group are equivalent.

Proof. Composition series are normal series and from Schreier’s theorem they have equivalent refinements.But composition series are already refined (by the maximality condition).

Exercises:

7.1. Show that for every group G, Z(G) is a characteristic subgroup of G. Find an example where Z(G) is not fullyinvariant (i.e. φ(Z(G)) ⊂ Z(G) where φ is an endomorphism of G).

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7.2 Solvable groups

Definition 7.2. A group G is called solvable if it has a normal series

1 C G1 C . . . C Gn = G,

such that Gi+1/Gi is Abelian for each i.

Definition 7.3. The higher commutator subgroups of G are defined inductively:

G(0) = G, G(i+1) = G(i)

where G(i+1) is the commutator of G(i). The series

G = G(0)≥ G(1)

≥ . . .

is called the derived series of G.

Lemma 7.2. For each G we haveG(i) cC G, for all i

Proof. Homework

Exercise 7.1. Show that there exist solvable groups of arbitrary large derived lengths

Theorem 7.3. A group G is solvable if and only if G(n) = 1 for some n.

Proof. Let assume that G is solvable. Then, there is a series

1 = G0 C G1 C . . . C Gn = G

such that each Gi+1/Gi is Abelian. We will show that G(n) = 1.Claim: For all i ≤ n, G(i)

≤ Gn−i.We proceed induction. If i = 0 then G0 = G = G(0). Assume that it is true for i = s. So G(s)

≤ Gk−s. Then,

(G(s))′ ≤ (Gk−s)′.

Hence, G(s+1)≤ (Gk−s)′. But Gk−s/Gk−s−1 is Abelian (solvable series). Then,

(Gk−s)′ ≤ Gk−s−1

(see ??) which impliesG(i+1)

≤ Gk−(s+1).

Hence, the claim is proved.Thus, G(n)

≤ G1 which implies that G(n) = 1. This completes the proof.The converse follows from the fact that the G(i+1)/G(i) is Abelian for all i.

Lemma 7.3. Suppose that G is a finite solvable group. Then there is a chain

e CH1 CH2 C . . . CHm = G

of subgroups of G, so that each Hi+1/Hi is cyclic.

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7 Shaska T.

Proof. Since G is solvable, then there is a chain

e C G1 C G2 C . . . C Gn = G

For every step Gi C Gi+1 we get a composition series (Jordan - Hölder theorem)

Gi CHi,1 C . . . CHi,k C . . . C Gi+1

Doing this for all the steps we get a composition series

e CH1 CH2 C . . . CHm = G

where all the factors Hi,k/Hi,K+1 are simple. Since Hi,k+1/Hi,k ≤Gi+1/Gi then Hi,k+1/Hi,k is Abelian and simple.Hence, it is of prime order and therefore cyclic.

Lemma 7.4. The homomorphic image of a solvable group is solvable.

Proof. Let f : G→H be surjective homomorphism and

e C G1 C G2 C . . . C Gn = G

a composition series for G. One can show (by induction) that

f (G(i) ) = f (G)(i)

for all i. This gives a composition series for H.

Lemma 7.5. Every subgroup of a solvable group is solvable

Proof. Let G be a solvable group and H ≤ G. Since G is solvable then there exists a composition series

e C G1 C G2 C . . . C Gn = G

for G. Then, the seriese C (G1∩H) C (G2∩H) C . . . C (Gn∩H) = H

is a composition series.Indeed, let a ∈ Gi ∩H. Then, for every g ∈ Gi+1 ∩H we have that gag−1

∈ Gi since Gi C Gi+1. Also,gag−1

∈ H since both a, g ∈ H. Thus, for all a ∈ Gi ∩H and g ∈ Gi+1 ∩H we have gag−1∈ Gi ∩H. Hence,

(Gi∩H) C (Gi+1∩H).Consider now the factor group (Gi+1∩H)/(Gi∩H).

Lemma 7.6. Let G be a group and H C G. Then, G is solvable if and only if H and G/H are solvable.

Proof. If G is solvable then H is solvable as a subgroup of G, see Lemma 7.5. Also G/H is solvable as ahomomorphic image under the natural projection πG→ G/H.

Lemma 7.7. Sn is not solvable for n ≥ 5

Proof. A normal series of Sn is1 C An C Sn

Factor groups are Sn/AnC2 and An and have no normal subgroups. Hence, this is a composition series.From the Jordan-Hölder theorem, every composition series is equivalent to this series. But factor groupsof this series are not Abelian. Hence, Sn is not solvable.

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Theorem 7.4. Every finite p-group is solvable.

Proof. Use induction on |G|. Assume that the theorem is true for all groups of order < |G|. Since G/Z(G) isa p-group of order < |G|, then it is solvable. The same can be said for Z(G). Then the result follows as aconsequence of the previous theorem.

Theorem 7.5. (Hall) If G is a solvable group and G = ab, (a,b) = 1, then G contains a subgroup of order a. Moreover,any two subgroups of order a are conjugate.

Theorem 7.6. Let G be a finite group.i) (Burnside) If |G| = paqb for some primes p,q then G is solvable.ii) (Hall)iii) (Feit-Thompson) If |G| is odd then G is solvableiv) (Thomspon) If for every pair of elements x, y ∈ G, 〈x, y〉 is a solvable group, then G is solvable.

Exercises:

7.2. Let G and H be solvable groups. Prove that G×H is also solvable.

7.3. If G has a composition series (main) and if N is a proper normal subgroup of G, prove that there exists acomposition series (main) which contains N.

7.4. Prove that G is a solvable group if and only if when G has a series subgroups

G = Pn ⊃ Pn−1 ⊃ · · · ⊃ P1 ⊃ P0 = e

where Pi is normal in Pi+1 and the order of Pi+1/Pi is prime.

7.5. Prove that Dn is solvable for all n.

7.6. Assume that G has a composition series. If N is a normal subgroup of G, prove that N and G/N have also acomposition series.

7.7. Let G be a p -cyclic group which has subgroups H and K. Prove that H is contained in K or K is contained in H.

7.8. Assume that G is a solvable group with order n ≥ 2. Prove that G has a nontrivial Abelian subgroup.

7.9. Let p and q be distinct primes and G a group of order |G| = p2·q. Prove that G is solvable.

7.10. Let |G| = 495. Prove thata) G has a normal subgroup of order 55b) G is solvable.

7.11. Let G = 520. Prove thata) G has a normal cyclic subgroup of order 65b) G is solvable.

7.12. Let G = 36. Prove that G is solvable.

7.13. Let G = 108. Prove that G is solvable.

7.14. Prove that a solvable group having a composition series must be solvable.

7.15. If p and q are primes with p < q, then every group of order pqn is solvable.

7.16. If G is a group with |G| < 60, then G is solvable.

7.17. Prove that the following two statements are equivalent:

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• every group of odd order is solvable

• every finite simple group has even order

7.18. A group G is called supersolvable if it has a chain of subgroups

e = G0 ≤ g1 ≤ · · · ≤ Gn = G

such that every i = 1, . . . ,n we have Gi C G and Gi+1/Gi is cyclic. Find an example of a group which is solvable, butnot supersolvable.

7.19. Prove that S4 is not supersolvable.

7.20. Prove that every p-group is supersolvable.

7.21. If G has a composition series and if H C G, then G has a composition series one of whose terms is H.

7.22. If H and K are solvable subgroups of G with H C G, then HK is solvable.

7.23. Every finite subgroup has a unique maximal normal subgroup F (G) . Moreover, G/F (G) has no nontrivialnormal solvable subgroups.

7.24. Find a composition series for A4, S4.

7.25. For a group G the Frattini subgroup of G, denoted by Φ(G), is defined to be the intersection of all maximalsubgroups of G. Find Φ(S3), Φ(A4), Φ(S4), Φ(A5), and Φ(S5).

7.26. Assume that G is a finite solvable group. Prove that we can find a chain G = G0 ≥G1 ≥ · · · ≥Gk = e of G suchthat every Gi+1 is normal in Gi and Gi/Gi+1 is cyclic.

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Igor Shafarevich (1923-2017)

Igor Rostislavovich Shafarevich (3 June 1923 – 19 February 2017)was a Russian mathematician who contributed to algebraic numbertheory and algebraic geometry. He wrote books and articles thatcriticize socialism, and was an important dissident during the Sovietregime.Shafarevich died on 19 February 2017 in Moscow, at the age of 93.

Shafarevich made fundamental contributions to several parts ofmathematics including algebraic number theory, algebraic geometryand arithmetic algebraic geometry. In algebraic number theory theShafarevich–Weil theorem extends the commutative reciprocity mapto the case of Galois groups which are extensions of abelian groupsby finite groups. Shafarevich was the first to give a completelyself-contained formula for the pairing which coincides with the wildHilbert symbol on local fields, thus initiating an important branchof the study of explicit formulas in number theory. Another famousresult is Shafarevich’s theorem on solvable Galois groups giving therealization of every finite solvable group as a Galois group overthe rationals. Another fundamental result is the Golod–Shafarevichtheorem on towers of unramified extensions of number fields.

Shafarevich and his school greatly contributed to the study of algebraic geometry of surfaces. Heinitiated a Moscow seminar on classification of algebraic surfaces that updated around 1960 the treatment ofbirational geometry, and was largely responsible for the early introduction of the scheme theory approachto algebraic geometry in the Soviet school. His investigation in arithmetic of elliptic curves led himindependently of John Tate to the introduction of the most mysterious group related to elliptic curves overnumber fields, the Tate–Shafarevich group. He introduced the Grothendieck–Ogg–Shafarevich formulaand the Néron-Ogg-Shafarevich criterion. He also formulated the Shafarevich conjecture which stated thefiniteness of the set of Abelian varieties over a number field having fixed dimension and prescribed set ofprimes of bad reduction. This conjecture was proved by Gerd Faltings as a step in his proof of the Mordellconjecture.

Shafarevich was a student of Boris Delone, and his students included Yuri Manin, A. N. Parshin, I.Dolgachev, Evgeny Golod, A. I. Kostrikin, Igor A. Kostrikin, S. Y. Arakelov, G. V. Belyi, Victor Abrashkin,Andrey N. Tyurin, and V. A. Kolyvagin. He did major work in collaboration with Ilya Piatetski-Shapiro onK3 surfaces.

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7.3 Nilpotent Groups

7.3.1 Central series

Let G be a group and H, K subgroups of G. Define

[H,K] := 〈 [h,k] |h ∈H and k ∈ K 〉

where [h,k] is as usual the commutator [h,k] = hkh−1k−1.

Definition 7.4. For any group G define the following subgroups

G0 = G, G1 = [G,G], Gi+1 = [G,Gi]

The chain of groupsG0 ≥ G1 ≥ . . .

is called the lower central series

LetZ0(G) = 1, Z1(G) = Z(G)

and consider the mapεi : G→ G/Zi

Define inductively

Zi+1(G) := ε−1i (Z(G/Zi)).

Then, we have the chainZ0 ≤ Z1 ≤ . . .

which we call the upper central series.

Definition 7.5. A group G is called nilpotent if Zn(G) = G for some n. The smallest such n is called the nilpotenceclass of G.

Lemma 7.8. a) Prove that a finite p-group is nilpotent.b) The direct product of a finite number of nilpotent groups is nilpotent.

Proof. LetZ0 = 1 ≤ Z1 ≤ · · · ≤Zi ≤ . . .

be the upper central series of G. We want to show that Zn = G for some n ≥ 0. For any i, we can assume thatZi(G) ,G, otherwise we are done. Then, Z(G/Zi(G)) , 1, since G/Zi(G) is a p-group. Thus, Zi(G) is a propersubgroup of Zi+1(G). Since G is finite, then it must be a n ≥ 0 such that Zn(G) = G

The proof of part b) comes straight from the properties of the direct product.

Lemma 7.9. a) Prove that if G is nilpotent, and H is any proper subgroup, then H is a proper subgroup of itsnormalizer.

b) Prove that G is nilpotent if and only if it is isomorphic to a direct product of its Sylow subgroups.

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Proof. a) LetZ0 = 1 ≤ Z1 ≤ · · · ≤Zi ≤ . . .

be the upper central series of G and n be the largest index such that Zn 1H. Take a ∈ Zn+1 \Zn. Then a <H.We will show that a ∈NG(H). As above, let εn : G→ G/Zn. Then, εn(a) ∈ Z(G/Zn). Thus, for all h ∈H,

Zna ·Znh = Znh ·ZnaZn · ah = Zn ·hab1 · ah = b2 ·ha, for some b1,b2 ∈ Zn

ah = b ·ha, b ∈ Zn ⊂H

aha−1 = bh ∈H

(7.1)

Thus, a ∈NG(H).b) Assume that G is nilpotent. Let P ∈ Sylp(G). If NG(P) = G then P C G and GP×G/P. If NG(P) is

proper in G, then by part a) it is proper in NG(Ng(P)). But, this is a contradiction; see Lemma 5.10.

Theorem 7.7. A group G is nilpotent if and only if Gn = 1 for some n ≥ 0

Proof.

Lemma 7.10. The following is trueGi≤ Gi

Proof. Exercise

Theorem 7.8. Prove that every nilpotent group is solvable.

Proof. Use the above lemma

Exercises:

7.27. Let p be a prime number and G be the group of all invertible n by n matrices which are lower triangular overthe field Fp of p elements.

a) Let U = a ∈ G : aii = 1, f or all i = 1,2,3... Prove that U is nilpotent.b) Show that G is solvable, but if n > 1, G is not nilpotent.

7.28. Let H < G and assume that Z(H) = e. Show that the following are equivalent:a) There is a subgroup J of G such that G = H× Jb) For every g ∈ G, g induces by conjugation an inner automorphism of H.

7.29. Let G be a group; call g ∈G a non-generator if, for each subset X of G so that X∪g generates G, then, in fact,X itself generates G. Let Fr(G) denote the set of all non-generators of G.

a) Prove that Fr(G) is a subgroup of G.b) Show that Fr(G) is the intersection of all maximal (proper) subgroups of G. (Careful with Zorn’s Lemma!)

7.30. a) Prove that a finite p-group is nilpotent.b) The direct product of a finite number of nilpotent groups is nilpotent.

7.31. a)Prove that if G is nilpotent, and H is any proper subgroup, then H is a proper subgroup of its normalizer.b) Prove that G is nilpotent if and only if it is isomorphic to a finite direct product of p-groups.

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138

Chapter 8

Extension and Cohomology

8.1 Extensions

The extension problem.

Example 8.1. Automorphism group of a hyperelliptic curve:

Definition 8.1. If K,Q are groups, then an extension of K by Q is a group G having a subgroup K1 isomorphic toK such that G/K1Q.

Example 8.2. S3 is an extension of Z3 by Z2: Indeed, Z3〈(123)〉 → S3 and S3/〈(123)〉Z2.

Example 8.3. Z6 is an extension of Z3 by Z2: Indeed, Z3 →Z6 and Z6/Z3Z2. In this case, Z6Z2×Z3.

Example 8.4. Direct products: Obviously any direct product GK×Q is an extension of K by Q.

Definition 8.2. Let K CG and G/KQ. Then, the degree of the extension G of K is called the cardinality |Q| of Q.

8.1. Given the group K. Find all possible degree n extensions of K (up to isomorphism).

Or we can narrow it down to the following:

8.2. Given the group K and Q such that |Q| = n. Find all possible extensions of K by Q (up to isomorphism).

Hölder Program:

8.2 More on automorphism groups

Theorem 8.1.

Definition 8.3. A group G is called complete if Z(G) = 1 and Aut (G) = Inn(G).

Lemma 8.1. If G is a cyclic group of order |G| = n, then Aut (G)U(Zn).

The following theorem states some results on the automorphism groups:

Theorem 8.2. The following are true:i) Aut (Z2) = 1ii)

Lemma 8.2. Let ϕ ∈ Aut (Sn). Then, φ preserves the transpositions if and only if ϕ ∈ Inn(Sn).

Theorem 8.3. If n , 2 and n , 6 then Sn is complete.

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Theorem 8.4. If G is a non-Abelian simple group, then Aut (G) is complete.

Definition 8.4. Automorphism tower

Theorem 8.5. (Wielandt) Let G be finite and Z(G) = 1. Then the automorphism tower is finite.

Definition 8.5. The holomorphic group of a group K, denoted by Hol(K), is the subgroup of SK generated by KL

and Aut (K).Hol(K) := 〈KL,Aut (K)〉 ≤ SK

Theorem 8.6. If K C G is a direct factor of G, then K is complete.

8.3 Semidirect Products

8.4 Cocycles and coboundaries

8.5 The second cohomology group and the Schreier theorem

8.6 Schur-Zassenhaus lemma

8.7 Projective Representations and the Schur Multiplier

Definition 8.6. A central extension of K by Q is an extension G of K by Q such that K ≤ Z(G).

Example 8.5. If G = KoθQ is a central extension then G is the direct product K×Q.

Lemma 8.3. Let (Q,K,θ) be given. Then, θ is trivial if and only if KoθQ is a central extension.

Corollary 8.1. There is a bijection from the set of all equivalence classes of central extension realizing (Q,K,θ) withθ trivial to H2(Q,K).

Definition 8.7. The Schur multiplier of a Q is the Abelian group

M(Q) := H2(Q,C?)

Definition 8.8. Let G be a finite group. The exponent of eG is the smallest integer e such that for all x ∈ G, xe = 1.

Lemma 8.4. Let Q be a finite group. Then the following are true:i) M(Q) is a finite Abelian group.ii) e|M(Q)|

| |Q|.

8.7.1 Projective Representations

Theorem 8.7 (Schur). Every finite group Q has a cover U which is a central extension of M(Q) by Q.

Exercises:

8.3. Show that the group Qn of generalized quaternions is not a semidirect product.

8.4. Prove that if K and Q are solvable, then KoθQ is solvable.

8.5. Find H2(Z2,Zn).

8.6. Find all central extensions of Z2 by Zn.

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8.7. Given a finite group G. Write a computer program in GAP which checks if G has a complement in Aut (G).

Final ExamApril 2017

Do the following problems according to your ticket:

1. 1.15, 1.45, 1.59, 2.18, 2.22, 3.17, 3.38, 3.39, 5.21, 5.34, 6.13,6.14, 7.23, 7.5,

2. 1.28, 1.46, 1.60, 2.17, 2.23, 3.16, 3.36, 3.40, 5.22, 5.35, 6.12,6.15, 7.24, 7.6,

3. 1.17, 1.47, 1.61, 2.16, 2.24,3.15, 3.33, 3.41, 5.23, 5.36, 6.11,6.16, 6.29, 7.7,

4. 1.18, 1.48, 1.62, 2.15, 2.25, 3.14, 3.32, 3.42, 5.24, 5.37, 6.10,6.17, 6.30, 7.8,

5. 1.19, 1.49, 1.65, 2.14, 2.26, 3.13, 3.31, 3.43, 5.25, 5.38, 6.9,6.18, 6.31, 7.9,

6. 1.20, 1.50, 1.66, 2.13, 2.27, 3.12, 3.25, 3.44, 5.49, 5.39, 6.8,6.19, 6.32, 7.18,

7. 1.21, 1.58, 1.67, 2.12, 2.28,3.11, 3.24, 3.45, 5.27, 5.46, 6.7,6.20, 6.33, 7.11,

8. 1.22, 1.52, 1.68, 2.11, 2.29, 3.10, 3.23, 5.13, 5.28, 5.45, 6.6,6.21, 6.34, 7.19,

9. 1.23, 1.53, 1.69, 2.10, 2.31, 3.6, 3.22, 5.14, 5.29, 5.47, 6.5,7.22, 7.26, 7.20,

10. 1.24, 1.54, 1.70, 2.20, 2.32, 3.4, 3.21, 5.18, 5.30, 5.48, 6.4,6.23, 7.1, 7.14,

11. 1.25, 1.55, 1.71, 2.8, 2.33, 3.3, 3.20, 5.19, 5.31, 5.50, 6.3,6.24, 7.2, 7.15,

12. 1.26, 1.56, 1.72, 2.7, 2.34, 3.7, 3.19, 5.17, 5.32, 5.51, 7.21,6.25, 7.3, 7.16,

13. 1.27, 1.57, 1.73, 2.6, 2.37, 3.5, 3.18, 5.20, 5.33, 5.52, 6.1,6.26, 7.4, 7.17,

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142

Part I

Ring theory

143

Chapter 9

Rings

9.1 Introduction to rings

The set R with two algebraic operation (R,+, ·) (addition and multiplication) that satisfies the followingconditions;

1. (R,+) is Abelian group

2. multiplication is associative property

(a ·b) · c = a · (b · c), ∀a,b,c ∈ R

3. it is true distributive property

(a + b)c = ac + bca(b + c) = ab + ac

for every a,b,c ∈ R, is called a ring .

A ring R in which multiplication is commutative is called commutative ring or Abelian ring. The ringR has identity when there exists the element eR ∈ R such that

∀a ∈ R, a · eR = a.

A ring R with identity (eR , 0) in which every element a ∈ R\ 0 has inverse with multiplication is called adivision ring. An Abelian ring which is also a division ring is called a field . In this book we usually willdeal with rings with identity.

Lemma 9.1. Let be given ring R. Then the following hold:a) 0a = a0 = 0b) (−a)b = a(−b) = −abc) the identity is a unique element and

−a = (−eR) · a

Proof. a) 0a = (0 + 0)a = 0a + 0a. Hence, 0a = 0a + 0a, 0a = 0.b) ab + (−a)b = (a− a)b = 0b = 0c) If R has two identities ea and eb then, ea · eb = ea, ea · eb = eb. Hence, ea = eb.

Let R be a ring. A nonzero element a ∈ R is called a zero divisor if there is an nonzero element b ∈ R

such that ab = 0 or ba = 0. A element u ∈ R is called unit in R if there exists v ∈ R that uv = vu = 1.An Abelian ring with identity is called an integral ring if has not zero divisors.

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Lemma 9.2. The integral rings have the cancellation property

ab = ac⇒ a = 0 or b = c

Proof. ab = ac⇒ a(b− c) = 0⇒ a = 0 or b = c

Example 9.1. The ring (Z,+, ·) is an integral ring since ab = 0 implies that a = 0 and b = 0. However, (Z,+, ·) isnot a field since most elements have no multiplicative inverses. The only elements with multiplicative inverses are 1and -1.

Example 9.2. Together with the usual addition and multiplication the following integers Z, rational numbers Q,real numbers R, complex numbers C are rings. Moreover, the reader can check that they are also fields, other than Z.

Example 9.3. We studied the set Zn =Z/nZ with addition and showed that it was an Abelian group. Define themultiplication in Zn as follows:

Z 7→Zn

a ·b = ab mod n

For example, inZ12, 5 ·7 ≡ 11 (mod 12). Then (Zn,+, ·) is a ring. Obviously, Zn is a commutative ring, but not anintegral ring. For example, 3 ·4 ≡ 0 (mod 12) in Z12

Example 9.4. The set of continuous functions on a fixed interval, for example

f : [a,b] 7→R,

together with addition and composition of functions form a commutative ring.

Example 9.5. The 2× 2 matrices with terms in Z form a ring with addition and multiplication of matrices. Thisring is not commutative, because multiplication of matrices is not commutative.

Example 9.6. We consider as a division ring the following

1 =

(1 00 1

)i =

(0 1−1 0

)

j =

(0 o fi 0

)k =

(i 00 −i

),

where i2 = −1. These elements satisfy the following relations

i2 = j2 = k2 = −1ij = k

jk = iki = jji = −k

kj = −iik = −j.

Let H be the set of elements of the forma + bi + cj + dk,

where a,b,c,d are real numbers. Equivalently, H can be thought of as the set of 2×2 matrices of the form(α β−β α

),

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where α = a + di and β = b + c i are complex numbers.We can define addition and multiplication in H or with addition as the usual addition of matrices or with

generators 1, i, j, and k :

(a1 + b1i + c1j + d1k) + (a2 + b2i + c2j + d2k) =

(a1 + a2) + (b1 + b2)i + (c1 + c2)j + (d1 + d2)k

and multiplication as(a1 + b1i + c1j + d1k)(a2 + b2i + c2j + d2k) = α+βi +γj +δk,

where

α = a1a2−b1b2− c1c2−d1d2

β = a1b2 + a1b1 + c1d2−d1c2

γ = a1c2− b1d2 + c1a2−d1b2

δ = a1d2 + b1c2− c1b2−d1a2.

The ring H is called the quaternion ring.

Exercise 9.1. Prove that H is a ring and then a divisor ring.

Notice that(a + bi + cj + dk)(a−bi− cj−dk) = a2 + b2 + c2 + d2.

Thus, the element (a + bi + cj + dk) can to be zero only if a, b, c, and d are all zero. Hence, if a + bi + cj + dk , 0,

(a + bi + cj + dk)(

a− bi− cj−dka2 + b2 + c2 + d2

)= 1.

Let K be a given field. Denote with K∗ the set K \ 0. A function

ν : K∗ −→Z

which has properties

1. ν(ab) = ν(a) +ν(b),∀a,b ∈ K∗

2. ν is surjective

3. ν(x + y) ≥minν(x),ν(y), for every x, y ∈ K∗,x + y , 0

is called discrete value.The set R ⊂ K∗, R := x ∈ K∗ : ν(x) ≥ 0∪ 0 is called the value ring of ν. A ring is called discrete value

ring, denoted by DVR, if there is a field K and a discrete valuation ν from K, such that R is the value ringof ν.

Example 9.7. Take K =Q and p a prime number. Define

νp :Q∗ −→Zab

= pαcd−→ α,

where p is relatively prime with c and d. The reader to prove that νp is a discrete value. What is the value ring of νp?It is the set of elements of Q for which α ≥ 0. Hence all those elements which have denominators not divisible by

p. The units of this ring are exactly those elements for which α = 0. Prove each of the above statements.

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A subring S of a ring R is a subset S of R such that S is also a ring with operations if R. For example, Zis subring of Q, Q is subring of R. The following result gives us necessary and sufficient condition that S isa subring of R.

Exercise 9.2. Let R be a ring and S a subset of R. Then, S is a subring of R if and only if the following hold:

1. S , ∅.

2. rs ∈ S for every r,s ∈ S.

3. r− s ∈ S for every r,s ∈ S.

Example 9.8. Let R =M2(Z) ring of matrices 2×2 with terms in Z. If U is the set of upper triangular matrices inR, for example,

U =

(a b0 c

): a,b,c ∈Z

,

then U is a subring of R. If

A =

(a b0 c

)and B =

(a′ b′

0 c′

)are in U, then A−B is also in U. Also,

AB =

(aa′ ab′+ bc′

0 cc′

)is in U.

9.1.1 Polynomial rings

Let be given a Abelian ring R with unity. Take a variable x and consider all polynomials with coefficientsfrom R,

p(x) = an xn + · · ·+ a1 x + a0

for n ≥ 0 and ai ∈ R. The number n is called that degree of the polynomial and an is called leading termof the polynomial. We denote the degree of a polynomial p(x) with degp. If an = 1, then polynomial iscalled monic. The set of all polynomials with coefficients from R is called the ring of polynomials of R anddenoted by R[x].

Exercise 9.3. Prove that R[x] is an Abelian ring with identity with addition and multiplication of polynomials.

Similarly we we prove that polynomials with many variables form a ring. We denote this ring byR[x1, . . . ,xn].

Lemma 9.3. Let be given R an integral ring. Then,a) units of R[x] are units of R.b) R[x] is an integral ring.

Proof. a) If p(x) is a unit then there exists q(x) such that p(x)q(x) = 1. Thus, degp(x) = degq(x) = 0. Thus, p(x)and q(x) are constant polynomials, hence they are in R.

b) Assume that R[x] is not an integral ring. Hence there exist p(x) and q(x)

p(x) = an xn + . . .a0

q(x) = bm xm + . . .b0

such that p(x)q(x) = 0. Then, p(x)q(x) = an bmxm+n + . . .a0 b0 = 0. Hence, an,bm ∈ R and an bm = 0. Thus, R isnot integral ring, which contradicts the hypothesis of the theorem.

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9.1.2 The rings of matrices

Another important type of rings is the rings of matrices. Take all n×n matrices with terms from a ring R.We denote this set with Mn(R). With addition and multiplication of matrices this set forms a non Abelianring.

Exercise 9.4. Prove that Mn(R) together with addition and multiplication of matrices forms a ring.

Matrices are presented as A = (ai j),1 ≤ i ≤ m,1 ≤ j ≤ n. A matrix A = (ai j) is called scalar if aii = a ∈ Rand all other entries are zero. The unit elements of Mn(R) form a multiplicative group that we denote it byGLn(R) and called it the general linear group. We have seen this group before.

Assume that R is field F, matrices with determinant 1 form a subgroup of GLn(F), that is called thespecial linear group and denoted by SLn(F). Let Z(SLn(F)), the center of SLn(F). Then,

PSLn(F) := SLn(F)/Z(SLn(F))

is called the projective special linear group .Does Mn(R) have zero divisor? We take R a ring and two elements a,b ∈ R such that ab , 0. Consider

the case n = 3, but the following is true for every n. Take matrices

A =

a 0 00 0 00 0 0

, B =

0 b 00 0 00 0 0

Then,

AB =

0 ab 00 0 00 0 0

, BA =

0 0 00 0 00 0 0

This shows that for every ring R, the ring Mn(R) has zero divisors for n ≥ 2.

Exercises:

9.1. Let R be a ring with identity and S a subring of R containing the identity. Prove that if u is a unit in S then u isa unit in R. Show that the converse is false.

9.2. Prove that the intersection of any nonempty collection of subrings of a ring is also a subring.

9.3. The center of a ring R isZ(R) =

x ∈ R |xy = yx, for all y ∈ R

prove that the center of a ring is a subring which contains the identity. Prove that the center of a division ring is afield.

9.4. Prove that if R is an integral domain and x2 = 1 for some x ∈ R, then x = ±1.

9.5. An element x ∈ R is called nilpotent if xm = 0 for some m ∈Z+.

a) Show that if n = akb for some integers a and b then ab is an nilpotent element of Z/nZ.

b) If a ∈Z is an integer, show that a ∈Z/nZ is nilpotent if and only if every prime divisor of of n is also a divisorof a.

9.6. Let R be a commutative ring and x nilpotent in R.

a) Prove that x is either zero or a zero divisor

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b) Prove that rx is nilpotent for every r ∈ R.

c) Prove that 1 + x is a unit in R.

d) Prove that the sum of a nilpotent and a unit is always a unit.

9.7. A ring is called a Boolean ring if a2 = a for all a ∈ R. Prove that every Boolean ring is commutative.

9.8. Prove that a Boolean ring which is an integral domain is Z/2Z.

9.9. Let D be an integer, which is not a complete square in Z. Take the set

Z[D] := a + b√

D : a,b ∈Z

a) Prove that Z[D] is a ring.b) Define the function

N :Z[D] −→Z

a + b√

D −→ a2−Db2.

This function is called a norm of Z[D]. Prove that ∀x, y ∈Z,N(xy) = N(x)N(y). Also prove that if u is a unit inZ[D] then N(u) = ±1

9.10. Let be given an integer that is not complete square and define

Q[D] := a + b√

D : a,b ∈Q.

Prove that Q[D] is field. Prove that the valuation ring is a ring.

9.11. Let be given the field K, discrete valuation ν in K and R the valuation ring of ν in R.a) Prove that for every nonzero element x ∈ K, x ∈ R or x−1

∈ Rb) Prove that x is a unit in R if and only if ν(x) = 0

9.12. Let R be the ring Z/6Z. How many polynomials are in R[x]?

9.13. Let R be a ring with identity 1 , 0, n a positive integer, and A ∈Mn(R) such that

A =[ai, j

]Let Ei, j =

[er,s

]be the element of Mn(R) such that

er,s =

1 if r = i, s = j0 otherwise

Prove the following:

a) Ei, jA is the matrix whose i-th row equals the j-th row of A and all other rows are zero.

b) AEi, j is the matrix whose j-th column equals the i-th column of A and all other columns are zero.

c) Ep,qAEr,s is the matrix whose p, s entry is aq,r and all other entries are zero.

9.14. Prove that the center of rings Mn(R) is the set of scalar matrices.

9.15. Let be given the field Fq with q elements. Prove that:|Mn(Fq)| = qn2

|GLn(Fq)| = (qn−1)(qn

− q)(qn−q2) . . . (qn

− qn−1)

|SLn(Fq)| = qn(n−1)

2∏i=n

i=2(qi−1)

|PSLn(Fq)| = 1(n,q−1) |SLn(Fq)|

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9.2 Ring homomorphisms and quotient rings

A ring homomorphism a function which preserves both ring operations.Let R and S be rings. A ring homomorphism is called the functionϕ : R−→ S that satisfies the following

properties:a) ϕ(a + b) = ϕ(a) +ϕ(b)b) ϕ(ab) = ϕ(a)ϕ(a)c) ϕ(eR) = ϕ(eS)ker(ϕ) is called the set

ker(ϕ) := x ∈ R : ϕ(x) = 0S

A bijective homomorphism is called an isomorphism. We will see a few examples of isomorphisms.

Exercise 9.5. Let be given the function

ϕ :Q[x] −→Q

anxn + · · ·+ a1x + a0 −→ a0

Prove that ϕ is homomorphism.

Example 9.9. For every integer n we can define a ring homomorphism

φ : Z→Zn,

where a 7→ a (mod n). This is a ring homomorphism because

φ(a + b) = (a + b) (mod n) = a (mod n) + b (mod n) = φ(a) +φ(b)

andφ(ab) = ab (mod n) = a (mod n) · b (mod n) = φ(a)φ(b).

The kernel of the homomorphism φ is nZ.

Example 9.10. Let C[a,b] ring of real valued functions continuous in an interval [a,b]. For a fixed α ∈ [a,b], we candefine a ring homomorphism

φα : C[a,b]→R,

where φα( f ) = f (α). This is indeed a ring homomorphism, because

φα( f + g) = ( f + g)(α) = f (α) + g(α) = φα( f ) +φα(g)φα( f g) = ( f g)(α) = f (α)g(α) = φα( f )φα(g).

The following is an easy exercise.

Lemma 9.4. Let ϕ : R→ S a ring homomorphism. Then

1. The image ϕ(R) is a subring of S

2. The kernel kerϕ is a subring of R.

3. If R is a commutative ring, then φ(R) is a commutative ring.

4. φ(0) = 0.

5. Let eR and eS, be respectively the identities of R and S. If φ is surjective, then φ(eR) = eS.

6. If R is a field and φ(R) , 0, then φ(R) is a field.

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Next, we study ideals of ring. A left ideal of the ring R is called the subgroup (I,+) ≤ (R,+) such that

∀x ∈ R, ∀y ∈ I, we have that,xy ∈ I.

Hence, RI ⊂ I. A right ideal is called I ⊂ R such that Ix ⊂ I for every x ∈ R. A ideal that is a left and rightideal is called an ideal. In commutative rings we have simply ideals. A ideal I , 0 and I ,R is called properideal.

Every ring R has at least two ideals, 0 and R. Let R a ring with identity and assume that I is a ideal inR such that 1 is in R. Since for every r ∈ R, r1 = r ∈ I from the definition of ideal, I = R.

Example 9.11. If a is some element in a commutative ring R with unity, then the set

〈a〉 = ar : r ∈ R

is a ideal in R. Obviously, 〈a〉 is nonempty, since together 0 = a0 and a = a1 are in 〈a〉. The sum of two elements in〈a〉 is again in 〈a〉 since ar+ar′ = a(r+ r′). The opposite of ar is −ar = a(−r) ∈ 〈a〉. Finally, if we multiply an elementar ∈ 〈a〉 with an any element s ∈ R, we get s(ar) = a(sr). Thus, 〈a〉 is an ideal.

If R is a commutative ring with identity, then a ideal 〈a〉 = ar : r ∈ R is called principal ideal.

Theorem 9.1. Every ideal in the ring of integers Z is a principal ideal.

Proof. The zero ideal 0 is a principal ideal, since 〈0〉 = 0. If I is a nonzero ideal in Z, then I must containsome positive integer n. From the well ordering principal we find the smallest n in I. Let a an element in I.Using the division algorithm we know that there exist integers q and r, such that

a = nq + r

where 0 ≤ r < n. This equation shows that that r = a−nq ∈ I. However, r must be 0 since n is the smallestpositive integer in I. Thus, a = nq and I = 〈n〉.

Example 9.12. The set nZ is an ideal in the ring Z. If na is in nZ and b is in Z, then nab is in nZ. Thus, nZ isan ideal, for any n. Indeed, these are the only ideals of Z.

Proposition 9.1. The kernel of a ring homomorphism φ : R→ S is a ideal in R.

Proof. From the theory of groups we know that kerφ is a additive subgroup of R. Assume that r ∈ R anda ∈ kerφ. Then, we must prove that ar and ra are in kerφ. However,

φ(ar) = φ(a)φ(r) = 0φ(r) = 0

andφ(ra) = φ(r)φ(a) = φ(r)0 = 0.

Theorem 9.2. Let I a ideal of R. The quotient group R/I is a ring, where multiplication is of defined as

(r + I)(s + I) = rs + I.

Proof. We know that R/I is Abelian group under addition. Let r + I and s + I in R/I. We want to show thatthe product (r + I)(s + I) = rs + I is independent from the choice of cosets representatives. Thus, if r′ ∈ r + Iand s′ ∈ s + I, then r′s′ must be in rs + I. Since r′ ∈ r + I, there exists a element a in I such that r′ = r + a.Similarly, there exists a b ∈ I such that s′ = s + b. Notice that

r′s′ = (r + a)(s + b) = rs + as + rb + ab

and as+ rb+ab ∈ I since I is an ideal. Hence, r′s′ ∈ rs+ I. To verify the associative and distributive propertieswe leave it as an exercise for the reader.

The ring R/I in Theorem 9.2 is called quotient ring. Next we will study properties of quotient rings.

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Theorem 9.3. Let I a ideal of R. The map

ψ : R −→ R/Ir −→ r + I

is a homomorphism of rings with kernel I.

Proof. The map ψ : R→ R/I is a surjective homomorphism of Abelian groups. Remains to prove that ψpreserves the multiplication of rings. Let r and s in R. Then,

ψ(r)ψ(s) = (r + I)(s + I) = rs + I = ψ(rs),

which concludes the theorem. The map ψ : R→ R/I is often called the natural projection or the canonical homomorphism. In the

theory of rings we have the Ring Isomorphism Theorems similar to those for groups.

Theorem 9.4 (First Isomorphism Theorem). Let φ : R→ S be a ring homomorphism. Then, kerφ is a ideal of R.If

ψ : R → R/kerφ

is the canonical homomorphism then there exists a unique isomorphism

η : R/ker φ → φ(R),

such that φ = ηψ.

Proof. Let K = kerφ. From the First Isomorphism Theorem for groups, there exists a well defined grouphomomorphism

η : R/K→ ψ(R),

such that η(r+K) = ψ(r) for groups Abelian of additions R and R/K. To prove that this is a homomorphismof rings, it is enough to prove that η((r + K)(s + K)) = η(r + K)η(s + K).

Thus,η((r + K)(s + K)) = η(rs + K) = ψ(rs) = ψ(r)ψ(s) = η(r + K)η(s + K).

Theorem 9.5 (Second Isomorphism Theorem). Let I a subring of a ring R and J a ideal of R. Then, I∩ J is a idealof I and

I/I∩ J (I + J)/J.

Proof. Exercise

Theorem 9.6 (Third Isomorphism Theorem). Let R a ring and I and J ideals of R where J ⊂ I. Then,

R/I R/JI/J

.

Theorem 9.7 (Correspondence Theorem). Let I a ideal of ring R. Then, there is a correspondence S→ S/I betweenthe set of subrings S, that contain I and the set of subrings of R/I. Moreover, ideals of R, that contain I correspondwith ideals of R/I.

Proof. Exercise

Exercises:

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9 Shaska T.

9.16. Let R be the ring of continuous, integrable, real valued functions on [0,1]. Prove that φ : R→R given by

φ( f ) =

∫ 1

0f (t)dt,

is a homomorphism of additive groups but not a ring homomorphism.

9.17. Let φ : R→ S be a surjective ring homomorphism. Prove that the image of the center of R is contained in thecenter of S.

9.18. If I and J are ideals in R, prove that I∩ J is an ideal of R. Prove that the collection of an arbitrary nonemptycollection of ideals in R is again an ideal in R.

9.19. Let I be an ideal of R and S a subring of R. Prove that I∩S is an ideal in S. Show by example that not everyideal of S is of the form I∩S, for some ideal I in R.

9.20. Let φ : R→ S be a ring homomorphism. Prove that if x is an nilpotent element of R then φ(x) is nilpotent in S.

9.3 Ideals, nilradical, Jacobson’s radical

In this section we study some important ideals in the study of rings.

Lemma 9.5. Let be given the ideal I ⊂ R.a) IR = Ib) I = R if and only if I contains an element a unitc) R is field if and only if the only ideals of R are 0 and R.

Proof. a) From the definition of ideals IR = Ib) If I = R then eR ∈ R, so I contains an element a unit. If u ∈ I is an element a unit then there exists v ∈ R

such that vu = eR. Then, ∀r ∈ R,r = reR = rvu = (rv)u ∈ I.

Thus, R ⊂ I, which means that I = R.c) We prove that if R is a field then the only ideals are 0 and R. We know that ring R is field if and only

if every nonzero element is a unit. Thus, every ideal nonzero has an element a unit. Therefore it is equal toR.

Let R a ring with ideals’s only S 0 and R. We prove that it is a field. Assume se there is an element r ∈ Rthat s’is a unit in R. Then, (r) , R, which is a contradiction. Thus, for every element in R is a unit. Then, Ris field.

A ideal m ⊂ R is called maximal ideal if the only ideals that contain M are M and R.

Lemma 9.6. In a ring with identity every proper ideal is contained in a maximal ideal.

Proof. Let be given ring R with unity and the proper ideal I ⊂ R. Let S the set of all proper ideals thatcontain I. S is nonempty because I ∈ S. Also S is ordered from set inclusion. Take an increasing chain C inS. Prove that has it has an upper bound. Take

J := ∪A∈CA

Prove that J is ideal. Obviously J is nonempty, because 0 ∈ J. If a,b ∈ J then there exist ideals A,B ∈ Csuch that a ∈ A and b ∈ B. From the definition of chain A ⊂ B or B ⊂ A. Hence a− b ∈ J, which implies J isclosed with addition. Since A is closed from left and right multiplication with the elements of R then and Jis closed. Thus, J is an ideal.

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J is proper ideal. If not then 1 ∈ J. Thus, there exists some A ∈ C, that 1 ∈A. this is contradiction, becauseA ∈ C ⊂ S. Finally, the hypothesis of Zorn’s lemma are satisfied which implies that that S has an maximalelementM.

Theorem 9.8. Every commutative ring A , 0 with identity contains a maximal ideal.

Proof. The proof is similar to that of the above lemma.

Corollary 9.1. If a , (1) is a ideal of A, there is a ideal of A that contains a.

Corollary 9.2. Every non unit element of A is contained in some maximal ideal.

Lemma 9.7. In a commutative ring, an ideal M is maximal if and only if R/M is field.

Proof. From the Fourth Isomorphism Theorem, ideals of R/I correspond one to one with ideals of R thatcontain M. If M is maximal, then does not exist any ideal that contains M. Thus, there are no ideals of R/Iother than 0 and R/I. Thus, R/I is field. If R/I is field, then does not have other ideals other than 0 and R/I.Thus, R does not have ideals that contain M.

Definition 9.1. A ideal is called a prime ideal if P , R and

ab ∈ P⇒ a ∈ P or b ∈ P.

Lemma 9.8. Let be given R a commutative ring. I is prime if and only if R/I is integral ring.

Proof. Assume that I is prime. Take the natural projection π : R −→ R/I. then, r ∈ I if and only if π(r) = 0 inR/I. If π(a)π(b) = 0 in R/I then ab ∈ I. Then, a ∈ I ose b ∈ I, so π(a) = 0 or π(b) = 0 in R/I. Thus, R/I does nothave zero divisors. The converse is similar.

Corollary 9.3. Every maximal ideal is prime.

Proof. If M is maximal ideal, then R/M is field. Every field is integral ring. From Lemma 9.8, M is a primeideal.

Example 9.13. Principal ideals generated from prime numbers in Z are prime and maximal ideals.

Example 9.14. The ideal (x) is prime in Z(x), because Z[x]/(x) Z. This ideal is not maximal. The ideal 0 is aprime ideal in Z[x], but not maximal ideal.

Proposition 9.2. The set ℵ of all nilpotent elements of a ring R is an ideal.

Proof. If x ∈ ℵ, then there is a n ∈Z that xn = 0. Thus, ∀a ∈R, (ax)n = anxn = a0 = 0. Thus, ax ∈ ℵ. Take x, y ∈ ℵ,xn = 0, ym = 0. Then,

(x + y)m+n−1 = (xn)m−1 + a1xm+n−2 y + . . .an xm−1 yn + . . . (ym)n−1 = 0

Thus, for every x, y ∈ ℵwe have that x + y ∈ ℵ and for every x ∈ ℵ,a ∈ R, ax ∈ ℵ. Thus, ℵ is ideal.

The ideal ℵ is called the nilradical of R. Another definition of ℵ is given from the following proposition.

Proposition 9.3. The nilradical ℵ is the intersection of all prime ideals of R.

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Proof. Denote with ℵ the nilradical of R and ℵ′ the intersection of all prime ideals of R. If x ∈ ℵ, thenxn = 0 ∈ p, for every prime ideal p. Then, x ∈ p, because p is prime. Thus, x ∈ ℵ′.

Take now x ∈ ℵ′. Assume , se x is not nilpotent. Let Σ the set of all ideals awith property that for n > 0,xn < a. Σ is not empty, because 0 ∈ Σ. From Zorn’s Lemma we have that Σ has an maximal element. Let pthe maximal element of Σ. If we prove that p is prime ideal, then the proof is complete. To prove that p isprime, we must of prove that

a < p,b < p⇒ ab < p

ideals p+ (a) and p+ (b) contain p, hence are not elements of Σ. Thus, there exist M and n that

xm∈ p+ (a), xn

∈ p+ (b)

Then, xm = p1 + r1 a and xn = p2 + r2 b, hence xm+n = p1p2 + p1 r2b + p2 r1a + ab ∈ p+ (ab). Thus, p+ (ab) is not inΣ, which implies ab < p.

The intersection of all maximal ideals of R is called the Jacobson ideal of R and denoted by<.

Proposition 9.4. x ∈<⇐⇒ 1−xy is a unit in R for every y ∈ R.

Proof. ⇒ Assume that 1− xy is not a unit. Then, it is contained in some maximal ideal m. However,x ∈< ⊂m, hence x y ∈m. Thus, 1 ∈mwhich is a contradiction.⇐ Assume that x <<. Then, there is a maximal ideal m, such that x < m. Thus, m and x generate

R. Hence, they generate the ideal (1). Thus we have 1 = u + xy for u ∈ m and y ∈ R, which implies thatu = 1−xy ∈m is not a unit in R.

If a is an ideal of A then the radical of a is

r(a) := x ∈ A : xn∈ a,n ∈Z+

.

Proposition 9.5. Let A be a given ring and a an ideal of A. The radical r(a) of a is the intersection of all prime idealsof A that contain a.

Proof. Prime ideals of A that contain a correspond with prime ideals of A/a; see Correspondence Theorem.If xn

∈ a then xn = 0 in A/a, so x is in the nilradical of A/a. Thus, x is in every prime ideal, that contains a.Conversely, if x is in every ideal that contains a, then x is in the nilradical of A/a. Thus, there exists n suchthat xn = 0 in A/awhich is equivalent with xn

∈ a.

Example 9.15. Let pZ be an ideal inZ, where p is a prime. Then, pZ is a maximal ideal sinceZ/pZ Zp is a field.

Example 9.16. Every ideal in Z is of the form nZ. The quotient ring Z/nZ Zn is a integral ring only when nis prime number. In this case Zn is a field. Thus, nonzero prime ideals in Z are ideals pZ, where p is prime. Thisexplains the use of the term ’prime’ for such ideals.

Exercises:

9.21. Let a and b be ideals of R such that a+b = R. Prove that

ab = a∩b

9.22. Let be given an integral ring R. Prove that, (a) = (b) for a,b ∈R if and only if when a = ub for some unit elementu of R.

9.23. Let x a nilpotent element of a commutative ring A. Prove that, the element 1+x is a unit element in A.Concludethat the sum of an nilpotent element with a unit, is a unit.

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9.24. Let be given a ring A and N its nilradical. Prove that the following are equivalent:i) A has only a prime ideal.ii) for every element of A is or element a unit or nilpotent.iii) A/N is a field

9.25. Let R be a finite commutative ring with identity. Prove that every prime ideal is a maximal ideal.

9.26. Let R be a commutative ring. Prove that the following are equivalent:

a) R has exactly one prime ideal

b) every element of R is either nilpotent or a unit

c) R/N(R) is a field

9.4 Ring of fractions

Let be given a commutative ring R. Let D ⊂ R, be the set of all elements of R that are not zero divisors.Also, 0 <D and D is a closed multiplicative set. We take

F := (r,d) : r ∈ R,d ∈D.

Define a relation in F as follows(r,d) ∼ (s,e)⇔ re = sd

Prove that this relation is an equivalence relation. For example, show the following1) (r,d) ∼ (r,d)⇔ rd = rd2) (r,d) ∼ (s,e)⇔ (s,e)(r,e)3) Transitive property

(r,d) ∼ (s,e)⇒ re = sd⇒ f re− f sd = 0(s,e) ∼ (t, f )⇒ s f = te⇒ ds f −dte = 0

Thus, f re−dte = 0 which implies that e( f r−dt) = 0. Since e ∈D, e is not a zero divisor and e , 0 we haver f − td = 0. This implies that (r,d) ∼ (t, f ).

The equivalence class of (r,d) is denoted by r

d . Denote with Q the set of all equivalence classes of therelation above. Notice that, r

d = rcdc in Q for all c ∈D, (dc ∈D because D is closed multiplicative set). Define

addition of multiplication in Q as follows:

ab

+cd

=ad + bc

bdab·

cd

=acbd

The reader can prove that

1) these are algebraic operations2) Q is a commutative group with addition, with zero 0

d , where d is every element from D and theopposite of a

b is − ab

3) multiplication is associative and has the distributive property with addition.4) Q has an identity.

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Thus, Q is commutative ring with identity. The ring Q is called the ring of fractions of R and denotedby D−1R. If R is integral ring, then D = R \ 0 and D−1R is field because every element has multiplicativeinverse. In this case D−1R is called the field of fractions of R. In any case, R ⊂ D−1R. Thus, D−1R is aextension of R.

Example 9.17. Show that Q is the field of fractions of Z.

Example 9.18. Since Q is a field, Q[x] is a integral ring. The field of fractions of Q[x] is the set of all rationalexpressions p(x)/q(x), where p(x) and q(x) are polynomials overQ and q(x) is not the zero polynomial. We will denotethis field with Q(x).

9.5 Chinese remainder theorem

Let us assume that by a ring we always mean a commutative ring with unity 1 , 0.For any two rings R1 and R2 we denote by R1×R2 their Cartesian product. Define on R1×R2 the addition

and multiplication as follows:

(r1,r2) + (s1,s2) = (r1 + s1,r2 + s2)(r1,r2)(s1,s2) = (r1s1,r2s2).

It is an easy exercise to show that R1×R2 is a ring which we call direct product of R1 and R2. Two ideals Iand J of a ring R are called co-maximal if I + J = R.

Theorem 9.9 (Chinese remainder theorem). Let I1, I2, ..., Ik ideals in R and the natural projection

R→ R/I1×R/I2× · · ·×R/Ik

r 7→ (r + I1,r + I2, . . . ,r + Ik)

with kernel I1∩ I2∩· · ·∩ Ik. If for every i, j ∈ 1,2, . . . ,k, where i , j, ideals Ii and I j are co-maximal, then this map issurjective and

I1∩ I2∩· · ·∩ Ik = I1I2 . . . Ik,

soR/(I1I2 . . . Ik) = R/(I1∩ I2∩ ...∩ Ik) R/I1×R/I2× ...R/Ik.

Proof. First, we prove the theorem for k = 2. The general case follows from induction. Let A = I1 and B = I2,consider the map

ϕ : R→ R/A×R/Br→ (r mod A, r mod B) ,

where mod A means the coset in R/A that contains r ( so r+A). This map is a ring homomorphism, becauseis simply the natural projection of R onto R/A and onto R/B for both components.

The kernel of ϕ consists in all the elements r ∈ R such that r ∈A∩B. To finish the proof we need to showthat when A and B are co-maximal, then ϕ is surjective and A∩B = AB.

If ϕ(x) = (0,1) and ϕ(y) = (1,0), then x ∈A and x = 1− y ∈ 1+B. If (r1 mod A,r2 mod B) is an element ofR/A×R/B, then r2x + r1y is mapped this element. So

ϕ(r2x + r1y) = ϕ(r2)ϕ(x) +ϕ(r1)ϕ(y) =

= (r2 mod A,r2 mod B)(0,1) + (r1 mod A,r1 mod B)(1,0)= (0,r2 mod B) + (r1 mod A,0)= (r1 mod A,r2 mod B).

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This shows that ϕ is surjective.Finally, the ideal AB is contained in A∪B. If A and B are co-maximal and x and y are as above, then for

every c ∈ A∪B,c = c1 = cx + cy ∈ AB. This shows A∪B ⊆ AB and completes the case when k = 2.In general, let A = A1 and B = A2 . . .Ak. Show that A1 and A2 . . .Ak are co-maximal. From assumption for

every i ∈ 2,3, . . . ,k there are the elements xi ∈Ai and yi ∈Ai, such that x1 + y1 = 1. Since xi + yi = yi mod A1,we have that 1 = (x2 + y2) . . . (xk + yk) is an element of A1 + (A2 . . .Ak). This completes the proof.

This theorem has taken this name from the special case of isomorphism of rings

Z/mnZ (Z/mZ)× (Z/nZ)

where m and n are integers relatively prime.In the case of Z/mnZ the theorem gives an isomorphism of the groups of units as follows:

(Z/mnZ)× (Z/mZ)×× (Z/nZ)×.

In general we have

Corollary 9.4. Let n a positive integer andpα1

1 pα22 . . .pαk

k

its factorization into powers of primes. Then,

Z/nZ (Z/pα11 Z)× (Z/pα2

2 Z)× · · ·× (Z/pαkk Z),

as rings. In particular we have the isomorphism between the multiplicative groups

(Z/nZ)× (Z/pα11 Z)×× (Z/pα2

2 Z)×× · · ·× (Z/pαkk Z)×.

If we compare orders of the groups on different sides of the above isomorphism we have

ϕ(n) = ϕ(pα11 )ϕ(pα2

2 ) . . .ϕ(pαkk )

for the Euler functionϕ. In number theory this is known as the multiplicative property of the Euler function.Thus,

ϕ(ab) = ϕ(a)ϕ(b)

where a and b are positive integers which are relatively prime. This corollary also implies the decompositionof the Abelian group (Z/nZ)× into a direct product of cyclic groups.

Exercises:

Let R a ring with identity 1 , 0.

9.27. A element e ∈ R is called idempotent if e2 = e. Assume that e is a idempotent in R and er = re,∀r ∈ R. Showthat e and 1− e are respectively units of subrings re and R(1− e).

9.28. Let R a finite Boolian ring with unity 1 , 0. Prove that R Z/2Z×Z/2Z× · · ·×Z/2Z.

9.29. Let R and S two rings with identity. Prove that every ideal of R×S is of the form I× J, where I is an ideal in Rand J an ideal in S.

9.30. Prove that, if R and S are nonzero rings, then R×S is not a field.

9.31. Let n1,n2, . . . ,nk integers, which are pairwise relatively prime (ni,n j) = 1, for all i , j.

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1. Show that Theorem 9.9 implies that for every a1,a2, . . . ,ak ∈Z the systemx ≡ a1 mod n1

x ≡ a2 mod n2

. . .

x ≡ ak mod nk

has a solution x ∈Z and this solution is unique mod n for n = n1n2 . . .nk.

2. Let n′

i = n/ni be the ratio of n with ni, which is relatively prime with ni from the hypothesis. Let ti be inverse ofn′

i mod ni. Prove that the solution x of the above is given by

x = a1t1n′

1 + a2t2n′

2 + · · ·+ aktkn′

k mod n.

3. Solve the system x ≡ 1 mod 8x ≡ 2 mod 25x ≡ 3 mod 81

9.32. Let f1(x), f2(x), . . . , fk(x) be polynomials with integer coefficients of same degree d. Let n1,n2, . . . ,nk integers,which are pairwise relatively prime. Prove that there is a polynomial f (x) with integer coefficients and degree d, suchthat

f (x) ≡ f1(x) mod n1

f (x) ≡ f2(x) mod n2

f (x) ≡ fk(x) mod nk

Hence coefficients of f (x) agree with all coefficients of fi(x) mod ni. Prove that, if all fi(x) are monic, then f (x) canbe chosen monic.

9.33. Let m and n integers, where n |m. Prove that natural projection of rings Z/mZ→Z/nZ is also surjective onthe units

(Z/mZ)×→ (Z/nZ)×.

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Chapter 10

Euclidean rings, PID’s, UFD’s

10.1 Integral domains and fields

Let’s recall briefly some definitions. If R is a ring and r is an nonzero element in R, then r is called zerodivisor if there exists some nonzero element s ∈ R, such that rs = 0. A commutative ring with identity iscalled an integral ring if it does not have zero divisor.

If an element a in a ring R with unity has a multiplicative inverse we say that a is an element a unit. Iffor every nonzero element in a ring R is a unit, then R is called division ring. A commutative division ringis called a field.

Proposition 10.1 (Cancellation property). Let D a commutative ring with identity. D is a integral ring if andonly if when for all the nonzero elements a ∈D,

ab = ac⇒ b = c.

Proof. Let D a integral ring. Then, D does not have zero divisor. Let ab = ac, where a , 0. Then, a(b− c) = 0.Thus, b− c = 0 and b = c.

Conversely, let ’s assume that cancellation property holds in D. Thus, assume that, ab = ac, implies b = c.Let ab = 0. If a , 0, then ab = a0 and b = 0. Thus, a can not be a zero divisor.

Example 10.1. If i2 = −1, then the setZ[i] = m + ni : m,n ∈Z

forms a ring which is called the Gaussian integers.

It is easily proved that Gaussian integers form a subring of complex numbers, since they are closedunder addition addition and multiplication.

Let α = a+bi a a unit in Z[i]. Then, α = a−bi is also a a unit, since if αβ = 1, then αβ = 1. If β = c+di, then

1 = α ·β ·α ·β = (a2 + b2)(c2 + d2).

Thus, a2 + b2 must be 1 or −1. Equivalently, a + bi = ±1 or a + bi = ±i. Thus, units of this ring are ±1 and ±i.Thus, Gaussian integers are not field. The reader should prove that Gaussian integers are an integral ring.

Example 10.2. The set of matrices

F =

(1 00 1

),

(1 11 0

),

(0 11 1

),

(0 00 0

)me elements in Z2 forms field.

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10 Shaska T.

Example 10.3. The setQ(√

2) = a + b√

2 : a,b ∈Q

is field. The inverse of the element a + b√

2 in Q(√

2) is

aa2−2b2 +

−ba2−2b2

√

2.

The following theorem was proven by Wedderburn.

Theorem 10.1 (Wedderburn). Every finite integral ring is field.

Proof. Let D a integral ring finite and D∗ the set of elements nonzero of D. We must of we prove that forevery element in D∗ has inverse.

For every a ∈D∗ can of define a function λa : D∗→D∗ where λa(d) = ad. This function is of well definedbecause if a , 0 and d , 0, then ad , 0.

The map λa is injective, since for d1,d2 ∈D∗,

ad1 = λa(d1) = λa(d2) = ad2

implies d1 = d2 from cancellation from the left. Since D∗ is a finite set , the map λa must be surjective. Thus,for some d ∈ D∗, λa(d) = ad = 1. Thus, a has a left inverse. Since D is commutative, d must be also a rightinverse for a. Hence, D is field.

For every non negative integer n and for every element r in a ring R, We write r + · · ·+ r (n times) as nr.

We define the characteristic of a ring R to be the smallest positive integer n such that nr = 0 for all r ∈ R. Ifthere is no such integer, then the characteristic of R is defined to be 0.

Example 10.4. For for every prime number p, Zp is field with characteristic p. Every nonzero element in Zp hasinverse. Thus, Zp is field. If a is a nonzero element in field, then pa = 0, since the order of for every the nonzeroelement in the Abelian group Zp is p.

Theorem 10.2. The characteristic of an integral ring is a prime number or zero.

Proof. Let D a integral ring and assume that the characteristic of D is n, where n , 0. If n is not primenumber, then n = ab, where 1 < a < n and 1 < b < n. Since

0 = n · eR = (ab) · eR = (a · eR)(b · eR)

and a,b are not zero divisor in D. Then, a · eR = 0 or b · eR = 0. Thus, characteristic of D must be smaller thann, which is contradiction. Thus, n must be prime.

Exercises:

10.1.

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10.2 Euclidean domains

Let be given an integral ring R. The function

N : R −→Z+∪0

such that N(0) = 0 is called a norm for R. A ring can have more than a norm.

Definition 10.1. The integral ring R is called a Euclidean domain if there is a norm N in R, such that for everytwo elements a of b , 0 in R, there exist elements q,r ∈ R such that:

a = qb + r, where N(r) <N(b).

The element q is called quotient of a of b and r is called remainder.

Example 10.5. The ring of integers Z is a Euclidean domain. Take as norm

N :Z −→Z+∪0

a −→ |a|

From elementary arithmetic we we know that in Z the Euclidean algorithm is valid. Hence, for every a,b ∈ Z, thereexist q,r ∈Z such that

a = q · b + r

where |r| < |b|.

Example 10.6 (Fields). Every field is a Euclidean domain. For example is given the field F. Take N : F −→Z+∪0,

such thatN(a) = 0,∀a ∈ F.

Then, ∀a,b ∈ F,a = (ab−1)b + 0, so q = ab−1,r = 0.

Example 10.7. Let be given the field F and the ring of polynomials F[x]. Then, F[x] is Euclidean domain. As a normwe take N(p(x)) = degp(x). We will see later that F[x] is a Euclidean domain.

Example 10.8. Every discrete valuation ring (DVR) is a Euclidean domain. Let be given R a DVR. Then, there is afield K, a valuation ν : K∗ −→Z, that R is valuation ring of ν. Take N : R −→Z∪0 such that ∀a ∈ R∗,N(a) = v(a),N(0) = 0. Then, ∀a,b ∈ R.

i) If N(a) <N(b), then a = 0 ·b + aii) If N(a) >N(b), then N(a)−N(b) > 0,N(ab−1) > 0⇒ ab−1

∈ R, so a = (ab−1)b + 0.

Lemma 10.1. Every ideal in a Euclidean domain R is principal.

Proof. Take the ideal I , 0 in R. If I = R, then I = (1). Assume that I is proper. Consider the set

A = N(a) : a ∈ I ⊂Z+∪0

This set has an minimum element M (Zorn’s lemma). Denote with d ∈ I the element such that N(d) = m.Prove that I = (d).

Take a ∈ I. From Euclidian ’s algorithm we have a = qd + r where N(r) <N(d). Thus, r = a− qd ∈ I, henceN(r) = 0 because d has minimal norm in I. which implies that a = qd. Thus, a ∈ (d).

The above Lemma says that in Z every ideal is principal. However, Z[x] is not Euclidean domain,because (2,x) is not principal ideal.

Definition 10.2. Let be given ring (commutative) R and a,b ∈ R,b , 0. The greatest common divisor of a and bis called the nonzero element d, such that

1) d|a and d|b2) If d1|a and d1|b, then d1|d.

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10 Shaska T.

The greatest common divisor of a and b is denoted by (a,b) or by gcd (a,b).

Theorem 10.3. Let be given an Euclidean domain R and the nonzero elements a,b ∈ R. Applying the Euclideanalgorithm

a = q0b + r0

b = q1r0 + r1

r0 = q2r1 + r2

. . . . . . . . .

rn−2 = qnrn−2 + rn

rn−1 = qn+1rn

The sequence rn is decreasing and finite and rn = (a,b) := d. Also, there exist x, y ∈ R such that that d = xa + by.

Proof. Since R is Euclidean domain there exists norma N : R −→ Z+∪ 0. Using Euclid ’s algorithm we

haveN(b) >N(r0) > · · · >N(rn).

Thus, the sequence rn is decreasing. Since ri ∈Z+ this sequence has a minimal element , so is finite. Prove

that rn|a and rn|b. From the last equality rn|rn−1. Then, rn|rn and rn|rn−1 therefore rn|rn−2. From inductionwe get rn|a and rn|b. If d1|a and d1|b, then from first equality d1|r0. Thus, d1|r0 and d1|b and therefore d1|r1.Again with induction we prove that d1|rn. Thus, rn = (a,b).

To prove that rn is a linear combination of a and b it is enough to prove that rn belongs to the idealI = (a,b). We have r0 ∈ I,r1 ∈ I, . . . . Thus, rn ∈ I.

If a,b ∈ R such that (a,b) = eR then a and b are called relatively prime.

Example 10.9. Gaussian integers are defined as

Z[i] = a + bi : a,b ∈Z.

Prove that they form a Euclidean ring.

Proof. We define the following norm,

ν :Z[i]→Z

a + bi→ a2 + b2

Then, ν(a + bi) = a2 + b2 is a Euclidean norm in Z[i]. Let z,w ∈Z[i]. Then,

ν(zw) = |zw|2 = |z|2|w|2 = ν(z)ν(w).

Since ν(z) ≥ 1 for every nonzero element z ∈ Z[i], then ν(z) = ν(z)ν(w). The reader can verify all theseproperties.

We have to show that for every z = a+bi and w = c+di in Z[i], where w , 0, there are the elements q andr in Z[i] such that

z = qw + r,

where r = 0 or ν(r) < ν(w).Consider, z and w as elements in Q(i) = p + qi : p,q ∈Q the field of fractions of Z[i]. Notice that

zw−1 = (a + bi)c−di

c2 + d2 =ac + bdc2 + d2 +

bc− adc2 + d2 i =

(m1 +

n1

c2 + d2

)+

(m2 +

n2

c2 + d2

)i

= (m1 + m2i) +( n1

c2 + d2 +n2

c2 + d2 i)

= (m1 + m2i) + (s + ti)

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Shaska T. 10

in Q(i).Finally we will write the real and the imaginary part as a sum of an integer and a proper fraction. Thus,

take the integer closest to mi such that the fractional part satisfies |ni/(a2 + b2)| ≤ 1/2.Take s and t as fractional parts of zw−1 = (m1 + m2i) + (s + ti). We know that s2 + t2

≤ 1/4 + 1/4 = 1/2.Multiplying with w, we get

z = zw−1w = w(m1 + m2i) + w(s + ti) = qw + r,

where q = m1 + m2i and r = w(s + ti). Since z and qw are in Z[i], r must be in Z[i]. Finally, we must provethat r = 0 or ν(r) < ν(w). However,

ν(r) = ν(w)ν(s + ti) ≤12ν(w) < ν(w).

This completes the proof.

Exercises:

10.2.

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10 Shaska T.

10.3 Principal ideal domains

In this section we study rings in which every ideal is principal.

Definition 10.3. A ring A is called a principal ideal domain, denoted by PID, if every ideal of A is principal.

From Lemma 10.1 we have that every Euclidean domain is principal. There exist principal ideal domainswhich are not Euclidean.

Lemma 10.2. Every prime ideal in a principal ideal domain is maximal.

Proof. Take an ideal (p) in the principal ideal domain R. We we know that there exists the proper ideal M,such that (p) ⊂M. Since R is a PID, there exists m ∈ R that m = (m). Thus, (p) ⊂ (m), which implies p = rm,for some r ∈ R. Since (p) is prime, then r ∈ (p) or m ∈ (p). If m ∈ (p), then (m) = (p) and this completes theproof. If r ∈ (p), then r = ps. Thus, p = rm = psm or sm = 1. Thus, M is a unit which implies that (m) = R. Thiscontradicts the hypothesis that m = (m) is a proper ideal.

Corollary 10.1. If R is commutative ring such that R[x] is principal ideal domain then R is field.

Proof. Since R ⊂ R[x], then R is integral ring, because R[x] is integral ring as principal ideal domain.However R = R[x]/(x), hence (x) is prime ideal. From the above lemma (x) is maximal,which implies thatR is field.

Exercises:

10.3.

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Shaska T. 10

10.4 Unique factorization domains

In this section we will study rings in which every element can be factored in a unique way as a product ofirreducible factors.

Definition 10.4. Let be given a integral ring R.1) A element r ∈ R, that is not a unit is called irreducible in R if

r = ab⇒ a ose b is unit.

2) A element r ∈ R is called reducible if r is written as product two elements r = ab, where a and b are not units.3) The element p ∈ R is called prime if (p) is a prime ideal.4) Two elements a,b are called associated if a = bu for some a unit u in R.

Lemma 10.3. A prime element is irreducible.

Proof. Let p an prime element in R. If p = ab then ab ∈ (p). Thus, a ∈ (p) ose b ∈ (p). Assume that a ∈ (p) thisimplies a = pr. Then, p = ab = prb therefore rb = 1. This implies that b is a unit. Finally (p) is irreducible.

Not all irreducible elements are prime. For example, take the ring

Z[√

−5] = a + b√

−5 : a,b ∈Z

Recall that the ring Z[√

D], has a norm

N :Z[√

D] −→Z

a + b√

D −→ a2−Db2

and N(xy) = N(x) ·N(y). Also u is a unit in R if and only if N(u) = ±1.Take α = 2 +

√−5 ∈ Z[

√−5]. Then, N(α) = −1. If α = ab then N(ab) = −1. Thus, a or b is a unit and

therefore α is irreducible. However α is not prime because

32 = (2 +√

−5)(2−√

−5

Thus, 32∈ (α) but 3 < (α).

Lemma 10.4. In a principal ideal domain R an nonzero element is prime if and only if it is irreducible.

Proof. ⇒ Directly from the above Lemma.⇐ Take an irreducible element P. We want to show that (P) is prime. Assume that I is a ideal that

contains (P). Since we are in a principal ideal domain I = (m) for some m ∈ R. Thus, p ∈ (m) and thereforep = rm, r ∈ R. P is irreducible so R or M is a unit. Thus, (p) = (m) or (m) = (1). which implies the only idealsthat contain (p) are (p) and R = (1). Thus, (p) is maximal and therefore prime.

Definition 10.5. A integral ring R is called unique factorization domain, denoted by UFD, if every element r ∈R,that is not a unit can be written as a product of irreducible elements pi

r = pα11 . . .pαk

k

and ifr = qβ1

1 . . .qβss ,

then s = k and pi is of associated with qi.

Example 10.10. A field F is UFD because every nonzero element is a unit.

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10 Shaska T.

Lemma 10.5. In a UFD an nonzero element is prime if and only if is irreducible.

Proof. Form the above statement a prime element is irreducible. Let ’s prove the converse.Take an irreducible element p ∈ R. Assume that ab ∈ (p). Thus, ab = pc for c ∈ R. Since R is UFD then

a = pα11 . . .pαk

k , and b = qβ11 . . .q

βss ,

where p1, . . . ,pk,q1, . . . ,qs are irreducible elements. Thus,

pc = pα11 . . .pαk

k qβ11 . . .q

βss .

Thus , p must be associated with some prime from pα11 . . .pαk

k qβ11 . . .q

βss . Assume that p is associated with p1.

Thus, p = p1u, where u is a unit. Thus,a = (up)pα1−1

1 . . .pαkk .

Thus, α is in (P). Above we proved that α = 2 +

√−5 is irreducible but not prime in Z[

√−5]. Thus, Z[

√−5] is not UFD.

This is the first example of an integral ring that is not UFD.

Theorem 10.4. Every principal ideal domain R is UFD.

Proof. Take an element r ∈ R, r , 0 and r is not a unit. We want to prove that r can not be written as productof irreducible elements of R. If r is irreducible, we are done. If r is reducible, then r can be written asproduct r = r1r2 where r1 and r2 are not units. If these elements are irreducible we are done, otherwise theyare written as product of other elements. We need to show that this process ends.

We know that,(r) ⊂ (r1) ⊂ · · · ⊂ R.

This increasing chain ends by Zorn ’s lemma.To prove that this factorization is unique we use induction on the number of irreducible factors n of r.

For n = 1 it is clear. Ifr = p1p2 . . .pn = q1q2 . . .qm m ≥ n

then p1 divides the right hand side, hence one of qi, i = 1,2, . . .m. Assume that, p1|q1. Then, q1 = p1u. Then,u must be a unit because q1 is irreducible. which implies p1 of q1 are of associated. Thus, we have

p1p2 . . .pn = p1uq2 . . .qm.

We can cancel p1 since the cancellation property holds on integral rings, and we get

p2 . . .pn = uq2 . . .qm.

Now we have (n−1) factors. From induction hypothesis pi and q j are of associated. Since and p1 and q1 areassociated, then this completes the proof.

Theorem 10.5 (Fundamental Theorem of Arithmetic). The ring of integers Z is UFD.

Proof. Z is a PID and from the above lemma it is a PID. We summarize some of the inclusions of these classes of rings.

Fields ⊂ ED ⊂ PID ⊂UFD ⊂ IntegralDomains

Every inclusion is proper. For example

1. Z is Euclidean domain but not a field.

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Shaska T. 10

2. Z[θ], where θ = 1+√−19

2 is principal ideal domain but not a Euclidean domain.

3. Z[x] is UFD but not PID.

4. Z[√−5] is integral ring but not UFD.

Exercises:

10.4. A is every integral ring finite field?

10.5. Prove, that m2(Q) does not have proper ideals.

10.6. Prove, that for every integral ring can is contained in a field.

10.7. Let be given A a commutative ring with identity. Prove, that in the ring A[x] the Jacobson radical is equal withthe nilradical of A[x].

10.8. Let be given R a ring in which x3 = x. Prove, that R is commutative.

10.9. Let be given R a ring in which x4 = x. Prove, that R is commutative.

10.10. Let be given I, J ideals of R and R1 = R/I, R2 = R/J. Prove that,

ϕ : R→ R1⊕R2

r→ (r + I,r + J)

is a homomorphism, such that kerϕ = I∩ J.

10.11. Let be given m,n ∈ such that (m,n) = 1. Prove, the isomorphism of rings

Zmn =Zm⊕Zn

10.12. Let be given x an element nilpotent of a ring A. Prove that, 1 + x is a a unit of A. Show that the sum of anilpotent element with a a unit is a unit.

10.13. Let be given ring A and N its nilradical. Prove that, the following statements are equivalent:i) A has exactly only a prime ideal.ii) Every element of A is a unit or nilpotent.iii) A/N is field.

10.14. Let be given A a ring and f ∈ A[x], such that

f = a0 + a1x + · · ·+ anxn.

Prove thati) f is a unit in A[x] if and only if when a0 is a unit in A and a1, . . . ,an are nilpotent.ii) f is nilpotent if and only if when a0,a1, . . . ,an are nilpotent.iii) f is a zero divisor if and only if there is an a , 0 in A, such that a f (x) = 0.iv) For every f , g ∈ A[x], f g is irreducible if and only if f and g are irreducible.

10.15. Let z = a+b√

3 i inZ[√

3 i]. If a+3b2 = 1, prove that, z must be a unit. Prove that, the only units ofZ[√

3o f ]are 1 and −1.

10.16. Gaussian integers, Z[i], are UFD. Factor each from the following elements in Z[i], in a product of irreducibleelements; i) 5, ii) 1 + 3i, iii) 6 + 8i, iv) 2.

10.17. Let D a integral ring.

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10 Shaska T.

1. Prove that, FD is a commutative group under with addition.

2. Prove that multiplication is well defined in the field of fractions FD.

3. Verify properties associative and commutative for multiplication in FD.

10.18. Prove or disprove: Every subring of a field F, that contains 1 is an integral ring.

10.19. Let F a field with characteristic zero. Prove that, F contains a subfield isomorphic to Q.

10.20. Le to be F field.

1. Prove that, field of fractions of F[x], of denoted by F(x), is isomorphic to the set of all rational expressionsp(x)/q(x), where q(x) is not polynomial zero.

2. Let p(x1, . . . ,xn) and q(x1, . . . ,xn) be polynomials in F[x1, . . . ,xn]. Prove that, the set of all rational expressionsp(x1, . . . ,xn)/q(x1, . . . ,xn) is isomorphic to field of fractions of F[x1, . . . ,xn]. Denote the field of fractions ofF[x1, . . . ,xn] by F(x1, . . . ,xn).

10.21. Let p be prime and denote the field of fractions of Zp[x] by Zp(x). Prove that Zp(x) is an infinite field ofcharacteristic p.

10.22. Prove that field of fractions of Gaussian integers, Z[i] is

Q(i) = p + qo f : p,q ∈Q.

10.23. A field F is called simple field if it has no proper subfield. If E is a subfield of F and E is simple field, then Eis a prime subfielde F.

1. Prove that, every field contains a unique prime subfield.

2. If F is field with characteristic 0, show that prime subfield of F is isomorphic to field of rational numbers, Q.

3. If F is field with characteristic p, show that prime subfield of F is isomorphic to Zp.

10.24. Let Z[√

2] = a + b√

2 : a,b ∈Z.

1. Prove that, Z[√

2] is a integral ring.

2. Find all units of Z[√

2].

3. Determine the field of fractions of Z[√

2].

4. Prove that, Z[√

2o f ] is a Euclidean domain with Euclidean norm ν(a + b√

2 i) = a2 + 2b2.

10.25. Let D be a UFD, d ∈ D is the greatest common divisor of a and b in D if d | a and d | b and d is divisiblefrom every other element, that divides together a and b.

1. If D is PID and a and b are together elements nonzero of D, show that there is the greatest common divisor,unique, of a and b. We write gcd (a,b) for the greatest common divisor of a and b.

2. Let D a PID and a and b elements nonzero of D. Prove that, there exist elements s and t in D, such thatgcd (a,b) = as + bt.

10.26. Let D a integral ring. Define a relation in D, where a ∼ b if a and b are associative in D. Prove that, ∼ isequivalence relation in D.

10.27. Let D a Euclidean domain with Euclidean norm ν. If u is a unit in D, prove that ν(u) = ν(1).

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10.28. Let D a Euclidean domain with Euclidean norm ν. If a and b are associative in D, show that ν(a) = ν(b).

10.29. Prove that Z[√

5 i] is not UFD.

10.30. Prove or disprove: Every subring of a UFD is again UFD.

10.31. A ideal of commutative ring R is called finitely generated if there are elements a1, . . . ,an in R such that everyelement r ∈ R can be written as a1r1 + · · ·+ anrn for some r1, . . . ,rn in R. Prove that, R satisfies the decreasing chaincondition if and only if,every ideal of R is finitely generated.

10.32. Let D a integral ring with a decreasing chain of ideals I1 ⊃ I2 ⊃ · · · . Prove that, there exists a N, such thatIk = IN for every k ≥N. A ring that satisfy this condition is called a ring with decreasing chain condition, or DCC.The rings which satisfy the DCC are called Artinian rings.

10.33. Let R a commutative ring with identity. We define the multiplicative subset of R to be a subset S, such that1 ∈ S and ab ∈ S if a,b ∈ S.

1. Define a relation ∼ in R×S, where (a,s) ∼ (a′,s′), if there exists a s ∈ S such that s(s′a− sa′) = 0. Prove that ∼is equivalence relation in R×S.

2. Let a/s the equivalence class of (a,s) ∈ R×S and let S−1R the set of all equivalence classes with ∼. Defineaddition and of multiplication on S−1R respectively as

as

+bt

=at + bs

stas

bt

=abst,

Prove that these operations are of well defined in S−1R and that S−1R is ring with identity with these operations.The ring S−1R is called ring of fractions of R related to with S.

3. Prove that, the map ψ : R→ S−1R of defined from ψ(a) = a/1, is homomorphism rings.

4. If R does not have zero divisor and 0 < S, prove that ψ is injective.

5. Prove that, P is prime ideal of R, if and only if S = R\P is multiplicative subset of R.

6. If P is prime ideal of R and S = R\P, prove that ring of fractions S−1R has a maximal ideal of only. Every ring,that has a maximal ideal of only is called local ring .

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172

Chapter 11

Polynomial rings

In this chapter we will review some of the basic properties of polynomial rings. Further, we will study theirreducibility criteria for polynomials, symmetric functions, resultants, and discriminants. Even thoughmost of the results can be extended to polynomials in several variables we will focus mainly on polynomialsin one variable.

11.1 Polynomials

In this chapter, R is a commutative ring with identity. An expression of the form

f (x) =

n∑i=0

aixi = a0 + a1x + a2x2 + · · ·+ anxn,

where ai ∈ R and an , 0, is called a polynomial over R with variable x. The elements a0,a1, . . . ,an are calledcoefficients of f (x). The coefficient an is called the leading coefficient. A polynomial is called monic if itsleading coefficient is 1.

If n is the largest non negative integer for which an , 0, then we say that the degree of f (x) is n and writedeg f (x) = n. If such an n does not exist, then we have f = 0 and degree of f (x) is∞.

The set of all polynomials , with coefficient in a ring R is denoted by R[x]. Two polynomials are equal iftheir corresponding coefficients are equal, so if we have

p(x) = a0 + a1x + · · ·+ anxn

q(x) = b0 + b1x + · · ·+ bmxm,

then p(x) = q(x) if and only if ai = bi for every i ≥ 0.To prove that the set of all polynomials forms a ring, we must first define addition and multiplication.

The sum of two polynomials we define as follows. Let p(x) and q(x) be as follows

p(x) = a0 + a1x + · · ·+ anxn

q(x) = b0 + b1x + · · ·+ bmxm.

Then, the sum of p(x) with q(x) isp(x) + q(x) = c0 + c1x + · · ·+ ckxk,

where ci = ai + bi for every i.

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The product of p(x) with q(x) is defined as

p(x)q(x) = c0 + c1x + · · ·+ cm+nxm+n,

where

ci =

i∑k=0

akbi−k = a0bi + a1bi−1 + · · ·+ ai−1b1 + aib0,

for every i. Notice that in all cases some of the coefficients can be zero.

Theorem 11.1. Let R a commutative ring with identity. Then, R[x] is commutative ring with identity.

Proof. First we want to show that R[x] is an Abelian group with additions of polynomials. The zeropolynomial, f (x) = 0 is the zero of the group. Let be given a polynomial p(x) =

∑ni=0 aixi, then the opposite

of p(x) is −p(x) =∑n

i=0(−ai)xi = −∑n

i=0 aixi. Commutativity and associativity come from the definition ofaddition.

To prove that multiplication of polynomials is associative, take p(x),q(x),r(x) as follows

p(x) =

m∑i=0

aixi, q(x) =

n∑i=0

bixi, r(x) =

p∑i=0

cixi.

Then,

[p(x)q(x)

]r(x) =

m∑

i=0

aixi

n∑

i=0

bixi

p∑

i=0

cixi

=

m+n∑i=0

i∑j=0

a jbi− j

xi

p∑

i=0

cixi

=

m+n+p∑i=0

i∑j=0

j∑

k=0

akb j−k

c j

xi

=

m+n+p∑i=0

∑j+k+l=i

a jbkcr

xi

=

m+n+p∑i=0

i∑j=0

a j

i− j∑k=0

bkci− j−k

xi

=

m∑i=0

aixi

n+p∑i=0

i∑j=0

b jci− j

xi

=

m∑i=0

aixi

n∑

i=0

bixi

p∑

i=0

cixi

= p(x)[q(x)r(x)

]The proof for the commutativity and distribution are left as an exercise for the reader.

Lemma 11.1. Let p(x) and q(x)polynomials in R[x], where R is a integral ring. Then,

deg(p ·q) = deg p + deg q.

Moreover, R[x] is a integral ring.

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Proof. Assume that we have two polynomials

p(x) = amxm + · · ·+ a1x + a0

q(x) = bnxn + · · ·+ b1x + b0

where am , 0 and bn , 0. The degrees of p and q are m and n, respectively. The leading term of p(x)q(x) isambnxm+n, which can not be zero since R is a integral ring; thus, degree of p(x)q(x) is m + n and p(x)q(x) , 0.Since p(x) , 0 and q(x) , 0 this means se p(x)q(x) , 0, so R[x] is an integral ring.

Next we have the following important result.

Theorem 11.2. Let be given a field F. Then, F[x] is a Euclidean domain with norm

N :F[x] −→Z+∪0

p(x) −→ deg p.

Proof. Take a(x),b(x) ∈ F[x], where a(x),b(x) , 0. We will prove the theorem by induction on the degreen = deg(a(x)). For n = 0,1 the theorem is simply the Euclid’s algorithm for Euclidean domains.

Assume that the theorem is true for k < n−1. If n <m, then a(x) = 0 ·b(x)+b(x) and the proof is complete.If n ≥m then we get

a(x) = anxn + · · ·+ a1x + a0

b(x) = bmxm + · · ·+ b1x + b0

Denote with a′(x) := a(x)− anbm

xn−mb(x). Thus,

a′(x) = anxn−1 + . . .+ a1x + a0−an ·bm−1

b0xn−m−1 + · · ·+

an

bmb0xn−m

Hence a′(x) is a polynomial with degree n−1. From induction hypothesis there exist q′(x) and r(x) that

a′(x) = q′(x)b(x) + r(x),

where r(x) = 0 or degr(x) < degb(x). Take q(x) = q′(x) + anbm

xn−m and we have:

a(x) = q(x)b(x) + r(x),

where r(x) = 0 or degr(x) < degb(x), because

q(x)b(x) + r(x) = (q′(x) +an

bmxn−m)b(x) + r(x)

= q′(x)b(x) +an

bmxn−mb(x) + r(x) =

= a′(x) +an

bmxn−m)b(x) = a(x).

To prove the uniqueness assume thata(x) = q1(x)b(x) + r1(x).

Then,

r(x) = a(x)− q(x)b(x)r1(x) = a(x)− q1(x)b(x)

deg(r(x)− r1(x)) = degb(x)(r(x)− r1(x) <m.

However, the degree of b(x) is m. Hence q1(x)− q(x) = 0. From this we get that r(x) = r1(x).

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11.1.1 Division algorithm

Recall that se division algorithm for integers says that if a and b are integers, where b > 0, then thereexist unique numbers q and r such that a = bq + r, where 0 ≤ r < b. The algorithm of finding q and r iscalled the Euclidean algorithm. For polynomials there exists a similar theorem. The division algorithm forpolynomials is similar of that of integers.

Theorem 11.3 (Division Algorithm). Let f (x) and g(x) be two nonzero polynomials in F[x], where F is a field andg(x) is a non-constant polynomial. Then, there exist unique polynomials q(x),r(x) ∈ F[x] such that

f (x) = g(x)q(x) + r(x),

where degr(x) < deg g(x) and r(x) is a nonzero polynomial.

Proof. First let’s study the existence of q(x) and r(x). Let S = f (x)− g(x)h(x) : h(x) ∈ F[x] and assume that

g(x) = a0 + a1x + · · ·+ anxn

is a polynomial with degree n. This set is nonempty, since f (x) ∈ S. If f (x) is the zero polynomial, then0 = f (x) = 0 · g(x) + 0. Hence, q and r, are zero polynomials.

Assume that the zero polynomial is not in S. In this case, the degree of for every polynomial in S isnon-negative. We pick a polynomial r(x) with smallest degree in S ; hence, there exists a q(x) ∈ F[x] suchthat

r(x) = f (x)− g(x)q(x),

orf (x) = g(x)q(x) + r(x).

We must show that degree of r(x) is smaller than degree of g(x). Assume that deg g(x) ≤ degr(x). Letr(x) = b0 + b1x + · · ·+ bmxm and m ≥ n. Then,

f (x)− g(x)(q(x)−

(bm

an

)xm−n

)= f (x)− g(x)q(x) +

(bm

an

)xm−ng(x)

= r(x) +

(bm

an

)xm−ng(x) = r(x) + bmxm + terms of lower degree

is contained in S. This is a polynomial with smaller degree than r(x), which contradicts the fact that r(x) isa polynomial with smallest degree in S ; so degr(x) < deg g(x).

To prove that q(x) and r(x) are unique, assume that there exist two other polynomials q′(x) and r′(x) suchthat f (x) = g(x)q′(x) + r′(x) and degr′(x) < deg g(x) or r′(x) = 0 ; hence,

f (x) = g(x)q(x) + r(x) = g(x)q′(x) + r′(x),

andg(x)[q(x)− q′(x)] = r′(x)− r(x).

If g is not the zero polynomial, then

deg(g(x)[q(x)− q′(x)]) = deg(r′(x)− r(x)) ≥ deg g(x).

However, degrees of r(x) and r′(x) are strictly less than degree of g(x). Hence, r(x) = r′(x) and q(x) = q′(x).

Let p(x) be a polynomial in F[x] and α ∈ F. We say that α is a zero or roots of p(x), if p(x) is in the kernelof the homomorphism φα or we say α is a zero of p(x) if p(α) = 0.

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Corollary 11.1. Let F be a field. An element α ∈ F is a zero of p(x) ∈ F[x], if and only if x−α is a factor of p(x) inF[x].

Proof. Assume that α ∈ F and p(α) = 0. From the division algorithm, there exist polynomials q(x) and r(x)such that

p(x) = (x−α)q(x) + r(x)

and the degree of r(x) must be smaller than degree of x−α. Since the degree of r(x) is smaller than 1, thenr(x) = a for a ∈ F ; hence,

p(x) = (x−α)q(x) + a.

However0 = p(α) = 0 ·q(x) + a = a.

Hence, p(x) = (x−α)q(x) and x−α is a factor of p(x).Conversely, assume that x−α is a factor of p(x) ; say p(x) = (x−α)q(x). Then, p(α) = 0 · q(x) = 0.

Corollary 11.2. Let F a field. A nonzero polynomial p(x) with degree n in F[x] must have at most n distinct zeroesin F.

Proof. We will use mathematical induction on the degree of p(x). If degp(x) = 0, then p(x) is a constantpolynomial and does not have a zero. Let degp(x) = 1. Then, p(x) = ax + b for some a and b in F, where α1and α2 are zeroes of p(x), we have aα1 + b = aα2 + b or α1 = α2.

Now assume that degp(x) > 1. If p(x) does not have some zero in F, then this completes the proof. Also,ifα is a zero of p(x), then p(x) = (x−α)q(x) for some q(x) ∈ F[x]. The degree of q(x) is n−1. Let β another zeroof p(x), which is different from α. Then, p(β) = (β−α)q(β) = 0. Since α , β and F is a field, q(β) = 0.

From induction hypothesis, p(x) can have at most n− 1 zeroes in F which are different from α. Thus,p(x) has at most n distinct zeroes in F.

Let F a field. A monic polynomial d(x) is called greatest common divisor of polynomials p(x),q(x) ∈ F[x]if d(x) divides p(x) and q(x) ; and if for every other polynomial d′(x) that divides p(x) and q(x), d′(x) | d(x).We write

d(x) = gcd (p(x),q(x)).

Two polynomials p(x) and q(x) are relatively prime if gcd (p(x),q(x)) = 1. Similarly as for the greatestcommon divisor of integers, we have the following:

Proposition 11.1. Let F be a field and assume that d(x) is the greatest common divisor of two polynomials p(x) andq(x) in F[x]. Then, there exist polynomials r(x) and s(x) such that

d(x) = r(x) ·p(x) + s(x) · q(x).

Thus, greatest common divisor of two polynomials is unique.

Proof. Let d(x)monic polynomial with smallest degree in the set

S = f (x)p(x) + g(x)q(x) : f (x), g(x) ∈ F[x].

We write d(x) = r(x)p(x) + s(x)q(x), for two polynomials r(x) and s(x) in F[x]. We must show that d(x)divides p(x) and q(x). First must prove that d(x) divides p(x). From division algorithm , there existpolynomials a(x) and b(x) such that p(x) = a(x)d(x) + b(x), where b(x) is the zero polynomial or degb(x) <degd(x). Thus,

b(x) = p(x)− a(x)d(x)= p(x)− a(x)(r(x)p(x) + s(x)q(x))= p(x)− a(x)r(x)p(x)− a(x)s(x)q(x)= p(x)(1− a(x)r(x)) + q(x)(−a(x)s(x))

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is a linear combination of p(x) and q(x) therefore is contained in S. However, b(x) must be the zeropolynomial, since d(x) was picked with smallest degree, hence d(x) divides p(x). A similar argument showsthat d(x) divides q(x). Thus, d(x) is a common divisor of p(x) and q(x).

To prove that d(x) is greatest common divisor of p(x) and q(x), assume that d′(x) is a other commondivisor of p(x) and q(x). We must prove that d′(x) | d(x). Since d′(x) is a other common divisor of p(x) andq(x), then there exist polynomials u(x) and v(x) such that p(x) = u(x)d′(x) and q(x) = v(x)d′(x). Thus,

d(x) = r(x)p(x) + s(x)q(x)= r(x)u(x)d′(x) + s(x)v(x)d′(x)= d′(x)[r(x)u(x) + s(x)v(x)].

which implies that d′(x) | d(x), d(x) is greatest common divisor of p(x) and q(x).Finally, we must prove that the greatest common divisor of p(x) and q(x)) is unique. Assume that d′(x)

is another common divisor of p(x) and q(x). It is enough to show that there exist polynomials u(x) and v(x)in F[x] such that

d(x) = d′(x)[r(x)u(x) + s(x)v(x)].

Sincedegd(x) = degd′(x) + deg[r(x)u(x) + s(x)v(x)]

and d(x) and d′(x) are both the greatest common divisor, then degd(x) = degd′(x). Since d(x) and d′(x) aretwo monomial polynomials with the same degree, then we have that d(x) = d′(x).

Theorem 11.4. If k is a field then k[x] is a PID.

Proof. Every Euclidean domain is a PID

Corollary 11.3. The ring k[x] is a unique factorization domain.

A polynomial f (x) ∈ k[x] is called irreducible if it has degree ≥ 1 and can not be written as

f (x) = g(x) ·h(x)

for some g,h ∈ k[x] and both g,h < k. Elements of k are called constant polynomials.Let A be a commutative ring and f ∈A[x]. Let B be an extension ring of A. Then α ∈ B is called a root of

f (x) if f (α) = 0.

Theorem 11.5. Let k be a field and f ∈ k[x] a polynomial of degree n. Then f (x) has at most n roots in k, and if x = ais a root of f (x) in k, then (x− a) divides f (x).

Proof. We prove the second part of the Theorem first. Assume that α ∈ k and f (α) = 0. From the divisionalgorithm, there exist polynomials q(x) and r(x) such that

f (x) = (x−α)q(x) + r(x)

and the degree of r(x) must be smaller than degree of x−α. Thus the degree of r(x) is smaller than 1, thenr(x) = a for some a ∈ k; hence,

f (x) = (x−α)q(x) + a.

However0 = f (α) = 0 ·q(x) + a = a.

Hence, f (x) = (x−α)q(x) and x−α is a factor of f (x).Conversely, assume that x−α is a factor of f (x); say f (x) = (x−α)q(x). Then, f (α) = 0 ·q(x) = 0.

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To prove the first part of the theorem we will use mathematical induction on the degree of f (x). Ifdeg f (x) = 0, then f (x) is a constant polynomial and does not have a zero. Let deg f (x) = 1. Then, f (x) = ax+bfor some a and b in k, where α1 and α2 are zeroes of f (x), we have aα1 + b = aα2 + b or α1 = α2.

Now assume that deg f (x) > 1. If f (x) does not have some zero in k, then this completes the proof. Also,if α is a zero of f (x), then f (x) = (x−α)q(x) for some q(x) ∈ k[x]. The degree of q(x) is n−1. Let β another zeroof f (x), which is different from α. Then, f (β) = (β−α)q(β) = 0. Since α , β and k is a field, q(β) = 0.

From induction hypothesis, f (x) can have at most n− 1 zeroes in k which are different from α. Thus,f (x) has at most n distinct zeroes in k.

Lemma 11.2. Let f (x) be a polynomial in F[x]. Then, F[x]/〈 f (x)〉 is a field if and only if f (x) is irreducible.

Proof. Exercise

Theorem 11.6. Every finite subgroup of the multiplication group of a field is cyclic

Proof. Let F be a field and F? denote the set of its nonzero elements. Let G ≤ F? such that |G| = n. Then Gis Abelian (since F? is Abelian) and from the fundamental theorem of Abelian groups G has an invariantfactor decomposition

GZm1 × · · ·×Zmk

such that mi |mi+1 and mi ≥ 2, for all i ≤ k. Take x ∈G. Then x ∈Zmi for some i. Hence, xmi = 1 which impliesthat xmk = 1. Thus, the polynomial xmk −1 has n roots (since G has n elements). But a polynomial can’t havemore roots then its degree, see Theorem 11.5. Hence, n = mk and GZmk .

Corollary 11.4. If F is finite then F? is cyclic.

An element ε in a field k such that εn = 1 is called the n-th root of unity. The set of n-th roots of unityare roots of the polynomial f (x) = xn

−1. This set forms a subgroup of k? of order n and by Theorem 11.6such subgroup is cyclic. Any generator of this group is called a primitive n-th root of unity. If k = C thenthe n-th roots of unity are

αr = er 2πin

for 1 ≤ r ≤ n. One of the primitive roots in this case is e2πin

Let A be a commutative ring and f (x) ∈ A[x], such that

f (x) = anxn + an−1xn−1 + · · ·a1x + a0.

The derivative of f (x) is defined as follows

f ′(x) =

n∑r=1

rarxr−1

One can easily verify that for every f , g ∈ A[x] and a ∈ A the following are true

( f + g)′ = f ′+ g′

( f · g)′ = f ′g + f g′

(a f )′ = a f ′

Hence, this definition of the derivative matches the definition from Calculus. Moreover, we have a map

A[x]→ A[x]f → f ′

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Let k be a field, f a non-zero polynomial in k[x] and α ∈ k a root of f (x). Then,

f (x) = (x−α)m· g(x)

such that (x−α) 6 | g(x). We call m the multiplicity of α in f (x). The root α is called a multiple root if m > 1.

Lemma 11.3. Let k be a field and f ∈ k[x]. Let α ∈ k be a root of f (x). Then, α is a multiple root if and only iff ′(α) = 0.

Proof. Homework.

Exercises:

11.1. Assume that R and S are isomorphic rings. Prove that R[x] is isomorphic to S[x].

11.2. Let F a field and a ∈ F. If p(x) ∈ F[x]. Prove that p(a) is the remainder obtained from the division of p(x) withx− a.

11.3. Let Q∗ the multiplicative group of non-negative rational numbers. Prove that Q∗ is isomorphic to (Z[x],+).

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11.2 Polynomials over UFD’s

Let A be a UFD and k its field of fractions. We take a ∈ k such that a = rs , where (r,s) = 1. For any prime

element p ∈ A, we can writea = pm a′

where m is a integer and a′ ∈ k such that p does not divide numerator or denominator of a′. The order of ain p is defined as m, say ordp(a) = m.

Let f (x) ∈ k[x] be given byf (x) = anxn + an−1xn−1 + · · ·+ a1x + a0.

Defineordp ( f ) = min ordp (ai) | ai , 0.

The content of f (x), which is denoted cont ( f ), is defined as the product (up to multiplication to a unit in A)

cont ( f ) :=∏

pordp ( f ), (11.1)

taking all p such that ordp ( f ) , 0. If cont ( f ) = 1, then f (x) is called a primitive polynomial. Thus, everypolynomial f (x) ∈ k[x] can be written as

f (x) = cont ( f ) · f1(x),

where f1(x) is primitive and f1(x) ∈ A[x].The height of f (x) is defined as

h( f ) := max ordp(ai) |ai , 0

The following result is known as Gauss’ lemma.

Theorem 11.7 (Gauss Lemma). Let A be a UFD, k its field of fractions and f , g ∈ k[x]. Then,

cont ( f g) = cont ( f ) · cont (g)

Proof. Let f (x) and g(x) be

f (x) = anxn + an−1xn−1 + . . .+ a1x + a0

g(x) = bmxm + bm−1xm−1 + . . .+ b1x + b0

Corollary 11.5. Let f , g ∈ A[x]. Then, f g is primitive if and only if f and g are both primitive.

Proof. Assume that f (x)g(x) is not primitive and we will show that f (x) or g(x) are not primitive. Thatf (x)g(x) is not a unit means that gcd of coefficients of this polynomial is not 1. Let p be an irreducible factorof this gcd . So f (x)g(x) looks like

f (x)g(x) = pcnxn + pcn−1xn−1 + . . .+ pc1x + pc0.

Consider the the natural homomorphism π : A→ A/pA and extend it to the ring of polynomials.

π : A[x]→(A/pA

)[x]

Since A is a integral ring, A/pA is an integral ring as well. Hence, (A/pA)[x] is a integral ring.We have that π( f (x)g(x)) = 0, since p divides all coefficients of f (x)g(x). Hence π( f (x)) = 0 or π(g(x)) = 0.

Hence, f (x) or g(x) is divisible from p, so one of these polynomials is not primitive.

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Corollary 11.6. If f (x) is reducible in k[x], then f (x) is reducible in A[x].

Proof. Let be given f (x) = g(x) ·h(x), where g(x),h(x) ∈ k[x]. Then,

f (x) = cont (g) · cont (h) · g1(x) ·h1(x)

where g1,h1 ∈ A[x] are primitive. Thus, f (x) is reducible in A[x].

Example 11.1.

Theorem 11.8. Let be given A a UFD. Then, A[x] is a UFD and its primes are or primes in A, or primitivepolynomials, irreducible in A[x].

Exercises:

11.4. Let be given a ring A and f ∈ A[x] such that

f = a0 + a1x + · · ·+ anxn.

Prove that

i) f is an element a unit in A[x] if and only if a0 is an element a unit in A and a1, . . . ,an are nilpotent.ii) f is nilpotent if and only if a0,a1, . . . ,an are nilpotent.iii) f is a zero divisor if and only if there exists a , 0 in A such that a f (x) = 0.iv) For every f , g ∈ A[x], f g is primitive if and only if f and g are primitive.

11.5. Prove that the ideals I = (x) and J = (x, y) are prime ideals in Q[x, y], but only J is maximal.

11.6. Prove that the ideals I = (x, y) and J = (2,x, y) are prime ideals in Z[x, y], but only J is maximal.

11.7. Prove that I = (x, y) is not a principal ideal in Q[x, y].

11.8. Show that the radical of the ideal I = (x, y2) in Q[x, y] is the ideal J = (x, y). Moreover, prove that I is a primaryideal that is not a power of a prime ideal.

11.9. Prove that the rings F[x, y]/(y2−x) and F[x, y]/(y2

−x2) are not isomorphic for any field F.

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11.3 Irreducibility of polynomials

Next we see another criteria of irreducibility for polynomials.

Lemma 11.4. Let be given field F and p(x) ∈ F[x]. Then, p(x) has a factor with degree 1 if and only if p(x) has a rootsin F (there exists α ∈ F that p(α) = 0 )

Proof. If p(x) has a factor (ax + b) = a(x− (− ba )) = a(x−α), then α = a

b ∈ F. Hence assume that p(x) has a factor(x−α). Then, p(α) = 0. On the contrary, assume that ∃α ∈ F that p(α) = 0. Then, p(x) = q(x)(x−α) + r fromEuclid’s algorithm. Substituting x = α we have r = 0.

11.3.1 Integer root test

Theorem 11.9 (Integer root test). Let R be a UFD, F the field of fractions of R, and p(x) ∈ R[x] as follows,

p(x) = anxn + · · ·+ a1x + a0

Let r,s,∈ R, such that rs ∈ F, (r,s) = 1. If r

s is root of p(x), then r |a0 and s |an.

Proof. Assume that p( rs ) = 0 = an( r

s )n + · · ·+ a0. Multiplying by sn we have;

anrn + an−1rn−1s + . . .+ a1x + a0sn = 0

Hence anrn = −s(an−1rn−1 + . . .+ a1x + a0sn−1), hence s|an. Also a0sn = −r(anrn−1 + . . .a1), so r|a0.

Corollary 11.7. If p(x) ∈Z[x] is given as follows

p(x) = xn + an−1xn−1 + . . .+ a1x + a0

and p(r) , 0, for every r|a0 then p(x) has no root in Q.

Proof. If rs ∈Q and p( r

s ) = 0, then s|1 so s = ±1. which implies r|a0 and p(r) = 0, that is contradiction.

Example 11.2. Prove thatg(x) = x4 + x3 + x2 + x + 1

is irreducible over field with two elements F2.

Proof. The field F2 has two elements 0 and 1. We see that no one of them is a root. Hence g(x) does nothave linear factors. Then, if g(x) is factored it will have only quadratic factors. Assume that there exist suchquadratic factors, say

g(x) = (x2 + ax + 1)(x2 + bx + 1)

Leading coefficients must be 1 that x4 to have coefficient 1. Also the constant terms must be 1 so that theirproduct is 1. By multiplying through and equaling coefficients we get b+c = 1 and bc = 1. This system doesnot have a solution in F2 (check 0,1). Hence g(x) is irreducible.

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11.3.2 Eisenstein Criteria

Theorem 11.10 (Eisenstein criteria). Let be given A a UFD and K its field of fractions. Let f ∈ A[x] such that

f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0

and p a prime number in A, such thati) p | ai for every i ≤ n−1ii) p2 - a0iii) p - an.

Then, f (x) is irreducible in K[x].

Proof. We can assume that cont ( f ) = 1. If there is a factorization in factors with degree ≥ 1 in K[x] then fromcorollary of Gauss Lemma there exists a factorization in A[x], for example f (x) = g(x)h(x), where

g(x) = bdxd + · · ·+ b0

h(x) = cmxm + · · ·+ c0,

where d,m ≥ 1 and bdcm , 0.Since b0c0 = a0 is divisible from p, but not from p2, we have that one of them isnot divisible from p, say b0. Then, p | c0.Since bdcm = an is not divisible from p, then p does not divide cm.Letcr the farthest coefficient on the right such that p | cr. Then, r ,m and

ar = b0cr + b1cr−1 + . . . .

Since p does not divide b0cr and divides all other terms in this sum, we conclude that p does not divide ar.This completes the proof.

Example 11.3. Using Eisenstein criteria the reader to prove that polynomials that follow are irreducible,

f (x) = x4 + 10x + 5, f (x) = xn−p.

Let be given a prime number p. The polynomial

φp(x) =xp−1

x−1= xp−1 + xp−2 + · · ·+ x + 1

is called cyclotomic polynomial in p.

Lemma 11.5. φp(x) is irreducible in Z[x]

Proof. If φp(x) is reducible, then φp(x+1) is also reducible. Thus, instead of φp(x) we consider φp(x+1). Wehave

φp(x + 1) =(x + 1)p

−1x

= xp−1 + pxp−2 + · · ·+p(p−1)

2x + p.

Then, from Eisenstein criteria φp(x + 1) is irreducible. Therefore, φp(x) is irreducible

Theorem 11.11 (Extension of Eisenstein criteria). Let be given A a UFD and K its field of fractions. Let be givenf ∈ A[x] such that

f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0

and p a prime number in A such that:

1. there is a r (0 ≤ r ≤ n ) such that p - ar

2. p | ai for all 0 ≤ o f ≤ r−1

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3. p2 - a0

4. f (x) = h(x) · g(x), such that h, g ∈ A[x].

Then, deg(h) ≥ r or deg(g) ≥ r.

Proof.

Example 11.4. Let p a prime number. Prove that

f (x) = x5 + 2x4 + 3x3 + 3

is irreducible in Q[x].

Proof. We will use the above theorem. 3 divides a0, . . . ,a3, but does not divide a4. Hence, r = 4. Thus, if f (x)is reducible, then it is the product of polynomials of degrees 4 and 1. Thus, f x has a rational root. Frominteger root test we can show that this doesn’t happen.

11.3.3 Reduction modulo a prime

Theorem 11.12 (Reduction criteria). Let be given A,B integral rings, a ring homomorphism

ϕ0 : A→ B,

and ϕ : A[x] 7→ B[x] its extension to polynomial rings. Let be given K,L the field of fractions of A,B respectively andf ∈ A[x] such that ϕ ( f ) , 0 and degϕ( f ) = deg( f ). If ϕ( f ) is irreducible in L[x], then f is irreducible in K[x].

Proof. Assume that f has a factorization f (x) = g(x) ·h(x) in K[x]. Then,

ϕ( f ) = ϕ(g) ·ϕ(h).

Since deg ϕ(g) ≤ deg g and degϕ(h) ≤ degh, then we have equality. From irreducibility in L(x) we concludethat one of ϕ(g) or ϕ(h) is a constant. Thus one of g or h is in A, which contradicts our assumption. Thiscompletes the proof.

Corollary 11.8 (Modulo p irreducibility test). Let be given p a prime number and f (x) ∈Z[x], such that deg f ≥ 1and p does not divide the leading coefficient of f (x). Let f (x) ∈ Zp[x] be the reduction of f (x) mod p. If f (x) isirreducible in Zp[x], then f (x) is irreducible in Q[x].

Example 11.5. Prove thatf (x) = x5

−5x4−6x−1

is irreducible in Q[x].

Proof. Let be given f (x) = f (x) mod 5 = x5 + 4x + 1. It can be easily shown that f (x) does not have root inZ5. Thus, if f (x) is reducible in Z5[x], then it is a product of polynomials with degree 2 and 3. Thesepolynomials are

f (x) = x5 + 4x + 1 = (x2 + ax + b) (x3 + mx2 + nx + r).

The reader can show that this doesn ’t happen.

Remark 11.1. Notice that the converse of the above statement is not true. There are polynomials which are reduciblein Zp for every prime number p, but irreducible in Z[x].

Example 11.6. The polynomialf (x) = x4

−10x2 + 1

is reducible in Zp[x], for every prime number p, but is irreducible in Z[x].

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Proof. Indeed, we check that

f (x) mod 2 = (x + 1)4

f (x) mod 3 = (x2 + 1)2

f (x) mod 5 = (x2 + 3)(x2 + 2)

f (x) mod 7 = (x2 + 6x + 6)(x2 + x + 6),

(11.2)

see Milne [?Mi] (pg. 9) for the proof that this is reducible for any prime p.To prove that f (x) is irreducible in Z[x] we use the above example. We prove that f (x) does not have

rational root. Thus, it is product polynomials with degree 2. By elementary arithmetic we can show thatthis doesn ’t happen.

Exercises:

11.10. Let f (x) irreducible. If f (x) | p(x)q(x), prove that f (x) | p(x) or f (x) | q(x).

11.11. Prove that f (x) = x3−3x−1 is irreducible in Z[x].

11.12. For every prime number pprove that x2−p and x3

−p are irreducible in Q.

11.13. Let be given α ∈ Z such that α is divisible from some prime number p, but p2 - α. Prove that xn−α is

irreducible.

11.14. Prove that f (x) = x4 + 1 is irreducible in Q.

11.15. Prove that the following polynomials are irreducible in Z[x].

f (x) = x4 + 10x + 5.

11.16. Prove that the following polynomials are irreducible in Z[x].

f (x) = x4 + 10x2 + 1.

11.17. Prove that the following polynomials are irreducible in Z[x].

f (x) = x4−4x3 + 6.

11.18. Prove that the following polynomials are irreducible in Z[x].

f (x) = x6 + 30x5−15x3 + 6x−120.

11.19. Prove that the following polynomials are irreducible in Z[x].

f (x) = x4 + 4x3 + 6x2 + 2x + 1.

11.20. Is the polynomialf (x) = x7 + 3x6 + 12x5 + 6x4 + 2x3

−4x2 + 6x + 2

irreducible in Q[x]?

11.21. Prove that the following polynomials are irreducible over Qi) x4−4x3 + 6

ii) x6 + 30x5−15x3 + 6x−120

iii) x4 + 4x3 + 6x2 + 2x + 1

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11.22. Is the following polynomial

f (x) = x7 + 3x6 + 12x5 + 6x4 + 2x3−4x2 + 6x + 2

irreducible?

11.23. Prove that polynomials

f (x) =

n∏i=1

(x− i)±1

are irreducible in Z for all n ≥ 1.

11.24. If a is rational and x− a divides a monic polynomial f (x) ∈Z[x], prove that a is a integer.

11.25. Find which of the following is reducible (irreducible) in Z2[x]. Justify your answeri) x2 + 1ii) x2 + x + 1iii) x3 + x + 1

11.26. Prove or disprove: xp + a is irreducible, for every a ∈Zp, where p is prime number.

11.27. Find a factorization off (x) = x4 + 1

in Z5[x].

11.28. Find a factorization forf (x) = x4 + 1

in Z5[x].

11.4 Symmetric polynomials and discriminant

Let’s recall Vieta’s formula from elementary algebra. Assume that x1, . . . ,xn are n roots of a polynomial

f (x) = xn + a1xn−1 + · · ·+ an−1x + an.

Then,

s1(x1, . . . ,xn) =

n∑i=1

xi = −a1

s2(x1, . . . ,xn) =∑

1≤i1<i2≤n

xi1xi2 = a2

. . . . . . . . . . . .

sm(x1, . . . ,xn) =∑

1≤i1<..<im≤n

xi1 . . .xim = (−1)mam

. . . . . . . . . . . .

sn(x1, . . . ,xn) = x1x2 . . .xn = (−1)nan.

Polynomials s1, . . .sn are called symmetric polynomials of polynomial of degree n. The polynomialsm(x1, . . . ,xn) is called the m -th elementary symmetric polynomial in x1, . . . ,xn. This polynomial has thefollowing property

sm(xσ(1), . . . ,xσ(n)) = sm(x1, . . . ,xn)

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for every permutation σ of 1, . . . ,n. Recall that a permutation of 1, . . . ,n is a correspondence 1-1

σ : 1, . . . ,n → 1, . . . ,n.

The above property leads to this definition

Definition 11.1. A polynomial is called symmetric polynomial if it satisfies

p(xσ(1), . . . ,xσ(n)) = p(x1, . . . ,xn)

for every permutation σ ∈ 1, . . . ,n.

Let F be field. A polynomial p(x1, . . . ,xn) ∈ F[x1, . . . ,xn] is called symmetric if it is fixed from everypermutation of its variables.

Example 11.7. The sum x1 + · · ·+ xn and the product x1x2 . . .xn are symmetric as and sums xr1 + . . .xr

n for r ≥ 1.

We define an action of Sn in F[x1, . . . ,xn]

(σp)(x1, . . . ,xn) = p(xσ−1(1), . . . ,xσ−1(n)).

Theorem 11.13. Every symmetric polynomial in k[t1, . . . , tn] can be written in unique way with elementary symmetricpolynomials s1, . . . ,sn.

Proof. Let p(x1, . . . ,xn) a Sn -invariant. Le to be

q :Z[x1, . . . ,xn−1,xn]→Z[x1, . . . ,xn−1],

the map which drops xn. Hence, q(xi) = xi where (1 ≤ o f < n), q(x) = 0 where (i = n).If p(x1, . . . ,xn) is Sn -invariant, then

q(p(x1, . . . ,xn−1,xn)) = p(x1, . . . ,xn−1,0)

is Sn−1 -invariant, we we get a copy of Sn−1 in Sn that fixes n.Notice that

q(si(x1, . . . ,xn)) = si(x1, . . . ,xn−1

for (1 ≤ o f < n) andq(si(x1, . . . ,xn)) = 0,

for i = n.Me induction in the number of variables, there is a polynomial P with n−1 variables, such that;

q(p(x1, . . . ,xn)) = P(s1(x1, . . . ,xn−1), . . . ,sn−1(x1, . . . ,xn−1)).

Now use the same polynomial P, but with elementary polynomials from introduction of xn with

g()x1, . . . ,xn) = P(s1(x1, . . . ,xn−1), . . . ,sn−1(x1, . . . ,xn−1)).

From the way we picked P we have that q(p(x1, . . . ,xn))− g(x1, . . . ,xn) = 0. Hence the map xn → 0 sendsthe difference p− g in 0. Using factorization of only in Z[x1, . . . ,xn] this implies that xn divides p− g. Sn-invariance of p− g implies that every xi divides p− g. Thus, from uniqueness of factorization sn(x1, . . . ,xn)divides p− g.

The total degree of monomial cxe11 . . .x

enn is the sum of exponents

deg(c ·xe11 . . .x

enn ) = e1 + · · ·+ en.

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The total degree of a polynomial is the maximum of total degrees of its monomials. Consider thepolynomial

p− gsn

=p(x1, . . . ,xn)− g(x1, . . . ,xn)

sn(x1, . . . ,xn).

It has total degree smaller than p. With induction on total degree, p−gsn

is expressed in terms of symmetricelementary polynomials.

Definition 11.2. Let be given f (x) ∈ k[x] such that

f (x) = (x−α1) · · · (x−αn).

Discriminant of f (x) , that denoted by D( f ,x), is defined as:

D( f ,x) :=∏i< j

(αi−α j)2

Lemma 11.6. Let be given f (x) ∈ C[x]. Then, f (x) has a double root if and only if D( f ,x) = 0.

Proof. The proof is left as an exercise. Discriminant is a symmetric polynomial of α1, . . . ,αn, therefore it can be expressed as a polynomial in

s1, . . . ,sn.

Example 11.8. Let be given f (x) as belowf (x) = ax2 + bx + c.

It is easily proved that D( f ,x) = b2−4ac.

Example 11.9. Let be given f (x), the cubic function

f (x) = ax3 + bx2 + cx + d.

Then, D( f ,x) = −27a2d2 + 18adbc + b2c2−4b3d−4ac3.

Example 11.10. Let be given the quartic function

f (x) = ax4 + bx3 + cx2 + dx + e.

Then,

D( f ,x) = 256a3e3−128a2e2c2

−4b3d3 + 16ac4e−4ac3d2−6aeb2d2 + 144ae2cb2

+ 144a2ecd2 + 18abd3c + c2b2d2−4c3b2e−192a2e2bd−80abdc2e

+ 18b3dce−27b4e2−27a2d4

Exercise 11.1. Discriminant of a general polynomial of degree n,

f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0

is a homogenous polynomial in a0, . . .an degree 2n−2.

Example 11.11 (Quadratics). Here p(X) = x2 + bX + c = (x−α1) (x−α2). Then we get ∆p = b2−4c.

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Example 11.12 (Cubics). Take p in the form

p(x) = x3 + ax + b

Then we get ∆p = −4a3− 27b2. Thus Cardano’s formulas become

xi =

−b2

+

√−∆p

108

1/3

+

−b2−

√−∆p

108

1/3

Exercise 11.2. Suppose a,b ∈ Q. We know that Dp = 0 iff p has a multiple root. Show that Dp > 0 iff p has threedistinct real roots. Thus Dp < 0 iff p has one real root and two non-real complex conjugate roots.

Lemma 11.7. Let be given f , g ∈ k[x], deg f = l > 0, deg g = m > 0. Then, f and g have a common factor if and onlyif there exist A,B ∈ k[x] such that:

1) A and B are of different from zero.2) degA ≤m−1, degB ≤ l−1.3) A f + Bg = 0.

Proof. Left to the reader Let f , g be as follows

f (x) = alxl + . . .+ a1x + a0

g(x) = bmxm + . . .+ b1x + b0

A(x) = cm−1xm−1 + . . .+ c1x + c0

B(x) = dl−1xl−1 + . . .+ d1x + d0

Substituting in the equationA f + Bg = 0,

we have

(alcm−1 + bmdl−1)xl+m−1+

+ (al−1cm−1 + alcm−2 + bm−1dl−1 + bmdl−2)xl+m−2+

+ . . . . . . · · ·+

+ (a0c0 + b0d0) = 0

Let these coefficients equal zero we get the system of (l + m) equations. Consider c0, . . . ,cm−1,d0, . . . ,dl−1 asvariables. The system is linear and its coefficient matrix is:

Syl( f , g,x) =

al bmal−1 al bm−1 bmal−2 al−1 . bm−2 bm−1 .. al−2 . . . . . .. . . . al . . . . bm

a1 . . . al−1 b0 . . . bm−1a0 a1 . . al−2 b0 . . .

a0 . . . . . .. . . .. . .

a0 b0

(11.3)

which is called Sylvestre’s matrix for f (x) and g(x).

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Definition 11.3. The resultant of f (x) and g(x), which denoted by Res ( f , g,x), is

Res ( f , g,x) := det(Syl( f , g,x)).

Lemma 11.8. Let be given f (x) and g(x) :

f (x) =∏

i

(x−αi), g(x) =∏

j

(x−β j)

Then,Res ( f , g,x) =

∏i, j

(αi−β j)

Proof. Exercise.

Corollary 11.9. Polynomials f (x), g(x) ∈ k[x] have a common factor in k[x] if and only if

Res ( f , g,x) = 0.

Exercise 11.3. Let be givenf (x) = anxn + . . .a1x + a0

and f ′(x) its derivative. Then,

D( f ,x) =(−1)

n(n−12

anRes ( f , f ′,x).

Exercise 11.4. Let f (x) for every monic polynomial of degree n. Then,

D( f ,x) = (−1)n(n−1)

2

n∏i=1

f ′(αi)

where α1, . . . ,αn are roots of f (x) and f ′(x) its derivative.

The following lemma is quite useful.

Lemma 11.9 (Product formula for discriminants). Let f , g ∈ k[x]. Then,

∆( f g,x) = ∆( f ,x) ·∆(g,x) ·Res ( f , g,x)2

Proof. Exercise

Exercises:

11.29. Prove that f ∈ k[x1, . . . ,xn] is symmetric, if and only if when

f (x1, . . . ,xn) = f (x2,x1,x3, . . . ,xn) = f (x2,x3, . . . ,xn,x1).

11.30. Find the discriminant of polynomials :i) x3 + px + qii) x3 + px2 + qiii) x3 + x2

−4x + 1iv) x4 + px2 + qx + r

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11 Shaska T.

11.31. Let be givenf (x) = xn + px + q.

Prove that∆( f ,x) = (−1)

n(n−1)2 nnqn−1 + (−1)

(n−1)(n−2)2 (n−1)(n−1)pn

11.32. Let p ∈Q[x] be a cubic polynomial. We know that that ∆p = 0 if and only if p(x) has a multiple root. Provethat,

i) ∆p > 0 if and only if p(x) has three distinct real roots.ii) ∆p < 0 if and only if p(x) has exactly one real root and two complex (non-real) roots.

11.33. Expressn∑

i=1

t3i +

∑1≤i< j≤n

(t2i t j + t2

j ti)+

∑1≤i< j<k≤n

tit jtk

as a polynomial of the elementary symmetric polynomials si(t) of ti’s.

11.34. Let be given polynomials f , g ∈Q[t],

f (t) = u(1 + t2)− t2 and g(t) = v(1 + t2)− t3

Find Res ( f , g, t).

11.35. Let be given

f (x) = x5−3x4

−2x3 + 3x2 + 7x + 6

g(x) = x4 + x2 + 1(11.4)

Find Res ( f , g,x).

11.36. Does the polynomialf (x) = 6x4

−23x3−19x + 4

have multiple roots in C?

11.37. Find b such thatf (x) = x4

− bx + 1

has a double root in C.

11.38. Find p such thatf (x) = x3

−px + 1

has a double root in C.

11.39. Let be givenf (x) = xn + px + q.

Prove thatD( f ,x) = (−1)

n(n−1)2 nnqn−1 + (−1)

(n−1)(n−2)2 (n−1)(n−1)pn

11.40. Let be given

f (x) = x5−3x4

−2x3 + 3x2 + 7x + 6

g(x) = x4 + x2 + 1(11.5)

Find Res ( f , g,x).

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11.41. Find an algebraic relation between a and b such that

xn + ax + b

has a double root in C.

11.42. i) Solve the equationx5−2x4 + 5x3

−10x2 + 3x−6 = 0

in complex numbers. Find all solutions.

ii) Find the discriminant of the polynomial

f (x) = ax2 + bx + c

using the definition.iii) Is the following polynomial reducible in Q

f (x) = x7 + 5x6−15x5

−3x4 + 6x3 + 9x2 + 12x−21

Prove your answer.

11.43. Find if the following polynomials

f (x) = x3 + x2−5x−5

g(x) = x3 + 2x2−5x−10

(11.6)

have common factors in Q[x].

11.44. i) Prove that the polynomial

f (x) =

n∏i=1

(x− i)−1

is irreducible in Z for all n ≥ 1.ii) Prove that the polynomial

f (x) =

n∏i=1

(x− i) + 1

is irreducible in Z for all numbers odd n ≥ 1.

11.45. If a ∈Q and x− a divides a monic polynomial f (x) ∈Z[x], prove that a is an integer.

11.5 Formal power series

In this section we give a brief introduction to power series.Let xi denote a symbol and x a singleton set. Let G = Map(x,N) be the set of all maps x →N. For

some f ∈ G, we write xn for f (x).Then G = x0,x1, . . .xn, . . . . Denote x0 = 1 and define a multiplication on G as

xn·xm = xn+m

Let A[[x]] denote the set of maps from G to A,

A[[x]] := Map(G,A)

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Then for f ∈ A[[x]] we have

f (1) = a + 0 ∈ Af (x) = a1 ∈ A

. . .

f (xn = an ∈ A

We writef = a0 + a1x + a2x2 + · · ·anxn + · · ·

For f , g ∈ A[[x]] as

f = a0 + a1x + a2x2 + · · ·anxn + · · ·

g = b0 + b1x + b2x2 + · · ·bnxn + · · ·

define the sum and product in A[[x]] as (f + g

)(x) = f (xi) + g(xi)

and( f g)(x) =

∑cixi,where ci =

∑k+l=i

akbk

Exercise 11.5. Prove that A[[x]] is a commutative ring.

A[[x]] is called the formal power series over A.Before we prove the following theorem we need some material covered in Appendix A.

Theorem 11.14. k[[x1, . . . ,xn]] is local and complete.

Proof.

Theorem 11.15. Let O be a local and complete ring with m the maximal ideal of O. Let f ∈ O[[x]] and not allcoefficients are in m, such that

f =

∞∑i=0

aixi

where an is the first coefficient not in m. Then, for every g ∈ O[[x]], there exist q,r ∈ O[[x]] such that

g(x) = q(x) · f (x) + r(x),

where degr ≤ n−1.

Manin.

Theorem 11.16 (Weierstrass preparation theorem). Let O be a local and complete ring with m the maximal idealof O. Let f ∈ O[[x]] and not all coefficients are in m, such that

f =

∞∑i=0

aixi, a0, . . . ,an <m.

Then, there exists a unique unit u(x) and b0, . . . ,bn−1 ∈m such that

f (x) =(xn + b0 + b1x + b2x2 + · · ·bn−1xn−1

)·u(x)

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Proof. Complete this

The integer n is called the Weierstrass degree of f (x).

Exercises:

11.46. Suppose that O is a complete local ring. Show that O[[x]] is also a complete local ring.

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196

Chapter 12

Local and Notherian rings

12.1 Introduction to local rings

Let A be a commutative ring. A set S ⊂ A is called multiplicative subset or multiplicatively closed if

∀x, y ∈ S =⇒ xy ∈ S

The set S−1A is called the quotient ring of A by S or the ring of fractions of A by S. We review quicklyits construction.

TakeF := (r,d) : r ∈ A,d ∈ S

Define a relation in F as follows:(r,d) ∼ (s,e)⇔ re = sd

Prove that this is an equivalence relation and denote the equivalence class of (r,d) with rd . Let S−1A be

the set of all equivalence classes of this relation. Note that rd = rc

dc in S−1A for all c ∈ S, (dc ∈ S because S is aclosed multiplicative set).

Define addition and multiplication in S−1A as follows:

ab

+cd

=ad + bc

bd

ab·

cd

=acbd

Prove that (S−1A,+, ·) is a commutative ring with identity. We call S−1A the ring of fractions of A anddenote it by S−1A.

Now let p be a prime ideal of A. Obviously S = A \p is a multiplicative set of A. The ring of quotientsS−1A is called the localization of A at p and denoted by Ap.

Definition 12.1. A local ring is a commutative ring with identity which has a unique maximal ideal.

Proposition 12.1. For every prime ideal p the ring Ap is a local ring.

Proof.

Exercise 12.1. Let S be a multiplicative set and 0 < S. Take P a maximal ideal such that P∩S = ∅. Prove that P isprime.

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Exercise 12.2. Let A be a local ring and m its maximal ideal. Prove that m is the set of all non-units of A and A\mis a multiplicative set.

Exercise 12.3. Let R be an integral domain and S a multiplicative set of R which does not contain 0. Show that Rcan be embedded in S−1R.

Let S be a multiplicative set of A and denote by J(A) the set of ideals of A. Define the following map asfollows:

ψS : J(A)→ J(S−1A)

a→ S−1a

where

S−1a =a

s| a ∈ a, s ∈ S

Exercise 12.4. Prove that

i) S−1a is an ideal of S−1A.ii) S−1 (a+b) = S−1a+ S−1b

iii) S−1 (ab) =(S−1a

)(S−1b

)iv) S−1 (a∩b) = S−1a∩S−1b

Exercises:

12.1. Let A = S−1Z where S is the set of all integers not divisible by a fixed prime p and I a unique maximal ideal ofA. Show that

i) ∩∞i=1In = 0ii) A is not complete

Exercises:

12.2. a) Suppose that A is a commutative ring with identity, and M is an A-module. Show that the following areequivalent:

a) M=0b) MP = 0 over AP, for each prime ideal P of A.c) Mm = 0 over AQ, for each maximal ideal m of A.[Note: Clearly state the facts about localization which are needed here.What is worth it to show is c)=⇒a)]b) Suppose R is a subring of a field K and S is a multiplicative set of R not containing 0. Show that if x ∈ K is

integral over S−1R then it may be written as x = bs where s ∈ S and b is integral over R.

12.3. If A is a commutative ring with identity which is Noetherian, then prove that A[[T]], the ring of formal powerseries, is also Noetherian.

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12.2 Introduction to Notherian rings

Let A be a commutative ring with identity. An ascending chain of ideals in A is called a chain

I1 ⊂ I2 ⊂ · · · ⊂ Ik ⊂ · · ·

A descending chain of ideals in A is called a chain

I1 ⊂ I2 ⊂ · · · ⊂ Ik ⊂ · · ·

A ring A is called Notherian if every ascending chain stabilizes after finitely many steps.

Proposition 12.2. A is Notherian if and only if every ideal in A is finitely generated.

Proof. Let us assume that every ideal in A is finitely generated. Let

M1 ⊂M2 ⊂ · · · ⊂Mn ⊂ · · ·

be an ascending chain of ideals. TakeM = ∪∞i=1Mi

Then M is finitely generated, sayM = 〈x1, . . . ,xr〉

and each generator is in some Mi. Hence, there exists some index s such that

x1, . . .xr ∈Ms

Then,〈x1, . . . ,xr〉 ⊂Ms ⊂M = 〈x1, . . . ,xr〉

Hence, for all j > s we have M j = Ms = M.Let I ⊂ A be any ideal. Pick some a0 ∈ I. If I = 〈a0〉 then we are done. Otherwise, let a1 ∈ I such that

a1 < 〈a0〉. We proceed inductively in the way to get

〈a0〉 ⊂ 〈a0,a1〉 ⊂ · · · ⊂ I

where each inclusion is proper. Then this ascending chain stabilizes after many steps. Hence, I is finitelygenerated.

Proposition 12.3. Let φ : A→ B be a ring homomorphism. If A is Notherian then φ(A) is Notherian.

Proof. Take an ideal a ⊂ φ(A). Obviously φ−1(a) is an ideal in A. Since A is Notherian then φ−1(a) is finitelygenerated. Let

φ−1(a) = 〈φ−1(a1),φ−1(a2), . . . ,φ−1(an)〉

Take b ∈ a. Then,φ−1(b) =

∑riφ−1(ai)

where i = 1, . . . ,n and ri ∈ A. Thenb = φ−1

(φ−1(b)

)=

∑riai

So a is finitely generated.

Proposition 12.4. Let A be Notherian. Then every subring and every quotient ring of A are Notherian.

Proof. Exercise

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Proposition 12.5. Let A be a commutative Notherian ring and S a multiplicative set of A. Then, S−1A is Notherian.

Proof. Indeed, the ideals of S−1A look like S−1awhere a is an ideal in A. Complete the proof.

The proof of the following theorem will be provided in the chapter of Notherian modules.

Theorem 12.1 (Cohn). Prove that A is Notherian if and only if every prime P is finitely generated.

Exercises:

12.4. (a) Let A be a commutative ring with identity, which satisfies the ascending chain condition on prime ideals.Must A be Noetherian? Prove, or else give a counterexample.

(b) Let A be a commutative ring with identity such that the local ring Ap is Notherian for every prime ideal P ofA. Is A necessary Notherian?

12.3 Hilbert’s basis theorem

Theorem 12.2 (Hilbert’s basis theorem). Let A be a Noetherian ring, then A[x] is Notherian.

Proof. Let I be an ideal of A[x]. Let ai be the set of leading coefficients of degree i polynomials in I and 0.

Claim: ai is an ideal in A.This proves the claim.

Then we havea0 ⊂ a1 ⊂ · · ·

This ascending chain stops after some r steps, since A is Notherian. So

a0 ⊂ a1 ⊂ · · ·ar = ar+1 = · · ·

Letai,1, . . . ,ai,ni

be the generators for each ai.Let fi, j be the corresponding coefficient in I with leading coefficient ai, j.

Claim: The set of polynomials fi, j generate I.We prove this by induction on the degree d of the polynomials in I.Let f ∈ I such that deg f = d. We want to show that f ∈ 〈 fi, j〉.If d > r then the leading coefficients of

xd−r fr,1, · · · ,xd−r fr,nr

generate ad. Hence, there exist elements such that

f −(c1xd−r fr,1 + · · ·+

)If d ≤ r thenThis completes the proof.

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Proof. Let I be an ideal in A[x] and a the set of all leading coefficients of polynomials in I.

Claim: a is a finitely generated ideal of A.

First we show that a is an ideal of A. Since I contains the zero polynomial then 0 ∈ a. Let

f = axd + . . .

g = bxe + · · ·

be polynomials in I with degrees d and e and leading coefficients a,b ∈ A. Then, for every r ∈ A we havera−b is zero or it is the leading coefficient of

rxe f −xdg.

Since this is in I then ra− b ∈ a, which means that a is an ideal of A. Since A is Notherian, then a is finitelygenerated. Say, a is generated by a1,a2, . . . ,an ∈ A.

For every i = 1, . . . ,n let fi be an element of I which has leading coefficient ai. Let ei be the degree of fi anddenote with N the maximum of e1,e2, . . . ,en. For every d ∈ 0,1, . . . ,N−1 denote by ad the set of all leadingcoefficients of elements in I with degree d together with 0.

Similarly we show that ad is an ideal of A. Since A is Notherian then ad is finitely generated.For every non-zero ideal ad, let

bd,1,bd,2, . . . ,bd,nd ∈ A,

be a generating set of ad and let fd,i a polynomial in I of degree d with leading coefficient bd,i.Let’s show that the polynomials f1, . . . , fn together with all polynomials fd,i for all ideals Ld form a

generating set for I. In other words,

I = ( f1, . . . , fn∪ fd,i|0 ≤ d <N,1 ≤ i ≤ nd).

From construction, the ideal I′ on the right side is contained in I, because all generators were selectedfrom I. If I , I′, there exists a non-zero polynomial f ∈ I with minimum degree such that f is not in I′. Letd = deg f and let a be the leading coefficient of f .

First we assume that d ≥N. Since a ∈ L we can write a as a linear combination of generators of L

a = r1a1 + · · ·+ rnan.

Then,g = r1xd−e1 f1 + · · ·+ rnxd−en fn

is an element of I′ of the same degree d and the same leading coefficient a as f . Th en, f − g ∈ I is apolynomial in I with degree less than f .Since f is of minimal degree we have f − g = 0, or f = g ∈ I′, whichis a contradiction.

Next, assume that d <N. In this case a ∈ Ld, for d <N. Hence, we have

a = r1bd,1 + · · ·+ rndbnd , per ri ∈ A.

Then,g = r1 fd,1 + · · ·+ rnd fnd

is a polynomial in I′ of the same degree d and the same leading coefficient a as f . Again we have acontradiction. Hence, I = I′. This completes the proof.

Theorem 12.3. If A is Notherian then A[[x]] is Notherian.

Proof.

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12.3.1 Primary decomposition on Notherian rings

An ideal a is called irreducible ifa = b∩ c =⇒ a = b or a = c

Proposition 12.6. In a Notherian ring A every ideal is a finite intersection of irreducible ideals.

Proof. Let I be an ideal in A. Suppose that I is not a finite intersection of irreducible ideals.Let S be the set of all ideals of A which are not a finite intersection of irreducible ideals. Then, S , ∅

since I ∈ S. By Zorn’s lemma, S has a maximal element. Let’s call it a. So a is reducible since a ∈ S. Then,there exist b,c such that

a = b∩ c =⇒ a = b or a = c

and a is proper in b and c.But since a is the maximal ideal which is not a finite intersection of irreducibles then b and c are a finite

intersection of irreducibles ideals. Then, so is a, which is a contradiction.

Proposition 12.7. Prove that in a Notherian ring every ideal has a primary decomposition.

Proof.

12.3.2 Artinian rings

Exercises:

12.5. Prove that an Artinian commutative ring A with identity has only a finite number of prime ideals, and thateach one is a maximal ideal.

12.6. Let A be a commutative ring with identity. Prove that:A is Artinian if and only if A Noetherian of dim 0.

12.4 Hilbert’s basis theorem

In this section we prove one of the most important results of the polynomial rings.

Theorem 12.4 (Hilbert). Let A a Notherian ring, then A[x] is Notherian.

Proof. Let I a ideal in A[x] and L the set of all leading coefficients all elements in I.First, let’s to show that L is a ideal of A. I contains the zero polynomial, 0 ∈ L. Let’s f = axd + . . . and

g = bxe be polynomials in I with degree d and e and with leading coefficients a, b respectively. Then, forevery r ∈ A or ra− b is the leading coefficient of rxe f −xdg. Since this polynomial is in I then and ra− b is inL, then L is ideal of A.

Since A is Notherian, the ideal L of A is finitely generated, for example generated from a1, a2, . . . , an ∈A.For every i = 1, . . . ,n let fi an element of I that has leading coefficient ai. Let ei degree of fi and denote with

N the maximum of e1,e2, . . . ,en. For every d ∈ 0,1, . . . ,N−1 denote with Ld the set of all leading coefficientsof elements of I with degree d together with 0. Similarly as for L, shows that Ld is an ideal of A and as suchis finitely generated, because A is Notherian.

For every ideal nonzero Ld, letbd,1,bd,2, . . . ,bd,nd ∈ A,

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a set of generators for Ld and let fd,i a polynomial in I with degree d with leading coefficient bd,i. To provethat polynomials f1, . . . , fn together with all polynomials fd,i all nonzero ideals Ld are a set generators for I,so

I = ( f1, . . . , fn∪ fd,i|0 ≤ d <N,1 ≤ o f ≤ nd).

From construction the ideal I′ is contained in I, because all generators were picked in I. If I , I′, thenthere is a nonzero polynomial h ∈ I with minimal degree such that h is not in I′. Let d = degh and let aleading coefficient of h.

Assume first that d ≥N. Since a ∈ L we can write a as a linear combination of generators of L

a = r1a1 + · · ·+ rnan.

Then,g = r1xd−e1 f1 + · · ·+ rnxd−en fn

is an element of I′ of the same degree d and with the same leading coefficient a as h. Then, h− g ∈ I is apolynomial in I with smaller degree than h. Since h is with minimal degree must of we have h− g = 0, soh = g ∈ I′, which is a contradiction.

Next assume that d <N. In this case a ∈ Ld, for d <N. Hence, we can write

a = r1bd,1 + · · ·+ rndbnd , for ri ∈ A.

Then,g = r1 fd,1 + · · ·+ rnd fnd

is a polynomial in I′ with of same degree d and with the same leading coefficient a as h, again we havecontradiction. Hence remains that I = I′ is finitely generated and since I is any ideal, this completes thetheorem.

Exercises:

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David Hilbert

(23 January 1862 – 14 February 1943) was a German mathemati-cian and one of the most influential and universal mathematicians ofthe 19th and early 20th centuries. Hilbert discovered and developeda broad range of fundamental ideas in many areas, including invari-ant theory, the calculus of variations, commutative algebra, algebraicnumber theory, the foundations of geometry, spectral theory of oper-ators and its application to integral equations, mathematical physics,and foundations of mathematics (particularly proof theory).

Hilbert adopted and warmly defended Georg Cantor’s set theoryand transfinite numbers. A famous example of his leadership inmathematics is his 1900 presentation of a collection of problems thatset the course for much of the mathematical research of the 20thcentury.

Hilbert’s first work on invariant functions led him to the demonstration in 1888 of his famous finitenesstheorem. Twenty years earlier, Paul Gordan had demonstrated the theorem of the finiteness of generatorsfor binary forms using a complex computational approach. Attempts to generalize his method to functionswith more than two variables failed because of the enormous difficulty of the calculations involved. Inorder to solve what had become known in some circles as Gordan’s Problem, Hilbert realized that it wasnecessary to take a completely different path. As a result, he demonstrated Hilbert’s basis theorem, showingthe existence of a finite set of generators, for the invariants of quantics in any number of variables, but inan abstract form. That is, while demonstrating the existence of such a set, it was not a constructive proof? it did not display "an object" ? but rather, it was an existence proof[28] and relied on use of the law ofexcluded middle in an infinite extension.

Hilbert sent his results to the Mathematische Annalen. Gordan, the house expert on the theory ofinvariants for the Mathematische Annalen, could not appreciate the revolutionary nature of Hilbert’stheorem and rejected the article, criticizing the exposition because it was insufficiently comprehensive. Hiscomment was:

Das ist nicht Mathematik. Das ist Theologie. (This is not Mathematics. This is Theology.)[29] Klein,on the other hand, recognized the importance of the work, and guaranteed that it would be publishedwithout any alterations. Encouraged by Klein, Hilbert extended his method in a second article, providingestimations on the maximum degree of the minimum set of generators, and he sent it once more to theAnnalen. After having read the manuscript, Klein wrote to him, saying:

Without doubt this is the most important work on general algebra that the Annalen has ever pub-lished.[30] Later, after the usefulness of Hilbert’s method was universally recognized, Gordan himselfwould say:

I have convinced myself that even theology has its merits.[31] For all his successes, the nature of his proofstirred up more trouble than Hilbert could have imagined at the time. Although Kronecker had conceded,Hilbert would later respond to others’ similar criticisms that "many different constructions are subsumedunder one fundamental idea" ? in other words (to quote Reid): "Through a proof of existence, Hilberthad been able to obtain a construction"; "the proof" (i.e. the symbols on the page) was "the object".[31]Not all were convinced. While Kronecker would die soon afterwards, his constructivist philosophy wouldcontinue with the young Brouwer and his developing intuitionist "school", much to Hilbert’s torment in hislater years.[32] Indeed, Hilbert would lose his "gifted pupil" Weyl to intuitionism ? "Hilbert was disturbedby his former student’s fascination with the ideas of Brouwer, which aroused in Hilbert the memory ofKronecker".[33] Brouwer the intuitionist in particular opposed the use of the Law of Excluded Middle overinfinite sets (as Hilbert had used it). Hilbert would respond:

Taking the Principle of the Excluded Middle from the mathematician ... is the same as ... prohibitingthe boxer the use of his fists.[34]

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Part II

Module theory

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Chapter 13

Introduction to modules

In this chapter we give a short introduction of modules and their basic properties.

13.1 Introduction to modules

First we give the definition of a module.

Definition 13.1. Let R a ring (not necessarily ring with identity or Abelian ring). A left R-module or a left moduleover the ring R is a set M together with

1. a binary operation + over M, with which the set M forms a Abelian group and

2. an operation of rings R over M (hence a map R×M→M ) which denoted by rm, for every m ∈M whichsatisfies conditions:

a) (r + s)m = rm + sm, for every r,s ∈ R and m ∈M,

b) (rs)m = r(sm), for every r,s ∈ R and m ∈M,

c) r(m + n) = rm + rn, for every r ∈ R and m,n ∈M.

If the ring R has identity 1R then we add the axion

d) 1R ·m = m, for every m ∈M.

The term "left" means that the multiplication is done on the left. Similarly we can define an R -rightmodule. If a ring R is an Abelian ring and M is a left module, we can think of it as a right module bymr = rm, for every m ∈M and r ∈R. When using the term "module" we always mean "left module". Moduleswhich satisfy the condition 2(d) are called module with identity and in this notes all modules will be withidentity.

When the ring R is a field F, axioms of a R module are exactly the same with those of a vector spacesover field F. Hence, we have

Lemma 13.1. Modules over a field F and vector spaces over F are the same.

Definition 13.2. Let R be a ring and M an R-module. A R-submodule of M is a subgroup N of M which is closedunder the multiplication by scalars o f R, which implies rn ∈N, for every r ∈ R and n ∈N.

Submodules of modules M are subsets of M which are also modules with operations on M. If R = F is afield, submodules are the same with subspaces. Every R-module M has at least two submodules M and 0.

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Example 13.1. Let R a ring. Then, M = R is an R -left module, where the multiplication is that of the ring R andaddition is also the same as that of R.

Every field F can be considered as a vector space (1 dimensional) over itself. When R is considered as a left moduleover itself, submodules of R are exactly left ideals of R.

Example 13.2. Let R = F be a field. As stated above every vector space over F is a F–module and conversely. Letn ∈Z+ and let

Fn = (a1,a2, · · · ,an)|ai ∈ F, for every i

(which is called the affine space over F ). Fn can be considered as a vector space defining addition and of multiplicationas follows:

(a1,a2, · · · ,an) + (b1,b2, · · · ,bn) = (a1 + b1,a2 + b2, · · · ,an + bn)α(a1,a2, · · · ,an) = (αa1,αa2, · · · ,αan), α ∈ F.

As in the case of Euclidean n -spaces (F =R ), the affine n -space is a vector space with dimension n over F.

Example 13.3. Let R be a ring with 1 and let n ∈Z+. Define,

Rn = (a1,a2, · · · ,an)|ai ∈ R, for every i

Rn can be made an R-module defining addition and of multiplication as above. The module Rn is called free modulewith rank n over R .

Example 13.4. Abelian groups can be modules for many different rings.For example, if M is a R-module and S is a subring of R, where 1S = 1R, then M is automatically also an S -module.

The field R is an R–module, a Q–module is a Z–module.

Example 13.5. If M is a R-module and for an ideal I of R, am = 0 for every a ∈ I and for every m ∈M, we say that Mis annihilated from I. In this case we can transform M to a (R/I)–module defining operations over the factor ringR/I over M in such that

(r + I)m = rm, for every m ∈M and for every coset r + I ∈ R/I .

Since am = 0 for every a ∈ I and for every m ∈M the above is well defined and it is easy to prove that M is aR/I–module. In particular when I is a maximal ideal in an Abelian ring R and IM = 0, then M is a vector space overfield R/I.

Next we describe a test for submodules which is similar to the subgroup test.

Proposition 13.1 (The submodule test). Let R a ring and M a R-module. A subset N of M is a submodule of M ifand only if

1. N , ∅ and

2. x + ry ∈N, for every r ∈ R and for every x, y ∈N.

Proof. If N is a submodule, then 0 ∈ N so N , ∅. Also N is closed under addition and under the action ofelements of R. Conversely, assume that (1) and (2) are true. Let’s have r = −1 and use the subgroup test tocheck that N is a subgroup of M. In particular, 0 ∈ N. Now let x = 0 and use (2) to check that N is closedunder the multiplication of R.

Definition 13.3. Let R a Abelian ring with identity. A R–algebra is a ring A with identity together with a ringhomomorphism f : R→ A, that maps 1R to 1A, such that the subring f (R) of A is contained in the center of A.

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Definition 13.4. If A and B are two R–algebras, an R–algebra homomorphism (resp. an isomorphism) is a ringhomomorphism (resp. isomorphism) ϕ : A→ B which maps 1A to 1B such that ϕ(ra) = rϕ(a) for every r ∈ R anda ∈ A.

Example 13.6. Let R a Abelian ring with identity.

1. Every ring with identity is a Z–algebra.

2. If A is a R–algebra, then A as a R-module depends only from the subring f (R).

Assume that A is a R–algebra. Then, A is a ring with identity so is a R –left module (with identity)which satisfies the property:

r · (ab) = (r · a)b = a(r · b)

for every r ∈ R and a,b ∈A. Conversely, over the ring A these conditions determine an R–algebra and oftenare used as the definition of an R–algebra.

Exercises:

In the following exercises R is ring with identity and M a left R-module.

13.1. Prove that 0m = 0 and (−1)m = −m for every m ∈M.

13.2. Suppose that rm = 0 for some r ∈ R and m ∈M where m , 0. Prove that r does not have a left inverse.

13.3. For every left ideal of I in R define

IM =

∑f inite

aimi|ai ∈ I,mi ∈M

to be the collection of all finite fields of elements of the form am, where a ∈ I and m ∈M. Prove that IM is a submoduleof M.

13.4. Prove that any intersection of not empty submodules of a R-module is a submodule.

13.5. Let z a element of the center of R, so zr = rz for every r ∈ R. Prove that zM is a submodule of M, where

zM = zm|m ∈M.

Prove that if R is ring of 2× 2 matrices over a field and e is a matrix with 1 in position (1, 1) and 0 in all otherpositions, then eR is n ot a R –left submodule.

13.6. If M is a finite Abelian group then M is aZ–module. Can we extend the action ofZ on M such that M becomesa Q -module?

13.2 Module homomorphisms and quotient modules

In this section we will give the basic theory of quotient modules and homomorphisms of modules.Let R a ring and let M and N be R-modules. A function ϕ : M→N is an R-module homomorphism if

it preserves the R-module structures M and N, so,a) ϕ(x + y) = ϕ(x) +ϕ(y) for every x, y ∈M andb) ϕ(rx) = rϕ(x), for every r ∈ R,x ∈M.

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A homomorphism R-modules is an isomorphism if it is injective and surjective. Modules M and Nare called isomorphic, denoted with M N, if there is an R-module isomorphism ϕ : M→N.

If ϕ : M→N is an R-module homomorphism, the kernel of ϕ is:

kerϕ = m ∈M |ϕ(m) = 0

and the image of ϕ is:ϕ(M) = y ∈N | y = ϕ(x) for some x ∈M

Let M and N be R-modules and define Hom R(M,N) to be the set of all homomorphisms of R-modules fromM in N.

Let M,N and L be R-modules.

Lemma 13.2. A map ϕ : M→N is an R-module homomorphism if and only if

ϕ(rx + y) = rϕ(x) +ϕ(y) for every x, y ∈M and for all r ∈ R.

Proof. If ϕ is a homomorphism R-modules then

ϕ(rx + y) = rϕ(x) +ϕ(y).

Conversely, if ϕ(rx+ y) = rϕ(x)+ϕ(y), we get r = 1 for of seen that ϕ is additive and we get y = 0 for of seenthat ϕ is commutative with operation in of R over M.

Let ϕ,ψ elements of Hom R(M,N). Define ϕ+ψ such that

(ϕ+ψ)(m) = ϕ(m) +ψ(m) for every m ∈M

Then, ϕ+ψ ∈HomR(M,N) and with this operation Hom R(M,N) is Abelian group. If R is commutative ringthen for r ∈ R define rϕ such that

(rϕ)(m) = r(ϕ(m)) for all m ∈M.

Then, rϕ ∈HomR(M,N) and with this action of R, the Abelian group Hom R(M,N) is a R-module.

Lemma 13.3. Hom R(M,N) is a R-module.

Proof. It is easy to prove that all axioms of Abelian groups and R-modules are satisfied with these definitions.Notice that commutative property of rings R is used to show that rϕ satisfies the second axiom of a R-modulehomomorphism, namely

(r1ϕ)(r2m) = r1ϕ(r2m) = r1r2(ϕ(m)) = r2r1ϕ(m) = r2(r1ϕ)(m)

Proposition 13.2. i) If ϕ ∈Hom R(M,N) and ψ ∈Hom R(M,N) then ψϕ ∈Hom R(M,N).ii) With the addition as above and with multiplication as composition of functions, Hom R(M,M) is ring with

identity. When R is commutative then Hom R(M,M) is an R–algebra.

Proof. i) Let given ϕ and ψ and r ∈ R, x, y ∈ L. Then, we have:

(ψϕ)(rx + y) = ψ(ϕ(rx + y))= ψ(rϕ(x) +ϕ(y))= rψ(ϕ(x)) +ψ(ϕ(y))= r(ψϕ)(x) + (ψϕ)(y)

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Hence, ψϕ is a homomorphism R-modules.ii) Since the set of definition and the set of values of elements of Hom R(M,M) are the same, composition

of functions is defined. From (3) it is a binary operation over Hom R(M,M). we know that compositionof functions satisfy the commutative property. The other properties of rings are easy to prove and left asexercises. The identity function, I, (I(x) = x, for every x ∈M ) is the identity of Hom R(M,M) under operationof multiplication. If R is Abelian, then (2) shows that the ring Hom R(M,M) is a R –left module and definedϕr = rϕ for every ϕ ∈Hom R(M,M) and r ∈ R is an R–algebra.

Definition 13.5. The ring Hom R(M,M) is called the endomorphism ring of M and we will denote by End R(M)or simply End (M). The elements of End (M) are called endomorphisms.

Let R be a ring, M an R-module and let N be a submodule of M. The factor group M/N can be made inan R-module defining operations

r(x + N) = (rx) + N, for every x ∈ R,x + N ∈M/N.

Proposition 13.3. M/N is an R-module. The natural projection

π : M→M/N,

such that π(x) = x + N is a homomorphism of R-modules with kernel N.

Proof. Since the group M is Abelian with addition +, the factor group M/N is an Abelian group. To see theoperation is well defined, assume that x + N = y + N, so x− y ∈ N. Since N is a left R-module, r(x− y) ∈ N.Thus, rx− ry ∈N and therefore rx+N = ry+N. Since action in M/N agrees with those in M, axioms to provethat M/N is a R-module are easy to prove. For example, axiom 2(b) is proved as follows: for every r1,r2 ∈ Rand x + M/N, from the definition of operation it of elements of rings over the elements of M/N

(r1r2)(x + N) = (r1r2x) + N= r1(r2x + N)= r1(r2(x + N)).

The other axioms are easy to prove and their proof is left as an exercise.Finally natural projection π of given above is natural projection of the Abelian group M over the

Abelian group M/N, therefore is a group homomorphism with kernel N. It is left only prove that π is ahomomorphism modules, so π(rm) = rπ(m). However,

π(rm) = rm + N = r(m + N) = rπ(m)

This completes the proof.

We call the module M/N a quotient module.

Definition 13.6. Let A,B submodules of an R-module M. The sum of A and B is the set:

A + B = a + b |a ∈ A,b ∈ B

Exercise 13.1. Prove that the sum of two submodules A and B is a submodule and is the smallest submodule thatcontains A and B.

The following theorem combines four isomorphism theorems for modules. Their proof is similar to theproofs for such theorems in the case of rings.

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Theorem 13.1 (Isomorphism Theorems for Modules). i) Let M and N,R–module and let ϕ : M→ N a homo-morphism R-modules. Then, kerϕ is a submodule of modules M and m/kerϕ ϕ(M).

ii) Let A,B submodules a R-module M. Then,

(A + B)/B A/(A∩B).

iii) Let M a R-module and Let’s are A and B submodules M where A ⊂ B. Then, (M/A)/(B/A) M/B.

iv) Let N a submodule of R-module M. There exists a bijection between submodules of M which contain N andsubmodules of M/N. The correspondence is given from A↔ A/N, for every A ⊇N.

Proof. Exercise

13.2.1 Local modules

Let M be an R module and S a multiplicative set on R. We follow the procedure of the construction of thering of fractions S−1R to construct a module of fractions S−1M. Define a relation on M×S as follows:

(m,s) ∼ (m′,s′) ⇐⇒ ∃t ∈ S, such that t(sm′− s′m) = 0

Exercise 13.2. Prove that the above is an equivalence relation.

Let m/s denote the equivalence class of (m,s) and S−1M the set of such fractions.

Exercise 13.3. Prove that S−1M is a S−1R-module.

Let p be a prime ideal in R. Then, S = R \p is a multiplicative set. We denote Rp:=S−1R and Mp:=S−1M.The module Mp is called the localization of M at p.

Lemma 13.4. Let N and Q be R-modules and f : N→ P an homomorphism. Then the following are equivalent:a) f is injectiveb) fp : Np→Qp is injective for every prime p.c) fm : Nm→Qm is injective for every maximal ideal m.

Proof. Exercise

Exercises:

13.7. Use the submodule criteria to prove that the kernel and the image of the homomorphism R-modules aresubmodule.

13.8. Prove that the relation "is R-module isomorphic " is a equivalence relation over for every set R-modules.

13.9. Give an example of a function from an R-module to another which is a group homomorphism but not ahomomorphism of R-modules.

13.10. Prove that Hom Z(Z/nZ,Z/mZ) Z/(m,n)Z.

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13.3 Direct sums and free modules

Let R a ring with identity. As above, by module we mean a left module.Let M be a module and N1, · · ·Nn submodules M. The sum of N1, · · · ,Nn is the set of all finite sums of

elements from the sets Ni. Say,a1 + a2 + · · ·+ an |ai ∈Ni for all i.

We denote this sum with N1 + · · ·Nn.For any subset A ⊂M let:

RA = r1a1 + r2a2 + · · ·+ rmam |r1, . . . ,rm ∈ R,a1, . . . ,am ∈ A,m ∈Z+

If A is the finite set a1, · · · ,an instead of RA we can write

Ra1 + Ra2 + · · ·+ Ran.

RA is called the submodule of M generated by A. If N is a submodule of M (possibly N = M ) and N = RA,for a subset A of M, we call A the set of generators for N and we say that N is generated by A.

A submodule N of a module M is finitely generated if there is a finite subset A of M such that N = RA,namely, if N is generated from a finite subset.

A submodule N of M (possibly N = M ) is cyclic if there is an element a ∈M such that N = Ra, namely, ifN is generated from an element:

N = Ra = ra |r ∈ R.

A submodule N of module M can have many generating sets. If the submodule N is finitely generated,then there is the smallest non-negative integer d such that N is generated from d elements. Every set ofgenerators which contains d elements is called a minimal set generators for the submodule N.

Example 13.7. Let’s have R =Z and let M be a R-module. Hence, M is some Abelian group.If a ∈M thenZa is simply the cyclic subgroup 〈a〉 of M generated from a. Moreover, M is generated ofZ–module

by a set A if and only if when M is generated as group by A.

Example 13.8. Let R a ring with 1R and let M = R be a R–left module. Notice that R is finitely generated, in fact itis a cyclic R-module because R = R1R.

Recall that the submodules of R are exactly left ideals of R, so when we say that I is a cyclic R–submodule of theleft R-module R is the same as saying that I is a principal ideal of R. Also, saying that I is a finitely generated R -submodule is the same as saying that I is a finitely generated ideal.

A principal integral domain is an integral domain (commutative) R with unity in which for every R-submoduleof R is cyclic.

Let M1, · · · ,Mn be a collection R-modules. As usual we denote

M1× · · ·×Mk := (m1,m2, · · · ,mk) |mi ∈Mi

Define addition and scalar multiplication in M1× · · ·×Mk as usual

(x1, . . . ,xk) + (y1, . . . , yk) = (x1 + y1, . . . ,xk + yk)

r(x1, . . . ,xk) = (rx1, . . . ,rxk)

Exercise 13.4. M1× · · ·×Mk with addition and scalar multiplication as above is an R-module.

We call this module a direct product of M1, · · · ,Mk and denote it by

M1× · · ·×Mk.

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Proposition 13.4. Let N1,N2, · · · ,Nn submodules a R-module M. Then, the following are equivalent:

1. The function

π : N1×N2× · · ·×Nk→N1 + N2 + · · ·+ Nk

(a1,a2, · · · ,an)→ a1 + a2 + · · ·+ ak

is a isomorphism(R-modules), in other words

N1 + N2 + · · ·+ Nk N1×N2× · · ·×Nk.

2. For every j ∈ 1,2, · · · ,k we have that

N j∩ (N1 + N2 + · · ·+ N j−1 + N j+1 + · · ·+ Nk) = 0.

3. Every x ∈N1 + N2 + · · ·+ Nk can be written uniquely in the form a1 + a2 + · · ·+ ak where ai ∈Ni.

Proof. Left to the reader.

If an R-module M = M1 + · · ·Mk is the sum of submodules M1, . . . ,Mk and one of the equivalent conditionsof Proposition 13.4 is satisfied then we that M is the direct sum of M1, . . . ,Mk and is written as

M = M1⊕ · · ·⊕Mk

A R-module F is called a free module on the subset A of F if for every nonzero element x ∈ F, there existunique nonzero elements r1,r2, · · · ,rn ∈ R and a1,a2, · · · ,an ∈ A such that

x = r1a1 + r2a2 + · · ·rnan

for some n ∈Z+. In this case we we say that A is a basis of free generators for F. If R is commutative ringcardinality of A is called the rank of F.

Theorem 13.2. i) For any set A there exists a free R-module F(A) on the set A and an embedding

ι : A → F(A)

ii) Moreover, F(A) satisfies the universal property, namely: If M is a R-module and

ϕ : A→M,

is any map of sets, then there is a unique R-module homomorphism φ : F(A)→M such that the following diagram iscommutative

A

ϕ!!

ι // F(A)

φ

M

Figure 13.1: Universal property of free modules

iii) If A is a finite set a1,a2, · · · ,an, then

F(A) = Ra1⊕Ra2⊕ · · ·⊕Ran Rn.

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Proof. If A = ∅ then we let F(A) = 0. Assume A , ∅. Let

F(A) = f : A→ R | f (a) = 0 for all but finitely many a ∈ A

Define addition and scalar multiplication on F(A) as follows

( f + g)(a) = f (a) + g(a)

(r f )(a) = r f (a)

Prove that this is an R-module. Define the function

ι : A→ F(A)

a→ fa

where fa : A→ R as follows f (a) = 1 and f (x) = 0 for all x , a. Thus, ι is an embedding.To prove part ii), think of F(A) as all finite linear combinations of elements of A, where f = r1a1 + · · ·+rnan

such that f (ai) = ri and is 0 everywhere else. By the definition of f ∈ F(A) each such expression is unique.Define

φ : F(A)→M

f =

n∑i=1

riai→

n∑i=1

riϕ(ai)

By uniqueness of the expression for f , φ is a well defined R-module homomorphism. By definition of themap φ the diagram commutes.

iii) By Proposition 13.4, part 3), we have that

F(A) = Ra1⊕ · · ·⊕Ran.

Notice that RRai, for all i, under that map r 7→ rai. By Proposition 13.4, part 1), we have that F(A)Rn.

Corollary 13.1. 1. If F1 and F2 are free modules over of same set A, there is a unique isomorphism between F1and F2 which is the identity function in A.

2. If F is a R free module with basis A, then FF(A). Moreover, F has the universal property with respect to A,as has F(A).

Proof. Left to the reader

When R =Z, a free module over the set A is a free Abelian group over A. If |A| = n, then F(A) is calleda free Abelian group with rank n and is isomorphic to Z⊕ · · ·⊕Z, n -times.

Lemma 13.5. Every module M is quotient of a free module.

Proof. Exercise

Theorem 13.3. M is a finitely generated R-module if and only if M is isomorphic to a quotient of Rn, for some n > 0.

Proof. Exercise. See [?AM], Prop. 2.3.

Lemma 13.6. Let R be a commutative ring and M a finitely generated R-module. If M = IM for some ideal I of R,then there exists an x ∈ I such that

(1−x)M = 0

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Proof. Exercise, see [?AM] Cor. 2.5.

Lemma 13.7 (Nakayama’s Lemma). Let R be a commutative ring, M a finitely generated R-module, and I an idealof R contained in the Jacobson radical J(R) of R. If IM = M then M = 0.

Proof. See [?AM], Prop. 2.6.

13.3.1 Irreducible modules

An R–module M is called irreducible if M , 0 and if it has no proper submodules.

Exercise 13.5. Show that M is irreducible if and only if M , 0 and M is a cyclic module with any nonzero elementsas a generator. Determine all irreducible Z-modules.

Exercise 13.6. Let R be a commutative ring. Prove that an R-module M is irreducible if and only if M is isomorphic(as an R-module) to R/I for some maximal ideal I.

Exercise 13.7. Show that if M and N are irreducible R-modules then every module homomorphism f : M→N is anisomorphism.

Lemma 13.8 (Schur). If M is an irreducible R-module then EndR(M) is a division ring.

Proof.

Exercises:

13.11. Suppose that A is a commutative ring with identity. Let F(m) and F(n) be the free modules on m and ngenerators, respectively. Prove that if F(m) F(n), then m = n.

(b) Give an example of a free A-module F such that two different basis of it have different cardinalities.[Hint: A has to be non-commutative, otherwise b) is true]

13.12. (a) Suppose that R is a principal ideal domain. Prove that any submodule of a free R-module is free.(b) Give an example of an ring R with 1 with an R-module F which is free but has a submodule which is not free.

13.13. Prove that if A and B are set with same cardinality, then free modules F(A) and F(B) are isomorphic.

13.14. Suppose that R is commutative. Prove that RnRm if and only if when n = m, so two free R-modules withfinite rank are isomorphic if and only if when they have same rank.

13.15. A R-module M is called a torsion module if for every m ∈M there is an nonzero element r ∈ R such thatrm = 0, i.e. M = Tor(M). Prove that every finite Abelian group is a torsion Z–module. Give a example a the infiniteAbelian group which is a torsion Z–module.

13.16. Let R a integral domain. Prove that for every finitely generated torsion R–module M there is a non-zeroelement r ∈ R such that rm = 0 for all m ∈M. In other words, Ann(M) , 0.

13.17. Let N a submodule of modules M. Prove that if both M/N and N are finitely generated then so is M.

13.18. Let R be a commutative ring and A, B, and M be R-modules. Prove that,i) Hom R(A×B,M)Hom R(A,M)×Hom R(B,M)ii) Hom R(M,A×B)Hom R(M,A)×Hom R(M,B)

13.19. Let R be a commutative ring and F a free R-module of finite rank. Prove that, Hom R(F,R)F.

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13.4 Tensor products

In this section we do of study the product n tensor of two modules M and N over ring R (not necessarilycommutative). For more details on tensor products we recommend Bourbaki [?bourbaki, pg. 243-266].

Let R be a ring (not necessarily commutative), M = MR a right R-module, and N =R N a left R-module.Let F(M×N) be the free Z-module generated by M×N. Hence, there is a canonical map

M×N→ F(M×N)

Let D be the Z-submodule of F(M×N) which is generated by elements of the following type

(x1 + x2, y)− (x1, y)− (x2, y)(x, y1 + y2)− (x, y1)− (x− y2)(x,λ, y)− (x,λy),

where x,x1,x2 ∈M, y, y1, y2 ∈N, and λ ∈ R.

Definition 13.7. The quotient Z-module F(M×N)/D will be called the tensor product of modules M and N anddenoted by M⊗R N.

Let B be any additive group. A mapϕ : M×N→ B

will be called Z-bilinear if

ϕ(x1 + x2, y) = ϕ(x1, y) +ϕ(x2, y)ϕ(x, y1 + y2) = ϕ(x, y1) +ϕ(x− y2)

for all x,x1,x2 ∈M, y, y1, y2 ∈N, and R-balanced if

ϕ(xλ, y) = ϕ(x,λy),

where x ∈M, y ∈N, and λ ∈ R.Define the canonical mapping

j : M×N −→M⊗R N(m,n) −→ (m,n) + D

and denote the symbol (m,n) + D by m⊕R n or simply m⊕n.

Exercise 13.8. Prove that the canonical map

j : M×N −→M⊗R N(m,n) −→m⊕n

is Z-bilinear and R-balanced.

Theorem 13.4. i) Let ψ : M⊗N→ G be a Z-linear mapping into a Z-module G. The mapping ψ j : M×N→ Gis Z-bilinear and R-balanced.

ii) Let G be any Abelian group and ϕ : M×N→ G, be any Z-bilinear, R-balanced mapping. Then, there exists aunique mapping ψ : M⊗N→ G such that ϕ = ψ j.

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Proof. i) See Exercise 13.8.ii) First we extend ϕ to a unique homomorphism ϕ? : F(M×N)→ G as follows

ϕ?(∑

αx,y(x, y))

=∑

αx,yϕ(x, y)

Then, ϕ?(D) = 0, since ϕ is bilinear. The reader must verify this. Hence, we can define a homomorphism

ψ : M⊗R N := F(M×N)/D→ Gx⊗R y := (x, y) + D→ ϕ(x, y)

Since, ϕ?(D) = 0 then such map is well-defined.The commutativity of the diagram is obvious from the definition. The uniqueness of the mapping ψ

follows from the fact that elements x⊗R y generate M⊗R N.

M×N

ϕ%%

j// M⊗R N

ψ

G

Figure 13.2: Universal property of the tensor product

The following properties are true in M⊗R N

(m1 + m2)⊗n = m1⊗n + m2⊗nm⊗ (n1 + n2) = m⊗n1 + m⊗n2

mr⊗n = m⊗ rn

Example 13.9. Notice thatZ⊗ZZ/2ZZ/2Z.

Indeed, z⊗ a = 1⊗ za.

Example 13.10. The tensor product of non-zero modules may be zero. For example, take the two Z-modulesM =Z/2Z and N =Z/3Z. Then, in M⊗N we have

x⊗ y = 3(x⊗ y)−2(x⊗ y) = x⊗ (3y)− (2x)⊗ y = 0

Hence,Z⊗ZZ/2Z 1Q⊗ZZ/2Z

Example 13.11. Notice that Q⊗ZZ/2Z = 0. Indeed,

q⊗ a =rs·2⊗ a =

rs⊗2a =

rs⊗0 = 0

The existence of the map ψ : M⊗N→ G, or the universal property, characterizes the tensor product. Inother words the following is true.

Proposition 13.5. Suppose that there is a R-module T such that for everyZ-bilinear, R-balanced mapϕ : M⊗N→Gthere exists a unique g : T→ G such that gjT = ϕ. Then,

TM⊗R N

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M×N

ϕ""

jT// T

g

G

Figure 13.3: Characterization of the tensor product

Proof. Homework

To summarize we state the two main properties of the tensor product:

1) The tensor product of M and N, R-modules exists and it is unique.2) Every bilinear, R-balanced mapping from M⊗R N to any Abelian group A factors through M⊗R N.

13.4.1 Tensor product of two homomorphisms

Let R be a ring, M, M′ two right R-modules, N, N′ two left R-modules and u : M→M′ and v : N→N′ twohomomorphisms. Define the following map

ϕ :M×N→M′⊗N′

(x, y)→ u(x)⊗v(y)

Exercise 13.9. Show that ϕ is Z-bilinear and R-balanced

The mapping ϕ is called canonical.By the property of the tensor product, there exists a unique map M⊗N→M′⊗N′ as in the following

diagram; see Fig. 13.4, whereψ(u⊗v)→ u(x)⊗v(x)

M×N

ϕ%%

j// M⊗N

u⊗v

M′⊗N′

Figure 13.4: Tensor product of mappings

This mapping is denoted by u⊗ v and is called the tensor product of the homomorphisms (or linearmaps) u and v. The canonical map (u,v)→ u⊗v induces the map

HomR(M,M′)×HomR(N,N′)→HomZ(M⊗R N,M′⊗R N′)

Theorem 13.5. Let M1,M2 be right R-modules and N a left R-module. If M = M1⊕M2, then

M⊗R NM1⊗R N⊕M2⊗N

Proof. Notes

Exercise 13.10. Let M =Q and M1 =Z. Why the above theorem does not work in this case? What about the caseM =Z and M1 = 2Z?

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13.4.2 Two sided modules

Let S and R rings. M is called an (S-R) two sided module if1) M is a left S-module2) M is a right R-module3) For every s ∈ S, m ∈M, and r ∈ R we have

(sm)r = s(mr)

A two-sided module is denoted as SMR. Given SMR and RN, we will define an S-module structure onM⊗R N.

Pick s ∈ S. Define the mapϕs : M×N→M⊗N

such that ϕs(m,n) = sm⊗n.

Exercise 13.11. Prove that ϕs is Z-bilinear and R-balanced.

Then, there exists a unique map ψs : M⊗N→M⊗N such that

ψs(m⊗n) = sm⊗n.

Hence, M⊗R N becomes an S-module via

M×N

ϕs%%

j// M⊗N

∃!ψs

M⊗N

Figure 13.5: Two sided modules

S×M⊗R N→M⊗R N(s,m⊗n)→ ψs(m⊗n) = sm⊗n

Exercise 13.12. Verify that the above is an S-module.

Theorem 13.6. Let M be a left R-module. Then, R⊗R MM, as R modules.

Proof. First we need to check that R is a two-sided module. Indeed,

(r1r2)r3 = r1(r2r3)

Using the last result we haver(r′⊗m) = rr′⊗m

Hence the map

R⊗R M→Mr⊗m→m

is well defined. Its inverse is

M→ R⊗R Mm→ 1⊗m

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is well defined.

Exercises:

13.20. Prove that(Z/mZ)⊗Z (Z/nZ) = 0

if m and n are coprime.

13.21. Let R a ring and I a ideal of R. Prove that

R/I⊗R MM/IM

for a R-module M.

13.22. Let M a Abelian group finitely generated. Prove that:

Q⊗Z M ⊕ri=1Q

where r is rank of M.

13.23. Let M,N,S, R-module. Prove that:

M⊗R (N⊗R S) (M⊗R N)⊗R S

13.24. Suppose that R is a ring with identity. Prove that

HomAb(B,Πi∈IGi) = Πi∈IHomAb(B,Gi),

as right R-modules, for all left R-modules B and all abelian groups Gi (i ∈ I). You may use resources from categorytheory; if so, outline your argument so that it is clear which theorems you are appealing to.

13.25. Adjoint Isomorphism Let R and S be rings and AR, CS modules, and B a bimodule. Then;

HomS(A⊗R B,C) HomR(A,HomS(B,C))

13.26. Suppose that R is a ring with identity. For each left R-module M, HomR(R,M) is naturally R-isomorphic toM. Prove this, and explain what the “natural” part is all about.

13.27. Suppose that R and S are rings with identity. Let SAR be an S-R-bimodule, and B be a left R-module. Provethat A⊗R B has a unique scalar multiplication making it a left S-module, so that s(a⊗b) = sa⊗b, for each s ∈ S, a ∈A,and b ∈ B.

13.5 Exact sequences

In this section all rings have 1. Letψ : A→ B,

be a monomorphism of R-modules. Hence, Aψ(A) ⊆ B ).To say that C is isomorphic to the factor group is the same as of we say that there is a surjective

homomorphism ϕ : B→ C with kerϕ = ψ(A). Hence this gives a pair of homomorphisms:

Aψ→ B

ϕ→ C,

where Img ψ = kerϕ. A pair of homomorphisms with this property is called exact.

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Definition 13.8. The pair of homomorphisms X α→ Y

β→ Z is said to be exact at Y if Img α = kerβ. A sequence of

homomorphisms· · · → xn−1→ xn→ xn+1→ ·· ·

is called exact sequence if it is exact in every xn between a pair of homomorphisms.

Proposition 13.6. Let A,B and C, be R-modules and 0 is the zero module. Then,i) The sequence

0→ Aψ→ B

is exact (in A ) if and only if ψ is injective.ii) The sequence

Bϕ→ C→ 0

is exact in C if and only if ϕ is surjective.

Proof. The homomorphism ψ : 0→ A has image 0 in A. This will be the kernel of ψ if and only if ψ isinjective. Similarly, the kernel of the homomorphism zero C→ 0 is all of C which is the image of ϕ if andonly if ϕ is surjective.

Corollary 13.2. The sequence

0→ Aψ→ B

ϕ→ C→ 0

is exact if and only if ψ is injective, ϕ is surjective and Img ψ = kerϕ, so B is a extension of C by A.

Definition 13.9. The exact sequence

0→ Aψ→ B

ϕ→ C→ 0

is called short exact sequence.

Example 13.12. For any direct sum A⊕C the sequence

O→ A t→ A⊕C π

→ C→O,

where t(a) = (a,0) and π(a,c) = C is a short exact sequence.

Example 13.13. As a special case of the above example we take two Z− modules A =Z and C =Z/nZ :

O→Zt→Z⊕ (Z/nZ)

ϕ→Z/nZ→O

which gives an extension of Z/nZ in Z.Another extension of Z/nZ in Z is given by the short exact sequence

O→Z[n]→Z

π→Z/nZ→O

where with [n] denote the function x→ nx and π denote the natural projection.

Notice that modules of the above short exact sequences are not isomorphic even though A and C areisomorphic. Thus, there exist at least two different ways to obtain non equivalent extension of degree nZ/nZ over Z.

Example 13.14. If ϕ : B→ C is a homomorphism we can always construct the short exact sequence

0→ kerϕ → Bϕ→ Img ϕ→O.

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LetO→ A→ B→ C→O

andO→ A′→ B′→ C′→O

be two short exact sequences.i) A homomorphism of exact short sequences is a triple α,β,γ of homomorphisms modules such that

the following diagram is commutative:

0 // A

α

// B

β

// C

γ

// 0

0 // A′ // B′ // C′ // 0

Figure 13.6: Homomorphism of exact sequences.

A homomorphism is an isomorphism of exact short sequences if α,β,γ are all isomorphisms. In thiscase extensions B and B′ are called isomorphic extensions.

ii) The two exact sequences are called equivalent if A = A′, C = C′ and there is a isomorphism betweenthem as in the above diagram. In this case B and B′ are called equivalent extension.

Lemma 13.9 (The short five lemma). Let α,β,γ be homomorphisms of short exact sequences as in the diagram

0 // A

α

// B

β

// C

γ

// 0

0 // A′ // B′ // C′ // 0

Figure 13.7: The short five Lemma.

Then the following are true:i) If α and γ are injective then so is βii) If α and γ are surjective then so is β.ii) If α and γ are isomorphisms then so is β.

Proof. Let R be a ring and

0→ Aψ→ B

ϕ→ C→ 0

be a short exact sequence of R-modules. We say that this sequence is split if there is an R-modulecomplement of ψ(A) in B. In this case B = ψ(A)⊕C′ for some submodule C′ of B. Since ψ is injective and ϕis surjective then BA⊕C.

Suppose now that we have a short exact sequence

1→ Aψ→ B

ϕ→ C→ 1

of groups. In this case we get B = ψ(A)oC′ or BAoC. We say that B is a split extension of C by A.

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Proposition 13.7. i) Let the following be an exact sequence of finite-dimensional vector spaces over a field k.

0 −→ V′ −→ V −→ V′′

−→ 0

Then,dimV′+ dimV

′′

= dimV

ii) Let the following be an exact sequence of finite-dimensional vector spaces over a field k.

0 −→ V1 −→ V2 −→ V3 −→ V4 −→ 0

Then,dimV4 = dimV3−dimV2 + dimV1

Proof.

Exercises:

13.28. Let X,X′,X” be A-modules. A sequence

X′ −→ X→ X”−→ 0

is exact if and only ifHom A(X′,Y)←−Hom A(X,Y)←−Hom A(X”,Y)←− 0

is exact for all A-modules Y.

13.29. Let R be a ring with 1 and A a right module. If the following is an exact sequence:

N −→M −→ P −→ 0

then show that:

A⊗R N −→ A⊗R M −→ A⊗R P −→ 0

is exact.

13.30. Let A be a commutative ring with identity, and S be a multiplicative system of A. Prove that S−1(·) is acovariant functor which carries short exact sequences of A-modules to short exact sequences of S−1A-modules.

13.31. Assume that R is a ring with identity, and that every short exact sequence of with unity R-modules splits.Prove that every with unity R -module is isomorphic to a direct sum of simple R-submodules.

13.32. i) Let N be a submodule of M andπ : M→M/N

the natural projection. Suppose that ϕ : M→M′ is a homomorphism of R-modules, and ϕ(N) = 0. Show that thereis a unique homomorphism

ϕ : M/N→M′

such that π ϕ = ϕ.ii)

13.33. If O is a local ring with maximal ideal m, there is a natural exact sequence of O-modules

0 −→mn/mn+1−→O/mn+1

−→O/mn−→ 0

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13.6 Projective, injective, and flat modules

In this section we give a brief introduction to projective, injective, and flat modules.

13.6.1 Projective modules

Let R be a ring an P an R-module. Then, P is called a projective module if the following propertyholds: given a homomorphism f : P→M, for every surjective homomorphism g : M→M′′ there exists ahomomorphism h : P→M such that the following diagram commutes.

P

f

h

~~M g

// M′′ // 0

Figure 13.8: Projective property

Theorem 13.7. Let P be an R-module. Then the following are equivalent.i) P is projectiveii) Every short exact sequence

0→M′→M′′→ P→ 0

of left R-modules splitsiii) P is a direct sumand of a free moduleiv) The functor

M 7→Hom R(P,M)

is exact.

Proof.

Example 13.15. Suppose that R is a ring with identity.

(a) Prove that each free left R-module is projective.(b) Prove that a left R-module P is projective if and only if each short exact sequence of left R-modules below splits.

0 −→ A −→ B −→ P −→ 0

You may use the fact that every left R-module is a homomorphic image of a free one.(c) Prove that: P is projective if and only if it is a summand of a free module.

13.6.2 Injective modules

A left module Q over the ring R is injective if it satisfies the following condition:If X and Y are left R-modules and f : X→ Y is an injective module homomorphism and g : X→Q is an

arbitrary module homomorphism, then there exists a module homomorphism h : Y→Q such that h f = g,i.e. such that the following diagram commutes:

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13 Shaska T.

0 // X

g

f// Y

h

Q

Figure 13.9: Injective property

Theorem 13.8. Let Q be an R-module. Then the following are equivalent.i) Q is injectiveii) Every short exact sequence

0→Q→M→N→ 0

of left R-modules splitsiii) If Q is a submodule of some other left R-module M, then there exists another submodule K of M such that M

is the internal direct sum of Q and K, i.e. Q + K = M and Q∩K = 0.iv) The contravariant functor

M 7→Hom R(M,Q)

is exact.

Proof.

Theorem 13.9 (Baer’s Criterion). A left R-module Q is injective if and only if any homomorphism g : I→Q definedon a left ideal I of R can be extended to all of R.

Proof. Homework

Divisible groups

An abelian group G is divisible if and only if for every positive integer n and every g ∈G, there exists y ∈Gsuch that ny = g.

An abelian group G is divisible if and only if G is an injective object in the category of abelian groups,so a divisible group is sometimes called an injective group.

Lemma 13.10. Prove that:(a) An abelian group is injective if and only if it is divisible.(b) Over a PID a module is injective if and only if it is divisible.

Proof. Homework

13.6.3 Flat modules

In Homological algebra, and algebraic geometry, a flat module over a ring R is an R-module M such thattaking the tensor product over R with M preserves exact sequences. A module is faithfully flat if taking thetensor product with a sequence produces an exact sequence if and only if the original sequence is exact.Vector spaces over a field are flat modules. Free modules, or more generally projective modules, are alsoflat, over any R. For finitely generated modules over a Noetherian local ring, flatness, projectivity, andfreeness are all equivalent.

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Consider the category of R-modules and homomorphisms N an R-module. The function F : M→M⊗R Nor (· ⊗R N) is a functor in this category. The R-module N is called flat if the functor F is exact. In otherwords, if for every exact sequence

M′→M→M”

we have thatM′⊗R N→M⊗R N→M”⊗R N

is exact.

Proposition 13.8. LetM′→M→M”→ 0

be an exact sequence of R-modules and N any R-module. Then,

M′⊗R N→M⊗R N→M”⊗R N→ 0

is exact.

Proof. See [?AM, Prop. 2.18, pg. 29].

Lemma 13.11. Every projective module is flat.

Proof. Hence, for every module, we have the following

free =⇒ projective =⇒ flat

Proposition 13.9. Let N be any R-module. Then the following are equivalenti) N is flatii) If

0→M′→M→M”→ 0

is exact then0→M′⊗R N→M⊗R N→M”⊗R N→ 0

is exact.iii) If f := M′→M is injective, then f ⊗1 : M′⊗N→M⊗N is injective.iv) If f := M′→M is injective and M,M′ are finitely generated, then f ⊗1 : M′⊗N→M⊗N is injective.

Proof. See [?AM, Prop. 2.19, pg. 29].

Lemma 13.12. Let f : A→ B be a ring homomorphism. If M is flat as an A-module, then B⊗A M is flat as aB-module.

Proof. Let0→ P→ K→Q→ 0

be an exact sequence of B-modules. We want to show that

0→ P⊗B (B⊗A M)→ K⊗B (B⊗AM M)→Q⊗B (B⊗A M)→ 0

is exact.Recall that for any module P the following is true

P⊗B (B⊗A M) (P⊗B B)⊗A MP⊗A M

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13 Shaska T.

Hence, it is enought to show that

0→ P⊗A M→ K⊗A M→Q⊗A M→ 0

is exact. This is true since M is flat.

Lemma 13.13. Suppose that A is a commutative ring with identity, and M is an A-module. Show that the followingare equivalent:

(a) M is flat over A.(b) MP is flat over AP, for each prime ideal P of A.(c) Mm is flat over Am, for each maximal ideal m of A.

Proof. a) =⇒ b): Let P be a prime in A. Then, there is an embedding

A → AP

Since M is flat, by the above Lemma we have that AP⊗A M is AP-flat. However,

AP⊗A M = S−1A⊗A MS−1M = MP

is flat as an AP-module.b) =⇒ a) : is obvious.c) =⇒ a) : Use Lemma 13.4.

Exercises:

13.34. (a) Give an example of a projective module which is not free. Explain.(b) Give an example of a ring with identity and a free module possessing a submodule which is not free.

13.35. Let R be a ring with identity. Prove that a direct sum of left R-modules is projective if and only if eachsummand is projective.

13.36. Assume that R is a ring with identity. Prove that every free R-module is projective.

13.37. Prove that every abelian group G can be embedded as a subgroup of a divisible abelian group.

13.38. Let G be an abelian group. Prove that G has a subgroup d(G) which is divisible and contains all divisiblesubgroups of G, and, moreover, that d(G) is a summand of G, such that G/d(G) has no nontrivial divisible subgroups.

13.39. Let R be a ring with identity. Prove that a direct product of left R-modules is injective if and only if each factoris injective.

13.40. Let R be a ring with identity, and J be an indecomposable injective left R-module, and S = EndR(J). Prove thatfor all s ∈ S at least one of s or 1− s is a unit.

[Hint: Recall that an injective module J is indecomposable if and only if every two nonzero submodules of J havenontrivial intersection.]

13.41. Let R be a ring with identity.(a) Define flat left R-module.(b) Prove that a direct is flat if and only if each module is flat.(c) Prove that a free left R-module is flat.(d) Prove that every projective module is flat.

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13.42. Suppose that R is a ring with identity, and that every short exact sequence of unital R-modules splits. Provethat every unital R-module is isomorphic to a direct sum of simple R-submodules.

13.43. Let

0 −→ A −→ B −→ P −→ 0

be an exact sequence of commutative groups with A and B of orders a and b respectively such that (a,b) = 1. LetB‘ = x ∈ E : bx = 0. The group E is the direct sum of A and B‘. Moreover, B‘ is the only subgroup of E isomorphicto B.

13.44. Let A be a commutative ring with identity. For each multiplicative set S of A, prove that S−1A is a flatA-module.

13.45. Let A be a commutative ring with identity. For each multiplicative system S of A, prove that the contractionQ 7→Q∩A is an order isomorphism from Spec(S−1A) onto the subset of Spec(A) consisting of all prime ideals P of Awhich are disjoint from S.

13.46. Suppose that R is a principal ideal domain (PID) and F its field of fractions. For every R-module torsion freeM, prove that M⊗R F is the injective hull of M. (You may use any results about injective and flat modules over aPID; please identify them clearly.)

13.47. Suppose that A is a commutative ring with identity. If P and Q are projective A-modules, prove that P⊗A Qis also projective.

13.48. Suppose that R is a principal ideal domain and F is its field of fractions. For any torsion-free R-module M,prove that M⊗R F is the injective hull of M. (You may use any results about injective and flat modules over a PID;please identify them clearly.)

13.7 The Snake Lemma

M

α

f// M′

β

g// M′′

γ

// 0

0 // N′f// N g

// N′′

Figure 13.10: The Snake diagram.

Lemma 13.14 (Snake Lemma). Given a snake diagram as above, the map

δ : kerγ→ Coker α

induced by δz′′ = f−1β g−1z′′ is well defined and the following sequence is exact

kerα→ kerβ→ kerγ→ Coker α→ Coker β→ Coker γ

Proof.

Exercises:

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13 Shaska T.

kerα

// kerβ

// kerγ

M′

α

f// M

β

g// M′′

γ

// 0

0 // N′f

//

N g//

N′′

Coker α // Coker β // Coker γ

Figure 13.11: The Snake diagram.

13.49. Assume that A is a commutative ring with identity. Let F(m) and F(n) be the free modules on M and ngenerators, respectively. Prove that if F(m) F(n), then m = n.

(b) Give an example of a free A-module F such that two different basis of it have different cardinalities.[Hint: A has to be non-commutative, otherwise b) is true]

13.50. Let R be a ring with identity, and J be an indecomposable injective left R-module, and S = EndR(J). Prove thatfor all s ∈ S at least one of s or 1− s is a unit.

[Hint: Recall that an injective module J is indecomposable if and only if every two nonzero submodules of J havenontrivial intersection.]

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Chapter 14

Modules over a Principal Ideal Domains

In this chapter we will study modules over principal ideal domains.

14.1 Notherian Modules

The left R-module M is called a Notherian R-module if it satisfies the ascending chain condition onsubmodules which says that for every increasing chain of submodules

M1 ⊆M2 ⊆M3 ⊆ · · ·

there is a positive number m such that for every k ≥m, Mk = Mm.The ring R is called Notherian if it is a left module over itself which implies if does not have an increasing

infinite chain of left ideals of R.

Theorem 14.1. Let R a ring and M a left R-module. Then, the following are equivalent:1) M is a Notherian R-module.2) Every nonempty set of submodules of M contains a maximal element under inclusion.3) Every submodule of M is finitely generated.

Proof. (1⇒ 2) Assume that M is Notherian and let Σ the collection of nonempty submodules of M. Letm1 ∈ Σ. If m1 is maximal then (2) is true, so assume that m1 is not maximal. Then, there is a m2 ∈ Σ suchthatm1 ⊂m2. Ifm2 is maximal in Σ, then (2) is true, otherwise we can assume that there is anm3 ∈ Σ whichcontainsm2. If we continue this way then we get an infinite strictly increasing chain of elements of Σ, whichcontradicts (1).

(2⇒ 3) Assume that 2) holds and let N be a submodule of M. Let Σ the collection of all finitely generatedsubmodules of N. Since 0 ∈ Σ, this collection is not empty. By 2) we have that Σ contains an maximalelement N′. If N′ , N, let x ∈ N \N′. Since N′ ∈ Σ, the submodule N′ is finitely generated by assumption.Therefore, the submodule generated by 〈N′,x〉 is finitely generated. This contradicts the maximality of N′.Hence, N = N′ is finitely generated.

(3⇒ 1) LetM1 ⊆M2 ⊆ · · ·

be a chain of submodules of M. Let

N =

∞⋃i=1

Mi

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14 Shaska T.

It can easily be proved that N is a submodule of M and by assumption it is finitely generated, say by

a1,a2, · · · ,an.

Since ai ∈N, for every i, ai is in one of submodules of chain, say M ji . Let

m = max j1, j2, · · · , jn.

Then, ai ∈Mm, for all i. Hence, the module that they generate is contained in Mm. Thus, N ⊆Mm. Thisimplies Mm = N = Mk for every k ≥m, which proves 1).

Corollary 14.1. If R is a PID then every nonempty set of ideals of R has an maximal element and R is a Notherianring.

Proof. Principal ideal domain satisfies condition 3) above with M = R.

Let M be a free R-module with rank n < ∞. The elements y1, y2, · · · , yn+1 ∈M are called R-linearlydependent if there exist elements r1,r2, · · · ,rn+1 ∈ R not all zero such that:

r1y1 + r2y2 + · · ·+ rn+1yn+1 = 0

Proposition 14.1. Let R be an a integral domain and let M be a free R-module with rank n <∞. Then, any n + 1elements of M are R-linearly dependent.

Proof. Let e1,e2, · · · ,en be a basis for M and y1, y2, · · · , yn+1, are distinct elements of M. For 1 ≤ i ≤ n + 1, wehave

yi = a1iei + a2iei + · · ·+ aniei

in terms of basis e1,e2, · · · ,en.Let A be a (n + 1)× (n + 1) matrix whose the i, j entry is ai j,1 ≤ o f ≤ n,1 ≤ j ≤ n + 1 and whose last row is

zero. Therefore, detA = 0. Since R is integral domain from Corollary 14.1 shows columns of A are R-linearlydependent. This implies dependence relations on the yi’s. This completes the proof.

14.2 Torsion modules over a PID

If R is a integral domain and M is a R-module then we define

Tor(M) := x ∈M | rx = 0 for a nonzero element r ∈ R.

Tor (M) is a submodule of M, which is called the torsion submodule of M. If N is a submodule of Tor (M),then N is called a torsion submodule of M. If Tor (M) = 0, then the module M is called torsion free.

For a submodule N of M, the annihilator of N is the ideal of R defined as:

Ann (N) = r ∈ R | rn = 0 for every n ∈N

Notice that if N is not a submodule torsion of M then Ann (N) = (0). It is easy to prove that if N and L aresubmodules M where N ⊆ L then Ann (L) ⊆Ann (N). If R is be a PID and N ⊆ L ⊆M with Ann (N) = (a) andAnn (L) = (b), then a | b. In particular, annihilator of for every the element x of M divides the annihilator ofM (this follows from Lagrange’s Theorem when R =Z.)

Definition 14.1. For every integral domain R the rank of an R-module M is the maximum number of R-linearlyindependent elements of M.

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Theorem 14.2. Let R a PID, let M be a free R-module with finite rank n, and N a submodule of M. Then, thefollowing hold:

1) N is free with rank m ≤ n,2) there exist a basis y1, y2, · · · , yn of M such that a1y1,a2y2, · · · ,amym is a basis for N where the elements

a1,a2, · · · ,am are nonzero elements of R such that

a1|a2| · · · |am.

Proof. Complete .... See [?DF]

14.3 Finitely generated modules over a Principal Ideal Domain

Recall that se a R-module C is a cyclic R-module if there is a x ∈ C such that C = Rx. Then, we can of de finea homomorphism R-modules

π : R→ Cπ(r) = rx

which is surjective from the assumption that C = Rx. From the First Isomorphism Theorem for ringsTheorem 9.4 we have an isomorphism of R-modules

R/kerπC

If R is a PID, kerπ is a principal ideal, (a), so modules cyclic are of the form R/(a) where (a) = Ann (C).

Theorem 14.3 (Fundamental Theorem, Existence: Invariant Factor Form). Let R be a PID and Mbe a finitelygenerated R-module.

1. Then, M is isomorphic to a direct sum of finitely many cyclic modules. More precisely,

MRr⊕R/(a1)⊕R/(a2)⊕ · · ·⊕R/(am)

for some integer r ≥ 0 and the elements nonzero a1,a2, · · · ,am ∈ R which are not units in R and which satisfy

a1|a2| · · · |am

2. M is torsion free if and only if when M is free.

3. In the decomposition of 1)Tor(M)R/(a1)⊕R/(a2)⊕ · · ·⊕R/(am)

In particular M is torsion module if and only if r = 0 and in this case the annihilator of M is the ideal (am).

Proof. The integer r in above theorem is called the free rank or the Betti number of M and the elements

a1,a2, · · · ,am ∈ R are called invariant factors of M.Using the Chinese Remainder Theorem modules we can further decompose the cyclic modules in

Theorem 14.3 so that M is a direct sum of cyclic modules whose annihilators are as simple as possible.

Theorem 14.4 (Elementary Divisors). Let R a principal ideal domain and let M a R-module finitely generated.Then, M is direct sum of a finite number cyclic modules annihilator of of cileve is (0) or generated from powers ofprime numbers in R, so,

MRr⊕R/(pα1

1 )⊕R/(pα22 )⊕ · · ·⊕R/(pαt

t )

where r ≥ 0 is a integer and pα11 pα2

2 · · ·pαtt are power pozitive of prime numbers in R.

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14 Shaska T.

Proof. Assume that a ∈ R and a , 0. Since R is also a UFD we can write

a = u ·pα11 pα2

2 · · ·pαss

where pi’s are distinct primes in R and u is a unit. This factorization is unique up to units, so ideals(pαi

i ), i = 1,2, · · · ,s are uniquely defined. For i , j, we have (pαii + p

α jj ) = 〈gcd (pi,p j〉 = 〈1〉 = R.

Also, the ideal 〈a〉 can be written as

〈a〉 = lcm (pα11 , . . . ,p

ass ) = ∩s

i=1〈pαii 〉

Then, from the Chinese Remainder Theorem

R/(a)R/(pα11 )⊕R/(pα2

2 )⊕ · · ·⊕R/(pαss )

as rings and also as R-module. Applying this to the decomposition from the Theorem 14.3 we completethe proof.

Let R be a PID and let M a R-module finitely generated as in above theorem. The prime powers

pα11 pα2

2 · · ·pαtt (up to multiplication by units) are called elementary divisors of M.

Assume that M is a finitely generated torsion module over R, where R is a PID. Then, for distinct primesp1,p2, · · · ,pn that appear in above theorem we group together all cyclic factors that correspond to the sameprime pi. Then, M can be written as direct sum:

M = N1⊕N2⊕ · · ·⊕Nn

where Ni contains all the elements of M which are annihilated from some power of pi. Then, we have thefollowing:

Theorem 14.5 (The Primary Decomposition Theorem). Let R be a PID, M be a nonzero torsion R-module,Ann (M) = 〈a〉, such that

a = u ·pα11 pα2

2 · · ·pαnn ,

where p1, . . . ,pn are distinct primes, and

Ni = x ∈M|pαii x = 0,1 ≤ o f ≤ n.

Then, Ni is a submodule of M, Ann (Ni) = 〈pαii and

M = N1⊕N2⊕ · · ·⊕Nn.

If M is finitely generated, then for every Ni is direct sum of a finitely many cyclic modules whose annihilators aredivisors of pαi

i .

Proof. complete [?DF] Thm. 7. pg. 465. The submodule Ni above is called the pi-primary component of M.Notice that the elementary divisors of a finitely generated module M are just the invariant factors of the

primary components of Tor(M).

Lemma 14.1. Let R be a PID, p a prime in R, and F = R/(p).i) Let M = Rr. Then, M/pMFr.ii) Let M = R/(a) where a is a nonzero element of R. Then,

M/pM

F if p divides a in R0 if p does not divide a in R.

iii) LetM = R/(a1)⊕R/(a2)⊕ · · ·⊕R/(ak)

where each ai is divisible by p. Then, M/pMFk.

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Shaska T. 14

Proof. See [?DF] Lemma 8, pg. 466.

Theorem 14.6 (Fundamental Theorem). Let R be a PID .

1. Two finitely generated R-modules M1 and M2 are isomorphic if and only if they have of the same free rank andthe same list of invariant factors.

2. Two finitely generated R-modules M1 and M2 are isomorphic if and only if they have of the same free rank andthe same list of elementary divisors.

Proof. [?DF]. pg. 466

Corollary 14.2. Let R be a PID and let M a finitely generated R-module.i) Elementary divisors of M are the prime power factors of the invariant factors of M.ii) The largest invariant factor is the product of the largest of the distinct prime powers among the elementary

divisors

Corollary 14.3. The Fundamental Theorem of Finitely Generated Abelian Groups holds.

Exercises:

14.1. Let M a module over integral domain R.a) Suppose that x is an nonzero element torsion in M. Prove that x and 0 are "linearly dependent". Conclude

that the rank of Tor (M) is 0, so every R-module torsion has rank 0.b) Prove that rank of M is the same with the rank of M/Tor (M).

14.2. Let M a module over integral domain R.a) Suppose that M has rank n and se x1,x2, · · · ,xn is a maximal set with linearly independent elements of M. Let

N = Rx1 + · · ·+ Rxn be a submodule generated from x1,x2, · · · ,xn. Prove that N is isomorphic to Rn and that M/N isan R-module torsion. (equivalently, the elements x1, · · · ,xn are linearly independent and for every y ∈M there is annonzero element r ∈ R such that ry is written as a linear combination r1x1 + · · ·+ rnxn of xi -ve.)

b) Conversely, prove that if M contains a submodule free N with rank n (NRn ) such that m/N is a R-moduletorsion then M has rank n.

14.3. Let R a integral domain and A and B be R-modules with rank m and n respectively. Prove that rank of A⊕B ism + n.

14.4 Endomorphisms of vector spaces

The main purpose of this section is classify the distinct linear transformations of a vector space or thesimilarity classes of matrices.

Let V be a n dimensional vector space over the field k and B a basis of V. Further, T : V→ V is alinear map and A = MB

B(T) is its associated matrix. Choosing a different basis B′ for V gives a new matrix

B = MB′

B′(T) associated with T, namely

B = P−1 AP

where P = MBB′

(id), see Chapter 4. Can we find B′ such that the matrix associated with T is as simple aspossible? The strategy is to pick B′ such that B is as close to a diagonal matrix as possible. We distinguishtwo cases:

i) k does not contain all the eigenvalues of A

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14 Shaska T.

ii) k contains all eigenvalues.

These cases lead respectively to the rational canonical form and the Jordan canonical form and will be studiedin sections 2 and 3.

As above k denotes a field and Matn×n(k) denotes the vector space of all n×n matrices with entries in k.Let A ∈Matn×n(k) and f ∈ k[x] given by

f (x) = an xn + · · ·+ a0.

We definef (A) := an An + · · ·+ a1 A + a0 I.

Then f (A) is an n by n matrix with entries in k.

Theorem 14.7. Let A ∈Matn×n(k). Then there exists a non-zero f ∈ k[x] such that

f (A) = 0.

Proof. The vector space Matn×n(k) is of dimension n2. Hence,

I,A,A2, . . . ,As

are linearly dependent for s > n2. Thus, there exist a0, . . . ,as such that

asAs + . . .aA + a0I = 0.

Take f (x) = asxs + . . .a1x + a0.

Definition 14.2. We call the minimal polynomial of A the unique monic polynomial m ∈ k[x] of minimal degreesuch that m(A) = 0. The minimal polynomial of A is denoted by mA(x).

Definition 14.3. Let f (x) be a monic polynomial in k[x] given by

f (x) = xn + an−1 xn−1 + · · ·+ a0.

The companion matrix of f (x) is the n×n matrix

C f :=

0 0 . . . . . . −a01 0 . . . . . . −a10 1 . . . . . . −a2

. . . . . . . . .

. . . . . . . . .0 0 . . . 1 −an−1

and we denote it by C f .

Lemma 14.2. Let f (x) ∈ k[x] and C f its companion matrix. The characteristic polynomial of C f is

char (C f ,x) = f (x).

Proof. Exercise.

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Shaska T. 14

For a given matrix A the characteristic polynomial char (A,x) = det(xI−A). The matrix (xI−A) can beconsidered as a matrix over the field k(x). Moreover, A is also in Matn×n(k(x) ). In the next theorem we showhow every matrix in Matn×n(k(x) ) can be transformed into a diagonal matrix by the elementary operations.These elementary operations consist of

i) Interchange of any two rows or columns (Ri←→ R j)

ii) Adding a multiple (in k[x]) of one row or column to another (Ri −→ q(x) ·Ri + R j).

iii) Multiplying any row or column by a non-zero element in k (Ri −→ u ·Ri, for u ∈ k)

Two matrices A and B, one of which can be obtained by a sequence of elementary operations on the other,are called Gaussian equivalent. For matrices whose entries are polynomials we have the following:

Theorem 14.8. Let M ∈Matn×n(k[x] ). Then, using elementary operations the matrix M can be put in a diagonalform

1·

·

1e1(x)

·

·

·

es(x)

where e1(x), . . . ,en(x) are monic polynomials such that

ei(x) | ei+1(x), for i = 1, . . . ,s−1.

Proof. We will use the elementary operations to transform M into a diagonal matrix. Among all matriceswhich are Gaussian equivalent to M pick the one which has the entry of smallest degree. Let such matrixbe A = [ai j(x)] and the entry with lowest degree is ai j =: m(x).

By an interchange of rows and columns bring this entry in (1,1)-position. All entries of the first columncan be written as (Euclidean algorithm)

a1 j = m(x)q j(x) + r j(x)

where degr j(x) < degm(x).By performing R j−m(x)q j(x)→ R j for j = 2, . . .n the first column of the matrix is

m(x)r2(x). . .. . .

rn(x)

.Choose the entry m′(x) with the smallest degree from the first column and by a row change move that

to the (1,1)-position. Perform the same process as above. Then degrees of r′j(x) will decrease by at leastone. Since k[x] is an Euclidean domain this process will end after finitely many steps and the first column

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14 Shaska T.

will look like m1(x)

0. . .. . .0

.Indeed, the maximum number of steps can be no bigger then degm(x).

Next we perform the same procedure for the first row to getm2(x) 0 . . . 0a′2,1(x) a′2,2(x) . . . a′2,n(x)a′3,1(x) a′3,2(x) . . . a′3,n(x). . .

a′n,1(x) a′n,2(x) . . . a′n,n(x)

Continuing again with the first column and so on, we get a sequence of operations

A→ A(1)→ A(2)

→ . . .

Let mi(x) denote the entry in the (1,1)-position after the i-th step. Then

degm(x) > degm1(x) > . . .

Thus, the procedure must stop and the matrix will be

e1(x) 0 . . . 0

0 a′′2,2(x) . . . a′′2,n(x)0 a′′3,2(x) . . . a′′3,n(x). . .0 a′′n,2(x) . . . a′′n,n(x)

where e1(x) has the smallest degree and divides all the entries a′′i, j(x).

Now we perform the same procedure focusing on the next row and column. Finally we will have

D :=

e1(x) 0 . . . 0

0 e2(x) . . . 00 0 . . . 0. . .0 0 . . . en(x)

such that ei(x) | ei+1(x), for i = 1, . . . ,n−1.

Remark 14.1. If any of ei(x) = 0 then it will occur in the last position since all other e j(x), j , i must divide ei(x).

Definition 14.4. Let A ∈Matn×n(k). Then by the above theorem the matrix xI−A can be put into the diagonal form

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1·

·

1e1(x)

·

·

·

es(x)

such that ei(x) are monic and ei(x) | ei+1(x), for i = 1, . . . ,s− 1. This is called the Smith normal form for A andelements ei(x) of nonzero degree are called invariant factors of A.

Lemma 14.3. The characteristic polynomial of A is the product of its invariant factors up to multiplication by aconstant.

Proof. We havechar (A,x) = det(xI−A).

Since (xI−A) ∼ Smith (A) thendet(xI−A) = c ·det(Smith (A)),

for some c ∈ k.

Lemma 14.4. Let e1(x), . . .es(x) be the invariant factors of A such that

ei(x) |ei+1(x), for i = 1, . . . ,s.

The minimal polynomial ma(x) is the largest invariant factor of A. In other words

es(x) = mA(x).

Proof. Exercise

Example 14.1. Find the Smith normal form of the matrix A given as follows:

A :=

2 -2 140 3 -70 0 2

Solution: We have

xI−A =

x - 2 2 - 140 x-3 70 0 x-2

We perform the following elementary operations

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14 Shaska T.

xI−A =

x−2 2 - 140 x−3 70 0 x−2

C1←→C2−→

2 x−2 - 14x−3 0 7

0 0 x−2

R2→(x−3)R1−2R2−→

2 x−2 - 140 (x−2)(x−3) −14(x−2)0 0 x−2

C2→(x−2)C1−2C2−→

2 0 - 140 −2(x−2)(x−3) −14(x−2)0 0 x−2

R1→12 R1, R2→−

12 R2

−→

1 0 - 70 (x−2)(x−3) 7(x−2)0 0 x−2

C3→7C1+C3−→

1 0 00 (x−2)(x−3) 7(x−2)0 0 x−2

C2←→C3−→

1 0 00 7(x−2) (x−2)(x−3)0 (x−2) 0

R3→R2−7R3−→

1 0 00 7(x−2) (x−2)(x−3)0 0 (x−2)(x−3)

C3→(x−3)C2−7C3−→

1 0 00 7(x−2) 00 0 −7(x−2)(x−3)

R2→17 R2, R3→−

17 R3

−→

1 0 00 (x−2) 00 0 (x−2)(x−3)

which is the Smith normal form Smith (A). The reader can check that the characteristic polynomial of Smith (A) and

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A are the same.

Exercises:

1. Find the companion matrix off (x) = x3

−x−1.

2. Find the companion matrix off (x) = (x−2)2(x−3).

3. Let A be a 2 by 2 matrix with entries in Q such that char (A,x) = x2 + 1. Find the minimal polynomialof A.

4. Let f (x) be an irreducible polynomial cubic in Q. For example

f (x) = ax3 + bx2 + cx + d.

Let A be a 3 by 3 matrix with entries in Q such that char (A,x) = f (x). Find the minimal polynomialmA(x) of A. Can you generalize to a degree n polynomial?

5. Find the Smith normal form of matrices in the previous two exercises.

6. Determine all possible minimal polynomials of a matrix A with characteristic polynomial

char (A,x) = (x−2)2(x−3)

7. Determine all possible Smith normal forms of a matrix A with characteristic polynomial

char (A,x) = (x−2)2(x−3)

8. Find all possible Smith normal forms of a matrix A with characteristic polynomial

char (A,x) = x3−1.

Programming exercises:

1) Write a computer program which finds the Smith normal form of a given matrix A.

14.5 The rational canonical form

Let f (x) be a polynomial with coefficients in a field k. As noted in the previous section not all roots of agiven polynomial are necessarily in k. For example, not all polynomials with rational coefficients factorinto linear factors over the rationals. Let A be a given matrix with entries in k. In this section we will seehow to find the "best" matrix D similar to A and with entries still in k. The reader can assume that in thissection k =Q.

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Let A ∈Matn×n(k) and D = Smith (A), its Smith normal form as in the previous section. Let e1(x), . . . ,es(x)be the invariant factors of A and C1, . . . ,Cs the corresponding companion matrices. The block-matrix

C1C2

·

·

·

Cs

is called the rational canonical form of A and is denoted by Rat (A). The word rational is used to indicatethat this form is calculated entirely within the field k. Notice that,

e1(x) · · ·es(x) = c · char (A,x)

implies thatdege1 + · · ·+ deges = deg char (A,x).

Hence, A and Rat (A) have the same dimensions.

Example 14.2. Find the rational canonical form of the matrix

A :=

2 -2 140 3 -70 0 2

Solution: We found the invariant factors of this matrix in Example 14.1 in the last section. They are e1(x) = x−2

and e2(x) = (x−2)(x−3). Then the rational form of A is

Rat (A) =

20 -61 5

Theorem 14.9. Let k be a field and A ∈Matn×n(k). Then the following hold:

i) Two matrices in Matn×n(k) are similar if and only if have the same rational form.ii) The rational form of A is unique.

Proof. Let A be similar to B. Then char A(x) = char B(x) as polynomials over k. Hence, the Smith normalform is the same for A and B. Thus, A and B have the same rational form.

If A and B have the same rational form, then they have the same invariant factors.

ii) There is only a unique choice of invariant factors. Hence a unique rational form.

Example 14.3. Let A be a 10 by 10 matrix such that its invariant factors are

e1(x) = x−2

e2(x) = (x−2)(x3 + x + 1)

e3(x) = (x−2)(x−3)(x3 + x + 1)

(14.1)

Find the rational canonical form of A.

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Solution: By multiplying through we have

e2(x) = x4−2x3 + x2

−x−2

e3(x) = x5−5x4 + 7x3

−4x2 + x + 6(14.2)

Hence, the rational canonical form of A is

Rat (A) =

20 0 0 21 0 0 10 1 0 -10 0 1 2

0 0 0 0 -61 0 0 0 -10 1 0 0 40 0 1 0 -70 0 0 1 5

Example 14.4. Let A be a 8 by 8 matrix such that its invariant factors are

e1(x) = x3 + x + 1

e2(x) = (x2 + 2)(x3 + x + 1) = x5 + 3x3 + x2 + 2x + 2(14.3)

Solution: Hence the rational canonical form is

Rat (A) =

0 0 -11 0 -10 1 0

0 0 0 0 -21 0 0 0 -20 1 0 0 -10 0 1 0 -30 0 0 1 0

Exercises:

1. Find the rational canonical form of this matrix over Q[1 23 4

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14 Shaska T.

2. Let A be the 8 by 8 matrix given by

A =

0 0 0 0 0 0 0 11 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 1 0 0 0 00 0 0 0 1 0 0 00 0 0 0 0 1 0 00 0 0 0 0 0 1 0

Find its eigenvalues. What about the eigenvalues of AT?

14.5.1 Caylay-Hamilton theorem

The Caylay-Hamilton theorem is one of the most recognized theorems of linear algebra. It can be quiteuseful at times to compute the rational canonical form of matrices.

Theorem 14.10. (Cayley - Hamilton) Let A ∈ Matn×n(k), mA(x) its minimal polynomial, and char A(x) thecharacteristic polynomial of A. Then,

mA(x) | char A(x).

Proof. Let e1(x), . . . ,es(x) be the invariant factors of A such that ei(x) |ei+1(x), for i = 1, . . .s. We know that

char A(x) = e1(x) · · ·es(x)

Since es(A) = mA(A) = 0 and es(x) | char A(x), then char A(A) = 0.Since m(x) is the minimal polynomial then

degmA(x) ≤ degcharA(x).

By the Euclidean algorithm,charA(x) = q(x)mA(x) + r(x)

such that degr(x) < degmA(x). Since charA(A) = 0, then r(A) = 0. Thus r(x) is the zero polynomial, otherwiser(x) would be the minimal polynomial.

14.5.2 Computing the rational canonical form

The previous section determines an algorithm for computing the Smith normal form of a matrix A. Thisgives us all the invariant factors of A. Once the invariant factors are known then it is easy to write downthe rational canonical form Rat (A) of A. However, there are techniques to directly compute the rationalform of a matrix by elementary operations or figure out the invariant factors without computing the Smithnormal form. In this section we illustrate some of these techniques through examples.

Example 14.5. Let A be the 3 by 3 matrix given below:

A =

233

703

203

- 43 - 11

3 - 43

-2 -7 -1

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Shaska T. 14

Find its rational canonical form.

Solution: The characteristic polynomial of A is

char (A,x) = (x−1)3.

Then, by Cayley-Hamilton theorem the minimal polynomial of A is one of the following:

mA(x) = (x−1), (x−1)2, (x−1)3

Furthermore, mA(A) = 0. We check that A− I , 0 and (A− I)2 = 0. Hence the minimal polynomial is

mA(x) = (x−1)2

Hence the Smith normal form is

Smith (A) =

1x−1

(x−1)2

and the rational form

Rat (A) =

10 -11 2

14.5.3 Computing the transformation matrix:

We know how to compute the rational form of a matrix A. Then, A is similar to its rational form Rat (A).Hence there exists an invertible matrix C such that

A = C−1 Rat (A)C

We would like to compute C. The strategy is to keep track of all elementary operations performed in xI−Aand to perform these operations on I in order to get C as a product of elementary matrices.

Algorithm 1. Input: A n×n matrix AOutput: The matrix C such that

A = C−1 Rat (A)C

1. 1) Create the matrix xI−A.

2. 2) Transform it to the Smith normal form and keep track of all the elementary operations.

3. 3) For each of the operations of step 2, perform the following operations on the identity matrix I by convertingto the following rules:

a) Ri←→ R j =⇒ Ci←→ C j

b) Ri −→ q(x) ·Ri + R j =⇒ Ci −→ q(x) ·Ci + C j

c) Ri −→ u ·Ri, for u ∈ k =⇒ Ci −→ u ·C j

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14 Shaska T.

4. 4) The matrix obtained after performing these operations on I is the sought matrix C.

Exercises:

1. Find the rational form of the 3 by 3 matrix with invariant factors

e1(x) = (x−1), e2(x) = (x−1), e3(x) = x−1.

2. Find the rational canonical form of matrices over Q

A =

0 -4 851 4 -300 0 3

, B =

2 2 10 2 -10 0 3

and determine if A and B are similar.

3. Find the invariant factors of 2 2 13 4 11 5 1

4. Prove that two non-scalar 2×2 matrices over k are similar if and only if they have the same charac-

teristic polynomial.

5. Find the rational canonical form of 0 -1 -10 0 0

-1 0 0

6. Determine all possible rational canonical forms for a matrix with characteristic polynomial

f (x) = x2 (x2 + 1)2

7. Determine all possible rational canonical forms for a matrix with characteristic polynomial

f (x) = xp−1

for an odd prime p.

8. The characteristic polynomial of a given matrix A is

char (A,x) = (x−1)2· (x + 1) · (x2 + x + 1).

What are the possible polynomials that can be minimal polynomials of A?

9. Find all similarity classes of 2×2 matrices with entries in Q and precise order 4 (i.e, A4 = I).

Programming exercises:

1) Write a computer program which finds the rational canonical normal form of a given matrix A.

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14.6 The Jordan canonical form

Let α ∈ k. Then a matrix of the form

Jα =

α 1α 1· ·

· ·

α 1α

is called a Jordan block.

Lemma 14.5. Let A be an s× s matrix with characteristic polynomial

char A(x) = (x−α)s.

Then, A is similar to the s× s Jordan block matrix Jα.

Proof. Let f (x) := (x−α)s. Then, the Cayley-Hamilton theorem implies that

f (A) = (A−αI)s = 0.

Hence, mA(x) = (x−α)r or equivalently mA−αI(x) = xr. Thus, (A−αI) is similar to the companion matrix D ofg(x) := xr, where

D =

0 10 1

0 ·

· ·

· ·

· 10

Thus, there is an invertible matrix P such that

P−1 (A−αI)P = D

which implies that P−1 AP = D +αI. In other words, A is similar to

D +αI =

α 1α 1

α 1· ·

· ·

α 1α

A matrix is in Jordan canonical form if it is a block diagonal matrix

J =

J1

J2.

.Jn

with Jordan blocks along the diagonal.

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14 Shaska T.

Theorem 14.11. Let A be a n×n matrix with entries in k and assume that k contains all eigenvalues of A. Then,

i) A is similar to a matrix in Jordan canonical form.ii) The Jordan canonical form of A denoted by J(A) is unique up to a permutation of blocks.

Thus, to find the Jordan canonical form of a n by n matrix A we first find its invariant factors e1(x), . . . ,es(x).Since the field k contains all eigenvalues of A and each ei(x) | char (A,x), then we factor invariant factors as

ei(x) = (x−α1)e1 · · · (x−αr)er

For each αi, i = 1, . . . ,er we have a Jordan block. Since the product of all the invariant factors equals thecharacteristic polynomial of A, the combination of all the Jordan blocks along the diagonal will create an nby n matrix (same dimensions as A).

Remark 14.2. The Jordan canonical form of a matrix A is diagonal if and only if A is diagonalizable.

Example 14.6. Both matrices

A =

0 1 1 11 0 1 11 1 0 11 1 1 0

, B =

5 2 -8 -8-6 -3 8 8-3 -1 3 43 1 -4 -5

,have the same characteristic polynomial

f (x) = (x−3)(x + 1)3.

Determine whether these matrices are similar and find their Jordan canonical forms.

Solution: The minimal polynomial for A and B is one of the following polynomials:

m1(x) =(x−3)(x + 1),

m2(x) =(x−3)(x + 1)2,

m3(x) =(x−3)(x + 1)3.

(14.4)

We check that (A− 3I) (A + I) = 0. In the same way we check that (B− 3I) (B + I) = 0. Hence, the minimalpolynomial of A and B is

m(x) = (x−3)(x + 1).

Its Smith normal forms are

Smith (A) = Smith (B) =

1

x + 1x + 1

(x−3)(x + 1)

Then the Jordan canonical forms are

J(A) = J(B) =

-1

-1-1

-3

Thus, A and B are similar. Further, A and B are diagonalizable matrices and we can diagonalize them using thetechniques of the previous chapter.

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Shaska T. 14

Example 14.7. Let A be a matrix such that its invariant factors are

e1(x) =(x−2)2(x2 + 1)

e2(x) =(x−2)3(x2 + 1)2 (14.5)

Find the rational and Jordan canonical form of A.

Solution: Multiplying out we have

e1(x) =x4−4x3 + 5x2

−4x + 4

e2(x) =x7 + 6x6 + 14x5−20x4 + 25x3

−22x2 + 12x−8(14.6)

The rational canonical form is

Rat (A) =

0 0 0 -41 0 0 40 1 0 -50 0 1 4

0 0 0 0 0 0 81 0 0 0 0 0 -120 1 0 0 0 0 220 0 1 0 0 0 -250 0 0 1 0 0 200 0 0 0 1 0 -140 0 0 0 0 1 -6

,

and the Jordan canonical form

J(A) =

2 10 2

-i+i

2 1 00 2 10 0 2

-i 10 -i

i 10 i

Example 14.8. Let A be a 3 by 3 matrix as below

A =

2 1 00 2 00 0 3

Find its Jordan canonical form.

Solution: Then char (A,λ) = (λ− 2)2(λ− 3). For the eigenvalue λ = 2, the algebraic multiplicity is 2 and the

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14 Shaska T.

eigenspace is given by

E2 = t

100

| t ∈QThe geometric multiplicity is 1, hence A is not similar to the diagonal matrix of eigenvalues.

We have

xI−A = x−2 1 00 x−2 00 0 x−3

C1←→C2−→

1 x−2 0x−2 0 0

0 0 x−3

R2=(x−2)R1−R2−→

1 x−2 00 (x−2)2 00 0 x−3

C2=(x−2)C1−C2−→

1 0 00 - (x−2)2 00 0 x−3

R2←→R3

C2←→C3−→

1 0 00 x−3 00 0 (x−2)2

−→ 1 0 0

0 1 00 0 (x−2)2(x−3)

Then its Jordan canonical form is

J(A) =

2 10 2

3

.Instead we could have recognized that A was already in the Jordan canonical form. Notice that the geometricmultiplicity for each eigenvalue is 1 and there is one Jordan block for each eigenvalue. Also the algebraic multiplicitiesof the eigenvalues are 2 and 1 and the corresponding Jordan blocks are of sizes 2 and 1 respectively. We will see thatthese facts are not a coincidence.

Exercises:

1. Let A be a matrix with characteristic polynomial

char (A,x) = x3 + x2 + x + 1

Find the rational form of A over Q and the Jordan canonical form of A over C.

2. Find the rational and Jordan canonical form of 2 1 11 2 01 1 3

.3. Compute the Jordan canonical form of the matrix with characteristic polynomial f (x) = xn

−1, for n≥ 2.

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Shaska T. 14

4. Show that if A2 = A then A is similar to a diagonal matrix which has only 0’s and 1’s along the diagonal.

5. Find the Jordan canonical form of 3 2 01 2 71 -2 3

.6. Find the Jordan canonical form of 1 0 0

0 0 -20 1 3

.7. Find the Jordan canonical form of matrices

A =

0 -4 851 4 -300 0 3

, B =

2 2 10 2 -10 0 3

and determine if A and B are similar.

8. Determine the Jordan canonical form for the n×n matrix over Q whose entries are all 1.

9. Let A be the 2× 2 matrix which corresponds to the rotation of the complex plane by 2π5 . Find the

Jordan canonical form of A. Explain in terms of complex numbers.

10. Let A be the 2×2 matrix which corresponds to the transformation of the complex plane T(z) = 1z . Find

the Jordan canonical form of A. Explain in terms of complex numbers.

Programming exercises:

1) Write a computer program which finds the Jordan canonical form of a given matrix A.

1. Find the rational canonical form of the 5 by 5 matrix A with characteristic polynomial

char (A,x) = x5 + 2x4−12x3 + 4x2

−6x + 10

2. Let A be a n by n matrix which has n distinct eigenvaluesλ1, . . . ,λn. Find the Jordan canonical form of A.

3. The characteristic polynomial of a 3 by 3 matrix A is

char (A,x) = (x−1)2(x−2).

Find all possibilities for the rational and Jordan canonical form of A.

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14 Shaska T.

4. Determine if the matrices A and B are similar

A =

-1 1 0 00 -1 0 00 0 -2 00 0 0 -2

, B =

-1 1 0 00 -1 0 00 0 -2 10 0 0 -2

5. Diagonalize the matrix or explain why it can’t be diagonalized.

A =

3 1 0 -14 0 0 3-4 2 2 -32 -4 0 7

6. Diagonalize the matrix or explain why it can’t be diagonalized.

A =

7 -1 0 2

-10 4 0 -45 -1 2 2

-15 3 0 -4

7. Let A be an n×n nilpotent matrix. Show that An = 0.

8. Let A be a strictly upper triangular matrix (all entries on the main diagonal and below are 0). Provethat A is nilpotent.

9. Let A be the 2× 2 matrix which corresponds to the rotation of the complex plane by 2πn . Find the

Jordan canonical form of A. Explain in terms of complex numbers.

10. Determine the set of similarity classes of 3×3 matrices A, over C, which satisfy A3 = 1.

11. Determine the set of similarity classes of 3×3 matrices A, over C, which satisfy A6 = 1.

12. Determine the set of similarity classes of 6×6 matrices A, over C, with characteristic polynomial:

char (A,x) = (x4−1)(x2

−1).

252

Part III

The theory of fields

253

Chapter 15

Field theory

15.1 Introduction to fields

Throughout this part a field is a commutative division ring. If we say that a field k is a subset of some ringR is is understood that it is a subring of R. First we review some basic properties of fields.

15.1.1 Characteristic of rings

Let R be a a commutative ring with identity. As usual, for a positive integer n, and any r ∈ R, we denote bythe symbol nr the following sum

nr := r + r + · · ·+ r

and for a negative n, we let nr := −(−n)r.The characteristic of R, denoted by char (R), is defined to be the smallest positive integer n ∈Z, such

thatnx = 0, for all x ∈ R

If no such integer exists then we say that the ring has characteristic zero.

Lemma 15.1. The characteristic of an integral domain R is either zero or a prime p.

Proof. Let n be the characteristic of R. Then, for all α ∈ R we have n ·α = 0. Assume that n is composite, sayn = ab. Then

n ·1 = (ab) ·1 = (a ·1) · (b ·1) = 0

Since R is an integral domain, then aα = 0 or bα = 0 where both a and b are smaller than n. Hence, thesmallest such integer is a prime.

The mapping σ : R→ R defined as follows

x→ xp

is called the Frobenius map.

Theorem 15.1. Let A be a commutative ring with prime characteristic p> 0. The mapping x→ xp is an endomorphismof A. In other words, we have the relations

i) : (a + b)p = ap + bp

ii) : (ab)p = ap bp,

for all a,b ∈ A.

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15 Shaska T.

Proof. The proof of the second part follows from commutativity. To prove the first part we use thebinomial formula and the fact that all coefficients

∑p−1i=1 are divisible by p. Hence, σ : R→ R is a ring

homomorphism. A ring A of characteristic p > 0 is called perfect if it is commutative and the Frobenius map

σ : x→ xp

is bijective.

15.1.2 Prime fields

The above results are valid for all commutative rings with identity; therefore, also for fields. We treat fieldsin more detail.

Let F be a field with char (F) = p and consider the homomorphism

ϕ :Z→ Fn→ n ·1F

Then, ker(ϕ) = pZ. From the First Isomorphism Theorem Theorem 9.4 we have that

Img (ϕ)Z/kerϕ =Z/pZ

Thus, F contains and isomorphic copy of Zp :=Z/pZ. This motivates the following definition:

Definition 15.1. The prime subfield of a field F is a subfield of F generated by 1F.

Now we summarize the previous discussion in the following theorem.

Theorem 15.2. Let F be a field. If char F = 0, the prime subfield of F is isomorphic to Q. If char F = p > 0, the primesubfield is isomorphic to Zp.

Proof. If char (F) = 0, then the mapϕ :Z→ F has kerϕ= 0 and therefore is injective. In this case, F containsan isomorphic copy of Z. Hence, it contains the field of fractions of Z as a subfield. Thus, F contains anisomorphic copy of Q.

If char F = p > 0, then ϕ(1) = 1F. Hence, 1 f ∈ Img (ϕ)Zp and Zp contains no proper subfields. Hence,the prime subfield of F is Img (ϕ). Hence, the prime subfield of F is isomorphic to either Q or Zp.

Exercise 15.1. Let F be a field. Then, there exists a unique subfield of F which is a prime field, and this is the leastsubfield of F.

Exercise 15.2. For a field to be prime it is necessary and sufficient that it contains no subfield other then itself.

15.1.3 Perfect fields

Let F be a field. The characteristic exponent of F is defined to be the integer

q =

1 if char F = 0p if char F = p > 0

Lemma 15.2. Let F be a field of characteristic exponent q. For every integer n ≥ 0 the mapping

ϕ :F→ F

x→ xqn

is an isomorphism of F on one of its subfields (denoted by Fqn).

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Proof. In the case q = 1 then this statement is trivial. If q, 1 then this is simply an extension of Theorem 15.1.Indeed,

(a + b)pn= apn

+ · · ·+

n−1∑i=2

(pn

i

)· aibpn

−i + · · ·bpn

All(pn

i)

are divisible by p. Hence(a + b)pn

= apn+ bpn

The formula(ab)pn

= apn· bpn

is easy to prove. A field F with characteristic exponent q is called perfect if Fq = F, otherwise is called imperfect.

Proposition 15.1. If F is a field of characteristic 0, or if F is a finite field, then F is perfect. In particular prime fieldsare perfect fields.

Proof. If F has characteristic 0 then q = 1 and obviously F1 = F. If F has characteristic p > 0 and is finite thenthe subfield Fp of F has the same cardinal as F. Thus, Fp = F and F is perfect. Since a prime field either hascharacteristic zero or is finite then prime fields are perfect.

Imperfect fields do exists.

Example 15.1. Give an example of an imperfect field.

Solution: Let F be a field of characteristic p > 0 and F(x) be the field of rational functions over F in the

indeterminate x. There exists no element f (x)g(x) in F(x) such that(

f (x)g(x)

)p

= x.

Indeed, if this were true then(

f (x))p = x

(g(x)

)p. Therefore, for r = deg f and s = deg g we have

rp = 1 + sp

which is impossible. Thus, F(x) is imperfect.

Exercises:

15.1. Let F be a field and p(x) ∈ F[x]. Show that F[x]/〈p(x)〉 is a field if and only if p(x) is irreducible in F[x].

15.2. Let F be a finite field of characteristic p > 0. Prove that |F| = pn, for some positive integer n.

15.3. Find the characteristic of Z, Q, R.

15.4. Find the characteristic of fields Fp :=Z/pZ.

15.5. In a field F with characteristic p > 0, prove the formula: for every two elements x, y ∈ F,

(x + y)p = xp + yp.

15.6. Prove that the characteristic subfield of Q and R is Q.

15.7. Prove that the characteristic subfield of Fp(x) is isomorphic to Fp.

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15 Shaska T.

15.8. Let F =Q(√

2). Prove thatσ : a + b

√

2 7→ a− b√

2,

is an automorphism of F.

15.9. Let C = a + bi | a,b ∈ R be the set of complex numbers. Prove directly the C/R is a field extension. Letφ : C 7→ C be defined as

a + bi 7→ a− bi

This is called the conjugation map. Prove that the conjugation map is an automorphism of C.

15.2 Field extensions

Let K, L be fields. If K ⊂ L then L is called a field extension of K, denoted by L/K. We already have seenexamples of field extensions.

Example 15.2. Let be given

F =Q(√

2) = a + b√

2 : a,b ∈Q, and E =Q(√

2 +√

3),

as the smallest fields which contain Q and√

2 and Q and√

2 +√

3.Both E and F are extension of Q. Notice also that E is an extension of F. For this it is enough to prove that

√2 is

in E. Since√

2 +√

3 is in E, then1

(√

2 +√

3)=√

3−√

2

must be also in E. The linear combination of√

2 +√

3 and√

3−√

2 gives that√

2 and√

3 must also be in E.

Exercise 15.3. Let F/K be a field extension. Show that F is a vector space over K, where the vector addition andscalar multiplication are the addition and multiplication of the field.

Let be given the extension L/K. The degree of the extension, which isdenoted by [L : K], is called the dimension of the vector space L over K. In otherwords, the cardinality of the basis of L over K. If [L : K] is finite (resp., infinite)then the field extension is finite (resp., infinite) extension. It is very useful toalso use diagrams to denote such extensions.

L

K

A ring homomorphism among fields is called a field homomorphism. The following result is obvious.

Lemma 15.3. Let be given a field homomorphism

ϕ : F −→ K

Then, ϕ is 0 or is injective. Thus, ϕ = 0 or F ϕ(F) ⊂ K.

Proof. We of we know that ker(ϕ) is an ideal of F. However F has as ideals only 0 and F. If ker(ϕ)= 0, then

ϕ is injective. If n ker(ϕ) = F, then ϕ = 0.

A field homomorphism σ : F→ F which is bijective is called an automorphism of F. The set of allautomorphisms of F is denoted by Aut (F).

Example 15.3. For any given field F prove that Aut (F) is a group.

Lemma 15.4. Let L/K/F be field extensions. Then,

[L : F] = [L : K] [K : F]

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Proof. Let B1 := αi | i ∈ A and B2 := β j | j ∈ B be the basis of K/F and L/K respectively. We denote by

B = αi β j | i ∈ A, j ∈ B.

It is enough to show that B is a basis of L over F.Let

∑fi, j (αi β j) = 0. Since β j’s are independent then fi, jαi = 0 for all i, j. Hence,

∑fi, j ai = 0. Since ai’s are

independent then fi, j = 0 for all i, j. Thus, B is independent.Let x ∈ L. Then x =

∑k j β j, where k j ∈ K for all j ∈ B. Since each k j is a linear combination of the αi’s, it

follows that x is a linear combination of αi β j.

15.2.1 Composite extensions and distinguished classes

Let E/K and F/K be field extensions. The intersection E∩F is a field and K ⊂ E∩F.Assume that both E and F are subfield of a larger field L. The smallest field in L containing E and F is

called the compositum of E and F and denoted by EF. Sometimes we say that EF/K is the lifting of E/K byF.

A tower of field extensions is called the following

F1 ⊂ F2 ⊂ · · · ⊂ Fi ⊂ . . .

Each Fi is called an intermediate field of the Fi+1/Fi1 extension.

Let C be a given class of field extensions F ⊂ E. We will call Ca distinguished class of extensions if the following properties aresatisfied:

i) Tower property: For every tower k ⊂ F ⊂ E we have: k ⊂ E is inC if and only if k ⊂ F is in C and F ⊂ E is in C.

ii) Lifting: C is closed under lifting: if k ⊂ E is in C and F is anyextension of k, and E,F are both contained in some larger field, thenF ⊂ EF is in C

iii) Composite: If k ⊂ F and k ⊂ E are in C. and E,F are subfieldsof a common field then k ⊂ EF is in C.

L

EF

E F

E∩F

K

In terms of diagrams we illustrate the above three properties in Fig. 15.1. Solid lines mean that thecorresponding extension belongs to C.

E

F

k

EF

F

E

k

EF

E F

k

Figure 15.1: Distinguished classes of extensions

Notice that property iii) follows from the first two conditions. EF/F is in C because of the ii) property.Then, F/k is in C and EF/F is in C implies that EF/k is in C by the first property.

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15 Shaska T.

Next we continue with some basic definitions about fields and field extensions.Let F/k be a field extension and α ∈ F. If there is a polynomial f (x) ∈ k[x] such that f (α) = 0 then α is

called an algebraic element over K. An element which is not algebraic is called transcendental.

Example 15.4. The number√

2 is algebraic over Q because is a root of the polynomial

x2−2 ∈Q[x].

The number π is a transcendental number over Q.

An extension F/k is called an algebraic extension if every element α ∈ F is algebraic over k. Otherwise,F/k is called a transcendental extension.

We apply some of the results above when the base field is Q. A complex number is called an algebraicnumber if it is algebraic over Q. A number that is not algebraic is called transcendental.

Remark 15.1. Transcendental numbers exist.

An algebraic number α is said to be an algebraic integer if it satisfies an equation of the form

αn + an−1αn−1 + · · ·+ a1α+ a0 = 0

where a0, . . . ,an−1 are integers.Let S ⊂ F be a finite set. By k(S) we denote the smallest subfield of F containing both S and k. If

S = α1, . . . ,αn, we usually write k(α1, . . . ,αn).The extension F/k is called a finitely generated extension if F = k(S) for some finite set S. If F = k(α) then

F/k is called a simple extension and α a primitive element.In the remaining chapters we will prove that the following classes of extensions are distinguished:

• finite extensions

• finitely generated extensions

• algebraic extensions

• separable extensions

and these classes are not distinguished

• simple extensions

• transcendental extensions

• normal extensions

Exercises:

15.10. Let C be a distinguished class of extensions. Prove that C is closed under taking a finite number of composites.

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15.3 Finitely generated and finite extensions

Let L be an extension of F and X a subset of L. The ring F[X], generated by F and X, is the intersection ofall subrings of L that contain F and X. The field F(X), generated by F and X, is the intersection of all thesubfields of L that contain F and X.

If X = α1, . . . ,αn, we writeF[X] = F[α1, . . . ,αn]

andF(X) = F(α1, . . . ,αn).

Proposition 15.2. Let L be a field extension of F and α ∈ L. Then,

F[α] =f (α) | f (x) ∈ F[x]

and

F(α) =

f (α)g(α)

∣∣∣∣∣ f , g ∈ F[x], g(α) , 0

Moreover, F(α) is the quotient field of F[a].

Proof. Homework In general we have

F ⊂ F(α1) ⊂ F(α1,α2) ⊂ · · · ⊂ F(α1, . . . ,αn)

It is clear that

Proposition 15.3. Let L be a field extension of F and α1, . . . ,αn ∈ L. Then

F[α1, . . . ,αn] =f (α1, . . . ,αn) | f ∈ F[x1, . . . ,xn]

and

F(α1, . . . ,αn) =

f (α1, . . . ,αn)g(α1, . . . ,αn)

∣∣∣∣∣ f , g ∈ F[x1, . . . ,xn], g(α1, . . . ,αn) , 0

Moreover, F(α1, . . . ,αn) is the quotient field of F[α1, . . . ,αn].

Proof. Homework

Theorem 15.3. The class of all finitely generated extensions is distinguished.

Proof. Let X and T be finite sets. Since,F ⊂ F(X) ⊂ F(X)(T)

then every step is finitely generated because F(X)(T) = F(X∪T).To show that for every tower F ⊂ K ⊂ F(X) the extension K/F is finitely generated we will discuss it in

the section of transcendental extension.To prove the lifting property, let E = F(X), where X is a finite set. If F ⊂ K, with EK defined, then

EK = K(F(X)) = K(X)

Hence, the compositum EK is finitely generated over K by X.

Theorem 15.4. An extension is finite if and only if it is generated by algebraic elements.

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15 Shaska T.

Proof. Let E/F be a finite extension and α1, . . . ,αn a basis for E over F. Then, for each αi the sequence ofpowers

αi,α2i ,α

3i ,α

4i , . . .

is linearly dependent over F. Hence, αi is algebraic over F. Thus, E/F is algebraic.

Theorem 15.5. Finite extensions form a distinguished class.

Proof. Homework

Lemma 15.5. Let E/F and K/F be given and EK well defined. If B a basis of E over F, then B spans EK over K.Moreover,

[EK : K] ≤ [E : F]

Proof. Let B = β1, . . . ,βn be a basis for E over F. Then, EK = K(β1, . . . ,βn). From the Theorem 15.4, eachβi is algebraic over K. Hence, EK is the set of polynomial expressions in β1, . . . ,βn with coefficients in K.However, any monomial in βi’s is in E and therefore a linear combination of β1, . . . ,βn over F. Hence, B

EK

E K

F

spans EK over K. If an extension E/F is a vector space with finite dimensions n, then we say that E is a finite extension of

degree n over F. We write [E : F] = n.

Theorem 15.6. Every finite extension E/F is algebraic.

Proof. Let α ∈ E. Since [E : F] = n, elements1,α, . . . ,αn

are not linearly independent. Thus, there are ai ∈ F, not all zero such that

anαn + an−1α

n−1 + · · ·+ a1α+ a0 = 0.

Thus,p(x) = anxn + · · ·+ a0 ∈ F[x],

is a nonzero polynomial where p(α) = 0.

Theorem 15.7. If E is a finite extension of F and K is a finite extension of E, then K is a finite extension of F and

[K : F] = [K : E][E : F].

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Proof. Let α1, . . . ,αn be a basis for E as vector space over F and β1, . . . ,βm a basis forK as vector space over E. We will show that αiβ j is basis for K over F.

First we will prove that these vectors generate K. Let u ∈ K. Then, u =∑m

j=1 b jβ j

and b j =∑n

i=1 ai jαi, where b j ∈ E and ai j ∈ F. Therefore,

u =

m∑j=1

n∑i=1

ai jαi

β j =∑i, j

ai j(αiβ j).

Thus, mn vectors αiβ j must generate K over F.

K

m

E

n

F

Figure 15.2: Finite ex-tensions

Next we must prove that αiβ j are linearly independent. Let

u =∑i, j

ci j(αiβ j) = 0,

for ci j ∈ F. We must of to show that all ci j are zero. We can rewrite u as

m∑j=1

n∑i=1

ci jαi

β j = 0,

where∑

i ci jαi ∈ E. Since β j are linearly independent in E, then

n∑i=1

ci jαi = 0

for all j. However, α j are also linearly independent over F. Thus, ci j = 0 for every i and j, which completesthe proof.

Corollary 15.1. If Fi is field, where i = 1, . . . ,k and Fi+1 is a finite extension of Fi, then Fk is a finite extension of F1and

[Fk : F1] = [Fk : Fk−1] · · · [F2 : F1].

Corollary 15.2. Let E/F be an extension. If α ∈ E is algebraic over F with polynomial minimal p(x) and β ∈ F(α)with minimal polynomial q(x), then degq(x) divides degp(x).

Proof. We know that degp(x) = [F(α) : F] and degq(x) = [F(β) : F]. Since F ⊂ F(β) ⊂ F(α) we have

[F(α) : F] = [F(α) : F(β)][F(β) : F].

Example 15.5. Find a basis for Q(√

3 +√

5)/Q.

Proof. We start with

[Q(√

3 +√

5) :Q] = 4.

We know that 1,√

3 is basis for Q(√

3) over Q. Thus,√

3 +√

5 cannot be in Q(

√3), since

√5 is not in Q(

√3). Thus, 1,

√5 is a basis for

Q(√

3,√

5) =Q(√

3)(√

5),

over Q(√

3) and1,√

3,√

5,√

3√

5 =√

15

is basis for Q(√

3,√

5) =Q(√

3 +√

5) over Q.

Q(√

3 +√

5)2

2

Q(√

3)2

Q(√

5)

2

Q

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15 Shaska T.

This example shows that it is possible for some extensions F(α1, . . . ,αn) to be a simple extensions of Feven though n > 1.

Example 15.6. Find a basis for Q( 3√5,√

5 i), where√

5 is the positive square root of 5 and 3√5 is the real cubic rootof 5.

Proof. We know that√

5i <Q( 3√5). Thus,

[Q(3√

5,√

5 i) :Q(3√

5)] = 2.

It is easy to show that 1,√

5i is basis for Q( 3√5,√

5 i) over Q( 3√5).We know that 1, 3√5, ( 3√5)2

is basis for Q( 3√5) over Q. A basis for Q(√

5, 3√5) over Q is

1,√

5 i,3√

5, (3√

5)2, (6√

5)5i, (6√

5)7i = 56√

5 i or6√

5 i.

Notice that 6√5 i is root of x6 +5. We can show that this polynomial is irreducible over Q, using Eisenstein’scriteria for p = 5. Therefore,

Q ⊂Q(6√

5) ⊂Q(3√

5,√

5 i).

However, it can’t happen that Q( 6√5 i) =Q( 3√5,√

5 i) since degree of extension is 6.

Theorem 15.8. Let E a extension field of F. Then, the following are equivalent.i) E is a finite extension of F.ii) There is a finite number of algebraic elements α1, . . . ,αn ∈ E, such that E = F(α1, . . . ,αn).iii) There is a tower of fields

E = F(α1, . . . ,αn) ⊃ F(α1, . . . ,αn−1) ⊃ · · · ⊃ F(α1) ⊃ F,

where each field F(α1, . . . ,αi) is algebraic over F(α1, . . . ,αi−1).

Proof. (1)⇒ (2). Let E/F be a finite extension. Then, E is vector space of finite dimension over F. Hence,there is a basis with elements α1, . . . ,αn in E such that E = F(α1, . . . ,αn). Every αi is algebraic over F fromTheorem 15.6.

(2) ⇒ (3). Assume that E = F(α1, . . . ,αn) where every αi is algebraic over F.Then,

E = F(α1, . . . ,αn) ⊃ F(α1, . . . ,αn−1) ⊃ · · · ⊃ F(α1) ⊃ F,

where every field F(α1, . . . ,αi) is algebraic over F(α1, . . . ,αi−1).

(3)⇒ (1). Let be given

E = F(α1, . . . ,αn) ⊃ F(α1, . . . ,αn−1) ⊃ · · · ⊃ F(α1) ⊃ F,

where every field F(α1, . . . ,αi) is algebraic over F(α1, . . . ,αi−1). Since

F(α1, . . . ,αi) = F(α1, . . . ,αi−1)(αi)

is a simple extension and αi are algebraic over F(α1, . . . ,αi−1) then we have that

[F(α1, . . . ,αi) : F(α1, . . . ,αi−1)]

is finite for every i. Thus [E : F] is finite.

E = F(α1, . . . ,αn)

F(α1, . . . ,αn−1)

...

F(α1)

F

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Exercises:

15.11. Let L/F and K/F be field extensions with degrees p and q respectively. Assume that (p,q) = 1. What is thedegree of LK/F? What about L∩K/F?

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15 Shaska T.

15.4 Simple extensions

An extension F(α)/F is called simple. Simple extensions could be algebraic or transcendental. In thissection we will briefly describe some of their properties. They both will be discussed in more detail whenwe study algebraic and transcendental extensions in coming chapters.

Simple extensions are non a distinguished class. For example, take x and y as independent variables.Then,

F ⊂ F(x) ⊂ F(x)(y) = F(x, y),

each step of the tower is simple, but F(x, y)/F is not.

15.4.1 Simple algebraic extensions

Let L/K be a field extension and α ∈ L. If there is a polynomial f (x) ∈ K[x] such that f (α) = 0 then α is calledan algebraic element over K. The ideal

Iα = h(x) ∈ K[x] | h(α) = 0

is principal in K[x] and generated by a unique monic polynomial which is called the minimal polynomialof α over K and denoted by min (α, K,x).

Lemma 15.6. Let L/K, α ∈ L be algebraic over K. Then, p(x) = min (α,K,x) is the unique monic irreduciblepolynomial such that p(α) = 0.

Proof. Suppose p(x) can be written asp(x) = f (x) g(x).

Then f (α) = 0 or g(α) = 0. Say, f (α) = 0. Then, Iα is not generated from p(x), which is a contradiction. An algebraic extension is a field extension L/K such that every element of L is algebraic over K.

Lemma 15.7. Let L be a finite extension of K. Then L/K is an algebraic extension.

Proof. Let α ∈ L. Since L/K is finite then the powers 1,α,α2, . . . cannot be algebraically independent. So itexists a polynomial for which α is a root. Hence, α is algebraic over K.

Let L/K be a field extension and [L : K] = n. Let

B = α1, . . . ,αn

be a basis of L over K. We denote this by L = K(α1, . . . ,αn). Any extension of the form L = K(α) is called asimple extension and α is called a primitive element.

Lemma 15.8. Let L/K be a field extension and α ∈ L be algebraic over K. Then,

K(α)K[x]/〈min (α,K,x)〉.

Proof. From previous lemma, the polynomial p(x) := min (α,K,x) is irreducible. Hence the ideal Iα =〈min (α,K,x)〉 is maximal and therefore

K[x]/〈min (α,K,x)〉,

is a field. We define the map

ψ : K[x]→ Lf (x)→ f (α)

(15.1)

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Shaska T. 15

The kernel of ψ is the ideal Iα = 〈p(x)〉. Thus,

ψ(K[x])K[x]/Iα

which implies that ψ(K[x]) is a field. It is left to show that ψ(K[x]) = K(α). First, ψ(K[x]) ⊂ K(a) sinceevery y ∈ ψ(K[x]) is y = h(a) ∈ K(a), for some h(x) ∈ K[x]. Secondly, K ⊂ ψ(K[x]) and ψ(x) = a ∈ ψ(K[x]). Thiscompletes the proof.

Thus, K(α) is the set of all expressions in α of degree

d < deg(min (α,K,x)).

Hence, the basis of K(α) over K is1,α, . . . ,αn−1

where n = deg(min (α,K,x)). Moreover

[K(α) : K] = deg(min (α,K,x)).

Thus, we have the following,

Lemma 15.9. Let K(α)/K be an algebraic extension. Then

[K(α) : K] = deg(min (α,K,x)).

Theorem 15.9 (Primitive Element Theorem). Let K be a field and L/K a finite extension. Then, L = K(α) if andonly if there are only finitely many intermediate fields.

Proof. Assume that L = K(α) and F an intermediate field.

L = K(α)

F

E = K(a0, . . . ,an)

K

Let p(x) = min (α,K,x) and pF(x) = min (α,F,x). Since p(x) can be considered as a polynomial in F[x] thenpF(x) | p(x). But p(x) is monic and F[x] is a UFD, so p(x) has finitely divisors. So it is left to show that everydivisor of p(x) determines only one intermediate field. For a fixed pF(x) given as

pF(x) = anxn + · · ·+ a0,

let E := K(a0, . . . ,an). Then, pF(x) is irreducible in E(x) and is satisfied by α. Hence,

[L : E] = deg pF(x) = [L : F]

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15 Shaska T.

which shows that F = E.For the converse we first deal with the case char (L) = 0. Let α,β ∈ L. We will show that there is a w ∈ L

such that K(α,β) = K(w). Consider all fields of type K(α+ mβ) for m ∈ K. They are all intermediate fields ofthe extension L/K. Since we have only finitely of them, there is m1,m2 ∈ K such that

K(α+ m1β) = K(α+ m2β).

Since α+ m1β and α+ m2β are in the same field then (m1−m2)β ∈ K(α+ m1β) = K(α+ m2β). In the same way,α ∈ K(α+ m1β) = K(α+ m2β). Hence, K(α,β) = K(α+ m1β).

When char (L) = p > 0 then L? is cyclic. Say L? = 〈α〉. Then L = K(α).

15.4.2 Simple transcendental extensions

Consider now a simple extension F(t)/F where t is transcendental over F. Then F(t) is the field of all rationalexpressions

F(t) =

f (t)g(t)

∣∣∣∣∣ f , g ∈ F[x], g(t) , 0

Theorem 15.10. Let L/F and t ∈ L be transcendental over F. Then F(t) is isomorphic to the field of all rationalfunctions F(x) in a single variable x.

L

F(x)φ // F(t)

F

Proof. The evaluation homomorphismφ : F(x)→ F(t)

defined by

φ

(f (x)g(x)

)=

f (t)g(t)

is an isomorphism.Indeed, f (x)

g(x) =f (t)g(t) = 0 implies that f (x) ≡ 0 polynomial, otherwise t would be the root of a polynomial

and therefore algebraic. Hence, kerφ = 0 and φ is injective. Obviously, φ is surjective.

Exercise 15.4. Prove that simple transcendental extensions do not form a distinguished class.

Theorem 15.11. Let F(t)/F be a transcendental extension and s ∈ F(t)\F. Then the following hold:i) F(s)/F is transcendentalii) F(t)/F(s) is algebraic and

[F(t) : F(s)] = max deg f ,deg g

iii) F(t) is algebraic over any intermediate field K other than F itself.

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Shaska T. 15

Proof. Since s ∈ F(t)\F means that s is a non-constant rational function in t, say

s =f (t)g(t)

,

where we can assume that f (t) and g(t) are co-prime. Then t is algebraic over F(s) because it satisfies thepolynomial

p(x) = g(x)s− f (x) ∈ F(s)[x]

From Theorem 15.4 it is also finite.

F(t)

F(s) = F( f (t)

g(t)

)

F

If F(s)/F is algebraic then it is a finite extension; see Theorem 15.4. Then by multiplicative property ofextensions we have F(t)/F is a finite extension. However, this is a contradiction since t is transcendental.

To determine the degree [F(t) : F(s)] we need to find the min (t,F(s), y). By Lemma 15.7,

[F(t) : F(s)] = degmin (t,F(s), y).

Let p(y) := f (y)− s · g(y). We will show that p(y) = min (t,F(s), y).Obviously, t is a root of p(y) since p(t) = f (t)− s g(t) = 0. By Gauss’ Lemma p(y) is irreducible in F(s)[y] if

and only if it is irreducible in (F[s])[y]. However, (F[s])[y] = (F[y])[s] = F[y,s]. So it is enough to show thatp(y) is irreducible over F[s, y]. Suppose not, then

p(y,s) = a(y)(b(y) · s + c(y)

),

where a(y),b(y),c(y) ore in F[y]. But, f (y) and g(y) are co-prime. Hence, a(y) must be a unit in F[y], whichimplies that p(y,s) is irreducible in F[y,s].

The above theorem motivates the following definition. The degree of a rational function

s =f (t)g(t)

,

where where f (t) and g(t) are co-prime is defined as follows

degs = max deg f ,deg g

Exercises:

15.12. Let f (x), g(x) be irreducible in F(x) and(deg f ,deg g

)= 1. Let α be a root of g(x) and L = F(α). Is f (x)

necessarily irreducible in L[x]?

15.13. Show that Q(√

2,√

3) =Q(√

2 +√

3).

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15 Shaska T.

15.14. Determine the minimal polynomial of the following extensions: Q(√

2)/Q,Q(√

3)/Q),Q(√

5)/Q,Q(√

2,√

3)/Q.

15.15. Determine the minimal polynomial of the following extensions: Q(√

2 +√

3)/Q(√

2), Q(√

2 +√

3)/Q(√

3),Q(√

5)/Q(√

2 +√

3).

15.16. Let F/Q be a field extension such that [F : Q] = 2. Show that F =Q(√

d), where d is an integer not divisiblebe the square of any prime.

15.17. Let a and b be rational numbers. Show that Q(√

a,√

b) =Q(√

a +√

b).

15.18. Let p be a prime integer and εp be a p-th primitive root of unity. Find [Q(εp) :Q].

15.19. Let F = k(α) where α is algebraic over k and [F : k] is odd. Show that F = k(α2).

15.20. If α,β are algebraic over k then α±β, αβ, and αβ for β , 0) are also algebraic over k.

15.21. Find an element α such that Q(√

2, 3√5) =Q(α).

15.22. Let α ∈Q and α is an algebraic integer. Prove that α ∈Z.

15.23. If α and β are algebraic integers satisfying α3 +α+ 1 = 0 and β2 + β− 3 = 0, then both α+ β and αβ arealgebraic integers.

15.24. Prove that:i) The sum of two algebraic integers is an algebraic integer.ii) The product of two algebraic integers is an algebraic integer.

15.25. For any integer m prove that sin m is an algebraic number.

15.26. Let m be an integer which is not a square and α+ β√

m be a root of some polynomial f (x) ∈ Q[x], whereα,β ∈Q. Show that α−β

√m is also a root of f (x).

15.27. Find the minimal polynomial of√

5 +√

3,√

5 +√

7 over Q.

15.28. Design an algorithm that computes the minimal polynomial of√

a +√

b over Q.

15.29. Find an extension which is algebraic, but not finite.

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15.5 Finite fields

A field is called finite when has number finite elements.

Lemma 15.10. Let be given a finite field F and a E/F a finite extension such that [E : F] = n. Then, |E| = |F|n.

Proof. Let α1, . . . ,αnbe a basis of E over F. Then, every element of E can be expressed as a linear combination

a1α1 + · · ·+ anαn

where ai ∈ F. We have |F| possibilities for ai and n possibilities for αi. Hence, in total |F|n possibilities.

Lemma 15.11. The order of a finite field F is pn, where p = char F and n any positive integer.

Proof. The characteristic of fields is a prime number p or 0. Every field with characteristic 0 is isomorphicto Q, so is infinite. Thus, characteristic of F is p > 0. Take the field homomorphism

ϕ :Z/pZ −→ F

1 −→ e f

This homomorphism is 0 or injective. Since 1 −→ eF then ϕ , 0, so it is injective. Thus, F has an isomorphiccopy of Z/pZ. Since F is finite then it is a finite extension of Z/pZ. Assume that [F :Z/pZ] = n. Then fromthe above lemma |F| = |Z/pZ| = pn.

Finite fields usually are denoted with Fq, where q = pn for p a prime number and n ∈Z. Next we will

study the existence of fields with pn elements.

Theorem 15.12. For every prime number p and a positive integer n ∈Z there is a field with q = pn elements. Allfields with pn elements are isomorphic.

Proof. To prove the first part of the theorem consider the polynomial

f (x) = xq−x

Let S be the splitting field of the polynomial f (x) over field Fp :=Z/pZ. Let R be the set of roots of f (x) inS. The set R is a field because contains 0 and 1 and

α ∈ R⇒ αq−α = 0⇒ αq = α

α,β ∈ R⇒ (α+β) = αq +βq = (α+β)q

because q is multiple of p. Also,α ∈ R⇒−α ∈ R

α,β ∈ R⇒ αβ−1 = αq(β−1)q = (αβ−1)q⇒ αβ−1

∈ R

Since S is the smallest field that contains roots of f (x) and R ⊂ S, then R = S. Hence S is the set of roots off (x). To prove that f (x) does not have double root we take the derivative of f (x)

f ′(x) = qxq−1−1 = pnxq−1

−1 = −1 , 0.

Since f ′(x) , 0 then we have no double root. Hence all factors of f (x) are linear. Therefore |S| = q becausedegree of f (x) is q.

To prove the second part recall that for every finite field F, F∗ = F \ 0 is a cyclic group. Let F be anotherfield, distinct from S, such that |F| = q. Then, ∀α ∈ F we have αq−1 = 1⇒ αq

−α = 0.Hence every α ∈ F is root of f (x) = xq

−x overZ/pZ. Since |F| = q and f (x) has q roots (all distinct) then Fis splitting field of f (x). We we know that every two splitting fields of a polynomial are isomorphic. Thus,S F.

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15 Shaska T.

Example 15.7. Let p some prime number and D a integral ring with characteristic p. Then,

apn+ bpn

= (a + b)pn

for all integers n.

Proof. We use induction on n. By the binomial formula for the case when n = 1 we have

(a + b)p =

p∑k=0

(pk

)akbp−k.

If 0 < k < p, then (pk

)=

p!k!(p− k)!

is divisible by p,since p does not divide k!(p− k)!. We know that D is a integral ring with characteristic p,hence all terms other than the first and the last are zero. Thus, (a + b)p = ap + bp.

Assume that the result holds for all 1 ≤ k ≤ n. By induction hypothesis

(a + b)pn+1= ((a + b)p)pn

= (ap + bp)pn= (ap)pn

+ (bp)pn= apn+1

+ bpn+1.

Thus, the result holds for n + 1.

The unique finite field with q = pn elements is called Galois field with order q = pn. This field is denotedby GF(pn) or Fq.

Theorem 15.13. Every subfield of GF(pn), has q = pm elements, where M divides n. Conversely, if m | n for m > 0,then there exists a unique field of GF(pn) isomorphic to GF(pm).

Proof. Let F a subfield of E = GF(pn). Then, F must be a extension field of K that contains pm elements,where K is isomorphic to Zp. Then, m | n, since [E : K] = [E : F][F : K].

To prove the converse assume that m | n for some m > 0. Then, pm− 1 divides pn

− 1. Thus, xpm−1− 1

divides xpn−1−1. Thus, xpm

−x must divide xpn−x, and every zero of xpm

−x is also a zero of xpn−x. Thus,

GF(pn) contains as a subfield, a splitting field of xpm−x, which is isomorphic to GF(pm).

Example 15.8. The lattice of subfields of GF(p24) is given in Fig. 15.3.

For every field F we have a group of nonzero elements of F which we denote by F∗ and call it themultiplicative group of F. The multiplicative group of every field is a cyclic group.

Theorem 15.14. If G is a finite subgroup group of F∗, then G is cyclic.

Proof. Let G a finite subgroup of F∗ with n = pe11 · · ·p

ekk elements, where pi are primes. From the Fundamental

Theorem of Abelian Groups,G Zp

e11× · · ·×Zp

ekk.

Let m be the least common multiple of pe11 , . . . ,p

ekk . Then, G has an element with order m. Since every α ∈ G

satisfies xr− 1 for some r that divides m, α is a root of xm

− 1. However, xm− 1 has at most m roots in F,

n ≤m. Since m ≤ |G|, then m = n. Thus, G contains an element with order n and hence is cyclic.

Corollary 15.3. The multiplicative group of all nonzero elements of a finite field is cyclic.

Corollary 15.4. Every finite extension E of a finite field F is a simple extension of F.

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Shaska T. 15

GF(p24 )

GF(p8 ) GF(p12 )

GF(p4 ) GF(p6 )

GF(p2 ) GF(p3 )

GF(p )

""""

""""

""

bb

bb

""

Figure 15.3: Subfields of GF(p24)

Proof. Let α generate the group cyclic E∗ of nonzero elements of E. Then, E = F(α).

Example 15.9. The finite field GF(24) is isomorphic with the fieldZ2/〈1+x+x4〉. Thus, the elements of GF(24) can

be obtained asa0 + a1α+ a2α

2 + a3α3 : ai ∈Z2 and 1 +α+α4 = 0.

Proof. Recall that 1 +α+α4 = 0. We add and multiply the elements of GF(24) as we add and multiplypolynomials. The multiplicative group of GF(24) is isomorphic to Z15 with generator α :

α1 = α α6 = α2 +α3 α11 = α+α2 +α3

α2 = α2 α7 = 1 +α+α3 α12 = 1 +α+α2 +α3

α3 = α3 α8 = 1 +α2 α13 = 1 +α2 +α3

α4 = 1 +α α9 = α+α3 α14 = 1 +α3

α5 = α+α2 α10 = 1 +α+α2 α15 = 1.

Exercises:

15.30. Let f (x) a irreducible polynomial in k[x]. Prove that, the following are equivalent:i) char(k) = p > 0 and f (x) = g(xp), for some g(x) ∈ k[x].ii) All of f (x) are multipleNotice: In a finite field Fq, with characteristic p, for every element β ∈ Fq can be written as β = αp, for some

α ∈ Fq.

15.31. Let E be an algebraic extension of a field F, and let σ a automorphism of E, which fixes F. Let α ∈ E. Provethat, σ induces a permutation of the set se all roots of the minimal polynomial of α, which are in E.

15.32. Prove or disprove: Let be given a polynomial p(x) in Z6[x], can you construct a ring R, such that, p(x) has aroot in R.

15.33. Let F a field with characteristic p. Prove that, p(x) = xp− a or is irreducible over F or splits in F.

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15 Shaska T.

274

Chapter 16

Algebraic Closure

In this chapter we study in more detail algebraic extensions which will be our focus for the remainingchapters. We will develop one of the very important concepts of algebra, that of algebraic closure, and willprove the Fundamental Theorem of Algebra.

16.1 Algebraic extensions revisited

The following theorem, from Kronecker, often is known as the Fundamental Theorem of field theory.

Theorem 16.1 (Kronecker). Let F field and p(x) a non constant irreducible polynomial in F[x]. Then, there is anextension field E of F and an element α ∈ E such that p(α) = 0.

Proof. We want to find an extension field E for F which contains a element α, such that p(α) = 0. The ideal〈p(x)〉 generated from p(x) is a maximal ideal in F[x] because p(x) is irreducible. Thus, from Lemma 11.2F[x]/〈p(x)〉 is field. We claim that E = F[x]/〈p(x)〉 is the field as claimed in the theorem.

First, we prove that E is extension field for F. We define the map

ψ : F→ F[x]/〈 p(x) 〉a→ a + 〈p(x)〉

for a ∈ F.It is easy to check that ψ is well defined ring homomorphism. Notice that, for a,b ∈ F

ψ(a) +ψ(b) = (a + 〈p(x)〉) + (b + 〈p(x)〉) = (a + b) + 〈p(x)〉 = ψ(a + b)

andψ(a)ψ(b) = (a + 〈p(x)〉)(b + 〈p(x)〉) = ab + 〈p(x)〉 = ψ(ab).

To show that ψ is injective assume that

ψ(a) = ψ(b) =⇒ a + 〈p(x)〉 = b + 〈p(x)〉 =⇒ a− b ∈ 〈p(x)〉

Thus, a−b is a multiple of p(x), since it is contained in the ideal 〈p(x)〉. Since p(x) is a non constant polynomialthen a− b = 0.

Since ψ is injective we can identify F with the subfield ψ(F) of E. Then, E is an extension field of F.It remains to be shown that p(x) has a root α ∈ E. Take α = x + 〈p(x)〉, then α is in E. If

p(x) = a0 + a1x + · · ·+ anxn,

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16 Shaska T.

then,

p(α) = a0 + a1(x + 〈p(x)〉) + · · ·+ an(x + 〈p(x)〉)n

= a0 + (a1x + 〈p(x)〉) + · · ·+ (anxn + 〈p(x)〉)= (a0 + a1x + · · ·+ anxn) + 〈p(x)〉= 0 + 〈p(x)〉.

Thus, we found an element α ∈ E = F[x]/〈p(x)〉 such that α is root of p(x).

Example 16.1. Let be given the polynomial

p(x) = x2 + x + 1 ∈Z2[x].

Construct the extension of Z2 that contains a root of p(x).

Solution: Since 0 and 1 are not roots of this polynomial then p(x) is irreducible overZ2. Hence,Z2[x]/〈p(x)〉

is a field; see Lemma 11.2. Let f (x) + 〈p(x)〉 be an element of Z2[x]/〈p(x)〉. From the division algorithm wehave

f (x) = (x2 + x + 1)q(x) + r(x),

where degr(x) < 2. Hence, r(x) is one of the following polynomials 0, 1, x, and x + 1. Therefore, E =Z2[x]/〈x2 + x + 1〉 is field with four elements, namely

E :=0 + p(x),1 + p(x),x + p(x), (x + 1) + p(x)

From the above theorem, E must be an extension field for Z2 which contains a root α of p(x). Notice that

p(α) = α2 +α+ 1 = 0.

Hence, the field Z2(α) consists of elements

Z2(α) =0 + p(α),1 + p(α),α+ p(α), (α+ 1) + p(a)

= 0,1,α,α+ 1

Thus, if we compute (1 +α)2 we have,

(1 +α)(1 +α) = 1 +α+α+ (α)2 = α.

The other computations are done similarly. We summarize such computations in the following tables

+ 0 1 α 1 +α0 0 1 α 1 +α1 1 0 1 +α αα α 1 +α 0 1

1 +α 1 +α α 1 0

· 0 1 α 1 +α0 0 0 0 01 0 1 α 1 +αα 0 α 1 +α 1

1 +α 0 1 +α 1 α

Example 16.2. Let be givenp(x) = x5 + x4 + 1 ∈Z2[x].

Then, p(x) has irreducible factors x2 + x + 1 and x3 + x + 1. To construct the extension E of Z2 such that p(x) has aroot in E, take E to be,

Z2[x]/〈x2 + x + 1〉 or Z2[x]/〈x3 + x + 1〉.

In the previous example we considered Z2[x]/〈x2 + x + 1〉.It is left as an exercise for the reader to prove that Z2[x]/〈x3 +x+1〉 is field with 23 = 8 elements and list all these

elements.

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Shaska T. 16

An extension E/F is called an algebraic extension of F if every element in E is algebraic over F. IfE = F(α) for some α ∈ E, then E/F is called a simple extension.

Example 16.3. Prove that√

2 +√

3 is algebraic over Q.

Proof. If α =

√2 +√

3, then α2 = 2 +√

3. Thus, α2−2 =

√3 and (α2

−2)2 = 3. Since α4−4α2 + 1 = 0 is true se

α is a root of the polynomial x4−4x2 + 1 ∈Q[x].

The following Lemma characterizes the transcendental extensions.

Theorem 16.2. Let E/F be an extension and α ∈ E. Then, α is transcendental over F if and only if F(α) is isomorphicto F(x).

Proof. Define the evaluation map at α as

φα : F[x]→ Ef (x)→ f (α)

Then, α is transcendental over F if and only if

φα(p(x)) = p(α) , 0,

for all polynomials p(x) ∈ F[x]. This is true if and only if kerφα = 0. Thus, is true only when φα is injective,which implies that the field E must contain an isomorphic copy of F[x]. The smallest field that containsF[x] is the field of fractions F(x). Thus, E must contain an isomorphic copy of this field.

In the case of algebraic extensions we have.

Theorem 16.3. Let E/F be an extension and α ∈ E algebraic over F. There exists a unique monic irreduciblepolynomial p(x) ∈ F[x] of smallest degree such that p(α) = 0.

If f (x) is another monic polynomial in F[x] such that f (α) = 0, then p(x) divides f (x).

Proof. The proof goes similarly to that of the above theorem. Define the evaluation map at α as

φα : F[x]→ Ef (x)→ f (α)

Since F[x] is a PID then the kernel of φα is a principal ideal generated by some polynomial p(x) ∈ F[x], wheredegp(x) ≥ 1. The ideal 〈p(x)〉 consists exactly of those elements of F[x] that of have α as a root. If f (α) = 0and f (x) is not a zero polynomial, then f (x) ∈ 〈p(x)〉 and p(x) divides f (x). Thus p(x) is a polynomial withminimal degree, which of has α as a root. Every other polynomial with of same degree, which of has α asa root must be of the form βp(x), for some β ∈ F.

Assume that p(x) = r(x)s(x) is a factorization of p(x) in polynomials with lower degree. Since p(α) = 0,then r(α)s(α) = 0 and as a consequence r(α) = 0 or s(α) = 0, which contradicts the fact that p(x) is of minimaldegree. Thus p(x) must be irreducible.

Let E a extension field of F and α ∈ E algebraic over F. The unique monic polynomial p(x) from the last

theorem is called the minimal polynomial of α over F and is denoted by min (α,F,x). The degree of p(x) isthe degree of α over F.

Example 16.4. Let f (x) = x2−2 and g(x) = x4

−4x2 + 1. These polynomials are respectively minimal polynomials

of√

2 and√

2 +√

3.

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16 Shaska T.

Proposition 16.1. Let E/F be an extension and α ∈ E algebraic over F. Then,

F(α) F[x]/〈min (α,F,x)〉

Proof. Define the evaluation map at α as

φα : F[x]→ Ef (x)→ f (α)

Its kernel is the ideal generated by the minimal polynomial min (α,F,x) of α. From the First IsomorphismTheorem for rings (Theorem 9.4) the image of φα in E is isomorphic to F(α).

Theorem 16.4. Let E = F(α) be a simple extension of F, where α ∈ E is algebraic over F. Assume that degree of αover F is n. Then, every element β ∈ E can be expressed in a unique way in the form

β = b0 + b1α+ · · ·+ bn−1αn−1

for bi ∈ F.

Proof. Since φα(F[x]) = F(α), every element in E = F(α) must be of the form φα( f (x)) = f (α), where f (α) is thepolynomial in α with coefficients in F. Let

p(x) = xn + an−1xn−1 + · · ·+ a0

be minimal polynomial of α. Then, p(α) = 0. Thus,

αn = −an−1αn−1− · · ·− a0.

Similarly we get

αn+1 = ααn

= −an−1αn− an−2α

n−1− · · ·− a0α

= −an−1(−an−1αn−1− · · ·− a0)− an−2α

n−1− · · ·− a0α.

Continuing in this way we can express every monomial αm for m ≥ n as of linear combination of powers ofα which are less than n. Thus, every β ∈ F(α) can be written as

β = b0 + b1α+ · · ·+ bn−1αn−1.

To prove uniqueness assume that

β = b0 + b1α+ · · ·+ bn−1αn−1 = c0 + c1α+ · · ·+ cn−1α

n−1,

for bi and ci in F. Then,g(x) = (b0− c0) + (b1− c1)x + · · ·+ (bn−1− cn−1)xn−1,

is in F[x] and g(α) = 0. Since deg g(x) = n−1 < degp(x) then g(x) must be polynomial zero. Hence,

b0− c0 = b1− c1 = · · · = bn−1− cn−1 = 0,

or bi = ci for i = 0,1, . . . ,n−1.

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Example 16.5. Since x2 + 1 is irreducible over R, then 〈x2 + 1〉 is a maximal ideal in R[x]. Thus E =R[x]/〈x2 + 1〉is an extension field of R which contains a root of x2 + 1. Let α = x + 〈x2 + 1〉.

Then, E is isomorphic to R(α) = a + bα : a,b ∈R. We we know that α2 = −1 in E since

α2 + 1 = (x + 〈x2 + 1〉)2 + (1 + 〈x2 + 1〉) = (x2 + 1) + 〈x2 + 1〉 0.

Thus, we have a isomorphism of R(α) with C of defined from the map which takes a + bα→ a + bi.

Exercises:

16.1. Let K/F be an algebraic extension such that: [K : F] = p, where p is prime number. Prove that, there is nointermediate field F ⊂ E ⊂ K.

16.2. Prove that, if [F(α) : F] is an odd number then F(α) = F(α2).

16.3. Prove that, Q(√

3,√

7) =Q(√

3 +√

7). Generalize this proof to show that Q(√

a,√

b ) =Q(√

a +√

b ).

16.4. Let α,β transcendental over Q. Prove that, or αβ, or α+β is also transcendental.

16.5. Let E a extension field of F and α ∈ E transcendental over F. Prove that, for every element in F(α), i cili is notin F, is also transcendental over F.

16.6. Prove or disprove: Q(√

2) Q(√

3).

16.7. Prove that, the fields Q( 4√3) and Q( 4√3 i) are isomorphic, but not equal.

16.8. Let K a algebraic extension of E, and E a algebraic extension of F. Prove that, K is algebraic over F.

16.9. Prove that, each from the following numbers is algebraic over Q, by finding the minimal polynomial over Q.√3√

2− i,√

1/3 +√

7,√

3 +3√

5,√

3 +√

2 i.

16.10. Determine all subfields of Q( 4√3,o f ).

16.11. Prove that, Z2[x]/〈x3 + x + 1〉 is a field with eight elements. Construct a multiplication table for the group ofmultiplication of the field.

16.12. Prove that, Q(√

3, 4√3, 8√3, . . .) is a algebraic extension of Q, but not a finite extension.

16.13. Prove or disprove: π is algebraic over Q(π3).

16.14. Prove that, the set of all elements in R, which are algebraic over Q, form a extension field for Q, which is notfinite.

16.15. Let E a extension field of F and α ∈ E. Determine [F(α) : F(α3)].

16.16. Prove or disprove: Z[x]/〈x3−2〉 is field.

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16 Shaska T.

16.2 Splitting fields

Let be given a polynomial p(x) ∈ F(x), where F is a field. From Kronecker’s theorem Theorem 16.1 thereexists a field K in which p(x) has a root α. Then,

p(x) = (x−α)q(x).

which implies that there is a extension of K where q(x) has a root. Hence, there is a extension of F thatcontains all roots of p(x). The smallest of such fields is called the splitting field of the polynomial p(x).

Definition 16.1. Let be given a polynomial p(x) ∈ F[x] with degree n, where F is field. Let α1, . . . ,αn be roots of thispolynomial. Then, the field F(α1, . . . ,αn) is called the splitting field of the polynomial p(x).

Theorem 16.5. For every field F and every polynomial f (x) ∈ F[x] there is a extension K of F which is splitting fieldof f (x).

Proof. We prove the existence of splitting fields with induction on the degree n of the polynomial. If apolynomial has degree n = 1, then we get K = F. Assume that the statement is true for k ≤ n−1. If f (x) isfactored in linear factors again we get K = F. On the contrary, there is a factor with degree ≥ 2. We knowthat there is a field E1 in which this factor has a root α. The polynomial f (x) has a root α in E1. The degreeof the other factor f1(x) of f (x) is ≤ n−1. By induction hypothesis there is a extension of E that contains allroots of f1(x). However α ∈ E1 ⊂ E. Thus, E contains all roots of f (x). Take as K as the intersection of allsubfields of E, that contain all roots of f (x).

Example 16.6. The splitting field for x2−2 is Q(

√2) because roots of x2

−2 are α12 = ±√

2 and Q(√

2,−√−2) =

Q(√

2). Then, [Q(√

2 :Q] = 2.

Example 16.7. The splitting field of x4 + 4 over Q is Q(i), because

x4 + 4 = (x2 + 2x + 2)(x2−2x + 2)

where of two factors are irreducible. Their roots are

±1,±i.

Thus, the splitting field is Q(i) and [Q(i) :Q] = 2 because i has as minimal polynomial x2 + 1.

Lemma 16.1. The splitting field of a polynomial with degree n over F has at most degree n! over F.

Proof. Take a polynomial f (x) with degree n. Let α be one of its roots. Then,

f (x) = (x−α) f1(x).

The degree of F(α) over F is at most n. The polynomial f1(x) has degree (n−1). Let β be one of its roots. Thedegree of F(α)(β) is at most (n−1). Continuing this way for all roots we get the desired result.

Example 16.8. Let p(x) = x4 + 2x2− 8 in Q[x]. Then, p(x) has irreducible factors x2

− 2 and x2 + 4. Thus fieldQ(√

2, ı) is splitting field for p(x).

Example 16.9. Let p(x) = x3−3 in Q[x]. Then, p(x) has a root in field Q( 3√3). However this field is not splitting

field for p(x) since complex cubic roots of 3−

3√3± ( 6√3)5o f2

,

are not in Q( 3√3).

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Next we study the question if two splitting fields of the same polynomial f (x) over some field F aresomehow related.

Lemma 16.2. Let φ : E→ F be a field isomorphism. Let K/E be an extension, α ∈ K algebraic over E, andp(x) = min (α,E,x).

Assume that L/F is an extension such that β is root of the polynomial φ(p(x)) in F[x]. Then, φ can be extendedto a unique isomorphism ψ : E(α)→ F(β) such that ψ(α) = β and ψ agrees with φ in E.

E(α)ψ−→ F(β)yσ yτ

E[x]/〈p(x)〉φ−→ F[x]/〈q(x)〉y y

Eφ−→ F

Proof. If p(x) has degree n, then from Theorem 16.4 we can write every element of E(α) as linear combinationof 1,α, . . . ,αn−1. Thus, we define

ψ(a0 + a1α+ · · ·+ an−1αn−1) = φ(a0) +φ(a1)β+ · · ·+φ(an−1)βn−1,

wherea0 + a1α+ · · ·+ an−1α

n−1

is an element of E(α). The fact that ψ is isomorphism can be easily checked.We can extend φ so that it becomes an isomorphism from E[x] to F[x], which we also denote with φ, by

takingφ(a0 + a1x + · · ·+ anxn) = φ(a0) +φ(a1)x + · · ·+φ(an)xn.

This extension agrees with the original isomorphism φ : E → F, since constant polynomials are mappedto constant polynomials. From the assumption φ(p(x)) = q(x). Thus, φ maps 〈p(x)〉 in 〈q(x)〉.

Hence, we have an isomorphism φ : E[x]/〈 p(x)〉 → F[x]/〈 q(x)〉 . The isomorphisms

σ : E[x]/〈p(x)〉 → F(α) and τ : F[x]/〈q(x)〉 → F(β),

are of defined respectively for values of α and β. Thus ψ = τ−1φσ is as claimedThe proof of uniqueness is left as an exercise.

Theorem 16.6. Let φ : E → F be a field isomorphism, p(x) a non-constant polynomial in E[x], and q(x) thecorresponding polynomial in F[x] related to this isomorphism. If K is a splitting field for p(x) and L is a splitting fieldfor q(x), then φ is extended to an isomorphism ψ : K→ L.

Proof. We use mathematical induction on the degree of p(x). Assume that p(x) is irreducible over E. Thusq(x) is also irreducible over F. If degp(x) = 1, then from the definition of splitting fields K = E and L = F.

Assume that the theorem is true for all polynomials with smaller degree than n. Since K is a splittingfield of E, all roots of p(x) are in K. Pick one from these roots, say α such that E ⊂ E(α) ⊂ K. Similarly we canfind a root β of q(x) in L such that F ⊂ F(β) ⊂ L.

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16 Shaska T.

From Lemma 16.2, there is an isomorphism φ : E(α)→ F(β) such that φ(α) = β and φ agrees with φ in E.

Kψ−→ Ly y

E(α)φ−→ F(β)y y

Eφ−→ F

Then, p(x) = (x−α) f (x) and q(x) = (x−β)g(x), where degrees of f (x) and g(x) are respectively less thandegrees of p(x) and q(x). The field extension K is splitting field for f (x) over E(α) and L is splitting field forg(x) over F(β). From induction hypothesis there is a isomorphism ψ : K→ L such that ψ agrees with φ inE(α). Thus, there is a isomorphism ψ : K→ L such that ψ agrees with φ in E.

Corollary 16.1. Let p(x) a polynomial in F[x]. Then, there is a splitting field K for p(x) which is unique up to aisomorphism.

Example 16.10. Find the degree of the splitting field of

x8−2

over Q.

Solution: Let be given α =8√2 an eighth root of 2. Since f (x) is irreducible then [Q(α) : Q] = 8. Then,

splitting field is Q(α,ε8), where ε8 is a primitive root of unity. Let’s say,

e8 =

√2

2(1 + ı).

Thus, splitting field is Q(α, ı). It is clear that, ı < Q(α), since α ∈ R. Thus, [Q(α, ı) : Q(α)] = 2. Therefore,[Q(α, ı) :Q] = 16.

Example 16.11. Let a and b be square free distinct integers such that, [Q(√

a +√

b) :Q] = 4. Find

min(√

a +√

b,Q,x).

Use this result to find the minimal polynomial for√

2 +√

3,√

2 +√

7,√

3 +√

5,

over Q.

Solution: Let be given a and b square free, distinct integers such that, [Q(√

a +√

b) :Q] = 4. Recall that we

have proved that Q(√

a +√

b) =Q(√

a,√

b) ).

If√

a+√

b is root and a, b are such that, [Q(√

a+√

b) :Q] = 4, then ±√

a±√

b are all roots. Then, minimalpolynomial is

f (x) = x4−2(a + b)x2 + (a− b)2.

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Shaska T. 16

It is easily proved that f (x) is irreducible over Q, since its roots are not in Q and all quadratic factors alsoare not polynomials in Q[x].

For a = 2, b = 3, we havef (x) = x4

−10x2 + 1.

Similarly if a = 2,b = 7, or a = 3,b = 5, we get

f (x) = x4−18x2 + 25, f (x) = x4

−16x2 + 4.

Let f (x) ∈K[x] be given. We say that f (x) splits over a field L if f (x) factors completely into linear factors

in L[x].

Definition 16.2. The extension L of K is called the splitting field of the polynomial f (x) ∈ K[x] if it is the smallestfield containing K where f (x) splits completely.

Theorem 16.7 (Kronecker). For any field K and f (x) ∈ K[x] there exists the splitting field of f (x) over K.

Proof. The proof goes by induction on the degree of f (x). If f (x) has degree one then this is obviously true.If the deg f > 1 then f (x) = (x−α) g(x) for some α. Then K(α)K[x]/〈 f 〉 contains α. By induction hypothesisthere is an extension of K(α) which contains all roots of g(x).

Example 16.12. The splitting field of x2−2 over Q is Q(

√2). Indeed, the two roots are ±

√2 and −

√2 ∈Q(

√2).

Example 16.13. Find the splitting field of x4 + 4 over Q. Then

f (x) = (x2−2x + 2)(x2 + 2x + 2)

where each factor is irreducible. The roots are ±1± i. Hence the splitting field is Q(i).

Lemma 16.3. A splitting field E f of a polynomial f (x) of degree n over K is of at most of degree n! over K. Moreover,if f (x) is irreducible over K then n | [E f : K].

Proof. Let f (x) ∈K[x] be a polynomial of degree n. Then adjoining one root α of f (x) to K we get an extensionof degree at most n. Then, f (x)

x−α has degree at most (n− 1). Adjoining a second root to K(α) the degree ofthe extension will be at most (n−1). Since the degrees of the extensions are multiplicative this proves thelemma.

If f (x) is irreducible then [K(α) : K] = deg f = n. Thus, n | [E f : K].

16.2.1 Splitting field of quadratics

Let f (x) ∈Q[x] be given asf (x) = ax2 + bx + c

Then the splitting field if f (x) is Q(λ), where λ =√

b2−4ac. Indeed, both roots can be expressed as linearcombinations of λ and scalars in Q.

Example 16.14. We continue with the previous example of Q(√

2,√

3)/Q. We showed that Q(√

2,√

3) =Q(√

2 +√

3). Let’s call α =√

2 +√

3. What is the minimal polynomial of α?Notice that

α2 = (√

2 +√

3)2 = 5 +√

6

andα4−10α2 + 1 = 0

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16 Shaska T.

Since we know that the polynomialp(x) = x4

−10x2 + 1

is irreducible over Q and [Q(α) :Q] = 4 then

min (α,Q,x) = x4−10x2 + 1.

What is α−1? We get α(10α−α3) = 1. So α−1 = 10α−α3 =√

3−√

2.

The roots of f (x) = x4−10x + 1 are

√

3±√

2, −√

3±√

2

So Q((√

2 +√

3) is the splitting field of f (x).

16.2.2 Splitting field of cubics

Let f (x) ∈Q[x] be irreducible and given as

f (x) = x3 + ax2 + bx + c

Let α1,α2,α3 be the roots of the cubic. Then, Q(α1,α2,α3) =Q(α1,α2) since α3 = −a− (α1 +α2). Since f (x) isirreducible then [Q(α1) : Q] = 3. The polynomial f (x)

x−α1has degree 1 or 2 and so [Q(α1,α2) : Q(α1)] = 1 or 2.

Hence, [Q(α1,α2) :Q] = 3 or 6.

Lemma 16.4. The splitting field of a cubic has degree 3 if and only if the discriminant of f (x) is a square in Q.

Proof. Every cubic can be transformed into a cubic of the form

g(x) = x3 + ax + b

So the splitting field of f (x) and g(x) are the same. The roots of g(x) are given by Cardano’s formulas

xi = (−b2

+

√b2

4+

a3

27)1/3 + (−

b2−

√b2

4+

a3

27)1/3

and the discriminant of g(x) is∆(g,x) = −4a3

−27b2

So the roots are

xi =

(−

b2

+1

18

√

−3∆

)1/3

+

(−

b2−

118

√

−3∆

)1/3

Now the result is obvious. We will discuss the solution of the cubic in detail later in the book.

Example 16.15. Letf (x) = x3

−3x + 1 ∈Q[x]

Its discriminant is∆( f ,x) = 34

Then the splitting field of f (x) has degree 3 over Q.

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Shaska T. 16

Q( 3√2,√−3)

32

Q( 3√2)

3

Q(√−3)

2

Q

Example 16.16. Let f (x) ∈Q[x] such thatf (x) = x3

−2

Its roots are3√

2,3√

2[−

12

+I2

√

3],

3√

2[−

12−

I2

√

3]

and the splitting field is Q( 3√2,√−3). We have the following lattice So the splitting field of f (x) = x3

−2 has degree6 over Q.

Example 16.17. Consider the polynomial

f (x) = x3−x2−2x + 1.

We leave it as an exercise to show that f (x) is irreducible over Q. Using Cardano’s formulas we find that its roots are

α1 :=16

d +14

3d +13

; (16.1)

where d = (−28 + 84√−3)

13 .

αi := −d

12−

73d

+13±

12

√

−3(

d6−

143d

)(16.2)

where i = 2,3. The first root can be written as

αi := −d

12−

d2

42+

d2√−3

84+

d√−3

12+

13

(16.3)

Hence, the splitting field of f (x) over Q has degree 6.

16.2.3 Higher degree polynomials

We will discuss in detail the higher degree polynomials through the Galois theory. As we will see for arandom polynomial of degree n the splitting field has degree n! or n!

2 . The next example provides a degreen polynomial f (x) such that its splitting field has degree n! over Q, for any given N ≥ 2.

Example 16.18. Take the polynomialf (x) = xr(x−1)n−r

−λ ∈Q[x]

where λ is an integer such that (r,λ) = 1. Then the degree of the split extension of f (x) is always n!, see Serre, pg. 85.

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16 Shaska T.

16.2.4 Uniqueness of splitting fields

In this section we will prove that the splitting field of a polynomial is unique up to isomorphism.

Lemma 16.5. Let σ : F→ F′ be a field isomorphism. Let f (x) ∈ F[x] be irreducible and α a root of f (x) in someextension K of F, and let α′ be a root of σ( f ) in some extension K′ of F′. Then there is an isomorphism τ : F(α)→ F′(α′)with τ(α) = α′ and τ |F= σ.

Proof. Since f (x) is irreducible and f (α) = 0 then f (x) is a constant multiple of the minimal polynomial of α.Thus, F(α)F[x]/〈 f (x)〉 via

ϕ : F[x]/〈 f (x)〉 → F(α)g(x) + 〈 f (x)〉 → g(α)

Similarly we have

ϕ : F′[x]/〈 f (x)〉 → F′(α)g(x) + 〈 f ′(x)〉 → g(α′)

We also have the extension of σ : F→ F′ as follows

φ : F[x]/〈 f (x)〉 → F′[x]/〈 f ′(x)〉g(x) + 〈 f (x)〉 → σ

(g(x)

)+ 〈 f ′(x)〉

It is an easy exercise to check that this is and isomorphism and an extension of σ. Thus, the composition

F[x]/〈 f (x)〉

ϕ

φ // F′[x]/〈 f ′(x)〉

ψ

F(α) τ F′(α′)

τ : F(α)→ F′(α) as τ := ψφϕ−1 is an isomorphism. It obviously extends σ and σ(α) = α′.

K K′

F(α) τ F′(α′)

Fσ

F′

The following lemma considers the case of a family of polynomials.

Lemma 16.6. Let σ : F→ F′ be a field isomorphism, K a field extension of F and K′ a field extension of F′. Supposethat K is a splitting field of fi over F and that τ : K→ K′ is a homomorphism with τ |F= σ. Then τ(K) is the splittingfield of σ( fi) over F′.

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Shaska T. 16

Proof. Since K is the splitting field of the set of polynomials fi then for each fi there exist a,α1, . . . ,αn ∈ Ksuch that fi(x) = a

∏j(x−α j). Thus,

f ′i (x) = σ(

fi(x))

= τ(

fi(x))

= τ(a)∏

j

(x−τ(α j))

Hence, fi(x) splits over τ(K).K is generated over F by the roots of fi; hence, τ(K) is generated over F′ by the roots of f ′i . Thus, τ(K)

is a splitting field over F′ for f ′i . The next theorem proves the uniqueness up to isomorphism of the splitting field and it is one of the

main results of Galois theory.

Theorem 16.8 (Isomorphism extension theorem). Let σ : F→ F′ be a field isomorphism. Let S = fi(x) be a setof polynomials over F and let S′ = σ( fi) be the corresponding polynomials over F′. Let K be the splitting field for Sover F and K′ be the splitting field for S′ over F′. Then there is an automorphism τ : K→ K′ with τ |F= σ.

Furthermore, if α ∈ K and α′ is any root of the polynomial σ(min (α,F,x)), then τ can be chosen so that τ(α) = α′.

Proof. Let A be the set of all pairs (L,ϕ) as below

A := (L,ϕ) | L a subfield of K and ϕ : L→ K′extends σ

Then, A , ∅ since (F,σ) ∈A. Also, A is partially ordered by

(L,ϕ) ≤ (L′,ϕ′) if L ⊂ L′ and ϕ′ |L= ϕ

Let (Li,ϕi) be a chain in A. This chain has an upper bound given by L = ∪iLi and ϕ : L→ K′ given by

ϕ(a) = ϕi(a), if a ∈ Li.

The reader should check that ϕ is a homomorphism extending σ. By Zorn’s lemma there is a maximalelement (M,τ) in A. We will show that M = K and τ(M) = K′.

Kτ

K′

F σ F′

If M , K, then there is an f (x) ∈ S which does not split over M. Let α ∈ K be a root of f (x) that is not in M.Denote by p(x) = min (x,F,α) and let p′ = σ(p) ∈ F′[x]. Let α′ ∈ K′ be a root of p′(x). Such α′ exists since p′ | f ′

and f ′ splits over K′/ By Lemma 16.5, there is a ρ : M(α)→ τ(M)(α′) that extends τ. Then, (M(α),ρ) ∈A islarger than (M,τ), which is a contradiction. Hence, M = K.

The equality τ(K) = K′ follows from Lemma 16.6, because τ(K)⊂ K′ is a splitting field for S′ over F′. Thiscompletes the proof.

Corollary 16.2. Let f (x) ∈ k[x]. Any two splitting fields of f (x) are k-isomorphic. In particular, every two algebraicclosures of k are k-isomorphic.

Proof. For the first part take σ = id in the above theorem. The second statement follows from the first sinceevery algebraic closure is the splitting field on the set of all nonconstant polynomials in k[x].

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16 Shaska T.

E f ooτ // E′f

k

ka1oo τ // ka

2

k

Example 16.19. Let f (x) = xn− a ∈Q[x]). Its splitting field is

Q( n√a,εnn√a,ε2

nn√a, . . . , εn−1

nn√a)

where εn is the primitive root of unity. The degree of this extension over Q will depend on a and n as we will see inthe section of cyclotomic extensions. If n√a ∈Q then the splitting field of f (x) is Q(εn).

Exercises:

16.17. Let K be the splitting field of x3 + x2 + 1 ∈Z2[x]. Prove or disprove that K is a radical extension.

16.18. Let F a field such that char F, 2. Prove that splitting field of f (x) = ax2 +bx+c is F(√α ), where α= b2

−4ac.

16.19. Let K a splitting field of a polynomial over F. If E is a extension field of F which is contained in K and[E : F] = 2, then E is splitting field of a polynomial in F[x].

16.20. Let be given f (x) irreducible in k[x], such that, deg f = m and let K a extension field of k, ku [K : k] = n. Provethat, if, gcd (m,n) = 1, then f (x) is irreducible over K.

16.21.

Determine splitting field over Q of polynomialsi) x4−2.

ii) x4 + 2.iii) x4 + x2 + 1.iv) x6

−4.

16.22. Let E be a finite extension of a field F. If [E : F] = 2, prove that, E is a splitting field of F.

16.23. Let p(x) a non-constant polynomial with degree n in F[x]. Prove that, there is a splitting field E for p(x), suchthat, [E : F] ≤ n!.

16.24. Compute the splitting fields of the following polynomials, over Q and find the degree of such field:

1. x5−2;

2. x3−3x + 3;

3. x3 + 2,

4. x4−3 or x4

−2,

5. x5−4x + 2

6. x4−10x2 + 4

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Shaska T. 16

7. x4 + 4x2 + 2

8. x3 + 3x2−x−1,

9. x3−3x + 1,

16.25. Let Q(√

1 +√

5)/Q be given. Find the minimal polynomial of√

1 +√

5 and the degree of the extension.

16.26. Let p be a prime integer and

Φp(x) = xp−1 + xp−2 + · · ·+ x2 + x + 1

Find the splitting field of Φp(x) over Q and its degree.

16.27. Let E be a finite extension of F. Prove that E is a splitting field for some polynomial if and only if everyirreducible polynomial over F, having a root in E, factors completely over E.

16.3 Normal extensions

Let L/K be an algebraic extension such that L is the splitting field of a collection of polynomials f (x) ∈ K[x].Then L is called a normal extension of K.

Example 16.20. Every degree 2 extension is normal. Let K/F be a field extension such that [K : F] = 2. Take

α ∈ K \F

then the minimal polynomial of α has a root in K thus it splits in K.

Proposition 16.2. Let K/F be an algebraic extension. Then the following are equivalent:i) K is normal over Fii) For any irreducible polynomial f (x) ∈ F[x], if f (x) has a root in K then it splits over K.

Proof.

Example 16.21. Letf (x) = x4

−2

Its solutions are±

4√

2, ±i4√

2

Thus Q( 4√2, i)/Q is a normal extension. We have the lattice

Q( 4√2, i)

42normal

Q( 4√2)

4not normal

Q(i)

2

Q

The following fact is rather obvious but nevertheless very important.

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16 Shaska T.

Remark 16.1. Let L/F/k be a tower of field extensions such that L/k is normal. Then L/F is normal.

L

normal

normal~~

F

K

Proof. If L is the splitting field of a family of polynomials in k[x] then it is the splitting field of the samefamily in F[x].

Example 16.22. Give an example such that K ⊂ E is normal, E ⊂ F is normal, but K ⊂ F is not normal.

Take Q ⊂ Q(√

2) ⊂ Q( 4√2). Each extension is normal because it is degree 2 but Q ⊂ Q( 4√2) is not normal asshown above.

Remark 16.2. Normal extensions don’t form a distinguished class.

The proof of the following can be found in Lang, pg. 238.

Remark 16.3. If K1,K2 are normal over k and are contained in some field L, then K1K2 is normal over k, and so isK1∩K2.

Exercises:

16.28. Let F ⊂ K ⊂ E be fields. If E is a normal extension of F, prove that E is also a normal extension of K.

16.29. Let α be a real number such that α4 = 5. Show that:

i) Q(iα2) is normal over Q.ii) Q(α+ iα) is normal over Q(iα2)iii) Q(α+ iα) is not normal over Q.

16.4 Algebraic closure

Let E a extension field of fields F. Define algebraic closure of a field F in E to be field of all elements of Ewhich are algebraic over F.

A field F is algebraically closed if every non-constant polynomial in F[x] has a root in F.

Theorem 16.9. A field F is algebraically closed if and only if every non-constant polynomial in F[x] is factored inlinear factors in F[x].

Proof. Let F a field algebraically closed. If p(x) ∈ F[x] is a non-constant polynomial, then p(x) has a root inF, say α. Thus x−α must be a factor of p(x) and hence p(x) = (x−α)q1(x), where degq1(x) = degp(x)−1.

Continue this process with q1(x) to find a factorization

p(x) = (x−α)(x−β)q2(x),

where degq2(x) = degp(x)−2. The process will stop because the degree of p(x) is finite.

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Shaska T. 16

Conversely, assume that every non-constant polynomial p(x) in F[x] is factored in linear factors. Letax−b a factor such. Then, p(b/a) = 0. Thus, F is algebraically closed.

Corollary 16.3. A algebraically closed field F does not have any proper algebraic extensions.

Proof. Let E a algebraic extension of F, then F ⊂ E. For α ∈ E minimal polynomial of α is x−α. Thus α ∈ Fand F = E.

Lemma 16.7 (Artin). Let F be a field. Then there is an extension of F which is algebraically closed.

Proof. First we construct an F1 such that every polynomial in F[x] has a root in F1. Then, construct anotherfield F2 such that every polynomial in F1[x] has a root in F2 and continue. We get a tower of fields

F ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fi ⊂ Fi+1 ⊂ . . .

LetF = ∪i∈I Fi,

for some index set I. Then F is algebraically closed, since for every polynomial f ∈ F, f has a root in one ofthe Fi’s.

Thus, it is left to construct F1 such that every polynomial in F has a root in F1. For every f ∈ F[x]introduce a new variable x f . Let S be the set of all new variables x f . Let R := F[S] and

I = 〈 f (x f )〉 f∈F[x]

I , R, otherwise1R = a1 f1(x f1) + · · ·+ an fn(x fn )

where ai ∈ F. Since for each of the fi exist a field extension of F such that fi has a root (see Kronecker’sTheorem Theorem 11.5), then there is a field K such that 1R = a1 ·0 + · · ·+αn ·0 = 0 which is a contradiction.Then, I is contained in a maximal idealM.

F[x]

$$

α // R[x]

π

(R/M) [x]

Let π : R[x]→ R/M [x] be the quotient map and α : F[x] → R[x] the inclusion map. Let σ := πα. Forevery f ∈ F[x], σ( f ) has a root in R/M, namely x f +M. Take F1 = R/M. This completes the proof.

The algebraic closure of a field F will be called a field F if it is an algebraic extension of F and it isalgebraically closed.

Theorem 16.10. Every field has an algebraic closure.

Proof. Let k be a field. From the previous theorem there exists an extension E of F which is algebraicallyclosed. Let ka be the union of all Fi such that k ⊂ Fi ⊂ E and Fi/k is algebraic. We claim that ka is the algebraicclosure of k.

Indeed, let α ∈ ka. Then α ∈ Fi for some i. Hence, α is algebraic over k. Thus, ka/k is algebraic.Let f (x) ∈ ka[x]. Then, f (x) has a root α in E. Thus, α is algebraic over ka. Hence, α is algebraic over k

(algebraic extensions form a distinguished class). So α ∈ Fi for some i, which implies that α ∈ ka. Thereforeka is algebraically closed. This completes the proof.

Usually the algebraic closure of a field k is denoted by k or ka.

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Example 16.23. It will be proved later in these notes thatC is algebraically closed. Indeed,Ra =C. However,Qa ,C

Exercises:

16.30. Let F/k be a field extension, f (x) ∈ k[x], and σ an isomorphism of F fixing every element of k. Prove that if αis a root of f (x) then σ(α) is also a root of f (x).

16.31. Prove that Q( 3√2) has no automorphisms other then the identity.

16.32. Prove that if α ∈ C is a root of f (x) ∈R[x] then α is also a root of f (x).

16.33. Let E be the algebraic closure of a field F. Prove that, every polynomial p(x) in F[x] splits in E.

16.34. If for every irreducible polynomial p(x) in F[x] is linear, prove that, F is a field algebraically closed.

16.5 Some classical problems

16.5.1 Geometric Constructions

Some of the most important problems of classical mathematics deal with geometric constructions. In thissection something is constructible if it can be constructed by a ruler and compass. This material can befound in most introductory books in algebra. We start with the following 4 classical questions:

i) Is it possible to trisect an angle?ii) Is it possible to square the circle?iii) Is it possible to double the cube?iv) For what n the regular n -gon is constructible?

A number α is called constructible if we can construct a line segment of length α by a ruler and compass.The following theorem is the link of geometric constructions to algebra:

Lemma 16.8. If c > 0 is constructible then√

c is also constructible

Proof. Take the right triangle with hypotenuse 1 + c and the height from the right angle splitting thehypotenuse in segments of length 1 and c. Then the height has length

h2 = 1 · c =⇒ h =√

c

Lemma 16.9. It is possible to construct similar triangles.

Proof. Exercise.

Corollary 16.4. If a,b are constructible then so is a + b, ab, ab .

Proof. Exercise.

Theorem 16.11. i) Constructible numbers form a field.

ii) A real number α is constructible if and only if it is contained in a field of the form

Q(√α1, . . .

√αs), αi ∈Q(

√α1, . . .

√αi−1)

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Shaska T. 16

y

x0 1–1

–i

i

ω

ω7ω5

ω3

Figure 16.1: Construction of roots

Proof. Part i) is an immediate consequence of the above corollary. From i) it follows that the field ofconstructible numbers is an extension of Q. Since every number in Q(

√α1, . . .

√αi−1) is constructible it is

left to show that every constructible number is in Q(√α1, . . .

√αi−1).

This follows from the fact that the intersection of a line and a circle contains only square roots.

Corollary 16.5. If α is constructible then α algebraic over Q and

[Q(α) :Q] = 2r

for some r.

Proof. If α is constructible then α ∈Q(√α1, . . .

√αn), for some α1, . . . ,αn. Then,

[Q(α) :Q] | [Q(√α1, . . .

√αn) :Q] = 2r

for some r ≥ 0.

We now have the following results:

Theorem 16.12. It is impossible to trisect any angle.

Proof. Trisecting an angle 3α is the same as constructing cosα when cos3α is given. The equation

cos3α = 4cos3α−3cosα

gives us the polynomialf (x) = 4x3

−3x− cos3α

for which cosα is a root of. For some values of α the polynomial f (x) is irreducible and [Q(cosα) : Q] = 3,which is not a power of 2.

For example, take α = 20. Then,f (x) = 8x3

−6x−1

is irreducible over Q. Hence, if α is a root of f (x) then [Q(α) :Q] = 3 which is not a power of 2.

Theorem 16.13. It is impossible to square the circle.

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16 Shaska T.

Proof. Let r be constructible and a circle of radius r is given. We want to construct a square of radius x suchthat

x2 = πr2

Since π is not even an algebraic number then the roots of the above equation are not even algebraic andtherefore can not be constructible.

Theorem 16.14. It is impossible to double the cube.

Proof. Take a cube of volume 1. To double the cube would mean to construct an x such that

x3 = 2.

The polynomialf (x) = x3

−2

is irreducible over Q and therefore for each root α of f (x) we have [Q(α) :Q] = 3, which is not a power of 2.

Theorem 16.15. The regular n -gon is constructible if and only if

n = 2k·p1 · · ·ps

where pi are distinct Fermat primes.

Proof. To construct an n -gon is equivalent with constructing cos 2πn . Let εn denote

en = cos2πn

+ isin2πn

Then,

cos2πn

=12

(εn +ε−1n )

Hence, Q(εn) is an extension of Q( 2πn ). We complete the proof only for n a prime p, the rest will be proven

Q(εn)

2

Q(cos 2πn )

n−12

Q

in the chapter of cyclotomic fields. So cos 2πn is constructible if

p−12

= 2r

for some r ≥ 0. Thus, this is possible only for primes p of the form p = 2k + 1. But these are exactly theFermat’s primes and they are in the form

p = 22r+ 1.

This completes the proof.

Exercises:

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Shaska T. 16

16.35. Give a geometric way of constructing a regular n -gon for

n = 3,4,5,6,8,10,12,15,16,17,20,24

16.5.2 Algebraic equations

Solving algebraic equations has always been a central question of mathematics. In this chapter we give abrief introduction to the solutions of the quadratic and cubic equations.

Equations of degree 2 and 3

All polynomials have coefficients in some field K of characteristic 0. When dealing with an equation in onevariable

Xn + an−1Xn−1 + . . . + a1X + a0 = 0

the first simplification is to substitute

X = Y −an−1

nwhich results in an equation

Yn + bn−2Yn−2 + . . . + b0 = 0

with zero Yn−1 term. This already solves the quadratic equation ("completing the square"): For n = 2 wejust get Y2 = −b0.

For the rest of this subsection we study the case n = 3, i.e., the equation

(1) Y3 + aY + b = 0

Substituting Y = u + v gives

u3 + v3 + 3 (uv +a3

) (u + v) + b = 0

which holds if u and v satisfy

(2) u3 + v3 = −b, uv = −a3

The latter gives

(Z−u3) (Z−v3) = Z2 + bZ −a3

27so without loss

u3 =−b +

√b2 + 4a3

27

2

v3 =−b −

√b2 + 4a3

27

2

Conversely, we can choose u and v as suitable cube roots of the expressions on the right hand side suchthat (2) holds. Then u+v is one solution of (1), and the others we get by different choice of cube roots. Moreprecisely, if ε is a primitive third root of unity then the solutions of (1) are

Y1 = u + v, Y2 = εu + ε2v, Y3 = ε2u + εv

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16 Shaska T.

which can be checked by expanding (Y−Y1)(Y−Y2)(Y−Y3). These are Cardano’s formulas, which areusually (but less precisely) written as

Yi = (−b2

+

√b2

4+

a3

27)1/3 + (−

b2−

√b2

4+

a3

27)1/3

We see that these formulas have a lot of symmetries, coming from the various choices of square rootsand cube roots, and also from the choice of a third root of unity. The crucial point is that the pattern ofsymmetries between the solutions can be defined without using the explicit formula for the solutions: Itis given by the Galois group of the equation. The fundamental idea of Galois theory can now simply bestated as follows: For general n, replace the explicit formula for the solutions (it doesn’t exist anyway) byusing the Galois group.

Starting point is the observation that adjoining any root of an irreducible equation gives a field extensionthat depends only on the equation, up to isomorphism. This yields field isomorphisms interchanging thedifferent roots of the equation. These isomorphisms yield the basic symmetries between the roots, whichform the Galois group.

16.5.3 Newton’s identities and the discriminant

Before we get to general Galois theory, we collect some facts on the discriminant. The discriminant Dp ofa polynomial p(X) vanishes iff p has a multiple root. We have seen last semester that Dp is a polynomialexpression in the coefficients of p. (Follows from the main theorem on symmetric polynomials). Now wewant to get an explicit formula for Dp.

Recall that if we write p(X) in the form

p(X) = Xn + an−1Xn−1 + . . . + a0 = (X−x1) . . . (X−xn)

thenDp =

∏i, j

(xi−x j)

The matrix

X :=

1 . . . 1x1 . . . xn. . . . .. . . . .. . . . .

xn−11 . . . xn−1

n

has

det(X) =∏i> j

(xi−x j)

by the well-known Vandermonde determinant formula.

Problem 1: Prove this formula.

Thus

(3) Dp = det(X)2 = det(XXt) = det

S0 S1 . . . Sn−1S1 S2 . . . Sn. . . . . .. . . . . .. . . . . .

Sn−1 Sn . . . S2n−2

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Shaska T. 16

whereSµ := xµ1 + . . . + xµn

It remains to express the power sums Sµ in terms of the elementary symmetric functions

σν =∑

i1<i2<···<iν

xi1xi2 . . . xiν

(whose basic property is that σν(x1, . . . ,xn) = (−1)ν an−ν ). This is provided by Newton’s identities (c.f. Coxet al., p. 317):

Sµ − σ1Sµ−1 + . . . + (−1)µ−1σµ−1S1 + (−1)µµσµ = 0, for 1 ≤ µ ≤ n

Sµ − σ1Sµ−1 + . . . + (−1)n−1σn−1Sµ−n+1 + (−1)nσnSµ−n = 0, for µ > n

Proof. Let z be a new variable and define

σ(z) =

n∏i=1

(1 − xiz)

Then

−zσ′(z)σ(z)

=z

∑ni=1 xi

∏j,i (1 − x jz)

σ(z)=

n∑i=1

xiz1 − xiz

=

n∑i=1

∞∑ν=1

xνi zν =

∞∑ν=1

(n∑

i=1

xνi ) zν =

∞∑ν=1

Sν zν(16.4)

Thus we get the following identity between formal power series in z:

σ(z)∞∑ν=1

Sν zν = −zσ′(z)

The basic property of the elementary symmetric functions yields

σ(z) =

n∑µ=0

(−1)µσµzµ

Thusn∑

j=0

(−1) jσ jz j·

∞∑ν=1

Sν zν =

n∑µ=1

(−1)µ+1µσµzµ

Comparing coefficients yields the claim.

Exercises:

16.36. Determine whether the following angles can be trisected.i) The angle β such that cosβ = 1

3 .ii) β = 120

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16 Shaska T.

16.37. Find the degree of the splitting field ofx8−2

over Q.

16.38. Let a and b be distinct square free integers such that [Q(√

a +√

b) :Q] = 4. Find

min(√

a +√

b,Q,x)

Use the result to write down the minimal polynomial for√

2 +√

3,√

2 +√

7,√

3 +√

5

over Q.Recall: we have shown that Q(

√a +√

b) =Q(√

a,√

b).

16.39. Let f (x) be an irreducible polynomial in k[x]. Prove that the following are equivalent:

i) char (k) = p > 0 and f (x) = g(xp) for some g(x) ∈ k[x].ii) all roots of f (x) are multiple.

Recall: In a finite field Fq of characteristic p, every element β ∈ Fq can be written as β = αp for some α ∈ Fq.

16.40. Let f (x) be irreducible in k[x] such that deg f = m and let K be a field extension of k with [K : k] = n. Provethat if gcd (m,n) = 1 then f (x) is irreducible over K.

16.41. Trisecting an angle β = 3α is the same as constructing cosα when cos3α is given. The equation

cos3α = 4cos3α−3cosα

gives us the polynomialf (x) = 4x3

−3x− cos3α

for which cosα is a root of.i) If cos3α = 1

3 then cosα is a root of the polynomial

f (x) = 4x3−3x−

13

which is irreducible. Then, [Q(cosα) :Q] = 3 which is not a power of 2. Hence, this angle can not be trisected.ii) If β = 120 then cosβ = − 1

2 . Then

f (x) = 4x3−3x +

12

which is irreducible over Q and as above the angle can not be trisected.

16.42. Let α =8√2 be a 8-th root of 2. Since f (x) is irreducible (Eisenstein) then [Q(α) : Q] = 8. Then the splitting

field is Q(α,ε8), where ε8 is a primitive root of unity. Say

e8 =

√2

2(1 + ı)

hence the splitting field isQ(α, ı). Clearly, ı<Q(α) sinceα∈R. Thus, [Q(α, ı) :Q(α)] = 2. Therefore, [Q(α, ı) :Q] = 16.

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Shaska T. 16

16.43. Let a and b be distinct square free integers such that [Q(√

a +√

b) : Q] = 4 (recall: we have shown thatQ(√

a +√

b) =Q(√

a,√

b) ).

Findmin(

√a +√

b,Q,x)

Use the result to write down the minimal polynomial for√

2 +√

3,√

2 +√

7,√

3 +√

5

over Q.

As in the lecture, if√

a +√

b is a root and a and b are such that [Q(√

a +√

b) : Q] = 4, then ±√

a±√

b are allroots. Then, the minimal polynomial is

f (x) = x4−2(a + b)x2 + (a− b)2.

It is easy to show that it is irreducible overQ since its roots are not inQ and all possible quadratic factors are also notpolynomials in Q[x].

For a = 2, b = 3 we havef (x) = x4

−10x2 + 1.

Similarly, if a = 2,b = 7 or a = 3,b = 5 we get

f (x) = x4−18x2 + 25, f (x) = x4

−16x2 + 4

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300

Chapter 17

Galois theory

In this chapter we will study one of the most elegant parts of mathematics. The beauty of Galois theoryis in its power to connect some powerful mathematics with some elementary classical problems as that ofroots of polynomials.

17.1 Automorphisms of fields

Our first task is to establish a link between group theory and field theory by examining automorphisms offields.

Proposition 17.1. The set of all automorphisms of a field F forms group with composition of functions.

Proof. If σ and τ are the automorphism of E, then such are στ and σ−1. The identity also is a automorphism.Thus, the set of all automorphisms of a field F is group.

Proposition 17.2. Let E a extension field of F. Then, the set of all automorphisms of E that fix every element of F isgroup. Hence the set of all automorphisms σ : E→ E such that σ(α) = α for every α ∈ F is group.

Proof. We must only prove that the set of all automorphisms of E that fix every element of F forms asubgroup of the group of automorphisms of E. Let σ and τ two the automorphism of E such that thatσ(α) = α and τ(α) = α for every α ∈ F. Then, στ(α) = σ(α) = α and σ−1(α) = α. Since identity fixes for everyelement of E, the set of automorphisms of E that fix the elements of F is a subgroup of the group ofautomorphisms of E.

Let E a extension field of F. The group of automorphisms of E we will denote by Aut(E). The Galois

group of E over F is the group of automorphisms of E that fix F -in in every element. Thus,

Gal (E/F) = σ ∈ Aut(E) : σ(α) = α for every α ∈ F .

If f (x) is a polynomial in F[x] and E is splitting field of f (x) over F, then the Galois group of f (x) is Gal (E/F).

Example 17.1. Complex conjugation, of defined as:

σ : a + bi 7→ a−bi

is a automorphism of complex numbers. Since

σ(a) = σ(a + 0i) = a−0i = a,

the automorphism defined from complex conjugation must be in Gal (C/R).

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17 Shaska T.

Example 17.2. Let be given field Q ⊂Q(√

5) ⊂Q(√

3,√

5). Then, for a,b ∈Q(√

5),

σ(a + b√

3) = a− b√

3

is a automorphism of Q(√

3,√

5) which fixes Q(√

5). Similarly

τ(a + b√

5) = a− b√

5

is a automorphism of Q(√

3,√

5) that fixes Q(√

3). The automorphism µ = στ moves√

3 and√

5. We will see nextthat id,σ,τ,µ is the Galois group of Q(

√3,√

5) over Q. The following table shows se this group is isomorphic toZ2×Z2.

id σ τ µid id σ τ µσ σ id µ ττ τ µ id σµ µ τ σ id

The field Q(√

3,√

5) can be thought as a vector field with basis 1,√

3,√

5,√

15 . It is not a coincidence the fact that|Gal (Q(

√3,√

5)/Q)| = [Q(√

3,√

5) :Q)] = 4.

Proposition 17.3. Let E/F be a field extension and f ∈ F[x]. Then, every automorphism in Gal (E/F) determines apermutation of roots of f (x) in E.

Proof. Let f (x) be given byf (x) = anxn + an−1xn−1 + . . .a1x + a0

and α ∈ E such that f (α) = 0. Then, for any automorphism σ ∈ Gal (E/F) we have σ(α) is also a root of f (x)since

f (σ(α)) = anσ(α)n + an−1σ(α)n−1 + · · ·+ a1σ(α) + a0

= σ(anαn + an−1α

n−1 + . . .a1α+ a0)= σ(0) = 0

Let E a algebraic extension of a field F. Two elements α,β ∈ E are called conjugate over F if they have

the same minimal polynomial. For example, in field Q(√

2) the elements√

2 and −√

2 are conjugated overQ since that of two they are root of the polynomial irreducible x2

−2.

Proposition 17.4. If α and β are conjugate over F, then there is a isomorphism

σ : F(α)→ F(β),

such that σ is identity when restricted over F.

Proof. Exercise.

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Shaska T. 17

Theorem 17.1. Let f ∈ F[x] and E f its splitting field over F. If f (x) does not have multiple roots, then

|Gal (E/F)| = [E : F].

Proof. We will use induction over the degree of f (x). If the degree of f (x) is 0 or 1, then E = F. Assumethat this is true for all polynomials with degree k where 0 ≤ k < n. Let p(x) an irreducible factor of f (x) withdegree r. Since all roots of p(x) are in E we can choose one of them, say α, such that F ⊂ F(α) ⊂ E. If β is anyroot of p(x), then F ⊂ F(β) ⊂ E. From the above there is a unique isomorphism σ : F(α)→ F(β) for every βthat fixes every element of F.

Since E is a splitting field of F(β), then are exactly r isomorphisms such that. We can factor p(x) in F(α)as p(x) = (x−α)p1(x). Degrees of p1(x) and q1(x) are less than r. Since we we know that E is a splitting fieldof p1(x) over F(α) we can of apply the hypothesis of induction to prove that

|Gal (E/F(α))| = [E : F(α)].

Hence[E : F] = [E : F(α)][F(α) : F]

are possible automorphisms of E that fix F, or |Gal (E/F)| = [E : F].

Corollary 17.1. Let F a finite field with a extension of finite E such that [E : F] = k. Then, Gal (E/F) is cyclic.

Proof. Let p characteristic of E and F and assume that the order of E and F is respectively pm and pn. Then,nk = m. Also assume that E is splitting field of xpm

− x over subfield with order p. Thus E is also splittingfield of xpm

−x over F. Applying the above theorem we get |Gal (E/F)| = k.To prove that Gal (E/F) is cyclic we must find a generator for Gal (E/F). Let σ : E→ E such σ(α) = αpn

.Assume that σ is the desired element in Gal (E/F). First we must prove that σ is in Aut (E). If α and β arein E, then

σ(α+β) = (α+β)pn= αpn

+βpn= σ(α) +σ(β).

Also it is easy to prove that σ(αβ) = σ(α)σ(β). Since σ is a nonzero homomorphism of fields it is injective;see ??. It is also surjective because E is a finite field. We we know that σ must be in Gal (E/F) since F issplitting field of xpn

−x over field basis with order p. This means se σ fixes for every element in F. Finallymust of prove that the order of σ is k. We we know that σk(α) = αpk

= α is identity of Gal (E/F). However σr

can not be identity for 1 ≤ r < k otherwise xprk−x is pm root which is impossible.

Example 17.3. Now we can confirm that the Galois group of Q(√

3,√

5) over Q is isomorphic to Z2 ×Z2. Thegroup H = id,σ,τ,µ is a subgroup of Gal (Q(

√3,√

5)/Q). However H must be of all Gal (Q(√

3,√

5)/Q) because

|H| = [Q(√

3,√

5) :Q] = |Gal (Q(√

3,√

5)/Q)| = 4.

Example 17.4. Find the Galois group of

f (x) = x4 + x3 + x2 + x + 1

over Q.

Proof. We we know that f (x) is irreducible. Moreover, since (x−1) f (x) = x5−1, we can use the De Moivre

formula to express roots ωi, for i = 1, . . . ,4 of f (x). We have

ω = cos(2π/5) + ısin(2π/5).

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17 Shaska T.

Thus, splitting field of f (x) must be Q(ω). We can define an automorphisms σi of Q(ω) with σi(ω) = ωi fori = 1, . . . ,4. It is easy to see that these automorphisms are different in Gal (Q(ω)/Q). Since

[Q(ω) :Q] = |Gal (Q(ω)/Q)| = 4,

σi’s must all be from Gal (Q(ω)/Q). Thus Gal (Q(ω)/Q)Z4 implying that ω is a generator for the Galoisgroup.

17.2 Separable Extensions

A polynomial f (x) ∈ F[x] with degree n is separable if it has n distinct roots in splitting field of f (x). Thus,f (x) is separable when it is factored in linear factors over splitting field of f . A extension E of F is called aseparable extension of F if every element in E is root of a splitting polynomial in F[x].

Example 17.5. The polynomial x2−2 is separable overQ since it is factored as (x−

√2)(x+

√2). Moreover,Q(

√2)

is a separable extension of Q. Let α = a + b√

2 be an element in Q. If b = 0, then α is root of x− a. If b , 0, then α isroot of a splitting polynomial

x2−2ax + a2

−2b2 = (x− (a + b√

2))(x− (a− b√

2)).

We have a test for determining if a polynomial is separable or not. Let’s have

f (x) = a0 + a1x + · · ·+ anxn

a polynomial in F[x]. Define the derivative of f (x) to be

f ′(x) = a1 + 2a2x + · · ·+ nanxn−1.

Lemma 17.1. Let F a field and f (x) ∈ F[x]. Then, f (x) is separable if and only if when f (x) and f ′(x) are relativelyprime, so gcd ( f , f ′) = 1.

Proof. Let f (x) separable. Then, f (x) is factored over extension field of F as f (x) = (x−α1)(x−α2) · · · (x−αn),where αi , α j for i , j. Then,

f ′(x) = (x−α2) · · · (x−αn)+ (x−α1)(x−α3) · · · (x−αn)+ · · ·+ (x−α1) · · · (x−αn−1).

Thus, f (x) and f ′(x) can not have common factors.To prove the converse assume that f (x) = (x−α)kg(x), where k > 1. Find the derivative

f ′(x) = k(x−α)k−1g(x) + (x−α)kg′(x).

Thus, f (x) and f ′(x) have a common factor.

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Shaska T. 17

17.2.1 Multiple roots of polynomials

Let k be a field of f (x) ∈ k[x]. Let α be a root of f (x). The multiplicity of α is the largest positive integer nsuch that (x−α)n divides f (x). If n = 1 then α is called a simple root, otherwise a multiple root.

We want to determine conditions when f (x) has multiple roots. Let f be factored into a product ofirreducibles as follows:

f = f m11 · · · f

mrr .

Obviously then f has multiple roots if some mi > 1 for i = 1, . . . ,r. So the question becomes whetherirreducible polynomials have multiple roots.

Lemma 17.2. Let f (x) ∈ k[x] be a non-constant irreducible polynomial. The the following are equivalent:

i) f (x) has a multiple rootii) ∆ f (x) = 0iii) Res( f , f ′, x) = 0iv) gcd ( f , f ′) , 1v) char (k) = p > 0 and f (x) = g(xp) for some g(x).vi) all roots are multiple.

Proof. The equivalence of i), ii) iii), and iv) has been shown in the chapter on polynomials.

iv) =⇒ v). Since f is irreducible then deg f ′ < deg f , then gcd ( f , f ′) , 1 implies that f ′ = 0. Hence,char (k) = p > 0 and f is a polynomial in xp.

v) =⇒ vi). Let f (x) = g(xp) whereg(x) =

∏(x−αi)mi

in some splitting field. Then

f (x) = g(xp) =∏

(xp− ai)mi =

∏(x−αi)pmi

where αp1 = ai. Thus every root of f (x) has multiplicity at least p.

vi) =⇒ i) is obvious and the proof is complete.

Definition 17.1. An irreducible polynomial f (x) ∈ k[x] is called a separable polynomial if it has no multiple rootsin any extension of k. An irreducible polynomial that is not separable is called inseparable.

Definition 17.2. A field k is called perfect if all irreducible polynomials in k[x] are separable.

Corollary 17.2. All fields of characteristic 0 are perfect.

Corollary 17.3. A field k such that char (k) = p > 0 is perfect if and only if every element of k is a p-th power (i.e.,k = kp)

Proof. Assume that k contains an element a which is not a p-th power. Then

f (x0 = xp− a ∈ k[x]

is not separable sincef (x) = (x−α)p, αp = a

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17 Shaska T.

in some splitting field of f (x), which is a contradiction. If every element in k is a p-th power then everypolynomial in xp is a p-th power in k[x]. For example,

f (xp) =∑

ai(xp)i =∑

bpi (xi)p =

∑(bixi)p =

(∑bixi

)p

Hence, f (x) is not irreducible. This completes the proof. We know that every field of characteristic 0 is perfect. The following is also true.

Remark 17.1. Every finite field is perfect (chapter on finite fields)

Then what fields are not perfect?

Example 17.6. Let Fq be a field of characteristic p > 0 (i.e., q = pn). Then the field Fp[x] is not perfect since x is nota p-th power.

Definition 17.3. Let F/k be a field extension. Then α ∈ F is called separable over k is min (α,k,x) is separable. Thefield extension F/k is called a separable extension if every α ∈ F is separable over k.

Let E splitting field of a polynomial f (x) in F[x]. Assume that f (x) is factored over E as

f (x) = (x−α1)n1 (x−α2)n2 · · · (x−αr)nr =

r∏i=1

(x−αi)ni .

We say that multiplicity of roots αi of f (x) is ni. A root with multiplicity 1 is called simple root. Recall thatse a polynomial f (x) ∈ F[x] with degree n is separable if it has n distinct roots in E. Thus, f (x) is separableif ai is factored in linear factors over E[x]. A extension E of F is a separable extension of F if every elementin E is root of a splitting polynomial F[x]. Also recall that f (x) is separable if and only if gcd ( f (x), f ′(x)) = 1.

Proposition 17.5. Let f (x) a irreducible polynomial over F[x]. If characteristic of F is 0, then f (x) is separable. Ifcharacteristic of F is p and f (x) , g(xp) for a g(x) in F[x], then f (x) is also separable.

Proof. First assume that char F = 0. Since deg f ′(x) < deg f (x) and f (x) is reducible, then gcd ( f (x), f ′(x)) , 1only if f ′(x) is the zero polynomial, which is impossible in a field with characteristic zero. If char F = p,then f ′(x) can to be polynomial zero if for every coefficient of f (x) is a multiple of p. This can happen onlyif we have a polynomial of the form f (x) = a0 + a1xp + a2x2p + · · ·+ anxnp.

Some extension fields of F of the form F(α) are easier to study. Let be given a extension field E of F.

lHow is it possible to find an element α ∈ E such that E = F(α). In this case α is called primitive element.We already have seen examples of extension of fields generated by primitive elements. For example,

Q(√

3,√

5) =Q(√

3 +√

5)

andQ(

3√5,√

5 ı) =Q(6√

5 ı).

There exists an element primitive for every finite, separable extension as we can prove next.

Theorem 17.2 (Primitive Element Theorem). Let E a finite separable extension of fields F. Then, there is anelement α ∈ E such that E = F(α).

Proof. Assume that E is a finite extension of an infinite field. We will prove the theorem for F(α,β). Thenthe general case is easily obtained by induction. Let f (x) and g(x) be minimal polynomials of α and βrespectively . Let K be the splitting field of f (x) and g(x). Assume that f (x) has roots α = α1, . . . ,αn in K and

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g(x) has roots β = β1, . . . ,βm in K. All of these roots have multiplicity 1 since E is separable over F. Since F isinfinite we can find a ∈ F such that

a ,αi−αβ−β j

for every i and j where j , 1. Thus a(β−β j) , αi−α. Let γ = α+ aβ. Then,

γ = α+ aβ , αi + aβ j.

Thus, γ−aβ j ,αi for every i, j where j, 1. Define h(x) ∈ F(γ)[x] such that h(x) = f (γ−ax). Then, h(β) = f (α) = 0.However h(β j) , 0 for j , 1. Thus, h(x) and g(x) have a common factor in F(γ)[x]. Hence irreduciblepolynomial of β over F(γ) must be linear because β is a common root of g(x) and h(x), which implies β ∈ F(γ)and α = γ− aβ is in F(γ). Thus, F(α,β) = F(γ).

Exercises:

17.1. Separable extensions form a distinguished class.

17.2. Let F be a field of characteristic p and

f (x) = xp−x− a ∈ F[x]

Prove that if f (x) is reducible in F[x], then it splits in F[x].

17.3. Find the splitting field off (x) = xq

−x

over Fp, where q = pn. What is the cardinality and the degree of the splitting field?

17.3 Galois extensions

Let F be a field and G its group of automorphisms. The fixed field of the group G is:

FG = α ∈ F | σ(α) = α,∀σ ∈ G

The reader can to prove that FG is subfield of F. Let F/k a algebraic extension. Recall that an automorphismof F over k is a automorphism σ of F such that σ(α) = α for every α ∈ k.

Definition 17.4. Let F/k a algebraic extension. We say that F/k is a extension Galois if it is normal and separable.

The group Aut k(F) of automorphisms of F over k is called the Galois group of F over k and denoted byGal (F/k).

Proposition 17.6. Let F/k a algebraic extension. The following are equivalent:

i) F/k is a extension Galois.ii) There exists a group G of automorphisms of F over k such that

k = FG

iii) F is splitting field of a separable family of polynomials over k.

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17 Shaska T.

Proof. ii) =⇒ i) : Let α ∈ F andf (x) = min (α,F,x)

To prove that F/k is normal we have to prove that all roots of f (x) are in F. Denote byOα, G -the orbit of α :

Oα := σ(α) | σ ∈ G = α = α1,α2, . . .αr

since each σ(α) is root of f (x) then Oα has cardinality ≤ deg f , and also Oα ⊂ F. Define h(x) such that:

h(x) :=r∏

i=1

(x−αi)

Then, for every σ ∈ G we have:

σ(h(x)) :=r∏

i=1

(x−σ(αi)) =

r∏i=1

(x−αi) = h(x)

Thus, h(x) ∈ k[x]. Then, we have thatdegh ≤ deg f

and h(x) is a multiple of f (x). Thus, h(x) = f (x) and F/k is normal.i) =⇒ ii) : All roots of h(x) are distinct and therefore h(x) = f (x) is separable.

o f ⇐⇒ iii) : First we assume se F/k is Galois. Let α ∈ F and

f (x) = min (α,k,x).

Since F/k is normal then F is splitting field of f (x). Hence it is separable.Assume that F = k(α1,α2, . . . ) is splitting field of a separable family of polynomials. Then, F/k is normal.All αi are roots of separable polynomials over k. Let an element α ∈ F then we have:

α =p(α1, . . . ,am)q(α1, . . . ,αm)

.

Hence α is separable over k. Therefore F/k is separable and Galois.

Lemma 17.3. Let L/k a finite extension Galois and G = Gal (L/k). If F is an intermediate subfield k ⊂ F ⊂ L, thenL/F is Galois and the function

Φ : F −→ Gal (L/F)

is injective.

Proof. From above, if L/F is normal and separable, then it is a Galois extension. Let F and F′ two intermediatefield and H := Gal (L/F), H′ = Gal (L/F′) corresponding groups. Then,

F = KH and F′ = KH′

If H = H′ then F = F′. Hence, Φ is injective.

Corollary 17.4. Let L/k a finite extension Galois and G = Gal (L/k). Also let F and F′ two intermediate fields andH := Gal (L/F), H′ = Gal (L/F′) corresponding groups. Then, the following statements are true:

i) Φ(FF′) = H∩H′

ii) Φ(F∩F′) is the subgroup the smallest of G that contains H and H′.

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L

FF′

F F′

F∩F′

k

//

eG

H∩H′

H H′

M

G

Figure 17.1: The Galois correspondence

Proof. Exercise.

The following Lemma determine the degree of extensions L/LG.

Lemma 17.4. Let G a finite subgroup of Aut (L) with order |G| = n. Then, degree of extensions L/LG is:

[L : LG] = n = |G|

Proof. Now we state the main result of Galois theory.

Theorem 17.3. Let L/k a finite extension Galois and G = Gal (L/k). There exists a bijection between the set ofintermediate subfields F of L/k and subgroups H < G, that is given from

F = LH←→H

and G(L/F)H. Also F/k is Galois if and only if H E G. In this case G(F/k)G/H.

Proof. The function Φ of Lemma 17.3 is injective. This function is also surjective because for every subgroupH ≤ G there is a intermediate field LH.

Assume that HEG. To prove that F/k is Galois we must show that

i) F/k is normalii) F/k is separable.

Let α ∈ F = LH and β a root of min (α,k,x). If we show that β ∈ F then we have proved that F/k is normal.From the theorem of isomorphic extensions Theorem 16.8 there is a σ ∈ G such that:

σ(α) = β

Let τ ∈H. Then,τ(β) = σ(σ−1τσ(α))

Since H is normal then σ−1τσ ∈H. Also since every element of H fixes the elements of F we have that

σ−1τσ(α) = α

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17 Shaska T.

Henceτ(β) = σ(α) = β ∈ F

which shows se F/k is normal and is separable because separable form a distinguished class.

L

F = LH

k

oo //

1G

H

G

Figure 17.2: The Galois correspondence

Conversely assume that F/k is Galois with Galois group Gal (F/k). Define the function

φ : G→Gal (F/k)

σ→ σ|F

Thus, σ|F is in Gal (F/k) and φ is of well defined. Then,

ker(φ) = σ ∈ G | σ|F = id = H

Thus, HEG. The function φ is surjective from Theorem 16.8. Hence, we have

Gal (F/k)G/H.

Let f (x) a irreducible polynomial in k[x] which is factored as follows:

f (x) = (x−α1) . . . (x−αn)

in a splitting field E f . Then, E f /k is Galois because is a normal extension and separable. The group Gal (E f /k)is called the Galois group of f (x) over k and do of denoted by Gal ( f ). The elements of Gal ( f )permuteroots of f (x). Thus, the Galois group of polynomial has a copy isomorphic in Sn, where n is degree of thepolynomial.

Example 17.7 (Cubic polynomials). Let f (x) be an irreducible cubic polynomial in k[x]. We have shown that[E f : k] = 3 or 6. Hence, the Galois group Gal ( f ) is a subgroup of S3 with order 3 or 6. Thus, Gal ( f )A3 if and onlyif ∆ f is a square in k, otherwise Gal( f )S3.

Exercises:

17.4. Suppose that E is a finite extension Galois of fields F. If Gal(E/F) has order pq, where p < q are two differentprimes and p does not divide q−1, then prove that E has two subfields Ep and Eq, which are fixed under the action ofGal(E/F), such that Ep∩Eq = F, Ep and Eq generate E and Gal(Ep/F) (resp. Gal(Eq/F) ) is cyclic with order p (resp.q ).

17.5. Let E a finite Galois extension of F and G = Gal (E/F). Denote with (·)′ the Galois correspondence L −→ L′

and H −→H′, which maps intermediate subfields to subgroups of G and conversely. Prove that(a) If L is an intermediate subfield which is invariant under all automorphisms of G, then L′ is normal in G.(b) If H is normal subgroup of G, then prove that σ(H′) = H′, for every σ ∈ G.

17.6. Prove that every automorphism of the real fields R is identity.

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17.4 Cyclotomic extensions

Let n a positive number n. A n-cyclotomic polynomial is a polynomial of the form

Φn(x) = (x−α1) . . . (x−αr)

where α1, . . . ,αr are root of n -te primitive of unity. Hence if fixes a primitive root of unity α then

Φn(x) =∏

(r,n)=1

(x−αr)

and degΦn(x) = ϕ(n) where ϕ(n) is the Euler’s function.Denote with Fn splitting field of xn

− 1 over field k. Then, Fn/k is called extension of n-te cyclotomicover k.

The main goal of this chapter is to determine Fn and Gal (Φn) = Gal (Fn/k). Let us see first some propertiesof cyclotomic polynomials:

Lemma 17.5. Let Φn(x) a cyclotomic polynomial over k. Then,

i) degΦn(x) = ϕ(n)ii) Φn(x) is a monic with coefficients from subfield of kiii) If k =Q then

Φn(x) ∈Z[x]

iv) The following is truexn−1 =

∏d|n

Φd(x)

Proof. Left to the reader

Example 17.8. Prove that:

Φ1(x) = x−1Φ2(x) = x + 1

Φ3(x) = x2 + x + 1

Φ4(x) = x2 + 1

Φ6(x) = x2−x + 1

Φ8(x) = x4 + 1

Φ10(x) = x4−x3 + x2

−x + 1

(17.1)

and in general for a prime number p we have:

Φp(x) = xp−1 + xp−2 + · · ·+ x + 1

Theorem 17.4. All cyclotomic polynomials Φn(x) ∈Q[x] are irreducible in Q[x].

Proof. If Φn(x) is irreducible in Q, then by Gauss’ lemma is also irreducible in Z[x] Hence assume that

Φn(x) = f (x) · g(x)

where f (x) and g(x) are monic and at least one of them is irreducible over Z. Suppose that f (x) is irre-ducible overZ. Let α root of f (x). Then, α is root of unity and αp is a root of n -te primitive of unity since p - n

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17 Shaska T.

: f (αp) = 0, for every prime number p - n.

Proof:Let f (αp) , 0. Then, αp must be root of g(x). Hence α is a root of g(xp).Since f (x) is a monic and irreducible then f (x) | g(xp) so, say

g(xp) = h(x) f (x)

where h(x ) is a polynomial monic h(x) ∈ Z[x]. Reduce mod p and denote b f the remainder mod p off ∈Z[x]. Then, in Fp we have:

Φn(x) = f · g.

Notice that Φn(x) | (xn−1) and it does not have root in some extension of Fp. Since in Fp we have that ap = a

then for every a ∈ Fp we have:

g(xp) = g(x)p.

Hence f | (g)p and as a consequence every factor q(x) in Fp[x] of f divides also g. Hence q2 divides Φn whichimplies that Φn(x) has multiple roots. This is a contradiction and this completes the proof.

Hence all primitive roots of unity are roots of f (x). Thus, f (x) = Φn(x) and Φn(x) is irreducible.

Let εn be an n-th primitive root and denote by Fn the splitting field of Φn(x). Then, we have that:

Corollary 17.5. If Fn is n-th cyclotomic extension of Q then Fn =Q(εn). Moreover, we have:

G(Q(εn)/Q) (Z/nZ)∗, and [Q(εn) :Q] = ϕ(n)

The main result of cyclotomic extensions is the Kronecker-Weber theorem.

Theorem 17.5 (Kronecker-Weber). Let F a finite extension Abelian of Q. Then, F is contained in some extensioncyclotomic of Q.

Proof. The proof is outside the scope of these notes.

Exercises:

17.7. Determine Φn(x) for n = 12, . . . ,20.

Let n > 1 a number odd. Prove thatΦ2n(x) = Φn(−x)

Let n a number odd. Prove that splitting field of Φn(x) acts the same as the splitting field of Φ2n(x).

Te of are n and M two positive numbers with

d = gcd (m,n), l = lcm (m,n).

17.8. Denote the n-th cyclotomic extension of Q by Sn. Prove that:i) If n |m then Sm is a extension of Sn.ii) SnSm = Sliii) Sn∩Sm = Sl

17.9. If d ∈Q show that Q(√

d) is in a cyclotomic extension Sn of Q (do not use the Kronecker-Weber theorem).

17.10. Determine which are roots of unity in the following separable fields: Q(i), Q(√

2, Q(√−2, Q(

√−3, Q(

√3.

17.11. For what integers n we have that [Q(εn) :Q] = 2?

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17.5 Norm and trace

In this section we study two very important concepts in the theory of fields, that of norm and trace. Let begiven F/k a extension field with order [F : k] = n. Fix α ∈ F and consider the linear map

Lα : F −→ Fx −→ ax

(17.2)

Recall that F is a vector space n -dimensional over k and Lα a function linear over this vector space. Let Mα

be the matrix associated with the linear map Lα.

Definition 17.5. Let be given F/k a finite field extension. The norm NFk and trace TrF

k of for every element α ∈ F aredefined as follows:

NFk (α) = det(Mα) TrF

k (α) = tr (Mα) (17.3)

Recall from linear algebra that the change of basis of a vector space changes the matrix Mα to a similarmatrix A−1MαA. The determinant and trace are the same since

det(A−1MαA) = det(Mα)

tr (A−1MαA) = tr (Mα)(17.4)

Thus, norm and trace of an element are well-defined.

Example 17.9. Let be given F = k(√

d) for a d ∈ F such that d is not a complete square in k and α ∈ F such α= a+b√

d.We want to find NF/k(α) and TrF/k(α).

Choose a basis B = 1,√

d. Then,

Lα(1) = a + b√

d

La(√

d) = (a + b√

d) ·√

d = bd + a√

d(17.5)

Then, the associated matrix is

Mα =

[a b

bd a

]t

=

[a bdb a

]Thus we have:

NFk (α) = det(Mα) = a2

− b2d

TrFk (α) = tr (Mα) = 2a

(17.6)

Lemma 17.6. Let F = k(α) be an algebraic extension where the minimal polynomial of α is

min (α,k,x) = xn +βn−1xn−1 + · · ·+β1x +β0.

Then,NF

k (α) = (−1)nβ0, TrFk (α) = −βn−1

Proof. We we know that a basis for k(α) is

B = 1,α,α2, . . . ,αn−1.

Then,Lα(1) = α = (0,1,0,0, . . .0)

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17 Shaska T.

Lα(α) = α2 = (0,0,1,0, . . .0)

Lα(α2) = α3 = (0,0,0,1, . . .0)

. . . . . . . . .

Lα(αn−1) = αn = (−β0,−β1,−β2,−β4, . . . ,βn−1)

and the matrix Mα is given nga:

C f :=

0 0 . . . . . . −β01 0 . . . . . . −β10 1 . . . . . . −β2

. . . . . . . . .

. . . . . . . . .0 0 . . . 1 −βn−1

For details see [?lin-alg, Chapter 4]. Then,

tr (Mα) = −βn, and det(Mα) = (−1)nβ0.

this completes the proof.

Remark 17.2. The matrix Mα is associated matrix of min (α,k,x).

Norm and trace are given as follows:

Theorem 17.6. Let be given F/k a finite field extension and σ1, . . . ,σn distinct embeddings of F to an algebraic closureka of k. For α ∈ F we have:

NFk (α) =

r∏j=1

σ j(α)

[F:k]i

and TFk (α) = [F : k]i

n∑j=1

σ j(α)

Proof. This theorem will be proved only for Galois extensions in the following corollary.

Corollary 17.6. Let be given F/k a extension Galois finite with group Galois G. Then, for every α ∈ F,

NFk (α) =

∏σ∈G

σ(α) and TFk (α) =

∑σ∈G

σ(α)

Proof. Let be given α ∈ F, f (x) = min (α,k,x) and

G = 1,σ, . . . ,σn−1.

We know that all σi(α), for i ≤ n are also root of f (x). The result now follows.

Example 17.10. Let be given F = k(√

d). Then, F/k is Galois since for every extension with degree two is Galois.The group Galois is G = id,σ ku

σ :√

d −→−√

d

Then, for α = a + b√α ∈ F, we have:

NFk (α) = α ·σ(a) = (a + b

√

d)(a− b√

d) = a2− b2d.

Similarly, TrFk (α) = 2a.

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Lemma 17.7. Let be given L/F/k finite extension of fields. Then,

NLk = NF

k NLF , TrL

k = TrFk TrL

F

Proof. Exercise.

Exercises:

17.12. Let be given p a prime number and K :=Q(εp). Prove that

NKQ(1−εp) = p

17.13. Let be given n ≥ 3 a integer, εn a root of n -te primitive of unity and K :=Q(εn). Prove that NKQ

(εn) = 1.

17.14. Let be given F =Q(√

3) and L =Q( 4√3). Determine NFQ

(√

3), NLF(√

3), TrFQ

(√

3), TrLF(√

3).

17.15. Let be given [K :Q] = n and α ∈Q. Prove that

NKQ(α) = αn, and TrK

Q(α) = nα.

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17 Shaska T.

17.6 Cyclic extensions

Definition 17.6. Let be given F/k a extension Galois and Gal(F/k) a cyclic group, Gal(F/k) = 〈σ〉. Then, F/k is calleda cyclic extension.

Theorem 17.7 (Hilbert’s 90 Theorem). Let be given K/k a extension cyclic finite with Galois group Gal(K/k) = 〈σ〉

and α ∈ K. Then, N(α) = 1 if and only if there exists a β ∈ K such that α =βσ(β) .

Proof. See notes ....

Theorem 17.8. Let be given k a field which contains a root of n -te of unity. Assume that if char k = p > 0 where(n,p) = 1. Then, the following are equivalent:

i) F/k is cyclic with degree d | nii) F = k(α) where

min (α,k,x) = xd− a

for d | n and a ∈ k.iii) F is a separable field of a irreducible polynomial

f (x) = xd− b

where d | n and b ∈ k.iv) F is a separable field of

f (x) = xn−b

for b ∈ k.

Proof. Use Hilbert’ 90.

Theorem 17.9 (Hilbert’s 90 Theorem, (additive version)). Let be given K/k a finite extension cyclic with Galoisgroup Gal(K/k) = 〈σ〉 and α ∈ K. Then, we have that Tr(α) = 0 if and only if when there is a β ∈ K such thatα = β−σ(β)

Proof. Exercise. The following will be accepted without proof.

Theorem 17.10 (Artin-Schreier). Let be given char (k) = p > 0. The polynomial

f (x) = xp−x− a ∈ k[x]

either is in k or is irreducible over k. Moreover, the following are equivalent:

i) F/k is cyclic and [F : k] = pii) F = k(α) where

min (α,k,x) = xp−x− a

for some α ∈ k.ii) F is splitting field of an irreducible polynomial

f (x) = xp−x− a

for a ∈ k.

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Exercises:

17.16. Let be given F which is a extension of k of generated from all n-roots of unity, for every n ≥ 1. Prove that F/kis Abelian.

17.17. Let be given F a field and σ ∈ Aut (F) such that |σ| = s > 1. Prove that there is a α ∈ F such that

σ(α) = α+ 1

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17 Shaska T.

17.7 Fundamental theorem of Galois theory

In this section we state and prove what is commonly known in literature as the Fundamental Theorem ofGalois theory. Most of the material is a review of what was covered in previous sections.

Proposition 17.7. Let σi : i ∈ I a collection the automorphisms of fields F. Then,

Fσi = a ∈ F : σi(a) = a for every σi

is a subfield of F.

Proof. Let σi(a) = a and σi(b) = b. Then,

σi(a±b) = σi(a)±σi(b) = a± b

andσi(ab) = σi(a)σi(b) = ab.

If a , 0, then σi(a−1) = [σi(a)]−1 = a−1. Finally σi(0) = 0 and σi(1) = 1 since σi is a automorphism.

Corollary 17.7. Let F a field and G a subgroup of Aut (F). Then,

FG = α ∈ F : σ(α) = α for every σ ∈ G (17.7)

is a subfield of F.

Subgroup Fσi of F is called the fixed field of σi. The fixed field for a subgroup G of Aut(F) do ofdenoted by FG.

Example 17.11. Let σ : Q(√

3,√

5)→ Q(√

3,√

5) the automorphism that maps√

3 to −√

3. Then, Q(√

5) issubfield of Q(

√3,√

5) which is fixed from σ.

Proposition 17.8. Let E a splitting field over F of an irreducible polynomial. Then, EGal (E/F) = F.

Proof. Let G = Gal (E/F). Thus, F ⊂ EG ⊂ E. Also E is splitting field of EG and Gal (E/F) = Gal (E/EG). Fromabove we have

|G| = [E : EG] = [E : F].

Thus [EG : F] = 1. therefore EG = F.

Exercise 17.1. Let G a finite group of automorphisms of E and let F = EG. Then, [E : F] ≤ |G|.

Let E be an algebraic extension of F. If for every irreducible polynomial in F[x] with a root in E of hasall its roots in E, then E is called a normal extension of F. Thus, for every polynomial irreducible in F[x]that contains a root in E is a product of linear factors in E[x].

Theorem 17.11. Le to be E a extension field of F. Then, the following are equivalent.

1. E is a finite extension , normal and separable of F.

2. E is a splitting field of an irreducible polynomial over F.

3. F = EG for a finite automorphism group of E.

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Shaska T. 17

id,σ,τ,µ

id,σ id,τ id,µ

id

@@@

@@@

Q(√

3,√

5)

Q(√

3) Q(√

5) Q(√

15)

Q

@@@

@@@

Figure 17.3: Gal (Q(√

3,√

5)/Q)

Proof. (1)⇒ (2). Let E a finite extension , normal and separable of F. From Theorem 17.2 we can find anelement α in E such that E = F(α). Let f (x) minimal polynomial of α over F. Then E must contain all rootsof f (x) since it is a normal extension of F. Thus, E is splitting field for f (x).

(2) ⇒ (3). Let E be the splitting field an irreducible polynomial over F. We know that EGal (E/F) = F.Since |Gal (E/F)| = [E : F], this is a finite group.

(3)⇒ (1). Let F = EG for a finite automorphism group G of E. Since [E : F] ≤ |G|, E is a finite extension ofF. To prove that E is a normal extension and finite of F, let f (x) ∈ F[x] a polynomial monic and irreduciblewhich has a root α in E. We must show that f (x) is product of linear factors in E[x].

We we know that automorphisms in G permute roots of f (x) that are in E. Thus, if G acts over αwe takethe distinct roots α1 = α,α2, . . . ,αn in E. Let g(x) =

∏ni=1(x−αi). Then, g(x) is separable over F and g(α) = 0.

Any σ in G permutes factors of g(x) since it permutes these roots. Thus, when σ acts over g(x) it must fixcoefficients of g(x). Thus coefficients of g(x) must be in F. Since deg g(x) ≤ deg f (x) and f (x) is minimalpolynomial of α, then f (x) = g(x).

Corollary 17.8. Let K a extension field of F such that F = KG for a finite automorphism group G of K. Then,G = Gal (K/F).

Proof. Since F = KG, then G is a subgroup of Gal (K/F). Thus,

[K : F] ≤ |G| ≤ |Gal (K/F)| = [K : F]

From this we conclude that G = Gal (K/F) since they must have the same order.

Example 17.12. Above we studied the automorphisms of Q(√

3,√

5) which fix Q. Fig. 17.3 shows the lattice ofsubfields ofQ(

√3,√

5)/Q and the lattice of subgroups of the group Gal (Q(√

3,√

5)/Q). The Fundamental Theoremof Galois Theory determines the correspondence between those two lattices.

Now we are ready to state and prove the Fundamental Theorem of Galois Theory.

Theorem 17.12 (Fundamental Theorem of Galois Theory). Let E/F be a Galois extension with Galois groupGal (E/F), then the following are true.

1. The function K 7→ Gal (E/K) is a bijection of subfields K of E that contains F with subgroups of Gal (E/F).

2. If F ⊂ K ⊂ E, then[E : K] = |Gal (E/K)| and [K : F] = [Gal (E/F) : Gal (E/K)].

3. F ⊂ K ⊂ L ⊂ E if and only if id ⊂ Gal (E/L) ⊂ Gal (E/K) ⊂ Gal (E/F).

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17 Shaska T.

4. K is a normal extension F if and only if Gal (E/K) is a normal subgroup of Gal (E/F). In this case

Gal (K/F) Gal (E/F)/Gal (E/K).

Proof. (1) Assume that Gal (E/K) = Gal (E/L) = G. Both K and L are fixed fields of G. Thus, K = L andthe function such K 7→ Gal (E/K) is bijective. To prove that the function is surjective let G a subgroup ofGal (E/F) and K to be the fixed field of G. Then, F ⊂ K ⊂ E. Therefore, E is a normal extension of K. Thus,Gal (E/K) = G and the function K 7→Gal (E/K) is a bijection.

(2) From above |Gal (E/K)| = [E : K]. Thus,

|Gal (E/F)| = [Gal (E/F) : Gal (E/K)] · |Gal (E/K)| = [E : F] = [E : K][K : F].

Thus, [K : F] = [Gal (E/F) : Gal (E/K)].(3) Prop. (3) is illustrated in Fig. 17.4. The proof is left as an exercise.

E

L

K

F Gal (E/F)

Gal (E/K)

Gal (E/L)

id

-

-

-

-

Figure 17.4: Subgroups of Gal (E/F) and subfield of E

(4) Let K a normal extension of F. If σ is in Gal (E/F) and τ is in Gal (E/K) must to prove that σ−1τσ isin Gal (E/K). Thus, It is enough we prove that σ−1τσ(α) = α for every α ∈ K. Assume that f (x) is minimalpolynomial of α over F. Then, σ(α) is also a root of f (x) which is contained in K since K is a normal extensionof F. Thus, τ(σ(α)) = σ(α) or σ−1τσ(α) = α.

Conversely, let Gal (E/K) a normal subgroup of Gal (E/F). We must show that F = KGal (K/F). Letτ ∈Gal (E/K). For every σ ∈Gal (E/F) there is a τ ∈Gal (E/K) such that τσ = στ. therefore for every α ∈ K

τ(σ(α)) = σ(τ(α)) = σ(α).

Thus, σ(α) must be in the fixed field of Gal (E/K). Let σ restriction of σ in K. Then, σ is a automorphismof K that fixes F since σ(α) ∈ K for every α ∈ K. Thus, σ ∈ Gal (K/F). Also we do prove that fixed field ofGal (K/F) is F. Let β be an element in K which is fixed from all automorphisms in Gal (K/F). In particularσ(β) = β for every σ ∈Gal (E/F). Thus, β is in the fixed subfield F of the group Gal (E/F).

Finally must prove that when K is a normal extension of F then

Gal (K/F) Gal (E/F)/Gal (E/K).

For σ ∈Gal (E/F), let σK the automorphism of K obtained from the restriction of σ in K. Since K is a normalextension the above paragraph shows se σK ∈ Gal (K/F). Therefore the function φ : Gal (E/F)→Gal (K/F)is defined as σ 7→ σK. This function is a group homomorphism since

φ(στ) = (στ)K = σKτK = φ(σ)φ(τ).

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Shaska T. 17

The kernel of φ is Gal (E/K). From (2),

|Gal (E/F)|/|Gal (E/K)| = [K : F] = |Gal (K/F)|.

Thus, the image of φ is Gal (K/F) and φ is surjective. Applying Theorem 9.4 we have: Gal (K/F) Gal (E/F)/Gal (E/K).

Next we see some examples of the galois correspondence.

Example 17.13. Construct the lattice of the field extension E f /Q, where E f is the splitting field of the polynomial

f (x) = x4−2

Q

Q(i)Q(√

2) Q(√

2 i)

Q(√

2, i)Q( 4√2) Q( 4√2 i) Q((1 + i) 4√2) Q((1− i) 4√2)

Q( 4√2)

HHHH

HHH

H

HHH

H

HHHH

PPPPPP

(a)

Figure 17.5: The group of Galois of x4−2

Solution: We will compare this latice with the lattice of subfields of the splitting field E f of f (x) = x4− 2

over Q. The splitting field E f of f (x) is Q( 4√2, ı). Notice that f (x) is factored as (x2 +√

2)(x2−√

2). Thus,roots of f (x) are ± 4√2 and ± 4√2 ı. First we add the root 4√2 to Q and and then ı to Q( 4√2). Thus, the splittingfield of f (x) is Q( 4√2)(ı) =Q( 4√2, ı).

Since [Q( 4√2) :Q] = 4 and ı is not in Q( 4√2) then [Q( 4√2, ı) :Q( 4√2)] = 2. Thus, [Q( 4√2, ı) :Q] = 8. The set

1,4√

2, (4√

2)2, (4√

2)3, ı, ı4√

2, ı(4√

2)2, ı(4√

2)3

is a basis for Q( 4√2, ı) over Q. The lattice of subfields of Q( 4√2, ı)/Q is presented in Fig. 17.5(a).The Galois group G of f (x) must have order 8. Let σ be the automorphism defined by

σ(4√

2) =4√

2

and σ(ı) = ı. Let τ the automorphism defined by complex conjugation. Thus, τ(ı) = −ı. Then, G has anelement of order 4 and an element of order 2. The elements of G are

G = id,σ,σ2,σ3,τ,στ,σ2τ,σ3τ,

with relations τ2 = id, σ4 = id and τστ = σ−1. Thus, G must be isomorphic to D4. The lattice of subgroups ofG is given in Fig. 17.5(b).

Exercises:

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17 Shaska T.

17.18. Prove that the Galois group of an irreducible polynomial of degree two is isomorphic to Z2.

17.19. Prove that the Galois group of irreducible polynomial of degree three is isomorphic to S3 or Z3.

17.20. Use the derivative to prove that if f (T) is irreducible polynomial over field F, then it has a root that again is insplitting field if and only if when characteristic of F is p > 0 and f (T) = g(Tp), for g(T) ∈ F[T].

17.21. Give the definition of the discriminant ∆ of a polynomial f (x) ∈ Q[x]. Prove that ∆2∈ Q. Also prove that

Gal ( f ) < An if and only if ∆2 is a complete square in Q.

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Shaska T. 17

17.8 Solvable extensions

In this section we introduce solvable extensions. Recall that our initial goal was to determine what algebraicequations can be solved by radicals. In this section we will determine how the should the extension E f /k befor a given polynomial f ∈ k[x] such that this polynomial is solvable by radicals. The reader should reviewall the definitions from Section 7.2.

A Galois extension F/k is called solvable if Gal (F/k) is a solvable group.

Exercise 17.2. Solvable extensions form a distinguished class.

We say that a extension field F/k has a radical series when there is a tower of fields

k = F0 < F1 < · · · < Fn = F

such that for every every step is one of the following types:i) Fi+1 = Fi(εn), where εn is an n-th root of unity.ii) Fi+1 = Fi(αi), where αn is a root of

f (x) = xn− a

for n > 1, a ∈ k and (n,p) = 1 if p > 0iii) If char (k) > 0 then Fi+1 = Fi(αi) where αn is a root of

f (x) = xn−x− a

for a ∈ k.A extension F/k is of solvable by radicals if it has a radical series. Then we have the following:

Theorem 17.13. If F/k is solvable by radicals then such is and Fn/k.

Theorem 17.14. A separable finite extension F/k is solvable by radicals if and only if it is solvable.

A extension field E of fields F is a radical extension if there exist the elements α1, . . . ,αr ∈ K and positiveintegers n1, . . . ,nr such that

E = F(α1, . . . ,αr),

where αn11 ∈ F and

αnii ∈ F(α1, . . . ,αi−1)

for i = 2, . . . ,r. A polynomial f (x) is solvable by radicals over F if splitting field K of f (x) over F is containedin a extension of F with radicals.

Example 17.14. The polynomial xn−1 is solvable with radicals overQ. Roots of this polynomial are 1,ω,ω2, . . . ,ωn−1,

where

ω = cos(2π

n

)+ o f sin

(2πn

).

splitting field of xn−1 over Q is Q(ω).

Lemma 17.8. Let F a field with characteristic zero and E f splitting field of the polynomial

f (x) = xn− a ∈ F[x].

Then, Gal (E f /F) is solvable group.

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17 Shaska T.

Proof. First assume that F contains all n -th roots of unity. Roots of xn−a are n√a,ω n√a, . . . ,ωn−1 n√a, whereω is

the n -th primitive root of unity. If ζ is one from these roots, then the other roots of xn−1 are ζ,ωζ, . . . ,ωn−1ζ

and E = F(ζ). Since Gal (E/F) permutes roots of xn−1 the elements in Gal (E/F) must be defined from their

action over these roots. Let σ and τ in Gal (E/F) and assume that σ(ζ) = ωiζ and τ(ζ) = ω jζ. If F containsroots of unity, then

στ(ζ) = σ(ω jζ) = ω jσ(ζ) = ωi jζ = ωiτ(ζ) = τ(ωiζ) = τσ(ζ).

Thus, στ = τσ and Gal (E/F) is Abelian and Gal (E/F) is solvable.Assume that F does not contain an n -th root of unity. Let ω a generator of the cyclic group of roots of n

-ta of unity. Let α a zero of xn− a. Since α and ωα are in splitting field of xn

− a, then ω = (ωα)/α is also inE. Let K = F(ω). Then, F ⊂ K ⊂ E. Since K is splitting field of xn

−1, then K is a normal extension of F. Anautomorphism σ in Gal (F(ω)/F) is determined from σ(ω). Then, σ(ω) = ωi for a integer i since all zeroes ofxn−1 are powers of ω. If τ(ω) = ω j is in Gal (F(ω)/F), then

στ(ω) = σ(ω j) = [σ(ω)] j = ωi j = [τ(ω)]i = τ(ωi) = τσ(ω).

Thus, Gal (F(ω)/F) is Abelian. From Theorem 17.12 the series

id ⊂Gal (E/F(ω)) ⊂Gal (E/F)

is a normal series. Since Gal (E/F(ω)) and

Gal (E/F)/Gal (E/F(ω)) Gal (F(ω)/F)

are both Abelian, then Gal (E/F) is solvable.

Lemma 17.9. Let F a field with characteristic zero and E/F a radical extension. Then, there is a normal radicalextension K/F which contains E.

Proof. Since E is a extension with radicals of F, there exist the elements α1, . . . ,αr ∈ K and positive integersn1, . . . ,nr such that

E = F(α1, . . . ,αr),

where αn11 ∈ F and

αnii ∈ F(α1, . . . ,αi−1)

for i = 2, . . . ,r. Let f (x) = f1(x) · · · fr(x), where fi is minimal polynomial of αi over F and let K splitting field ofK over F. Every root of f (x) in K is of the form σ(αi), where σ ∈Gal (K/F). Thus, for for every σ ∈Gal (K/F)we have [σ(α1)]n1 ∈ F and [σ(αi)]ni ∈ F(α1, . . . ,αi−1) for i = 2, . . . ,r. Thus, if Gal (K/F) = σ1 = id,σ2, . . . ,σk, thenK = F(σ1(α j)) is a extension with radicals of F.

We will prove the Fundamental Theorem for solvability with radicals.

Theorem 17.15. Let f (x) in F[x], where char F = 0. Then, f (x) is of solvable with radicals if and only if Gal f issolvable.

Proof. Let K splitting field of f (x) over F. Since f (x) is solvable with radicals, there is a radical extension Esuch that

F = F0 ⊂ F1 ⊂ · · ·Fn = E.

Since Fi is normal over Fi−1, then E is a normal extension of for every Fi. From Theorem 17.12 we have thatGal (E/Fi) is a normal subgroup of Gal (E/Fi−1). Thus, we have a subnormal series subgroups Gal (E/F)such that

id ⊂Gal (E/Fn−1) ⊂ · · · ⊂Gal (E/F1) ⊂Gal (E/F).

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Shaska T. 17

Again from Theorem 17.12 we know that

Gal (E/Fi−1)/Gal (E/Fi) Gal (Fi/Fi−1).

Hence, Gal (Fi/Fi−1) is solvable. Thus, Gal (E/F) is also solvable.

Exercises:

17.22. Prove that polynomial x5−6x−2 is not solvable with radicals.

17.23. Let be given p a prime number.(a) Prove that if a subgroup H of Sp contains a p -cycle and a transposition, then H = Sp.

(b) If p(T) is a irreducible polynomial over Q with order p which has exactly two root not real, then the Galoisgroup of splitting field of p(T) is Sp.

17.24. Prove that for every group Sn, (n ≥ 2 ) there exists a polynomial f (x) ∈Q[x] such that

Gal( f )Sn

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17 Shaska T.

17.9 Fundamental theorem of Algebra

Now we are ready to prove one of the most important theorems of Algebra.

Theorem 17.16 (Fundamental Theorem of Algebra). The field of complex numbers is algebraically closed. Thus,every polynomial with a root from C[x] has all roots in C.

Proof. Assume that E is a finite extension of C. Since for every finite extension field with characteristic zerois primitive then there is a α ∈ E such that E = C(α) where α is the root of some irreducible polynomial f (x)in C[x].

Splitting field L of f (x) is a normal separable extension of C which contains E. We must prove that it isimpossible for L to be a proper extension of C.

E = C(α)

C

R

Figure 17.6: The Galois correspondence

Assume that L is a proper extension of C. Since L is splitting field of f (x) = x2 + 1 over R, then L is anormal and separable, finite extension ofR. Let K fixed field of Sylow 2-subgroups of G of Gal (L/R). Then,R ⊂ K ⊂ L and |Gal (L/K)| = [L : K]. Since,

[L :R] = [L : K][K :R],

then [K :R] must be odd. Thus, K =R(β), where β has a minimal polynomial f (x) with odd degree. Thus,K =R.

We know that Gal (L/R) is a 2-group. Follows that Gal (L/C) is a 2-group. We have accepted that L ,C,then |Gal (L/C)| ≥ 2. From Sylow’s Theorem and Fundamental Theorem of Galois theory (Theorem 17.12)there is a subgroup Gal (L/R) of Gal (L/C) with index 2 and a field E which is fixed by Gal (L/R). Then,[E : C] = 2 and there exists an element γ ∈ E with minimal polynomial x2 + bx + c in C[x]. This polynomialhas roots

−b±√

b2−4c2

which are in C since ∆ = b2−4c is in C. This is impossible. Hence, L = C.

Exercises:

17.25. Let G be the Galois group of polynomial with degree n. Prove that |G| |n!.

17.26. Let F ⊂ E. If f (x) is solvable over F, prove that f (x) is also solvable over E.

17.27. Construct a polynomial f (x) in Q[x] with degree 7 which is not solvable with radicals.

17.28. Let p be a prime number. Prove that there is a polynomial f (x) ∈ Q[x] with degree p and Galois groupisomorphic to Sp. Generalize that for every prime p with p ≥ 5 there is a polynomial with degree p which is not issolvable with radicals.

326

Shaska T. 17

17.29. Let p be a prime number and Zp(t) the field of rational functions over Zp. Prove that f (x) = xp− t is a

irreducible polynomial in Zp(t)[x]. Prove that f (x) is not splitting.

17.30. Let E be an extension field of F. Assume that K and L are two intermediate fields. If there is an elementσ ∈ Gal (E/F) such that σ(K) = L, then K and L are called conjugate fields. Prove that K and L are conjugate if andonly if Gal (E/K) and Gal (E/L) are conjugated subgroups Gal (E/F).

17.31. Let σ ∈ Aut (R). If a is a positive real number prove that σ(a) > 0.

17.32. Prove or disprove that: Two different subgroups of a Galois group have different fixed fields.

17.33. We know that the cyclotomic polynomial

Φp(x) =xp−1

x−1= xp−1 + xp−2 + · · ·+ x + 1

is irreducible over Q for every prime number p. Let ω be a root of Φp(x). Determine Q(ω).

17.34. Prove that ω,ω2, . . . ,ωp−1 are different zeroes of Φp(x) and conclude that they are all zeroes of Φp(x).

17.35. Prove that Gal (Q(ω)/Q) is Abelian with order p−1.

17.36. Prove that fixed field of Gal (Q(ω)/Q) is Q.

17.37. Let F a finite field or a field of zero characteristic. Let E be a finite normal extension of F with Galois groupGal (E/F). Prove that F ⊂ K ⊂ L ⊂ E if and only if id ⊂ Gal (E/L) ⊂ Gal (E/K) ⊂ Gal (E/F).

17.38. Let F a field with characteristic zero and let f (x) ∈ F[x] be a splitting polynomial with degree n. If E is thesplitting field e f (x), let α1, . . . ,αn be roots of f (x) in E. Let ∆ =

∏i, j(αi−α j). Prove that the discriminant of f (x) is

∆2.

17.39. If σ ∈ Gal (E/F) is a transposition of two roots of f (x), prove that σ(∆) = −∆.

17.40. If σ ∈ Gal (E/F) is an even permutation of roots of f (x), prove that σ(∆) = ∆.

17.41. Prove that Gal (E/F) is isomorphic to a subgroup of An if and only if ∆ ∈ F.

17.42. Determine the Galois group of x3 + 2x−4 and x3 + x−3.

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17 Shaska T.

Evaroiste Galois

Carl Gustav Jacob Jacobi (10 December 1804 – 18 February 1851)was a German mathematician, who made fundamental contributionsto elliptic functions, algebraic geometry, dynamics, differential equa-tions, and number theory. His name is occasionally written as CarolusGustavus Iacobus Iacobi in his Latin books, and his first name is some-times given as Karl.

One of Jacobi’s greatest accomplishments was his theory of ellipticfunctions and their relation to the elliptic theta function. This wasdeveloped in his great treatise Fundamenta nova theoriae functionumellipticarum (1829), and in later papers in Crelle’s Journal. Thetafunctions are of great importance in mathematical physics becauseof their role in the inverse problem for periodic and quasi-periodicflows. The equations of motion are integrable in terms of Jacobi’selliptic functions in the well-known cases of the pendulum, the Eulertop, the symmetric Lagrange top in a gravitational field and the Keplerproblem (planetary motion in a central gravitational field).

He also made fundamental contributions in the study of differen-tial equations and to rational mechanics, notably the Hamilton–Jacobitheory.

It was in algebraic development that Jacobi’s peculiar power mainly lay, and he made importantcontributions of this kind to many areas of mathematics, as shown by his long list of papers in Crelle’sJournal and elsewhere from 1826 onwards. One of his maxims was: ’Invert, always invert’ (’man mussimmer umkehren’), expressing his belief that the solution of many hard problems can be clarified byre-expressing them in inverse form.

In his 1835 paper, Jacobi proved the following basic result classifying periodic (including elliptic)functions: If a univariate single-valued function is multiply periodic, then such a function cannot havemore than two periods, and the ratio of the periods cannot be a real number. He discovered many of thefundamental properties of theta functions, including the functional equation and the Jacobi triple productformula, as well as many other results on q-series and hypergeometric series.

The solution of the Jacobi inversion problem for the hyperelliptic Abel map by Weierstrass in 1854required the introduction of the hyperelliptic theta function and later the general Riemann theta functionfor algebraic curves of arbitrary genus. The complex torus associated to a genus g algebraic curve, obtainedby quotienting Cg by the lattice of periods is referred to as the Jacobian variety. This method of inversion,and its subsequent extension by Weierstrass and Riemann to arbitrary algebraic curves, may be seen as ahigher genus generalization of the relation between elliptic integrals and the Jacobi, or Weierstrass ellipticfunctions

Jacobi was the first to apply elliptic functions to number theory, for example proving of Fermat’s two-square theorem and Lagrange’s four-square theorem, and similar results for 6 and 8 squares. His otherwork in number theory continued the work of Gauss: new proofs of quadratic reciprocity and introductionof the Jacobi symbol; contributions to higher reciprocity laws, investigations of continued fractions, andthe invention of Jacobi sums.

He was also one of the early founders of the theory of determinants; in particular, he invented theJacobian determinant formed from the n2 differential coefficients of n given functions of n independentvariables, and which has played an important part in many analytical investigations. In 1841 he reintro-duced the partial derivative ∂ notation of Legendre, which was to become standard. Students of vectorfields and Lie theory often encounter the Jacobi identity, the analog of associativity for the Lie bracketoperation.

328

Chapter 18

Computing Galois groups ofpolynomials

We only consider only fields of characteristic 0. While we will develop the theory for polynomials over afield K, all our examples will be for f ∈Q[x].

18.1 The Galois group of a polynomial

Let f (x) ∈ K[x] a polynomial of degree n with discriminant ∆ f , 0. Then

f (x) = (x−α1) · · · (x−αn)

with distinct roots α1, . . . ,αn in the splitting field L of f . The Galois group of f over K, denoted by GK( f ), isthe group Gal (L/K), viewed as a permutation group of the roots α1, . . . ,αn. Thus GK( f ) is a subgroup of Sn,determined up to conjugacy by f .

Proposition 18.1. (i) Let G = GK( f ) and H = G∩An. Then H = Gal (L/K(√

∆ f )). In particular, G is contained inthe alternating group An iff the discriminant ∆ f is a square in K.

(ii) The irreducible factors of f in K[x] correspond to the orbits of G. In particular, G is a transitive subgroup ofSn iff f is irreducible.

Proof. (i) We have ∆ f = d2f , where d f =

∏i> j (αi−α j). For g ∈ G we have g(d f ) = sgn(g)d f . Thus H = G∩An

is the stabilizer of d f in G. But this stabilizer equals Gal (L/K(d f )). Hence the claim.

(ii) G acts transitively on the roots of each irreducible factor of f , by Lemma ??.

Example 18.1. n = 2 : Letf (x) = x2 + bx + c

Then ∆ f = b2− 4c. Hence G = A2 = 1 if and only if ∆ f is a square.

18.1.1 Cubics

Let f (x) ∈Q[x] be given byf (x) = x3 + ax2 + bx + c

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18 Shaska T.

The group S3 has two transitive subgroups A3 and S3. Thus for irreducible f , we have G = A3 if and only if

∆ f = −4a3c + a2b2 + 18abc−4b3−27c2

is a square in Q.

Lemma 18.1. Let f (x) ∈Q[x] be given by

f (x) = x3 + ax2 + bx + c

We know that ∆ f = 0 if and only if f has a multiple root. Show that

i) ∆ f > 0 if and only if f has three distinct real roots.ii) ∆ f < 0 if and only if f has one real root and two non-real complex conjugate roots.

Proof. Let α1,α2,α3 be the roots of f (x). Then

∆ f = [(α1−α2)(α2−α3)(α2−α3)]2

i) If α1,α2,α3 are real roots and distinct then obviously ∆ f > 0. Suppose that ∆ f > 0. If one of the roots isnon-real then its complex conjugate is also a root. Thus, we can assume that the roots are

a + bi, a− bi, r

for a,b,r ∈R. Then∆ f =

(2bi ·β · β

)2 ,

where β = (a− r) + bi. Hence,

∆ f = −4 ·b2· ||β||4 < 0,

which contradicts our hypothesis that ∆ f > 0. Part ii) is an immediate consequence of i).

Remark 18.1. Recall that for a degree n irreducible polynomial f (x) over k with splitting field E f we have n | [E f : k],see Lemma 16.3.

Exercises:

18.1. Compute the Galois groups of the splitting fields of the following polynomials, over Q:i) p(x) = x3

−3x + 3,ii) x3 + 2,iii) x3 + 3x2

−x−1,iv) x3

−3x + 1

18.2. Determine the Galois group of the field Q(√

2 +√

3)

over Q.

18.3. (i) Compute the Galois group of x3−2 over Q.

(ii) Compute the Galois group of x9−1 over Q and primitive elements for all subfields of its splitting field L.

(iii) Show that x3−2 is irreducible over L.

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18.2 Galois groups of quartics

Let f (x) ∈Q[x] be an irreducible polynomial. Then G := Gal ( f ) is a transitive subgroup of S4. Further 4 | |G|,see Lemma 16.3. So the order of G is 4, 8, 12, or 24. It is an exercise in group theory to check that transitivesubgroups of S4 of such orders are isomorphic to one of the following Z4, D4, V4, A4, S4. We would like tofind conditions on the coefficients of f (x) which determines the Galois group of f (x).

We consider the normalized polynomial

(4) f (x) = x4 + ax2 + bx + c = (x−α1) . . . (x−α4)

with a,b,c ∈ k. Let E f = k(α1, . . . ,α4) be the splitting field of f over k. Since f has no x3-term, we haveα1 + · · ·+α4 = 0. We assume ∆ f , 0, so α1, . . . ,α4 are distinct. Let G = Gk( f ), viewed as a subgroup of S4 viapermuting α1, . . . ,α4.

There are 3 partitions of 1, . . . ,4 into two pairs. S4 permutes these 3 partitions, with kernel

V4 = (12)(34), (13)(24), (14)(23), id.

Thus S4/V4 S3, the full symmetric group on these 3 partitions. Associate with these partitions the elements

β1 = α1α2 +α3α4, β2 = α1α3 +α2α4, β3 = α1α4 +α2α3

of L. If β1 = β2 then α1(α2 −α3) = α4(α2 −α3), a contradiction. Similarly, β1,β2,β3 are 3 distinct elements.Then G acts as a subgroup of S4 on α1, . . . ,α4, and as the corresponding subgroup of S3 S4/V4 on β1, . . . ,β3.Thus the subgroup of G fixing all βi is G∩V4. This proves

Lemma 18.2. The subgroup G∩V4 of G corresponds to the subfield k(β1,β2,β3) of E f . This subfield is the splittingfield over k of the cubic polynomial ("cubic resolvent")

(4) g(x) = (x−β1) (x−β2) (x−β3) = x3− ax2

− 4cx + −b2 + 4ac

The rootsβi of the cubic resolvent can be found by Cardano’s formulas. The extension k(α1, . . . ,α4)/k(β1,β2,β3)has Galois group ≤ V4, hence is obtained by adjoining at most two square roots to k(β1,β2,β3).

L := k(α1,α2,α3,α4)

G=G∩V4

F := k(β1,β2,β3)

d

k

Figure 18.1: The Galois extension of quartics

Example 18.2. Show that ∆( f ,x) = ∆(g,x)

We denote by d := [k(β1,β2,β3) : k]. Then we have the following:

Lemma 18.3. The Galois group of f (x) is one of the following:

i) d = 1⇐⇒ GV4.ii) d = 3⇐⇒ GA4.

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18 Shaska T.

iii) d = 6⇐⇒ GS4.iv) If d = 2 then we have

a) f (x) is irreducible over F⇐⇒ GD4b) f (x) is reducible over F⇐⇒ GZ4

Proof. i) Let d = 1, then G = G∩V4 and 4 | |G|. Hence GV4. The converse is obvious.Parts ii) and iii): If d = 3,6 then 3 | |G|. Thus, G is either A4 or S4 which forces G∩V4 = V4. Hence,

[L : k] = 4d. So, if d = 3 (resp., d = 4) we have GA4 (resp., GS4).Conversely if G is A4 or S4 and F/k is the splitting field of a cubic, then G = V4. Hence, d = 3,6 respectively.iv) Let d = 2. Since the order of G must be a multiple of 4 then G =Z2,V4. f (x) is irreducible over F iff

G = V4 which is equivalent to GD4. If f (x) is reducible then G = Z2 which is equivalent with |G| = 4 orGZ4.

Example 18.3. Letf (x) = x4 + ax3 + bx2 + cx + d

be an irreducible quartic with coefficients in Q. Give necessary and sufficient conditions in terms of a,b,c,d whichcharacterize Gal ( f ).

One can first eliminate the coefficient of x3 by the substituting x with x− a4 . Then using the formula for the

resolvent given above we compute g(x) which is

g(x) := x3− bx2 + (ac−4d)x− a2d + 4bd− c2

The discriminant of f (x) is the same as the discriminant of g(x) and is given below:

∆ = −27a4d2 + 18a3bcd−4a3c3−4a2b3d + a2b2c2 + 144a2bd2

−6a2c2d−80ab2cd

+ 18abc3 + 16b4d−4b3c2−192acd2

−128b2d2 + 144bc2d−27c4 + 256d3

18.2.1 Solving quartics

The element (α1 +α2)(α3 +α4) is fixed by G∩V4, hence lies in K(β1,β2,β3). We find

−(α1 +α2)2 = (α1 +α2)(α3 +α4) = β2 +β3

By this and symmetry we get Ferrari’s formulas

α1 +α2 =√−β2−β3

α1 +α3 =√−β1−β3

α1 +α4 =√−β1−β2

or

α1 =

√−β1−β2 +

√−β1−β3 +

√−β2−β3

2

α2 =−

√−β1−β2 −

√−β1−β3 +

√−β2−β3

2

α3 =−

√−β1−β2 +

√−β1−β3 −

√−β2−β3

2

α4 =

√−β1−β2 −

√−β1−β3 −

√−β2−β3

2This completes the case for the quartics. We next give a few examples how to compute the galois groups ofa few quartics.

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Example 18.4. Let f (x) = x4−3. Recall from the section on splitting fields that [E f :Q] = 8. Then Gal ( f )D4.

Example 18.5. Let f (x) = x4−10x2 + 4. Find Gal ( f ).

The first method would be to compute the discriminant of f (x). We get

∆ f = 210·32·72

Since this a square in Q then Gal ( f ) ≤ A4. The resolvent is

g(x) = x3 + 10x2−16x−160

and splits in Q asg(x) = (x + 10)(x−4)(x + 4)

Thus, Gal ( f ) = V4.

The second method would be to compute the Galois action explicitly. The roots of f (x) are

α1 =

√142

+

√6

2, α2 =

√142−

√6

2,α3 = −

√142

+

√6

2,α4 = −

√142−

√6

2

Then we have the following automorphisms;

σ1 : α1→ α2

α3→ α3

α4→ α4

σ2 : α3→ α4

α1→ α1

α2→ α2

σ1σ2 : α1→ α2

α3→ α4

Then the Galois group isGal ( f ) = id,σ1,σ2,σ1σ2

which is isomorphic to V4.

Example 18.6. Letf (x) = x4 + 4x3 + 3x + 2

The resultant cubic isg(x) = x3

−41 + 4x

which is irreducible over Q. Then∆ f = ∆g = −13 ·3511

which is not a square. Hence, d = 6 and Gal ( f )S4.

Exercises:

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18 Shaska T.

18.4. Compute the Galois groups of the splitting fields of the following polynomials, over Q:i) x4−3 or x4

−2,ii) x4

−10x2 + 4iii) x4 + 4x2 + 2iii) x4 + x3 + 3x + 1iii) x4

−x2−x−12

18.3 Galois groups of quintics

In this section we want to find the list of groups that occur as Galois groups of quintics, and find methodsfor determining such groups.

Lemma 18.4. Let f (x) ∈ k[x] be an irreducible quintic. Then its Galois group is one of the following: Z5,D5,F5 = AGL(1,5), A5, or S5

Proof. G is transitive, hence its 5-Sylow subgroup is isomorphic to Z5 (generated by a 5-cycle). If Z5is not normal, then G has at least 6 of 5-Sylow subgroups; then |G| ≥ 6 · 5 = 30, hence [S5 : G] ≤ 4 whichimplies G = S5,A5. If Z5 is normal in G then G is conjugate either Z5, D5 (dihedral group of order 10) orF5 = AGL(1,5), the full normalizer of Z5 in S5, of order 20 (called also the Frobenius group of order 20).

Remark 18.2. If the discriminant of the quintic is a square in k then Gal ( f ) is contained in A5. Hence, it is Z5,D5,or A5.

In the next few examples we give a polynomial for each of the above groups. However proofs is someof the cases are difficult and will be shown in the next section. The reader can check the Galois group ofeach polynomial in Maple by using the command "galois(f, x)".

Example 18.7. Let f (x) = x5 + x4−4x3

−3x2 + 3x + 1. Then Gal ( f ) over Q is isomorphic to Z5

Example 18.8. Let f (x) = x5 + 11x + 44. Then Gal ( f ) over Q is isomorphic to D5

Example 18.9. Let f (x) = x5−2. Show that Gal ( f ) over Q is isomorphic to F5.

We will give a direct proof without making use of the previous lemma. First, f (x) is irreducible by Eisenstein’scriteria. Its roots are

5√

2,ε ·5√

2,ε2·

5√

2,ε3·

5√

2,ε4·

5√

2

where ε5 = 1. Hence E f =Q( 5√2,ε). Since x5−2 is irreducible overQ(ε) and we know that [Q(ε) :Q] = 4 (cyclotomic

extension), then [E f :Q] = 20. Hence Gal( f ) = F5.

Example 18.10. Let f (x) = x5 + 20x + 16. Show that Gal ( f ) over Q is isomorphic to A5

The discriminant is ∆ f = 216·56. Thus, Gal ( f ) is contained in A5 and is either Z5,D5, or A5. We will see later

how to determine which group it is.

Example 18.11. Let f (x) = x5−4x + 2. Show that Gal ( f ) over Q is isomorphic to S5

Notice that the derivative of f (x) is f ′(x) = 5x4−4. Study the sign of the derivative and show that f (x) intersect

the real line in only three points. Thus, it must have to non-real roots. Then, Gal ( f ) = S5.

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18.3.1 Solvable quintics

If G = S5,A5 then the equation f (x) = 0 is not solvable by radicals. We want to investigate here the caseG 6 S5, A5.

Let f (x) be an irreducible quintic in k[x] given by

(5) f (x) = x5 + c4x4 + · · · + c0 = (x−α1) · · · (x−α5)

Let G = Gal ( f ), viewed as a (transitive) subgroup of S5 via permuting the (distinct) roots α1, · · · ,α5. Asbefore E f = k(α1, · · · ,α5) denotes the splitting field.

A 5-cycle in S5 = Sym(1, . . . ,5) corresponds to an oriented pentagon with vertices 1, . . . ,5. A 5-cycle andits inverse correspond to a (non-oriented) pentagon, and the full Z5 corresponds to a pentagon togetherwith its "opposite". Thus F5, the normalizer of CZ5 in S5, is the subgroup permuting the pentagon and itsopposite. D5 is the subgroup of F5 fixing the pentagon (symmetry group of the pentagon), and C5 is thesubgroup of rotations. Thus if G ≤ F5 then G fixes

δ1 = (α1−α2)2(α2−α3)2(α3−α4)2(α4−α5)2(α5−α1)2

− (α1−α3)2(α3−α5)2(α5−α2)2(α2−α4)2(α4−α1)2 (18.1)

where the first (resp., second) term corresponds to the edges of the pentagon (resp., its opposite).Let δ1, . . . ,δ6 be the elements associated in this way to the six 5-Sylow’s of S5, i.e., to the six pentagon-

opposite pentagon pairs on five given letters. Clearly, G permutes δ1, . . . ,δ6. If G is conjugate to a subgroupof F5, it fixes one of δ1, . . . ,δ6; this fixed δi must then lie in k.

Thus, a necessary condition for the (irreducible) polynomial (5) to be solvable by radicals is that one δilies in k, i.e., that the polynomial

g(x) = (x−δ1) . . . (x−δ6) ∈ k[x]

has a root in k. It is also sufficient: If G fixes one δi then G is conjugate to a subgroup of F5, provided thatδ1, . . . ,δ6 are all distinct. To check this is:

Problem 5: Show δ1, . . . ,δ6 are mutually distinct (under the hypothesis D f , 0).

The coefficients of g(x) are symmetric functions in α1, . . . ,α5, hence are polynomial expressions inc0, . . . ,c4. The goal is to find these expressions explicitly. This gives an explicit criterion to check whetherf (x) = 0 is solvable by radicals.

Expressing the coefficients of g(x) explicitly in terms of α1, . . . ,α5 yields expressions that are too big(even for a computer). A better way to proceed is by noting that these expressions have certain symmetries(invariance properties). E.g., the substitution x′ = x + a in f (x) doesn’t change the δ′i s. To see the fullinvariance properties, we need to "projectivize".

18.3.2 Invariants of binary quintics

Consider a binary form of degree n, i.e., homogeneous polynomial F ∈ K[x, y] of degree n. Such F can bewritten in the form

(6) F(x, y) = bnxn + bn−1xn−1y + . . . + b0yn = (β1x − γ1y) · · · (βnx − γny)

with βi,γi ∈ K. To prove this, we may assume bn , 0. (If bn = 0 apply induction to F/y). Then

F = bnyn ( (xy

)n +bn−1

bn(xy

)n−1 + · · ·+b0

bn) = bnyn (

xy−α1) · · · (

xy−αn)

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18 Shaska T.

for certain αi ∈ K, which proves the claim.The group GL(2,k) acts on binary forms in the following way: The matrix

h =

(a bc d

)maps F to

h(F) = F((x, y)h) = F(ax + cy, bx + dy) = (β′1x − γ′1y) · · · (β′nx − γ′ny)

From

F = det(

x yγ1 β1

)· · · det

(x yγn βn

)we get

(γ′i ,β′

i ) = (γi,βi) h−1det(h)

From now on n = 5, so F is a binary quintic. Our present set-up reduces to that of the preceding sectionby setting y = 1 = βi. The generalized version of the δi’s are elements δ1, . . . , δ6, formed by replacing αi−α jby

Di j = det(γi βiγ j β j

)in the formulas defining the δi’s. In particular,

δ1 = D212 D2

23 D234 D2

45 D251 − D2

13 D235 D2

52 D224 D2

41

Lemma 18.5. Let σν(X1, . . . ,X6), ν = 1, . . . ,6, be the elementary symmetric polynomial

σν =∑

i1<i2<···<iν

Xi1Xi2 . . . Xiν .

Thendν := σν(δ1, . . . , δ6)

is a homogeneous polynomial expression in b0, . . . ,b5 of degree 4ν. These polynomials are invariant under the actionof SL(2,k) on binary quintics: For any h ∈ SL(2,k) the quintic h(F) has the same associated dν’s.

Proof. For α j := γ j/β j we have δi = (β1 · · ·β5)4 δi = b45 δi. Thus dν = b4ν

5 σν(δ1, , . . . ,δ6). But the σν(δ1, , . . . ,δ6)are polynomial expressions in the c j = b j/b5, j = 0, . . . ,4 (see the last section). Thus dν is a rational function inb0, . . . ,b5, where the denominator is a power of b5. Switching the roles of x and y yields that the denominatoris also a power of b0. Thus it is constant, i.e., dν is a polynomial in b0, . . . ,b5. If we replace each β j by cβ j fora scalar c then each δi gets multiplied by c4, so dν gets multiplied by c4ν. Thus dν is homogeneous of degree4ν. The rest of the claim is clear.

There are four basic invariants of quintics, denoted by J,K,L, I, of degrees 4,8,12 and 18, such that everySL(2,k)-invariant polynomial in b0, . . . ,b5 is a polynomial in J,K,L, I (see e.g. I. Schur, Vorlesungen ueberInvariantentheorie, Springer 1968). To define J,K,L, we need auxiliary quantities

A =20b4−8b1b3 + 3b2

2

100

B =100b5−12b1b4 + 2b2b3

100

C =20b1b5−8b2b4 + 3b2

3

100

(18.2)

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Shaska T. 18

and D,E,F,G defined by

11000

det

10u + 2b1v 2b1u + b2v b2u + b3v2b1u + b2v b2u + b3v b3u + 2b4vb2u + b3v b3u + 2b4v 2b4u + 10b5v

=Du3 + Eu2v + Fuv2 + Gv3

(18.3)

Then

J = 53 (B2− 4AC)

K = 25·56

[2A(3EG−F2)−B(9DG−EF) + 2C(3FD−E2)

]L = −210

·59·3−1

[4(3EG−F2)(3FD−E2) − (9DG−EF)2

] (18.4)

By using special quintics one gets linear equations for the coefficients expressing the dν’s in terms ofJ,K,L. The result is: (Due to Berwick 1915, see B. King, Beyond the quartic equation, Birkhaeuser 1996)

d1 = −10 J

d2 = 35 J2 + 10 K

d3 = −60 J3− 30 JK − 10 L

d4 = 55 J4 + 30 J2K + 25 K2 + 50 JL

d5 = −26 J5− 10 J3K − 44 JK2

− 59 J2L − 14 KL

d6 = 5J6 + 20 J2K2 + 20 J3L + 20 JKL + 25 L2

(18.5)

These are the coefficients of the polynomial g(x) from the previous section. This g(x) has a root in thebase field iff the Galois group of f (x) is solvable. We summarize in the following:

Lemma 18.6. Let f (x) be a irreducible quintic over k and d1, . . . ,d6 defined in terms of the coefficients of f (x) asabove. Then f (x) is solvable by radicals if and only if

g(x) = x6 + d1x5 + · · ·d5x + d6

has a root in k.

Project: Investigate whether or not for each group that occurs as a Galois group of a quintic you can char-acterize the case in terms of algebraic relations of invariants of cubics (I have not seen this done anywhereand would not be surprised if it has never been done).

Exercises:

18.5. Compute the Galois groups of the splitting fields of the following polynomials, over Q:1. p(x) = x5

−2,2. x5

−4x + 2

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18 Shaska T.

18.4 Determining the Galois group of higher degree polynomials

In this section we discuss several techniques of computing Galois groups of higher degree polynomials.Let f (x) ∈Q[x] be an irreducible polynomial such that deg f = n. Let G := GalQ( f ). Recall that the followingare true:

i) G is isomorphic to a transitive subgroup of Sn.ii) n divides |G|,iii) G is a subgroup of An if and only if ∆ f is a square in Q.

18.4.1 Reduction mod p

The reduction method is very powerful and uses the fact that once a every polynomial with rationalcoefficients can be transformed into a monic polynomial with integer coefficients without changing thesplitting field.

Let f (x) ∈Q[x] be given byf (x) = xn + an−1xn−1 + · · ·+ a1x + a0

Let d be the common denominator of all coefficients a0, · · · ,an−1. Then g(x) := d f ( xd ) is a monic polynomial

with integer coefficients. Clearly the splitting field of f (x) is the same as the splitting field of g(x). Thus,without loss of generality we can assume that f (x) is a monic polynomial with integer coefficients.

Theorem 18.1. (Dedekind) Let f (x) ∈ Z[x] be a monic polynomial such that deg f = n, Gal Q( f ) = G, and p aprime such that p - ∆ f . If fp := f (x) mod p factors in Zp[x] as a product of irreducible factors of degree

n1,n2,n3, · · · ,nk,

then G contains a permutation of type(n1) (n2) · · · (nk)

Proof. van der Warden section 8.10 The Dedekind theorem can be used to determine the Galois group in many cases. Consider for example

polynomials of degree 5. Then it is an easy example in group theory to determine the cycle types for allgroups that occur as Galois groups of quintics.

(2) (2)2 (3) (4) (3)(2) (5)S5 10 15 20 30 20 24A5 15 20 24F5 5 10 4D5 5 4Z5 4

Table 18.1: Cycle types for Galois groups of quintics

Example 18.12. Find the Galois group off (x) = x5

−4x + 2

over the rationals.

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Shaska T. 18

Solution: We see that ∆ f = −24·13259. Let p = 3. Then,

f3(x) = x5 + 2x + 2.

Hence there is a 5-cycle in the Galois group of f (x). For p = 5 we have

f5(x) = (x2 + 2x + 3)(x + 1)(x2 + 2x + 4)

and for p = 7 we havef7(x) = (x3 + 3x2 + 3x + 5)(x2 + 4x + 6)

From the above table we conclude that Gal( f ) = S5.

Below we display the table for the type of elements in S6.

() (2) (2)(2) (2)(2)(2) (3) (3)(2) (3)(3) (4) (4)(2) (5) (6) OrderS6 1 15 45 15 40 120 40 90 90 144 120 720A6 1 0 45 0 40 0 40 0 90 144 0 360S5 1 0 15 10 0 0 20 30 0 24 20 120

(S3×S3)oC2 1 6 9 6 4 12 4 0 18 0 12 72A5 1 0 15 0 0 0 20 0 0 24 0 60

C2×S4 1 3 9 7 0 0 8 6 6 0 8 48(C3×C3)oC4 1 0 9 0 4 0 4 0 18 0 0 36

S3×S3 1 0 9 6 4 0 4 0 0 0 12 36S4 1 0 3 6 0 0 8 6 0 0 0 24S4 1 0 9 0 0 0 8 0 6 0 0 24

C2×A4 1 3 3 1 0 0 8 0 0 0 8 24C3×S3 1 0 0 3 4 0 4 0 0 0 6 18

A4 1 0 3 0 0 0 8 0 0 0 0 12D12 1 0 3 4 0 0 2 0 0 0 2 12S3 1 0 0 3 0 0 2 0 0 0 0 6C6 1 0 0 1 0 0 2 0 0 0 2 6

Table 18.2: Cycle types for Galois groups of sextics

Exercises:

18.6. Compute the Galois group Gal ( f ) of the polynomial

f (x) = x5 + 20x + 16

18.7. Compute the Galois group Gal ( f ) of the polynomial

f (x) = x5 + 11x + 44

18.8. Compute the Galois group Gal ( f ) of the polynomial

f (x) = x5 + 15x + 12

18.9. Compute the Galois group Gal ( f ) of the polynomial

f (x) = x6 + 24x−20

18.10. Compute the Galois group Gal ( f ) of the polynomial

f (x) = x7 + 7x4 + 14x + 3

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18 Shaska T.

18.5 Polynomials with non-real roots

Let f (x) ∈ Q[x] be an irreducible polynomial of degree n > 5. Denote by r the number of non-real roots off (x). Since the complex conjugation permutes the roots then r is even, say r = 2s. By a reordering of theroots we may assume that if f (x) has r non-real roots then

α := (1,2)(3,4) · · · (r−1,r) ∈ Gal( f ).

Since determining the number of non-real roots can be very fast, we would like to know to what extent thenumber of non-real roots of f (x) determines Gal( f ). The complex conjugation assures that m(G) ≤ r. Theexistence of α can narrow down the list of candidates for Gal( f ). However, it is unlikely that the group canbe determined only on this information unless p is "large" enough. In this case the number of non-real rootsof f (x) can almost determine the Galois group of f (x), as we will see in the next section. Nevertheless, thetest is worth running for all p since it is very fast and improves the algorithm overall.

18.5.1 Polynomials of prime degree

The next theorem determines the Galois group of a prime degree polynomial f (x) with r non-real rootswhen the degree of f (x) is large enough with respect to r.

Theorem 18.2. Let f (x) ∈ Q[x] be an irreducible polynomial of prime degree p ≥ 5 and r = 2s be the number ofnon-real roots of f (x). If s satisfies

s (s logs + 2logs + 3) ≤ p

then Gal( f ) = Ap,Sp.

For a fixed p the above bound is not sharp as we will see below. However, the above theorem can beused successfully if s is fixed. We denote the above bound on p by

N(r) :=[s (s logs + 2logs + 3)

]for r = 2s. Hence, for a fixed number of non-real roots, for p ≥N(r) the Galois group is always Ap or Sp.

Corollary 18.1. Let a polynomial of prime degree p have r non-real roots. If one of the following holds:(i) r = 4 and p > 7,(ii) r = 6 and p > 13,(iii) r = 8 and p > 23,(iv) r = 10 and p > 37,

then Gal( f ) = Ap or Sp.

Remark 18.3. The above results gives a very quick way of determining the Galois group for polynomials with non-realroots. Whether or not the discriminant is a complete square can be used to distinguish between Ap and Sp.

18.5.2 Polynomials of prime degree p with Galois group Ap

Let f (x) be a polynomial in Q(t) as below

f (x) = (n−1)xn−nxn−1 + t.

The discriminant of f (x) with respect to x is

∆ f = (−1)n(n−1

2 nn(n−1)n−1tn−2(t−1).

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Shaska T. 18

∆ f is a complete square in Q if (−1)(n−1

2 nt(t− 1) is a complete square; see [?Se] (pg. 44) for more on thisfamily of polynomials. Let n = 23. Then

∆ f = −222·1122

·2323· t21(t−1).

Hence, ∆ f is a complete square in Q if G(t) = −23t(t−1) is a complete square. In other words, for all thoserational points on the curve

y2 = G(t).

This is a genus 0 curve and can be parameterized as follows:

(y, t) =

(−

23mm2 + 23

,23

(m2 + 23

)Consider f (x) for t = 23

(m2+23 . Since we prefer to work with polynomials with integer coefficients then take

f (x) = (22m2 + 506)x23− (23m2 + 529)x22 + 23.

It is easily checked that f (x) is irreducible over Q and its discriminant is

∆ f = 222·1122

·2344·m2 (23 + m2)22

which is a complete square in Q. Thus, Gal( f ) is inside A23. It is an simple calculus exercise to show thatthe number of real roots of these polynomials is ≤ 3. Hence, the Galois group is A23.

We conclude with the following open problem:

Problem: Find a degree 23 polynomial f (x) ∈Q[x] with exactly 7 real roots such that ∆ f is a complete square in Qbut Gal( f ) is not isomorphic to A23.

Exercises:

18.11. Letf (x) = x4 + ax3 + bx2 + cx + d

be an irreducible quartic with coefficients in Q. Give necessary and sufficient conditions in terms of a,b,c,d whichcharacterize Gal ( f ).

18.12. Using Maple or some other computer algebra package express d1, . . . ,d6 of Lemma 18.6 in terms of thecoefficients of f (x).

18.13. From Lemma 18.6, an irreducible quintic f (x) ∈ k[x] is solvable by radicals if and only if

g(x) = x6 + d1x5 + · · ·+ d5x + d6

has a root in k. Find such formulas when f (x) is solvable.

18.14. Let f (x) ∈ Q[x] be an irreducible polynomial of degree 7. What are the possible groups that can be Galoisgroups of f (x).

18.15. For small degree polynomials (i.e, 3,4,5) the reduction method can be used quite effectively to determine theGalois group of a polynomial. However, there is no guarantee that two non-isomorphic groups have different cycle-typeelements. Find an example that this happens (i.e., two non-isomorphic groups, transitive in the same Sn, with thesame cycle-type elements). How can you distinguish the Galois groups in this case.

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18 Shaska T.

18.16. Using GAP we can easily compile a list of groups which are candidates for Galois groups of a degree npolynomial (i.e., transitive in Sn and the order is a multiple of n). Compile such list for all polynomials of degree ≤ 10.

18.17. For each group G from the list in the above problem, provide a polynomial f (x) such that Gal ( f )G.

Final ExamDecember 2017

Do the following problems according to your ticket:

1. 9.16, 9.21, 11.5, 11.18, 15.1, 16.30, 18.6,

2. 9.17, 9.22, 11.6, 11.19, 15.12,16.31, 18.7,

3. 9.18, 9.23, 11.7, 11.20, 15.19,16.32, 18.8,

4. 9.19, 9.24, 11.8, 11.21, 15.22,16.33, 18.9,

5. 9.20, 9.25, 11.9, 11.22, 15.29,16.34, 18.10,

342

Chapter 19

Abelian Extensions

19.1 Abelian extensions and Abelian closure

19.2 Roots of unity

19.3 Cyclotomic extensions

Let n be a positive integer n. The n-th cyclotomic polynomial is

Φn(x) = (x−α1) . . . (x−αr)

where α1, . . . ,αr are the n-th primitive roots of unity. Hence, if we fix a primitive root of unity α then

Φn(x) =∏

(r,n)=1

(x−αr)

and degΦn(x) = ϕ(n), where ϕ(n) is the Euler function.Let Fn denote the splitting field of xn

− 1 over some field k. Then Fn/k is called the n-th cyclotomicextension over k.

The main goal of this section is to determine Fn and Gal (Φn) = G(Fn/k). We start by some properties ofcyclotomic polynomials:

Lemma 19.1. Let Φn(x) be the cyclotomic polynomial over k. Then,

i) degΦn(x) = ϕ(n)ii) Φn(x) is monic and has coefficients in the prime subfield of kiii) If k =Q then

Φn(x) ∈Z[x]

iv) The following holds:

xn−1 =

∏d|n

Φd(x)

Proof. Exercise

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19 Shaska T.

Example 19.1. It is easy to check that

Φ1(x) = x−1Φ2(x) = x + 1

Φ3(x) = x2 + x + 1

Φ4(x) = x2 + 1

Φ6(x) = x2−x + 1

Φ8(x) = x4 + 1

Φ10(x) = x4−x3 + x2

−x + 1

(19.1)

and in general for a prime p we have

Φp(x) = xp−1 + xp−2 + · · ·+ x + 1

Theorem 19.1. All cyclotomic polynomials Φn(x) over Q are irreducible in Q[x].

Proof. Assume that Φn(x) is reducible over Q. By Gauss’ lemma it is reducible in Z[x]. Say

Φn(x) = f (x) · g(x)

where f (x) and g(x) are monic and at least one of them irreducible over Z. We assume f (x) is irreducibleover Z. Let α be a root of f (x). Then α is a root of unity and αp is a primitive n-th root of unity since p - n

Claim: f (αp) = 0, for all primes p - n.

Proof: Say f (αp) , 0. Then αp must be a root of g(x). Hence α is a root of g(xp).Since f (x) is monic and irreducible then f (x) | g(xp), say

g(xp) = h(x) f (x)

for some monic polynomial h(x) ∈Z[x]. We reduce mod p. For f ∈Z[x] let f be the residue mod p of f .Thus in Fp we have

Φn(x) = f · g.

Notice that Φn(x) | (xn−1) therefore it has no repeated roots in any extension of Fp. Since ap = a in Fp for all

a ∈ Fp we have

g(xp) = g(x)p.

Thus, f | (g)p and therefore every factor q(x) in Fp[x] of f divides also g. Hence q2 divides Φn which impliesthat Φn(x) has multiple roots. This is a contradiction and completes the proof of the claim.

Thus, from the Claim, we get that all primitive roots of unity are roots of f (x). Hence f (x) = Φn(x) andΦn(x) is irreducible.

Let εn be the n-th primitive root of unity and Fn denote the splitting field of Φn(x). Then we have thefollowing:

Corollary 19.1. If Fn is the n-th cyclotomic extension of Q then Fn =Q(εn). Moreover,

G(Q(εn)/Q) (Z/nZ)∗, and [Q(εn) :Q] = ϕ(n)

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Shaska T. 19

Proof. The first part is obvious since Φn(εn) = 0 and Φn(x) is irreducible. Hence, [Fn :Q] = ϕ(n). A basis forQ(εn)/Q is 1,εi

n) for (i,n) = 1. All σ ∈Gal (Fn/Q) look like

εn→ εin, where (i,n) = 1

Define

f : G(Q(εn)/Q) −→ (Z/nZ)∗

(εn→ εin) −→ i

(19.2)

It is easily shown that this is an isomorphism. The highlight of the cyclotomic extensions is the celebrated Kronecker-Weber theorem.

Theorem 19.2. Let F be a finite Abelian extension of Q. Then F is contained is some cyclotomic extension of Q.

Proof. The proof is beyond the scope of this book. The interested reader can see [?Kr].

Exercises:

19.1. Compute Φn(x) for n = 12, . . . ,20.

19.2. Let n > 1 be an odd integer. Show that

Φ2n(x) = Φn(−x)

19.3. Let n be an odd integer. Show that the splitting field of Φn(x) is the same as the splitting field of Φ2n(x).

19.4. Let n and m be positive integers with

d = gcd (m,n), l = lcm (m,n).

Denote the n-th cyclotomic extension over Q by Sn. Show that

i) If n |m then Sm is an extension of Sn.ii) SnSm = Sliii) Sn∩Sm = Sl

19.5. If d ∈Q show that Q(√

d) lies in some cyclotomic polynomial Sn of Q (don’t use Kronecker-Weber theorem).

19.6. Determine which roots of unity are in the following: Q(i), Q(√

2, Q(√−2), Q(

√−3), Q(

√3).

19.7. For what integers n does [Q(εn) :Q] = 2 ?

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19 Shaska T.

19.4 Cyclic Extensions

Definition 19.1. Let F/k be a Galois extension and Gal(F/k) be a cyclic group, say Gal(F/k) = 〈σ〉. Then F/k is calleda cyclic extension.

The goal of this section is to determine cyclic extensions. This can be done when the ground field hasenough roots of unity. The following theorem plays a central role in determining such extensions.

Theorem 19.3 (Hilbert’s Theorem 90). Let K/k be a finite cyclic extension with Galois group Gal(K/k) = 〈σ〉 andα ∈ K. Then, N(α) = 1 if and only if there is a β ∈ K such that α =

βσ(β)

Proof. See notes.

Theorem 19.4. Let k be a field containing a primitive n-th root unity. Assume that if char k = p > 0 then (n,p) = 1.Then the following are equivalent:

i) F/k is cyclic of degree d | nii) F = k(α) where

min (α,k,x) = xd− a

for d | n and a ∈ k.iii) F is a splitting field of an irreducible polynomial

f (x) = xd− b

where d | n and b ∈ k.iv) F is a splitting field of

f (x) = xn−b

for b ∈ k.

Proof. Roman, pg. 210

Theorem 19.5 (Hilbert’s Theorem 90, additive version). Let K/k be a finite cyclic extension with Galois groupGal(K/k) = 〈σ〉 and α ∈ K. Then, Tr(α) = 0 if and only if there is a β ∈ K such that α = β−σ(β)

Proof. Similarly to the multiplicative version. Exercise.

Exercises:

Theorem 19.6 (Artin-Schreier). Let char (k) = p > 0. The polynomial

f (x) = xp−x− a ∈ k[x]

either splits in k or is irreducible over k. Moreover, the following are equivalent:

i) F/k is cyclic and [F : k] = pii) F = k(α) where

min (α,k,x) = xp−x− a

for some α ∈ k.

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Shaska T. 19

ii) F is the splitting field of the irreducible polynomial

f (x) = xp−x− a

for a ∈ k.

Proof. Roman, pg. 213.

Exercises:

19.8. Let F be an extension of k generated by all n-th roots of unity, for all n ≥ 1. Show that F/k is Abelian.

19.9. Let F be a field and σ ∈ Aut (F) such that |σ| = s > 1. Show that there is an α ∈ F such that

σ(α) = α+ 1

19.5 Kumer extensions

Theorem 19.7. Let k be a field containing a primitive n-th root of unity and F a finite extension of k. Then F/k is ann-Kummer extension if and only if F = k( n√a1, . . . , n√ar) for some a1, . . . ,ar ∈ k.

Proof. Morandi pg. 105

Exercises:

19.10. Let p1, . . . ,pn be distinct primes. Show that

[Q(√

p1, . . . ,√

pn) :Q] = 2n

19.6 Artin-Schreier theory

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19 Shaska T.

348

Chapter 20

Finite Fields

20.1 Basic definitions

Let Fp be a finite field of p-elements. We call the characteristics of Fp the smallest n ∈Z greater than zerosuch that for all α ∈ Fp, pα=0. Let F be a finite field with char (F) = p. Throughout this chapter p denotes aprime.

The Frobenious map is defined as

σ :Fp −→ Fq

x −→ xp

The proof of the following lemma is elementary and we leave it as an exercise:

Lemma 20.1. i) The Frobenious map is a monomorphism.ii) Fq is a vector space over Fp.

Example 20.1. In any characteristic p > 0 field, show that

(α+β)p = αp +βp

(α−β)p = αp−βp

Lemma 20.2. The order of a finite field is pn for some prime p.

Proof. We have shown that F has a copy of Fp inside. Thus, F is a finite extension of Fp. Let [F : Fp] = n.Hence, F = Fp(α1, . . . ,αn) for some α1, . . . ,αn. Each element of F is a linear combination

r1α1 + · · ·+ rnαn,

where r1, . . . ,rn ∈ Fp. Thus there are p-choices for each ri. Hence, F has pn elements.

Theorem 20.1. For every q = pn there is, up to isomorphism, a unique field Fq of size q.

Proof. Let L be the splitting field of f (x) = xq− x over Fq. Let S be the set of roots of f (x) in L. It is easily

verified that S is a field. Thus S = L. Since

f ′(x) = qxq−1−1 = −1 , 0

then f (x) has no multiple roots. Hence, |L|= q. Uniqueness comes from the uniqueness of splitting fields.

Lemma 20.3. Let Fq be a field of size q. Then, F∗q is a cyclic group under multiplication.

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20 Shaska T.

Proof.

Theorem 20.2. Fq is the splitting filed of f (x) = xq−x. (In other words, elements of Fq are the root of f (x) = xq

−x.)

Proof.

Theorem 20.3. (i) Every finite field has size pn for some p > 0, and n ∈Z+.(ii) For every q = pn there is up to isomorphism a unique fieldFq of size q, which is the splitting field of f (x) = xq

−xover Fq.

Proof.

20.2 Separable extensions

Theorem 20.4. Let F be a field such that char F = p and f (x) ∈ F[x] and irreducible. Then, f (x) has multiple rootsif and only if f (x) = g(xp) for some g(x) ∈ F[x].

Proof.

Corollary 20.1. All irreducible polynomials over a finite field are separable.

Proof.

20.3 Constructing Finite Fields

In this section we give a brief review of different methods of constructing finite fields. Let’s first recall afew facts:

1. F is a field F [x] is a UFD.

2. R is a ring, I is an ideal.

(a) R/I is a field if and only if I is maximal.

(b) R/I is a integral domain if and only if I is prime.

3. Let f ∈ F[x]. If f (x) is irreducible then 〈 f 〉 is a maximal ideal.

Lemma 20.4. Let f (x) ∈ Fq[x] s.t. deg f = n and f (x) is irreducible. Then Fp[x]/〈 f (x)〉 is a field of pn elements.

Proof.

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Shaska T. 20

20.4 Irreducibility of polynomials over finite fields

20.5 Artin-Schreier extensions

20.6 The algebraic closure of a finite field

Exercises:

20.1. Prove that any finite subgroup of the multiplicative group of nonzero elements of a field is cyclic.

20.2. Prove that for each prime number p and positive integer n there is (up to isomorphism) one field of order pn.Your proof should include an argument which shows that the order of a finite field is necessarily the power of a primenumber.

20.3. Prove the Wedderburn’s Theorem that every finite division ring is a field.

20.4. Let F be a field and p(x) ∈ F[x]. Show that F[x]/〈p(x)〉 is a field if and only if p(x) is irreducible in F[x].

20.5. Let Fq be a finite field, where q = pn. Show that Aut (Fq) is cyclic of order n. Show that Fq/Fp is normal andseparable.

20.6. Let k ⊂ E ⊂ F be a tower of fields such that F/E and E/k are separable. Show that F/k is separable.

20.7. Let k ⊂ E ⊂ F be a tower of fields such that F = k(α), where α is algebraic over k. Let

p(x) = a0 + a1x + a2x2 + · · ·+ an−1xn−1 + xn = Irr (α,E,x)

Show that E = k(α0, . . . ,αn−1).

20.8. i) Let E = k(α), α is algebraic over k with deg Irr (α,k,x) an odd number. Show that E = k(α2).ii) Let xn

− a ∈ k[x] be irreducible and θ be a root of xn− a in k. Suppose m|n. Show that [k(α) : k] = n

m .iii) In ii) do not assume m|n (m is an arbitrary positive integer). Find [k(θm) : k].

351

20 Shaska T.

352

Chapter 21

Transcendental Extensions

21.1 Transcendental Extensions

21.2 Lüroth and Castelnuovo theorem

Theorem 21.1. Let x be transcendental over k and F an intermediate field of the extension k(x)/k. Then:

i) F/k is purely transcendental and F = k(s) for some

s =p(x)q(x)

, p(x),q(x) ∈ k[x]

such that (p(x),q(x)) = 1.

ii) [k(x) : k(s)] = maxdegp,degq.

Proof.

Theorem 21.2 (Castelnuovo).

21.2.1 Automorphisms of k(x)

Lemma 21.1. Aut (k(x)/k)PGL2(k)

Proof.

21.2.2 Finite subgroups of PGL2(C)

Theorem 21.3. Let G be a finite subgroup of PGL2(C). Then G is isomorphic to one of the following: Zn,Dn,A4,S4,A5.

Proof.

where ω = −1+√

52 , ω = −1−

√5

2 , ζ is a primitive nth root of unity, ε is a primitive 5th root of unity, and i is aprimitive 4th root of unity.

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21 Shaska T.

21.3 Noether Normalization Lemma

Theorem 21.4 (Noether Normalization Lemma). If k is an infinite field and A is a finitely generated k-algebra,then A is integral over k or we can choose x1,x2, · · · ,xn and an index 1 ≤ r ≤ n, such that A = K[x1,x2, · · · ,xn] and

(a) the set x1,x2, · · · ,xr is algebraically independent over k and(b) A is integral over k[x1,x2, · · · ,xr].

21.4 Linearly disjoint extensions

21.5 Separable and Inseparable extensions

354

Chapter 22

Field Extensions

355

22 Shaska T.

356

Chapter 23

Norms and Traces

23.1 Introduction

In this section we define the very useful concepts of norm and trace. We will make use of them in the nextsection to prove the Hilbert 90 theorem on cyclic extensions.

Let F/k be a field extension with [F : k] = n. Fix α ∈ F and consider the linear map

Lα : F −→ Fx −→ ax

(23.1)

Recall that F is a n-dimensional vector space over k and Lα linear map on this vector space. Let Mα be theassociated matrix to the map Lα.

Definition 23.1. Let F/k be a finite field extension. The norm NFk and trace TrF

k of any α ∈ F are defined by

NFk (α) = det(Mα)

TrFk (α) = tr(Mα)

(23.2)

Remark 23.1. Recall from linear algebra that changing the base of the vector space would change the matrix Mα toa matrix similar to it, say A−1MαA. Hence the trace and the determinant would still be the same, since

det(A−1MαA) = det(Mα)

tr(A−1MαA) = tr(Mα)(23.3)

Thus, norm and trace are well defined.

The following example is part of the folklore in the theory of quadratic extensions.

Example 23.1. Let F = k(√

d) for some d ∈ F which is not a square in k. Let α ∈ F, say α = a+b√

d. We want to findNF/k(α) and TrF/k(α).

Pick a basis B = 1,√

d. Then

Lα(1) = a + b√

d

La(√

d) = (a + b√

d) ·√

d = bd + a√

d(23.4)

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23 Shaska T.

Then the associated matrix is

Mα =

[a b

bd a

]t

=

[a bdb a

]Thus we have

NFk (α) = det(Mα) = a2

−b2d

TrFk (α) = tr(Mα) = 2a

(23.5)

The reader hopefully recognizes these formulas from elementary number theory.

Lemma 23.1. Let F(α)/k be an algebraic extension where the minimal polynomial of α is

min (α,k,x) = xn +βn−1xn−1 + · · ·+β1x +β0.

Then,NF

k (α) = (−1)nβ0, TrFk (α) = −βn−1

Proof. We know that a basis for k(α) is

B = 1,α,α2, . . . ,αn−1.

Then,Lα(1) = α = (0,1,0,0, . . .0)

Lα(α) = α2 = (0,0,1,0, . . .0)

Lα(α2) = α3 = (0,0,0,1, . . .0)

. . . . . . . . .

Lα(αn−1) = αn = (−s0,−s1,−s2,−s4, . . .sn−1)

and the matrix Mα is given by

C f :=

0 0 . . . . . . −s01 0 . . . . . . −s10 1 . . . . . . −s2

. . . . . . . . .

. . . . . . . . .0 0 . . . 1 −sn−1

Then

tr(Mα) = −sn, and (−1)ns0.

This completes the proof.

Remark 23.2. Note that the matrix Mα is the companion matrix of the min (α,k,x).

The norm and trace can be characterized as follows:

Theorem 23.1. Let F/k be a finite field extension and σ1, . . . ,σn the distinct embeddings of F in an algebraic closureka of K. For α ∈ F we have

NFk (α) =

r∏j=1

σ j(α)

[F:k]i

and TFk (α) = [F : k]i

n∑j=1

σ j(α)

Proof. We will prove this theorem only for Galois extensions in the following corollary.

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Shaska T. 23

Corollary 23.1. Let F/k be a finite Galois extension with Galois group G. Then for each α ∈ F,

NFk (α) =

∏σ∈G

σ(α) and TFk (α) =

∑σ∈G

σ(α)

Proof. Let α ∈ F, f (x) = min (α,k,x) andG = 1,σ, . . . ,σn−1

.

We know that all σi(α), for i ≤ n are also roots of f (x). The result follows.

Example 23.2. Let F = k(√

d). Then F/k is Galois since every degree 2 extension is Galois. The Galois group isG = id,σ where

σ :√

d −→−√

d

Then for α = a + b√α ∈ F, we have

NFk (α) = α ·σ(a) = (a + b

√

d)(a−b√

d) = a2−b2d.

Similarly, TrFk (α) = 2a.

Lemma 23.2. Let L/F/k be finite field extensions. Then

NLk = NF

k NLF , TrL

k = TrFk TrL

F

Proof. Exercise

Exercises:

23.1. Let p be an odd prime and K :=Q(εp). Show that

NKQ(1−εp) = p

23.2. Let n ≥ 3 be an integer, εn a primitive n-th root of unity, and K :=Q(εn). Show that NKQ

(εn) = 1.

23.3. Let F =Q(√

3) and L =Q( 4√3). Compute NFQ

(√

3), NLF(√

3), TrFQ

(√

3), TrLF(√

3).

23.4. Let [K :Q] = n and α ∈Q. Show that

NKQ(α) = αn, and TrK

Q(α) = nα.

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23 Shaska T.

360

Chapter 24

Solutions

1.1 Let us start fresh ...

1.2 Let us

1.3 hjbkj

1.4 jhvjkh

1.5 gjg

1.6

1.7

1.8

1.9

1.10

1.11

1.12

1.13

1.14

1.15 Suppose that x2≡−1 ( mod p), then p - a, therefore by Fermat’s Little Theorem xp−1

≡ x4n+2≡ x2(2n+1)

≡

(−1)2n+1≡ −1(mod p). Thus 1 ≡ −1 (mod p), so 2 ≡ 0 (mod p) implies p|2 which in turn implies p = 2. Since

p = 4n + 3, p , 2 then there is no solution to the equation x2≡ −1 (mod p) .

1.16

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24 Shaska T.

1.17 By Corollary 1.2, we know φ(mn) = φ(m) ∗φ(n). We also know by Lemma 1.9 that φ(p) = p−1 and thatφ(pα) = pα−pα−1 = pα(1− 1

p )

φ(100) = φ(52∗22)

= φ(52) ∗φ(22)

= (52−5) ∗ (22

−2)= 20 ∗2= 40

φ(101) = 101−1 = 100 since 101 is prime

φ(102) = φ(2) ∗φ(3) ∗φ(17)= 1 ∗2 ∗16= 32

φ(103) = 103−1 = 102 since 103 is prime

φ(104) = φ(23) ∗φ(13)

= (23−22) ∗ (13−1)

= 4 ∗12= 48

φ(105) = φ(5) ∗φ(3) ∗φ(7)= 4 ∗2 ∗6= 48

φ(106) = φ(2) ∗φ(53)= 1 ∗52= 52

φ(107) = 107−1 = 106 since 107 is prime

φ(108) = φ(22) ∗φ(32)

= (22−2) ∗ (32

−3)= 2 ∗6= 12

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Shaska T. 24

φ(109) = 109−1 = 108 since 109 is prime

φ(110) = φ(2) ∗φ(5) ∗φ(11)= 1 ∗4 ∗10= 40

1.18 Let us consider the dihedral group D4 and look at Caley’s Table for it. Let

1. e = 0/360 rotation,

2. σ,σ2,σ3 = 90,180,270 (respectively),

3. τ,τ2 = Horizontal axis flip, Vertical axis flip (respectively),

4. ρ,ρ2 = 45 Diagonal Flip,135 Diagonal Flip (respectively)

Symmetries for D4

∗ e σ σ2 σ3 τ τ2 ρ ρ2

e e σ σ2 σ3 τ τ2 ρ ρ2

σ σ σ2 σ3 e ρ2 ρ τ τ2

σ2 σ2 σ3 e σ τ2 τ ρ2 ρσ3 σ3 e σ σ2 ρ ρ2 τ2 ττ τ ρ τ2 ρ2 e σ2 σ σ3

τ2 τ2 ρ2 τ ρ σ2 e σ3 σρ ρ τ2 ρ2 τ σ3 σ e σ2

ρ2 ρ2 τ ρ τ2 σ σ3 σ2 e

Let g = τ and h = ρ and we’ll choose 2 for n.

(gh)n = (τρ)2 = (σ)2 = σ2

gnhn = τ2ρ2 = e · e = e

Thus, e , σ2. This follows with Lemma 1.3 saying that if a group i s Abelian, then (gh)n = gnhn. Since D4is not Abelian, then (gh)n , gnhn.

1.19

1.20 Since the group is finite an cannot be distinct for all n ∈ Z. Let m1,m2 ∈ Z where m1 ,m2 such thatam1 = am2 Without loss of generality, let m1 >m2 then, m1−m2 ∈ Z+ and therefore am1−m2 = e

1.21 U(n) = m ∈Zn : gcd (m,n) = 1Note that the gcd(n−1,n) = 1, therefore n−1 ∈U(n).Let k = n−1, Then

k2 =(n−1)2

=n2−2n + 1 ≡ 1 mod n

→ k2 = 1

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24 Shaska T.

and k ∈U(n)

I f k = 1,n−1 ≡ 1 mod n→−1 ≡ 1 mod n→ 0 ≡ 2 mod n

This is possible only when n = 2. but from the assumption n > 2. Hence, n− 1 . 1 mod n therefore k isgreater than 2.

1.22 Let a,b ∈ G. By our assumption, (ab)2 = a2b2. Now, (ab)2 = (ab)(ab) = a(ba)b = a2b2. Now, by leftcancellation on a we have that (ba)b = ab2. Applying right cancellation on b yield ba = ab. Thus, G is abelian.

1.23 Let x, y ∈ G. x−1, y−1∈ G as well, since G is a group, so inverses exist.

xy = x−1y−1

(xy)−1 = (x−1y−1)−1

y−1x−1 = ((yx)−1)−1

y−1x−1 = yx

1.24 Let there exist an element c such that (ab)c = e is in fact c = b−1a−1 :

(ab)b−1a−1 = a(bb−1)a−1

= aea−1

= aa−1

= e

1.25 We know that (Zp,+) = 〈1〉. Suppose to the contrary that (Zp,+) has a proper subgroup H. We alsoknow that (Zp,+) is cyclic and since H ≤ (Zp,+) it is known that H is also cyclic. Let H = 〈x〉. Since x = x ·1,therefore the order of x is p

gcd (x,p)=

p1 = p. This makes H =Zp. So, the only subgroups ofZp are 0 and the

entire set. Therefore, (Zp,+) does not have a proper subgroup.

1.26 Let g,h ∈G such that |g| = 15 and |h| = 16. By definition of the order of an element we know that g15 = eand h16 = e, where e is the identity of the group G.

The group generated by g is

〈g〉 = e, g, g2, g3, g4, g5, g6, g7, g8, g9, g10, g11, g12, g13, g14

and the group generated by h is

〈h〉 = e,h,h2,h3,h4,h5,h6,h7,h8,h9,h10,h11,h12,h13,h14,h15

We know gcd(15,16) = 1. Therefore 〈g〉∩ 〈h〉 = e and |〈g〉∩ 〈h〉| = 1.

1.27 Let a be an element of a group G. Let 〈am〉 be the generator of G1 and 〈an

〉 be the generator of G2. Sowe want to find the generator of G1∩G2, this will be 〈ai

〉 where i is an integer. Since i must divide both mand n. Therefore, i = lcm(m,n).

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Shaska T. 24

1.28 Let x ∈Zn, where n > 2 generates Zn. Since 〈x〉 = 〈x〉−1, −x is also a generator of Zn. If we assume thatx = −x, then 2x = 0 (i.e. |x| = 2) then we have a contradiction because |x| , n. Therefore, we know that x , −xwhich means that all generators come in pairs. From this, it can be said that there are y pairs of generatorsin Zn where n > 2, meaning that there are 2y generators. Therefore Zn for n > 2 has an even number ofgenerators as 2y is an even number.

1.46 Let be given σ ∈ Sn. If σ is not a cycle, prove that σ can be written as product of at most (n− 2)transpositions.

For any σ ∈ Sn, σ can be expressed as a product of cycles. So, σ = (a1, ...,an1 )... ((ans−1+1, ...,ans ) where ns ≤ n.Since (a1, ...,an1 ) = (a1,an1 )...(a1,a2) it can be written as n1−1 transpositions. By doing the same to all of thesecycles, we obtain n1−1+ ...+ns−ns−1−1 = ns−s≤ n−s. As it has been given that σ is not a cycle, n−s≤ n−2and so there are at most n−2 transpositions.

1.29

1.30

1.31

1.32

1.33

1.34

1.35

1.36

1.37

1.38

1.39

1.40

1.41 It is obvious thatα = (12345) = (15)(14)(13)(12)

andβ = (1632)(457) = (12)(13)(16)(47)(45).

However this is not the only way to express α and β. Indeed, one can prove that

α = (12345) = (54)(53)(52)(51)

orα = (12345) = (54)(52)(21)(25)(23)(13)

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24 Shaska T.

1.42 Indeed

α =

(1 2 3 4 54 3 2 5 1

)Then,

α2 =

(1 2 3 4 55 2 3 1 4

)or

α4 = α2α2 =

(1 2 3 4 54 2 3 5 1

)and finally we have that:

α6 = α4α2 =

(1 2 3 4 51 2 3 4 5

)1.42

1.43

1.44

1.45 Let σ ∈ Sn, by theorem 1.6 then σ = τ1τ2 · · ·τr with τi (where i = 1,2, · · ·r) is the transposition beingdisjoint r-cycle of length ni and 1,2, · · · ,r is the number of transpositions.Since τ is a cycle, say τi = (a1,a2, · · ·ani ) then every cycle can be written as a product of transpositions,such that (a1,a2, · · ·ani ) = (a1ani )(a1ani−1) · · · )(a1,a3)(a1,a2), this implies that each τi can be written as ni − 1transposition. So σ can be written as n1−1 + n2−1 + · · ·nr−1 = n− r ≤ n−1 transposition.

1.46

1.47 Let σ be a cycle of odd length, then:

σ = (a1a2...a2ka2k+1)

σ2 = (a1a2...a2ka2k+1)(a1a2...a2ka2k+1)= (a1a3a5...a2k−1a2k+1a2a4...a2k)

We justify the last line by observing that ai is sent to ai+1 in the original cycle, and then multiplying itby the next cycle it is sent to ai+2, so for odd i, where i < 2k + 1, ai goes to the succeeding odd index, andlikewise for even i, it goes to the succeeding even index. a2k+1 is sent to a1, but is then sent to a2 in themultiplication. This results in a single cycle of equal length to the original σ, so σ2 is still odd.

1.48 Let us consider a 3-cycle (αβγ), for some α,β,γ ∈Z. To show that it is an even permutation, we need toshow that it can be written as a product of an even amount of transpositions. The simplest example wouldbe:

(αβγ) = (αβ)(αγ)

This shows that α can go to either β or γ, β can go to α or to γ through α, and finally γ can go to α orto β through α. Therefore, since there are two transpositions in the product, then the 3-cycle is an evenpermutation.

1.49

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Shaska T. 24

1.50 |An| =n!2 for n ≥ 2. So |A4| =

4!2 = 12 Here are the transpositions

(12)(34), (13)(24), (14)(23)

and 8 3-cycles(123), (132), (142), (234), (124), (134), (143), (243)

Since there are 8 elements of 3-cycles then there cannot exist a subgroup of order 6.

1.51

1.52 Consider two elements of An such that n ≥ 4, σ and β where β = (234) and σ = (123). Composing βσand then σβ we see that σβ = (234)(123) = (13)(24) and βσ = (123)(234) = (12)(34). (13)(24) , (12)(34). Hence,σβ , βσ. Thus, An for n ≥ 4 isn’t abelian.

1.53 Let αn = e and β2 = e. Since the dihedral groups represent the symmetries present in a regular n-gon,all Dn with n ≥ 3 are non-Abelian because there is no diagonal that the elements can be reflected over intheir Cayley tables.

1.54 This doesn’t compile. The latex source makes no sense and the solution has absolutely no ideas.

1.55 a) We know every element of Sn can be written as a product of transpositions. For transpositions (i j)where i, j , 1 can be written (i j) = (1i)(1 j)(1i). So 1→ i→ 1 and i→ 1→ j and j→ 1→ i. Therefore i→ j andj→ i. So (12), (13), · · · , (1n) generates all of Sn.

b) We must prove by induction that (1k) can be written as (12), (23), ..., (i i+1), ..., (n−1 n) for k = 2,3, ...,n.The base case is clear: (12) = (12). By the inductive step (1 k + 1) = (1k)(k k + 1)(1k). By part (a), the set(12), (13), ..., (1n) generates Sn, and so (12), (23), ..., (n−1 n) does as well.

c) By induction we must show that (k−1 k) can be written as (12), (12...n) for k = 2,3, ...,n. The base caseis again clear with (12) = (12). The inductive step gets k k + 1) = (12...n)(k−1 k)(n...n1). By part (b), the set(12), (23), ..., (n−1 n) generates Sn, and so (12), (12...n) does as well.

1.56

F20 = 〈(2354), (12345)〉= (1), (12)(35), (13)(45), (14)(23), (15)(24), (25)(34), (1243), (1254), (1325), (1342),

(1435), (1452), (1523), (1534), (2354), (2453), (12345), (13524), (14523), (15432)

Therefore|F20| = 20

and the lattice of F20 is:

〈(2354), (12345)〉

〈(12345)〉〈(2354)〉

(1)

Where 〈(2354)〉 = (1), (25)(34), (2354), (2453) and 〈(12345)〉 = (1), (12345), (13524), (14253).

1.57

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24 Shaska T.

1.58 D8 can be written as D8 = 1,r,r2,r3,r2s,r3swhere r is a rotation of 45 degrees and s is a 180 degree fliprsr = s, therefore r and s do not commute therefore they are not in the center of D8, but rsr = s mean thatsr = r3s so r3 and s do not commute therefore they are not in the center of D8. Then (rs)s = r but srs = (sr)s =r3, so rs is not in the center. (r2s)r = rs but r(r2s) = r3s so r2s is not in the center either. finally (r3s)s = r3 buts(r3) = r so r3s is not in the center. This leaves 1 and r2 of which both commute with every other elementtherefore they are in the center. Thus the center of D8 = 1,r2

. The center of D10 is trivial, meaning that thecenter of D10 = 1. in General, in the group Dn, sr = rn−1s, so s will not commute with rk unless rk = rn−k

which can only happen if n is even. Therefore the center of Dn is 1 when n is odd. When n is even, thecenter is 1,r(n/2)

.

1.59 1) Multiply both sides of the equation by σ we get

στ = (σ(a1),σ(a2, · · · ,σ(ak)σ

Let A = στ and B = (σ(a1),σ(a2), · · · ,σ(ak))σ. We want to show A = B by proving A(x) = B(x) for somex =

1,2,3, · · · ,n

. We consider cases:

Case 1: x = a j for any j <1,2,3, · · · ,k

, then x is fixed in τ, thus A(x) = στ(x) = στ(a j) = σ(τ(a j)) = σ(a j).

Similarly for the right hand side: B(x) = (σ(a1),σ(a2), · · · ,σ(ak))σ(x) = (σ(a1),σ(a2), · · · ,σ(ak))σ(a j) = σ(a j). HenceA(x) = B(x).Case 2: If x = ai for some i ∈

1,2,3, · · · ,k

Sub case 2.1: 1 ≤ i ≤ k−1. Then

A(x) = στ(x) = στ(ai) = σ(τ(ai)) = σ(ai+1)

.B(x) = (σ(a1),σ(a2), · · · ,σ(ak))σ(ai) = (σ(a1),σ(a2), · · · ,σ(ak))(σ(ai)) = σ(ai+1)

. Hence A(x) = B(x).Subcase 2.2: i = k. Then

A(x) = στ(x) = στ(ak) = σ(a1)

B(x) = (σ(a1),σ(a2), · · · ,σ(ak))σ(ak) = σ(ai)

From case 1 and 2, we conclude that A(x) = B(x) for all x. Thus A = B implies Aσ−1 = Bσ−1. Therefore, theresult holds. 2) Suppose µ = (b1,b2, · · · ,bk). Now let σ be the permutation that satisfies σ(ai) = bi for somei = 1,2, · · · ,k. By part 1) we have:

στσ−1 = (σ(a1),σ(a2, · · · ,σ(ak)= (b1,b2, · · · ,bk)= µ

(24.1)

1.60 Assume that αβ = βα and α , e. Let c,d, f be three distinct elements such that α(c) = d and β = d f .Therefore, we can say that αβ(c) = α(c) = d and also that βα(c) = β(d) = f . Therefore, αβ = d , f = βα and soαβ , βα. Therefore, α is the identity permutation if αβ = βα for every β ∈ Sn.

1.61 Let α be even. Then it can be expressed by the sequence of disjoint transpositions

σ1σ2...σ2k

Then the inverse is simply

σ−12k σ2k−1−1...σ−1

2 σ−11

since each transposition is disjoint. So the inverse is trivially even. The same argument is made for an oddpermutation, resulting in an odd permutation for the inverse.

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Shaska T. 24

1.62 First, it is necessary to prove this lemma.

Lemma. If α is an even cycle, then α−1 is also even.

Proof. We know that an even permutation can be written as a product of an even amount of transpositions.Let α = α1α2α3 . . .αn−1αn where α is a cycle and αn (n ∈ Z+) is a transposition of α. from this we take theinverse.

(α)−1 = (α1α2α3 . . .αn−1αn)−1

= (αnαn−1 . . .α3α2α1)

= α−1

Since both α and α−1 have the same number of transpositions, then α−1 is also even.

Using the lemma above, Let

α = α1α2α3 . . .αn α−1 = αn . . .α3α2α1

β = β1β2β3 . . .βn β−1 = βn . . .β3β2β1

If then we multiply them, we get

α−1β−1αβ = (αn . . .α3α2α1)(βn . . .β3β2β1)(α1α2α3 . . .αn)(β1β2β3 . . .βn)

Thus, α−1,β−1,α and β are all of even transpositions. Therefore, since we know that any amount of evensmultiplied together yield an even number, then we can say that α−1β−1αβ is also even.

1.66 Let matrix A =

[a1 b1c1 d1

]where ad−bc , 0 and matrix B =

[a2 b2c2 d2

]where ad− bc , 0 and A,B ∈ G

Now, [a1 b1c1 d1

]∗

[a2 b2c2 d2

]=

[a1a2 + b1c2 a1b2 + b1d2c1a2 + d1c2 c1b2 + d1d2

]Since A,B ∈ G then, a1,a2,b1,b2,c1,c2,d1,d2 ∈ G So addition and multiplication of these matrices are closedand therefore associative. Let the identity matrix be

I =

[1 00 1

]Then, [

a1 b1c1 d1

]∗

[1 00 1

]=

[a1 b1c1 d1

]So the identity exists. Since det(A) , 0 and det(B) , 0 then,

det(AB) = det(A)det(B)

and

det(A−1) =1

det(A)

therefore the inverse exists. Thus G forms a group under matrix multiplication.

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24 Shaska T.

1.67 First to show G is a group, the identity must exist within the Group. Let a = 1,d = 1,b = 0→[1 00 1

]∈

Mat2(R) and ad , 0. Thus the identity element is in G.

Second, associativity must hold for the supposed group. Let A,B,C be matrices and A =

[a b0 d

],B =[

x y0 z

],C =

[r k0 m

]such that (AB)C = A(BC). First consider (AB)C.

(AB)C =

[a b0 d

][x y0 z

]C =

[ax ay + bz0 dz

]C =

[ax ay + bz0 dz

][r k0 m

]=

[r(ax) (ax)k + m(ay + bz)

0 dzm

]Now consider A(BC).

A(BC) = A[x y0 z

][r k0 m

]= A

[xr xk + ym0 zm

]=

[a b0 d

][xr xk + ym0 zm

]=

[a(xr) a(xk + ym) + bzm

0 dzm

]Since (AB)C = A(BC) G is associative.

Third, there must exist an inverse in G such that gg−1 = e. Consider A−1 = 1ad

[d 0−b a

]∈ G since ad , 0 and

all entries are real numbers. 1ad

[d −b0 a

][a b0 d

]=

[1 00 1

]which is the identity. Therefore ∀g ∈ G ∃g−1 such

that gg−1 = e.Now consider AB = BA to check if G is abelian.

AB =

[a b0 d

][x y0 z

]=

[ax ay + bz0 zd

]BA =

[x y0 z

][a b0 d

]=

[xa xb + yd0 zd

]Thus AB , BA and G is not abelian.

1.68 closure: Let A,B ∈ G such that A =

(a 00 a−1

)and B =

(b 00 b−1

)where a,b , 0, we see that AB =(

a 00 a−1

)(b 00 b−1

)=

(ab 00 (ab)−1

)∈ G since ab , 0.

Associativity: Matrix multiplication is associative in general, and so it holds in G as well.

Identity: e =

(1 00 1

)is the identity in G as CI = IC = C, ∀C ∈ G.

Inverses: A =

(a 00 a−1

)where a , 0. Well,

(a 00 a−1

)−1

=

(a−1 00 a

)where a−1 , 0. Thus, G is a group under

the operation of matrix multiplication.

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Shaska T. 24

To see that G is abelian, we have the following:

AB =

(a 00 a−1

)(b 00 b−1

)=

(ab 00 (ab)−1

)=

(b 00 b−1

)(a 00 a−1

)Hence, G is abelian.

1.69

1.70 |G| ≤ 34 = 81, since a,b,c,d can each take one of 3 values (mod 3).1.) Suppose ad = bc = 0(mod3). Then (a = 0) or (d = 0) and (b = 0) or (c = 0), leading to 25 possible values

for a,b,c,d.2.) Suppose ad = bc = 1(mod3). Then (a = d = 1) or (a = d = 2) and (b = c = 1) or (b = c = 2), leading to 4

possible values for a,b,c,d.3.) Suppose ad = bc = 2(mod3). Then (a,d) = (1,2) or (a,d) = (2,1) and (b,c) = (1,2) or (b,c) = (2,1), leading

to 4 possible values for a,b,c,d.So there are in total 25 + 4 + 4 = 33 such matrices where ad− bc = 0(mod3). That means there are at most

81 - 33 = 48 such matrices where ad− bc , 0.

1.71 By the formula |SLn(Fp)| = pn(n−1)

2∏i=n

i=2(pi−1) we can find |SL2(3)| by plugging n = 2 and p = 3 into the

formula. We get 32(2−1)

2∏i=2

i=2(3i−1) = 3

22 (32−1) = 3(9−1) = 24. Therefore, |SL2(3)| = 24.

1.72 From the construction of GL2(p), we know the entries of the matrices (a,b,c, and d) are from Fp. Recallthat Fp = [0], [1], . . . , [p−1] and |Fp| = p. Thus each entry has p possibilities when not restricted; however,in our construction of GL2(p) we have that the determinant cannot equal zero.

Let’s treat the matrices M ∈ GL2(p) as two vectors u and v such that

M = [uv] where u =

[ac

]and v =

[bd

].

Clearly, u , 0 and v , 0 since ad− bc , 0. WLOG, we will say that u , 0 is the only restriction on u. Thenthe possibilities for u are everything but when both a = 0 and b = 0 which equates to (p2

−1). Next we willlook at the possibilities of v, taking into account we only put one restriction on u. If v is any scaled versionof u, then ad− bc could equal 0. Therefore v , tu for some t ∈ Fp. Since |Fp| = p, there are p possibilities fort that we must take into account. And so the possibilities for v are everything but when v can be writtenas tu which equates to (p2

−p). Multiplying the possibilities of the two vectors u and v together gives thepossibilities of matrices in GL2(p).

1.73

1.77

1.81

1.82

2.1 First, we prove that this subset is not empty. Since, e2 = e we have that e ∈H. Let a and b be two elementsfrom H. We have a2 = e and b2 = e and want to show that (ab−1)2 = e. Since G is an Abelian group we have:

(ab−1)2 = (ab−1)(ab−1) = ab−1ab−1 = (aa)(b−1b−1) = (a2)(b−1)2 = e(b2)−1 = e.

Thus, (ab−1) ∈H. Therefore, H is subgroup of G.

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24 Shaska T.

2.2 First, we prove that H , ∅. Since e2 = e, we have that e ∈H.Let a,b ∈ G such that a2 and b2 are in H. Notice that

a2(b2)−1 = (ab−1)2∈H.

From the first subgroup test H is a subgroup of the group G.

2.3 The identity of Rtimes is 1. Since 1 = 1/1 is the ratio of two nonzero integers then the identity of Rtimesis inQ×. If two elements p/q and r/s are given inQ×, then their product pr/qs is also inQ×. Also the inverseof every element p/q ∈Q× is again in Q× because (p/q)−1 = q/p. Finally, since multiplication in Rtimes hasthe association property then this property is true also in Q×.

2.5 A matrix

A =

(a bc d

)is in SL2(R) when ad− bc = 1.

The 2×2 identity matrix I2 is in SL2(R) and so is the inverse of the matrix A

A−1 =1

ad−bc·

(d −b−c a

),

sincedetA−1 =

1ad− bc

· (ad−bc) = 1.

Finally we have to show that the multiplication is closed. The product of two matrices with determinant 1is again a matrix with determinant 1.

2.6 Let G be an Abelian group and Cube(G) = g3| g ∈G. WTS: Cube(G) is a subgroup of G using the Second

Subgroup TestSince G is a group eG ∈ G. (eG)3 = e3

∈ Cube(G).Since G is a group, g1g2 ∈ G, so (g1g2)3

∈ Cube(G). And since g1, g2 ∈ G, (g1)3, (g2)3∈ Cube(G). Then,

(g1)3(g2)3 = (g1g2)3. Since (g1g2)3∈ Cube(G), Cube(G) is closed.

Since G is a group g−1∈ G, (g−1)3 = (g3)−1, so (g3)−1

∈ Cube(G).

2.7 We will use the second subgroup test so we must show that Gn is a subset of G, the identity of G is inGn, closure in Gn under the operation, and the elements in Gn have inverses that are also in Gn.

i) Subset:By our construction of Gn, we are taking elements from G and performing the operation n times oneach element. Since G is a group, it is closed under the operation with its elements. Therefore, Gn isa subset of G.

ii) Identity:Let e be the identity in G. Then en

∈ Gn but en = e. Thus, e ∈ Gn.

iii) Closure:Let g,h ∈ G. Then gn,hn

∈ Gn such that

gn = g1g2 . . . gn−1gn and hn = h1h2 . . .hn−1hn

where gi = g and hi = h, ∀ i = 1,2, . . . ,n (used to depict an order).

And sognhn = (g1g2 . . . gn−1gn)(h1h2 . . .hn−1hn)

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Shaska T. 24

Since G is an Abelian group, we know ∀ a,b ∈ G, ab = ba. We can use this to rearrange:

gnhn = g1g2 . . . gn−1(h1gn)h2 . . .hn−1hn

= g1g2 . . . (h1gn−1)(h2gn) . . .hn−1hn

...

= g1h1g2h2 . . . gn−1hn−1gnhn

= (gh)n

Since gh ∈ G, (gh)n∈ Gn and thus Gn is closed.

iv) Inverses:Let a,a−1

∈ G such that aa−1 = a−1a = e, where e is the identity in G.

Then an, (a−1)n∈ Gn where

an = a1a2 . . .an−1an and (a−1)n = a−11 a−1

2 . . .a−1n−1a−1

n

where ai = a and a−1i = a−1, ∀ i = 1,2, . . . ,n (used to depict an order).

And soan(a−1)n = (a1a2 . . .an−1an)(a−1

1 a−12 . . .a−1

n−1a−1n )

We will again use the fact that G is an Abelian group to rearrange:

an(a−1)n = a1a2 . . .an−1(a−11 an)a−1

2 . . .a−1n−1a−1

n

= a1a2 . . . (a−11 an−1)(a−1

2 an) . . .a−1n−1a−1

n

...

= a1a−11 a2a−1

2 . . .an−1a−1n−1ana−1

n

= (aa−1)n

= en = e

Therefore ∀ an∈ Gn, its inverse is (a−1)n and (a−1)n

∈ Gn.

Therefore, by the second subgroup test, Gn is a subgroup of G.

2.8 By the third subgroup test we will show that H is closed under multiplication which is the operationassigned to the groupQ. Let h1,h2 ∈H such that h1 = 2k1 and h2 = 2k2 such that k1,k2 ∈Z. Want to show thath1h2 ∈H. So, h1h2 = 2k12k2 = 2k1+k2 . Since k1,k2 ∈Z, the sum (k1 + k2) ∈Z and therefore h1h2 ∈H. Since H isclosed under multiplication H ≤Q∗.

2.9

2.10 To prove that a group is a subgroup four properties must be satisfied: G ⊂RX, the identity of G is thesame as that of R, G is closed under multiplication, and every element of G has an inverse.

• Subset: Since both a and b ∈Q and Q ⊆R , and G , 0, since a and b both don’t equal 0.

• Identity: 1 is the identity of (RX,×) because 1 is the identity of multiplicative groups. So,

1(a + b√

2) = a + b√

2

(a + b√

2)1 = a +√

2

Since both equal a +√

2, which was the initial input, 1 is the identity.

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24 Shaska T.

• Closed under multiplication: Let a,b,c,d ∈G, where a and b, both don’t equal zero and c and d , bothdon’t equal zero. So a,b,c,d ∈RX, since G ⊂RX

(a + b√

2)(c +√

2)

ac + ad√

2 + cb√

2 + 2bd

ac + 2bd + ad√

2 + cb√

2

ac + 2bd + (ad + cb)√

2

Since ac + 2bd + (ad + cb)√

2 ∈RX, G is closed under multiplication.

• Inverse:a + b√

21

·1

a + b√

2

=a + b√

2

a + b√

2

= 1

Therefore, G is a subgroup of (RX,×)

2.11 Closure: Let A,B ∈H such that A =

(a bc d

)and B =

(e fg h

)where a + d = 0 and e + h = 0.

A + B =

(a bc d

)(e fg h

)=

(a + e b + fc + g d + h

)Now, (a + e) + (d + h) = a + e + d + h = a + d + e + h = 0 + 0 = 0 Hence, H is closed under the group operation ofmatrix addition.

Associativity: Matrix addition is associative in G, hence it is also associative in H. Hence, the associativeproperty holds in H.

Identity: Consider eG =

(0 00 0

). a + d = 0 + 0 = 0 Hence, eG ∈H.

Inverses: Consider A =

(a bc d

). Then,

A−1 =

(a bc d

)−1

=1

ad− bc

(d −b−c a

)=

d

ad− bc−b

ad− bc−c

ad− bca

ad− bc

Now,

dad− bc

+a

ad−bc=

d + aad−bc

=0

ad−bc= 0. Hence, A−1

∈H. So, H is closed under inverses.

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Shaska T. 24

Thus, H ≤ G.

2.12 Clearly since Z ⊂ R→ SL2(Z) ⊂ SL2(R). Since SL2(Z) ⊂ SL2(R) we can apply the 2nd subgroup test.

Assume that SL2(Z) ≤ SL2(R). First we must verify that the identity is in SL2(Z). Since[1 00 1

]∈ SL2(Z) we

can confirm that the identity is in the group. Next, since SL2(Z) is an established group we know that theclosure property is true. Finally let A ∈ SL2(Z). A−1 exists since the determinant is 1 the matrix is just of the

form A−1 =

[d −b−c a

]where a,b,c,d ∈Z and AA−1 = e Thus SL2(Z) ≤ SL2(R).

2.13 Let Q8 = I,−I, J,−J,K,−K,L,−Lwhere I is the identity matrix and J,K,L are matrices.

J =

[i 00 i

]K =

[0 1−1 0

]L =

[0 ii 0

]I =

[1 00 1

]By the third subgroup test, the subgroups are S1 = I,−I with generator −I S2 = I,−I, J,−J with generatorJ or −J S3 = I,−I,K,−K with generator K or −K and S4 = I,−I,L,−L with generator L or −L Q8 is also asubgroup.

2.14

2.15 To disprove this claim, we need to prove that HK fails a subgroup test. We will look at the secondsubgroup test. Specifically the second part where it states:

ii) if h1,h2 ∈H, then h1h2 ∈H

In this case, let H = HK and h1,h2 = x, y, such that x, y ∈HK. We would make x = h1k1 and y = h2k2. If welook at xy = h1k1h2k2, we cannot move the elements in anyway since we do not know if G is Abelian. ThusHK fails the second subgroup test and is not a subgroup of G.

Now, if G is Abelian, then we can re-approach the problem and show that HK ≤ G.Let us use the second subgroup test:

1. Identity e of G is in HKSince H and K both are subgroups of G, then they both contain the identity e. If we take e and applyit to HK,

ee = e : e ∈H and e ∈ K

Thus, e ∈HK

2. If x, y ∈HK, then xy ∈HK.From earlier, let x, y ∈HK, such that x = h1k1 and y = h2k2. Since G is Abelian, then

xy = (h1k1)(h2k2)= h1k1h2k2

= h1h2k1k2

= (h1h2)(k1k2)

Thus, since h1h2 ∈H and k1k2 ∈ K, then xy ∈HK.

3. If x ∈HK, then x−1∈HK

Since H and K are subgroups of G, let h−1∈H and k−1

∈ K. Then, since H and K are abelian and we’veshown that xy ∈HK,

h−1k−1 = (hk)−1.

Thus, by hk ∈HK, then hk−1∈HK. Therefore, x−1

∈HK.

375

24 Shaska T.

∴ from the Second Subgroup Test, HK ≤ G.

2.16 Let the matrix in question be A. Let an element in the centralizer of A be called B. We want to find allB such that AB = BA. So let

B =

[a bc d

]So we want

[1 10 1

][a bc d

]=

[a bc d

][1 10 1

]In other words:[

a + c b + dc d

]=

[a a + bc c + d

]So then we have a system of equations:

a + c = a (24.2)b + d = a + b (24.3)

d = c + d (24.4)

From (1) and (3) we can see that c = 0. From (2) we can see that a = d. So we get the matrix

B =

[a b0 a

]where a and b are any real number. This comprises the centralizer of A.

2.17 Let |G| = 2n and x ∈ G such that |x| , 2 and x , e. Then, |x−1| , 2. So for every such x of order not equal

to two, x−1 has also order not equal to 2.Thus, we have an even number of elements x , e of order different from 2, say this number is 2t. Thus,

the number of elements of order 2 is 2n−2t−1. Therefore, it is an odd number.

2.18 To show that s : G→ G is a surjective we have to show that ∀b ∈ G, there exists x ∈ G such that x2 = b.In other words, every element of G can be represented as a square. Let b ∈ G. Since |G| = 2n + 1 for someinteger n, then b2n+1 = e. Then (bn+1)2 = b2n+2 = b2n+1

· b = b. Thus, (bn+1)2 = b. So it does exist an x = bn+1

such that x2 = b.Next we show when this is a homomorphism? If s is an homomorphism then s(xy) = s(x) · s(y) = x2

· y2.Also, s(xy) == (xy)2. Hence, the map is a homomorphism if and only if (xy)2 = x2

· y2. We have shownbefore that this is equivalent with G being Abelian.

2.19 We have φ : (Z,+)→ G, such that φ(1) = g. Then

φ(n) = φ(1) · · · · ·φ(1) = gn.

So φ(n) is uniquely determined.

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Shaska T. 24

2.20 Indeed,φ(ϕ+β) = cos(ϕ+β) + isin(ϕ+β)

(cos(ϕ)cos(β) + sin(ϕ)sin(β)) + i(sin(ϕ)cos(β) + cos(ϕ)sin(β))

(cos(ϕ) + isin(ϕ) + (cos(β) + isin(β)))

φ(ϕ)φ(β).

2.21

2.22 False. Counterexample:Klein 4-group: V4 =

e,a,b,c

is not cyclic, but subgroups of V4 : 〈e〉 =

e;〈a〉 =

e,a

since a0 = e,a1 = a,a2 =

e,a3 = a2a = a,a4 = e; similarly 〈b〉 =e,b

;〈c〉 =

e,c

are cyclic.

2.23 Suppose group G is infinite. Then, take some g ∈ G and consider the subgroup 〈g〉. This subgroupwould then have infinitely many subgroups (it is isomorphic toZ). So for G to be finite, it must not containthis infinite cyclic subgroup. So, every element generates a cyclic subgroup that is finite and G is then theunion of all of the cyclic subgroups. Since all of these subgroups are finite and there are in effect finitelymany as well we can say that G too, much be finite.

2.24 Given that the quaternion group consists of 2×2 matrices, we can observe that the identity for othersuch matrices is still the identity for this group, that is, the identity is

1 =

[1 00 1

]So any cyclic subgroup must be generated by an element, say k, such that for some integer n, we have

kn = 1. Consider first our elements I, J,K:

I =

[0 1−1 0

]J =

[0 ii 0

]K =

[i 00 −i

]Very quickly it is apparent that I2 = J2 = K2 = −1 and we can observe that −12 = 1, so I, J and K are all

generators of cyclic subgroups of order 4, these sets being 〈I〉,〈J〉,〈K〉. Furthermore, −I,−J,−K are distinctelements as well. However, these elements are generated by the previous elements, and we see

−I = I3

−J = J3

−K = K3

So (−I)2 = I6 = I2∗ 1 = −1 and (−I)3 = I9 = (I4)2I = I, and a similar observation is made for −J and −K,

so they do not produce a distinct group. We also have the element −1, which we can observe produces acyclic group of order 2. Having covered all the elements of Q8 we can see that there are 4 cyclic subgroupsin the quaternion group:

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24 Shaska T.

〈I〉〈J〉〈K〉〈−1〉

2.25 Let G be a group where |G| = pq and gcd(p,q) = 1. Assume a,b ∈ G such that, |a| = p and |b| = q.Suppose |ab| = n for some n. Then (ab)n = e and since G is Abelian, we have that anbn = eIf n = pq, then

apq = (ap)q = eq = ebpq = (bq)p = ep = e

Thus, |ab| = n =⇒ n | pq. This means n = 1,pq,p,orq, since gcd(p,q) = 1If n = 1, then (ab)1 = e and thus ab = e. From there we can see that |a|= |b| and thus p = q, which contradicts

gcd(p,q) = 1.If n = p, then we’d have

(ab)p = apbp = ebp =⇒ bp = e

Then, |b| = p which then means that q | p which again contradicts gcd(p,q) = 1. We can similarly observe thesame for when n = q.

We are left then with n = pq. Which then |ab| = pq. Therefore 〈ab〉 = G and thus, G is cyclic

2.26

2.27 Let p,q be distinct prime numbers so gcd(p,q) = 1. Since Zpq has pq elements and any element of Zpq canbe a generator then let n be the generator of Zpq where gcd(n,pq) = 1. Since gcd(n,p) = p then p,2p, ..., (q−1)pcannot be generators. Since gcd(n,q) = q then q,2q, ..., (p−1)q cannot be generators as well and the identityis not a generator. Then the number of generators is

pq− (q−1)− (p−1)−1 = pq− q + 1−p + 1−1

= pq−q + 1−p

= q(p−1)− (p−1)

= (p−1)(q−1)

2.28 We know that an integer k ∈Zn is a generator of Zn iff gcd(k,n) = 1the number of generators of

Zn = |Un|

= φ(n)

So the number of generators of

Zpr = φ(pr)

= pr(1−1p

)

= pr−1(p−1)

thus the number of generators of Zpr = pr−1(p−1)

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Shaska T. 24

2.29 Since G is a finite group, ∃a ∈ G such that a has maximal order. Consider < a >, where | < a > | = n. IfG =< a >, then it must be the case that G is cyclic. Suppose that G ,< a >. That is, ∃b ∈G such that b << a >.Denote |b|= k. Now, it must be the case that k ≤ n, as n is the order of a, which has maximal order in G. Now,since G is abelian, ∃r ∈ G such that |r| = lcm(n,k). Since n is the maximal order of an element in G, we havethat |r| = n, since clearly lcm(n,k) ≥ n. This implies that lcm(n,k) = n. Which is true only when k divides n.Hence, n = qk for some q ∈Z. Hence, bn = bqk = (bk)q. Since |b| = k, we have that (bk)q = eq = e. Hence, b is asolution to xn = e. Now, since all elements in < a > satisfy this equation, we have found n+1 such elementsthat do, which is a contradiction against our assumption that only n such solutions exist. So, it must be thecase that G =< a >. And thus, G is cyclic.

2.30

2.31 The group of integers is always over the binary relation of addition. ∃ a subgroup of (Z,+) such thatthe subgroup has an infinite index. The trivial subgroup 0 has an infinite index because ∀n ∈Z 0+n = nand since n is in the integers and are infinite n is infinite. Therefore, 0 is infinite, and not every subgroupof the integers has finite index.

2.32 absolutely useless

2.33 Want to prove that [Q : Z], that is |Q||Z| is infinite. Let x, y ∈ Q with 0 ≤ x < 1 and 0 ≤ y < 1. Then,

x− y ∈ [0,1). Now look at x +Z = y +Z⇒ x− y +Z = Z⇒ x− y = 0. Since x− y ∈ [0,1), x− y = 0⇒ x = y.Therefore, the coset x +Z, for 0 ≤ x < 1, form infinite distinct cosets of Z in Q. Therefore, [Q :Z] is infinite.

2.34 We want to find the index of the additive group of real numbers in the additive group of complexnumbers. This index is the number of cosets ofR in C. These cosets look like z+R = z+n | n ∈R for somez ∈ C.

Let x, y ∈ C such that x = a + bi and y = c + di where a,b,c,d ∈R and i =√−1. Then

x +R = y +R⇔ (x− y) ∈R⇔ b = d⇔ b−d = 0

We know b−d = 0 has the same amount of possibilities as the order of the real numbers and thus, b−d = f ,for some f ∈ R where f , 0, has infinitely many more possibilities than b− d = 0. Therefore, there areinfinitely many cosets ofR in C and so the additive group of real numbers has infinite index in the additivegroup of complex numbers (i.e. [(C, +) : (R, +)] =∞).

2.35

2.36

2.37 Let G be a group such that |G| = 2n.WTS: the number of elements with order 2 is odd.Since the order of an element of a finite group has to divide the order of the group, there exists an

element with order two since the order of the group is even. Consider the set G−e. |G−e| is odd. LetA be the set of all elements in G with order 2. Let S be the set of all nonidentity elements a , a−1. So, |S| iseven. Since |G| is even, |e| is odd, and |S| is even, |A|must be odd. So the set of all elements with order 2 isodd.

WTS: G contains a subgroup with order 2.Pick a ∈ G where |a| = n. Then, a ∗ a = e. Consider the group, e,a. The identity of G is the identity of the

new group. e ∗ e = e, e ∗ a = a, a ∗ e = a, and a ∗ a = e, so the new group is closed. The inverse of e is e and theinverse of a is a, so the inverse of each element exists in the new group. Therefore, by the second subgrouptest, the group e,a is a subgroup in G with order 2.

2.38

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24 Shaska T.

2.39

2.40

2.41

2.42

3.1 Indeed the function exponential has inverse the function lnx, so it is a bijection. Also we have that:

f (x + y) = ex+y = ex ey = f (x) f (y) for every x, y ∈R

Thus, we proved that (R,+) (R+, ·).

3.1

3.2 Define a map

φ :Z→Q×

n→ 2n.

Then,φ(m + n) = 2m+n = 2m 2n = φ(m)φ(n).

From the definition, the map φ is surjective in the subset

φ(Z) = 2n : n ∈Z ⊂Q×

To prove that φ is injective, assume that m, n. If we can prove that, φ(m),φ(n), then of we have completedthe proof. Assume, that m > n. If φ(m) = φ(n), then 2m = 2n or 2m−n = 1, which is impossible becausem−n > 0.

3.3 First we will show the identity map eG : G −→ G is an isomorphism for every a,b ∈ G. So eG(ab) = ab =eG(a)eG(b). Therefore the bijection eG is a homomorphism of G.

Next suppose f : G −→ G′ and g : G′ −→ G′′ are homomorphisms. We will show g f : G −→ G′′ isa homomorphism. We have (g f )(ab) = g( f (ab)) = g( f (a) f (b)) = g( f (a))g( f (b)) = (g f )(a)(g f )(b) for alla,b ∈ G.

Finally suppose f : G −→ G′ is an isomorphism and we will show the comosition inverse f−1 : G′ −→ Gis an isomorphism. Since f and f−1 are inverse functions f−1( f (a)) = a for all a ∈ G and f ( f−1(b)) = b for allb ∈ G′. Therefore

f−1(ab) = f−1( f ( f−1(a)))( f ( f−1(b)))

= f−1( f ( f−1(a) f−1(b)))

= f−1(a) f−1(b)

for all a,b ∈ G′.Since eG ∈Aut(G), the set Aut(G) is closed under composition, and every element in Aut(G) has an inverse

in Aut(G) we have shown that Aut(G) is a subgroup of the group permutations of G (i.e Aut(G) ≤ SG).

3.4 absolutely useless

380

Shaska T. 24

3.5 Let H /G and K /G.WTS: H∩K /G.First, want to show that H∩K ≤ G. Since H /G and K /G, eG ∈ H and eG ∈ K, so eG ∈ H∩K. So H∩K is

nonempty. Since a ∈H∩K, a ∈H and a ∈ K. Since H is a group, a−1∈H. Similarly, a−1

∈ K. So, a−1∈H∩K.

Then, a∗b−1∈H since a,b−1

∈H. Similarly, a,b−1∈K. So, a∗b−1

∈H∩K. By the first subgroup test, H∩K ≤G.Second, want to show that H∩K /G. By definition, since H /G, ∀n ∈ H∩K∀g ∈ G, g ∗n ∗ g−1

∈ H. Thisholds since n ∈H∩K⇒ n ∈H. Similarly for K. So, ∀n ∈H∩K∀g ∈ G, g ∗n ∗ g−1

∈H∩K. Sp, H∩K /G.

3.6 First prove HK is a subgroup of G.

• Subset: Since G is a closed group and H and K are subgroups of G, so HK ⊂ G

• Identity: e is the identity of G. Since H and K are both subgroups of G, so e ∈H and e ∈ K. ee = HK, soe = HK.

• Closed Under Multiplication(h1k1)(h2k2)

∃h1(h3k1)k2 because h2,h3 ∈H, g ∈ G such thatgh2 = h3g.in this case k1,k2 ∈ K are present instead ofg

(h1h3)(k1k2) because H and K are groups.

(h1h3)(k1k2) = HK since h1h3 ∈Hsince H is a closed group and k1k2 ∈ K since K is a closed group.

• Inverse: Let hk ∈HK h−1∈H and k−1

∈ K because both H and K are subgroups.

(hk)−1 = k−1h−1

k−1h−1 = h−1k−1 since both Hand K are both normal inG

Let H /G such that H = gHg−1, and K /G such that K = gKg−1 both ∀g ∈ G. Show that gHKg−1 = HK

HK = HK

(gHg−1)(gKg−1) = HK

gH(g−1g)Kg−1 = HKSince H and K are groups they are associative

gHKg−1 = HK Because gg−1 = e

Therefore, HK /G

3.7 Let H be a subgroup of G, N a normal subgroup of G, and H∩N = M. It is enough to show that hMh−1

is a subset of M, ∀ h ∈H because of Theorem 3.4.Fix some h ∈H, with inverse h−1

∈H, and let m1 ∈M. Then m1 ∈H and m1 ∈N by our construction of M.Since h,h−1,m1 ∈H, H is a group, and H is closed under the operation, we know hm1h−1

∈H.Since m1 ∈N, hm1h−1

∈ hNh−1. Since N is normal in G and h,h−1∈H⇒ h,h−1

∈ G,

(hN)h−1 = (Nh)h−1 = N(hh−1) = N(e) = N

and so hm1h−1∈N.

Therefore hm1h−1∈H and hm1h−1

∈N. Thus hm1h−1∈H∩N = M. Since h is an arbitrary element in H,

hMh−1 is a subset of M, ∀ h ∈H. Therefore M is a normal subgroup of H (i.e. H∩N is a normal subgroupof H).

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24 Shaska T.

3.8

3.9

3.10 Since |G| = 2p, by Lagrange’s theorem, the orders of all possible proper subgroups are divisors of 2p,namely 2 and p. By Cauchy’s theorem, ∃a ∈ G such that |a| = p. That is, | < a > | = p. And so, G has a

subgroup of order p. Now, since [G :< a >] =|G|| < a > |

=2p2

= p. To show that < a > is normal, we prove that

all subgroups of index 2 are normal.

Claim: let G be a finite group. And let H be a subgroup of index 2 in G. Then, H is normal in G.

Consider g ∈ G such that g ∈ H as well. Then, gH = H = Hg. If g ∈ G such that g < H, then gH = G/H,as the two cosets partition G. Likewise, if g ∈ G such that g < H, then Hg = G/H. Hence, gH = Hg. Hence,H = gHg−1. Thus, H is normal in G.

Now, returning back to our original problem, we see that < a > is normal in G. And so, any group oforder 2p where p is a prime, has a normal subgroup of order p.

3.11

3.12 Let A =

[x y0 2

]∈ G and B =

[1 b0 1

]∈H Since A ∈ G then xz , 0, therefore

ABA−1 =

[x y0 2

]∗

[1 b0 1

]∗

1xz

[z −y0 x

]=

[x xb + y0 z

]∗

1xz

[z −y0 x

]

ABA−1 =1xz

[xz x2b0 xz

]=

[1 xb

z0 1

]∈H

Since xz , 0 then H /G. Then corresponding matrices are

A =

[a1 a20 a3

],B =

[b1 b20 b3

]where A,B ∈ G a1,a2 , 0 and b1,b2 , 0. Then,

A−1 =1

a1a3

[a3 −a20 a1

],B−1 =

1b1b3

[b3 −b20 b1

]Then,

B−1A−1 =1

a1a3b1b3

[b3a3 −b3a2− b2a1

0 b1a1

]and then,

BA =

[b1a1 b1a2 + b2a3

0 b3a3

]Therefore,

B−1A−1BA =1

a1a3b1b3

[b3a3 −b3a2−b2a1

0 b1a1

]∗

[b1a1 b1a2 + b2a3

0 b3a3

]B−1A−1BA =

1a1b1a3b3

[b1a1 a3b3(b1a2− b2a1 + b2a3− b3a2)

0 b1a1b3a3

]382

Shaska T. 24

So then,

B−1A−1BA =

[1 H0 1

]∈H

where h =a2b1−a1b2+a3b2−a2b2

a1b1. Therefore G/H is Abelian.

3.13

3.14 Assume G/Z(G) is cyclic. Then G/Z(G) = 〈gZ(G)〉 be some generator for some g ∈ G. Let α,β ∈ G.

αZ(G) = (gZ(G))i = giZ(G), for some i ∈Z

=⇒ α = giz1, for some z1 ∈ Z(G)

and,

βZ(G) = (gZ(G)) j = g jZ(G), for some j ∈Z

=⇒ β = giz2, for some z2 ∈ Z(G)

We then observe αβ = (giz1)(g jz2). Since the center commutes with all elements we than can proceed assuch,

αβ = giz1g jz2

= gig jz1z2

= gi+ jz1z2

= g j+iz2z1

= g jgiz2z1

= g jz2giz1

= (g jz2)(giz1)= βα.

Therefore, since αβ = βα, then G is Abelian.

3.15 Let G = GL2(R) and H = SL2(R). We can see that H is entirely contained within G, and since it is agroup, so we want to show that H is a group. From linear algebra, we know that for two matrices A,B,det(AB) = det(A)det(B), and if A is invertible (as it is in the case of SL2(R)) then det(A−1) = det(A)−1. Soconsider h,k ∈H:

det(hk) = det(h)det(k) = 1 ∗1 = 1

det(h−1) = det(h)−1 = 1−1 = 1

So then hk ∈ H and h−1∈ H, and we have shown closure and inverseness. Associativity and identity

naturally extend from G, so then H ≤ G.It remains to be shown that H C G. That is, we want to show that for each g ∈ G, gHg−1

∈H. So, take anelement g ∈ G and let h be any element in H. Then det(h) = 1 and det(g) = n for some n ∈R. Now consider

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24 Shaska T.

det(ghg−1) = det(g)det(h)det(g−1)

= det(g)det(h)det(g)−1)

= n ∗1 ∗n−1

= n ∗n−1

= 1

So then ghg−1 describes a matrix with determinant 1, that is, a matrix in H. But this applies to allmatrices in H, so then H C G, which is the same as SL2(R) C GL2(R).

3.16 Let h ∈ H and k ∈ K. Since H and K are normal subgroups, kh−1k−1∈ H and hkh−1

∈ K. Furthermore,it can be said that h(kh−1k−1) ∈ H and (hkh−1)k−1

∈ K since H and K are closed. It follows that h(kh−1k−1) =(hkh−1)k−1 = hkh−1k−1 by associativity. Therefore hkh−1k−1

∈H and hkh−1k−1∈ K. Since H∩K = e, it is clear

that hkh−1k−1 = e

hkh−1k−1 = e

hkh−1k−1k = ek

hkh−1 = k

hkh−1h = khhk = kh

Thus, it has been proved that hk = kh for every h ∈H and k ∈ K

3.17 Let n ≥ 1 and let σ = (n + 1n + 2). We define the map:

φ : Sn 7−→ An+2

by φ(τ) = τ if τ is even and φ(τ) = τσi fτ is odd.i) φ is a homomorphism: we have the following cases:

• If τ1 even, τ2 even, then φ(τ1τ2) = τ1τ2 = φ(τ1)φ(τ2)

• If τ1 odd, τ2 odd then φ(τ1τ2) = τ1στ2σ = φ(τ1)φ(τ2).

• If τ1 odd, τ2 even, then φ(τ1τ2) = τ1στ2 = φ(τ1)φ(τ2)

• If τ1 even, τ2 odd, then φ(τ1τ2) = τ1τ2σ = φ(τ1)φ(τ2)Thus φ is a homomorphism.ii) Injective: Suppose τ ∈ ker(φ) then we want to show that φ(τ1kerφ) = φ(τ2kerφ) implies that τ1kerφ=τ2kerφ.Since the identity e is even and φ(τ) = e, we have τ =1. So the kernel of φ is trivial and that φ isinjective. Thus it defines a isomorphism with its image, a subgroup of An+2.

3.18 Let G be a group with order 35. Since |G| = 3 ∗5. By example discussed in notes, If |G| = p ∗ q, where pand q are primes where q > p, p 6 |q−1 =⇒ G is cyclic. So, since 3 and 5 are primes and 3 6 |5−1 = 4, a groupwith order 35 is cyclic.

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3.19 Let G be a group with order pnm where p is prime and p,m are relatively prime. Let P be a subgroup ofG with order pn. Then, by Sylow’s Theorems we get that the number of groups with order pn is congruentto 1 modulo p and divides m. Thus, the possibilities are 1 and m.

If m < p, then m is not congruent to 1 modulo p and thus the number of subgroups of order pn is oneand it must be normal in G and we are done.

Assume m < p. If np = m, then [G : NG(P)] = m. From Lagrange’s Theorem (Theorem 2.12), we know thatsince P is a subgroup of G,

[G : P] =|G||P|

=pnmpn = m

Thus,|NG(P)| = |P|

and P is normal in G follows. Similarly to above, P is also normal in G if np = 1.Therefore, P is a normal subgroup of G.

3.20

3.21 Since we know that G has an element of order 21, by Cauchy, you also know that there’s at least onesubgroup of order 21, let’s call it H.

Suppose that K ≤ G is another subgroup of order 21, then we can consider the subset HK that has order

|HK| = |H||K|/|H∩K||HK| = |H||K|/|H∩K|

.If H and K were distinct then H∩K should be a proper subgroup of both of them, but since they have

order the prime 21 this is possible iff H∩K = (id) and so |HK| = 21 ·21 = 441 which is clearly bigger then 42.We arrived to an absurd we have to conclude that H is the only subgroup of order 21 and so it’s

characteristic, hence normal.

3.22 Let G = eG, g1, g2, g3, g4, g5, g6, g7, g8, g9, g10, g11, g12, g13, g14, g15, g16, g17, g18, g19, g20 and H = eH,h1,h2,h3,h4,h5,h6,h7,h8,h9,h10,h11,h12,h13,h14,h15,h16,h17,h18,h19,h20

Each element of G can be mapped to each element of H and every element of H can be mapped to everyelement of G.

φ : G 7→H

eG, g1, . . . , g20 7→ eH,h1, . . . ,h20

To prove that groups are isomorphic they must be: well-defined, a bijection between the groups, and ahomorphism. The construction of φ is well-defined, and there is a bijection between G and H.

Homomorphism: Let φ(gi) = hi∀i = 1,2, . . . ,20, and where φ(eG) = eH

φ(g1g2) = h1h2

h1+2 = φ(g1+2)

(h1)(h2) = φ(g1)φ(g2)

3.23 We begin by showing that any group of order 9 is abelian. Let H be a group with |H| = 9. If ∃b ∈ Hsuch that |b| = 9, then H is cylic, hence abelian. So suppose that all elements of H, excluding the identity,have order 3. Consider x ∈ H such that x , e. Denote < x >= K. Also, Consider z ∈ H such that z < K. So,elements in H are of the form xazb where a,b ∈ 0,1,2. To show that H is abelian, we demonstrate thatxz = zx. Examining zx, we see that it must be of the form xazb for some a,b ∈ 0,1,2. That is zx = xazb. Bycancellation, we see that both a,b , 0. Now, if zx = x2z =⇒ zxz−1 = x2. So, z3xz−3 = x8 = x2 , x, which isimpossible as y3 = e. Similarly, zx , xz2. Now, if zx = x2z2 =⇒ zx = x−1z−1 = (zx)−1 =⇒ the order of zx is 1or 2, which is impossible. Thus it must be the case that zx = xz, and that H is abliean.

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24 Shaska T.

Now, back to our original problem. According to Cauchy’s theorem, since |G| = 99 = 9 ∗ 11, and 11 isprime, ∃b ∈ G such that |b| = | < b > | = 11. Denote < b >= N. Additionally, [G : N] = 9, by the fact that 11 - 9.Hence, N is normal in G. Next, we will demonstrate that x ∈ N =⇒ x ∈ Z(G). That is, N is a subgroupof the center of G. Now, by the normality of N in G, we have that gng−1 = nm =⇒ g11ng−11 = (nm)11, forsome n ∈ N and some m ∈Z+. Now, Fermat’s little theorem states that given prime p and some integer awhere p - a, then ap−1

≡ 1 mod p. Hence, m11≡m mod 11 =⇒ (nm)11 = nm. Thus, (nm)11 = g11ng−11 = gng−1,

hence g10n = ng10. Now, since |G| = 99 and 10 - 99, we see that ng = gn. Hence, n commutes with G, and

so n ∈ Z(G). Thus, N ⊂ Z(G). Now, [G : N] = |G|/|N| =9911

= 9. So, N is indeed abelian. Now, if p,q ∈ G, we

have that pq = rqp for some r ∈ N. Hence, r ∈ Z(G), hence r11 = e as all elements of N have order 11. So,rqp2 = (rp)(qp) = (pr)(qp) = p(pq) = p2q, by the fact that r ∈ Z(G). So we see that pmq = rqpm. Additionally, ifm = 11, then p11q = qp11. Hence, if |p| = 3, we have that p11 = p2 = p−1, hence qp−1 = p−1q. Hence, p ∈ Z(G).And so, |Z(G)| = 33 since there are two elements in Z(G), one with order 11 and the other with order 3=⇒ Z(G) contains an element with order lcm(11,3) = 33. Hence, [G : Z(G)] = 1 or 3. If [G : Z(G)] = 1, thenG = Z(G), and so G is abelian. On the other hand, if [g : Z(G)] = 3, since 3 is prime, G/Z(G) is cyclic, andhence abelian. In either case, we have that our group G, is, in fact, Abelian.

3.24 If G is a non-abelian group, then |G| = pq. Also ∃ H a normal subgroup of G with |H| = p (by SylowTheorem). Let K be the other subgroup of order q. Then, HK ≤ G since H is normal in G and HK=G since|KH| = |H| · |K|/|H∩K| = pq = Gso since HCG, K ≤ G, HK=G, H∩K = 1, this implies thatG is ismorphic to HxK thus there exists a unique non-abelian group of order pq.

3.25 Let G,H be a group with |G| = pq and |H| = pq by Cauchy’s corollary, let a ∈ G then < a > /G. and Letb ∈H then < b > /H. Now G = e,a,a2

and H = e′,b,b2. Take the map

f : G→H

where f (e) = e′, f (a) = b, and f (a2) = b2. Since any two cyclic groups of same order are isomorphic, and G iscyclic and H is cyclic, then G H.

3.7

3.27

3.28

3.29 Every element of C is a p-cycle other than the identity and any two conjugates of C intersect triviallyat the identity or coincide since |C| = p, a prime number. Since p-cycles are conjugate, every p-cycle iscontained in a conjugate of C. Every permutation α, that normalizes C, α : a 7→ (1, ...,p), a total of p choicesand α : b 7→ (1, ...,p), a total of (p−1) choices because α cannot map different elements to the same outcome.Therefore |NSp (C)| = p(p−1).

3.30 Let be given H, the set of elements with order odd and α ∈ G is an element with order 2n. FromCayley’s Theorem Theorem 3.14 we have

G → Sr

g→ Lg(24.5)

such that Lg(x) = gx and r = 2n m. Then, Lα is an element with order 2n in Sr and therefore product of 2n

-cycles. Moreover, Lα does not fix any point, otherwise Lα(x) = gx = x would imply that α = 1. Thus, Lα isa product of m cycles such that. Thus, Lα is an odd permutation.

Then, in the group G half of elements are even permutations. Let’s denote with An−1 the set of suchelements. Then, |An−1| = 2n−1m. Moreover, H ≤ An−1 and L2

α has order 2n−1. If HAn−1 then the proof is

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Shaska T. 24

complete, otherwise use the same argument for An−1 then take An−2 and so on. Finally, HAn−i for somei ≤ n and [G : H] = 2n−i. The proof is completed by induction.

3.30

3.31

3.32 The quarterion group Q8 is defined as such

Q8 = ±1,±I,±J,±K,

such that

I =

[1 00 1

]I =

[0 1−1 0

]J =

[0 ii 0

]K =

[i 00 −i

]The first subgroup is the trivial one: S0 = 1Then we have the center: S1 = ±1with a generator of 〈1〉Next, using the Third Subgroup Test, we find all the subsets that are closed under the operation of Q8. Weget:

S2 = ±1,±I, with a generator of 〈I〉S3 = ±1,±J, with a generator of 〈J〉S4 = ±1,±K, with a generator of 〈K〉S5 = ±1,±I,±J,±K = Q8

An example to find S2 = ±1,±I, we multiply all the elements with one another and see that

1J = J1 = J 1(−J) = −J1 = −J−1J = J(−1) = −J −1(−J) = −J(−1) = J1 ·1 = 1 −1 ·1 = 1 · (−1) = −1 −1 · (−1) = 1J · J = −1 −J · J = J · (−J) = 1 −J · (−J) = 1

Since all the products are still in the set, S2 ≤Q8.Since the subgroups are cyclical, then they are all normal. The factor groups would then be as follows:

Q8/S0 = 1S0,−1S0, IS0,−IS0, JS0,−JS0,KS0,−KS0

Q8/S1 = 1S1, IS1, JS1,KS1

Q8/S2 = 1S2, IS2

Q8/S3 = 1S3, JS3

Q8/S4 = 1S4,KS4

Q8/S5 = S5,

where they are each the sets of cosets of each normal subgroup of Q8

3.33

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24 Shaska T.

3.34 For (a), consider two elements a,b ∈U. Then

a ∗b

=

[1 x0 1

][1 y0 1

]=

[1 + 0 y + x0 + 0 0 + 1

]which is clearly of the same form, since x + y ∈ Q as Q is a group with addition, so we have closure in U.Now we need to show the existence of an inverse. So again consider a and b from above. If y = −x then

a ∗b =

[1 00 1

]which is the identity, so b is the inverse of a. Since these are arbitrary elements, then we can construct aninverse for any given u ∈U, so we have established inverseness. Since associativity is inherited from T andthe identity matrix exists in U, then U ≤ T.

For (b), to prove U is Abelian, we must show for any two elements a,b ∈ U, ab = ba. So consider thesame elements a and b. We have ab, so consider ba:[

1 y0 1

][1 x0 1

]=

[1 x + y0 1

]Since y + x = x + y, then ab = ba, so U is Abelian.For (c), let t ∈ T, we want to show that tUt−1. First observe that for

t =

[a b0 c

]det(t) = ac and ac , 0. Then for the inverse we get

t−1 = (1/ac)[c −b0 a

]=

[1a −

bac

0 1c

]So then with this, we take a ∈U to be a general element in U, then consider tat−1 =[

a b0 c

][1 x0 1

][1a −

bac

0 1c

]=

[a ax + b0 c

][1a −

bac

0 1c

]=

[1 ax+b

c −bc

0 1

]=

[1 ax

c0 1

]

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Shaska T. 24

And axc describes all the rational numbers when x is variable. So gUg−1 = U. Therefore, U C T.

For (d), we know that T/U forms a group with the operation by Theorem 3.5, but we need to show thatthis is an Abelian group. We know that an element A ∈ T/U is of the form tU for some t ∈ T. To show this isAbelian, we want to show that ∀t1U, t2U ∈ T/U, that t1Ut2U = t2Ut1U. Since U is normal, though, we knowthat

t1U = Ut1

t2U = Ut2

So then we get

t1Ut2U = t1(t2U)U= t1t2U

t2Ut1U = t2(t1U)U= t2t1U

So from this we get that T/U is abelian if and only if t1t2U = t2t1U, for all t1, t2 ∈ T. So consider t1t2U

[a1 b10 c1

][a2 b20 c2

][1 x0 1

]=

[a1a2 a1b2 + b1c2

0 c1c2

][1 x0 1

]=

[a1a2 a1a2x + a1b2 + b1c2

0 c1c2

]and now look at t2t1U

[a2 b20 c2

][a1 b10 c1

][1 y0 1

]=

[a2a1 a2b1 + b2c1

0 c1c2

][1 y0 1

]=

[a2a1 a2a1y + a2b1 + b2c1

0 c1c2

]So then t1t2U = t2t1U if and only if[

a1a2 a1a2x + a1b2 + b1c20 c1c2

]=

[a2a1 a2a1y + a2b1 + b2c1

0 c1c2

]for some x, y ∈Q. And we can easily construct a linear system for finding y in relation to x, namely, let

a1a2x + a1b2 + b1c2 = a2a1y + a2b1 + b2c1

a1a2x + a1b2 + b1c2− a2b1 + b2c1 = a2a1y

(a1a2x + a1b2 + b1c2− a2b1 + b2c1

a2a1) = y

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24 Shaska T.

So we know we can find a y for any x since a2,a1 , 0 by our definition of the group. Therefore, T/U isabelian.

For (e), consider an element in GL(Q), let it be g. Then we want to show that gT = Tg, or that gTg−1 = T,so consider this equation.

[e fg h

][a b0 c

](

1eh− f g

)[

h − f−g e

]=

[ea eb + f cga gb + bc

](

1eh− f g

)[

h − f−g e

]= (

1eh− f g

)[eah− g(eb + f c) −ea f + e(eb + f c)gah− g(gb + bc) −ga f + e(gb + bc)

]which, in general, is not equal to T, so T is not abelian in GL(Q).

3.35

3.36 Let αH,βH ∈ G/H. Therefore (αH)(βH) = (αβ)H via multiplication in factor groups. Furthermoresince G is abelian, we are able to write (αβ)H = (βα)H. From the previous statement, we can say that(βα)H = (βH)(αH). Hence, (αH)(βH) = (βH)(αH) and so G/H is therefore abelian as well.

3.37

3.38 We define φ(gH1) = φ(g)H2 for all g ∈ G1. We show that this is well defined. If g′H1 = gH1 theng′g−1

∈H1, so φ(g′g−1) ∈ φ(H1) ⊆H2. Thus φ(g′)φ(g)−1∈H2, so φ(g′H1) = φ(g′)H2 = φ(gH1).

It is also a homomorphism since

φ(g1H1)(g′H1) = φ(gg′H1)= φ(gg′)H2

= φ(g)φ(g′)H2

= φ(g)φ(g′)H2

= (φ(g)H2)(φ(g′)H2)

= φ(gH1)φ(g′H1)

(24.6)

3.39 Claim: every automorphism of A4 is not an inner automorphism. Elements of A4 are:e of order 1;(12) (34), (13) (24), (14) (23) of order 2; (?)(123) (132), (124) (142), (134) (143), (234) (243) of order 3.

Moreover, we know that (123) is not conjugate to (132). Because if (123) were conjugate to (132) = (123)−1

then we would have (123) ∈ (?), but clearly we do not.

3.40

3.41 First note that S3 is generated by 〈s〉 and 〈t〉, where s has order 2 and 〈t〉 has order 3, so s is a reflectionand t is a rotation. We can also note that (st)2 = e and (st2)2 = e where e is the identity. Furthermore, t3 = eand (t2)3 = e. Also note that ts = st2 and t2s = st, so we have covered all elements in the set. Then we havethree elements of order 2: s,st,st2, and two elements of order 3: t, t2.

Since an automorphism itself needs to retain the order of the elements (in order to be a homomorphism)then elements of order 2 must map to elements of order 2 and likewise for elements of order 3. So then wehave the following automorphisms:

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Shaska T. 24

s 7→ s t 7→ t (24.7)s 7→ st t 7→ t (24.8)

s 7→ st2 t 7→ t (24.9)

s 7→ s t 7→ t2 (24.10)

s 7→ st t 7→ t2 (24.11)

s 7→ st2 t 7→ t2 (24.12)

And e 7→ e for all automorphisms. The maps from st,st2, and t2 are implicitly defined here by the factthat these are homomorphisms, so a mapping f (xy) = f (x) f (y). This then describes Aut(S3).

Now we can look at Inn(S3), which are all the automorphisms brought on by conjugacies in the group.Note that any element can be described as a conjugate of others. That is, knowing that each element oforder 2 is its own inverse, and t and t2 are inverses:

s = s(s)s = st(st2)st = st2(st)st2 = t(st2)t2 = t2(st)t

st = s(st2)s = st(st)st = st2(s)st2 = t(s)t2 = t2(st2)t

st2 = s(st)s = st(s)st = st2(st2)st2 = t(st)t2 = t2(s)t

t = s(t2)s = st(t2)st = st2(t2)st2 = t(t)t2 = t2(t)t

t2 = s(t)s = st(t)st = st2(t)st2 = t(t2)t2 = t2(t2)t

which shows there is a conjugacy for each possible mapping that we have shown in the automorphisms.Therefore, Inn(S3) = Aut(S3).

3.42 From earlier . Let D4 = e,σ,σ2,σ3,τ,τ2,ρ,ρ2. To find the Aut(D4) and Inn(D4) we can look at the

Caley’s Table again.

Symmetries for D4

∗ e σ σ2 σ3 τ τ2 ρ ρ2

e e σ σ2 σ3 τ τ2 ρ ρ2

σ σ σ2 σ3 e ρ2 ρ τ τ2

σ2 σ2 σ3 e σ τ2 τ ρ2 ρσ3 σ3 e σ σ2 ρ ρ2 τ2 ττ τ ρ τ2 ρ2 e σ2 σ σ3

τ2 τ2 ρ2 τ ρ σ2 e σ3 σρ ρ τ2 ρ2 τ σ3 σ e σ2

ρ2 ρ2 τ ρ τ2 σ σ3 σ2 e

It is easy to see that the Aut(D4) = e

3.43

3.44 Let a1,a2 ∈ Zn then I need to showφ(a1a2) = φ(a1)φ(a2)

since a→ ka thenφ(a1a2) = k(a1a2)

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24 Shaska T.

φ(a1a2) = ka1ka2

φ(a1a2) = φ(a1)φ(a2)

3.45 Let G = a,a2, ...,an−1where a = 11/n Define a map

φ : G 7→Zn

φ(ak) = k

Let x, y ∈ G then φ(x · y)

= φ(ak1ak2 )

= φ(a(k1 + k2)= k1 + k2

= φ(ak1) +φ(ak2 )

= φ(x) +φ(y)

Thus φ is a homomorphism.Let φ(n) = φ(y)→ k1 = k2→ ak1 = ak2 → n = yThus φ is one-to-one’For each t ∈Zn∃at

∈ G such that φat = t.Thus φ is onto.Therefore φ is an isomorphism and G is isomorphic to Zn.

3.46

4.1

5.1

5.2

5.3 We know that the center of a p-group is nontrivial. Since G is non Abelian then Z(G) , G. Hence theorder of Z(G) is p or p2.

If |Z(G)| = p2, then |G/Z(G)| = p. Hence, G/Z(G) is cyclic and so G is Abelian. Thus, |Z(G)| = p.Then, |G/Z(G)| = p2 and G/Z(G) is Abelian. From the property of the commutator in Eq. (3.1) (G/N is

abelian, then G′ ≤N) we have that G′ ≤ Z(G).So either G′ = e or G′ = Z(G). But if G′ = e, then G/G′ = G and G is abelian. Thus, G′ = Z(G)

5.4

5.5

5.6

5.7

5.8

5.9

5.10

5.11

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Shaska T. 24

5.12

5.13 We know that GLn(Fp) = (pn−1)(pn

−p)(pn−p2)∗ ...∗(pn

−pn−1) =

n−1∏i=0

(pn−pi) =

n−1∏i=0

pi(pn−i−1) =

n−1∏i=0

pin−1∏i=0

pn−1−1.

Manipulatingn−1∏i=0

pi, we have thatn−1∏i=0

pi = p

n−1∑i=0

i= p

n(n−1)2 which is the formula for the sum of integers from 0

to n−1. Hence, a sylow p-subgroup of GLn(Fp) has pn(n−1)

2 elements in it. Now, by the form of an upper uni-triangular matrix, we know that |UTn(Fp)| = pk where k represents the number of elements above the main

diagonal. So, k =n∑

i=1(n− i) =

n−1∑i=0

i =n(n−1)

2. Thus, by the sylow theorems, UTn(Fp) is a sylow p-subgroup

of GLn(Fp) with ordern(n−1)

2.

5.14 Let G = GLn(Fp), where p is prime and n is an integer greater than 1. Fp is the set of all integers mod p.GLn is the set of all n×n matrices with a determinant not equal to zero. p-Sylow groups don’t overlap sosince there are at least two subgroups in G, H1

⋂H2 = e.

5.15 We have that HCp ×Cp which implies that |H| = p2 and p2| |G|. Thus |G| = pα ·m, where α ≥ 2 and

(p,m) = 1. We want to show that α > 2.Suppose that α = 2. Then, |G| = p2

·m. Then |SylpG| = p2 and so H is a Sylow p− subgroup and K is aSylow p−subgroup. Thus, H is conjugate to K, say H = x−1Kx, for some x ∈ G. But Kx−1Kx implies thatHK. Hence, Cp×CpCp2 which is a contradiction.

5.16 We want to show first that G has a prime power order. Suppose not, then

|G| = pα11 · · ·p

ann

From the Sylow’s theorem there exists a Pi such that |Pi|= pαii . Each of Pi is contained in the unique maximal

group M. Hence, pαii | |M| for all i = 1, . . .n. Thus |G| | |M|which is a contradiction because a maximal subgroup

is proper by definition.Next we want to show that G is cyclic. Suppose not, say a1, . . .an is a generating set for G. Thus,

G = 〈a1, . . . ,an〉

ConsiderH := 〈a1, . . . ,an−1〉

Then, H ≤M1 for some maximal subgroup M1. For the same reason 〈a2, . . . ,an〉 ≤M2 for some maximalsubgroup M2. Since G has a unique maximal subgroup then M1 = M2, which is a contradiction since an <M1but an ∈M2.

5.17 Assume p,q are primes such that p , q. Let G be a group where |G| = p2q. We will take two cases:

case 1: p > q

We know from Sylow’s Theorems (Theorem 5.2) that np ≡ 1 mod p, np = [G : NG(P)], and np | q. Thusnp = 1 or q. However, q < p so q . 1 mod p. Therefore np = 1 which is equivalent to saying P, a Sylowp−subgroup, is normal in G. Thus, G has a normal subgroup and is not simple.

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case 2: p < q

We know from Sylow’s Theorems (Theorem 5.2) that nq ≡ 1 mod q, nq = [G : NG(Q)], and nq | p2. Thus,nq = 1,p, or p2. However, p < q so p . 1 mod q which leaves 1 and p2. If nq = p2, then there would bep2 Sylow q−subgroups having order q. These Sylow q−subgroups all intersect at the identity whichgives us p2(q−1) distinct elements in the Sylow q−subgroups and thus in G. This leaves p2 elementsin G, including the identity.

We know from Sylow’s Theorems (Theorem 5.2) that np ≡ 1 mod p, np = [G : NG(P)], and np | q. Thus,np = 1 or q. We know that Sylow p−subgroups have order p2 and from above there are only p2 elementsleft in G for Sylow p−subgroups. Therefore, np = 1 which means there is only one Sylow p−subgroupand it is normal in G. Therefore G has a normal subgroup and is not simple.

In either case, we get that G has a normal subgroup and thus cannot be a simple group.

5.18 doesn’t compile

5.19

5.20

5.21G = S4,H :=

σ ∈ S4|σ(4) = 4

≤ G

P = 〈(124)〉 =(1, (124), (142))

Apparently H∩P < Sylp(H).

5.22 Proof by induction. Base case, let G have order pn for some prime p and integer n, then G equals itssylow p-subgroup and it is trivial that it is the product of its sylow subgroups. Induction case, Supposethat |G| = pn1

1 ...pnkk of p j, j ≤ k distinct primes...

5.23 From Exercise 1.25 we found that

|GLn(Fp)| =n∏

i=1

(pn−pi−1)

And from that we can get

|GLn(Fp)| =n∏

i=1

(pi−1pn−i−pi−1)

=

n∏i=1

pi−1(pn−i−1)

So let G = GLn(Fp). Then we know for certain that |G| = pαm for some α,m ∈Z such that (p,m) = 1. Thenwe know that G has at least one Sylow p-subgroup. Consider the set of upper triangular matrices U suchthat an element u ∈U looks like

[1 x0 1

]Now, it is clear that there are p elements in U as there are p unique choices for our one variable element

x. Closure on this subset with our operand is shown identically to how we found it for question 3.33

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earlier in this paper, as we know there is closure for addition in a finite group Fp, so x + y ∈ Fp if x, y ∈ Fp.Additionally, for any element u ∈U, we can find the inverse u−1 simply by finding the additive inverse ofx in Fp, and let that be our y in u−1. So then U ≤ G such that |U| = p, so then U is a Sylow p-subgroup.

It would be sufficient to find one more Sylow p-subgroup of G such that its only common element withU is the identity matrix. We in fact have a good candidate for this, that is L, the set of lower triangularmatrices, so an element l ∈ L looks like

[1 0x 1

]It is easy to see again that there are p such elements, and for the sake of clarity, note that two elements

l1, l2 ∈ L multiply so to give

l1l2 =

[1 0x 1

][1 0y 1

]=

[1 0

x + y 1

]So L has closure for the same reason as U did, as well as inverseness. So L ≤ G. Furthermore, it is a

Sylow p-subgroup as it has p elements. Moreover it’s clear that U∩L = e, since their only common elementis the identity matrix. Thus, we have shown two Sylow p-subgroups of G such that their intersection istrivial.

5.24 Let G be a group where |G| = pqr. p,q,r are primes, and p < q < r. Since r - q and r - p, then r - pq. UsingSylow’s Theorem, we can observe that

nr ≡ 1 mod r

and

nr | pq.

Then nr = 1 or r. But since r - pq, then nr = 1. And from Corollary 5.2, since there is only one Sylow r-group,that group is a normal subgroup.

5.25

5.26

5.27 We know that |S4| = 4! = 24 = 23∗3. Since the exponent on 3 is 1, the order of the sylow 3-subgroups

must be 3. Hence, they are cyclic as 3 is prime, hence there exists a generator for each one. Next, we provethat all sylow 3-subgroup of S4 are generated by 3-cycles.

Claim: All sylow 3-subgroup of S4 are generated by 3-cycles.

Proof: Consider a sylow 3-subgroup of S4. Denote this subgroup as A = e, f , g. Well, f 2 = g, henceg2 = f =⇒ f 3 = f 2 f = g f . Hence, g f = e =⇒ A =< f >.

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24 Shaska T.

Now, back to our original problem. Observing the set of all 3-cycles of S4 we have that C3 =(123), (124), (132), (134), (142), (143), (234), (243). So there are 8 3-cycles in S4. The cyclic subgroups oforder 3 are < 123 >, < 124 >=< (34)(123)(34) >, < 134 >=< (24)(123)(24) >, and < 234 >=< (14)(123)(14) >.Now, we know that n3 ≡ 1 mod 3 and must divide 8, so n3 = 1 or n3 = 4. We have already found 4 suchsubgroups, so these 4 must account for all sylow 3-subgroups in S4. To see that they are all conjugate, sincefor any (i jk) and (pqr), at most one element can differ between them. WLOG, we have (i jk) and (i jl). So,(kl)(i jk)(kl) = (i jl), hence all 3-cycles are conjugate. This can also be seen by the fact that all permutations ofthe same cycle type belong to the same conjugacy class in Sn in general. Hence, all sylow 3-subgroups areconjugate in S4.

5.28

5.29 Let a group, G, have the order 45|G| = 45 = 32

·5

n3 = 1

n5 = 1

Since both n3 and n5 only have the solution of 1, they both are normal subgroups because primes only havefactors of 1 and themselves, and 1 is always a normal group in Sylow groups, by 5.6 in the book. n3 has theorder of 9, so 9 is a normal subgroup of every group with order 45.

5.30 Copied from another book.

5.31 Let G be a finite group such that |96|. Since 96 = 25·3 then by the Sylow’s Theorem we get

n2 = 1,3n3 = 1,4,16.

Assume that G is simple. Therefore n2 = 3 otherwise G would not be simple. But by the index theorem[G : NG(P2)] = 3 where G is a finite group and NG(P2) ≤ G. If 196 - 3! then G is not simple. This is acontradiction to our assumption so therefore there is no simple group with order 96.

5.32 Let G be a group and |G| = 160 = 25·5.

We know from Sylow’s Theorems (Theorem 5.2) that n2 ≡ 1 mod 2, n2 = [G : NG(P2)], and n2 | 5. Thus,n2 = 1 or 5. We also know from Sylow’s Theorems (Theorem 5.2) that n5 ≡ 1 mod 5, n5 = [G : NG(Q5)], andn5 | 25 = 16. Thus, n5 = 1 or 16.

If G is simple, then n2 = 5 and n5 = 16. Thus [G : NG(P2)] = 5 and [G : NG(Q5)] = 16 where P2 is aSylow 2-subgroup and Q5 is a Sylow 5-subgroup. The Index Theorem (Theorem 3.16) allows us to use thefollowing:

i) G is a finite groupii) P2 is a subgroup of G which means the normalizer, NG(P2), is also a subgroup of G such

that [G : NG(P2)] = 5iii) 160 - 5! = 120

to conclude that G is not a simple group.

5.33 Let H /G such that |H| = pk for a prime number p.WTS: H is contained in every Sylow p-subgroup of G.Since H /G, any conjugate of H is H. Also, for a given prime p, each Sylow p-subgroup is a conjugate of

each other. By Sylow’s theorem, any p-subgroup is contained in a Sylow p-subgroup. Let P1 be a maximalp-subgroup. So, |P1| = pn. Let P2 be a second Sylow p-subgroup of G. Let H be contained in P1, the maximalSylow p-subgroup of G. Then, P2 is a conjugate of P1. Since the conjugate of H is H, H is contained in P2.This shows that H is contained in every Sylow p-subgroup of G.

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5.34 i) Let G be the group and |G| = p2g2. By Sylow first theorem, G has Sylow p-subgroups and Sylowq-subgroups. Let np and nq be the numbers of Sylow p-subgroups and Sylow q-subgroups respectively. Soby Sylow third theorem,

np = kp + 1,k ≥ 0,np|q2 and nq = tq + 1, t ≥ 1,nq|p2. Thus gcd(p,q) = 1. We claim: np = 1,nq = 1If np , 1, then np|q2 or kp + 1|q2 implies q2

≡ 1 (mod p) which is not possible. Thus np must be 1.Similarly np must be 1.Therefore, G has a unique Sylow p-subgroup H and a unique Sylow q- subgroup K. Thus |H| = p2, |K| = q2

and H and K are both normal, so H Zp2 Zp×Zp and K Zq2 Zq×Zq

Now H∩K =e, hence |HK| =

|H||K||H∩K|

= p2q2. Thus H and K are two normal subgroups of G such that

G = HK and H∩K = e. Hence G H×K Zp2 ×Zq2 .G is an internal direct product of two cyclic groups H,K. It is Abelian.ii) Examples: p2,p2q,p2q2 where p,q are primes

5.35 Since |G|= 33 and 33 = 3×11, we consider sylow subgroups for p = 3 and p = 11. By the Sylow Theorem,the number of subgroups satisfies np 1 mod p and also n3|11 and n11|3 therefore n3 = n11 = 1. Therefore,each p-subgroup must be unique and a group with order 33 must have a unique Sylow 3-subgroup.

5.36 If a group G has order |G| = 108, then

|G| = 22∗33

Then with Sylow’s theorem, for the Sylow p-subgroups we have possibilities

n2 = 1,3,9,27n3 = 1,4

First consider the requirements for this group to have no normal subgroups. Then it is necessary thatn3 = 4. Additionally, a Sylow 3-subgroup has 27 elements in this group, so excluding the identity, we have26 ∗4 = 104 elements. Then there are only 4 remaining elements for the Sylow 2-subgroups, and each onehas 4 elements in this group, so n2 = 1, and |G| has a normal subgroup

Otherwise, n3 = 1, so |G| still has a normal subgroup. Therefore, a group of order 108 must have anormal subgroup.

5.37 Let G175 be an arbitrary group with |G175| = 175. If we break up the prime factors then we have|G175| = 52

·7. From Sylow’s Theorem, we have

P : n5 ≡ 1 mod 5, where n5 | 7

and

Q : n7 ≡ 1 mod 7, where n7 | 25

Since n5 = 1 and n7 = 1, then there are only one P and one Q Sylow normal subgroup. So, PCG175 andQCG175 and G175 is not simple.

From Corollary 5.1, we know that a group with an order of a prime squared (p2), is Abelian. ThereforeP is Abelian. Also, from Corollary 2.8, since Q has a prime order, then it is Abelian. Thus, G175 is Abelian.

From Lemma 6.6, then

G175 C175

C25×C7

C5×C5×C7

This classifies all the groups of order 175 up to isomorphism.

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5.38

5.39 |S5| = 120 = 23∗3 ∗5. Let K be a Sylow 2 -subgroup of S5 then, |K| = 23.

Let a = (1234),b = (12)(34) and let K = e,a,a2,a3,b,ba,ba2,ba3.

ab = (1234)(12)(34) = (13) = (12)(34)(1432) = ba3 so K is closed under multiplication andtherefore is a subgroup of S5 and |K| = 8.Now D4 = e,a,a2,a3,ab,ab2,ab3

. Let T be a map such that

T(bkat→Hkat

with 0 ≤ k ≤ 1 and 0 ≤ t ≤ 3. Then T is an one to one map and T(e) = e.

T(bkatbman) = T(bkbmatm+n)

T(bkatbman) = Hk+ma3tm+n

T(bkatbman) = Hk+man−tm

T(bkatbman) = Hk(Hma−tm)an

T(bkatbman) = HkatHman

So T is a homomorphism and one to one so therefore T is an isomorphism from S5 to D4

5.40 From Sylow’s theorem, the group G contains one or more Sylow 5 -subgroups. The number of these 5-subgroups is congruent to 1 (mod 5) and also must divide 20. Hence, n5 = 1. Thus, P5 C G.

5.41 Let G be a finite group with order pn, n > 1 and p a prime number. From Lemma 5.4, G has nontrivialcenter. Since the center is a normal subgroup, then G can not be a simple group.

Thus, groups with order 4, 8, 9, 16, 25, 27, 32, 49, 64 and 81 are not simple. In fact, groups with order 4,9, 25 and 49 are Abelian from Corollary 5.1.

5.42 Let’s compute the number of Sylow subgroups for p = 2,7.We have n2 = 1,7 and n7 = 1,8. Assume that n2 = 7 and n7 = 8. By counting the elements of the group

we have1 + 8 ·6 = 49 elements of order 7 and the identity

In the 2-subgroups we have at least 2 such subgroups, which have at most 4 elements in common (countingthe identity). Hence, they have at least 11 elements of order 2. Adding to theses 11 elements the 49 elementsthat we got above we get a total of 60 > 56. Hence, n2 = 1 or n7 = 1. Therefore, one of the Sylow subgroupsis normal.

5.43 To show that a group G with order 48 is not simple we need to prove that G contains a normal subgroupwith order 8 or a normal subgroup with order 16. From Sylow theorem, the group G has either one or threeSylow 2-subgroups with order 16. If has only a subgroup then it is normal.

Assume we have 3 Sylow 2-subgroups with order 16 and two of them are H and K. Assume that|H∩K| = 8. If |H∩K| ≤ 4, then from Lemma 2.10

|HK| =16 ·16

4= 64,

that is impossible. Hence H∩K is normal in of two subgroups H and K since has index 2. Normalizer ofH∩K contains as H and K and also |H∩K|must be a multiple of 16 bigger than se 1 and also divide 48. Theonly possibility is that |N(H∩K)| = 48. Thus, N(H∩K) = G.

5.44

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5.45

5.46 Note that 3159 = 35·13 Let G be a simple group of order 3159. From Sylow’s theorem we have n3 = 13

and n3 = 27. Note that n3(G) = 13 . 1mod 9. There exists P3,Q3 ∈ Syl3(G) such that |NG(P3∩Q3)| is divisibleby 35 and 13. Thus NG(P3∩Q3) = G, a contradiction. Thus no group of order 3159 is simple.

5.47 a)

G = 168 = 23·3 ·7

n2 = 1,3,7,21

n3 = 1,4,7,28

n7 = 1,8

n2, n3, and n7 are not equal to 1 because G is a simple group, so it contains no normal groups other thetrival subgroup. n7 = 8. Using the index theorem to check n2 = 3,7,21 and n3 = 4,7,28, n2 = 3 and n3 = 4 arealso not possible solutions.

168 6 |3!

168 6 |6

This is should only be true if G is not a simple subgroup, but since it is n2 , 3 Similarly,

168 6 |4!

168 6 |24

Since, G is a simple group n3 , 4. The amount of element of the Sylow group can be found by adding thetrivial subgroup with the amount of the order of n2 groups with the amount of the order of n3 groups andwith the amount of the order of n7.

1 + 7(21) + 2(7) + 6(8) = 210

7 is used instead of 1 for the n2 location because 2 is to the 3rd power.b) Sylow 3-subgroups and 7-subgroups are cyclic because they are p-groups where p is raised only to

the power of 1. The p is the generator of the cyclic group.

5.48 We have|G| = (23

−1)(23−2)(23

−22) = 168 = 23·3 ·7.

Let v,w ∈ V be non-zero and let v1 = v,v2,v3 and w1 = w,w2,w3 be two basis of V. Then the function F suchthat f (i) = wi, i = 1,2,3 extends uniquely to an invertible linear transformation of V in V, i.e. to an elementx of G. Since v(x) = w, G is transitive on the 7 non-zero vectors of V. Let N , 1 be a normal subgroupof G. N is transitive so 7||N|. The stabilizer Gv of a vector v , 0 has index 7, and therefore order 24. Theaction on the cosets of Gv yields a homomorphism of G in S7, whose kernal K is contained in Gv. If K , 1,and normal, its order is divisible by 7. But then 7||Gv| = 24, impossible. Hence, K = 1, and G imbeds inS7. A subgroup of order 7 cannot be normal in G. If it is, its image in S7 would have a normalizer of atleast 168, but a subgroup of order 7 in S7 is generated by a 7-cycle and therefore its normalizer has order7 ·ϕ(7) = 7 ·6 = 42. It follows that the number of 7-Sylows is 8, and since 7||N| and NEG, the eight 7-Syloware all contained in N. Thus, |N| is divisible by 8 and 7, and so by 56. It contains 8 · (7−1) = 48 7-elements,and in addition the eight elements of a 2-Sylow, and so at least 56 elements. If |N| −56, the 2-Sylow is uniquein N, therefore characteristic in N and so normal in G. However, a normal subgroup of G must have orderdivisible by 7. Hence, |N| > 56, |N| = 168 and N = G. Hence, it is shown that a transitive subgroup of S7 oforder 168 is simple.

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5.49 Let G = 2 ∗3 ∗7 ∗11 by Sylow’s theorem,

n2 = 1,3,7,11,21,33,77,231

n3 = 1,7,22,154

n7 = 1,22

n11 = 1

Since n11 = 1 there exists a normal subgroup of G and therefore G is not simple.

5.50 Let |G| = 132. Since 132 = 22·3 ·11 then by the Sylow’s Theorem we get

n2 = 1,3,11,33n3 = 1,4,22

n11 = 1,12.

Assume G is simple then it will not have any proper normal subgroups. Therefore, n2 = 3,n3 = 4 andn11 = 12. So 1+3(x)+4 ·2+12 ·10 = 3(x)+129 where 3(x) is at least 5 elements. Therefore, 3(x)+129≥ 134> 132.So, |G| = 132 is not simple.

5.51 Let G be a simple group such that |G| = 168 = 23·3 ·7.

We know from Sylow’s Theorems (Theorem 5.2) that n7 ≡ 1 mod 7 and n7 | 23·3 = 24. Thus n7 = 1 or 8.

If n7 = 1, then P7 is a normal subgroup in G and thus G is not simple. However we are assuming that G isa simple group and thus n7 = 8. We know that the Sylow 7−subgroups have order 7 and their intersectionis the identity so we will count their order as (7-1)n7 = 6(8) = 48.

We know from Sylow’s Theorems (Theorem 5.2) that n3 ≡ 1 mod 3, n3 = [G : NG(P3)], and n3 | 23·7 = 28.

Thus n3 = 1,4,7, or 28. If n3 = 1, then P3 is a normal subgroup in G and thus G is not simple but we areassuming G is simple. Therefore n3 = 4,7 or 28. If we assume n3 = 4, the Index Theorem (Theorem 3.16)allows us to use the following:

i) G is a finite group

ii) P3 is a subgroup of G which means the normalizer, NG(P3), is also a subgroup of G suchthat [G : NG(P3)] = 4

iii) 168 - 4! = 24

to conclude that G is not a simple group. However we are assuming G is a simple group and thusn3 = 7 or 28. We know that the Sylow 3−subgroups have order 3 and their intersection is the identity sowe will count their order as (3-1)n3 = 2n3. Therefore the number of elements in the Sylow 3−subgroups iseither 2(7) = 14 or 2(28) = 56.

We know from Sylow’s Theorems (Theorem 5.2) that n2 ≡ 1 mod 2, n2 = [G : NG(P2)], and n2 | 3 ·7 = 21.Thus n2 = 1,3,7, or 21. If n2 = 1, then P2 is a normal subgroup in G and thus G is not simple but we areassuming G is simple. Therefore n2 = 3,7 or 21. If we assume n2 = 3, the Index Theorem (Theorem 3.16)allows us to use the following:

i) G is a finite group

ii) P2 is a subgroup of G which means the normalizer, NG(P2), is also a subgroup of G suchthat [G : NG(P2)] = 3

iii) 168 - 3! = 6

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to conclude that G is not a simple group. However we are assuming G is a simple group and thusn2 = 7 or 21. We know that the Sylow 2−subgroups have order 23 = 8 and their intersection is the identityso we will count their order as (8-1)n2 = 7n2. Therefore the number of elements in the Sylow 2−subgroupsis either 7(7) = 49 or 7(21) = 147.

Thus the possibilities for the amount of elements in the Sylow p−subgroups is the sum of the identity,|n2|, |n3|, and |n7|. We know n7 = 8 and the identity is only one element. If n2 = 21 and n3 = 7, then the sum is

1 + 147 + 14 + 48 = 210 > 168

and so this is not possible.If n3 = 7 and n7 = 8 then these are p−groups and thus are cyclic by Corollary 2.8.

5.52

6.1 Let G be an inner direct product of subgroups H and K.WTS: the function φ : G→H×K such that φ(g) = (h,k) for g = hk, where h ∈H and k ∈ K, in injective and

surjective.Injectivity: Let h1,h2 ∈H and k1,k2 ∈ K such that h1k1 = h2k2. WTS: g1 = g2. If φ(g1) = φ(g2)⇐⇒ (h1,k1) =

(h2,k2)⇐⇒ h1k1 = h2k2⇐⇒ g1 = g2. so φ is injective.Surjectivity: Since H and K are subgroups of G, if (h,k) ∈ H ×K, then h,k ∈ G. So, hk→ (h,k). So

φissurjective.

6.2 The statement is not true. ConsiderG =Z2×Z2× · · ·

Obviously G is a group of infinite order.If a ∈ G, then a = (· · · ,xi, · · · ), where xi ∈ Z2. Then 2a = (· · · ,2xi, · · · ). But 2xi = 0 because xi ∈ Z2. Thus,

2a = 0⇒ |a| = 2, for all a ∈ G, and G has infinite order.

6.3 Let ni and n j be relatively prime. The order of Cni ×Cn j is nin j. Let Cni = 〈x〉 and Cn j = 〈y〉 then (x,e)and (e, y) are elements of Cni ×Cn j of orders ni and n j respectively. Since ni and n j are relatively prime(e, y)× (x,e) = (x, y) = (x,e)× (e, y). Therefore (x, y) has order nin j. So (x, y) is an element of Cni ×Cn j of ordernin j, which generates the whole group. So Cni ×Cn j is the cyclic group of order nin j and Cnin j is cyclic withorder nin j. Any two cyclic groups of the same order are isomorphic. Therefore∏k

i=1 Cni Cn1···nk

when gcd(ni,n j) = 1 for i , j.

6.4 Since the greatest common divisor of peii and p

e jj is 1 for i , j, the proof follows from the problem above.

6.5 Let (g1, g2) ∈ G1×G2, where g1 ∈ G1 and g2 ∈ G2. So,

(g1, g2)× (g2, g1) = (g1 · g2, g2 · g1)

(g2, g1)× (g1, g2) = (g2 · g1, g1 · g1)

Since multiplication is commutative, G1×G2 G2×G2

6.6 #1.) We prove that this is true by the fundamental theorem of homomorphisms. Define

φ : G×G→ G

where φ(g,h) = gh−1.

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24 Shaska T.

Homomorphism: φ((g,h)(u,v)) = f (gu,hv) = (gu)(hv)−1 = guv−1h−1. Now, since G is abelian, guv−1h−1 =gv−1uh−1 = gh−1uv−1 = φ(g,h) = φ(u,v). Hence, φ is a homomorphism.

Onto: For any g ∈G,we have that ∃(h, g−1h) ∈G×G such that φ(h, g−1h) = h(g−1h)−1 = hh−1g = g. Hence,φ is onto.

Now,

Ker(φ) = (g,h) ∈ G×G : f (g,h) = e = (g,h) ∈ G×G : gh−1 = e = (g,h) ∈ G×G : g = h = (g, g) : g ∈ G = D

Hence, by the fundamental theorem of homomorphisms, we have thatG×GKer(φ)

≈ Img(φ). However, Img(φ) =

G since φ is onto. Thus,G×G

D≈ G.

#2.) If G is abelian, then as is G×G. Likewise, every subgroup of an abelian group is normal. Now,suppose that D is a normal subgroup of G×G, and let x, g ∈ G. Then, (g,x)(g, g)(g−1,x−1) = (g,xgx−1) ∈ D.Now, by the structure of D, we have that g = xgx−1, hence gx = xg. And so, G is abelian.

6.7 Assume G is the innner direct product of H and K. Defineφ : H1xH2x...xHn 7→G such thatφ(h1,h2, ...,hn) =h1h2....hnFirst we verify this is a homomorphism. φ(h1)φ(h2).... = h1h2...hn thus it is a homomorphism. and everyelement commutes with one another.

6.8 Letf : H1x...xHn→ G

f (a1, ...,an)(b1, ...,bn) = a1...an

where this is a bijection. This is also a homomorphism because

f (a) f (b) = (a1...an) ∗ (b1...bn)

f (a) f (b) = a1b1 ∗ a2...an ∗b2...bn

f (a) f (b) = a1b1 ∗ a2b2 ∗ a3...anb3...bn

f (a) f (b) = a1b1 ∗ a2b2...anbn

f (a) f (b) = f (ab)

Since ab = (a1b1...,anbn) then f is an isomorphism of groups.

6.9 here

6.10 here

6.11 To construct such a group, it must have some element g such that g is not uniquely written as g = g1...gnfor gi ∈Ni.

Using Lemma 6.4, we can simplify this to finding a gi ∈Ni and g j ∈N j such that gig j , g jgi, so if we findsome non-Abelian group with the requisite properties, we are done.

6.12 here

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6.13 By Structure Theorem for Finitely Generated Abelian Groups, the finite group G is a direct productof cyclic groups G1,G2, · · · ,Gk. We have: G = G1 ×G2 × · · · ×Gk. We have: G = G1 ×

P× · · · ×

e

ande× · · ·×

e×

Gk

intersect trivially, so H must be trivial.

So we see that a non trivial H ≤G is contained in every subgroup of G, then necessarily k = 1, since otherwiseby the argument above we should have H trivial, a contradiction. So G = G1, which is a cyclic group.Next, the order of G: the order of G is the power of prime, since the argument above, we could have takenthe cyclic groups Gi to be prime order. But conversely, G =Z/pkZ has a unique subgroup of order p, whichis contain in all the non trivial subgroup of G. So nothing else can be said about the order of G.

6.14 Let |G| = Pα11 Pα2

2 · · ·Pαll where each of the Pi. By theorem 6.6 (“Every finite Abelian group G is the direct

product of p-groups”) we get G G(P1)×· · ·×G(Pe). Then by lemma 6.1 (“Let G be a finite Abelian p-groupand assume that g ∈ G has maximal order. Then the group G can be written as 〈g〉×H for sone subgroupH of G”) each of the G(Pi) can be decomposed further such that G(Pi) CP

n1i×CP

n2i× · · ·×C

Pntii

where Cx is

the a cyclic group of order x. Therefore, we have that G is isomorphic to a direct product of cyclic groupsof prime power order.

6.15 Let |G| = p2 and p be prime. By Cauchy’s Theorem, there exists an element of p2 or every element is oforder p. If the former is the case, then G Cp2 because the element z of order p2 generates the cyclic groupof order p2. If the case is the the latter and G is not cyclic, every element in G is of order p and G Cp×Cp.Thus, G Cp2 or G Cp×Cp.

6.16 An automorphism of Cp ×Cp is an isomorphism to itself. As these elements are 2-tuples, we canrepresent these isomorphisms as 2×2 matrices multiplying 2-tuples as vectors. This would then result ina vector, or 2-tuple. Furthermore, these matrices must be invertible because isomorphisms are bijective, sothey must all have determinants not equal to 0. Aside from this, we know that matrix mutliplication on atuple (a vector) is homomorphic. So then, this implies that the automorphism group is the entire generallinear group across the elements of Cp. In otherwords

Aut(Cp×Cp) = GL2(Fp)

6.17 Let G = H1×H2× · · ·×Hn. From Theorem 6.1, if G H1×H2× · · ·×Hn, then G = H1H2 . . .Hn. Thus,

Z(G) = Z(H1H2 . . .Hn)

. And since Hi is normal, then we can further notice that

Z(G) = Z(H1)Z(H2) . . .Z(Hn)

Z(G) Z(H1)×Z(H2)× · · ·×Z(Hn)

Therefore, the center of G is isomorphic to the direct product of the centers of the the normal subgroups.

6.18 here

6.19 Suppose G is generated by finitely many elements k. Since each element in G has order 2, the group< g1, ...gk > has order of at most 2k. This is a contradiction since G has infinite order, and such a set couldnot generate G. Therefore, G is not finitely generated.

6.20

6.21

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24 Shaska T.

6.22 Consider rational numbers Q with addition. Assume that Q is finitely generated by generatorsp1q1, . . . ,

pnqn

, where each one piqi

is a simplified ratio. Let p be a prime number of such that it does not divide any

of the denominators q1, . . . ,qn. Then, 1/p can not be in the subgroup Q, which is generated from p1q1, . . . ,

pnqn

,because p does not divide the denominators of any element in this group. This can be seen easily, since thesum of any two generators is

pi

qi+

p j

q j=

piq j + p jqi

qiq j.

6.23 By the Third Sylow Theorem, G has only one subgroup H1 of order 17. So G/H1 has order 35 andmust be abelian. Hence, the commutator subgroup of G is contained in H which tells us that |G′| is either1 or 17. If |G′| = 1, we are done. Suppose that |G′| = 17. The Third Sylow Theorem tells us that G has onlyone subgroup of order 5 and one subgroup of order 7. So there exist normal subgroups H2 and H3 in G,where |H2| = 5 and |H3| = 7. In either case the quotient group is abelian; hence, G′ must be a subgroup ofHi, i = 1,2. Therefore, the order of G′ is 1, 5, or 7. However, we already have determined that |G′| = 1or17.So the commutator subgroup of G is trivial, and consequently G is abelian.

6.24 Let G be Abelian and |G| = 40. Since 40 = 23·5 we get G ≈ P23 ×P5. We have |P2| = 8 = 23 and |P5| = 5.

Therefore

P2 3 −→3 ≈ C23

2 + 1 ≈ C22 ×C2

1 + 1 + 1 ≈ C2×C2×C2

So the Abelian groups up to order of 40 are

C23 ×C5

C22 ×C2×C5

C2×C2×C2×C5

6.25 Let G be a group with order 200 = 23·52.

From Theorem 6.2 we know that every finite Abelian group is isomorphic to the direct product of itsSylow subgroups,

G ≈ P2×P5

The Sylow 2-subgroup P2 has order 23. The partitions of the exponent 3 are

3,3 = 2 + 1, and 3 = 1 + 1 + 1.

Hence, P2 is isomorphic toC8,C4×C2, and C2×C2×C2

The Sylow 5-subgroup P5 has order 52. The partitions of the exponent 2 are

2 and 2 = 1 + 1.

Hence, P5 is isomorphic toC25 and C5×C5

Putting all cases together we have that G is isomorphic to:

• C8×C25

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Shaska T. 24

• C8×C5×C5

• C4×C2×C25

• C4×C2×C5×C5

• C2×C2×C2×C25

• C2×C2×C2×C5×C5

6.26 |G| = 720 = 24∗32∗5

Since 4 = 4,4 = 3+1,4 = 2+2,4 = 2+1+1,4 = 1+1+1+1, we get C16,C8×C2,C4×C2×C2,C2×C2×C2×C2,respectively. Since 3 = 3,3 = 2 + 1,3 = 1 + 1 + 1, we get C27,C9×C3,C3×C3×C3, respectively.

So,G C27×C16×C5

G C16×C9×C5×C3

G C16×C5×C3×C3×C3

G C27×C8×C5×C2

G C9×C8×C5×C3×C2

G C8×C5×C3×C3×C3×C2

G C27×C5×C4×C2×C2

G C9×C5×C4×C3×C2×C2

G C5×C4×C3×C3×C3×C2×C2

G C27×C5×C2×C2×C2×C2

G C9×C5×C3×C2×C2×C2×C2

G C5×C3×C3×C3×C2×C2×C2×C2

6.27 Let G be a group such that |G| > 1. We will prove it by induction. Assume the statement is true allgroups of order less than |G| = n = p ·m.

We want to show that if it is true for |G| = n. If |G| = p, then we are done (Every element x of G has theproperty that xp = e). Let |G|> p. There exists x ∈G,x, e. If p | |X|, then let |x|= p ·r. Hence, xpr = e⇒ (xr)p = e,which implies |xr

| = 1. Note that xr , e because then |x| = r , prSo suppose p - |x|. Let N = 〈x〉 , e. Hence, |N| , 1. Also N C G because G is abelian. By Lagrange’s

Theorem,∣∣∣G/N∣∣∣ = |G|/|N|. Since |N| , 1, then |G/N| < |G|. But since p - |x|, then p - |N|, and p | |G|>. Therefore,

p | |G/N|. So |G/N| < |G| = n and p | |G/N|. By assumption of induction hypothesis we have that G/N has anelement of order p.

Let’s call that element y = yN. Note that y < N (otherwise y = N implies |y = 1|). Also yp = N and soyp∈N.So we have y < N and yp

∈ N. Hence 〈yp〉 , 〈y〉. Therefore, |yp

| < |y|, and so p | |y| which brings thesituation in the previous case.

6.28 Since |G| = n is square free, then |G| = p1p2 · · ·pk. Hence, GCp1 ×Cp2 × · · ·Cpk . Since (pi,p j) = 1, for alli, j ∈ 1,2, · · · ,k, then

Cp1 ×Cp2 × · · ·Cpk Cp1p2···pk = Cn.

Thus GCn.

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24 Shaska T.

6.29 We can note that a group G such that |G| = 108 = 22∗3 ∗5 ∗7 has the following property by Theorem 6.3:

G G1×G2×G3×G4

where

|G1| = 22, |G2| = 3, |G3| = 5, |G4| = 7

Furthermore, for each of these Gn groups,

G1 = C2×C2 or G1 = C4

G2 = C3

G3 = C5

G4 = C7

and each composition is unique. Then since we only have on choice for G2,G3, and G4, and two for G1,then we have two possible compositions for a group of order 420. Therefore we have

G = C2×C2×C3×C5×C7

G = C4×C3×C5×C7

And these are the only possible Abelian groups of order 420.

6.30 D6 is the symmetries of a hexagon, such that

D6 = e,σ,σ2,σ3,σ4,σ5,τ,τσ,τσ2,τσ3,τσ4,τσ5

where, e = identity, σ = 60 rotaions, and τ = flip about the x-axis.Let M = e,σ2,σ4,τ,τσ2,τσ4

and N = e,σ3. N is normal since it is the center (Z(D6)). M is also normal

by the Third Subgroup Test. It’s clear to see that M⋂

N = e. We can also observe that since

|MN| =|M||N||M

⋂N|

=6 ·2

1= 12

So, MK = D6 and thus D6 M×N. From Theorem 3.3, since M,N,S3, and C2 are cyclic groups, |M| = |S3|

and |N| = |C2|, thenM S3 and N C2.

Therefore, we can conclude that D6 S3×C2From above, we can make a guess that D2n Sl ×C2.. There exists subgroups X,Y ∈ D2n such that

|Sl| = |X| = 2n2(where l! = 2n2) and |C2| = |Y| = 2 and X⋂

Y = e. From that we can see that

|XY| =|X||Y||X

⋂Y|

=2n2· (2)

1= 4n2

. Since |D2n| = 4n2, we can similarly conclude that D2n Sl×C2.

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6.31

6.32 Disproof: A4 is a subgroup of S4 where S4 is a symmetric group of n elements. The order of A4 = 4!2 = 12.

Since 6 divides 12, but A4 has no subgroups of order 6 this is a counterexample.

6.33 Let G,H,K be groups. we have that (GxK) is isomorphic to (HxK). Since (GxK) is isomorphic to (HxK)we have a map call it f defined as

f : (GxK) 7→ (HxK)f (g,k) = (h,k)

and this map is one-to-one and a homomorphism.Now define the function

φ : (G) 7→ (H)φ(g) = h

where f(g,k)=(h,k).1.Let G1, g2 ∈ Gfor this we have f (g1,k) = (h1,k) and f (g2,k) = (h2,k)now consider

φ(g1) = φ(g2)→ h1 = h2

(h1,k) = (h2,k)f (g1,k) = (g2,k)( f → one− to− one)g1 = g2

Thus φ is one-to-one.

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24 Shaska T.

2.Let h ∈H. Then (h,k) ∈HxKSince we can find (g,k) ∈ GxK such thatf((g,k))=(h,k)for this g ∈ G, we have φ(g) = hTherefore φ is onto.3. Let g1, g2 ∈ G→ g1g2 ∈ GWith f (g1,k) = (h1,k) and f (g2,k) = (h2,k)

(g1g2,k) = (g1,k)(g2,k)f (g1g2,k) = f ((g1,k)(g2,k))

= f (g1,k) f (g2,k)( f = homomorphism)= (h1,k)(h2,k)= (h1h2,k)

so we have φ(g1g2) = h1h2 = φ(g1)φ(g2)Thus φ is a homomorphism. Therefore φ is an isomorphism from G to H.Thus G is isomorphic to H.

6.34

7.1 We have to show that σ(Z(G)) = Z(G).i) Let’s show first that σ(Z(G)) ⊂ Z(G).Let a ∈ Z(G). Want to show that σ(a) · g = g ·σ(a) for any g ∈ G. Since σ is an automorphism, then exists

c ∈ G such that σ(c) = g. So σ(a) · g = σ(a) ·σ(c) = σ(ac) = σ(ca) = σ(c) ·σ(a) = gσ(a).ii) Z(G) ⊂ σ(Z(G)).Want to show that if a ∈ Z(G) then a ∈ σ(Z(G)) which is equivalent to say that there exists b ∈ Z(G) such

that σ(b) = a.Let b = σ−1(a), where σ is a bijection. Want to show that b ∈ Z(G). Let g ∈ G. Then g ·b = σ−1(g′) ·σ−1(a) =

σ−1(g′ · a) = σ−1(ag′) = σ−1(a) ·σ−1(g′) = b · g. Note that since g ∈ G, then there exists g′ such that σ−1(g′) = g.So g · b = b · g and so b ∈ Z(G)⇒ a ∈ Z(G).

7.2 Since G is solvable, it has a normal series

1 /G1 / · · · /Gn = G,

such that Gi+1/Gi is Abelian for each i. Likewise, since H is solvable, it has a normal series

1 /H1 / · · · /Hm = H,

such that H j+1/H j is Abelian for each j. G×H creates the normal series

1× 1 /G1×H1 / · · · /Gn×Hm = G×H,

such that (Gi+1×H j+1)/(Gi×H j) is Abelian for each i and j. Therefore, G×H is a solvable group.

7.3 Let G be a group and N be a proper normal subgroup of G.Let

e /H1 / · · · /Hn = G (24.13)

be a composition series of G. Since N is a proper normal subgroup of G,

e /N /G (24.14)

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Shaska T. 24

is a normal series of G containing N. If this normal series is also a composition series then we are done.If it is not a composition series, then you can get to a composition function by refining it. By the Jordan -Hölder Theorem (Theorem 7.2), any two composition series of a group are equivalent and thus the abovecomposition series (1) must be isomorphic to a refinement of (2) since composition functions are alreadyrefined.

7.4 Let G be a solvable group. Then, by the definition of being solvable, G has a normal series such thate= P0 /P1 /.../Gn = G where Gi +1/Gi is Abelian for all i. Pi+1/Pi being Abelian implies that it is cyclic. Sinceit is cyclic, |Pi+1/Pi| is prime. It is also true that, by the normal series defined, Pn ⊃ Pn−1 ⊃ ... ⊃ P1 ⊃ P0 = e.The forward implication has been proven.

Let G have a series of subgroups such that G = Pn ⊃ Pn−1 ⊃ ... ⊃ P1 ⊃ P0 = ewhere Pi is normal in Pi+1and the order of Pi+1/Pi is prime. Then, since the order of Pi+1/Pi is prime, Pi+1/Pi is cyclic and thereforeAbelian. Also, since Pi is normal in Pi+1 and the series of subgroups is defined as above, these Pi subgroupsform a normal series of G. By definition, G is a solvable group.

7.5 Let n be a positive integer, D2n =r,s : s2 = rn = 1;rs = sr−1

, take H =

1,r,r2, · · · ,rn−1 so that H is cyclic

subgroup of D2n of order n. And the seriese/H /D2n. We want to show it’s a composition series.

H is isomorphic to H/e

and H is Abelian and cyclic group of order n generated by r. Also note that[D2n : H

]=|D2n|

|H|=

2nn

= 2, so H is normal in D2n (proven in previous exercise). Also D2n/H is isomorphic

to Z2 (theorem 3.3) which is Abelian. So we showede/H /D2n is the composition series for Dn, so Dn is

solvable.

7.6

7.7 If G is a p-cyclic group, then it is generated by an element x with order pn for some n. Then any subgroupof G must be generated by an element xpa

where 0 ≤ a ≤ n. So let H,K be subgroups of G. Without loss ofgeneralization, say

H = 〈xpa〉

K = 〈xpb〉

where 0 ≤ a ≤ b ≤ n. Then we know that pa|pb, so pb = cpa for some integer c. Then xpb

= (xpa)c and H ≤ K.

We can similarly say K ≤H if b ≤ a.

7.8 Let G be a solvable group of order n ≥ 2. If we look at the group as the chains

1CG1CG2C · · ·CGn = G,

Gn≤ · · · ≤ G2

≤ G1≤ G0 = G.

where Gi+1/Gi is Abelian and Gn = 1 . Since we know that since G0 = G and that G1 is the commutatorsubgroup of G0, then G1 is Abelian. Then, following the steps from the proof of Theorem 7.3, we can seethat G1

≤ Gn−1 and, since the series is Abelian, that Gn−1 ≤ G1. Therefore the subgroup Gn−1 is abelian andnon-trivial.

7.9 here

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24 Shaska T.

7.10 a) From Example 5.10 we know that n5 = 1 or n11 = 1.

Case i) n11 = 1. Then, P11 C G. Hence, G = G/P11 has order 45. Thus, |G| = 45 = 32·5 implies that

n3 = 1 and n5 = 1

Therefore, we have that P5 C G.From the Theorem 3.9 we have that there exists K C G such that |K/P11| = |P5|. Hence, |K| = 55.

G // G

K P5

oo

P11 // 1G

Case ii): Assume that n5 = 1. Then, P5 C G and G := G/P5 has order 99. By the Sylow’s theorem we get

n3 = 1 and n11 = 1

Thus, P11 C G.G // G

K P11

oo

P5 // 1G

From Theorem 3.9 we have that there exists K C G such that |K/P11| = |P5|. Hence, |K| = 55.b) From a) there exists K C G such that |K| = 55. K is solvable since it is a direct product of two solvable

groups. The quotient group G/K has order 9 and it is solvable since every p-group is solvable. FromLemma 7.6 we have that G is solvable.

7.11 From the Sylow’s Theorem we get

n2 = 1,5,13,65n5 = 1,26

n13 = 1,40

By a counting argument we easily get that

n5 = 1 or n13 = 1

Assume that n5 = 1. Then, P5 C G. Let G = G/P5. Then, |G| = 23·13. By Sylow’s theorem P13 C G. From

Theorem 3.9 there exists K C G such that [K : P5] = |P13| = 13. Hence |K| = 65.Assume now that n13 = 1. Then, P11 C G. Let G = G/P11. Thus, |G| = 23

·5. Applying Sylow’s theoremin G we have n5 = 1. Hence, P5 C G. From the correspondence theorem there exists K C G such that[K : P11] = |P5|. Hence, |K| = 65.

b) Since |K| = 65 = 5 ·13 then K is solvable (in fact H is even cyclic). Then G/K has order 8 and it is alsosolvable as a 2-group. Hence, from Lemma 7.6 we have that G is solvable.

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7.12 Since |G| = 22·32 then from Sylow’s theorem

n2 = 1,3, and n3 = 1,4.

If n3 = 1 then P3 CG and G/P3 has order 4. Both P3 and G/P3 are solvable as p-groups. Hence G is solvable.If n3 = 1 then since np is the index of the normalizer of the Sylow p-subgroup then [G : NG(P3)] = 4. Let’s

denote H := NG(P3). Then from Theorem 3.15 there is a homomorphism

φ : G→ S4

such that kerφ ≤H. Since |G| - 4! then φ is not injective. Hence K = kerφ has order 3 or 9. In both cases Kis solvable (as a p-group) and K C G (as a kernel). It is enough to show that G/K is solvable.

If |K| = 3 then G/K has order 2 ·32 = 18. The reader can easily show directly that this is solvable or usethe result 7.9. If |K| = 4 then G/K has order 9 and it is therefore solvable as a p-group.

7.13 Since |G| = 22·33, then from the Sylow’s Theorem we get

n2 = 1,3,9,27, and n3 = 1,4.

There exists a Sylow 3-group H := P3 < G such that [G : P3] = 4. From Theorem 3.15 there exists a homo-morphism

φ : G→ S4

such that kerφ ⊂H. Let K := kerφ. Since |G| - 4! then φ is not an embedding.Then |K| = 3,9,27. Since kernels are normal then G has a normal subgroup of order 3, 9, or 27. Notice

that K is solvable as a p-group. It is enough to show that G/K is solvable.If |K| = 27 then G/K is solvable as as p-group (it has order 4). Hence G is solvable.If |K| = 9 then G/K has order 12 = 22

·3. Every group of order 12 is solvable from a direct application ofSylow’s Theorem or the following result 7.9. Thus K and G/K are solvable. Hence, G is solvable.

If |K| = 3 then G/K has order 36. Every group of order 36 is solvable; see 7.12.

7.14 A subgroup of a solvable group is solvable (Define Hi = H∩Gi, then Hi is a series of H with abelianfactors), and a quotient group of a solvable group is solvable (Define Qi = GiN/N, then Qi is a series ofQ = G/N with abelian factors).

If Ki is a composition series of G, then each Ki/Ki+1 is a quotient of a subgroup of G, and so also solvable.A solvable simple group F is abelian, since [F,F] is a proper normal subgroup of the simple group F, and somust be the identity. Hence, each composition factor has finite (prime) order. Hence G itself is finite, andits order being the product of the orders of its finitely many composition factors.

7.15 Let G be a finite group and let |G| = pqn where p and q are primes and p < q. Want to show that G issolvable. By the Sylow’s Theorem we can get

np = 1,qn

nq = 1,p.

np = q is not possible since p. 1modq. Assume np = qn and nq = p but 1+p(q−1)+qn(p−1)> pqn. Thereforenp = 1 and nq = 1 so G is solvable.

7.16 We will use some theorems to help prove this. First, we know that the group of just the identity hasorder one and is solvable. From Theorem 7.4, we know that all finite p−groups are solvable. This leaves uswith

6,10,12,14,15,18,20,21,22,24,26,28,30,32,33,34,35,38,39,20,42,44,45,46,48,50,51,52,54,55,56,57,58

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24 Shaska T.

From Burnside Theorem, we know that a group of order paqb where p,q are primes and a,b ∈Z is solvable.This leaves us with

30 and 42

We must show that groups of order pqr where p,q,r are primes such that WLOG p < q < r are solvable. LetH be such a group with order pqr. Then r - p, r - p, and thus r - pq. Using Sylow’s Theorem, we know thatnr = 1 or r. However r - pq and so nr = 1 and thus R, a Sylow r− subgroup is simple. Thus we can create thenormal series

e /R /G

Since the order of R is r, a prime number, it is cyclic and therefore G/R and R/e are Abelian.Therefore, G has a normal series with Abelian factor groups and thus is solvable.

7.17 WTS: (1) Every group of odd order is solvable⇐⇒ (2) every finite simple group has even order.(1) =⇒ (2). If every group with odd order is solvable, then, consider every group that is not solvable,

simple groups. These simple groups cannot have odd order, else they would be solvable. Therefore, everyfinite simple group has even order.

(2) =⇒ (1). If every finite simple group has even order, then, consider groups of odd order. If thesegroups of odd order were simple, they would be of even order. A contradiction. Therefore, all groups ofodd order are solvable.

7.18 Take S4, Since e /V4 /A4 /S4, then S4 is a solvable group because it has a norma series but it is notsuper - solvable because S4 has no subgroups which are cyclic.

7.19 We know that if a group is supersolvable, it has a sylow tower. So, we will show that S4 has no sylowtower. |S4| = 24, so a 3-sylow subgroup will have order 3, and a 2-sylow subgroup will have order 8. Thesubgroup K of S4 generated by (13) and (1234) has order 8, and is thus a sylow-2 subgroup of S4, potentially≈ to D4. The subgroup M of S4 generated by (123) has order 3, hence it is a sylow-3 subgroup of S4. Now,since neither K nor M are normal in S4, it must be the case that S4 doesn’t have a sylow tower. And so, S4isn’t super solvable.

7.20 Let a group be supersolvable if G has a chain of subgroups:

e = G0 ≤ G1 ≤ · · · ≤ Gn = G

such that every i = 1, . . . ,n , there is Gi /G and Gi+1/Gi is cyclic. All p-groups have cyclic factor groupsbecause each element of p gets raised to the next power of p, so the factor group is always of the order pwhich is cyclic because it is the group generated by the element p.

7.21 Let G be a group with a composition series. Assume that H is a normal subgroup of G. If H = G thenwe are done so assume that H , G.

Lete /M1 / · · · /Mn = G (24.15)

be a composition series of G. Since we know H is normal in G,

e /H /G

is a normal series of G with H as a term. If H is maximal normal in G, then leave the series to the right ofH alone. However, if H is not maximal normal in G, then there is a maximal normal subgroup of G thatcontains H. Find the maximal normal subgroup of G and repeat this process of finding maximal normalsubgroups between G and H. This will refine the right side of the series until we can no longer refine itwithout adding unnecessary terms.

If H is simple, then we are done because we have a composition series of G with H as a term. If H is notsimple, then there is at least one normal subgroup of H. Find the maximal normal subgroup as before and

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Shaska T. 24

repeat this process until arriving at a simple maximal normal subgroup. This will refine the left side of theseries and ends when we start adding unnecessary terms.

Finally, we will have a composition series of G that has H as a term. This composition series will beequivalent to the composition series (3) by the Jordan - Hölder Theorem (Theorem 7.2).

7.22

7.23 Suppose M and N are normal solvable subgroup of G. For MN/N M/(M∩N), and M/(M∩N) issolvable as quotient of solvable M, so MN/N is solvable, N is solvable. Thus MN must be solvable. Thisshow that in every group, there is a unique maximal normal subgroup (it could be trivial). As we call thisgroup F (G).Next, we assume that K/F (G) is non-trivial normal solvable subgroup of G/F (G). Then F (G) < K /G withthe commutator subgroup K′ ⊆ F (G). Since F (G) is solvable, K′ is solvable and hence K must be solvable,thus K ⊆F (G). But we prove thatF (G) is the unique maximal normal subgroup of G, this is a contradiction.

7.24 eE (1 2)(3 4)E (1 2)(3 4), (1 3)(2 4)EA4

eE e, (1 2)(3 4)E e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)EA4ES4

7.25

7.26 This chain is the definition of a solvable group. p-group would be cyclic in G.

7.27 here

7.28 here

7.29

7.30

7.31

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24 Shaska T.

414

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