Lectures on Modern Algebra - homepage.fudan.edu.cnhomepage.fudan.edu.cn/jgyang/files/2012/02/algebra.pdf · iii is indispensable. Due to its importance, modern algebra (or abstract

Lectures on Modern Algebra

Author Jin-Gen Yang

Foreword

For a sufficiently educated person, the word “ algebra” often reminds

him or her a great deal of high school mathematics, such like factor-

ization of quadratic polynomials, solving an equation or a system of

equations, exponential functions, logarithmic functions and so forth.

These subjects are known as precalculus algebra or elementary alge-

bra.

This course is quite beyond the precalculus algebra. It emphasizes

the inner structures of groups, rings, fields and vector spaces as well

as the maps between algebraic structures. For this reason this branch

of mathematics is often refered to as “abstract algebra” or “modern

algebra”. The ideas of algebra evolved through many generations of

mathematicians around the turn of 20-th century. Among many promi-

nent mathematicians we mention Emmy Noether and Emil Artin, who

laid the foundation of modern algebra.

In the freshmen linear algebra, the students have encountered the

algebraic structure of vector spaces over real number field or complex

number field, as well as the structures of Euclidean spaces or unitary

spaces after the introduction of inner product. If we look at the set

of integers and the set of real polynomials in one variable closely, we

may find their properties strikingly similar. From the algebraic point of

view, their structures share many common properties. Roughly speak-

ing, a so-called algebraic structure is a set with one or more operations

satisfying certain conditions. This is one of the basic target of study in

mathematics. Another important basic structure is topological struc-

ture, which is not in the scope of this course. The development of

mathematics in the recent century has confirmed that modern algebra

ii

iii

is indispensable.

Due to its importance, modern algebra (or abstract algebra) has

become a basic standard course for math major students, usually of-

fered for sophomores or juniors. An ideal duration of the course is

one whole year. In recent years, only one semester of modern algebra

is available for many major universities of China. It is not easy for

instructors of this course to cover Galois theory in one semester. But

Galois theory is recognized as a milestone of algebra. It will be regret-

ful if a student does not know Galois theory after taking the course of

algebra. Through many years of teaching the author has made careful

selection of the materials for groups, rings and fields so that the stu-

dents can reach the main theorem of Galois theory and the proof of

the insolvability by radicals of equations of degrees greater than four.

Among all algebraic structures, the group is a basic structure. To

shorten the first chapter on groups, I postpone some less fundamental

but more technical subjects such like Sylow’s groups, finitely gener-

ated abelian groups and solvable groups to a later chapter. The first

three chapters aim at basic knowledge on groups, rings and fields. For

ring theory, emphasis is put on the residue ring Z/nZ and polynomial

rings, of one variable and of several variables. Through these concrete

rings students can understand more abstract concepts such like princi-

pal ideal domain and unique factorization ring. The non-commutative

rings are restricted to basic knowledge and a few standard examples

such like matrix rings and quaternions.

The fourth chapter is a brief account of linear algebra over an ar-

bitrary field. Since the students are assumed to have taken the first

course of linear algebra already, the account is often sketchy. The main

purpose of this chapter is to emphasize the difference of vector spaces

over fields of different characteristics and prepare for the theory of ex-

tensions of fields. Students should establish a point of view to regard

the extension of a field as a vector space over the base field.

Chapter 5 covers Sylow groups, finitely generate abelian groups

and solvable groups as I mentioned before. The last three chapters

are devoted to the field theory exclusively. By author experience, it is

reasonable to finish all eight chapters in 15 or 16 weeks.

iv FOREWORD

The appendices are for more enthusiastic students. A proof of

quadratic reciprocity law displays elegant techniques of finite fields.

The proof of the theorem concerning the structure of finite skew fields

displays a clever use of group actions. The proofs of these two theo-

rems are adapted from the ones of J.-P. Serre and A.Weil respectively.

The author wishes that the interested readers can feel the beauty of a

mathematical proof by reading these two appendices.

Due to the limit of time, in the last section only the insolvability

of equations of degree greater than four is proved. Using Galois theory

to deduce the formulas for the solutions of cubic and quartic equations

are put in appendices.

The sections marked with an asterisk can be skipped on the first

reading.

There are adequate amount of exercises throughout the book. The

degree of difficulty varies. Some exercises are chosen from Ph.D qual-

ifying exams. Hints or solutions are provided as an appendix. Most

problems have many different proofs, which are impossible to be in-

cluded. Students are strongly encouraged to find their own solutions

and do not rely on the appendix too heavily. Only by laying one’s

own hand on hard problems, one can feel the charm of mathematical

deduction.

In summary, the author has tried to give a concise and easy to

understand account of basic knowledge and methods in algebra without

loss of rigor. The main perspective readers are sophomores or juniors

of math major.

Professor Li Kezheng offered many valuable comments and pointed

out many mistakes. Many students of Fudan University who have taken

this course in the past several years have provided useful comments

too. I regret that I am not able to list their names. I thank all those

people who have made contributions to this book. The assistance of

Ms. Yao Lili of Science Press in the publication of this book is greatly

appreciated.

J.-G. Yang, Sept., 2011

Contents

Foreword ii

Preliminaries and Notations vi

1 Elements of Groups 1

1.1 Definitions and Examples . . . . . . . . . . . . . . . . 1

1.2 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Permutation Groups . . . . . . . . . . . . . . . . . . . 11

1.4 Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5 Normal Subgroups and Quotient Groups . . . . . . . . 23

1.6 Alternating Groups . . . . . . . . . . . . . . . . . . . . 28

1.7 Homomorphisms of Groups . . . . . . . . . . . . . . . . 32

1.8 Direct Product of Groups . . . . . . . . . . . . . . . . 39

1.9 * Automorphisms of Finite Cyclic Groups and the Euler

Function . . . . . . . . . . . . . . . . . . . . . . . . . . 43

1.10 Group Action . . . . . . . . . . . . . . . . . . . . . . . 45

2 Elements of Rings and Fields 51

2.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . 51

2.2 Ideals and Quotient Rings . . . . . . . . . . . . . . . . 55

2.3 Homomorphisms of Rings . . . . . . . . . . . . . . . . 60

2.4 Elementary Properties of Fields . . . . . . . . . . . . . 63

3 Polynomials and Rational Functions 71

3.1 Polynomials in One Variable . . . . . . . . . . . . . . . 71

3.2 Division Algorithm . . . . . . . . . . . . . . . . . . . . 72

3.3 Polynomials in Several Variables . . . . . . . . . . . . . 76

v

vi CONTENTS

3.4 Factorization . . . . . . . . . . . . . . . . . . . . . . . 78

3.5 * Polynomial Functions . . . . . . . . . . . . . . . . . 87

4 Vector Spaces 90

4.1 Vector Spaces and Linear Transformations . . . . . . . 90

4.2 Quotient spaces . . . . . . . . . . . . . . . . . . . . . . 95

5 Topics in Group Theory 98

5.1 The Orbit Formula of an Action by a Finite Group . . 98

5.2 Sylow subgroups . . . . . . . . . . . . . . . . . . . . . 101

5.3 * Structure of finitely generated abelian groups . . . . 106

5.4 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . 116

6 Field Extensions 120

6.1 Definitions and First Properties of Field Extensions . . 120

6.2 Algebraic Extensions . . . . . . . . . . . . . . . . . . . 123

6.3 Constructions of Field Extensions . . . . . . . . . . . . 128

6.4 Algebraically Closed Field . . . . . . . . . . . . . . . . 131

6.5 * Ruler and Compass Construction . . . . . . . . . . . 136

7 Finite fields 142

7.1 Basic Theory . . . . . . . . . . . . . . . . . . . . . . . 142

7.2 The structure of Multiplicative Group of a finite field . 144

8 Finite Galois Theory 148

8.1 Basic theory . . . . . . . . . . . . . . . . . . . . . . . . 150

8.2 * Solvable Extension and Solvability of Algebraic Equa-

tions by Radicals . . . . . . . . . . . . . . . . . . . . . 162

A Quadratic residues 169

B Every finite skew field is a field 174

C Solutions of a cubic equation and Hilbert theorem 90 178

D Solutions of quartic equations 183

E Hints or solutions for exercises 187

CONTENTS vii

Bibliography 232

Index 234

Preliminaries and Notations

Since this course is intended for the undergraduate students in math

major, we assume that the readers have already studied linear alge-

bra and calculus (at least calculus in one variable). The readers are

assumed to have the basic knowledge on sets and maps.

Some conventions and notations are given below.

A map f from one set S to another set T is an injection if f(x) 6=f(y) wherever x and y are two distinct elements in S. The map f is a

surjection if there exists x ∈ S such that f(x) = z for every z ∈ T. A

map is a bijection if it is injective and surjective. Let U be a subset of

S. Denote by f |U the restriction of f on U.

A map f from S to T carrying x to z is denoted by f : S → T, x 7→z. For example x 7→ ex is the exponential function. Sometimes the

identity map from a set S into itself is denoted by id or 1.

Let f be a map from S to T and g be a map from T to U. The map

g f : S → U, x 7→ g(f(x)) is the composite of f and g. It can also be

denoted by gf.

A subset of a set S is often defined by the notation x ∈ S|P, where

P is the condition that x should satisfy. For example x ∈ R|0 ≤ x ≤1 is the closed interval [0, 1]. The union and the intersection of sets

are denoted by ∪ and ∩ respectively. The difference x ∈ A|x /∈ B of

the sets A and B is denoted by A−B or A\B, and we prefer to use the

latter. A set with a few elements is usually denote by · · · , in which

· · · is the list of all elements. For example a is a set consisting of a

single element a, 0, 1 is the set consisting of the elements 0 and 1.

The empty set is denoted by ∅.The knowledge of equivalence relations and equivalence classes is

viii

ix

helpful, but is not indispensable. We will explain them when they are

first used in the text.

Readers are required to know the number fields well. By definition,

a number field is a subset of the complex number field closed under

addition, subtraction, multiplication and division. The most frequently

used number fields are complex number field, real number field and

rational number field.

The equality x = ±a means that x is equal to either a or −a.Some common notations are listed as follows:

N — set of natural numbers

Z — set of integers

Q — set of rational number

R — set of real numbers

C — set of complex numbers

x PRELIMINARIES AND NOTATIONS

Chapter 1

Elements of Groups

Many objects we encounter in mathematics are sets equipped with one

or more operations. Such objects are usually called algebraic struc-

tures. In this course the algebraic structures we will study include

groups, rings and vector spaces. In terms of the number of operations

the group theory can be considered to be a natural starting point for

this course, since a group involves only one basic operation. In this

chapter the definition, basic properties of groups and homomorphisms

between groups will be explained. More advanced topics on groups will

be covered in later chapters.

1.1 Definitions and Examples

Let S be a set. Denote the Cartesian product of S with itself by S×S.It consists of all pairs (a, b) with a, b ∈ S. Note that (a, b) and (b, a)

are different elements in S × S if a 6= b. A map f from the set S × S

to S is called a (binary) operation.

For example, let S be the set of all real numbers. Defined the map

f : S × S → S by f(a, b) = a + b. Then f is a binary operation.

We may use all sorts of ways to define operations such as f(a, b) =

a2 + b3, f(a, b) = ab, . . . .

Since it is awkward to use the notation f(a, b) for an operation,

there are more convenient notations such as a + b, [A,B],u × v, . . . ,

depending on occasions. The simplest notation is ab. We will use this

notation when no confusion will be caused. In practice, it is better

1

2 CHAPTER 1. ELEMENTS OF GROUPS

to use the notations that we are already familiar with. For instance,

the addition of numbers is denoted by “+”, the cross product of two

vectors in R3 is denoted by “×”.

Let’s look at some more examples.

• Let S be a 3-dimensional Euclidean space. The addition u + v

and cross product u× v are binary operations. The dot product

u · v is not a binary operation since the result of the dot product

is not a vector.

• Let S be the set of all n × n matrices. Then A + B,AB,AB −BA are three different operations on S (the last one is a basic

operation in Lie Algebra).

• Let S be the set of all everywhere well-defined real functions.

Then the composite f g is a binary operation on S.

Unitary operations (involving one variable) appear quite often too,

such as −a (the negative of a number), a2 (the square of a number),

a (the conjugate of a complex number), AT (the transpose of a square

matrix), etc..

We say that a binary operation f on a set S satisfies the law of

associativity if

f(a, f(b, c)) = f(f(a, b), c)

holds for any a, b, c ∈ S. By our convention this condition can be written

simply as

a(bc) = (ab)c,

which is exactly the form we are familiar with.

In the previous examples many binary operations satisfy the law of

associativity. The cross product of vectors and the operation AB−BAfor square matrices A,B do not satisfy the law of associativity. It is

easy to verify that the operation f(a, b) = a2 + b3 on real numbers does

not satisfy the law of associativity.

An element e in a set S equipped with a binary operation is called

an identity element if ea = ae = a for any a ∈ S.

Examples:

1.1. DEFINITIONS AND EXAMPLES 3

• 0 is an identity element in the set of real numbers under addition.

• 1 is an identity element in the set of real numbers under multi-

plication.

• The identity matrix In is an identity element in the set of n × n

matrices under multiplication.

Proposition 1.1.1. A set S with a binary operation has at most

one identity element.

Proof. Let e and e′ be two identity elements of S. Since e is an

identity element, ee′ = e′ holds. On the other hand, ee′ = e holds since

e′ is also an identity element. Therefore e = ee′ = e′.

Definition 1.1.2. A nonempty set S with a binary operation is

called a semigroup if the law of associativity is satisfied. A semigroup

containing an identity element is called a monoid.

According to Proposition 1.1.1 there is a unique identity element in

a monoid.

Example 1.1.3. The set of natural numbers under addition is a

semigroup but not a monoid.

Definition 1.1.4. Let a be an element in a monoid S and let e

be the identity element of S. If there is an element b ∈ S such that

ba = ab = e then a is called an invertible element and b is called the

inverse of a.

Note that the article in front of the word “inverse” is “the” instead

of “an”. This will be justified by the following proposition.

Proposition 1.1.5. There is a unique inverse for an invertible

element in a monoid.

Proof. Let b, b′ be inverses of a. Then b = be = b(ab′) = (ba)b′ =

eb′ = b′.

It is obvious that the relation “inverse” is symmetric. That is to

say, if a is invertible with b as its inverse, then b is also invertible and

a is the inverse of b.

The inverse of an invertible element a is commonly denoted by a−1.


Definition 1.1.6. A nonempty set G with a binary operation is

defined to be a group if the following three conditions are satisfied:

• The law of associativity is satisfied;

• The identity element exists;

• Every element in G is invertible.

The first two conditions in the definition mean that every group is

a monoid. Although the concepts of semigroup and monoid are more

general than group, but we are primarily interested in groups.

The unitary operation a 7→ a−1 in a group is called the inverse

operation. Since it is totally determined by the binary operation of the

group, we do not consider the inverse as a basic operation of the group.

As we have mentioned before, the binary operation in a group uses

different notations depending on occasions. The default notation is ab,

or a · b occasionally. For this reason, the operation can also be referred

to as “multiplication” and ab is called the product of a and b. In this

case the inverse of a is denoted by a−1 and the identity element e can

also be denoted by 1, or 1G to avoid confusion. For a natural number

n, the n-th power an is defined to be the product of a with itself n

times. The power a−n with negative integral exponent is defined to be

(a−1)n.

Let G be a group. If ab = ba for any a, b ∈ G, then G is called a

commutative group, or abelian group.

For many abelian groups, the binary operation is usually called

“addition” and denoted by a + b. In this case the identity element is

often called the zero (element) of the group and denoted by 0. The

inverse of an element a is denoted by −a and the n-th power of an

element a becomes na = a+ · · ·+ a.

A groups consisting of a finite number of elements is called a finite

group, otherwise it is called an infinite group. The number of ele-

ments in a group G is denoted by |G|, called the order of G. Hence

|G| is ∞ if G is an infinite group and |G| is a natural number if G is a

finite group.

1.1. DEFINITIONS AND EXAMPLES 5

Let g be an element of a group G. If there is no natural number n

such that gn = 1, then g is called an element of infinite order, otherwise,

define the order of g to be the smallest natural number n such that

gn = 1, denoted by o(g). Hence an element in G has order one if and

only if it is the identity element. By convention, o(g) = ∞ if g is an

element of infinite order.

• Every number field under addition forms an abelian group, called

the additive group of that number field. One thing to keep in mind

is that the multiplication operation of that number field is neglected

when it is regarded as a group, since by definition a group involves only

one basic operation.

• The set Z of all integers under addition is an abelian group, called

the additive group of integers.

• All n × n invertible matrices over a number field K under the

(matrix) multiplication is a group called general linear group and

denoted by GLn(K). It is not an abelian group when n > 1.

• A number field K under multiplication is a monoid, but not a

group since the element 0 in not invertible. Denote K∗ = K\0.Then K∗ is an abelian group under multiplication, which is essentially

the same as GL1(K). Although this group is abelian, but it would be

absurd to use “+” to denote the multiplication.

• Any vector space is an abelian group under addition.

• 3-dimensional Euclidean space R3 is not a semigroup under cross

product, since the law of associativity is not valid.

• The smallest group has only one element. Such a group is called

a trivial group.

• Empty set is not a group.

Proposition 1.1.7 (law of cancelation). If the element a, b, c in a

group satisfy ab = ac or ba = ca, then b = c.

Proof. Assume that ab = ac. By left multiplying a−1 to both sides

of ab = ac the equality a−1(ab) = a−1(ac) is obtained, so (a−1a)b =

(a−1a)c by the law of associativity. Hence eb = ec by the definition of

inverse. Finally b = c follows from the definition of the identity element

e.


The same argument shows that ba = ca implies b = c.

Corollary 1.1.8. Let a, b be two elements of a group G.

1) If ab = a or ba = a then b = 1G;

2) If ab = 1G or ba = 1G then b = a−1.

Proof. 1) Apply the law of cancelation to ab = a1G.

2) Apply the law of cancelation to ab = aa−1.

Proposition 1.1.9. Let a, b be two elements of a group G. Then

(ab)−1 = b−1a−1.

Proof. The equality (b−1a−1)(ab) = b−1(a−1a)b = b−1b = 1 and

Corollary 1.1.8.2) imply b−1a−1 = (ab)−1.

This property is the generalization of the well-known formula (AB)−1 =

B−1A−1 in linear algebra, where A and B are two invertible n×n ma-

trices.

As we have seen, any number field is a group under addition. For

this reason we may consider addition as the first important operation

of this number field. In fact, this is the first operation taught in ele-

mentary school. The subtraction is not regarded as a basic operation.

Since a − b = a + (−b), the subtraction can be considered to be the

composite of the inverse and the addition.

Exercises

1. In the set G = a ∈ R|a > 0, a 6= 1 define a binary operation

a∗ b = aln b. Is G a group under this operation? Here ln b is the natural

logarithm of b.

2. Let A be the set of all strictly increasing continuous functions

on [0, 1] satisfying f(0) = 0, f(1) = 1. For any f, g ∈ A define fg to

be the composite of f and g, i.e., (fg)(x) = f [g(x)] for any x ∈ [0, 1].

Prove that A is a group under this operation.

3. Let G be a semigroup satisfying the following two conditions:

1) there is some e ∈ G such that ea = a for any a ∈ G;

2) for every a ∈ G, there is some b ∈ G such that ba = e.

1.2. SUBGROUPS 7

Prove that G is a group.

4. Prove that every element of a finite group has finite order.

5. Let G be a group and a, b ∈ G. Prove that o(ab) = o(ba).

6. Assume that every element a of a group G satisfies a−1 = a.

Prove that G is an abelian group.

1.2 Subgroups

Definition 1.2.1. Let H be a nonempty subset of a group G sat-

isfying the following two conditions

1) (closure under the multiplication) ab ∈ H for any a, b ∈ H;

2) (closure under the inverse operation) a−1 ∈ H for any a ∈ H.Then H is called a subgroup of G.

By the closure of multiplication, a subgroup H of G inherits the

binary operation from G. Naturally this inherited operation preserves

the law of associativity. Since H is nonempty, there is some element

a ∈ H. Hence a−1 ∈ H by the closure of inverse operation, which

implies that 1 = a−1a ∈ H. So H is a group by itself. The readers

may compare the concept of subgroup with that of subspace in linear

algebra.

It is easy to see that any subgroup of an abelian group is abelian.

Example 1.2.2. • Every group G has two subgroups for free. They

are 1 and G. (They become the same if |G| = 1.) They are called

trivial subgroups of G. A subgroup H of G is called a proper subgroup

if H 6= G.

• Let n be a natural number. Let nZ denote the set of all integers

divisible by n. Then nZ is a subgroup of Z.• Let SLn(K) be the set of all n×n matrices with determinant one

over a number field K. It is a subgroup of GLn(K), called the special

linear group.

• The set of matrices consisting of[1 0

0 1

],

[−1 0

0 −1

]


is a subgroup of SL2(K) with |H| = 2.

• The real number field R is a group under addition and R∗ =

R\0 is a group under multiplication. Although the latter is a subset

of the former, but R∗ = R − 0 is not a subgroup of R because the

binary operations of these two groups are different.

Proposition 1.2.3 (a criterion for subgroup). Let H be a nonempty

subset of a group G. If ab−1 ∈ H for any a, b ∈ H, then H is a subgroup

of G.

Proof. Let c be an arbitrary element in H. Then 1 = cc−1 ∈ H

by the hypothesis of the proposition. For any a ∈ H, we have a−1 =

1a−1 ∈ H. Hence H is closed under the inverse operation.

Since ab = a(b−1)−1 ∈ H, H is also closed under multiplication.

Proposition 1.2.4. Let Hii∈I be a set (not necessarily finite) of

subgroups of a group G. Then H =⋂

i∈I Hi is a subgroup of G.

Proof. Since 1 ∈ ∩i∈IHi the set H is not empty. Let a, b ∈ H.

Then ab−1 ∈ Hi for every i ∈ I, since everyHi is a subgroup ofG. Hence

ab−1 ∈ H. It follows from Proposition 1.2.3 that H is a subgroup of

G.

Definition 1.2.5. Let g be an element of a group G. The set

C(g) = a ∈ G|ag = ga is called the centralizer of g in G. For any

nonempty subset S of G, the set C(S) = a ∈ G|ag = ga for all g ∈ Sis called the centralizer of S in G. In particular, C(G) is called the cen-

ter of G.

It is easy to see that C(g) and C(S) =⋂

g∈S C(g) are subgroups of

G and C(G) is an abelian group. The group G is abelian if and only if

G = C(G).

Let S be a nonempty subset of a group G. Let 〈S〉 denote the in-

tersection of all subgroups of G containing S. Then 〈S〉 is the smallest

subgroup of G containing S in the sense that 〈S〉 ⊆ H for every sub-

group H of G with S ⊆ H. The subgroup 〈S〉 is called the subgroup

generated by S. If S is a finite set consisting of the element a1, . . . , an,

1.2. SUBGROUPS 9

then 〈S〉 can also be denoted by 〈a1, . . . , an〉. If G = 〈S〉, then we say

that G is generated by S.

Example 1.2.6. • Z = 〈1〉 = 〈−1〉.

• GLn(K) is generated by all n× n elementary matrices.

A group generated by a finite set is called a finitely generated

group. In particular, a group generated by a single element is called a

cyclic group. For instance, Z is a cyclic group while GLn(K) is not.

It is obvious that every cyclic group is abelian.

Let a be an element of a group G. Then 〈a〉 is a subgroup of G,

called a cyclic subgroup of G. It is easy to verify that o(a) = |〈a〉|. In

fact, if o(a) = ∞ then

〈a〉 = . . . , a−2, a−1, 1, a, a2, . . .

and if o(a) = n <∞ then

〈a〉 = 1, a, a2, . . . , an−1.

Example 1.2.7. In GLn(K) the matrix[1 1

0 1

]

generates an infinite cyclic subgroup while the matrix[−1

2−

√3

2√3

2−1

2

]

generates a cyclic subgroup of order 3.

Proposition 1.2.8. Let S be a nonempty subset of a group G. Then

〈S〉 = ae11 a

e22 · · · aen

n |a1, . . . , an ∈ S, e1, . . . , en = ±1,

n is an arbitrary nonnegative integer.

Proof. First note that the elements a1, a2, . . . , an involved in the

expression are not necessarily distinct.


Denote the set of the right hand side by T. Then T is a subgroup

containing S. Hence 〈S〉 ⊆ T.

Let H be a subgroup of G such that S ⊆ H. By the definition

of subgroup every element that can be expressed as ae11 a

e22 · · · aen

n (ai ∈S, ej = ±1) is in H. Hence T ⊆ H. Therefore T = 〈S〉.

At this point we want to determine all subgroups of Z. First of all,

as we have already seen before, 0,Z, 2Z, 3Z, . . . are all subgroups of Z.We show that there are no other subgroups.

Let H be a nontrivial subgroup of Z. Then there is some nonzero

integer a inH. Since −a is also inH by the definition of subgroup, there

exists a natural number in H. Let n be the smallest natural number

in H. Then nZ ⊆ H. For any m ∈ H, there are integers q, r such that

m = qn + r in which 0 ≤ r ≤ n − 1. Since r = m − qn ∈ H, r cannot

be a natural number by the minimality of n. Hence r = 0. This implies

that m ∈ nZ. It follows that H ⊆ nZ. Therefore every subgroup of Zis an infinite cyclic group.

Exercises

1. Let G be the set of all 3 × 3 real upper triangular matrices

with all diagonal elements equal to 1. Show that G is a group under

multiplication and determine the center of G.

2. Let X and Y be two subsets of a group G. Prove that

1) if X ⊆ Y then C(X) ⊇ C(Y );

2) X ⊆ C(C(X));

3) C(X) = C(C(C(X))).

Here C(X) denotes the centralizer of X.

3. Let H be a subgroup of a group G such that H is contained in

every nontrivial subgroup of G. Prove that H is contained in the center

of G.

4. An element a of a group G is called a perfect square if there

exists b ∈ G such that a = b2. Assume that G is a cyclic group and

a, b ∈ G are not perfect squares. Show that ab is a perfect square.

1.3. PERMUTATION GROUPS 11

Give an example to show that this statement is not true for non-cyclic

groups.

5. Let H be a nonempty subset of a finite group G. Show that H

is a subgroup of G if ab ∈ H for any a, b ∈ H.

6. Let A be an n × n real invertible matrix and let G be the set

consisting of all n×n real matrices P such that P TAP = A. Show that

G is a subgroup of GLn(R). Here P T is the transpose of P.

1.3 Permutation Groups

Permutations and combinations have already been studied in high

school mathematics. For instance, 2, 1, 4, 5, 3 is a permutation (or re-

arrangement) of the sequence 1, 2, 3, 4, 5. It is known that the number

of permutations of 1, 2, 3, 4, 5 is equal to 5! = 120.

Let’s look at the permutations from another point of view. The re-

arrangement 2, 1, 4, 5, 3 is treated as a bijection σ of the set 1, 2, 3, 4, 5into itself, which carries 1 to 2, 2 to 1, 3 to 4, 4 to 5 and 5 to 3. We

can use the following table to specify this bijection[1 2 3 4 5

2 1 4 5 3

].

This interpretation suggests the following definition.

Definition 1.3.1. Let n be a natural number. A bijection from

the set 1, 2, 3, . . . , n to itself is a permutation of n objects.

As in the example, a permutation can be represented by a table[1 2 3 · · · n

σ(1) σ(2) σ(3) · · · σ(n)

].

Here the numbers 1, 2, 3, . . . , n do not have numerical meaning what-

soever. They are merely convenient labels for distinct objects. Denote

the set of all permutations of n objects by Sn. Then Sn contains n! (the

factorial of n) elements.


Introduce a binary operation in Sn in the following way. For any

σ, τ ∈ Sn define στ to be the composite of σ and τ, the map τ followed

by σ. Thus στ(i) = σ[τ(i)] for every i ∈ 1, 2, . . . , n. Evidently this is

a well-defined binary operation of the set Sn.

Proposition 1.3.2. The set Sn is a group under the binary oper-

ation of composite.

Proof. By the rule of composite of maps σ(τπ) = (στ)π holds for

any σ, τ, π ∈ Sn. Hence the law of associativity is valid.

Let id be the identity map of 1, 2, 3, . . . , n, i.e., id(i) = i for every

i ∈ 1, 2, 3, . . . , n. Then id σ = σ id = σ for any σ ∈ Sn. Hence id

is the identity element.

Since σ ∈ Sn is a bijection of 1, 2, . . . , n, its inverse map τ exists.

It means that στ = τσ = id.

The product of two permutations can be read off from their tables.

For example,[1 2 3 4 5

3 1 5 4 2

][1 2 3 4 5

2 1 4 5 3

]=

[1 2 3 4 5

1 3 4 2 5

].

[1 2 3 4 5

3 1 5 4 2

]−1

=

[1 2 3 4 5

2 5 1 4 3

].

Definition 1.3.3. A subgroup of Sn is called a permutation

group. Sn is called the symmetric group of n objects.

There are two reasons for introducing permutation groups at this

point. First of all, they form a large class of very important finite

groups. Secondly, many non-abelian groups can be found in permuta-

tion groups.

Let’s look at some symmetric groups of low order. When n ≤ 2,

they are too simple to worth studying. The first nontrivial symmetric

group is S3. Its six elements can be enumerated as

σ0 =

[1 2 3

1 2 3

], σ1 =

[1 2 3

1 3 2

], σ2 =

[1 2 3

3 1 2

],


σ3 =

[1 2 3

3 2 1

], σ4 =

[1 2 3

2 1 3

], σ5 =

[1 2 3

2 3 1

].

It is easy to verify that σ1σ2 = σ4 and σ2σ1 = σ3. Thus S3 is not

abelian.

It is not hard to verify that S3 has following four nontrivial sub-

groups σ0, σ1, σ0, σ3, σ0, σ4, σ0, σ2, σ5.The relations between the subgroups of S3 can be described by the

following diagram.

S4

rrrrrrrrrrr

LLLLLLLLLLL

VVVVVVVVVVVVVVVVVVVVVVV

σ0, σ1

LLLLLLLLLLσ0, σ3 σ0, σ4

rrrrrrrrrrσ0, σ2, σ5

hhhhhhhhhhhhhhhhhhhhhh

σ0

If two subgroups are directly connected by a straight line, the sub-

group in the lower position is contained in the one in the upper position.

The diagram makes it easy to see the relations of subgroups at a glance.

Serious readers may try to write all elements of S4 and as many as

possible subgroups and draw a diagram of subgroups.

You may probably notice that this two line notation for permuta-

tions is not economic. The first line is really not necessary. To be

worse, in S9 a permutation of “swapping 1 and 2” is denoted by[1 2 3 4 5 6 7 8 9

2 1 3 4 5 6 7 8 9

],

which is hardly bearable. This suggests the following concepts.

Definition 1.3.4. Let i1, i2, . . . , id be d distinct objects in 1, 2, . . . , n.Let σ be an element in Sn such that

σ(i1) = i2, σ(i2) = i3, . . . , σ(id) = i1

and σ(i) = i for all i /∈ i1, i2, . . . , id. Then σ is called a d-cycle,

denoted by (i1i2 · · · id). The notation of a cycle is not unique. For

instance, (i2i3 · · · idi1) and (i1i2 · · · id) are the same cycle. Two cycles


are called disjoint if every object in the first cycle does not appear in

the second one. For instance, (142) and (36) are disjoint while (261)

and (3245) are not.

A 2-cycle is called a transposition.

It is not hard to see that every permutation can be expressed as the

product of mutually disjoint cycles. For example,[1 2 3 4 5 6

3 5 6 4 2 1

]= (136)(52).

Although the cycle notation is not as straightforward as the table

notation, we will soon see that in some occasions it is indispensable.

One thing we should mention is: (135)(52) can denote an element in S5

as well as an element in any Sn with n > 5. This ambiguity is tolerable,

since the value of n is usually clear from the context.

The multiplication of two permutations under the cycle notation can

be accomplished by changing them into the table notion first. This is

not necessary. One may use his (or her) favorite method to write out the

result directly from the cycle notation. For example, let σ = (152)(34),

τ = (35)(41). Then στ = (152)(34)(35)(41) = (132)(45).

Proposition 1.3.5. Every permutation can be written as the prod-

uct of transpositions, not necessarily disjoint.

Proof. It suffices to show that every cycle can be written as the

product of some transpositions.

It is easy to verify that (i1i2 · · · id) = (i1i2)(i2i3) · · · (id−2id−1)(id−1id).

Let σ ∈ Sn. Let r be the number of elements in the set

(i, j)|1 ≤ i < j ≤ n, σ(i) > σ(j).

In other words, r is the number of pairs out of order in the sequence

σ(1), σ(2), · · · , σ(n). The permutation σ is called an even permutation

if r is even, otherwise it is called an odd permutation. The set of all

even permutations in Sn is denoted by An.


Example 1.3.6. Let

σ =

[1 2 3 4 5

3 1 5 4 2

].

The sequence 3, 1, 5, 4, 2 has five pairs out of order. Hence σ is an odd

permutation.

Lemma 1.3.7. Let σ ∈ Sn and let τ = (ij) be a transposition with

1 ≤ i < j ≤ n. If σ is an even permutation then στ is an odd permu-

tation. If σ is an odd permutation then στ is an even permutation.

Proof. First assume that j = i+ 1. Express σ as[1 2 · · · i i+ 1 · · · n

σ(1) σ(2) · · · σ(i) σ(i+ 1) · · · σ(n)

].

Then

σ (i, i+ 1) =

[1 2 · · · i i+ 1 · · · n

σ(1) σ(2) · · · σ(i+ 1) σ(i) · · · σ(n)

].

One may observe that στ has one more (or less) pair out of order than

σ. Hence the multiplication of τ from the right changes the parity of σ.

For the general case 1 ≤ i < j ≤ n, it follows from the equality

(ij) = (i, i+ 1)(i+ 1, i+ 2) · · · (j − 2, j − 1)

(j − 1, j)(j − 2, j − 1) · · · (i+ 1, i+ 2)(i, i+ 1)

and what we have proved that σ (ij) and σ have different parity.

Corollary 1.3.8. For any natural number n > 1, the number of

even permutations in Sn(n > 1) is equal to that of odd permutations.

Proof. Let A and B denote the sets of all even permutations and

odd permutations respectively. By Lemma 1.3.7 the map σ 7→ σ · (12)

is a one to one correspondence from A to B. Therefore A and B contain

the same number of elements.


Corollary 1.3.9. A permutation is even (odd resp.) if and only

if it can be written as a product of even (odd resp.) number of trans-

positions.

Proof. This follows from Proposition 1.3.5 and 1.3.7 immediately.

Permutation groups appear in applications in various forms. One

important application is the symmetry of a geometric figure.

Let Γ be a geometric figure in a Euclidean space. The set of all rigid

transformations of Γ to itself under composite operation form a group,

called the group of symmetries of Γ. Here a rigid transformation is

a bijection from Γ to Γ which is a composite of translations, rotations

and reflections.

As plane figures, the groups of symmetries of the following figures

have order 2 and 1 respectively.

@@

@

@@

@

The group of symmetries of a geometric figure reflects how sym-

metric the figure is. The figures in the examples above are not quite

symmetric compared with a circle, whose group of symmetries is an

infinite group.

The group of symmetries of a regular polygon of n sides is denoted

by Dn, called a dihedral group.

We use S = 1, 2, . . . , n to denote the set of vertices of a regular

polygon of n sides, as illustrated in Figure 1. Since a rigid transforma-

tion carries vertices to vertices and once the images of all vertices are

known the rigid transform is determined, Dn is a subgroup of Sn. In

other words, Dn is a permutation group.

Let us denote by σ the rotation of the polygon by 2π/n in the

counterclockwise direction. Choose an axis of symmetry such as the

x-axis in Figure 1. Denote the reflection across this axis by τ. Then

σ, τ ∈ Dn. It is obvious that o(σ) = n and o(τ) = 2. Under cycle


notation of permutation groups, we may write σ = (123 · · ·n) and

τ = (1n)(2, n− 1) · · · .

-

6

@@

@@@

@@

@@@

OX

Y

1

23

4

6

5

7

8

Figure 1 Figure 2

!!!!!!!!!!

SS

SS

@@

@@

@@

@@@

1

2 3

4

It is easy to verify that

στ = τσn−1.

Hence Dn is not abelian.

Every rigid transformation carries the adjacent vertices 1, 2 into two

adjacent vertices. The transformation is determined by the images of

1 and 2, which can be enumerated as

(1, 2), (2, 3), . . . , (n− 1, n), (n, 1),

(2, 1), (3, 2), . . . , (n, n− 1), (1, n).

This tells us that |Dn| = 2n.

Let’s investigate the group of symmetries of a regular tetrahedron.

Let 1, 2, 3, 4 denote the vertices of the tetrahedron, as shown in Figure


2. It is easy to see that every permutation of the vertices determines a

rigid transformation of the tetrahedron. Hence the group of symmetries

of tetrahedron is S4.

Exercises

1. Find all natural numbers n such that there is an element of order

n in S7.

2. Find an element of order 20 in S9. Show that there is no element

of order 18 in S9.

3. Show that the group of symmetries of a non-square rectangle

contains four elements.

4. Without explicit enumeration show that the groups of symme-

tries of a cube and a regular octahedron have the same order. Do the

groups of symmetries for regular dodecahedron and icosahedron have

the same number of elements?

1.4 Cosets

Definition 1.4.1. Let H be a subgroup of a group G. Let a be an

element of G. Denote aH = ah|h ∈ H and Ha = ha|h ∈ H. The

sets aH and Ha are called a left coset and a right coset of H in G

respectively.

Left cosets and right cosets are subsets of the group G, but not

subgroups in general. The subgroup H itself is a left coset as well as a

right coset, since H = 1H = H1.

Example 1.4.2. Let G = GL2(R) and let H be the subgroup of G

consisting of all 2× 2 invertible upper-triangular real matrices. Let

g =

[1 0

1 1

].

Then

1.4. COSETS 19

gH =[1 0

1 1

][a b

0 c

] ∣∣∣a 6= 0, c 6= 0

=[a b

a b+ c

] ∣∣∣a 6= 0, c 6= 0

=[a b

a d

] ∣∣∣a 6= 0, d 6= b.

Hg =[a b

0 c

][1 0

1 1

] ∣∣∣a 6= 0, c 6= 0

=[a+ b b

c c

] ∣∣∣a 6= 0, c 6= 0

=[d b

c c

] ∣∣∣c 6= 0, d 6= b.

Note that gH 6= Hg in this example.

Example 1.4.3. H = e, (123), (132) is a subgroup of S3. (12)H =

(12), (23), (13) is a left coset of H in S3. H(12) = (12), (13), (23)is a right coset. Thus (12)H = H(12). In next section we will see that

this phenomenon is not coincidental.

Example 1.4.4. In the dihedral group Dn, let σ be the rotation by

2π/n in counterclockwise direction. Let τ be the reflection across one

axis of symmetry.

Since o(σ) = n, the subset

1, σ, σ2, . . . , σn−1

is a cyclic subgroup of Dn consisting of all rotations. The reflection τ

is not in H. Hence

τ, στ, σ2τ, . . . , σn−1τ

is a right coset of H which is different from H. Moreover, none of the

elements in this coset is a rotation. HenceH∩Hτ = ∅. Since |Dn| = 2n,


the dihedral group Dn is a disjoint union of H and Hτ. In particular

σ, τ generate Dn. In this way, the group Dn is better understood.

Now let us explore basic properties of cosets.

Proposition 1.4.5. Let H be a subgroup of a group G. For any

element a, b ∈ G, the map f : aH → bH, g 7→ ba−1g is a bijection from

aH to bH.

Proof. It is easy to see that this map is well-defined. Define a map

φ : bH → aH, g 7→ ab−1g. Then φ f and f φ are identity maps.

Proposition 1.4.6. Two cosets aH and bH are equal if and only

if a−1b ∈ H.

Proof. Assume that aH = bH. Then b ∈ aH. Hence b = ah for

some h ∈ H, which implies a−1b = h ∈ H.Conversely, assume that a−1b ∈ H. Then h = a−1b ∈ H. Hence

b = ah ∈ aH, which implies bH ⊆ aH. On the other hand, the equality

b−1a = (a−1b)−1 ∈ H implies aH ⊆ bH. Therefore aH = bH.

This proposition provides a convenient way to check whether two

left cosets are equal.

Corollary 1.4.7. If aH ∩ bH 6= ∅, Then aH = bH.

Proof. Assume that g ∈ aH ∩ bH. Then g = ah = bh′ for some

h, h′ ∈ H. Hence a−1b = hh′−1 ∈ H. It follows from Proposition 1.4.6

that aH = bH.

Since a ∈ aH for every a ∈ G, every element of G belongs to at least

one left coset of H. If follows from Corollary 1.4.7 that G is the disjoint

union of some left costs. In other words, the left cosets of H in G give

a partition of G. The number r of the left cosets in this partition is

called the index of H in G, denoted by (G : H). This number is a

natural number or ∞.

Similarly, the right cosets of H give another partition of G, which

might be different from the partition into left cosets. However the

number of right cosets in this partition is also equal to (G : H). In

1.4. COSETS 21

fact, it is easy to check that the map aH 7→ Ha−1 gives a one to one

correspondence from the set of left cosets of H to the set of right cosets

of H.

Theorem 1.4.8 (Lagrange). Let H be a subgroup of a finite group

G. Then

|G| = |H|(G : H).

Proof. This is because every left coset of H contains |H| elements

due to Proposition 1.4.5.

Lagrange’s theorem tells us that the number of elements in any

subgroup of a finite group G always divides the number of elements in

G.

Corollary 1.4.9. Let a be an element in a group G of order n.

Then o(a)||G|.

Proof. This is because |〈a〉| = o(a).

Corollary 1.4.10. If the order of a group is a prime number then

this group is a cyclic group.

Proof. Denote p = |G|. Choose any a ∈ G such that a 6= 1. Then

o(a) > 1. It follows from 1.4.9 that o(a)|p. Since p is a prime number,

o(a) = p. Hence G = 〈a〉.

Corollary 1.4.11. Let a be an element of a finite group G with

n = |G|. Then an = 1.

Proof. Corollary 1.4.9 implies n = o(a)m for some integer m.

Hence an = (ao(a))m = 1m = 1.

The concept of coset is a special case of the so called “equivalence

class”. In order to deepen our understanding of the partition of a group

into left cosets let us recall equivalence relations and equivalence classes

in the set theory.


Let S be a set. A relation ∼ between the elments of S is called an

equivalence relation if the following three conditions hold:

1. (reflexivity) a ∼ a for every a ∈ S;

2. (symmetry) a ∼ b if and only if b ∼ a;

3. (transitivity) a ∼ b and b ∼ c imply a ∼ c.

For example, although the relation “≤” in the set of real numbers

satisfies reflexivity and transitivity, but it does not satisfy symmetry.

Hence it is not an equivalence relation. The relation of similarity in the

set of all triangles in the Euclidean plane is an equivalence relation. The

relation of similarity in the set of n× n matrices is also an equivalence

relation.

Let ∼ be an equivalence relation on a set S. For any a ∈ S, denote

[a] = x ∈ S|x ∼ a, called an equivalence class represented by a.

Let b be another element in S. If [a]∩ [b] 6= ∅, then c ∈ [a]∩ [b] for some

c ∈ S, i.e.,c ∼ a, c ∼ b. It follows from symmetry and transitivity of ∼that a ∼ b, which implies [a] = [b]. This means that the equivalence

classes give a partition of S.

Let H be a subgroup G. For any a, b ∈ G, define a ∼ b if a−1b ∈ H.It is easy to verify that ∼ is an equivalence relation on G. According to

Proposition 1.4.6 an equivalence class under this relation is a left coset

of H in G.

Exercises

1. Let H = e, (12) ⊂ S3. List all left cosets and right cosets of H.

2. Let H be a subgroup of a group G and let n = (G : H) < ∞.

Determine whether the following statements hold. Prove it or disprove

it by a counterexample.

1) if a ∈ G then an ∈ H;

2) if a ∈ G then there is a natural number k with k ≤ n such that

ak ∈ H.

3. Let H and K be subgroups of a group G such that (G : H) <∞.

Show that (K : H ∩K) ≤ (G : H).

1.5. NORMAL SUBGROUPS AND QUOTIENT GROUPS 23

4. Let K be a nonempty subset of a group G. Denote gK = gk|k ∈K for any g ∈ G. Assume that aK and bK are either identical or

disjoint for any a, b ∈ G. Show that K is a left coset of some subgroup

of G.

5. LetH1, H2 be two different subgroups of a group G. Let a1H1 and

a2H2 be two left cosets of H1 and H2 respectively. Show a1H1 6= a2H2.

Give an example of two different subgroups H1, H2 in the symmetric

group S3 and a1, a2 ∈ S3 such that a1H1 = H2a2.

1.5 Normal Subgroups and Quotient Groups

Let H be a subgroup of a group G. Let Γ denote the set of all left

cosets of H in G. For an element aH ∈ Γ, a is a representative of aH.

It is not unique. Every element in aH can be a representative of aH.

Let us try to use the binary operation of G to equip Γ with an

operation under which Γ become a new group.

For aH, bH ∈ Γ, is it possible to define the product of aH and

bH to be (ab)H? This operation makes sense only if the left coset

(ab)H does not depend upon the choices of the representatives a and

b. Let a′ and b′ be two other representatives of aH and bH respec-

tively, i.e., a′ = ah, b′ = bk for some h, k ∈ H. In order that (ab)H =

(a′b′)H the element (ab)−1(a′b′) must be in H by Proposition 1.4.6, i.e.,

b−1a−1ahbk = b−1hbk ∈ H should hold for any b ∈ G and h, k ∈ H.

This is equivalent to saying the b−1hb ∈ H for any b ∈ G, h ∈ H.This suggests the following definition.

Definition 1.5.1. Let H be a subgroup of a group G. If b−1hb ∈ Hfor any b ∈ G, h ∈ H, then H is called a normal subgroup of G,

denoted by H G.

Assume that HG. By the analysis at the beginning of this section

the binary operation (aH)(bH) = (ab)H on the set Γ is meaningful. It

is straightforward to verify that Γ becomes a group under this oper-

ation, called the quotient group of G by H, denoted by G/H. It is

obvious that |G/H| = (G : H).


Similar discussion applies to the right cosets as well. Can we obtain

another “quotient group”? The following property of normal subgroups

answers this question.

Proposition 1.5.2. Let H be a subgroup of a group G. Then HG

if and only if aH = Ha for any a ∈ G.

Proof. Assume that H G. Then ah = (aha−1)a ∈ Ha for any

h ∈ H. Hence aH ⊆ Ha. For the same reason Ha ⊆ aH. Therefore

aH = Ha.

Conversely assume that aH = Ha for any a ∈ G. For any a ∈ G

and any h ∈ H, the element ha belongs to aH, i.e,. ha = ah′ for some

h′ ∈ H. Hence a−1ha = h′ ∈ H.

Thus when H G there is no distinction between left cosets and

right cosets. We can simply call them cosets.

Keep in mind that an element in the quotient G/H is a coset aH,

which is usually denoted by a or [a], called the coset represented by

the element a ∈ G.The operation in many abelian groups is denoted by “+”. In that

occasion a coset is denoted by a+H.

All elements of G/H can be listed as

a1H, a2H, . . .

in which a−1i aj /∈ H whenever i 6= j and for every a ∈ G there is

some i such that a−1ai ∈ H.

Example 1.5.3. • SLn(K) GLn(K). But the subgroup of

GLn(K) consisting of upper triangular matrices is not a normal

subgroup of GLn(K).

• H = cIn|c ∈ K, c 6= 0 is a normal subgroup of GLn(K). The

quotient group GLn(K)/H is denoted by PGLn(K), called pro-

jective general linear group.

• LetH = id, (123), (132) ⊂ S3.We know thatHS3 in Example

1.4.3. The quotient group S3/H contains two elements H and

(12)H, of which the first one is the identity element.


Example 1.5.4. K = id, (12) is a subgroup of S3. The left coset

of K in S3 are

K = eK,

(13)K = (13), (123),

(23)K = (23), (132).

The set ab|a ∈ (13)K, b ∈ (23)K = (132), (23), (12), id is not a left

coset of K.

This example shows that for a non-normal subgroup we can not

define a natural binary operation in the set of left cosets.

Example 1.5.5. • Trivial subgroups are always normal.

• All subgroups of an abelian group are normal.

• Assume that HG and K is a subgroup of G containing H. Then

H K.

Example 1.5.6 ( an important example). Let n be a natural num-

ber greater than 1. Let nZ be the set of all integers divisible by n. Then

Z/nZ is a cyclic group of order n, whose elements can be enumerated

as

0, 1, 2, . . . , n− 1

in which 1 can be chosen to be the generator. Since the group Z/nZwill be frequently used, it is abbreviated as Zn.

Example 1.5.7. The group Q of all rational numbers under addi-

tion is an abelian group. The quotient group Q/Z is an infinite group

of which every element has finite order.

Let H be a subgroup of a group G. Denote a−1Ha = a−1ha|h ∈H. It is easy to verify that a−1Ha is a subgroup of G, called a conju-

gate subgroup of H. Its order is equal to |H|. but a−1Ha = H does

not hold in general. Readers may verity that HG if and only if every

conjugate subgroup of H is equal to H.

In Example 1.4.3 we see that the left cosets id, (123), (132) in

S3 coincide with the right cosets. We may explain it without explicit

computation by the following observation.


Proposition 1.5.8. Let H be a subgroup of a group G such that

(G : H) = 2. Then H G.

Proof. It suffices to show that gH = Hg holds for every g ∈ G.

If g ∈ H, then gH = H = Hg. Hence we may assume that g /∈ H.

Then gH is a left coset different from H. Since (G : H) = 2, there are

only two left cosets of H in G. Hence gH = G\H. For the same reason

Hg = G\H. Therefore gH = Hg.

Example 1.5.9. The center of a group G is always a normal sub-

group of G.

Example 1.5.10. The intersection of arbitrarily many normal sub-

groups of a group is a normal subgroup.

Example 1.5.11. Let S be a subset of a group G. Assume that

g−1ag ∈ S for any g ∈ G, a ∈ S. Then 〈S〉G.

Proof. Let b be an arbitrary element in 〈S〉. Then

b = ae11 · · · aen

n ,

in which ai ∈ S, ej = ±1. Hence

g−1bg = (g−1a1g)e1 · · · (g−1ang)

en .

By the given condition we have g−1bg ∈ 〈S〉.

As a special case, let S = a−1b−1ab|a, b ∈ G. Since

g−1(a−1b−1ab)g = (g−1ag)−1(g−1bg)−1(g−1ag)(g−1bg) ∈ S,

S satisfies the above condition. Denote [G,G] = 〈S〉, called the com-

mutator subgroup of G. It is a normal subgroup of G.

Example 1.5.12. Let H be a subgroup of a group G. Let

NG(H) = g ∈ G|g−1Hg = H.


ThenNG(H) is a subgroup ofG containingH such thatHNG(H). For

any subgroupK ofG containingH andHK, the relationK ⊆ NG(H)

holds. The subgroup NG(H) of G is called the normalizer of H in G.

Proof. Assume that g ∈ NG(H). Then g−1Hg = H,Hence gHg−1 =

H, which implies that g−1 ∈ NG(H). Hence NG(H) is closed under the

inverse operation.

Let a, b ∈ NG(H). Then (ab)−1H(ab) = b−1(a−1Ha)b = b−1Hb = H.

Hence ab ∈ NG(H), i.e., NG(H) is closed under multiplication. Hence

NG(H) is a subgroup of G. Obviously it contains H.

It follows from the definition of NG(H) that H NG(H).

For an arbitrary a ∈ K, a−1Ha = H sinceHK. Hence a ∈ NG(H).

This shows that K ⊆ NG(H).

Caution: H1 H2 G does not imply H1 G.

Proposition 1.5.13. For any group G the quotient group G/[G,G]

is an abelian group. If H G and G/H is an abelian group then

[G,G] ⊆ H.

Proof. Let a be an element in G/[G,G] represented by a ∈ G.

Then

a−1b−1ab = a−1b−1ab = 1.

Hence ab = ba for any a, b ∈ G. This shows that G/[G,G] is abelian.

Denote by [a] and element in G/H represented by a ∈ G. The

equality [a−1b−1ab] = [a]−1[b]−1[a][b] = [1] holds for any a, b ∈ G. Hence

a−1b−1ab ∈ H, which implies that [G,G] ⊆ H.

Definition 1.5.14. A groups is called a simple group if it contains

at least two elements and does not have a nontrivial normal subgroup.

Exercises

1. Show that the intersection of arbitrarily many normal subgroups

of a a group is normal.

2. Show that e, (12)(34), (13)(24), (14)(23) is a normal subgroup

of S4.


3. Let r be a natural number. Assume that a group G has only one

subgroup H of order r. Show that H G.

4. Let G be a group of order 120 and let H be a subgroup of G with

|H| = 24. Assume that there exist a, b ∈ G such that aH = Hb 6= H.

Show that H G.

5. Give a counterexample to show that H1 H2 G does not imply

H1 G.

6. Let H,K be two subgroups of a group G. Let HK = ab|a ∈H, b ∈ K.

1. Assume that H G. Show that HK is a subgroup of G.

2. Let G = S3, H = e, (12), K = e, (13). Show that HK is not

a subgroup of G.

3. Assume that H G,K G. Show that HK G.

7. Let H be a normal subgroup of a finite group G and a ∈ G. Let

n be the order of aH in G/H and let m be the order of an in G. Show

that mn is the order of a in G.

8. Let N be a cyclic subgroup of a group G. Assume that N G.

Show that ab = ba for every a ∈ N and every b ∈ [G,G].

9. Let G be a non-abelian finite group. Let Z be the center of G.

Show that 4|Z| ≤ |G|.

10. Let H1, . . . , Hn be subgroups of a group G. Assume that aiaj =

ajai for any 1 ≤ i < j ≤ n and any ai ∈ Hi, aj ∈ Hj. If every element

of G can be expressed as b1b2 · · · bn, in which bi ∈ Hi. Show that Hi G

for every i.

11. Assume that a finite group G has exactly two subgroup H,K

of order n and G is generated by H ∪K. Show that H G,K G.

1.6 Alternating Groups

In this section we will study more properties of the symmetric group Sn

and its subgroups. Recall that An denotes the subset of Sn consisting

of all even permutations.

1.6. ALTERNATING GROUPS 29

Theorem 1.6.1. An is a normal subgroup of Sn and (Sn : An) = 2

for all n > 1.

Proof. Assume that σ, τ ∈ An. They can be expressed as products

of even number of transpositions by Corollary 1.3.9. Hence στ can

also be expressed as a product of even number of transpositions. This

implies στ ∈ An. By the same argument An is closed under the inverse

operation. Hence An is a subgroup of Sn.

Since the map σ 7→ (12)σ is a bijection from An into the set (12)An

of odd permutations and every permutation is either even or odd, we

have Sn = An ∪ (12)An. Hence (Sn : An) = 2. It follows from Proposi-

tion 1.5.8 that An Sn.

The groups An is called the alternating group of n objects. By

the way, the theorem implies that Sn is not a simple group for n > 2.

The usefulness of the cycle notation for permutations is justified by

the following proposition and its corollaries.

Proposition 1.6.2. Let σ = (i1 · · · ir) be an r-cycle and τ ∈ Sn.

Then τστ−1 is also an r-cycle. More precisely,

τστ−1 = (τ(i1) · · · τ(ir)).

Proof. If 1 ≤ i ≤ n does not appear in τ(i1), . . . , τ(ir) then τ−1(i)

does not appear in i1, . . . , ir. Hence

τστ−1(i) = ττ−1(i) = i,

for every i that does not appear in τ(i1), . . . , τ(ir).

The proposition follows from τστ−1(τ(ij)) = τ(ij+1) for all 1 ≤ j <

r and τστ−1(τ(ir)) = τ(i1).

Corollary 1.6.3. Let σ = (a1 · · · ar) · · · (c1 · · · cs) be a product of

cycles. Then

τστ−1 = (τ(a1) · · · τ(ar)) · · · (τ(c1) · · · τ(cs)).

The following proposition shows that Sn and An can be generated

by some simple permutations.


Proposition 1.6.4.

Sn = 〈(12), (13), . . . , (1n)〉 = 〈(12), (12 · · ·n)〉.

An = 〈(123), (124), . . . , (12n)〉.

Proof. The equality (1i)(1j)(1i) = (ij) and Proposition 1.3.5 im-

ply

Sn = 〈(12), (13), . . . , (1n)〉.

It follows from Proposition 1.6.2 that (12 · · ·n)(i, i + 1)(12 · · ·n)−1 =

(i+ 1, i+ 2) for all 1 ≤ i ≤ n− 1. Hence

Sn = 〈(12), (12 · · ·n)〉.

Every element in An can be written as

(1m1)(1m2) · · · (1m2r).

It follows from

(1j)(1i) = (1ij) = (12j)(12i)−1

that

An = 〈(123), (124), . . . , (12n)〉.

Theorem 1.6.5. The alternating group An is a simple group if n ≥5.

Proof. Let K be a normal subgroup of An with |K| > 1. We need

to show that K = An.

First assume that K contains a 3-cycle (rst). For any i > 2, Propo-

sition 1.6.2 implies that σ(rst)σ−1 = (12i), where σ is any permutation

satisfying σ(r) = 1, σ(s) = 2, σ(t) = i. If σ /∈ An, choose u, v such

that 1 ≤ u < v ≤ n and u /∈ r, s, t, v /∈ r, s, t. Replace σ by

σ (uv). Then this new σ is in An and σ(rst)σ−1 = (12i) still holds.

Hence there is always some σ ∈ An such that σ(rst)σ−1 = (12i). So K

1.6. ALTERNATING GROUPS 31

contains (123), (124), . . . , (12n). If follows from Proposition 1.6.4 that

K = An. Hence K = An if K contains a 3-cycle.

For any α ∈ K and any β ∈ An, the element βαβ−1 is in K since

K An. Hence βαβ−1α−1 ∈ K. It suffices to find suitable α ∈ K and

β ∈ An such that βαβ−1α−1 becomes a 3-cycle.

Since |K| > 1, there is an element α in K which is not the identity

element. If α is a 3-cycle, then the proof is done. Assume that α is not

a 3-cycle. Write α as a product of mutually disjoint cycles. At least

one of the following six cases occurs:

1) α contains a cycle of length greater than 4;

2) α contains a 4-cycle;

3) α contains two 3-cycles;

4) α contains four transpositions;

5) α is the product of two transpositions;

6) α is the product of one 3-cycle and two transpositions.

Let’s treat these cases one by one.

1) Without loss of generality assume that α = (12 · · ·m)α1, in

which m ≥ 5 and α1 is the product of mutually disjoint cycles that

do not contain any of the objects 1, 2, . . . ,m. Let β = (345). Then

βαβ−1α−1 = (134) (for m = 5) or (634) (for m > 5).

2) Assume that α contains a cycle (1234). Since α is an even permu-

tation, it contains at least one more cycle, say (56 · · · ). Let β = (345).

Then βαβ−1α−1 = (15346). This brings us back to Case 1).

3) Assume that α contains two cycles (123), (456). Let β = (124).

Then βαβ−1α−1 = (12534), reduced to Case 1).

4) Assume that α contains four transpositions (12), (34), (56), (78).

Let β = (135). Then βαβ−1α−1 = (135)(264), reduced to Case 3).

5) Assume that α = (12)(34). Let β = (135). Then βαβ−1α−1 =

(13542), reduced to Case 1).

6) Assume that α = (12)(34)(567). Let β = (135). Then βαβ−1α−1 =

(135)(264), reduced to Case 3.

Exercises

1. Let G be a subgroup of the symmetric group S6. Assume that G

has an element of order 6. Show that G contains a normal subgroup H


such that (G : H) = 2.

2. Show that A4 is not a simple group.

3. Let S be the group of all bijections from N to N under the

composite operation. Let

G = σ ∈ S| there exists n > 0 such that σ(i) = i ∀ i > n

and the number of pairs out of order is even .

Show that G is a simple group.

4. Let G be a subgroup of Sn with |G| > 2. Assume that G is a

simple group. Show that G ⊆ An.

5. Let G be a normal subgroup of Sn(n > 1) satisfying (Sn : G) = 2.

Show that G = An.

6. Compute the commutator subgroup [Sn, Sn] of Sn. Then compute

the commutator subgroup of [Sn, Sn]. (Hint: Discuss the cases n ≤2, n = 3, n = 4, n ≥ 5.)

7. Two elements x, y in a group G are called conjugate if there

exists g ∈ G such that g−1xg = y.

i) Show that the conjugacy in G is an equivalence relation;

ii) Show that every element in Sn is conjugate to its inverse;

iii) Find an element in A4 which is not conjugate to its inverse.

8. Let γ = (a1, a2, · · · , an) be an n-cycle in Sn and β ∈ Sn. Assume

that βγ = γβ. Show that there is some integer k such that β = γk.

1.7 Homomorphisms of Groups

Definition 1.7.1. Let G1, G2 two groups with identity elements e1

and e2 respectively. A map f : G1 → G2 is called a homomorphism

if f(ab) = f(a)f(b) for any a, b ∈ G1.

If f is a homomorphism, then f(e1) = e2 and f(a−1) = f(a)−1 for

any a ∈ G1.

For a homomorphism f, denote Ker(f) = a ∈ G1|f(a) = e2. This

is a normal subgroup of G1, called the kernel of f. The homomorphism

1.7. HOMOMORPHISMS OF GROUPS 33

f is injective if and only if Ker(f) = e1. An injective homomorphism

is called a monomorphism.

Denote Im(f) = f(a)|a ∈ G1. This is a subgroup of G2, called the

image of f. The homomorphism f is surjective if and only if Im(f) =

G2. A surjective homomorphism is called an epimorphism.

A bijective homomorphism is called an isomorphism.

If there is an isomorphism between two groups G1, G2 then we say

that G1 and G2 are isomorphic, denoted by G1∼= G2.

An isomorphism from G to itself is called an automorphism.

That two groups are isomorphic means that they have the same

group structure, which means that they are identical if all other proper-

ties of the underlying sets are ignored. Sometimes two different groups

are considered to be the same if they are isomorphic. For instance, all

groups with two elements are isomorphic, so we can say that there is

only one group of order two. If we say that there are only two groups

of order 6, it means that there exist two non-isomorphic groups G1 and

G2 with |G1| = |G2| = 6 and any group of order 6 is isomorphic to

either G1 or G2.

If G1∼= G2 then there is an isomorphism f : G1 → G2, but such an

isomorphism f is not necessarily unique.

Let j : H → G be a monomorphism. Then H is isomorphic to

the subgroup j(H) of G. We may regard H as a subgroup of G under

the monomorphism j, or simply say that H is a subgroup of G, if no

confusion will be caused.

Example 1.7.2. The map f : G1 → G2, g 7→ e2 for all g ∈ G1 is

a homomorphism with Ker(f) = G1, Im(f) = e2. Such a homomor-

phism is called a trivial homomorphism.

Example 1.7.3. Let H be a subgroup of a group G. Then the

injective map i : H → G, h 7→ h is a monomorphism.

Example 1.7.4. Assume that H G. Then π : G → G/H, a 7→ a

is an epimorphism, called the canonical homomorphism from G to

G/H.

Example 1.7.5. Let K be a number field. Denote by K∗ the group


of all nonzero elements in K under multiplication. Then

det : GLn(K) → K∗, A 7→ |A|

is an epimorphism. Here |A| is the determinant of A.

Example 1.7.6. Let

G =[1 x

0 1

] ∣∣∣x ∈ R.

Then G is a group under multiplication. The map

f : G→ R,

[1 x

0 1

]7→ x

is an isomorphism from G to the additive group of real numbers.

Example 1.7.7. Let g be an element in a group G. Then the map

f : G→ G, a 7→ g−1ag

is an automorphism of G, called an inner automorphism.

For example, if G = GLn(K) then the inner automorphism is the

similarity in linear algebra.

To show that two groups are isomorphic requires the construction

of a homomorphism which is both injective and surjective. The veri-

fication can be straightforward or difficult depending on the problem.

To show that two groups are not isomorphic, it is enough to find a

property that one group possesses but the other group does not. A

simple example is given below.

Example 1.7.8. Although Z/6Z and S3 have the same order 6, but

they are not isomorphic since the former is abelian while the latter is

not.

Theorem 1.7.9 (Fundamental Theorem of Homomorphisms). Let

f : G1 → G2 be a homomorphism of groups. Then

1) Ker(f) G1;


2) G1/Ker(f) ∼= Im(f).

Proof. We have mentioned before that Ker(f)G1. The verifica-

tion is easy and is left as an exercise.

In order to prove the second statement we construct a map p :

G1/Ker(f) → Im(f) as follows. For an arbitrary aKer(f) ∈ G1/Ker(f),

define p(aKer(f)) = f(a). We need to verify that this definition makes

sense, i.e., p(aKer(f)) does not depend upon the choice of the rep-

resentative a. Assume that a′ ∈ aKer(f) is another representative of

aKer(f). Then a′ = ah for some h ∈ Ker(f). Thus f(a′) = f(a)f(h) =

f(a)e2 = f(a). Hence the map p is well-defined.

It is evident that p is surjective. It remains to show that p is injec-

tive. Assume that aKer(f), bKer(f) ∈ G/Ker(f) satisfy f(a) = f(b).

Then f(a−1b) = e2. Hence a−1b ∈ Ker(f), which implies aKer(f) =

bKer(f).

The following example displays a typical application of Theorem

1.7.9.

Example 1.7.10. Let n ∈ N. Every cyclic group of order n is iso-

morphic to Zn. Hence two cyclic groups are isomorphic if and only if

they have the same order.

Proof. Let G be a cyclic group of order n with generator g. Define

a map

f : Z → G,m 7→ gm.

Then f is an epimorphism with nZ as its kernel. It follows from The-

orem 1.7.9 that

Zn∼= G.

Hence all cyclic finite groups of the same order are isomorphic. It is

easy to see that every infinite cyclic group is isomorphic to Z.

From this example we see that in order to prove that G/H is isomor-

phic to another group K we need to construct a surjective epimorphism

f : G → K such that H = Ker(f). The proof of this style is usually

concise and convincing.


Lemma 1.7.11. Let f : G1 → G2 be a homomorphism of groups.

Let H2 be a subgroup of G2. Then f−1(H2) = g ∈ G1|f(g) ∈ H2 is a

subgroup of G1 containing Ker(f).

Proof. Assume that a, b ∈ f−1(H2). Then f(a), f(b) ∈ H2. Thus

f(ab−1) = f(a)f(b)−1 ∈ H2. Hence ab−1 ∈ f−1(H2). Therefore f−1(H2)

is a subgroup of G1.

Assume that a ∈ Ker(f). Then f(a) = e2 ∈ H2. So a ∈ f−1(H2),

which implies Ker(f) ⊆ f−1(H2).

The lemma can be stated in plain English: the inverse image of any

subgroup is a subgroup containing the kernel.

Proposition 1.7.12. Assume that H G and f : G → G/H is

the canonical homomorphism. Let Γ be the set of all subgroups of G

containing H and let Γ′ be the set of all subgroups of G/H. Then the

map f−1 : Γ′ → Γ, K 7→ f−1(K) is a bijection from Γ′ to Γ.

Proof. By Lemma 1.7.11 the map f−1 : Γ′ → Γ is well-defined.

Construct a map φ : Γ → Γ′ as follows. For any T ∈ Γ define T ′ to be

the set of all cosets aH with a ∈ T. Then T ′ is a subset of G/H. It is

easy to verify that T ′ is a subgroup of G/H. Define φ(T ) = T ′.

The inclusion T ⊆ f−1(T ′) is obvious. If b ∈ f−1(T ′), then f(b) ∈T ′, which means that there is some a ∈ T such that bH = aH. Hence

a−1b ∈ H. Thus b ∈ T, since H ⊆ T. Hence f−1(T ′) = T, which implies

that f−1 φ is the identity map. It is easy to verify that φ f−1 is also

the identity map.

We have seen before that every subgroup of an infinite cyclic group

is an infinite cyclic group. Let us classify the subgroups of finite cyclic

groups. Assume that G = Zn, in which n is a natural number. By

Proposition 1.7.12 there is one-to-one correspondence between the set

of subgroups of G and the set of subgroups of Z containing nZ. A

subgroup mZ of Z contains nZ if and only if m|n. All subgroups of G

are thus determined.


Theorem 1.7.13 (First Isomorphism Theorem). Assume that H

G,N G and H ⊆ N. Then

(G/H)/(N/H) ∼= G/N.

Proof. It is easy to see that H N and N/H G/H. So all the

quotient groups involved in the theorem are well-defined. The inclusion

relations of these groups are described by the following diagram:

G G/H

N

H N/H

Let φ : G → (G/H)/(N/H) be the composite of the canonical

homomorphisms G → G/H and G/H → (G/H)/(N/H). Then φ is

evidently an epimorphism. It is obvious that N ⊆ Ker(φ). Let g ∈Ker(φ). Then gH ∈ N/H. Hence gH = sH for some s ∈ N. So g ∈ N.Hence N = Ker(φ). The theorem follows from Theorem 1.7.9.

Theorem 1.7.14 (Second Isomorphism Theorem). Assume that

H G and K is a subgroup of G. Let

KH = ab|a ∈ K, b ∈ H.

Then KH is a subgroup of G and

KH/H ∼= K/K ∩H.

Proof. It follows from Exercise 6 of Section 5 that KH is a sub-

group of G.

The inclusion relations of the five groups involved are described by

the following diagram:


G

KH

vvvvvvvvv

HHHHHHHHH

H

HHHHHHHHH K

vvvvvvvvv

K ∩HAssume that st ∈ KH. Then s = ab, t = cd in which a, c ∈ K and

b, d ∈ H. Hence st−1 = abd−1c−1 = (ac−1)(cbd−1c−1) ∈ KH. which

implies that KH is a subgroup of G.

It is clear that H KH and K ∩H K.

Define φ : K → KH/H, a 7→ aH. This is an epimorphism with

Ker(φ) = K ∩H. The theorem follows from Theorem 1.7.9.

Exercises

1. Let f : G1 → G2 be a homomorphism of groups. Verify

1) f(e1) = e2, where e1, e2 are the identity elements of G1, G2;

2) f(a−1) = f(a)−1 for any a ∈ G1;

3) Ker(f) G1.

2. Let H be a commutative subgroup of a group G such that n =

(G : H) <∞. Let

b1H, b2H, . . . , bnH

be all left cosets of H in G. Define a map τ : G→ H by the following

rule: For every a ∈ G and every 1 ≤ i ≤ n, the element abi belongs to

exact one bjH, i.e. there exists a unique hi ∈ H such that abi = bjhi.

Define

τ(a) =n∏

i=1

hi.

Show that τ is a homomorphism.

3. Show that the additive group C is not isomorphic to GL2(R).

4. Let G be a finite group. Let N be a normal subgroup of G and

let H be a subgroup of G. Assume that |N | and (G : H) are coprime.

Show that N ⊆ H.

1.8. DIRECT PRODUCT OF GROUPS 39

5. Let G = C−1. In G define a binary operation ab = a+b−ab.Verify that G is an abelian group under this operation and show that it

is isomorphic to the multiplicative group of nonzero complex numbers.

( The problems 6-10 involve automorphisms.)

Let G be a group and let Aut(G) denote the set of all automorphism

of G. Define a binary operation on Aut(G) in the following way. For

σ, τ ∈ Aut(G), define στ to be the composite of σ and τ, i.e., (στ)(a) =

σ(τ(a)) for every a ∈ G.

6. Show that Aut(G) is a group under the operation defined above.

This group is called the group of automorphisms of G.

7. Show that the set consisting of all inner automorphism is a

normal subgroup of Aut(G), denoted by Inn(G).

8. Show that Inn(G) ∼= G/C(G). Here C(G) is the center of G.

9. Let G be an infinite cyclic group. Show that Aut(G) is a group

of order 2.

10. Determine all automorphisms of the additive group Q of rational

numbers.

11. Let N be a normal subgroup of a group G. If a subgroup H of

G satisfies NH = G,N ∩H = e, then H is called a compliment sub-

group of N. Show that all compliment subgroups of N are isomorphic.

12. Let G = Q/Z as in Example 1.5.7. Show that G has a unique

subgroup of order n for every natural number n.

13. Let H be a maximal subgroup of a finite group G, i.e., H 6= G

and there is no subgroup between H and G. Assume that (G : H) is

not a prime number. Show that H is not a normal subgroup of G.

1.8 Direct Product of Groups

So far we have learned two methods to get new groups from old ones:

subgroups and quotient groups. Both methods yield smaller groups. In

this section we introduce the most common construction that combines

more than one groups into a single group.


Recall that in linear algebra the space Rn can be regarded as the

direct sum of n one-dimensional spaces.

Rn = R⊕ · · · ⊕ R= (x1, . . . , xn)|x1, . . . , xn ∈ R.

Let G1, . . . , Gn be groups. Let

G1 × · · · ×Gn = (x1, . . . , xn)|xi ∈ Gi.

For arbitrary (a1, . . . , an), (b1, . . . , bn) ∈ G1×· · ·×Gn, define the prod-

uct

(a1, . . . , an)(b1, . . . , bn) = (a1b1, . . . , anbn).

It is easy to see that G1 × · · · × Gn is a group under this operation,

called the direct product of G1, . . . , Gn, denoted by G1× · · ·×Gn or∏ni=1Gi. Obviously, the identity element of

∏ni=1Gi is (1G1 , . . . , 1Gn)

and the inverse of (a1, . . . , an) is (a−11 , . . . , a−1

n ).

When Gi’s are abelian groups and the operations are denoted by

“+”, the direct product is also called direct sum, denoted by G1 ⊕· · · ⊕Gn or ⊕n

i=1Gi. The addition is given by

(a1, . . . , an) + (b1, . . . , bn) = (a1 + b1, . . . , an + bn)

for any (a1, . . . , an), (b1, . . . , bn) ∈ G1⊕· · ·⊕Gn. An element (a1, . . . , an)

in G1 ⊕ · · · ⊕Gn can also be denoted by a1 ⊕ · · · ⊕ an.

Let G1, . . . , Gn be groups. For each index 1 ≤ i ≤ n there are two

homomorphisms:

ji : Gi → G1 × · · · ×Gn

a 7→ (1G1 , . . . , 1Gi−1, a, 1Gi+1

, . . . , 1Gn)

and

pi : G1 × · · · ×Gn → Gi

(a1, . . . , an) 7→ ai.

Obviously ji is a monomorphism such that ji(Gi) G1 × · · · ×Gn and

1.8. DIRECT PRODUCT OF GROUPS 41

pi is an epimorphism. It is clear that (a1, . . . , an) = j1(a1) · · · jn(an) for

any (a1, . . . , an) ∈ G1 × · · · ×Gn.

Due to the monomorphism ji, we may treat Gi as a subgroup of

G1 × · · · × Gn. For instance, an element in the form (a, 1, . . . , 1) in

G1×· · ·×Gn can be regarded as the element a in G1. As we have seen,

Gi G1 × · · · ×Gn for every 1 ≤ i ≤ n.

Once we know the meanings of direct product and direct sum, the

next task is the decomposition of a group into direct product or direct

sum, which is harder.

In order to understand the structure of a group, we usually start

with its decomposition into direct product (or direct sum) of some

smaller groups. For this purpose we need to find some groupsG1, . . . , Gn

and an isomorphism

φ : G1 × · · · ×Gn → G.

It will be more desirable if every Gi cannot be decomposed further. If

every component Gi is some group that we are familiar with, then we

may consider the structure of G is understood.

How to find these G1, . . . , Gn? By the analysis before, if there exists

an isomorphism

φ : G1 × · · · ×Gn → G

Then all subgroups H1 = φ j1(G1), . . . , Hn = φ jn(Gn) are normal

subgroups of G and aiaj = ajai for any ai ∈ Hi, aj ∈ Hj. So all normal

subgroups of G are the candidates of Gi.

Theorem 1.8.1. Let G,G1, . . . , Gn be groups. Then

G ∼= G1 × · · · ×Gn

if and only if there are subgroups H1, . . . , Hn of G satisfying the follow-

ing three conditions:

1) Hi∼= Gi for each 1 ≤ i ≤ n;

2) aiaj = ajai for any ai ∈ Hi, aj ∈ Hj with i 6= j;

3) the map ψ : H1 × · · · × Hn → G, (a1, . . . , an) 7→ a1 · · · an is

bijective.


Remark 1.8.2. Exercise 1.5.10 shows that Hi G is a consequence

of 2) and 3).

Proof. We first make a remark on the third condition. H1×· · ·×Hn is not a subset of G, although every Hi is a subgroup of G.

Assume that there is an isomorphism φ : G1 × · · · × Gn → G. Let

Hi = φ · ji(Gi). It is clear that the subgroups H1, . . . , Hn of G satisfy

all three conditions.

Conversely assume that the three conditions are satisfied forH1, . . . , Hn.

It suffices to show that ψ is a homomorphism. Let a = (a1, . . . , an), b =

(b1, . . . , bn) be two elements of H1 × · · · ×Hn. Then

ψ(ab) = a1b1a2b2 · · · anbn.

It follows from the second condition that

a1b1a2b2 · · · anbn = a1a2 · · · anb1b2 · · · bn = ψ(a)ψ(b).

Hence ψ is a homomorphism.

In application, the surjectivity of ψ is usually easy to verify. The in-

jectivity can be verified by showing thatHi∩(H1 · · ·Hi−1Hi+1 · · ·Hn) =

e for each i.

Example 1.8.3. A cyclic group of order 6 is the direct product of

a cyclic group of order 2 and a cyclic group of order 3.

Proof. Let G = Z/6Z. Then G has a subgroup H1 = 0, 3 of

order 2 and a subgroup H2 = 0, 2, 4 of order 3. Obviously, H1 and

H2 satisfy all three conditions in the theorem.

Example 1.8.4. The infinite cyclic group is not isomorphic to the

direct product of two proper subgroups.

Proof. Otherwise Z would have two proper subgroups nZ and mZsatisfying the three conditions of the theorem. According the third

condition nZ ∩ mZ = 0 would hold. This is absurd, since nm ∈nZ ∩mZ.

1.9. * AUTOMORPHISMS OF FINITE CYCLIC GROUPS AND THE EULER FUNCTION43

Exercises

1. Let G be a finite group and let G1 = G×G. Assume that G1 has

exactly four normal subgroups. Show that G is not an abelian group.

2. Let G be an abelian group whose operation is denoted by “+”.

Let u, v : G → G be homomorphisms. Define two maps from G to G

by

f(x) = x− v(u(x)), g(x) = x− u(v(x)).

Show that f is surjective if and only if g is surjective.

3. Let H be a subgroup of an abelian group G. Let G = G/H. Are

the following statements true? If true give a proof, otherwise give a

counterexample.

1) If G is a finite cyclic group, then G ∼= G×H;

2) If G is an infinite cyclic group, then G ∼= G×H.

4. Let N1, N2, N3 be normal subgroups of G. Assume that Ni∩Nj =

e, G = NiNj for any 1 ≤ i < j ≤ 3. Show that G is an abelian group

and N1, N2, N3 are mutually isomorphic.

5. Let S, T be two groups, G = S×T. Assume that a subgroup H of

G satisfies the condition SH = G = TH. (Here S and T are considered

to be the subgroups of G.)

i) Show that S ∩H G, T ∩H G;

ii) If S ∩H = e = T ∩H, show that S ∼= T.

iii) If S ∩H = e = T ∩H and H G, show that G is an abelian

group.

1.9 * Automorphisms of Finite Cyclic Groups and

the Euler Function

Let G = 〈a〉 be a cyclic group of finite order n. Then

G = 1, a, a2, . . . , an−1.


Let m be a natural number smaller than n and coprime with n. The

map

σm : G→ G, b 7→ bm

is a homomorphism. Assume ak ∈ Ker(σm). Then amk = 1. Thus n|mk.Since m and n are coprime, so n|k. Hence ak = 1, which implies that

σm is a monomorphism. Since every monomorphism from a finite group

to itself is bijective, σm is bijective. Hence σm ∈ Aut(G).

Conversely assume that σ ∈ Aut(G). Then σ(a) = am for some

m with 0 ≤ m ≤ n − 1. Let d be the greatest common divisor of m

and n. Suppose that d > 1. Let n′ = n/d. Then 1 ≤ n′ < n and

σ(an′) = amn′ = 1. Hence an′ ∈ Ker(σ). This would contradict the

assumption that σ is an isomorphism. Therefore d = 1, i.e., m and n

are coprime.

Let Z∗n denote the set of all natural numbers less than n and co-

prime with n. The above discussion shows that there is a one-to-one

correspondence between Aut(G) and Z∗n. For anym,m′ ∈ Z∗

n, The com-

posite of the automorphisms σm and σm′ is the isomorphism b 7→ bmm′.

Let r be the remainder of mm′ divided by n. Then σmσm′ = σr.

This means that the set Z∗n becomes a group under the multiplica-

tion mm′ = r and

Aut(G) ∼= Z∗n.

Z∗n is an abelian group, but not necessarily a cyclic group. For

example Z∗8 consists of four elements 1, 3, 5, 7. It is easy to verify that

it does not contain an element of order 4.

The function φ(n) = |Z∗n| is an important arithmetic function, called

the Euler function. A few of its values are

φ(2) = 1, φ(3) = 2, φ(4) = 2, φ(5) = 4, . . .

Lagrange’s theorem implies the celebrated Euler’s Theorem in ele-

mentary number theory:

If two natural numbers m and n are coprime, then the remainder

of mφ(n) divided by n is equal to 1.

1.10. GROUP ACTION 45

The following statement is a special case of Euler’s theorem.

Let p be a prime number and let a be an integer not divisible by p.

Then the remainder of ap−1 divided by p is equal to 1.

This result is known as Fermat’s little theorem.

1.10 Group Action

Let G be a group and let S be a set. If there is a map σ : G× S → S

satisfying the following two conditions:

1) σ(e, x) = x for any x ∈ S;

2) σ(g1g2, x) = σ(g1, σ(g2, x)) for any x ∈ S and any g1, g2 ∈ G,then σ is called a (left) action of G on S.

For a fixed g ∈ G, a map from S into itself is defined by x 7→ σ(g, a).

It is more convenient to denote σ(g, a) by ga. We should be aware

that gx is different from the multiplication in a group, since the two

elements g and x involved in the action are contained in two different

sets G and S. The element g is sometimes called an “operator”. Under

the notation gx, the two conditions in the definition can be rewritten

as

1) ex = x for any x ∈ S;

2) (g1g2)x = g1(g2x) for any x ∈ S and any g1, g2 ∈ G.The expression g1g2 on the left hand side of 2) is the product of g1

and g2 in G, while the right hand side is the result to two successive

actions. The second condition guarantees that the notation g1g2x is

meaningful.

A right action of G on S can be defined similarly. It satisfies the

following conditions:

1) xe = x for any x ∈ S;

2) x(g1g2) = (xg1)g2 for any x ∈ S and any g1, g2 ∈ G.The main difference between the left action and the right action is

the difference of the order in which g1g2 acts on x ∈ S. For the left

action, x is acted by g2 followed by g1 while the right action takes the

reverse order.


Proposition 1.10.1. Assume that G has a left action on a set S.

Then for any g ∈ G the map x 7→ gx is a bijection of S onto itself.

Proof. This follows from g−1(gx) = (g−1g)x = ex = x.

The following examples of group actions are easy to verify.

Example 1.10.2. • Every group has a trivial action on any set S.

That is, gx = x, ∀g ∈ G, ∀x ∈ S.• GLn(K) acts on the space of n-dimensional column vectors by

multiplying from the left. This is a left action. GLn(K) acts on the

space of n-dimensional row vectors by multiplying from the right. This

is a right action.

• An action of G on S induces an action of any subgroup of G on

S.

• G has a left action on itself (the latter is treated as a set instead

of a group):

G×G→ G

(g, a) 7→ ga,

called a left translation.

This is an important action, the underlying set of the operator group

is identical to the set on which G acts. The right translation is defined

similarly.

• There is another important left action of a group G on itself:

G×G→ G

(g, a) 7→ gag−1,

called a conjugation. It is not appropriate to denote this action by

ga. Similarly g−1ag is a right action of G on itself. The translation and

conjugation are two frequently used actions in group theory.

• Let H be a subgroup of a group G. Let S be the set of all left

cosets of H in G. Then (g, aH) 7→ (ga)H is a left action of G on S.

• Assume that H G. Then g, aH 7→ (gag−1)H is a left action of

G on the quotient group G/H.


Due to Proposition 1.10.1 a left action of a group G on a finite set

S induces a natural homomorphism φ : G 7→ Sn, where n is equal to

the number of elements in S. Conversely , it is easy to see that every

homomorphism from G to Sn is induced from a left action of G on a

set of n elements. In fancier terminology, a group action is nothing

but a “group representation”. This point of view is useful in dealing

with some problems. A typical example is the so called n! theorem as

follows.

Example 1.10.3. Let H be a subgroup of a finite group G with

n = (G : H) > 1. If |G| > n!, then G is not a simple group.

Proof. Let S be the set of all left cosets of H in G. Then |S| = n.

There is a left action of G on the set S : g(aH) = (ga)H, ∀g, a ∈ G.

This determines a homomorphism φ from G to Sn. Since H is a proper

subgroup of G, Ker(φ) 6= G. Since |Im(φ)| ≤ n!, the fundamental

theorem of homomorphisms implies that (G : Ker(φ)) ≤ n!. Hence

|Ker(φ)| ≥ |G|/n! > 1, which implies that Ker(φ) is a nontrivial normal

subgroup of G.

From now on, by default an action is a left action, unless otherwise

stated.

Definition 1.10.4. Assume that G acts on a set S. For an element

x ∈ S, the set

Gx = gx|g ∈ G

is called an orbit. The number of elements in an orbit is called the

length of that orbit. The stabilizer of an element x ∈ S is defined to

be

Stab(x) = g ∈ G|gx = x.

It is easy to check that Stab(x) is indeed a subgroup of G. The

stabilizer of x can also be denoted by Gx.

Lemma 1.10.5. For x, y ∈ S, if Gx ∩Gy 6= ∅ then Gx = Gy.

Proof. Assume that g1x = g2y. Then g−11 g2y = x. Hence gx =

gg−11 g2y ∈ Gy for any g ∈ G. Therefore, Gx ⊆ Gy. For the same

reason, Gy ⊆ Gx.


This lemma implies that the orbits give a partition of the set S,

which is useful in counting the number of elements in S.

Definition 1.10.6. An action with only one orbit is called a tran-

sitive action.

In other words, an action is transitive if and only if there exists

g ∈ G such that gx = y for any x, y ∈ S.For example the left translation of a group is transitive while the

conjugation of a nontrivial group is not transitive.

Assume that a G acts on a set S and x ∈ S. The orbit Gx is closely

related to the stabilizer Gx. Let Γ denote the set of all left cosets of Gx

in G.

Theorem 1.10.7. There is a one-to-one correspondence between the

set Γ and the orbit Gx.

Proof. Define a map

φ : Γ → Gx, gGx 7→ gx.

Let us check that this map is well-defined. Assume that g′ = gh is

another representative of the left coset gGx in which h ∈ Gx. Then

g′x = (gh)x = g(hx) = gx.

Hence φ is well-defined. It is evident that φ is surjective. It remains to

show that φ is injective.

Assume that φ(g1Gx) = φ(g2Gx). Then g1x = g2x. So g−11 g2x = x,

i.e., g−11 g2 ∈ Gx, which implies g1Gx = g2Gx. Hence φ is injective.

Corollary 1.10.8. Assume that a group G acts on a set S and

x ∈ S. Then

|Gx| = (G : Gx).

Example 1.10.9. Let U(2) be the 2× 2 unitary group. It consists

of all 2× 2 unitary matrices. Let

S3 = (u, v) ∈ C2| |u|2 + |v|2 = 1.


Then S3 is the unit sphere in the 4-dimensional Euclidean space, i.e.,

S3 = (x1, x2, x3, x4) ∈ R4|x21 + x2

2 + x23 + x2

4 = 1.

By the definition of unitary matrices,

A

[u

v

]∈ S3

holds for A ∈ U(2). This gives an action of U(2) on S3. By the Gram-

Schmidt process in linear algebra, this action is transitive. Hence the

map

f : U(2) → S3, A 7→ A

[1

0

]is surjective. By Theorem 1.10.7 the inverse image of any (u, v) ∈ S3

in U(2) is a right coset of

Stab

([1

0

]).

In particular,

f−1

([1

0

])= Stab

([1

0

]).

It is not hard to see that

Stab

([1

0

])= U(1),

which is the set of all complex numbers with modulus 1, which is a

unit circle on the complex plane.

This analysis enables us to visualize U(2) in some degree. Since

3-dimensional sphere is bounded as well as the unit circle. There is

reason to believe that U(2) is also bounded. In the terminology of

topology, U(2) is a compact manifold. In term of Lie theory, U(2) is a

compact Lie group.


Exercises

1. Assume that a group G acts on a set S. Define a relation on S

by x ∼ y if and only if there is g ∈ G such that gx = y. Show that

∼ is an equivalence relation and an equivalence class is an orbit of the

given action.

2. A subset S of a group G is called a conjugacy class if there is an

element a ∈ G such that

S = bab−1|b ∈ G.

Assume that G is a finite group.

1) Show that the number of elements in any conjugacy class divides

the order of G;

2) Do all conjugacy classes contain same number of elements ?

3) If G has only two conjugacy classes, show that |G| = 2.

3. Find all conjugacy classes of the special linear group SL2(C).

4. Assume that a finite group G acts on a finite set S with n =

|S| > 2 such that for any x1 6= x2, y1 6= y2 ∈ S there exists g ∈ G such

that gx1 = y1, gx2 = y2. show that n(n− 1) divides |G|.

5. Assume that a group G acts on a set S with |S| > 1 transitively.

Show that GxGy is a proper subset of G for any two distinct elements

x, y of S.

6. Regard S4 as a subgroup of S5 in the natural way ( i.e., the sub-

group of S5 consisting of all permutations fixing the object 5). Define

an action of S4 × S4 on S5 by

(σ1, σ2)g = σ1gσ−12 .

Find the total number of orbits and the length of each orbit.

Chapter 2

Elements of Rings and Fields

2.1 Basic Definitions

There are more than one operation in many sets that we are familiar

with. These operations obey certain common rules. Typical examples

are set of integers Z and set of all n×n real matrices Mn(R), on which

there are two basic operations — addition and multiplication.

Definition 2.1.1. Let R be a nonempty set. Assume that R has

two binary operations +, · satisfying the following conditions:

1) R is an abelian group under the addition +;

2) a · (b · c) = (a · b) · c for any a, b, c ∈ R;

3) (a+b) ·c = a ·c+b ·c and a ·(b+c) = a ·b+a ·c for any a, b, c ∈ R.Then R is called a pseudo-ring.

If there is an element e ∈ R such that e ·a = a ·e = a for any a ∈ R,then e is called a unity of R (or multiplicative identity). A psudo-ring

with unity is defined to be a ring.

If a ring F satisfies one more condition

4) a · b = b · a for any a, b ∈ R,then this ring is called a commutative ring, otherwise it is a non-

commutative ring.

Remark 2.1.2. The multiplication a · b in a ring is usually denoted

by ab. The identity element is usually called the zero element and is

denoted by 0, and the unity is usually denoted by 1. They can also be

denoted by 0R and 1R to avoid confusion.

51

52 CHAPTER 2. ELEMENTS OF RINGS AND FIELDS

Let a ∈ R and let n be a natural number. The sum and product of

n copies of a is denoted by na and an respectively.

Example 2.1.3. • By definition a ring contains at least one element.

Is there a ring with only one element? The answer is yes. Let R be a

set containing only one element a. Define a+ a = a, aa = a. Then R is

a ring in deed. The element a serves as both zero element and unity.

This ring is simply denoted by 0, called the zero ring.

• Z is a commutative ring under usual addition and multiplication,

called the ring of integers.

•Mn(R) is a ring under the addition and multiplication of matrices,

which is non-commutative when n ≥ 2. Note the difference between

Mn(R) and GLn(R).

• The set of all polynomial K[x] over a number field K is a com-

mutative ring under usual addition and multiplication, called the poly-

nomial ring over K. Later on we will see that the ring of integers and

the polynomial ring share many common properties.

Proposition 2.1.4 (Basic properties of rings). 1) The unity is

unique.

2) 0 · a = a · 0 = 0 for any a ∈ R.3) (−a) · b = a · (−b) = −(a · b) for any a, b ∈ R.4)

(a1 + · · ·+ am) · (b1 + · · ·+ bn) =∑

1≤i≤m,1≤j≤n

aibj

for any a1, . . . , am, b1, . . . , bn ∈ R.5) If ab = ba, then

(a+ b)n =n∑

i=0

(n

i

)aibj;

6) In a nonzero ring, i.e, a ring with more than one element, the

unity 1 is not equal to 0.

The proof is left as an exercise.

We have learned in Chapter 1 that na is the sum of n copies of a

for any a ∈ R and any natural number n. There are two interpretation

2.1. BASIC DEFINITIONS 53

of the notation na if n is negative. It can be understood either as the

sum of −n copies of −a or as −[(−n)a]. It is easy to verify that they

are the same. Hence the notation na is unambiguous for any integer n.

It is easy to verify the following properties:

• (n+m)a = na+ma for any integers n,m;

• n(a+ b) = na+ nb for any integer n;

• n(ma) = (nm)a for any integers n,m.

Definition 2.1.5. Let R be a ring. If a, b ∈ R satisfy ab = 0, then

a is called a left zero-divisor of b and b is called a right zero-divisor

of a. If a, b ∈ R satisfy ab = 1, then a is called a left inverse of b and

b is called a right inverse of a. Moreover if ab = ba = 1, then a is

called the (multiplicative) inverse of b, denoted by b−1. By symmetry,

b is also the inverse of a, i.e., b = a−1. A nonzero element a is called a

unit if its multiplicative inverse exists.

Let R be a nonzero commutative ring. If the product of any two

nonzero elements in R is not equal to zero, then R is called an integral

domain.

If every nonzero element in a nonzero ring R has an inverse, then

R is called a division ring or skew field. A commutative skew field

is called a field.

Example 2.1.6. • All number fields are fields.

• Mn(R) is not an integral domain if n ≥ 2.

• Z is an integral domain.

• The polynomial ring K[x1, . . . , xn] in n variables over a number

field is an integral domain.

• The set of all rational functions over a number field K is a field.

• The zero ring 0 is not an integral domain.

Let A be a nonzero ring such that the product of any two nonzero

elements is nonzero. The the set of all nonzero elements of R form a

monoid under the multiplication. The set of all units form a group,


denoted by A∗. When A is a division ring, then the group A∗ = A\0is called the multiplicative group of A. The multiplicative group of a

field is an abelian group.

Recall the introduction of a complex number a + bi in which i is

an imaginary number satisfying i2 = −1. A new kind of “number”,

calld quaternion, can be defined in a similar manner. A quaternion

is written as a+ bi+ cj + dk, in which a, b, c, d are numbers and i, j, k

are three distinct elements satisfying

i2 = j2 = k2 = −1, ij = k, ji = −k, jk = i, kj = −i, ki = j, ik = −j.

The addition of two quaternions are given by

(a+bi+cj+dk)+(a′+b′i+c′j+d′k) = (a+a′)+(b+b′)i+(c+c′)j+(d+d′)k

and the product is given by

(a+ bi+ cj + dk)(a′ + b′i+ c′j + d′k)

= (aa′ − bb′ − cc′ − dd′) + (ab′ + ba′ + cd′ − dc′)i

+(ac′ + ca′ + db′ − bd′)j + (ad′ + da′ + bc′ − cb′)k.

It is easy, though tedious, to verify that the set Q of all quaternions

is a ring under addition and multiplication with 1 as the unity. Since

ij 6= ji, it is not a commutative ring.

Since

(a+ bi+ cj + dk)(a− bi− cj − dk) = a2 + b2 + c2 + d2,

Every nonzero element of Q is a unit. That is to say that Q is a division

ring. This is the simplest non-commutative division ring.

The set a + bi + cj + dk ∈ Q|a, b, c, d ∈ Z is a non-commutative

ring.

The definition of the direct product of rings is similar to that of

groups.

Exercises

2.2. IDEALS AND QUOTIENT RINGS 55

1. Verify 1)-3) of Proposition 2.1.4.

2. Show that the inverse of a unit in a ring is unique.

3. Let R be a commutative ring and let u be a unit in R. Assume

that a ∈ R and there exists a natural number n such that an = 0.

Show that u+ a is a unit.

4. Let A be the set of maps from the set N of all natural numbers

to the complex number field, i.e., the set of all sequences of complex

numbers. For any f, g ∈ A, define

(f + g)(n) = f(n) + g(n), ∀n ∈ N,

and

(f ∗ g)(n) =∑d|n

f(d)g(nd

).

Show that A is a commutative ring under the addition and the multi-

plication “*”. What is the unity in this ring?

5. Show that the direct product of two fields is not a field.

6. Let A be a ring such that x2 = x for all x ∈ A. Show that 2x = 0

for all x ∈ A.

2.2 Ideals and Quotient Rings

Many mathematical structures contain substructures. For example,

sets have subsets, vector spaces have subspaces, groups have subgroups.

Naturally rings have subrings which are defined in the following obvious

way.

Definition 2.2.1. Let S be a nonempty additive subgroup of a

ring R. If S is closed under multiplication and contains the unity, then

S is a subring of R.

The subring S itself is a ring.

The subrings inherit most properties of their mother ring, such like

commutativity or being an integral domain, etc, but not all. For ex-

ample, a subring of a non-commutative ring can be commutative, a


subring of a field is not necessarily a field. If a subring of a field is a

field then this subring is called a subfield.

Let I be an additive subgroup of a ring R. Since R is an abelian

group under addition, I is a normal subgroup of R. Hence the additive

group R/I is well-defined. We want to know whether the multiplication

in R induces a well-defined multiplication in R/I to make it into a ring.

Write two elements in R/I as a + I, b + I, in which a, b ∈ I. A

reasonable multiplication should be (a+I)(b+I) = ab+I. An important

issue arises as in the discussion of quotient groups: does ab+ I depend

upon the choices of the representatives of a+ I and b+ I?

Change the representatives into a + u and b + v, in which u, v can

be any elements in I. Then

(a+ u)(b+ v) = ab+ av + ub+ uv.

In order that the multiplication (a + I)(b + I) = ab + I makes sense,

(a + u)(b + v) and ab must be in the same coset of I. This amounts

to saying that (a + u)v + ub ∈ I for any a, b ∈ R, u, v ∈ I, which is

equivalent to av ∈ I, ub ∈ I for any a, b ∈ R, u, v ∈ I. The following

definition is suggested:

Definition 2.2.2. Let I be an additive subgroup of a ring R. If

au ∈ I, ua ∈ I for any a ∈ R, u ∈ I then I is called a (two-sided) ideal

of R.

Proposition 2.2.3. Let I be an ideal of R. Then R/I forms a ring

under the multiplication (a + I)(b + I) = ab + I, called the quotient

ring of R over I.

Proof. We have already seen that the multiplication is well-defined.

It is evident that R/I is an abelian group under addition. It remains

to check the associative law, the distributive law and the existence of

unity.

Associative law:

(a+ I)[(b+ I)(c+ I)] = (a+ I)(bc+ I) = abc+ I.


[(a+ I)(b+ I)](c+ I) = (ab+ I)(c+ I) = abc+ I.

Distributive law:

(a+ I)[(b+ I) + (c+ I)] = (a+ I)(b+ c+ I)

= a(b+ c) + I

= (a+ I)(b+ I) + (a+ I)(c+ I).

The verification of

[(b+ I) + (c+ I)](a+ I) = (b+ I)(a+ I) + (c+ I)(a+ I)

is similar. Since (1 + I)(a + I) = 1a + I = a + I for every a ∈ R, the

coset 1 + I is the unity of R/I.

Remark 2.2.4. If the condition au ∈ I, ua ∈ I for all a ∈ R, u ∈ Iis replaced by au ∈ I for all a ∈ R, u ∈ I then I is called a left ideal

of R. The definition of right ideal is similar. For commutative rings,

there is no difference among ideal, left ideal and right ideal.

The role played by ideals in ring theory is similar to normal sub-

groups in group theory. But an ideal is not a subring, except for the

whole ring, since a subring contains unity by definition and an ideal is

the whole ring if and only if it contains the unity.

Two elements a, b in a ring R are called congruent with respect to

the ideal I if a− b ∈ I, denoted by

a ≡ b (mod I).

In particular a ≡ 0 (mod I) means a ∈ I.

Example 2.2.5. • Let n be a natural number greater than one.

Then nZ is an ideal of Z. The quotient ring Zn = Z/nZ is an

integral domain if and only if n is a prime number. In fact, Zn is

a field if n is a prime number. The notation a ≡ b (mod nZ) can

be abbreviated as a ≡ b (mod n), which is a standard notation

in elementary number theory.


• Any nonzero ring R is shipped with two ideals for free. They are

0 and R, called trivial ideals of R. A nonzero ring is called

a simple ring if it is not a division ring and does not have any

nontrivial ideals.

• Let I be the set of all 2×2 matrices whose first row is zero. Then

I is a left ideal of M2(R), but not an ideal.

Proposition 2.2.6. 1) Let Iλλ∈Λ be a collection of ideals of a

ring R. Then⋂

λ∈Λ Iλ is also an ideal of R.

2) Let I and J be ideals of a ring R. Let I+J = a+b|a ∈ I, b ∈ J.Then I+J is an ideal of R. More generally, for any collection of ideals

Iλλ∈Λ of R, let∑λ∈Λ

Iλ = ∑

λ

aλ|aλ ∈ Iλ, there are only finitely many nonzero terms .

Then∑

λ∈Λ Iλ is an ideal of R.

3) Let I, J be ideals of a ring R. Let

IJ = n∑

i=1

aibi|ai ∈ I, bi ∈ J, n <∞.

Then IJ is an ideal of R.

The proof is left as an exercise.

Example 2.2.7. In Z let I = nZ, J = mZ. Then

I ∩ J = [m,n]Z.

I + J = (m,n)Z.

Here [m,n] and (m,n) denote the least common multiple and greatest

common divisor of m and n respectively.

IJ = mnZ.

This suggests the following definition.


Definition 2.2.8. Let I, J be ideals of a ring R. If I+J = R, then

I, J are coprime.

Let S be a nonempty subset of a ring R. Let 〈S〉 be the intersection

of all ideals of R containing S. Since R contains S, the intersection

makes sense. Then 〈S〉 is an ideal of R, called the ideal generated by

S. It is not hard to see that

〈S〉 = a1u1b1 + · · ·+anunbn|a1, . . . , an, b1, . . . , bn ∈ R, u1, . . . , un ∈ S.

When R is commutative then

〈S〉 = a1u1 + · · · anun|a1, . . . , an ∈ R, u1, . . . , un ∈ S.

The left and right ideals generated by S are defined in a similar

manner.

Let I be an ideal of a ring R. If there is a finite subset S of R such

that I = 〈S〉 , then I is called a finitely generated ideal. Moreover,

if I can be generated by a single element, then I is called a principal

ideal.

If an ideal I is generated by a finite number of elements s1, . . . , sn,

the I can be denoted by (s1, . . . , sn). For example in the ring of poly-

nomials Q[x], the principal ideal generated by x2 + x is denoted by

(x2 + x).

Exercises

1. Prove Proposition 2.2.6

2. Let A be the ring of all real matrices in the form[a b

0 c

].

Find all ideals of A.

3. Let Mn(K) denote the ring of all n× n matrices over a number

field K. Show that it is a simple ring.

4. Let R = Zk where k is a natural number. Show that the following

two statements are equivalent


1) There is an element a ∈ R and a natural number n such that

a 6= 0 and an = 0.

2) k is divisible by the square of a prime number.

5. Let A be a commutative ring and let I1, . . . , In be mutually

coprime ideals of A. Prove the following statements:

1) For any 1 ≤ i ≤ n there exists bi ∈ A such that

bi ≡ 1 (mod Ii)

and

bi ≡ 0 (mod Ij)

for all j 6= i.

2) (The Chinese Remainder Theorem) For any a1, . . . , an ∈ A, there

exists b such that

b ≡ ai (mod Ii)

for all i.

6. Let I1, I2, . . . be a sequence of ideals of R such that I1 ⊆ I2 ⊆ · · · .Show that

⋃∞i=0 Ii is an ideal of R.

7. Let I be the ideal generated by three elements 12, 48, 30 in Z.Find a ∈ Z such that I = (a).

8. Let I be the left ideal generated by all elements in the form

ab− ba(a, b ∈ R) in a ring R. Show that I is an ideal of R.

2.3 Homomorphisms of Rings

Definition 2.3.1. Let R1 and R2 be two rings. A map f : R1 → R2

is called a homomorphism if the following conditions are satisfied:

1) f(a+ b) = f(a) + f(b) for any a, b ∈ R1.

2) f(ab) = f(a)f(b) for any a, b ∈ R1.

3) f(1R1) is equal to 1R2 .

The following example shows that the third condition cannot be

derived from the first two conditions.

2.3. HOMOMORPHISMS OF RINGS 61

Example 2.3.2. Let R = Z × Z. The map f : Z → R, a 7→ (a, 0).

satisfies 1) and 2), but not 3).

For a homomorphism f, denote Ker(f) = a ∈ R1|f(a) = 0. Then

Ker(f) is called the kernel of f. f is injective if and only if Ker(f) =

0. Denote Im(f) = f(a)|a ∈ R1. Then Im(f) is a subring of R2.

If Im(f) = R2, then f is an epimorphism. A bijective homomorphism

is called an isomorphism. If there is an isomorphism from a ring

R1 to another ring R2 then these two rings are isomorphic, denoted

by R1∼= R2. A homomorphism from a ring R to itself is called an

endomorphism. A bijective endomorphism is an automorphism.

Theorem 2.3.3 (The Fundamental Theorem of Homomorphisms).

Let f : R1 → R2 be a ring homomorphism. Then Ker(f) is an ideal of

R1 and

R1/Ker(f) ∼= Im(f).

Proof. Assume that a ∈ R1, h ∈ Ker(f). Then

f(ah) = f(a)f(h) = f(a)0 = 0,

f(ha) = f(h)f(0) = 0f(a) = 0.

Hence ah ∈ Ker(f), ha ∈ Ker(f). This implies that Ker(f) is an ideal

of R1. Denote I = Ker(f).

It follows from the fundamental theorem of group homomorphisms

that the group homomorphism

φ : R1/I → Im(f), a+ I 7→ f(a)

is bijective. It remains to verify φ((a + I)(b + I)) = φ(a + I)φ(b + I),

which is obvious from the explicit formula of φ.

Proposition 2.3.4. Let I be an ideal of a ring R. Let f : R→ R/I

be the canonical homomorphism. Let Γ be the set of all ideals of R

containing I. Let Γ′ be the set of all ideals of R/I. The map f−1 is a

one-to-one correspondence from Γ′ to Γ.

Proof. Define a map φ from Γ to Γ′ as follows: For any J ∈ Γ let


J ′ be the set of all coset of I in the form a + I(a ∈ J). Then J ′ is an

ideal of R/I. Define φ(J) = J ′.

Since f−1(J ′) = J the composite f−1 φ is the identify map. It is

easy to verify that φf−1 is also the identity map. Hence f−1 : Γ′ → Γ

is bijective.

Proposition 2.3.5. Let S be a subring of a ring R and let I be an

ideal of R. Then S + I = a + b|a ∈ S, b ∈ I is a subring of R, I is

an ideal of S + I, S ∩ I is an ideal of S and

S + I/I ∼= S/S ∩ I.

Proof. Obviously S+I is an additive subgroup of R. Assume that

a1, a2 ∈ S, b1, b2 ∈ I. Then (a1+b1)(a2+b2) = a1a2+(a1b2+b1a2+b1b2) ∈S + I. It is obvious that S + I contains the unity. Hence S + I is a

subring of R. It is easy to check that I is an ideal of S + I and S ∩ Iis an ideal of S.

Construct a map f : S → S + I/I, a 7→ a + I. Then f is an

epimorphism with Ker(f) = S ∩ I. It follows from the fundamental

theorem of homomorphisms that

S + I/I ∼= S/S ∩ I.

Exercises

1. Let a be a nonzero element of a commutative ring R and let I

be the principal ideal in R[x] generated by ax− 1. Let f : R→ R[x]/I

be the homomorphism defined by f(b) = b + I. Show that Ker(f) =

b ∈ R| there exists a natural number n such that anb = 0.

2. Let I, J be two ideals of a ring R. Define the map

f : R→ R/I ×R/J, a 7→ (a+ I, a+ J).

Show that

1) f is a ring homomorphism;

2.4. ELEMENTARY PROPERTIES OF FIELDS 63

2) f is injective if and only if I ∩ J = 0;

3) f is surjective if and only if I + J = R.

3. Denote by F [x, y, z] and F [t] the rings of polynomials in three

variables and in one variable respectively over a field F. Let φ : F [x, y, z] →F [t] be a homomorphism carrying f(x, y, z) to f(t2, t3, t4). Show that

Ker(φ) is the ideal generated by y2 − x3 and z − x2.

2.4 Elementary Properties of Fields

Fields are special commutative rings in which every nonzero element is

a unit. Recall the definition of a number field. A number field K is a

subset of C satisfying the following three conditions

1. a+ b, a− b, ab ∈ K for any a, b ∈ K;

2. 1/a ∈ K for any nonzero a ∈ K;

3. K contains at least one nonzero number.

The generalization is straightforward.

Proposition 2.4.1. A subset E of a field F is a subfield of F if

and only if the following conditions are satisfied:

1. a+ b, a− b, ab ∈ E for any a, b ∈ E;

2. 1/a ∈ E for any nonzero a ∈ E;

3. E contains at least one nonzero element.

The homomorphism from a field to a ring is relatively simple, as

stated in the following proposition.

Proposition 2.4.2. Let f : F → A be a homomorphism from a

field to a nonzero ring A. Then f is injective.

Proof. Since f(1F ) = 1A 6= 0, Ker(f) is a proper ideal of F, which

contains only one element 0. Hence f is injective.


The arithmetic of elementary schools starts with integers followed

by fractional numbers. We may imitate this process to construct a field

from a given integral domain R.

Immediately one may say that the set of all a/b with a, b ∈ R(b 6= 0)

is the desired field. On second thought, what does the expression a/b

mean? If a and b are natural numbers, a math teacher of elementary

school may interpret a/b graphically or verbally: cut a cake into b

pieces of equal size and then take a pieces. But for an abstract integral

domain R, such explanation does not make sense, since the elements in

R are not quantities any more. The expression a/b is merely an ordered

pair of elements in R. A better notation is (a, b) with a, b ∈ R, b 6= 0.

The set S of all such pairs are not the field we try to construct, because

there is redundancy in this set. Take fractional numbers as example,

(2, 3), (4, 6), . . . represent the same fraction number, although they are

different pairs. This reminds us to give a rule to equalize the elements in

S. In mathematical jargon, we need to establish an equivalence relation

and consider the set of equivalence classes.

More specifically, for any (a1, b1), (a2, b2) ∈ S, define (a1, b1) ∼(a2, b2) if and only if a1b2 = a2b1. Let us verify that “∼” is an equiva-

lence relation.

1) reflexivity: ab = ba implies (a, b) ∼ (a, b).

2) symmetry: obvious.

3) (a1, b1) ∼ (a2, b2), (a2, b2) ∼ (a3, b3). So b2(a1b3 − a3b1) = (a1b2 −a2b1)b3 − (a3b2 − a2b3)b1 = 0. Hence (a1, b1) ∼ (a3, b3).

Let a/b denote the equivalence class in S represented by (a, b). This

is the correct interpretation of the notation a/b. In our familiar terms,

a1/b1 = a2/b2 if and only if a1b2 = a2b1.

Define addition and multiplication between equivalence classes by

the following rules:

a1

b1+a2

b2=a1b2 + a2b1

b1b2,

a1

b1

a2

b2=a1a2

b1b2.


It is indispensable to check that these two operations do not depend

upon the choices of representatives. Assume that a1/b1 = a′1/b′1 and

a2/b2 = a′2/b′2. Then

(a1b2 + a2b1)b′1b′2 = a1b2b

′1b′2 + a2b1b

′1b′2

= a′1b1b2b′2 + a′2b2b1b

′1 = (a′1b

′2 + a′2b

′1)b1b2.

Hence the addition does not depend upon the choices of representatives.

The readers can verify the multiplication easily.

The remaining work is routine: check that the set of equivalence

classes is an abelian group under addition with 0/1 as its zero element,

the law of associativity and commutativity for multiplication and the

law of distributivity are satisfied, 1/1 is the unity and every nonzero

element is a unit.

The field F thus obtained is called the field of fractions of R.

Define a map j : R → F, a 7→ a/1. It is easy to check that j is

injective, which means that the integral domain can be treated as a

subring of its field of fractions.

Remark 2.4.3. This method of construction can be summarized as

follows.

In order to construct a new algebraic structure from a known alge-

braic structure, first construct a set which is larger than needed, then

define an equivalence relation in this large set and finally define a new

algebraic structure in the set of equivalence classes.

This method is very common in various branches of mathematics.

The method we have used to “enlarge” an integral domain to its field

of fractions is more or less what we are familiar with in elementary and

middle schools. Let’s consider the other direction. Can we “shrink” a

ring R into a field? The first natural candidate might be a subring.

Look at a simple example R = Z. It is easy to see that no subring

of Z is a field. Therefore only limited new fields can be produced as

subrings of known rings.

The correct way of thinking is to consider the quotient ring of a

given ring R. Choose an ideal I of R. Then the quotient ring R/I is


“smaller” than R. When I satisfies certain conditions R/I can be a

field.

Definition 2.4.4. Let I be a proper ideal of a ring R. If there is

no other ideal between I and R then I is called a maximal ideal of

R.

Proposition 2.4.5. Let I be an ideal of a commutative ring R.

Then R/I is a field if and only if I is a maximal ideal of R.

Proof. ⇒: Assume that R/I is a field. Let J be an ideal of R

containing I but not equal to I. It suffices to show that J = R. Choose

any a ∈ J\I. Then a + I is a nonzero element in R/I. So there is

b + I ∈ R/I such that (a + I)(b + I) = 1 + I, which means that

ab − 1 ∈ I. So 1 = ab + u for some u ∈ I. Hence 1 ∈ J, which implies

J = R.

⇐: Assume that I is a maximal ideal of R. For any a ∈ R\I let J

be the ideal of R generated by I and a. Then J = R by the definition

of maximal ideals. Hence 1 ∈ J, i.e., 1 = ba + u for some b ∈ R and

u ∈ I. This implies (a + I)(b + I) = 1 + I in R/I. Hence R/I is a

field.

It is easy to find all maximal ideals of Z since its every ideal has

the form nZ for some nonnegative integer n. The relation nZ ⊆ mZholds if and only if m|n or n = 0. It follows that nZ is a maximal ideal

if and only if n is a prime number.

Let p be a prime number. Denote Fp = Zp = Z/pZ. Then Fp is a

field containing p elements. This a very important field which must be

kept in mind.

Let F be an arbitrary field. Let

f : Z → F, n→ n · 1.

Recall that n·1 = 1+1+· · ·+1 n times. The map f is a homomorphism.

Let Ker(f) = mZ, in which m is a nonnegative integer. The integer m

is called the characteristic of the field F.

There are two cases:


1) the characteristic is 0.

In this case, f is a monomorphism. Thus Z can be treated as a

subring of the field F. More precisely, F contains a subring isomorphic

to Z. Since F is a field, all m/n with m,n ∈ Z, n 6= 0 also belongs to

F. This yields the following

Theorem 2.4.6. Every field of characteristic 0 contains a subfield

isomorphic to Q.

It is immediate that this subfield is the smallest subfield of F, which

is called the prime field of F.

2) the characteristic m > 0.

In this case we want to show that m is a prime number. Suppose

that m = ab in which a and b are natural numbers less than m. Then

f(a)f(b) = f(m) = 0. But both a and b are not in Ker(f) = mZ.Hence f(a) 6= 0, f(b) 6= 0, which contradicts the assumption that F is

a field. Hence m is a prime number, which is usually denoted by p.

It is evident that Im(f) ∼= Fp.

Theorem 2.4.7. Every field of characteristic p > 0 contains a sub-

field isomorphic to Fp.

It is natural to call Fp the prime field of F in this case.

Lemma 2.4.8. Let F be a field of characteristic p > 0 and let a ∈F. Let n,m be two integers such that p|(m − n). Then ma = na. In

particular, if p|n then na = 0 for any a ∈ F.

Proof. Since p|(m− n), m− n = pr for some r ∈ Z.First assume that a = 1. Then pr · 1 = 0 for any integer r. Hence

ma− na = m · 1− n · 1 = pr · 1 = 0.

For an arbitrary a ∈ F, it follows from ma−na = m(1·a)−n(1·a) =

(m · 1− n · 1)a that ma− na = 0.

The following formula is useful.

Proposition 2.4.9. Let a and b be two elements in a field F of

characteristic p > 0. Let r be a natural number. Then

(a+ b)pr

= apr

+ bpr

.


Proof. Apply induction to r.

For any 1 < i < p, the natural number(

pi

)is divisible by p. Hence(

pi

)aibp−i = 0 holds for all 0 < i < p. It follows from

(a+ b)p =

p∑i=0

(p

i

)aibp−i

that

(a+ b)p = ap + bp.

The remaining step of the proof is simple and left to the readers.

Fields are divided into two big classes: those of characteristic zero

and nonzero. These two classes have quite different properties. Among

the fields of prime characteristics, those of characteristic 2 are very

special, mainly because a quadratic polynomial x2 + bx + c cannot be

changed into the form y2 + d by a change of variable y = x+ e.

In terms of the number of elements, fields are divided into finite

fields and infinite fields. We have already met finite fields Fp, whose

characteristic is a prime number p. Later we will learn that there are

other finite fields. A field of prime characteristic is not necessarily a

finite field. In field theory, the classification of finite fields is totally

solved. But the classification of infinite fields is far from clear.

To conclude this chapter we prove an interesting theorem which

plays an important role in Galois theory.

Theorem 2.4.10 (E. Artin). Let χ1, . . . , χn be distinct homomor-

phisms from a group G to the multiplicative group F ∗ of a field F. Let

a1, . . . , an be a set of elements in F not all equal to zero. Then there

exists g ∈ G such that

a1χ1(g) + a2χ2(g) + · · ·+ anχn(g) 6= 0.

Proof. Apply induction to n. The statement is obviously true for

n = 1.

Assume that n > 1. If one among a1, . . . , an is equal to zero, then

the statement holds by induction hypothesis. Hence we may assume


that a1, . . . , an ∈ F ∗.

Since χ1, . . . , χn are distinct, there is h ∈ G such that χ1(h) 6=χ2(h). Hence a2(χ2(h)− χ1(h)) 6= 0. Since the n− 1 elements

a2(χ2(h)− χ1(h)), a3(χ3(h)− χ1(h)), . . . , an(χn(h)− χ1(h))

in F are not all equal to zero, by induction hypothesis there exists

g ∈ G such that

a2(χ2(h)−χ1(h))χ2(g)+a3(χ3(h)−χ1(h))χ3(g)+· · ·+an(χn(h)−χ1(h))χ(g) 6= 0.

Let

u = a1χ1(hg) + a2χ2(hg) + · · ·+ anχn(hg),

v = a1χ1(g) + a2χ2(g) + · · ·+ anχn(g).

Then

u− χ1(h)v

= a2(χ2(h)− χ1(h))χ2(g) + a3(χ3(h)− χ1(h))χ3(g) + · · ·+ an(χn(h)− χ1(h))χ(g)

6= 0.

So u, v cannot be zero at the same time. This shows that the theo-

rem is true for n too.

Exercises

1. Give an example to show that the condition of commutative ring

in Proposition 2.4.5 is indispensable.

2. Let R denote the field of real numbers. Find all maximal ideals

of the ring R× R.

3. Let C denote the field of complex numbers. Show that the ideal

in C[x, y] generated by x+y2 and y+x2 +2xy2 +y4 is a maximal ideal.

4. Let k be a field with infinitely many elements and let n be a

natural number greater than 1. Assume that (a + b)n = an + bn holds


for all a, b ∈ k. Show that the characteristic of k is a prime number

and n is the power of a prime number.

5. Let S be the ring consisting of all real matrices in the form[a b

−b a

].

1) Write out an isomorphism from A to the complex number field

C without proof;

2) Let M = M2(R) denote the ring consisting of all 2 × 2 real

matrices. Show that there exists a subring N of M containing

A =

[0 3

−4 1

]such that N ∼= S;

3) Show that there exists X ∈M2(R) such that X4 + 13X = A.

6. Let A be a commutative ring. An ideal P of A is called a prime

ideal if ab /∈ P for any a, b ∈ A\P.1) Show that an ideal of A is a prime ideal if and only if A/P is an

integral domain;

2)Show that every maximal ideal of A is a prime ideal.

Chapter 3

Polynomials and Rational

Functions

Polynomials over an arbitrary field are defined similar to those over

number fields (real polynomials, complex polynomials, · · · ). In fact

the coefficients of polynomials can be in an arbitrary ring. To avoid

zero-divisors we restrict the discussion to polynomials over an integral

domain.

3.1 Polynomials in One Variable

Definition 3.1.1. Let A be an integral domain. An expression in

the form

anxn + an−1x

n−1 + · · ·+ a1x+ a0

with

an, an−1, . . . , a1, a0 ∈ A.

is called a polynomial over A. For any i with ai 6= 0, aixi is called a

term of that polynomial and ai is called the coefficient of that term.

The terms of a polynomial can be exchanged at will. When an 6= 0 the

number n is called the degree of that polynomial. The term anxn is

the initial term. A polynomial whose coefficient of the initial term is

equal to one is called a monic polynomial. The symbol x is called

the indeterminate or variable. The term a0 is called the constant

71

72 CHAPTER 3. POLYNOMIALS AND RATIONAL FUNCTIONS

term. The zero polynomial 0 is a special polynomial, whose degree is

equal to −∞ by convention.

The set of all polynomials over A with indeterminate x is denoted

by A[x].

The degree of f(x) ∈ A[x] is denoted by deg(f(x)) or deg(f). The

degree is equal to zero if and only if f(x) ∈ A\0. In this case we may

say that f(x) is a nonzero constant.

Two polynomials are equal if and only if they have the same degree

and their coefficients of the degree i terms are equal for each i.

The addition, subtraction and multiplication between polynomials

are defined in usual way. They satisfy the laws of commutativity, as-

sociativity and distributivity. Hence A[x] is a commutative ring.

The composite f(g(x)) of f(x), g(x) ∈ A[x] is defined as usual. The

following properties are easily verified:

• deg(fg) = deg(f) + deg(g). Thus the product of two nonzero

polynomials is not equal to zero. Hence A[x] is an integral domain.

• Let c be a nonzero element in A. Then deg(cf(x)) = deg(f) for

any f(x) ∈ A[x].

• deg(f ± g) ≤ max(deg(f), deg(g)), and the equality holds when

deg(f) 6= deg(g).

3.2 Division Algorithm

Theorem 3.2.1. Let A be an integral domain and let f(x), g(x) ∈A[x]. If g(x) 6= 0 and its leading coefficient is an invertible element in

A then there exist a unique pair of polynomials q(x), r(x) ∈ A[x] such

that

1) f(x) = g(x)q(x) + r(x);

2) deg(r) < deg(g).

The polynomials q(x) and r(x) are called the quotient and re-

mainder of f(x) by g(x) respectively.

Proof. This can be proved by induction on the degree of f(x).

The details are left as an exercise.

3.2. DIVISION ALGORITHM 73

Definition 3.2.2. Let A be an integral domain and let f(x), g(x) ∈A[x]. If f(x) 6= 0 and there exists h(x) ∈ A[x] such that g(x) =

f(x)h(x), then f(x) divides g(x) or g(x) is divisible by f(x), denoted

by f(x)|g(x) or simply f |g.

Some basic facts are listed as follows:

• Every invertible element of A divides any polynomial;

• Every nonzero polynomial divides the zero polynomial 0;

• 0 does not divide any polynomial;

• If g(x)|f(x), g(x)|h(x), then g(x)|a(x)f(x) + b(x)h(x) for any

a(x), b(x) ∈ A[x];

• If g(x)|f(x) and f(x) 6= 0, then deg(g) ≤ deg(f).

• g(x)|f(x) implies g(h(x))|f(h(x)) for any h(x) ∈ A[x] such that

g(h(x)) 6= 0.

Lemma 3.2.3. Assume that f(x), g(x) ∈ A[x], g(x) 6= 0, the leading

coefficient of g(x) is an invertible element in A. Then g(x)|f(x) if and

only if the remainder of f(x) by g(x) is zero.

Proof. ⇒: there is some b(x) ∈ A[x] such that f(x) = g(x)b(x). So

f(x) = g(x)b(x) + 0. Since deg(0) < deg(g), the remainder of f(x) by

g(x) is equal to zero.

⇐: the remainder of f(x) by g(x) being zero means that f(x) =

g(x)q(x), i.e., g(x)|f(x). 2

Lemma 3.2.4. Let f(x), g(x) be nonzero polynomials. If f(x)|g(x)and g(x)|f(x), then f(x) = c · g(x), where c is an invertible element in

A.

Proof. Assume that f(x) = g(x)a(x), g(x) = f(x)b(x). Then f(x) =

f(x)b(x)a(x), so a(x)b(x) = 1. Hence both a(x) and b(x) are invertible

elements in A. 2

The following result is the central theorem in the theory of polyno-

mials in one variable.

Theorem 3.2.5. Every ideal in the polynomial ring F [x] over a

field F is a principal ideal.


Proof. Let I be an ideal of F [x]. If I = 0, then I is evidently a

principal ideal. It remains to consider that case I 6= 0.Among all nonzero elements in I choose an element g such that

deg(g) reaches the minimum degree. We show that I is generated by

g. For every f ∈ I there are q, r ∈ F [x] such that f = qg + r and

deg(r) < deg(g) by Theorem 3.2.1

Since r = f − qg ∈ I, r must be zero by the choice of g. This means

that f = qg, which implies that f is in the ideal generated by g. 2

We have mentioned that the rings F [x] and Z share many common

properties. The reason is that both of them have division algorithm

and every ideal is a principal ideal. Another interesting ring with this

property is the ring of Gauss integers (cf. Exercise 4). An integral

domain in which every ideal is principal is called a principal ideal

domain, abbreviated as PID.

Definition 3.2.6. Let F be a field and let f(x), g(x), h(x) ∈ F [x].

If h(x)|f(x), h(x)|g(x), then h(x) is called a common divisor of f(x)

g(x). A common divisor d(x) of f(x) and g(x) that divides every com-

mon divisor of f(x), g(x) is called a greatest common divisor of

f(x) and g(x).

Example 3.2.7. There is no greatest common divisor when f(x) =

g(x) = 0.

Example 3.2.8. If f(x) 6= 0, then f(x) is greatest common divisor

of f(x) and 0.

Lemma 3.2.9. If d1(x), d2(x) are greatest common divisors of f(x), g(x)

then d1(x) = c · d2(x) for some nonzero element c ∈ F.

Proof. The definition of greatest common divisor implies d1(x)|d2(x), d2(x)|d1(x).

Then apply Lemma 3.2.4. 2

This lemma shows that the greatest common divisor is unique up

to a constant factor. If the leading coefficient is required to be 1, then

it is unique. The monic greatest common divisor of f(x) and g(x) is

denoted by gcd(f(x), g(x)) or gcd(f, g).

3.2. DIVISION ALGORITHM 75

Theorem 3.2.10. If one of f(x), g(x) ∈ F [x] is nonzero then gcd(f(x), g(x))

exists and there are a(x), b(x) ∈ F [x] such that

gcd(f(x), g(x)) = a(x)f(x) + b(x)g(x).

Proof. Let I be the ideal of F [x] generated by f(x) and g(x).

Theorem 3.2.5 implies that I is a principal ideal. Let d(x) be a monic

generator of I. Then there are a(x), b(x) ∈ F [x] such that d(x) =

a(x)f(x) + b(x)g(x).

It remains to show that d(x) = gcd(f(x), g(x)). The definition

of principal ideal implies that d(x)|f(x), d(x)|g(x). Hence d(x) is a

common divisor of f(x) and g(x). Assume that h(x)|f(x), h(x)|g(x),i.e., there are u(x), v(x) ∈ F [x] such that f(x) = u(x)h(x), g(x) =

v(x)h(x). Then

d(x) = [a(x)u(x) + b(x)v(x)]h(x),

which implies that h(x)|d(x). 2

This proof is not constructive, since if does not tell us how to find

the greatest common divisor. An effective way to compute it is the

Euclidean algorithm, which is based on the division algorithm.

Definition 3.2.11. If gcd(f(x), g(x)) = 1 then f(x) and g(x) are

coprime or relatively prime.

Corollary 3.2.12. Two polynomials f(x) and g(x) over a field

F are coprime if and only if there are u(x), v(x) ∈ F [x] such that

f(x)u(x) + g(x)v(x) = 1.

Exercises

1. Give a detailed proof of Theorem 3.2.1.

2. Let F = F3, f(x) = 2x4 + 2x + 1, g(x) = x2 − 2x + 2. Find the

quotient and remainder of f(x) by g(x).

3. Prove Lemma 3.2.12

4. The integral domain Z[i] = m+ ni|m,n ∈ Z is called the ring

of Gauss integers.


1) For every m+ ni ∈ Z[i] let N(m+ ni) = m2 + n2. Show that for

any a, b ∈ Z[i], a 6= 0, there are q, r ∈ Z[i] such that b = qa + r and

N(r) < N(a).

2) Show that Z[i] is a principal ideal domain.

3) Find the greatest common divisor of 11 + 7i and 18 + i.

5. Show that Z[x] is not a principal ideal domain.

6. Show that the quotient ring Q[x]/(x2−4) is isomorphic to Q×Q.

3.3 Polynomials in Several Variables

Let A be an integral domain and let x1, x2, . . . , xn be variables (also

known as indeterminates). An expression in the form cxe11 x

e22 · · ·xen

n

with a nonzero c in A is called a monomial. Here e1, . . . , en are non-

negative integers.

For fixed variables x1, x2, . . . , xn a finite sum of monomials is called

a polynomial in these variables. Every nonzero monomial appearing

in a polynomial is called a term of this polynomial. Note the 0 is also

considered to be a polynomial. The set of polynomials in x1, . . . , xn is

denoted by A[x1, . . . , xn]. The addition,substraction and multiplication

of polynomials are performed in the usual way, keeping in mind that

the arithmetic on the coefficients adopts that in the integral domain

A. It is evident that A[x1, . . . , xn] in a commutative ring.

The polynomial ring A[x1, . . . , xn] can be regarded as B[xn] with

B = A[x1, . . . , xn−1]. Thus it is easy to prove that A[x1, . . . , xn] is an

integral domain by induction on n.

The degree of a monomial cxe11 x

e22 · · ·xen

n is defined to be e1+· · ·+en.

The degree of a nonzero polynomial is defined to be the maximum

degree of all its terms. The degree of the zero polynomial 0 is equal to

−∞ by convention. The degree of f(x1, . . . , xn) is denoted by deg(f).

The equality

deg(fg) = deg(f) + deg(g)

holds for any f(x1, . . . , xn), g(x1, . . . , xn) ∈ A[x1, . . . , xn].

3.3. POLYNOMIALS IN SEVERAL VARIABLES 77

Assume that f(x1, . . . , xn), g(x1, . . . , xn) ∈ A[x1, . . . , xn] with g(x1, . . . , xn) 6=0. If there is some q(x1, . . . , xn) ∈ A[x1, . . . , xn] such that

f(x1, . . . , xn) = g(x1, . . . , xn)q(x1, . . . , xn),

then we say that g divides f, or g is a factor of f, denoted by g|f.When n > 1 there is no division algorithm in F [x1, . . . , xn], even

when F is a field. This is a significant difference between the polynomial

rings in one variable and in several variables.

Let F be a field. The field of fractions of F [x1, . . . , xn] is denoted

by F (x1, . . . , xn), called the field of rational functions in variables

x1, . . . , xn. Be careful about the parenthesis and bracket in the nota-

tion. Every element in F (x1, . . . , xn) is written as the quotient of two

polynomials.

We have mentioned before that a field of prime characteristic p is not

necessarily a finite field. Here is an example: the field Fp(x1, . . . , xn) is

an infinite field of characteristic p.

A polynomial f(x1, . . . , xn) ∈ F [x1, . . . , xn] is called a symmetric

polynomial if one of the following equivalent conditions is satisfied:

1) f(x1, . . . , xi, . . . , xj, . . . , xn) = f(x1, . . . , xj, . . . , xi, . . . , xn) for any

1 ≤ i < j ≤ n;

2) f(x1, . . . , xn) = f(xσ(1), . . . , xσ(n)) for any permutation σ.

Recall the elementary symmetric polynomials:

σ1 = x1 + · · ·+ xn =n∑

i=1

xi,

σ2 =∑

1≤i≤j≤n

xixj,

· · ·

σn = x1 · · ·xn.

The following theorem is the most important theorem concerning

symmetric polynomials.

Theorem 3.3.1. For any symmetric polynomials f(x1, . . . , xn) ∈F [x1, . . . , xn] there is a unique g(y1, . . . , yn) ∈ F [y1, . . . , yn] such that


f(x1, . . . , xn) = g(σ1, . . . , σn).

The theorem will be used in the last chapter of this book.

3.4 Factorization

The fundamental theorem of arithmetic says that every natural number

greater than one can be factored uniquely (up to the order of the fac-

tors) into a product of prime numbers. For polynomials in one variable,

the irreducible polynomials play the same role as the prime numbers in

the ring of integers and the unique factorization theorem holds in the

ring of polynomials in one variable. One may ask whether the unique

factorization theorem holds for any integral domain. The answer is

negative.

To investigate this property, we need to understand the exact mean-

ing of “unique factorization”.

Definition 3.4.1. Let A be an integral domain. For any a, b ∈ A

with b 6= 0, if a = bc for some c ∈ A, then b is said to divide a (or b is

a factor of a), denoted by b|aIf a|b, b|a for two nonzero elements a and b in A, the a is said to be

associated with b.

The divisibility is the immediate generalization of the divisibility in

the ring of integers and the ring of polynomials.

Example 3.4.2. A natural number n is associated to −n in Z.

Lemma 3.4.3. 1) a is associated to b if and only if a = ub for some

unit u in A.

2) Being associated is an equivalent relation.

Proof. 1) ⇐: Let u′ = u−1. Then b = u′a and hence a|b, b|a.⇒: It follows from a|b, b|a that a = ub, b = u′a for some u, u′ ∈ A.

Thus a = uu′a. Since a 6= 0 and A is an integral domain, uu′ = 1,

which implies that u and u′ are units.

2) is a direct consequence of 1).

3.4. FACTORIZATION 79

Definition 3.4.4. A nonzero element p in an integral domain A is

called an irreducible element if

1) it is not a unit and

2) p = ab implies that either a or b is a unit.

In other words, any factor of an irreducible element p is either a

unit or associated with p.

Definition 3.4.5. An integral domain A is a unique factorization

domain (abbreviated as UFD) if the following two conditions are sat-

isfied:

1) Every nonzero element a of A can be expressed as

a = cp1 · · · pr,

in which c is a unit and p1, . . . , pr are irreducible elements.

2) If cp1 · · · ps = dq1 · · · qt, with units c, d and irreducible elements

p1, . . . , ps, q1, . . . , qt, then s = t and pi is associated with qi for 1 ≤ i ≤ r

after a suitable permutation of q1, . . . , qt.

It is easy to verify that if p is an irreducible element in a unique

factorization domain A such that p|(a1 · · · an) then p|aj for some 1 ≤j ≤ n.

Now we are going to show that the polynomial ring in one variable

over an arbitrary field is a UFD.

Let F be a field. According to definition 3.4.4 a polynomial p(x) ∈F [x] is irreducible if and only if deg(p) > 0 and p(x) is not a product of

two polynomials of degrees less than deg(p). A polynomial of positive

degree which is not irreducible is called a reducible polynomial.

Lemma 3.4.6. Let f(x), g(x) ∈ F [x]. If f(x) is irreducible and f(x)

does not divide g(x), then gcd(f(x), g(x)) = 1.

Proof. Let d(x) = gcd(f(x), g(x)). Then f(x) = d(x)h(x) for

some h(x) ∈ F [x]. Since f(x) is irreducible, either deg(d) = 0 or

deg(h) = 0. Suppose that deg(h) = 0. Then h(x) is a nonzero ele-

ment in F. It follows from d(x)|g(x) that f(x)|g(x), contradicting the

assumption that f(x) does not divide g(x). Hence deg(d) = 0.


Theorem 3.4.7. Assume that an irreducible polynomial p(x) di-

vides f(x)g(x). Then p(x)|f(x) or p(x)|g(x).

Proof. Assume that p(x) does not divide f(x). Then Lemma 3.4.6

implies that gcd(p(x), f(x)) = 1. So a(x)p(x) + b(x)f(x) = 1 for some

a(x), b(x) ∈ F [x] by Theorem 3.2.10. Thus

a(x)p(x)g(x) + b(x)f(x)g(x) = g(x).

It follows that p(x)|g(x).

Corollary 3.4.8. Assume that an irreducible polynomial p(x) di-

vides f1(x) · · · fn(x). Then p(x)|fi(x) for at least one i.

Proof. Apply induction on n.

With these preparations we can prove that F [x] is a UFD.

Theorem 3.4.9. Let f(x) be a polynomial of positive degree over a

field F. Then

f(x) = cp1(x) · · · pn(x),

in which c is a nonzero element in F, p1(x), . . . , pn(x) are irreducible

monic polynomials. This equality is called an irreducible decompo-

sition of f(x).

The element c and the sequence p1(x), . . . , pn(x) of irreducible fac-

tors are uniquely determined by f(x) up to a permutation.

Proof. The existence of the factorization is proved by induction

on deg(f). When deg(f) = 1, f(x) is irreducible. Let c be the leading

coefficient of f and let f1 = f/c. Then f(x) = cf1(x) is the required

irreducible decomposition.

Assume that n > 1 and that every polynomial of degree less than

n has an irreducible decomposition. For a polynomial f of degree n, if

f(x) is irreducible then it is obvious that f has an irreducible decompo-

sition, otherwise f(x) = f1(x)f2(x) with deg(f1) < deg(f), deg(f2) <

deg(f). By induction hypothesis we have

f1(x) = cp1(x) · · · pr(x), (3.1)


f2(x) = c′pr+1(x) · · · pn(x), (3.2)

in which c, c′ are nonzero elements of F, p1(x), . . . , pn(x) are irre-

ducible monic polynomials. After plugging (3.2) and (3.1) into f(x) =

f1(x)f2(x) an irreducible decomposition is obtained.

The uniqueness is proved by induction on the number n of irre-

ducible factors of f. When n = 1 the factorization of f is obviously

unique. Assume that n > 1. Let

f(x) = c′q1(x) . . . qm(x)

be another irreducible decomposition. Since all pi(x), qj(x) are monic

polynomials, both c and c′ are equal to the leading coefficient of f(x),

so c = c′.

Since p1(x)|f(x), Corollary 3.4.8 implies that p1(x) divides some

qj(x), say, p1(x)|q1(x). Since both p1(x) and q1(x) are irreducible monic

polynomials, so p1(x) = q1(x) and

p2(x) · · · pn(x) = q2(x) · · · qm(x).

The induction hypothesis implies thatm = n and the sequences p2(x), . . . , pn(x)

and q2(x), . . . , qm(x) differ by a permutation.

This theorem means that the polynomial ring in one variable over

an arbitrary field is a UFD.

Let

f(x) = cp1(x) · · · pn(x)

be an irreducible decomposition of f(x). The factors are not necessarily

distinct. By collecting the same factors it can be written as

f(x) = cp1(x)e1 · · · pr(x)

er , (3.3)

in which p1(x), . . . , pr(x) are distinct irreducible monic polynomials and

e1, . . . , er are natural numbers. Such an expression is referred to as the

standard irreducible decomposition of f(x). The number ei is called


the multiplicity of the factor pi(x) in f(x). A factor with multiplicity

greater than one is called a multiple factor.

Let f(x) = a0xn + a1x

n−1 + · · · + an−1x + an ∈ F [x]. Define the

formal derivative of f(x) = a0xn + an−1x

n−1 + · · ·+ an−1x+ an as

f ′(x) = na0xn−1 + (n− 1)a1x

n−2 + · · ·+ an−1.

It has no geometric or physical interpretation of the derivative as in

calculus.

The well-known formulae

[f(x) + g(x)]′ = f ′(x) + g′(x)

and

(f(x)g(x))′ = f ′(x)g(x) + f(x)g′(x)

still hold.

Over a field of characteristic zero, the degree of f ′(x) is is equal

to deg(f) − 1 as long as deg(f) > 0. However this is not true for

polynomials over a field of positive characteristic. Let p > 0 be the

characteristic of the field over which the polynomials are defined. Let

m be a natural number divisible by p. Then the derivative of xm is

zero!

The following result is well-known.

Proposition 3.4.10. Let f(x) = cp1(x)e1 · · · pr(x)

er be the standard

irreducible decomposition of a polynomial f(x) with positive degree over

a number field. Then

gcd(f(x), f ′(x)) =r∏

i=1

pi(x)ei−1.

This result can be generalized to the fields of characteristic zero but

not to those of positive characteristic. However, the following result

holds for arbitrary fields.

Proposition 3.4.11. If gcd(f(x), f ′(x)) = 1, then f(x) has no

multiple factors.


Proof. Assume that f(x) = g(x)rh(x), in which deg(g) > 0,

h(x) 6= 0, r > 1. Then

f ′(x) = rg(x)r−1g′(x)h(x) + g(x)rh′(x)

= g(x)r−1[rg′(x)h(x) + g(x)h′(x)].

Hence g(x)|gcd(f(x), f ′(x)).

So far, we have learned that the ring of integers Z and the ring

of polynomials in one variable over a field are UFD’s. We wish to

construct new UFD’s from known UFD’s.

In the remaining of this section A is always assumed to be a UFD

with F as its field of fractions. Then A[x] is a subring of F [x].

Definition 3.4.12. A nonzero polynomial f(x) = anxn+an−1x

n−1+

· · · + a1x + a0 ∈ A[x] is a primitive polynomial if the only nonzero

elements in A dividing all coefficients a0, a1, . . . , an are units.

The following theorem is also known as Gauss’s Lemma.

Theorem 3.4.13 ( Gauss ). The product of two primitive polyno-

mials is still primitive.

Proof. Let

f(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0

and

g(x) = bmxm + bm−1x

m−1 + · · ·+ b1x+ b0

be primitive polynomials. Expand f(x)g(x) as

f(x)g(x) = cn+mxn+m + cn+m−1x

n+m−1 + · · ·+ c1x+ c0.

Then

ck =∑

i+j=k

aibj

for any k.

Suppose that f(x)g(x) is not primitive. Then there is an irreducible

element p in A dividing every ci. Let r (s resp.) be the largest index


such that ar (bs resp.) is not divisible by p. Then

cr+s = arbs +∑i>0

ar−ibs+i +∑i>0

ar+ibs−i

is not divisible by p. This leads to a contraction.

Lemma 3.4.14. Let f(x), g(x) be primitive polynomials. Assume

that f(x) = cg(x) for some c ∈ F. Then c is a unit in A. In other

words, two primitive polynomials f(x), g(x) ∈ A[x] are associated in

F [x] if and only if they are associated in A[x].

Proof. Write c as c = a/b where a, b ∈ A satisfy gcd(a, b) = 1.

Then

bf(x) = ag(x). (3.4)

Let ci and di denote the coefficients of the terms of degree i in f(x)

and g(x) respectively. Then bci = adi holds for every i by comparing

the terms of degree i in both sides of (3.4). Suppose that b is not a

unit. Since b is coprime with a, the equality bci = adi implies that b|di,

so b divides every coefficient of g(x). This contradicts the assumption

that g(x) is primitive. Hence b is a unit in A. The element a is also a

unit in A for the same reason.

Proposition 3.4.15. If a primitive polynomial in A[x] is reducible

in F [x] then it can be factored into a product of two polynomials in

A[x] of lower degrees.

Proof. Let f be a primitive polynomial such that f(x) = g(x)h(x)

for some g(x), h(x) ∈ F [x] with deg(g) < deg(f), deg(h) < deg(f).

Choose r, s ∈ F such that rg(x), sh(x) are primitive. This is possible

since A is a UFD.

By Theorem 3.4.13 the right hand side of

rsf(x) = [rg(x)][sh(x)]

is primitive. Then Lemma 3.4.14 implies that rs is a unit in A.

The units and irreducible elements in A[x] are characterized by the

following lemma.


Lemma 3.4.16. 1) Every unit in A[x] is a unit in A.

2) A polynomial f(x) ∈ A[x] of positive degree is irreducible in A[x]

if and only if it is primitive and it is irreducible in F [x].

Proof. 1) Let g(x) be a unit in A[x]. Then g(x)h(x) = 1 for some

h(x) ∈ A[x]. Hence deg(g) = deg(h) = 0, which implies that both g(x)

and h(x) are units in A.

2) First assume that f(x) is an irreducible element in A[x]. Let a

be the greatest common divisor of all coefficients of f(x). Then f(x) =

af1(x), with f1(x) being primitive. f1(x) is not a unit since deg(f1) =

deg(f) > 0. Thus a is a unit in A since f(x) is irreducible in A[x]. This

implies that f(x) is primitive.

By Proposition 3.4.15 if f(x) is reducible in F [x] then it would

be decomposed into the product of two polynomials in A[x] of lower

degrees. This would contradict the assumption that f(x) is irreducible

in A[x]. Hence f(x) is irreducible in F [x].

Conversely assume that a primitive polynomial f(x) ∈ A[x] is irre-

ducible in F [x]. Assume that f(x) = g(x)h(x) with g(x), h(x) ∈ A[x].

Since f(x) is irreducible in F [x], either deg(g) = 0 or deg(h) = 0.

Assume that deg(g) = 0 without loss of generality. This amounts to

saying that g(x) = c is a nonzero element in A. Since f(x) is primitive,

c is a unit. This shows that f(x) is an irreducible element in A[x].

With all these preparation, we can prove the following theorem:

Theorem 3.4.17. A[x] is a UFD if A is a UFD.

Proof. We need to verify for A[x] the two conditions in the defi-

nition of UFD are satisfied.

1) Let f(x) be a nonzero element in A[x]. We show that f(x) can

be decomposed into a product of a unit and some irreducible elements

in A[x] by induction on deg(f).

If deg(f) = 0, then f ∈ A and f is decomposed into a product of a

unit and some irreducible elements, since A is a UFD by assumption.

Assume that deg(f) > 0. Then f = cf1 with c ∈ A and f1 prim-

itive. Let c = dc1c2 · · · cr be the irreducible decomposition of c in A

where d is a unit and c1, c2, . . . , cr are irreducible. If f1 is irreducible


then f = dc1c2 · · · crf1 is the required decomposition already, other-

wise Proposition 3.4.15 implies that f1 = gh, in which g, h ∈ A[x],

deg(g) < deg(f), deg(h) < deg(f). By induction hypothesis both g

and h have irreducible decompositions. Hence f has an irreducible

decomposition.

2) Assume that cp1 · · · ps = dq1 · · · qt in which c, d are units and

p1, . . . , ps, q1, . . . , qt are irreducible elements in A[x]. Assume that

p1, . . . , pu, q1, . . . , qv ∈ A

and

pu+1, . . . , ps, qv+1, . . . , qt /∈ A.

By Lemma 3.4.16 pu+1, . . . , ps, qv+1, . . . , qt are primitive and irreducible

in F [x]. By Theorem 3.4.13 the products pu+1 · · · ps and qv+1 · · · qt are

also primitive polynomials. Hence

p1 · · · pu = λq1 · · · qv (3.5)

and

pu+1 · · · ps = µqv+1 · · · qt, (3.6)

in which λ, µ are units. Since A is a UFD, u = v and after a suitable

permutation of indices pi is associated with qi for 1 ≤ i ≤ u by (3.5).

It follows from (3.6) and Theorem 3.4.9 that s = t and pi is associ-

ated with qi in F [x] after a suitable permutation for 1 < i ≤ s. Lemma

3.4.14 implies that pi and qi are associated in A[x]. The proof of the

uniqueness is concluded.

Corollary 3.4.18. If A is a UFD, then A[x1, x2, . . . , xn] is a UFD.

Proof. Apply the theorem n times.

Corollary 3.4.19. 1) The polynomial ring F [x1, . . . , xn] is a UFD

for any field F.

2) Z[x1, . . . , xn] is a UFD.

Exercises

3.5. * POLYNOMIAL FUNCTIONS 87

1. Let F3 be the field with three elements. Decompose the polyno-

mial x9 − x over F3 into a product of irreducible polynomials.

2. Let F be a field with four elements. Let α be a nonzero element

of F. Show that x3 + αx+ 1 is an irreducible polynomial over F.

3. Let A be the set of all complex numbers in the form m + n√

5i

with m,n ∈ Z.1) Show that A is a subring of the complex number field.

2) Show that 3, 2 +√

5i, 2−√

5i are irreducible elements in A.

3) Show that 3 and 2 +√

5i are not associated.

4) Show that A is not a UFD.

3.5 * Polynomial Functions

Definition 3.5.1. Let F be a field and

f(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0 ∈ F [x],

b ∈ F. Then

f(b) = anbn + an−1b

n−1 + · · ·+ a1b+ a0 ∈ F

is called the value of f(x) at b. If f(b) = 0, then b is called a zero of

f(x).

For any field F, a function on F with values in F is a map from F

into F by definition. Any polynomial f(x) over F determines a function

by b 7→ f(b). Such a function is called a polynomial function. Let

f(x) and g(x) be distinct polynomials. We may ask whether they

determine different functions. At the first glance the answer seems to

be affirmative. However this is an illusion.

Let F be a finite field containing q elements. Then xq and x are

two distinct polynomials over F.We show that they determine the same

function on F.

The set F ∗ = F\0 is a group of order q− 1 under multiplication.

Thus bq−1 = 1 for every b ∈ F ∗ by Lagrange’s theorem. Hence bq = b


for every b ∈ F. This means that the polynomials xq and x determine

the same polynomial function.

This example tells us that the concepts of polynomial and polyno-

mial functions over a finite field are different. In fact, over a finite field

there are only finitely many functions but there are infinitely many

polynomials. A wonderful result is that every function on a finite field

is a polynomial function (cf. Exercise 3).

Theorem 3.5.2. Assume that f(x) ∈ F [x], b ∈ F. Then f(b) is the

remainder of f(x) divided by x− b.

Proof. By the division algorithm there are q(x), r(x) ∈ F [x] such

that

f(x) = (x− b)q(x) + r(x), (3.7)

with deg(r) < deg(x− b) = 1. Hence r(x) is a constant c ∈ F.Replace x in (3.7) by b and f(b) = c is obtained.

Corollary 3.5.3. An element b ∈ F is a zero of f(x) if and only

if (x− b)|f(x).

Corollary 3.5.4. A nonzero polynomial of degree n over a field

F has at most n distinct zeros in F.

Proof. Suppose that a polynomial of degree n has n + 1 distinct

elements b1, . . . , bn+1. Then (x − bi)|f(x) for i = 1, . . . , n + 1. Hence

(x− b1) · · · (x− bn+1)|f(x), which is impossible.

Corollary 3.5.5. Let f(x), g(x) be nonzero polynomials over F

with degrees at most n. If there are n+1 distinct b1, . . . , bn+1 ∈ F such

that f(bi) = g(bi) for every i, then f(x) = g(x).

Proof. Let h(x) = f(x)− g(x). Suppose that h(x) 6= 0, then h(x)

would be a nonzero polynomial of degree at most n with n+ 1 distinct

zeros b1, . . . , bn+1, which is impossible. Hence h(x) = 0.

Definition 3.5.6. If (x − b)d|f(x) but (x − b)d+1 does not divide

f(x), then b is a zero of multiplicity d of f(x). A zero of f(x) of mul-

tiplicity greater than one is called a multiple zero of f(x).

3.5. * POLYNOMIAL FUNCTIONS 89

Theorem 3.5.7. Let f(x) be a nonzero polynomial of degree n and

let b1, . . . , br be the set of all distinct zeros of f(x) with multiplicities

d1, . . . , dr. Then

d1 + · · ·+ dr ≤ n.

Proof. Since

f(x) = (x− b1)d1 · · · (x− br)

drg(x)

where g(x) is a nonzero polynomial. Hence

d1 + · · ·+ dr ≤ d1 + · · ·+ dr + deg(g) = deg(f) = n.

Example 3.5.8. Let F be a finite field containing q elements. Then

xq − x =∏a∈F

(x− a)

in F [x].

Proof. Since both sides are monic polynomials of degree q, it suf-

fices to show that the right hand side divides the left hand side. This is

true since aq = a for every element a ∈ F as we have verified before.

Exercises

1. Let F be a field containing infinitely many elements. Let f(x), g(x) ∈F [x]. If f(x) and g(x) determine the same function on F, then f(x) =

g(x).

2. Let I = f(x) ∈ R[x]|f(2) = f ′(2) = f ′′(2) = 0. Show that I is

an ideal of R[x] and find a monic polynomial that generates I. Is the

set J = f(x) ∈ R[x]|f(2) = 0, f ′(3) = 0 an ideal of R[x]? Explain

the reason.

3. Calculate the number of polynomial functions on a finite field

containing q elements. Use this to show that every function on a finite

field is a polynomial function.

Chapter 4

Vector Spaces

The freshman linear algebra is restricted to the linear algebra over

number fields, especially the real and complex number fields. Now that

the concept of abstract fields is at our disposal, most results we have

known in linear algebra can be generalized to arbitrary fields. There

are a few materials that can not be generalized in straight forward way.

The concepts of vector space and its related concepts will be for-

mulated in the framework of algebraic structures. This point of view

will enable us to deepen our understanding of the subject.

4.1 Vector Spaces and Linear Transformations

Definition 4.1.1. Let F be a field and let V be an (additive)

abelian group. If there is an operation

F × V → V, (a, u) 7→ au,

called the scalar multiplication, satisfying the following four conditions

1) 1u = u for any u ∈ V ;

2) (ab)u = a(bu) for any a, b ∈ F, u ∈ V ;

3) (a+ b)u = au+ bu for any a, b ∈ F, u ∈ V ;

4) a(u+ w) = au+ aw for any a ∈ F, u, w ∈ V.then V is called a vector space over the field F. The elements in

V are called vectors.

90

4.1. VECTOR SPACES AND LINEAR TRANSFORMATIONS 91

A vector space is a structure of different nature from groups or rings.

The scalar multiplication is an operation different from those in groups

and rings. Instead of the multiplication of two elements in V, it is the

multiplication of two elements from two different sets F and V.

The concepts of linear independence and basis can be defined in the

same way as in elementary linear algebra.

Definition 4.1.2. 1) A set S of vectors in a vector space V over

a field F is linear independent if for any finite number of vectors

u1, . . . , un with n > 0 in S and any nonzero elements c1, . . . , cn in

F, c1u1 + · · ·+ cnun is not equal to zero.

2) A subset S spans the vector space V if for every w ∈ V there

exist c1, . . . , cn ∈ F, u1, . . . , un ∈ S such that w = c1u1 + · · ·+ cnun.

3) A linearly independent subset of a vector space is a basis of the

space if it spans the whole space. The number of elements in a basis is

defined to be the dimension of the space, denoted by dim(V ).

If a vector space is spanned by a finite number of vectors, then the

existence of the basis can be proved in the same way as in elementary

linear algebra. In that case, the number of elements in the basis is

independent of the choice of the basis. Hence the dimension is a well-

defined natural number. If a vector space cannot be spanned by a finite

number of vectors, then we say that it is an infinite dimensional space.

The exists a basis for an infinite dimensional space. The proof uses the

axiom of choice.

Let e1, . . . , en be a basis of a finite dimensional vector space V.

Every vector u in V can be uniquely expressed as a linear combination

of e1, . . . , en :

u = c1u1 + · · ·+ cnun.

This gives a one to one correspondence between V and the space of

column vectors (or row vectors), which depends on the choice of basis.

A subset W of a vector space V is called a subspace if it is closed

under addition and scalar multiplication.

Let V1 and V2 be two vector spaces over the same field F. A map

f from V1 to V2 is a linear map if it preserves the addition and scalar

multiplication, i.e., f(au+ bv) = af(u)+ bf(v) fo any a, b ∈ F and any

92 CHAPTER 4. VECTOR SPACES

u, v ∈ V1. A linear map is also refered to as a homomorphism. When

V1 = V2 = V, a linear map is called a linear transformation. In analysis

a linear map is also called a linear operator. In algebra a linear trans-

form is sometimes called an endomorphism. A bijective endomorphism

is an automorphism, which is an invertible linear transformation.

Let f be a linear transformation of a finite dimensional vector space

V. After fixing a basis e1, . . . , en of V the linear transformation f de-

termines an n× n matrix with entries in F. The rules of addition and

multiplication of matrices over an arbitrary field are the same as over

a number field. The concepts such like rank of a matrix, determinant

of a square matrix are defined without modification.

The set of all invertible linear transformations forms a group un-

der composite operation, denoted by GL(V ), which is isomorphic to

GLn(F ).

The theory of system of linear equations, characteristic polynomials,

eigenvalues and eigenvectors still holds for arbitrary field.

Besides groups, rings, fields, vector spaces, there is another useful

algebraic structure. Let A be a ring containing a field F as a subring.

Then A is called an F -algebra.

For example, the ring Mn(F ) consisting of all n×n matrices over a

field F is an F -algebra, the skew field of quaternions is an R-algebra.

Every F -algebra A is a vector space over F, in which the scalar mul-

tiplication coincides with the multiplication of A. If a field L contains a

subfield F, then L is an F -algebra. There is an underlying structure of

F -space. This point of view is useful, which we will discuss in greater

details in the future.

Let p be a prime number. An abelian group whose every nonzero

element has order p has a hidden structure of vector space over Fp, as

explained in the next lemma.

Lemma 4.1.3. Let p be a prime number and let G be an (additive)

abelian group. If pa = 0 for every nonzero a ∈ G, then G is a vector

space over the finite field Fp.

Proof. It is known that Fp = 0, 1, . . . , p− 1. For every i ∈ Fp

and u ∈ G, define the scalar multiplication iu = iu. Let us verify that

4.1. VECTOR SPACES AND LINEAR TRANSFORMATIONS 93

G become a vector space under this scalar multiplication

1) 1u = 1u = u is obvious.

2) For arbitrary i, j ∈ Fp, we have ij = k, where k = ij − rp for

some nonnegative integer r. Thus (ij)u = ku = (ij − rp)u. Since the

order of u divides p, we have rpu = 0. Therefore (ij)u = iju = i(ju).

3) Similarly one can easily show that (i+ j)u = iu+ ju

and

4) i(u+ v) = i(u+ v) = iu+ iv = iu+ iv.

The following example shows how useful this lemma is.

Example 4.1.4. Let p be a prime number and let A = Zp2 ⊕Zp2 ⊕Zp3 . Find the number of all non-cyclic subgroups of A with order p2.

Solution:

Every element of A is expressed as (a, b, c) in which a, b ∈ Zp2 and

c ∈ Zp3 . Its order is maxo(a), o(b), o(c). Recall that o(a) is the order

of a. A subgroup H of A with order p2 is non-cyclic if and only if the

order of every nonzero element in H has order p. Let V be the subset

of all elements in A with order less than or equal to p. Then V is a

subgroup of order p3. By Lemma 4.1.3 G is a 3-dimensional space over

Fp. Hence a subgroup H of A with order p2 is non-cyclic if and only if

it is a 2-dimensional subspace of V. Thus the problem is converted into

the counting of 2-dimensional subspaces in a 3-dimensional space over

Fp. This becomes a problem in linear algebra.

There are at least two methods of counting.

A 2-dimensional subspace is spanned by two linearly independent

vectors (a1, a2, a3) and (b1, b2, b3). The first vector is a nonzero vector.

Hence there are p3 − 1 choices for (a1, a2, a3). The second vector is

linear independent from the first if and only if (b1, b2, b3) 6= λ(a1, a2, a3)

for any λ ∈ Fp\0. Hence with fixed (a1, a2, a3), the second vector

(b1, b2, b3) has p3 − p choices. Thus the number of choices of ordered

pairs of linearly independent vectors is (p3 − 1)(p3 − p).

Two ordered pairs (a1, a2, a3), (b1, b2, b3) and (c1, c2, c3), (d1, d2, d3)of linearly independent vectors span the same 2-dimensional subspace


if and only if there is a nonsingular matrix

P =

[p11 p12

p21 p22

]

such that [a1 a2 a3

b1 b2 b3

]= P

[c1 c2 c3

d1 d2 d3

].

By using the same method as above, the number of 2× 2 nonsingular

matrices with entries in Fp is (p2− 1)(p2− p). Therefore the number of

2-dimensional spaces of F3p is

(p3 − 1)(p3 − p)

(p2 − 1)(p2 − p)= p2 + p+ 1.

The second method is more elegant. Every 2-dimensional subspace

is the space of solutions of a nonzero homogeneous linear equation.

Two homogeneous linear equations

a1x1 + a2x2 + a3x3 = 0

and

b1x1 + b2x2 + b3x3 = 0

have the same solutions if and only if there is a nonzero λ ∈ Fp such

that

(a1, a2, a3) = λ(b1, b2, b3).

Hence the total number of 2-dimensional subspaces is equal to

p3 − 1

p− 1= p2 + p+ 1.2

Exercises

1. Let V be the real vector space of all continuous real functions

on the closed interval [0, 1]. Show that x3, sin x and cos x are linearly

independent.

2. Let V be a vector space of finite dimension over a field F and

4.2. QUOTIENT SPACES 95

let f be a linear transformation of V. Show that f is surjective if and

only if f is injective.

3. Let R be an integral domain containing a field F. If R, regarded

as a vector space over F, has finite dimension, then R is a field.

4.2 Quotient spaces

Let W be a subspace of a vector space V over a field F. By disregarding

the scalar multiplication, V becomes an abelian group. Hence W is a

normal subgroup of V. Let V/W the the quotient group. We wish to

construct a scalar multiplication on V/W so that it becomes a vector

space over F.

For any a ∈ F and any x ∈ V/W, in which x ∈ V, define

ax = ax.

Let x′ ∈ V be another representative of x. Then x′ − x ∈ W. So

ax′ − ax = a(x′ − x) ∈ W, which means that ax′ = ax. Hence ax does

not depend upon the choice of x. It is easy to verify that V/W becomes

a vector space over F under this scalar multiplication. The space V/W

is called the quotient space of V over the subspace W.

Let w1, . . . , wr be a basis of the subspace W. Extend it to a basis

w1, . . . , wr, ur+1, . . . , un of the whole space V.

Proposition 4.2.1. ur+1, . . . , un form a basis of V/W. In particu-

lar, dim(V/W ) = dim(V )− dim(W ).

Proof. Assume that ar+1, . . . , an ∈ F satisfy

ar+1ur+1 + · · ·+ anun = 0.

Then

ar+1ur+1 + · · ·+ anun ∈ W.

Hence

ar+1ur+1 + · · ·+ anun = b1w1 + · · ·+ brwr.


It follows that

−b1w1 − · · · − brwr + ar+1ur+1 + · · ·+ anun = 0.

Since w1, . . . , wr, ur+1, . . . , un are linearly independent,

ar+1 = · · · = an = 0.

This proves that ur+1, . . . , un are linearly independent in V/W.

For any x ∈ V/W there exist a1, . . . , an ∈ F such that

x = a1w1 + · · ·+ arwr + ar+1ur+1 + · · ·+ anun.

Hence

x = ar+1ur+1 + · · ·+ anun.

This shows that every element of V/W is a linear combination of

ur+1, . . . , un.

An arbitrary linear transformation f of V generally does not induce

a linear transformation of the quotient space V/W.

Proposition 4.2.2. For an f -invariant subspaceW, which means

that f(u) ∈ W for any u ∈ W, define

f : V/W → V/W, x 7→ f(x).

Then f is a linear transformation of V/W.

Proof. An important task is to check that the map f is well-

defined, i.e.,f(x) does not depend upon the choice of the representative

x.

Let x′ be another representative of x. Then x′ − x ∈ W. Since W is

invariant under f, f(x′ − x) ∈ W. Hence f(x′) = f(x).

The remaining part of the proof is straightforward and is left to the

readers.

Let u1, . . . , un be a basis of V, in which u1, . . . , ur form a basis of

4.2. QUOTIENT SPACES 97

the f -invariant subspace W. The matrix of f under this basis is[A B

0 C

],

in which A and C are square matrices of orders r and n−r respectively.

It is easy to see that A is the matrix of f |W under the basis u1, . . . , ur

and C is the matrix of f under the basis ur+1, . . . , un.

Exercises

1. Let p be a prime number and let H be a cyclic group of order

p. Let G be the direct sum of n copies of H, i.e., G = Hn. Assume

that 1 ≤ n ≤ p. Show that the automorphism group Aut(G) does not

contain elements of order p2.

Chapter 5

Topics in Group Theory

In this chapter we choose three topics in group theory. The first one

is the Sylow theorem for finite groups. Since its proof involves clever

application of group actions, the basic technique of group actions is

discussed in the first section. The second topic is the structure theorem

of finitely generated abelian groups, which has many applications. It is

indispensable in some branches of mathematics such as number theory,

algebra and topology. The role of this theorem is similar to the Jordan

canonical form in linear algebra. The last topic is the solvable groups,

which will be used in the last chapter on Galois theory.

5.1 The Orbit Formula of an Action by a Finite

Group

The group action is a powerful tool in the theory of finite groups. Its

application is full of tricks. For basic definitions see the last section of

Chapter 1.

Assume that a finite group G acts on a finite set S. The S is the

union of mutually disjoint orbits. Let Gx be an orbit. Then

|Gx| = (G : Stab(x)).

This formula can also be expressed as |Gx| · |Stab(x)| = |G|, which

roughly means that the larger the orbit is the smaller the stabilizer is.

Unlike the coset decomposition of a group, different orbits may contain

98

5.1. THE ORBIT FORMULA OF AN ACTION BY A FINITE GROUP99

different number of elements.

Choose one representative in each orbit and they form a finite set

D = x1, . . . , xd. This means that S is the disjoint union of the orbits

Gx1, . . . , Gxd. Such a set D is called the complete system of represen-

tatives of the orbits. Thus we obtain the following formula

|S| =d∑

i=1

(G : Stab(xi)). (5.1)

which is referred to as the orbit decomposition formula or simply orbit

formula. This is the basis of a counting technique in finite groups. This

formula will be used to prove the celebrated Sylow theorem in the next

section. In order to get familiar with this technique, let us look at some

simple examples.

Example 5.1.1. The order of the group of symmetry of a cube.

This group acts transitively on the set of eight vertices of the cube.

Fix one vertex. The stabilizer of that vertex is a group of order 6.

Hence the order of the group of symmetry is equal to 48.

The orders of groups of symmetry for other regular polyhedra can

be computed in the same way.

Example 5.1.2. Let G be a finite group with |G| > 2 and the

conjugacy class of one element a ∈ G contains two elements. Then G

is not a simple group.

Proof. The group G acts on G itself by conjugacy. By the given

condition the orbit on which a is located has length 2. Thus (G :

Stab(a)) = 2. So Stab(a) / G by 1.5.8. Since 1 < |Stab(a)| < |G|,Stab(a) is a nontrivial normal subgroup of G.

The examples above only use the formular |Gx| = (G : Stab(x)).

The following example is the application of the orbit formula 5.1.

Example 5.1.3. Let p be a prime number and let G be a group with

|G| = pn(n > 1). Assume that H / G and |H| = p. Then H ⊆ C(G),

where C(G) is the center of G.

100 CHAPTER 5. TOPICS IN GROUP THEORY

Proof. Since H is a normal subgroup of G, the group G acts on

H by conjugacy. Let e = x1, . . . , xm be a complete system of represen-

tatives of the orbits. It follows from the orbit formula that

m∑i=1

(G : Stab(xi)) = |H| = p.

Since|G| = pn, every number (G : Stab(xi)) is a power of p. Since

Stab(e) = G, i.e., (G : Stab(x1)) = 1, so m = p and Stab(xi) = G for

each i. Hence xi ∈ C(G) for every i.

Let p be a prime number. A group of order pn with n > 0 is called

a p-group.

Proposition 5.1.4. Let G be a p-group for some prime number p.

Then the following statements hold:

1) The center Z of G contains more than one element;

2) (Normalizer grows) Let H be a proper subgroup of G. Then H 6=NG(H).

Proof. 1) Assume that |G| = pn with n > 0. Consider the action

of G on itself by conjugacy. Let Z denote the center of G and let

r = |Z|. Then Z = x ∈ G|Stab(x) = G. It follows from 1 ∈ Z that

r > 0. By the orbit formula the sum of the lengths of all orbits is equal

to pn. Since the length of every orbit containing more than one element

is divisible by p, |Z| is divisible by p. Hence r > 1.

2) Suppose that n is the smallest integer such that the statement

fails. By hypothesis there is a proper subgroup H of G such that

H = NG(H). Since Z ⊆ NG(H), the subgroup H = NG(H) contains

Z. Since |H| > 1 by the result of 1), the order of G/Z is strictly less

then pn. Since H/Z is a proper subgroup of G/Z, we have H/Z 6=NG/Z(H/Z). There is some g ∈ G such that g ∈ NG/Z(H/Z)\H/Z.Hence g ∈ NG(H)\H, leading to contradiction.

Example 5.1.5. Any group G of order p2 is abelian for any prime

number p.

5.2. SYLOW SUBGROUPS 101

Proof. If G is not abelian, then the center Z of G is a proper

subgroup of G. Then Proposition 5.1.4 implies that |Z| = p. Hence

G/Z is a cyclic group. Let y with y ∈ G be a generator of G/Z. Then

G = ayia∈Z,0≤i<p. It is clear ayi commutes with byj for any a, b ∈ Zand any i, j.

It follows that the smallest possible non-abelian p-subgroup has or-

der 8. Indeed such a group exists: The eight quaternions ±1,±i,±j,±kform a non-abelian group of order 8.

An elegant application of the orbit formula is the proof of Wedder-

burn’s theorem of finite skew fields. Refer to Appendix 2.

Exercises

1. Let n = 2m, where m is an odd integer greater than one. Let G

be a subgroup of Sn of order 2r in which r is a natural number. Show

that there are 1 ≤ i < j ≤ n such that σ(i) ∈ i, j, σ(j) ∈ i, j for

every σ ∈ G.

2. Let G be a finite group of order n. Denote by S(G) the set of all

bijections from G to G. Let στ denote the composite of σ, τ ∈ S(G).

Then S(G) becomes a group.

1) Show that S(G) is isomorphic to Sn;

2) Prove Cayley’s theorem: every finite group is isomorphic to a

permutation group.

3) Let G be a subgroup of the symmetric group Sn, in which n is

an odd integer. Assume that |G| is a power of 2. Show that there is

some i among 1, 2, . . . , n such that σ(i) = i for all σ ∈ G.

5.2 Sylow subgroups

In this section G is always a fixed finite group with |G| = prs, in which

p is a prime number, r > 0, and s is not divisible by p.

A subgroup H of G with |H| = pr is called a Sylow p-subgroup of

G, or simply a Sylow subgroup if p is fixed.

The Sylow’s theorem answers the following three questions:

1) Does a Sylow subgroup always exist?


2) How many Sylow subgroups are there?

3) What is the relation between Sylow subgroups?

First we need a preparatory lemma.

Lemma 5.2.1. For a natural number n = prs, in which r > 0 and s

is not divisible by p, the congruence(n

pr

)≡ s (mod p)

holds.

Proof. First of all(

npr

)is the coefficient of xpr

in the expansion of

(1 + x)n.

Over the finite field Fp the equality

(1 + x)n = [(1 + x)pr

]s = (1 + xpr

)s

holds. Hence the coefficient of xprof (1 + x)n over Fp is equal to s.

This means that (n

pr

)≡ s (mod p).

Now we state and prove the main theorem on Sylow subgroups.

Theorem 5.2.2 (Sylow). Let G be a finite group and let p be a

prime factor of |G|. For this fixed p the following statements hold:

1) Sylow subgroups exist in G.

2) Let m be the number of Sylow subgroups in G, then m ≡ 1

(mod p).

3) All Sylow subgroups are conjugate.

Proof. Assume that |G| = n = prs, in which r > 0 and p does

not divide s. Let Γ be the set of all subsets of G containing pr elements

and let N be the number of members in Γ. Then N =(

npr

). By Lemma

5.2.1 N ≡ s (mod p).

For any A ∈ Γ and any g ∈ G, let gA = ga|a ∈ A. Then gA ∈ Γ.

This gives an action of G on Γ.


Let A1, A2, . . . , Aq ∈ Γ be a complete system of representatives of

the orbits, i.e., they are in different orbits and every member of Γ is

in the same orbit with some Ai. For each 1 ≤ i ≤ q Let Hi denote the

stabilizer of Ai. This means that ha ∈ Ai for any a ∈ Ai, h ∈ Hi. Then

Ai is a union of some right cosets of Hi. Hence |Hi| divides the number

of elements in Ai, which implies that |Hi| = pdi for some integer di ≤ r.

The number of elements in the orbit of Ai is equal to |G|/pdi =

pr−dis. By Formula (5.1) N is equal to∑q

i=1 pr−dis. Since N is not

divisible by p, there is some i such that di = r. The corresponding Hi

is a Sylow p-subgroup. Thus 1) is proved.

Moreover, all orbits can be divided into two families: the first family

consists of all orbits of length s and the second family contains the

remaining orbits. This implies that the length of each orbit in the

second family is divisible by p. Assume that there are m orbits in the

first family. Then N ≡ ms (mod p) by the equality N =∑q

i=1 pr−dis.

It follows from N ≡ s (mod p) that m ≡ 1 (mod p).

Let A be a member in an orbit of the second family. Then A is not

a subgroup for otherwise A would be contained in its stabilizer, which

would contradict the inequality (G : Stab(A)) > s.

Every orbit of the first family contains an element A such that

1 ∈ A. Let H be the stabilizer of A. Then h = h ·1 ∈ A for each h ∈ H.Thus H ⊆ A. So H = A since |H| = pr = |A|. Hence A is a Sylow p-

subgroup of G. Moreover the orbit containing A consists all left cosets

of H in G. This shows that every orbit of the first family contains exact

one Sylow p-subgroup. Therefore there are m Sylow p-subgroups in G

and 2) is proved.

Finally we prove a statement stronger than 3):

Let P is a Sylow p-subgroup and let K be a subgroup of G with

|K| = pt for some natural number t. Then there is some g ∈ G such

that K ⊆ gPg−1.

Let Σ be the set of all subgroups of G conjugate to P. Then gBg−1 ∈Γ for any g ∈ G,B ∈ Σ. This gives a transitive action of G on the set

Σ. Since P is contained in its stabilizer, the number of elements in Σ

is a factor of s, which is not divisible by p.

This action induces an action of K on Σ. Let Ω ⊆ Σ be an orbit


of the action of K on Σ. For any B ∈ Ω, if the stabilizer of B is not

equal to K then |Ω| is divisible by p. However the sum of lengths of

all orbits is equal to |Σ|, which is not divisible by p. Thus there is

some P ′ = gPg−1 ∈ Σ whose stabilizer is equal to K. This implies that

K ⊆ NG(P ′). Since P ′ / NG(P ′), the conditions to apply the second

isomorphism theorem 1.7.14 are satisfied (see the figure below). So

KP ′/P ′ ∼= K/(K ∩ P ′).

NG(P ′)

KP ′

vvvvvvvvv

HHHHHHHHH

K

HHHHHHHHH P ′

vvvvvvvvv

K ∩ P ′

Hence |KP ′/P ′| = |K/K ∩ P ′] is a power of p, which implies that

|KP ′| is also a power of p. Hence |KP ′| ≤ pr = |P ′|,which implies

that KP ′ = P ′. Therefore K ⊆ P ′. This concludes the proof of the

statement.

It is easy to use this statement to prove 3). Let P and P ′ be

arbitrary Sylow p-subgroups. By the statement we have just proved

there is g ∈ G such that P ′ ⊆ gPg−1. It follows from |P ′| = |gPg−1|that P ′ = gPg−1.

Corollary 5.2.3. Let G be a finite group of order prs, where r > 0,

s is not divisible by the prime number p. Then the number of Sylow p-

subgroups in G divides s.

Proof. Let n be the number of Sylow p-subgroups in G. Since the

conjugate action of G on the set of all Sylow p-subgroups is transitive

by 3) of Sylow’s theorem, the number n divides prs. On the other hand,

n ≡ 1 (mod p) implies that n is not divisible by p. Hence n|s.

Corollary 5.2.4. Let p be a prime factor of the order of a finite

group G. Then there exists an element of order p.


Proof. Take any element a of a Sylow p-subgroup P of G which

is not the identity element. Then the order of a is pt for some natural

number t and apt−1is an element of order p.

Remark 5.2.5. In some textbooks this corollary appears as a prepara-

tory lemma to prove Sylow’s theorem.

For a finite group, Corollary 5.2.3 and 2) of Sylow’s theorem can

often be used to prove that this group has only one Sylow p-subgroup

for a prime factor p of the order of the group. This will imply that

the group is not a simple group if p is not the only prime factor of the

given group. Look at the following typical example.

Example 5.2.6. Let G be a group of order prs, in which p is a

prime number, r > 0. Assume that 1 < s < p. Then G is not a simple

group.

Proof. Let n denote the number of Sylow p-subgroups of G. Then

n ≡ 1 (mod p) by Sylow’s theorem and n divides s by Corollary 5.2.3.

It follows from the hypothesis s < p that n = 1, which means that G

has only one Sylow p-subgroup, which is a nontrivial normal subgroup

of G.

Exercises

1. Let p be a prime number and let Fp be the finite field containing

p elements. Let GLn(Fp) be the group of all invertible n× n matrices

over Fp and let U be the subgroup of GLn(Fp) consisting of all upper

triangular matrices whose elements on the diagonal are equal to 1.

Show that U is a Sylow p-subgroup of GLn(Fp) and find the number

of conjugates of U in GLn(Fp).

2. Let H be a subgroup of a finite group G such that every element

in G is conjugate to some element in H. Show that H = G.

3. Prove that a group of order 1225 is abelian.

4. Prove that a group of order 640 is not a simple group.

5. Let G be a simple group of order 168 and let S be the set of all

Sylow 7-subgroups of G. Let H ∈ S. Then H acts on S by conjugacy,


i.e., g ∈ H carries K ∈ S into gKg−1. Find the number of orbits and

the length of each orbit.

6. Find all Sylow 2-subgroups of S4.

7. Assume that a finite group G has 50 Sylow 7-subgroups and P

is a Sylow 7-subgroup. Let N = NG(P ) = g ∈ G|gPg−1 = P.1) Show that N is a maximal subgroup of G, i.e.,N 6= G and there

is no subgroup between N and G.

2) If N has a normal Sylow 5-subgroup Q, then QG. [ Hint Use

2) of Proposition 5.1.4.]

8. Let H be a proper normal subgroup of a finite group G and let p

be a prime factor of |G/H|. Show that the number of Sylow p-subgroups

of G/H divides the number of Sylow p-subgroups of G.

5.3 * Structure of finitely generated abelian groups

Let G be an abelian group, whose operation is always denoted by + in

this section. If there are finitely many elements g1, . . . , gn inG such that

every element g ∈ G can be expressed as g = k1g1 + · · ·+kngn for some

integers k1, . . . , kn, then G is called a finitely generated abelian group

and g1, . . . , gn is a set of generators. If furthermore the coefficients

k1, . . . kn are uniquely determined by g for every g ∈ G, then G is called

a finitely generated free abelian group and g1, . . . , gn is a basis of G.

LetG be a finitely generated free abelian group with basis g1, . . . , gn.It is easy to verify that the map

f : Z⊕ · · · ⊕ Z → G,

(k1, . . . , kn) 7→ k1g1 + · · ·+ kngn

is well defined and f is an isomorphism.

This tells us that the structure of a finitely generated free abelian

group is simple. It is the direct sum of a finite number of copies of Z. Itsstructure is totally determined by the number of elements in the basis.

This number is called the rank of the group, similar to the dimension

of a finite dimensional vector space.

5.3. * STRUCTURE OF FINITELY GENERATED ABELIAN GROUPS107

Next let us investigate the structure of a finite abelian group, which

is more complicated.

Lemma 5.3.1. Let k1, . . . , kn be integers, not all zero and gcd(k1, . . . , kn) =

1. Then there exists an n × n matrix A with integral entries of deter-

minant ±1 whose first row is k1, . . . , kn.

Proof. Apply induction on n. The lemma is obviously true for

n = 1. Assume that n > 1.

If k2 = · · · = kn = 0 then k1 = ±1, and replacing the first row of

the n × n identity matrix by k1, . . . , kn produces a matrix satisfying

the requirement. It remains to consider the case that ki 6= 0 for some

2 ≤ i ≤ n. Let d = gcd(k2, . . . , kn). Then there are integers s, t such

sk1 + td = 1.

Let qi = ki/d for 2 ≤ i ≤ n. Then q2, . . . , qn are integers such

that gcd(q2, . . . , qn) = 1. By induction hypothesis there are integers

bij(2 ≤ i ≤ n− 1, 2 ≤ j ≤ n) such that∣∣∣∣∣∣∣∣∣q2 q3 · · · qn

b22 b23 · · · b2n

· · ·bn−1,2 bn−1,3 · · · bn−1,n

∣∣∣∣∣∣∣∣∣ = ±1.

Without loss of generality we may assume that this determinant is

equal to 1. Then∣∣∣∣∣∣∣∣∣∣∣∣

k1 k2 k3 · · · kn

−t sq2 sq3 · · · sqn

0 b22 b23 · · · b2n

· · ·0 bn−1,2 bn−1,3 · · · bn−1,n

∣∣∣∣∣∣∣∣∣∣∣∣= sk1 + td = 1.

Lemma 5.3.2. Let A be an abelian group and let x, y ∈ A be two

elements of finite order n and m respectively. If gcd(m,n) = 1, then

the order of x+ y is mn.


Proof. It is evident that nm(x+ y) = 0. Assume that k(x+ y) =

0. Then kx = −ky. Since o(kx)|n and o(−ky)|m, we have o(kx) =

o(−ky) = 1. Hence kx = ky = 0, which implies mn|k.

Lemma 5.3.3. Let A be an abelian group and let x, y ∈ A be two

elements of finite orders n and m respectively. Let lcm(n,m) be the

least common multiple of n and m. There exist integers s and t such

that the order of sx+ ty is lcm(n,m).

Proof. The natural numbers n and m can be written as

n =r∏

i=1

peii ,m =

r∏i=1

pfi

i

where p1, . . . , pr are distinct prime numbers and e1, . . . , er, f1, . . . , fr

are non-negative integers. For each 1 ≤ i ≤ r, let

ui =

pei

i , if ei ≥ fi

0, otherwise

and

vi =

pfi

i , if ei < fi

0, otherwise.

Let u =∏r

i ui and v =∏r

i vi. Then u|n, v|m, gcd(u, v) = 1 and uv =

lcm(n,m). Since o((n/u)x) = u, o((m/v)y) = v, the order of (n/u)x+

(m/v)y is uv by Lemma 5.3.2.

Lemma 5.3.4. Let x and y be two elements of finite orders of an

abelian group A. Let n and m be the orders of x and y respectively.

Assume that m|n and dy ∈ 〈x〉 for some natural number d. Then dy ∈〈dx〉.

Proof. It is easy to verify the m/ gcd(m, d) divides n/ gcd(n, d).

Hence o(dy)|o(dx), which implies dy ∈ 〈dx〉.

Theorem 5.3.5. Let A be a finite abelian group with |A| > 1 gen-

erated by a1, . . . , an. Then A can be decomposed into a direct sum of

cyclic groups B1, . . . , Bm of orders d1, . . . , dm respectively, satisfying

the following conditions:


1) m ≤ n;

2) dm > 1 and di divides di−1 for 2 ≤ i ≤ m.

Proof. Apply induction on |A|. If |A| = 2 then A is a cyclic group

and the theorem holds. Assume that |A| > 2.

Take an element b′1 in A of highest order d1. Write

b′1 = s1a1 + · · ·+ snan,

where si ∈ Z. Let q = gcd(s1, . . . , sn), ki = si/q, and let

b1 = k1a1 + · · ·+ knan ∈ A.

Then b′1 = qb1 and the order of b1 is greater than or equal to d1. Since

d1 is an upper bound of orders of elements in A, the order b1 is equal

to d1. Note that gcd(k1, . . . , kn) = 1.

Lemma 5.3.3 implies that the order of every element in A divide d1.

Since gcd(k1, . . . , kn) = 1, Lemma 5.3.1 implies that there are inte-

gers cij(2 ≤ i ≤ n− 1, 1 ≤ j ≤ n) such that∣∣∣∣∣∣∣∣∣k1 · · · kn

c21 · · · c2n

· · ·cn1 · · · cnn

∣∣∣∣∣∣∣∣∣ = ±1.

Let

e2 = c21a1 + · · ·+ c2nan,

· · ·

en = cn1a1 + · · ·+ cnnan.

Then b1, e2, . . . , en generate A.

Let B1 be the cyclic subgroup of A generated by b1 and let A =

A/B1. If |A| = 1, then A = B1 and there is nothing more to prove.

Assume that |A| > 1. Denote by e2, . . . , en the images of e2, . . . , en

in A. Then e2, . . . , en generate A. By induction hypothesis there are

f2, . . . , fm ∈ A such that

1) m ≤ n;


2) A = 〈f2〉 ⊕ · · · ⊕ 〈fm〉;3) dm > 1 and di divides di−1 for 3 ≤ i ≤ m, where di is the order

of fi in A for 2 ≤ i ≤ m.

The element difi is in B1 for 2 ≤ i ≤ m. By Lemma 5.3.4 difi ∈〈dib1〉, since the order of fi divides that of b1. Hence difi = uidib1 for

some natural number ui.

Let bi = fi − uib1. Then bi = fi and the order of bi is equal to di.

For each 2 ≤ i ≤ m let Bi be the cyclic subgroup of A generated by

bi. We claim that A = B1⊕B2⊕· · ·⊕Bm. It is obvious that b1, . . . , bm

generate A. Assume that c1, . . . , cm ∈ Z satisfy

c1b1 + · · ·+ cmbm = 0.

Then

c2b2 + · · ·+ cmbm = 0,

i.e.,

c2f2 + · · ·+ cmfm = 0.

By the conditions that f2, . . . , fm satisfy, di|ci for 2 ≤ i ≤ m. Hence

cibi = 0 for 2 ≤ i ≤ m, which implies c1b1 = 0. This proves the

claim.

Corollary 5.3.6. Let A be a finite abelian group. Then A can

be decomposed into the direct sum of some cyclic groups B1, . . . , Bm of

orders d1, . . . , dm respectively such that

1 < dm|dm−1| · · · |d2|d1.

The number m and the decreasing sequence of natural numbers d1, d2, . . . , dm

are totally determined by A.

Proof. The first half is a consequence of Theorem 5.3.5. Assume

that A is decomposed into the direct sum of another set of cyclic groups

C1, . . . , Cn of orders e1, . . . , en respectively such that

1 < en|en−1| · · · |e2|e1.

Both d1 and e1 are equal to maxa∈A o(a). Hence d1 = e1. Suppose


that the decreasing sequences d1, d2, . . . , dm and e1, e2, . . . , en are dif-

ferent. Since∏m

i=1 di =∏n

i=1 ei, there is r > 1 such that di = ei for

1 ≤ i < r and dr 6= er. Without loss of generality we may assume

that dr < er. Let K = drA = dra|a ∈ A. It is obvious that K is a

subgroup of A. The decomposition A ∼= B1 ⊕ · · · ⊕Bm yields

|K| =r−1∏i=1

di

dr

,

while the second decomposition A ∼= C1 ⊕ · · · ⊕ Cn implies that

|K| =r∏

i=1

ei

dr

>

r−1∏i=1

ei

dr

.

This leads to contradiction.

The proof of Theorem 5.3.5 is constructive. It gives an algorithm

to find the explicit decomposition.

Example 5.3.7. Let

B = Z8 ⊕ Z12 ⊕ Z16

and let A be the subgroup of B generated by

u = 2⊕ 3⊕ 2,

v = 4⊕ 4⊕ 8,

w = 4⊕ 6⊕ 6.

Find the decomposition of A and generators of cyclic components sat-

isfying the conditions in Theorem 5.3.5.

Solution:

The orders of the elements u, v, w are 8, 6, 8 respectively. Hence

u+v = 6⊕ 7⊕10 is an element of highest order in A. Its order is equal


to 24. Due to ∣∣∣∣∣∣∣1 1 0

0 1 0

0 0 1

∣∣∣∣∣∣∣ = 1,

A is generated by u+ v, v, w.

Let z1 = u+ v. Then

z1 = 6⊕ 7⊕ 10,

2z1 = 4⊕ 2⊕ 4,

3z1 = 2⊕ 9⊕ 14,

4z1 = 0⊕ 4⊕ 8,

5z1 = 6⊕ 11⊕ 2,

6z1 = 4⊕ 6⊕ 12,

7z1 = 2⊕ 1⊕ 6,

8z1 = 0⊕ 8⊕ 0,

9z1 = 6⊕ 3⊕ 10,

10z1 = 4⊕ 10⊕ 4,

11z1 = 2⊕ 5⊕ 14,

12z1 = 0⊕ 0⊕ 8,

13z1 = 6⊕ 7⊕ 2,

14z1 = 4⊕ 2⊕ 12,

15z1 = 2⊕ 9⊕ 6,

16z1 = 0⊕ 4⊕ 0,

17z1 = 6⊕ 11⊕ 10,

18z1 = 4⊕ 6⊕ 4,

19z1 = 2⊕ 1⊕ 14,


20z1 = 0⊕ 8⊕ 8,

21z1 = 6⊕ 3⊕ 2,

22z1 = 4⊕ 10⊕ 12,

23z1 = 2⊕ 5⊕ 6.

Denote B = B/〈z1〉.Since

v = 4⊕ 4⊕ 8 /∈ B/Zz1, 2v = 0⊕ 8⊕ 0 = 8z1,

the element [v] ∈ B has order 2. Similarly [w] ∈ B has order 4 and

4w = 12z1.

Since 2w−v = 4⊕8⊕4 /∈ 〈z1〉, so 2[w] 6= [v]. Hence B = 〈[w]〉⊕〈[v]〉.Let v′ = v−4z1, w

′ = w−3z1. Then the orders of v′ and w′ in A are

equal to 2 and 4 respectively. Let z2 = w′ = 2⊕9⊕8, z3 = v′ = 4⊕0⊕0.

Then A = 〈z1〉 ⊕ 〈z2〉 ⊕ 〈z3〉 with o(z1) = 24, o(z2) = 4, o(z3) = 2. In

particular |A| = 24× 4× 2 = 192.

Lemma 5.3.8. Let A be a finitely generated abelian group. If every

element of A has finite order then A is a finite group.

Proof. Assume that A is generated by a1, a2, . . . , an of orders

d1, d2, . . . , dn respectively. Then every element of A can be written

as

k1a1 + k2a2 + · · ·+ knan

with 0 ≤ ki < di, although not necessarily uniquely. Hence |A| ≤d1d2 · · · dn.

Next we discuss the structure of a finitely generated abelian group.

Theorem 5.3.9. A finitely generated abelian group A is isomorphic

to Zn ⊕ Zd1 ⊕ · · · ⊕ Zdm , in which

1 < dm|dm−1| · · · |d2|d1.

The natural numbers n,m and the sequence d1, d2, . . . , dm are uniquely

determined by A.


Proof. Since Corollary 5.3.6 covers the case where A is finite, we

may assume that |A| = ∞.

Let T be the set of all elements of finite order in A. Then T is a

normal subgroup of A and K = A/T is also a finitely generated abelian

group. Due to Lemma 5.3.8, K 6= 0.

We claim that the only element of finite order is K is 0. Assume

that a ∈ A satisfies ma = 0 for some natural number m. Then ma ∈ T.So there is some natural number k such that kma = 0. This means

that a ∈ T and a = 0. The claim is proved.

Choose a set of generators x1, . . . , xn of K so that n reaches the

minimum value.

Assume that

m1x1 + · · ·+mnxn = 0

for some m1, . . . ,mn ∈ Z. Let d = gcd(m1, . . . ,mn),mi = dki. Then

d(k1x1 + · · ·+ knxn) = 0.

Since the only element of finite order in K is 0,

k1x1 + · · ·+ knxn = 0.

Since gcd(k1, . . . , kn) = 1 there is an integral n × n matrix A = (aij)

with determinant ±1 whose first row is (k1, . . . , kn).

Let yi = ai1x1 + · · ·+ ainxn(2 ≤ i ≤ n). Then

A

x1

...

xn

=

0

y2

...

yn

.Hence x1

...

xn

= A−1

0

y2

...

yn

,


which implies that K is generated by y2, . . . , yn, contradicting the min-

imality of n. Hence K is a free abelian group with basis x1, . . . , xn.

Choose zi ∈ A such that the image of in K = A/T is xi for 1 ≤ i ≤n.

For an arbitrary element a ∈ A, there is a unique set of integers

k1, . . . , kn such that

a = k1x1 + · · ·+ knxn.

This means that a− (k1z1 + · · ·+ knzn) ∈ T. Hence

A ∼= Zn ⊕ T.

In particular, T ∼= A/Zn. Hence T is finitely generated. It follows from

Lemma 5.3.8 and Corollary 5.3.6 that T has the required decomposi-

tion.

The uniqueness is obvious.

Remark 5.3.10. An element of finite order in a abelian group is

called a torsion element. The part T in the above theorem is called

the torsion part of A and n is called the rank of A.

Exercises

1. Show that the additive group Q of rational numbers is not a

finitely generated abelian group.

2. Determine the structure of the abelian group A = Z2/Z(6⊕ 30).

3. Determine all abelian groups of order 80.

4. Assume that an abelian group A is isomorphic to the direct sum

of three cyclic groups of orders 24, 6, 2 respectively. Determine the

structures of Sylow 2-subgroup and Sylow 3-subgroup of A.

5. Let

A = (a1, a2) ∈ Q2|4a1b1 + 2a1b2 + 2a2b1 + 4a2b2 ∈ Z ∀(b1, b2) ∈ Z2.

Determine the structure of the quotient group A/Z2.


6. Assume that a finite group G satisfies the following property:

For any pair of subgroups H,K either H ⊆ K or K ⊆ H. Show that

G is a cyclic group of order pn where p is a prime number.

7. Determine all finitely generated abelian groups G satisfying

|Aut(G)| <∞.

5.4 Solvable Groups

One cannot expect a structure theorem of non-abelian groups as simple

as that of finitely generated abelian groups.

Definition 5.4.1. Let G be a group. A chain

e = G0 / G1 / · · · / Gr = G

is called a subnormal series of G. The quotient groups Gi/Gi−1(1 ≤i ≤ r) are called the factor groups of that series. The subnormal series

is called a composition series if every quotient factor is simple.

In a subnormal series, Gi−1 is a normal subgroup of Gi for 1 ≤ i ≤ r,

but Gi is not necessarily a normal subgroup of Gj if j − i > 1.

Example 5.4.2. The set of eight quaternions G = ±1,±i,±j,±kis a non-abelian group under multiplication. Set G0 = 1, G1 =

±1, G2 = ±1,±i, G3 = G. Then G0 / G1 / G2 / G3 is a compo-

sition series.

Set G′0 = 1, G′

1 = ±1, G′2 = ±1,±j, G′

3 = G. Then G′0 / G

′1 /

G′2 / G

′3 is another composition series.

Lemma 5.4.3. Let

e = G0 / G1 / · · · / Gr = G

be a composition series of a group G. Then Gi−1 is a maximal normal

subgroup of Gi, which means that Gi−1 6= Gi and there is no proper

normal subgroup H of Gi such that Gi−1 ⊂ H ⊂ Gi.

5.4. SOLVABLE GROUPS 117

Proof. If H / Gi and Gi−1 ⊆ H, then H/Gi−1 / Gi/Gi−1. Since

Gi/Gi−1 is a simple group, H/Gi−1 is a trivial subgroup of Gi/Gi−1,

which means that H = Gi−1 or H = Gi.

Remark 5.4.4. A group may have more than one maximal normal

subgroup.

Proposition 5.4.5. Every finite group has a composition series.

Proof. The induction on the order of the group works.

Remark 5.4.6. This proposition is not valid for infinite groups.

For example, Z does not have composition series, since every nontrivial

subgroup of Z is isomorphic to Z.

Theorem 5.4.7 (Jordan-Holder). Let

e = H0 / H1 / · · · / Hs = G

and

e = G0 / G1 / · · · / Gr = G

be two composition series of a finite group G. Then r = s and there

exists a permutation σ of 1, . . . , r such that

Gi/Gi−1∼= Hσ(i)/Hσ(i)−1

for 1 ≤ i ≤ r.

Proof. Apply induction on |G|. When |G| = 1 the theorem holds

obviously. Assume that |G| > 1. Then r ≥ 1, s ≥ 1. If Gr−1∼= Hs−1

then the induction hypothesis implies the required result. Assume that

Gr−1 6= Hs−1.

Since Gr−1Hs−1 / G and Gr−1 is a maximal normal subgroup of G,

Gr−1Hs−1 = G.

Let K = Gr−1 ∩Hs−1. Theorem 1.7.14 implies

Gr−1/K ∼= G/Hs−1, Hs−1/K ∼= G/Gr−1.


Choose an arbitrary composition series

e = K0 / K1 / · · · / Km = K

of K. The theorem is true by induction hypothesis and by compareing

the following four composition series of G :

e = H0 / H1 / · · · / Hs−2 / Hs−1 / G,

e = G0 / G1 / · · · / Gr−2 / Gr−1 / G,

e = K0 / K1 / · · · / Km / Hs−1 / G,

e = K0 / K1 / · · · / Km / Gr−1 / G.

Definition 5.4.8. If a group G has a subnormal series whose all

factor groups are abelian groups, then G is called a solvable group.

By convention, the trivial group 1 is solvable.

Proposition 5.4.9. Every factor group of a composition series of

a finite solvable group is a cyclic group of prime order.

Proof. This is because a finite abelian group is simple if and only

if it is a cyclic group of prime order.

Proposition 5.4.10. Any subgroup or quotient group of a solvable

group is solvable.

Proof. Let H be subgroup of a solvable group G. Let

e = G0 / G1 / · · · / Gr = G

be a subnormal series such that all Gi/Gi−1 are abelian. Then

e = G0 ∩H / G1 ∩H / · · · / Gr ∩H = H

is a subnormal series. SinceGi∩H/Gi−1∩H is isomorphic to a subgroup

of Gi/Gi−1, it is abelian. Hence H is solvable.

5.4. SOLVABLE GROUPS 119

Let G/N be a quotient group of G. Then

e = G0/G0 ∩N / G1/G1 ∩N / · · · / Gr/Gr ∩N = G/N

is a subnormal series of G/N. Since (Gi/Gi ∩N)/(Gi−1/Gi−1 ∩N) is a

quotient group of Gi/Gi−1 for every i, G/N is solvable.

Theorem 5.4.11. The symmetric group Sn is solvable if and only

if n ≤ 4.

Proof. S1 and S2 are obviously solvable.

Since S3 has a normal subgroup of order 3, S3 is solvable.

As for S4, letG0 = e, G1 = 1, (12)(34), (13)(24), (14)(23), G2 =

A4, G3 = S4. Then

G0 / G1 / G2 / G3

is a subnormal series all whose factor groups are abelian.

When n ≥ 5 the alternating group An is simple by Theorem 1.6.5.

Hence

e / An / Sn

is a composition series and the factor group An is not abelian.

Exercises

1. Give a complete proof of Proposition 5.4.5.

2. Let F be a field an let G be the group of all 2 × 2 invertible

upper triangular matrices over F. Show that G is a solvable group.

Chapter 6

Field Extensions

6.1 Definitions and First Properties of Field Ex-

tensions

Definition 6.1.1. Let F be a subfield of a field E. Then E is an

extension of F. This relation is denoted by E/F. We may also say that

E/F is a field extension (or simply extension if the context makes its

meaning clear). A subfield E ′ of E containing F is called an interme-

diate field of E/F.

Let E/F be a field extension. Under the addition and multiplication

of E, E is a vector space over F, whose dimension is denoted by [E : F ],

called the degree of the extension E/F. If [E : F ] = ∞, then E/F

is called an infinite extension, otherwise a finite extension. A finite

extension E/F has degree one if and only if E = F.

Proposition 6.1.2. Let E/F, L/E be field extensions (or equiva-

lently, E is an intermediate field of L/F ). Then

[L : F ] = [L : E][E : F ].

Proof. First assume that E/F, L/E are finite extensions. Let

a1, . . . , an be a basis of the F -space E and let b1, . . . , bm be a basis

of the E-space L. It suffices to show that aibj1≤i≤n,1≤j≤m is a basis

of the F -space L.

120

6.1. DEFINITIONS AND FIRST PROPERTIES OF FIELD EXTENSIONS121

Assume that there are cij ∈ F, (1 ≤ i ≤ n, 1 ≤ j ≤ m) such that∑1≤i≤n,1≤j≤m

cijaibj = 0.

Thenm∑

j=1

(n∑

i=1

cijai

)bj = 0.

Since b1, . . . , bm are linearly independent over E,

n∑i=1

cijai = 0

holds for 1 ≤ j ≤ m. Hence cij = 0 for 1 ≤ i ≤ n, 1 ≤ j ≤ m

since a1, . . . , an are linearly independent over F. Hence themn elements

aibj1≤i≤n,1≤j≤m are linearly independent over F.

Let u be an arbitrary element in L. Then u can be expressed as

u =m∑

j=1

djbj,

in which dj ∈ E. Express dj as

dj =n∑

i=1

eijai,

in which eij ∈ F. Then

u =m∑

j=1

n∑i=1

eijaibj,

in other words, u is a linear combination of the elements aibj1≤i≤n,1≤j≤m.

Hence aibj1≤i≤n,1≤j≤m is a basis of L over F, which implies that

[L : F ] = mn = [L : E][E : F ].

If [E : F ] = ∞ or [L : E] = ∞ Then there are arbitrarily many

elements in L that are linearly independent over F, which means that

[L : F ] = ∞. Hence [L : F ] = [L : E][E : F ] holds in this case too.

122 CHAPTER 6. FIELD EXTENSIONS

The following lemma is obvious.

Lemma 6.1.3. Let Kλλ∈Λ be a collection of intermediate fields of

E/F in which Λ 6= ∅. Then⋂

λ∈ΛKλ is an intermediate field of E/F.

Proposition 6.1.4. Let E/F be a field extension and let S be a

subset of E. There exists an intermediate field F (S) of E/F such that

1) S ⊆ F (S);

2) F (S) ⊆ K for any intermediate field K of E/F containing S.

Such an intermediate field F (S) is uniquely determined by S.

Proof. Let Γ be the set of all intermediate field of E/F containing

S. Since E ∈ Γ, the set Γ is not empty. Let F (S) =⋂

E′∈ΓE′. Then

F (S) clearly satisfies the two conditions.

Assume that L is a field such that S ⊆ L and L ⊆ K for any

intermediate field K of E/F containing S. Then F (S) ⊆ L since F (S)

is an intermediate field of E/F containing S. For the same reason, L ⊆F (S). Therefore L = F (S). This proves the uniqueness of F (S).

Remark 6.1.5. The field F (S) is understood to be the smallest

intermediate of E/F containing S, called the subfield of E generated

by the set S over F.

Another presentation of F (S) is

F (S) =f(a1, . . . , an)

g(a1, . . . , an)

∣∣n ≥ 0, f, g ∈ F [x1, . . . , xn],

a1, . . . , an ∈ S, g(a1, . . . , an) 6= 0.

Definition 6.1.6. Let E/F be a field extension and let S be a

subset of E. If E = F (S), then we say that the extension E/F is

generated by S. If there is a finite subset S = t1, . . . , tn of E such

that E = F (S), then E/F is called a finitely generated extension,

denoted by E = F (t1, . . . , tn). In particular if E/F is generated by a

single element α ∈ E, i.e., E = F (α), then E/F is called a simple

extension.

Remark 6.1.7.

F (α) =f(α)

g(α)

∣∣f, g ∈ F [x], g(α) 6= 0.

6.2. ALGEBRAIC EXTENSIONS 123

Remark 6.1.8. Let E/F be a field extension, t1, . . . , tn ∈ E.Denote

F [t1, . . . , tn] = f(t1, . . . , tn)|f ∈ F [x1, . . . , xn].

It is a subring of E containing F and the elements t1, . . . , tn, but not

necessarily a subfield.

For example in the extension R/Q, the ring

Q[π] = a0 + a1π + · · ·+ anπn|a0, a1, . . . , an ∈ Q

is different from Q(π), since the element 1/π is in Q(π) but not in Q[π].

6.2 Algebraic Extensions

Let E/F be a field extension and let α ∈ E. There is a ring homomor-

phism

φ : F [x] → E, f(x) 7→ f(α).

If Ker(φ) = 0, then α is called a transcendental element over F. In

other words, a transcendental element is not a zero of any nonzero

polynomial with coefficients in F.

Any non-transcendental element in a field extension is called an

algebraic element (over F ). In other words, an algebraic element is the

zero of some nonzero polynomial over F. In particular, every element

in F is algebraic.

Let α ∈ E be an algebraic element over F. Then Ker(φ) 6= 0, i.e.,

Ker(φ) is a nonzero ideal of the polynomial ring F [x]. Since F [x] is a

principal ideal domain, the ideal Ker(φ) is generated by some monic

polynomial f(x). If follows from Ker(φ) 6= F [x] that deg(f) ≥ 1.

We claim that f(x) is irreducible. Otherwise f(x) = g(x)h(x) with

deg(g) < deg(f), deg(h) < deg(f). Then g(x) /∈ Ker(φ), h(x) /∈ Ker(φ).

Let g and h denote the images of g(x) and h(x) in E respectively. Then

g 6= 0, h 6= 0. However, gh = φ(gh) = φ(f) = 0. This would contradict

the fact that E is a field. Hence f(x) is an irreducible polynomial

over F and so Ker(φ) is a maximal ideal of F [x]. By Proposition 2.4.5

Im(φ) ∼= F [x]/(f(x)) is a field, which is nothing but F (α).


The polynomial f(x) above is called the minimal polynomial of

the algebraic element α, whose degree is defined to be the degree of the

element α over the base field F.

Proposition 6.2.1. Let f(x) be the minimal polynomial of an al-

gebraic element in a field extension E/F and let n = deg(f). Then

[F (α) : F ] = n and 1, α, α2, . . . , αn−1 form a basis the the F -space

F (α).

Proof. It is clear that the map

φ : F [x] → F (α), g(x) 7→ g(α)

is an epimorphism. Hence every element β ∈ F (α) can be written as

β = g(α) for some g(x) ∈ F [x]. Let

g(x) = q(x)f(x) + r(x),

where q(x), r(x) ∈ F [x] and deg(r) < n. Then β = g(α) = r(α). This

shows that β is a linear combination of 1, α, α2, . . . , αn−1 over F.

It remains to show that 1, α, α2, . . . , αn−1 are linear independent

over F. Suppose that there are not all zero elements c0, c1, . . . , cn−1 ∈ Fsuch that

c0 + c1α+ c2α2 + · · ·+ cn−1α

n−1 = 0.

Let g(x) = c0+c1x+c2x2+ · · ·+cn−1x

n−1. Then g(x) is a nonzero poly-

nomial over F such that g(α) = 0 and deg(g) < n, which contradicts

the assumption that f(x) is the minimal polynomial of α.

Let α ∈ E and let F [α] = f(α)|f(x) ∈ F [x]. Then F [α] is a

subring of E.

The proposition above shows that F [α] = F (α) when α is an alge-

braic element.

When α is transcendental, F [α] is isomorphic to the polynomial

ring F [x], which is not a field. Hence F [α] = F (α) if and only if α is

algebraic.

Definition 6.2.2. A field extension E/F is called an algebraic

extension if every element in E is algebraic over F.


Lemma 6.2.3. Let K be an intermediate field of an extension E/F

and let α ∈ E be an algebraic element over F. Then α is also an

algebraic element over K.

Proof. By definition there is a nonzero polynomial f(x) ∈ F [x]

such that f(α) = 0. Since F ⊆ K, f(x) ∈ K[x]. Hence α is also an

algebraic element over K.

Theorem 6.2.4. A field extension E/F is a finite extension if and

only if it is a finitely generated algebraic extension.

Proof. Assume that E/F is a finitely generated algebraic ex-

tension, i.e., there are algebraic elements α1, . . . , αn such that E =

F (α1, . . . , αn). Denote Ei = F (α1, . . . , αi) for 1 ≤ i ≤ n. Lemma

6.2.3 implies that αi is algebriac over Ei−1 for each i > 1. Hence

[Ei : Ei−1] <∞. It follows that

[E : F ] = [En : En−1] · · · [E2 : E1][E1 : F ] <∞.

Next assume that [E : F ] = d < ∞. For any α ∈ E the d + 1

elements

1, α, α2, . . . , αd

are linearly dependent over F, i.e., there are elements

c0, c1, . . . , cd ∈ F,

not all zero, such that

c0 + c1α+ c2α2 + · · ·+ cdα

d = 0.

Hence α is algebraic. This proves that E/F is an algebraic exten-

sion. Choose a basis b1, . . . , bd of the F -space E. It is clear that

E = F (b1, . . . , bd). Hence E/F is finitely generated.

Proposition 6.2.5. Assume that E/F, L/E are algebraic exten-

sions. Then L/F is an algebraic extension.

Proof. We need to show that every element γ ∈ L is an algebraic


element over F. For this purpose, it suffices to show that [F (γ) : F ] <

∞ by virtue of Theorem 6.2.4.

Since γ is algebraic over E, there is a nonzero f(x) ∈ E[x] such that

f(γ) = 0. Let

f(x) = bnxn + bn−1x

n−1 + · · ·+ b1x+ b0,

in which b0, b1, . . . bn ∈ E. By hypothesis, all coefficients b0, b1, . . . , bn

are algebraic elements over F. Hence F (b0, b1, . . . , bn) is a finite exten-

sion of F by Theorem 6.2.4.

Since f(γ) = 0, γ is algebraic over F (b0, b1, . . . , bn). Hence

[F (b0, b1, . . . , bn, γ) : F (b0, b1, . . . , bn)] <∞,

which implies that

[F (b0, b1, . . . , bn, γ) : F ]

= [F (b0, b1, . . . , bn, γ) : F (b0, b1, . . . , bn)][F (b0, b1, . . . , bn) : F ] <∞.

Therefore γ is algebraic over F.

Let E/F,K/F be two extensions of a field F and let f be a ho-

momorphism from E to K such that f(a) = a for all a ∈ F. Then

f is called a homomorphism from E/F to K/F, also referred to as a

homomorphism from E to K over F.

Proposition 6.2.6. Let E/F be an algebraic extension and let f

be a homomorphism from E/F to E/F itself. Then f : E → E is an

isomorphism.

Proof. The homomorphism is injective by Proposition 2.4.2. Hence

it suffices to show that f is surjective.

Let α ∈ E, By assumption α is algebraic over F. Let p(x) be the

minimal polynomial of α and let α = α1, α2, . . . , αn be all zeros of

p(x) in E. Let K = F (α1, α2, . . . , αn). Then K is a finitely generated

algebraic extension of F. Theorem 6.2.4 implies that K/F is a finite

extension. Thus K is an F -space of finite dimension.


Since f is a homomorphism,

p(f(αi)) = f(p(αi)) = f(0) = 0

for all 1 ≤ i ≤ n. Hence f(αi) ∈ K. This shows that f(K) ⊆ K. Since

any injective linear transformation of a vector space of finite dimension

is surjective, f : K → K is surjective. Hence there is some β ∈ K such

that f(β) = α). Therefore f : E → E is surjective.

Example 6.2.7. The condition of algebraic extension in the pro-

postion is indispensable. For example,

f : F (x) → F (x),a(x)

b(x)7→ a(x2)

b(x2)

is a monomorphism over F, but not an epimorphism.

Exercises

1. Show that u =√

2 +√

3 is an algebraic number, i.e., there is

f(x) ∈ Q[x] such that f(u) = 0. Find the ideal I of Q[x] such that

Q[u] ∼= Q[x]/I.

2. Let α is a complex root of the equation x3− 3x+ 4 = 0. Express

the multiplicative inverse of α2 +α+1 in Q[α] in the form a+ bα+ cα2

in which a, b, c are rational numbers.

3. Let F (α) be a simple extension of a field F and let β ∈ F (α)\F.Show that α is an algebraic element over F (β).

4. Let E/F be a field extension and α ∈ E such that [F (α) : F ] = 5,

Show that F (α2) = F (α).

5. Let E/F be a finite extension and let K be an intermediate

field. Let α ∈ E and let p(x) be the minimal polynomial of α over F.

Assume that deg(p) is relative prime to [K : F ]. Show that p(x) is an

irreducible polynomial in K[x].


6.3 Constructions of Field Extensions

Thus far we have discussed the properties of the intermediate fields of

a given extension E/F. To construct an intermediate field it suffices to

give a set of generators.

Now we may ask how to construct an extension E/F from a given

F without any knowledge of E. For example, we may ask what kind of

fields can be extensions of the finite field Fp. To answer these questions

we need to learn a new method.

Let F be a field. We consider to add a new element and the addition

and multiplication of this new element with the elements in F.Of course

a lot of other elements must also be thrown in to form a field.

For example, starting from the field Q of rational number, if we

want to add√

2 to obtain a larger field we must add all elements in

the form a+ b√

2, (a, b ∈ Q).

This example is easy, because the fields of real numbers and com-

plex numbers are available and the arithmetic involving√

2 is well-

known. This construction is essentially the construction of an interme-

diate field.

For an abstract field F, there is no larger field such like complex

number field available. This makes the construction of extensions of F

more difficult.

As the first step, we wish to construct a simple extension E/F, say

E = F (α) such that the element α satisfies some preassigned condi-

tions.

If we wish α to be transcendental over F, then the construction is

easy. Let E = F (x) be the field of all fractions over F in one variable

x. Then E is a simple extension of F generated by a transcendental

element x over F.

Suppose that we wish α to be algebraic over F. If E = F (α) has

been constructed then there will be a surjective homomorphism φ :

F [x] → E, f(x) → f(α), with Ker(φ) = (g(x)) where g(x) is the

minimal polynomial of α over F. Then

E ∼= F [x]/(g(x)).

6.3. CONSTRUCTIONS OF FIELD EXTENSIONS 129

The key observation that g(x) is an irreducible polynomial in F [x]

suggests us to start the construction of E from an arbitrary monic

irreducible polynomial g(x) ∈ F [x], and then define E = F [x]/(g(x)).

Since (g(x)) is a maximal ideal, E is a field by Proposition 2.4.5. Every

element in E can be expressed as f(x), in which f(x) ∈ F [x]. Recall

that f(x) is a coset of the principal ideal I = (g(x)) represented by

f(x). The addition and multiplication in E follow the rules

f1(x) + f2(x) = f1(x) + f2(x),

f1(x)f2(x) = f1(x)f2(x).

We are facing another issue: F and E are two different fields because

they contain elements of different nature. In what sense is F regarded

as a subfield of E?

The answer is that E contains a subfield isomorphic to F. This view

of point can hardly be over-emphasized.

Every element in F is also an element in F [x]. Let F ′ = a|a ∈ F.It is easy to verify that F ′ is a subfield of E.

The map

σ : F → F ′, a 7→ a

is obviously an isomorphism.

Example 6.3.1. x2 + x + 1 is an irreducible polynomial over the

finite field F2.

Proof. Every reducible polynomial of degree two can be decom-

posed into the product of two polynomials of degree one. There are

only two linear polynomials x and x+ 1 over F2. Hence there are only

three reducible polynomials x2, x2 + x, x2 + 1 of degree two. Hence

x2 + x+ 1 is irreducible over F2.

It follows that E = F2[x]/(x2+x+1) is a simple algebraic extension

of F2, with x2 +x+ 1 being the minimal polynomial of its generator x.

Hence E/F is an extension of degree 2, which implies that E is a field

containing four elements. Denote α = x. Then the four elements of E


are listed as 0, 1, α, α+ 1. The addition and multiplication tables in E

are

+ 0 1 α α+1

0 0 1 α α+1

1 1 0 α+1 α

α α α+1 0 1

α+1 α+1 α 1 0

and

× 0 1 α α+1

0 0 0 0 0

1 0 1 α α+1

α 0 α α+1 1

α+1 0 α+1 1 α

respectively.

By the same method it can be verified that x3+x+1 is an irreducible

polynomial over F2. So E = F2[x]/(x3 + x + 1) is a field containing 8

elements. Let α = x. The eight elements of E are

0, 1, α, α + 1, α2, α2 + 1, α2 + 1, α2 + α+ 1.

The product (α2 + 1)(α2 + α + 1) can be computed in the following

way. Let

f(x) = (x2 + 1)(x2 + x+ 1) = x4 + x3 + x+ 1.

Use division algorithm to compute the remainder of f(x) divided by

x3 + x+ 1 :

f(x) = (x+ 1)(x3 + x+ 1) + x2 + x.

Thus

(α2 + 1)(α2 + α+ 1) = f(α) = α2 + α.

Exercises

6.4. ALGEBRAICALLY CLOSED FIELD 131

1. Let E/F be an algebraic extension and let f(x) ∈ F [x] be an

irreducible polynomial over F. Assume that α, β ∈ E satisfy f(α) =

f(β) = 0. Show that there exists a homomorphism φ : F [α] → E such

that

1) φ(a) = a for all a ∈ F ;

2) φ(α) = β.

Show that φ induces an isomorphism from F [α] to F [β].

2. Show that x2 + x + 2 is an irreducible polynomial over F3. Let

E = F3[x]/(x2 + x+ 2). Write out the multiplication table of E.

6.4 Algebraically Closed Field

Let F be a field. According to the previous section, in order to con-

struct a simple algebraic extension F (α)/F of degree d it is enough to

find an irreducible polynomial of degree d over the field F. This exten-

sion F (α)/F is called trivial if F (α) = F, which is equivalent to d = 1.

Thus the field F has a nontrivial algebraic extension if and only if there

is an irreducible polynomial of degree greater than one over F. In other

words, the following three statements are equivalent.

• The only algebraic extension of F is F/F.

• A polynomial f(x) ∈ F [x] is irreducible if and only if deg(f) = 1.

• Every polynomial of positive degree over F has a zero in F.

A field satisfying the above equivalent conditions is called an alge-

braically closed field.

Our most familiar algebraically closed field is the complex number

field due to the Fundamental Theorem of Algebra.

Definition 6.4.1. Let L/F be an algebraic extension. If L is al-

gebraically closed then L is called the algebraic closure of F.

Proposition 6.4.2. Let E/F be an extension. Let E ′ be the set of

all algebraic elements (over F ) in E. Then

1) E ′ is an intermediate field of E/F, refered to as the algebraic

closure of F in E.

2) If E is algebraically closed then E ′ is the algebraic closure of F.


Proof. 1) Let α, β ∈ E ′. Then [F (α, β) : F ] < ∞, which implies

that F (α, β)/F is an algebraic extension. Hence F (α, β) ⊆ E ′. So

α + β ∈ E ′, α − β ∈ E ′, αβ ∈ E ′, and α/β ∈ E ′ if β 6= 0. This

shows that E ′ is a subfield of E. Obviously F ⊆ E ′. Hence E ′ is an

intermediate field of E/F.

2) Let f(x) be a polynomial of positive degree over E ′. Since E

is algebraically closed by assumption, there is some α ∈ E such that

f(α) = 0. This implies that E ′(α)/E ′ is an algebraic extension. Hence

E ′(α)/F is an algebraic extension, which implies that α is algebraic

over F, i.e., α ∈ E ′. This shows that E ′ is algebraically closed. It is

evident that E ′/F is an algebraic extension. Hence E ′ is the algebraic

closure of F by definition.

The condition of E being algebraically closed in 2) of Proposition

6.4.2 is indispensable. For example the algebraic closure of Q in R is

not an algebraically closed field.

For any subfield F of the complex number field C, the algebraic

closure of F in C is algebraically closed by Proposition 6.4.2. A typical

example is the algebraic closure of Q in C, commonly denoted by Q.The numbers in Q are called algebraic numbers while the other number

are transcendental numbers.

Now a fundamental problem we are facing is: For an arbitrary field

F, does its algebraic closure exist? If exists, is it unique?

Unlike the cases of number fields, there is no “universal” field for F

to play the role of the complex number field. The fields such like finite

fields or fields of rational functions are not contained in the complex

number field. A new method to resolve this problem is required.

We may try to solve the existence problem naively. If F is already

algebraically closed, then F itself is the algebraic closure of F. Other-

wise there exists a nontrivial simple algebraic extension F (α)/F using

the approach in the previous section. Then treat the new field F (α) in

the same way. If the same process is repeated again and again, prob-

ably after infinitely many times, an algebraically closure of F will be

obtained. This argument looks like induction. But this is not a solid

proof since the term “infinitely many times” is ambiguous.


Similar issues of this nature arise in many fundamental definitions

and theorems in mathematics. It forces us to take a few axioms as

granted, upon which the normal deduction in mathematics is based.

Here we need to use the celebrated Zorn’s lemma.

Definition 6.4.3. Let ≺ be a relation in a set S, i.e, a ≺ b holds

for some pairs a, b of elements in S. If the following conditions

1) a ≺ a for any a ∈ S;

2) a ≺ b, b ≺ a implies a = b;

3) a ≺ b, b ≺ c, implies a ≺ c,

then S is called a partially-ordered set under the partial order ≺ .

It is denoted by (S,≺), or simply S if no confusion arises.

Remark 6.4.4. A partially-ordered set (S,≺) is a totally-ordered

set if it satisfies the following additional condition

4) For any pair a, b of elements in S at least one of a ≺ b and b ≺ a

holds.

Example 6.4.5. • Any set of real numbers is a totally-ordered set

under the relation “≤”.

• For any set S let Ω be the set of all subsets of S. For any U, V ∈ Ω,

define U ≺ V if and only if U ⊆ V. Then Ω is a partially-ordered set

under “≺”, but not a totally-ordered set if S contains more than one

element.

Definition 6.4.6. Let (S,≺) be a partially-ordered set and let T

be a subset of S. An element b ∈ S is called an upper bound of T if

a ≺ b for any a ∈ T.An element b in S is called a maximal element of S if b ≺ a ∈ S

implies b = a.

It is possible that a partially-ordered set has no maximal element

or has more than one maximal elements.

Zorn’s Lemma

Let (S,≺) be a non-empty partially-ordered set. If every totally-

ordered subset of S has an upper bound in S then there exists a max-

imal element in S.


The following example is a typical application of Zorn’s lemma in

algebra.

Example 6.4.7. Let I be an ideal of a commutative ring A such

that I 6= A. Then there exists a maximal ideal m of A such that I ⊆ m.

Proof. Let Γ be the set of all ideals of A containing I but not

equal to A. Since I ∈ Γ, Γ is not empty.

For any J1, J2 ∈ Γ, define J1 ≺ J2 if and only if J1 ⊆ J2. Then

(Γ,≺) is a partially-ordered set.

Let T be a totally-ordered subset of Γ. Since 1 /∈ J for every J ∈ T,⋃J∈T J 6= A. Hence

⋃J∈T J ∈ Γ and J ≺

⋃J∈T J for every J ∈ T,

which means that the condition for Zorn’ lemma is satisfied. Hence

there is a maximal element m in Γ.

Let J be an ideal of A such that m ⊆ J and J 6= m. The maximality

of m implies that J /∈ Γ. This forces J = A. Hence m is a maximal

ideal of A.

Theorem 6.4.8. Let σ : E → L be a homomorphism of fields with

L algebraically closed. Let K/E be an algebraic extension. Then there

exist a homomorphism τ : K → L such that τ |E = σ.

Proof. Let Γ = (H, γ)| H is an intermediate field of K/E, γ :

H → L is a homomorphism such that γ|E = σ.. Define a partial order

in Γ by (H1, γ1) ≺ (H2, γ2) if and only if H1 ⊆ H2 and γ2|H1 = γ1.

Let T be a totally-ordered subset of Γ and let B =⋃

(H,γ)∈T H.

Then B is an intermediate field of K/E. Define a map β : B → L by

the following way: for an arbitrary x ∈ B, choose (H, γ) ∈ T such that

x ∈ H, let β(x) = γ(x). It is easy to verify that β(x) does not depend

upon the choice of (H, γ) and β is a homomorphism from B to L such

that β|H = γ for any (H, γ) ∈ T. This means that (B, β) ∈ Γ is an

upper bound of T.

By Zorn’s lemma, there is a maximal element (A, η) in Γ. It suffices

to show that A = K. Suppose that it were not the case, there would be

some element α ∈ K\A. Let f(x) ∈ A[x] be the minimal polynomial

of α over A. There is an isomorphism q : A[α] → A[x]/(f(x)). Identify

A and η(A) via η, so that f(x) is regarded as a polynomial over η(A).


Since L is assumed to be algebraically closed, f(x) has a zero a in L.

Thus there is an homomorphism p : A[x]/(f(x)) → L such that p(x) =

a. Let τ = p q : A[α] → L. Then (A[α], τ) ∈ Γ and (A, η) ≺ (A[α], τ).

This contradicts the hypothesis that (A, η) is a maximal element of Γ.

Hence A = K.

Theorem 6.4.9. Let F be a field. Then the following statements

hold.

1) The algebraic closure L of F exists.

2) Let L1, L2 be two algebraic closures of F. There exists isomor-

phism f : L1 → L2 such that f(a) = a for every a ∈ F.

Proof. 1) Let Ω be the set of all algebraic extensions of F. For

any E/F and K/F in Ω, define E/F ≺ K/F if E ⊆ K. This set Ω

becomes a partially-ordered set under the relation “≺”.

Let ∆ be a totally-ordered subset of Ω. In order to give an upper

bound of ∆, it is natural to consider the union of all members of ∆

as done in proofs of Example 6.4.7 and Theorem 6.4.8. But a new

difficulty pops up. The members of ∆ are not necessarily contained in

a common large set, so it does not make sense to take the union. We

need to be careful to construct an upper bound of ∆.

For any E/F ≺ K/F let jE,K be the injection of E into K. There

exists a set A satisfying the following two conditions:

1) For any E/F ∈ ∆ there is an injective map σE : E → A such

that σE = σK jE,K for any E/F ≺ K/F in ∆.

2) A =⋃

E/F∈∆ Im(σE). 1

For any x, y ∈ A there existE/F,K/F ∈ ∆ such that x = σE(a), y =

σK(b) for some a ∈ E, b ∈ K. Since ∆ is totally-ordered, E/F ≺ K/F

or K/F ≺ E/F. Without loss of generality we may assume that E/F ≺K/F. Define x+ y = σK(a+ b), xy = σK(ab). By the condition 1) that

A satisfies, x + y and xy do not depend upon the choices of E/F and

K/F. It is easy to verify that the set A under the addition and mul-

tiplication just defined is a field and σE : E → A is a monomorphism1The set A can be constructed in the following way: Let B = tE/F∈∆E be the disjoint

union of all members of ∆, i.e., B = (a, E/F )| E/F ∈ ∆, a ∈ E. Define a relation in Bby (a, E/F ) ∼ (b, K/F ) if and only if E/F ≺ K/F and b = jE,K(a) or K/F ≺ E/F andb = jK,E(a). This is an equivalence relation. Let A be the set of all equivalence classes.For any E/F ∈ ∆, define σE : E → A, a 7→ [(a, E/F )]. Then all conditions are satisfied.


for each E/F ∈ ∆. Hence we may identify the elements of E and those

of σE(E). The condition 2) guarantees that every element of A is alge-

braic over F, so A/F ∈ Ω. Since E/F ≺ A/F for each E/F ∈ ∆, A/F

is an upper bound of ∆.

By Zorn’s lemma, there exists a maximal element L/F in Ω. It

remains to show that L is algebraically closed. Let L′/L be an algebraic

extension. Then L′/F ∈ Ω and L/F ≺ L′/F. Since L/F is maximal in

Ω, L = L′. This proves that L is algebraically closed.

2) Let L1 and L2 be two algebraic closures of F. By Theorem 6.4.8

there is a monomorphism φ : L1 → L2 such that φ(a) = a for all a ∈ F.For the same reason there is a monomorphism ψ : L2 → L1 such that

ψ(a) = a for all a ∈ F. Since ψ φ is a homomorphism from L1/F to

L1/F, Proposition 6.2.6 implies that ψ φ is an automorphism. For

the same reason φ ψ is also an automorphism. Hence φ and ψ are

isomorphisms.

Exercises

1. Let A be a commutative ring and let S be a subset of A such

that 0 /∈ S and ab ∈ S for any a, b ∈ S. Use Zorn’s lemma to show the

existence of a prime ideal P of A such that P ∩ S = ∅.

2. Show that Q is not isomorphic to Q(x).

3. Let F = a ∈ R|a is algebraic over Q. Show that F is not

isomorphic to Q.

4. Let E/F be an algebraic extension. Show that E is also the

algebraic closure of F.

6.5 * Ruler and Compass Construction

A fascinating type of problems in plane geometry is ruler and compass

construction. One can only use a pair of compass and an unmarked

ruler to construct geometric figures with preassigned properties.

Many problems are impossible. A notorious example is the im-

possibility of trisection of an arbitrary angle. In this section the field

6.5. * RULER AND COMPASS CONSTRUCTION 137

extension theory will be used to settle this issue. This is a remarkable

example of the application of algebra.

First of all the construction problem should be converted into a

problem of algebra. Most problems can be reduced to the construction

of one or more points. For example, to construct a line or a circle it is

enough to construct two points on the line or the center of the circle

and a point on the circle. We concentrate on the construction of one

point. The rules of the construction are formulated as follows.

Let S0 be a finite set of points on the plane that contains at least two

points. These points are the initial data of the relevant construction

problem.

Then the set S0 is extended inductively to a finite sequence S1, S2, . . . , Sn

of sets of points such that

S0 ⊂ S1 ⊂ · · · ⊂ Sn,

where Si = Si−1 ∪ Pi and Pi is a point obtained from either one of

the following three methods:

1) Choose any four distinct points Q1, Q2, Q3, Q4 in Si−1 and draw

a line ` passing Q1 and Q2 by ruler and another line `′ passing Q3 and

Q4. Take the intersection point of ` and `′ to be Pi.

2) Choose any two distinct points Q1, Q2 in Si−1 and draw a line

` passing Q1 and Q2. Choose any three points A,B1, B2 in Si−1, not

necessarily distinct. Draw a circle C by compass with A as the center

and the distance from B1 to B2 as the radius. Take an intersection

point (if any) of ` and C to be Pi.

3) Draw two distinct circles by the rule in 2). Take an intersection

point of these two circles to be Pi.

Such a sequence is the process of construction.

A point P is said to be constructible from S0 if there exists a process

of construction

S0 ⊂ S1 ⊂ · · · ⊂ Sn

such that P ∈ Sn.

Example: Construct a square of a given side length.

This is a typical construction problem. We may assume that the


given length is 1 and one side lies on the x-axis from 0 to 1. Then S0 =

(0, 0), (1, 0). The object is to construct two points (0, 1) and (1, 1).

We may set P = (0, 1), since the other point (1, 1) can be constructed

in the similar way.

Draw a circle C with (0, 0) as its center and the distance from (0, 0)

to (1, 0) as its radius. It intersects the x-axis ( the line passing (0, 0)

and (1, 0)) at a new point P1 = (−1, 0). Let S1 = (0, 0), (1, 0), P1.Let d be the distance from (−1, 0) to (1, 0). Draw two circles of radius

d with centers at (1, 0) and (−1, 0) respectively. Take an intersection

point P2 of these two circles. Let S2 = (0, 0), (1, 0), P1, P2. Draw a

line L passing (0, 0) and P2. Take the intersection point P3 of L and

the circle C in the upper half plane. Let S3 = (0, 0), (1, 0), P1, P2, P3.Then P = P3. This means that P is a constructible point.

Here are some more examples.

• the construction of a regular polygon of n sides : S0 = (0, 0), (1, 0), P =

(cos(2π/n), sin(2π/n)).

• the trisection of an arbitrary angle: S0 = (0, 0), (1, 0), (cosα, sinα)

and P = (cos(α/3), sin(α/3)).

• the construction of a square having the same area as a given circle:

S0 = (0, 0), (1, 0) , P = (√π, 0).

• the construction of the edge of a cube having twice the volume of

a given cube: S0 = (0, 0), (1, 0) and P = ( 3√

2, 0).

Assume that S0 = (x1, y1), (x2, y2), . . . , (xm, ym). Let F0 be the

field Q(x1, y1, x2, y2, . . . , xm, ym). It is a subfield of R.

Theorem 6.5.1. A point P = (x, y) on the plane is constructible

from the given set S0 = (x1, y1), . . . , (xm, ym) only if [F0(x, y) : F0]

is a power of 2.

Proof. Let S0 ⊂ S1 ⊂ · · · ⊂ Sm be a process of construction

such that Si = Si−1 ∪ Pi for 1 ≤ i ≤ m, where Pi = (ui, vi). Let

Fi = F0(u1, v1, . . . , ui, vi). Then we obtain a chain of field extensions

F0 ⊆ F1 ⊆ · · · ⊆ Fm.

If P is constructible, then both x and y are in the field Fm, which


implies that F0(x, y) is an intermediate field of the extension Fm/F0.

Hence it suffices to show that [Fm : F0] is a power of 2. For this purpose

we need to show that [Fi : Fi−1] is a power of 2 for 1 ≤ i ≤ m. This

can be achieved by treating three separate cases according to how Pi

is constructed.

Case 1) Pi is the intersection of two lines.

In this case (ui, vi) is the solution of a system

ax+ by = c,

dx+ ey = f

where a, b, c, d, e, f ∈ Fi−1 such that ae − bd 6= 0. Hence ui, vi ∈ Fi−1,

which implies Fi = Fi−1.

Case 2) Pi is the intersection of a circle and a line.

Assume that the equations of the line and circle are ax + by = c

and

(x− e1)2 + (y − e2)

2 = r

where a, b, c, e1, e2, r ∈ Fi−1. Then a 6= 0 or b 6= 0. We may assume that

b 6= 0 without loss of generality. Then vi = (c− aui)/b and

(ui − e1)2 +

(cb− e2 −

a

bui

)= r.

Hence [Fi−1(ui, vi) : Fi−1] is less than or equal to 2. This implies that

[Fi : Fi−1] is a power of 2.

Case 3) Pi is the intersection of two circles. Assume that the equa-

tions of the two circles are

(x− a1)2 + (y − b1)

2 = c1 (6.1)

and

(x− a2)2 + (y − b2)

2 = c2 (6.2)

respectively. Subtracting (6.1) from (6.2) yields a linear equation.

Hence this case is reduced to Case 2).

This concludes the proof of the theorem.


Example 6.5.2. • If S0 = (0, 0), (1, 0) then the point P = ( 3√

2, 0)

is not constructible, since [Q( 3√

2) : Q] = 3 is not a power of 2. Hence

it is impossible to double the volume of a cube by ruler and compass.

• If S0 = (0, 0), (1, 0) then the point P = (√π, 0) is not con-

structible, since [Q(√π) : Q] = ∞ is not a power of 2. Hence it is

impossible to construct a square whose area is equal to that of a given

circle.

• Let S0 = (0, 0), (1, 0), (cosα, sinα) and P = (cos(α/3), sin(α/3)).

Let F0 = Q(cosα, sinα). It follows from(cos

α

3+ i sin

α

3

)3

= cosα+ i sinα

that

cos3 α

3− 3 cos

α

3(1− cos2 α

3) = cosα

and

− sin3 α

3+ 3 sin

α

3(1− sin2 α

3) = sinα.

Except for a few special values of α, the polynomial x3 − 3x(1− x2)−cosα is irreducible. Hence [F0(cos α

3, sin α

3) : F0] is not a power of

3. Hence the trisection of an arbitrary angle by ruler and compass is

impossible.

Remark 6.5.3. The power of 2 in Theorem 6.5.1 is only a nec-

essary condition of the constructibility. For the sufficient conditions,

interested readers may consult references such like [5].

Exercises

1. Solve the following problems:

1) Let C be a unit circle with its center at (0, 0). Let D be the circle

with (n, 0) as its center and a positive integer n as its radius. Find the

x-coordinate of the intersection points of C and D.

2) Design a method to subdivide a given line segment into three

equal parts.

3) Given two points (0, 0), (1, 0). For any rational number r show

that the point (r, 0) is constructible.


2. Let ζ = cos(2π/5) + i sin(2π/5).

1) Show that ζ + ζ = (√

5− 1)/2.

2) Construct a regular pentagon by ruler and compass.

Chapter 7

Finite fields

7.1 Basic Theory

Lemma 7.1.1. The number of elements of a finite field is a power

of a prime number.

Proof. Let E be a finite field. Its characteristic must be a prime

number p. Hence E contains a subfield isomorphic to Fp. Let n = [E :

F ]. Then |E| = pn.

Let p be a prime number. Denote by Fp the algebraic closure of Fp.

Theorem 7.1.2. Let n be an arbitrary natural number. Then there

is a unique subfield E of Fp containing pn elements.

Proof. Let E be the set of all zeros of the polynomial xpn − x in

Fp.

Let a1, a2 ∈ E. Then

(a1 − a2)pn − (a1 − a2) = apn

1 − apn

2 − a1 − a2 = 0.

(a1a2)pn − (a1a2) = 0.

If a is a nonzero element in E, then

(1/a)pn − 1/a = (1/apn

)− 1/a = 0.

Hence E is a subfield of Fp.

142

7.1. BASIC THEORY 143

Since the formal derivative of xpn −x is −1, the polynomial xpn −xhas no multiple zeros by Proposition 3.4.11. Hence E contains exactly

pn elements. The existence of E is proved.

Assume that K is a subfield of Fp containing pn elements. By

example 3.5.8 every element in K is a zero of xpn − x. Therefore K =

E.

Corollary 7.1.3. Fp is an infinite field.

Proposition 7.1.4. Let E,F be subfields of Fp containing pn and

pm elements respectively. Then F ⊆ E if and only if m|n.

Proof. Since E,F are zeros of xpn − x and xpm − x in Fp respec-

tively,

F ⊆ E ⇔

xpm − x|xpn − x⇔

xpm−1 − 1|xpn−1 − 1 ⇔

pm − 1|pn − 1 ⇔

m|n.

Remark 7.1.5. Since every finite extension of Fp can be embedded

into Fp, the finite field containing q = pn elements is unique up to

isomorphism. This field is usually denoted by Fq.

Exercises

1. How many monic irreducible polynomials of degree 2 are there

over F5? of degree 3?

2. Discuss the irreducibility of x4+1 over R,Q and F16 respectively.

3. Decompose the polynomials x4 − 4 and x3 − 2 into the products

of irreducible polynomials over the fields R,C and F3 respectively.

4. Show that every finite integral domain is a finite field. Give an

example of a finite commutative ring which is not a field.

144 CHAPTER 7. FINITE FIELDS

7.2 The structure of Multiplicative Group of a fi-

nite field

Lemma 7.2.1. Let G be a finite abelian group. If G has at most

one subgroup of order m for every natural number m then G is a cyclic

group.

Proof. Let n = |G| and let n = pe11 . . . per

r be the standard prime

decomposition of n in which p1, . . . , pr are distinct prime numbers.

For each 1 ≤ i ≤ r, let Gi be the Sylow pi subgroup of G. By the

structure theorem of finite abelian groups, every Gi is a direct product

of some cyclic subgroups, say, Gi = H1 × · · · × Ht, where each Hj is

cyclic. According to Lagrange’s theorem, the order of every Hj is a

power of pi. So every Hj contains a subgroup of order pi and G has

at least t subgroups of order p. Hence t = 1, which means that every

Sylow subgroup of G is a cyclic group. Lemma 5.3.2 implies that G is

cyclic.

Theorem 7.2.2. Let p be a prime number and let n be a natural

number. If p|n then there is no subgroup of order n in F∗p, otherwise

there is a unique subgroup H of order n in F∗p. Moreover, this subgroup

H is cyclic.

Proof. First assume that p|n. Suppose that H is a subgroup of

F∗p with |H| = n. Then there is an element α of order p in H. Hence

(α− 1)p = αp − 1 = 0, which leads to contradiction. Hence there is no

subgroup of order n in F∗p.Next assume that n is not divisible by p. Since the polynomial xn−1

is coprime to its formal derivative, it has no multiple zeros in Fp. Since

Fp is algebraically closed, xn − 1 has exactly n zeros in Fp. Let H be

the set of all zeros of xn − 1 in Fp. It is obvious that H is a subgroup

of F∗p with |H| = n.

LetK be a subgroup of F∗p with |K| = n. Lagrange’s theorem implies

that an = 1 for every a ∈ K, which implies that every element in K is

a zero of xn − 1 in Fp. So K = H. This proves the uniqueness of the

subgroup of order n in F∗p.

7.2. THE STRUCTURE OF MULTIPLICATIVE GROUP OF A FINITE FIELD145

In particular, H contains at most one subgroup of order m for any

natural number m. It follows from 7.2.1 that H is a cyclic group.

Corollary 7.2.3. Assume that q = pn, where p is a prime number

and n is a natural number. Then the multiplicative group F∗q = Fq\0is a cyclic group.

Corollary 7.2.4. Let F be a subfield of a finite field E. Then E/F

is a simple extension.

Proof. Take any generator α of the cyclic group E∗. Then E =

F [α].

Corollary 7.2.5. There exist irreducible polynomials of any given

degree over a finite field Fq.

Proof. For any natural number n, Corollary 7.2.4 implies that

Fqn = Fq[α] for some α ∈ Fqn . The minimal polynomial of α over Fq is

an irreducible polynomial of degree n.

Definition 7.2.6. Let F be a finite field. Any generator of the

cyclic group F ∗ is called a primitive element of F.

Example 7.2.7. Let p be a prime number and let m be a natural

number not divisible by p. Then there is a natural number n such that

m divides pn − 1.

Proof. According to Theorem 7.2.2 there is a cyclic subgroup of

order m in the multiplicative group F∗p. Let α be the generator of this

cyclic subgroup. and let E = Fp[α]. Then |E| = pn for some natural

number n. Hence m|pn − 1 by Lagrange’s theorem.

This example shows that some problems in elementary number the-

ory can be solved by using finite fields. The following example gives

one more simple application of finite fields in number theory. For more

advanced application see the proof of quadratic reciprocity in Appendix

1.

146 CHAPTER 7. FINITE FIELDS

Theorem 7.2.8 (Wilson). For any prime number p the congruence

(p− 1)! ≡ −1 (mod p)

holds.

Proof. The theorem holds evidently when p = 2. Assume that p

is an odd prime. Then 1 6= −1 in Fp and ±1 are all solutions of the

equation x2 = 1 in Fp. Hence t 6= t−1 for any t ∈ Fp\1,−1. This

implies that ∏t∈F∗p\1,−1

t = 1.

So ∏t∈F∗p

t = −1,

which can be rewritten as

p−1∏j=1

j = −1.

This is obviously equivalent to

(p− 1)! ≡ −1 (mod p).

Exercises

1. Let Fp be a field of p elements in which p is a prime number.

Give an irreducible polynomial of degree p over Fp.

2. Let a be a nonzero element in a finite field F. Show that there

are x, y ∈ F such that x2 + y2 = a.

3. Let p be a prime number and f(x) ∈ Fp[x], g(x) = xp − x. Show

that gcd(f, g) is the product of all distinct factors of degree one of f(x).

4. Decompose x4 +x3 +x+3 ∈ F5[x] into the product of irreducible

polynomials.

7.2. THE STRUCTURE OF MULTIPLICATIVE GROUP OF A FINITE FIELD147

5. Let q be a power of a prime number. Find the number of solutions

of the equation yqz + yzq = xq+1 in the finite field Fq2 .

Chapter 8

Finite Galois Theory

One classical problem in the history of mathematics is to find formulas

to solve a polynomial equation in one variable.

More precisely, let

f(x) = xn + a1xn−1 + · · ·+ an−1x+ an = 0 (8.1)

be an equation of degree n. Is there any formula which can give a

solution in terms of the coefficients a1, . . . , an involving addition, sub-

traction, multiplication, division and radicals ?

It is known that equations of degrees upto 4 has such formulas. The

solutions of x2 + a1x+ a2 = 0 are

x =−a1 ±

√a2

1 − 4a2

2

for arbitrary elements a1, a2. This solution is actually a function with

a1, a2 as its variables. More generally the solution of the equation 8.1

is a multi-valued function in the variables a1, . . . , an.

We may consider to approach the problem by using field extensions

as in the ruler and compass construction.

First we need to formulate the problem properly. What we are

given is the complex number field and the variables a1, . . . , an. So we

start with the field C(a1, . . . , an) in which a1, . . . , an are indeterminates.

Then we can add new elements by ordinary arithmetic “+,−,×, /”.

They are contained in C(a1, . . . , an).

148

149

However, adding an n-th root will extend the field. Let F be some

field. Let n be a natural number greater than 1 and let a ∈ F. The

zeros of xn − a = 0 are not necessarily in F. It is natural to consider a

finite extension E of F containing all zeros of xn − a = 0, preferably

minimal. Unlike the ruler and compass construction, the degree [E : F ]

is not necessarily a power of 2. It can be an arbitrary natural number.

To handle the problem in the way similar to the ruler and compass

construction we formulate a recursive process searching for a solution

of (8.1) as follows. Let L be the algebraic closure of F0 = C(a1, . . . , an).

For i ≥ 0 let Fi+1 = Fi(α1, . . . , αr), in which α1, . . . , αr are all zeros of

xni − a = 0 in L for some a ∈ Fi and some natural number ni. Such an

extension Fi+1/Fi is called a radical extension.

If there is a sequence of radical extensions

F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Fm

such that Fm contains all zeros of (8.1) then we say the the equation

(8.1) is solvable by radicals.

Since L is algebraically closed,

f(x) = (x− α1) . . . (x− αn)

where α1, . . . , αn ∈ L. Let E = F0(α1, . . . , αn). Hence the equation

f(x) = 0 is solvable by radicals if and only if there is a finite sequence

of radical extensions

F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Fm

such that E ⊆ Fm.

It is desirable to test the solvability by radicals directly from the

extension E/F0. This is not as simple as the ruler and compass con-

struction problem in which the degree of the extension can give a neg-

ative answer. The Galois theory relates the problem of solvability by

radicals to the group theory. The extension E/F0 above corresponds to

a finite group called the Galois group of E/F0. The equation f(x) = 0

is solvable by radicals if and only if the corresponding Galois group is

150 CHAPTER 8. FINITE GALOIS THEORY

solvable. Since the Galois group of every polynomial equation of degree

greater than or equal to five with general coefficients is not solvable,

there is no formula for the zeros of f(x) = 0 involving addition, sub-

traction, multiplication, division and radicals only.

This is only a typical application of Galois theory, which has promi-

nent significance in mathematics. It becomes an important tool in

modern algebra, number theory and other branches of mathematics.

In this chapter we will discuss the main theorem of Galois theory and

its classical application to the solution of polynomial equations by rad-

icals.

8.1 Basic theory

Definition 8.1.1. Let E/F be an extension. Denote by Aut(E/F )

the group of all automorphisms σ of E such that σ(a) = a ∀a ∈ F,

called the automorphism group E over F.

It is straightforward to check that Aut(E/F ) is a group under the

operation of composite.

Let K be an intermediate field of E/F. Then Aut(E/K) is a sub-

group of Aut(E/F ). This gives a map from the set of all intermediate

fields of E/F to the set of all subgroups of Aut(E/F ) by

K 7→ Aut(E/K).

On the other hand, for any subgroup H of Aut(E/F ) let

EH = b ∈ E|σ(b) = b for all σ ∈ H.

Then EH is an intermediate field of E/F, called the fixed subfield of

H. It gives a map from the set of subgroups of Aut(E/F ) to the set of

all intermediate fields of E/F by

H 7→ EH .

Lemma 8.1.2. 1) Let K1, K2 be intermediate fields of E/F with

K1 ⊆ K2. Then Aut(E/K1) ⊇ Aut(E/K2);


2) Let H1, H2 be subgroups of Aut(E/F ) with H1 ⊆ H2. Then EH1 ⊇EH2 ;

3) K ⊆ EAut(E/K) holds for any intermediate field K of E/F ;

4) H ⊆ Aut(E/EH) holds for any subgroup H of Aut(E/F ).

The proof is left to the readers.

1) and 2) of Lemma 8.1.2 mean that the corresponding subgroup

shrinks when the intermediate field grows and vice versa. But be aware

that it is possible that Aut(E/K1) and Aut(E/K2) can be the same

when K1 ⊂ K2.

Definition 8.1.3. Let E/F be a finite extension and let L/F be

an arbitrary extension of F. A homomorphism σ : E → L such that

σ(a) = a for all a ∈ F is called an embedding of E/F into L/F.

Proposition 8.1.4. The number of embeddings of a finite extension

E/F to L/F does not exceed [E : F ] for any extension L/F.

Proof. Let n = [E : F ]. Suppose that there are n + 1 distinct

embeddings σ1, . . . , σn, σn+1 from E/F to L/F. Let u1, u2, . . . , un be a

basis of E/F. Since the number of rows of the matrixσ1(u1) σ2(u1) · · · σn+1(u1)

· · ·σ1(un) σ2(un) · · · σn+1(un)

is less than the number of columns, there are not all zero elements

a1, . . . , an+1 ∈ L such thatσ1(u1) σ2(u1) · · · σn+1(u1)

· · ·σ1(un) σ2(un) · · · σn+1(un)

a1

...

an+1

= 0.

Hence

[c1 · · · cn

]σ1(u1) σ2(u1) · · · σn+1(u1)

· · ·σ1(un) σ2(un) · · · σn+1(un)

a1

...

an+1

= 0


for arbitrary c1, . . . , cn ∈ F. This implies that

a1σ1(c1u1 + · · ·+ cnun) + · · ·+ an+1σn+1(c1u1 + · · ·+ cnun) = 0

for arbitrary c1, . . . , cn ∈ F. Since u1, . . . , un form a basis of E/F,

a1σ1(x) + · · ·+ an+1σn+1(x) = 0

for any x ∈ E. This contradicts Theorem 2.4.10.

Corollary 8.1.5. For any finite extension E/F, the inequality

|Aut(E/F )| ≤ [E : F ] holds.

Proof. If follows from Proposition 6.2.6 that Aut(E/F ) consists

all embeddings of E/F into E/F. So the corollary is a consequence of

Proposition 8.1.4

This corollary tells us that [E : F ] is an upper bound of |Aut(E/F)|.The extension E/F should be of special interest when this upper bound

is reached. Thus it is natural to introduce the following definition.

Definition 8.1.6. Let E/F be a finite extension. If [E : F ] =

|Aut(E/F)| then E/F is called a Galois extension and the group

Aut(E/F) is called the Galois group of E/F, denoted by Gal(E/F)

or G(E/F).

Example 8.1.7. Assume the the characteristic of a field F is not

equal to 2. Then any quadratic extension E/F (i.e., [E : F ] = 2) is a

Galois extension.

Proof. Choose any α ∈ E\F. Then E = F [α] and the minimal

polynomial p(x) of α is a quadratic polynomial

p(x) = x2 + ax+ b ∈ F [x].

Since the characteristic of F is not equal to 2 by assumption there

exists c ∈ F such that a = 2c. Hence p(x) = (x + c)2 + b − c2. Since

f(x) is irreducible over F, b− c2 6= 0. Since p′(x) = 2(x+ c) 6= 0, p(x)

does not have multiple zeros. Hence p(x) = (x − α)(x − β) in which


β ∈ E and β 6= α. It is evident that 1, α and 1, β are two bases of E as

a vector space over F. Define a map

σ : E → E, c+ dα 7→ c+ dβ, (c, d ∈ F ).

Then σ is an invertible linear transformation of the F -space E. To verify

that σ ∈ Aut(E/F) it suffices to check that σ((c1 + d1α)(c2 + d2α)) =

σ(c1 + d1α)σ(c2 + d2α) for any c1, d1, c2, d2 ∈ F. It follows from

(c1 + d1α)(c2 + d2α) = (c1c2 − bd1d2) + (c1d2 + c2d1)α,

that

σ((c1+d1α)(c2+d2α)) = (c1c2−bd1d2)+(c1d2+c2d1)β = σ(c1+d1α)σ(c2+d2α),

So σ ∈ Aut(E/F) and Aut(E/F) contains at least two elements id and

σ. It follows from |Aut(E/F)| ≤ [E : F ] = 2 that |Aut(E/F)| = 2.

Example 8.1.8. The cubic extension Q[ 3√

2]/Q is not a Galois ex-

tension, in which 3√

2 is the real cubic root of 2.

Proof. Let σ ∈ Aut(Q[ 3√

2]/Q) and let α = σ( 3√

2). Since ( 3√

2)3 =

2, α3 = 2. But 3√

2 is the only number in Q[ 3√

2] ⊂ R whose cubic

power is equal to 2, so α = 3√

2. Hence σ is the identity map. Hence

|Aut(Q[ 3√

2]/Q)| = 1.

Proposition 8.1.9. Every finite extension of a finite field is a Ga-

lois extension with a cyclic group as its Galois group.

Proof. Let F = Fq be a finite field containing q elements and let

E/F be a finite extension with [E : F ] = n. Then E ∼= Fqn . Let

φ : E → E, a 7→ aq.

Then φ ∈ Aut(E/F) and φ generates a cyclic subgroup of order n of

Aut(E/F). Hence |Aut(E/F)| = n = [E : F ] by Corollary 8.1.5, which

implies that E/F is a Galois extension and G(E/F) is a cyclic group

of order n.


Proposition 8.1.10. Let K be an intermediate field of a finite

Galois extension E/F. Then E/K is a Galois extension.

Proof. Let n = [E : F ],m = [E : K], d = [K : F ]. Let A be the

set of all embeddings from K/F to E/F. For any g ∈ G(E/F), let g|Kdenote the restriction of g on K. Then g 7→ g|K gives a map

f : G(E/F) → A.

For an arbitrary h ∈ A such that f−1(h) is not empty, there is some

u ∈ G(E/F) such that u|K = h. It is easy to verify the f−1(h) =

uAut(E/K), i.e., f−1(h) is a left coset of the subgroup Aut(E/K) in

G(E/F). This means that f−1(h) contains at most |Aut(E/K)| ele-

ments for any h ∈ A. Since G(E/F ) is the disjoint union of f−1(h)

where h runs over all elements of A, the inequality

|G(E/F)| ≤ |Aut(E/K)| · |A| (8.2)

holds. It follows from Corollary 8.1.5 that |Aut(E/K)| ≤ [E : K].

Lemma 8.1.4 tells us that |A| ≤ [K : F ]. Hence the inequality (8.2)

implies |G(E/F)| ≤ [E : K][K : F ] = [E : F ]. Since E/F is a Galois

extension by assumption, |G(E/F )| = n. Hence |Aut(E/K)| = [E : K],

which means that E/K is a Galois extension.

Lemma 8.1.11. Let G be a finite subgroup of Aut(E) for a field E.

Let F = EG = a ∈ E|g(a) = a for all g ∈ G. Then E/F is a

finite Galois extension with G(E/F) = G.

Proof. Let g1, g2, . . . , gn be all elements of G. It suffices to show

that [E : F ] ≤ n. Suppose that there are n+1 elements u1, . . . , un+1 in

E linearly independent over F. The system of homogeneous equations

g1(u1)x1 + · · ·+ g1(un+1)xn+1 = 0,

· · · (8.3)

gn(u1)x1 + · · ·+ gn(un+1)xn+1 = 0


has a nonzero solution

x1 = a1, . . . , xn+1 = an+1

in the field E. Choose one solution with minimum number of nonzero

elements. Without loss of generality, assume that a1 = 1. If ai ∈ F for

1 ≤ i ≤ n+ 1, then

g1(a1u1 + · · ·+ an+1un+1) = 0,

so

a1u1 + · · ·+ an+1un+1 = 0,

contradicting the hypothesis that u1, . . . , un+1 are linearly independent

over F. Hence there is at least one ai that is not in F. Assume that

a2 /∈ F without loss of generality.

According to the definition of F there is some g ∈ G such that

g(a2) 6= a2. By applying g to each equation in (8.3) we obtain

gg1(u1)g(a1) + · · ·+ gg1(un+1)g(an+1) = 0,

· · ·

ggn(u1)g(a1) + · · ·+ ggn(un+1)g(an+1) = 0.

Since gg1, gg2, . . . , ggn are still all elements of G, these equations are

merely a rearrangement of the equations

g1(u1)g(a1) + · · ·+ g1(un+1)g(an+1) = 0,

· · ·

gn(u1)g(a1) + · · ·+ gn(un+1)g(an+1) = 0.

By subtracting the equations

g1(u1)a1 + · · ·+ g1(un+1)an+1 = 0,


· · ·

gn(u1)a1 + · · ·+ gn(un+1)an+1 = 0

one by one we obtain equations

g1(u1)(g(a1)− a1) + g1(u2)(g(a2)− a2) + · · · = 0,

· · ·

gn(u1)(g(a1)− a1) + gn(u2)(g(a2)− a2) + · · · = 0,

which means that g(a1)− a1, g(a2)− a2, . . . , g(an+1)− an+1 form a new

nonzero solution of the system (8.3). Since g(a1)−a1 = 0, g(a2)−a2 6= 0

and g(ai) − ai = 0 whenever ai = 0 for 3 ≤ i ≤ n + 1, the number

of nonzero elements in g(a1) − a1, g(a2) − a2, . . . , g(an+1) − an+1) is

less than that in a1, a2, . . . , an+1. A contradiction is reached. Hence

[E : F ] ≤ n.

Corollary 8.1.12. A finite extension E/F is Galois if and only

if for every α ∈ E\F there is some σ ∈ Aut(E/F) such that σ(α) 6= α.

Proof. ⇐: Let G = Aut(E/F). Then EG = F and E/F is a Galois

extension by Lemma 8.1.11.

⇒: Assume that E/F is a Galois extension. Let K = EG(E/F).

Then E/K is a Galois extension with G(E/F) as its Galois group.

Hence [E : K] = |G(E/F)| = [E : F ]. It follows from F ⊆ K that

K = F.

Remark 8.1.13. Due to Lemma 8.1.11 the item 4) of Lemma 8.1.2

can be improved as: H = Aut(E/EH) for every subgroupH of Aut(E/F).

Theorem 8.1.14 (The fundamental theorem of Galois theory). Let

E/F be a finite Galois extension with n = [E : F ]. Then the following

three statements hold:

1) The maps

φ : K 7→ Aut(E/K)


and

ψ : H 7→ EH

give a one-to-one correspondence between the set of all intermediate

fields of E/F and the set of all subgroups of G(E/F).

2) [K : F ] = (G(E/F) : G(E/K)) for any intermediate subfield K

of E/F.

3) For any intermediate field K, K/F is a Galois extension if and

only if G(E/K) /G(E/F). In this case, G(K/F) ∼= G(E/F)/G(E/K).

Proof. To prove 1) we need to verify that both ψ φ and φ ψare identity maps.

First of all, Lemma 8.1.11 means exactly that φ ψ is the identity

map.

According to Proposition 8.1.10 E/K is a Galois extension for any

intermediate fieldK. Hence |Aut(E/K)| = [E : K]. Let B = EAut(E/K).

Then B ⊇ K. By Lemma 8.1.11 E/B is a Galois extension, so [E :

B] = |Aut(E/K)| = [E : K] and B = K. This proves that ψ φ is the

identity map.

2) Since both E/F and E/K are Galois extensions, [E : F ] =

|G(E/F)|, [E : K] = |G(E/K)|. Hence [K : F ] = [E : F ]/[E : K] =

(G(E/F) : G(E/K)).

3) Assume that H/G(E/F). Let K = EH . and let g be an arbitrary

element in G(E/F). We claim that g(b) ∈ K for every b ∈ K. Otherwise

there would be some b ∈ K such that g(b) /∈ K. So there would be

some h ∈ H such that h(g(b)) 6= g(b). It follows from g−1hg ∈ H that

g−1hg(b) = b, i.e., hg(b) = g(b), contradiction.

Thus a homomorphism

φ : G(E/F) → Aut(K/F ), g 7→ g|K

is well defined. It is evident that Ker(φ) = G(E/K). By the funda-

mental theorem of homomorphism we have

|Imφ| = (G(E/F) : G(E/K)) = [E : F ]/[E : K] = [K : F ].


Since

[K : F ] = |Imφ| ≤ |Aut(K/F )| ≤ [K : F ],

the homomorphism φ is surjective and |Aut(K/F )| = [K : F ]. This

means thatK/F is a Galois extension and G(K/F) ∼= G(E/F)/G(E/K).

Conversely, assume that K is an intermediate field such that K/F

is Galois. Then |G(K/F)| = [K : F ]. Since there are at most [K : F ]

embeddings fromK/F into E/F,G(K/F) contains all embeddings from

K/F to E/F. For every σ ∈ G(E/F), σ|K is an embedding of K/F into

E/F. Hence σ|K ∈ G(K/F). This give a homomorphism

G(E/F) → G(K/F), σ 7→ σ|K

whose kernel is G(E/K). Hence G(E/K) /G(E/F).

Theorem 8.1.15. A finite extension E/F is Galois if and only if

there is some α ∈ E satisfying the following two conditions:

1) E = F [α];

2) the minimal polynomial f(x) ∈ F [x] of α is decomposed in E[x]

into

f(x) = (x− α1)(x− α2) · · · (x− αn)

for some distinct α1, α2, . . . , αn in E.

Proof. When F is a finite field the statement is true due to Propo-

sition 8.1.9 and Corollary 7.2.4. Hence we may assume that F is an

infinite field.

⇒: Assume that E/F is a Galois extension of degree n. For any

u ∈ E let

Hu = g ∈ G(E/F)|g(u) = u.

Then Hu is a subgroup of G(E/F).

For any u, v ∈ E we claim that there is some w ∈ E such that

Hw = Hu ∩Hv. Since G(E/F) has only finitely many subgroups but F

contains infinitely many elements, there are c1, c2 ∈ F, c1 6= c2 such that

Hu+c1v = Hu+c2v by pigeonhole principle. Let w = u+c1v. It is obvious

that Hu ∩Hv ⊆ Hw. Since g(u + c1v) = u + c1v, g(u + c2v) = u + c2v

for every g ∈ Hw, (c1 − c2)g(v) = (c1 − c2)v, which implies g(v) = v


and thus g(u) = u. So g ∈ Hu ∩Hv. This shows that Hw = Hu ∩Hv.

This concludes the proof of the claim.

By induction, for any u1, . . . , um ∈ E there is w ∈ E such that

Hw = Hu1 ∩ · · · ∩ Hum . In particular, let u1, . . . , un ∈ E be a basis of

E over F. Then

Hu1 ∩ · · · ∩Hun = g ∈ G(E/F)|g(x) = x ∀x ∈ E.

Hence

Hu1 ∩ · · · ∩Hun = 1

by the fundamental theorem of Galois theory. Therefore there exists

α ∈ E such that Hα = 1.Let f(x) be the minimal polynomial of α over F. Then deg(f) ≤ n.

Suppose that deg(f) < n. Since f(g(α)) = 0 for any g ∈ G(E/F), by

pigeonhole principle there exist two distinct elements g1, g2 ∈ G(E/F)

such that g1(α) = g2(α), so g−11 g2 ∈ Hα and thus g−1

1 g2 = 1. This

contradicts with g1 6= g2. Therefore E = F [α].

Let 1 = g1, g2, . . . , gn be all elements of G(E/F) and let αi =

gi(α), 1 ≤ i ≤ n. Then α1, . . . , αn are distinct elements in E and each

of them is a zero of f(x). Hence

f(x) = (x− α1)(x− α2) · · · (x− αn).

⇐: For every αi there is a surjective homomorphism

fi : F [x] → E, h(x) 7→ h(αi)

such that Ker(fi) = (f(x)). Hence fi induces an isomorphism φi :

F [x]/(f(x)) → E. Denote gi = φi φ−11 , (1 ≤ i ≤ n). Then g1, . . . , gn

are distinct elements of Aut(E/F). Hence |Aut(E/F)| ≥ n. It follows

from Corollary 8.1.5 that |Aut(E/F)| = [E : F ]. Therefore E/F is a

Galois extension.

Let E and K be subfields of a field L. The compositum of E and

K is the smallest subfield of L containing E and K, denoted by EK. It

is the extension of E generated by the set K or that of K generated by


E. The compositum of more than two subfields is defined in a similar

way.

Proposition 8.1.16. Let E,K and F be subfields of a field L such

that F ⊆ E and F ⊆ K. Assume that E/F is a finite Galois extension.

Then EK/K is a finite Galois extension and G(EK/K) ∼= G(E/E ∩K).

Proof. If E = F then the proposition is trivial. We may assume

that [E : F ] > 1.

According to Theorem 8.1.15 E = F [α] such that the minimal poly-

nomial f(x) of α over F is equal to

(x− α1) · · · (x− αn)

for distinct elements α1, . . . , αn ∈ E. We may assume that α1 = α.

Then KE = K[α] is a finite extension of K.

Let p(x) = xm + cm−1xm−1 + · · ·+ c1x+ c0 ∈ K[x] be the minimal

polynomial of α over K. Then p(x) divides f(x) in K[x]. Hence every

zero of p(x) in L is a zero of f(x). Without loss of generality, we may

assume that p(x) = (x − α1) · · · (x − αm). It follows from Theorem

8.1.15 again that EK/K is a finite Galois extension.

Since every coefficient ci of p(x) belongs to F [α1, . . . , αm], so ci ∈E ∩ K for 0 ≤ i < m. This means that p(x) ∈ E ∩ K[x]. It is clear

that p(x) is irreducible over E ∩ K. Hence p(x) is also the minimal

polynomial of α over E ∩K. Therefore [EK : K] = [E : E ∩K].

The restriction map g 7→ g|E gives a homomorphism φ : G(EK/K) →G(E/E ∩K). Assume that g ∈ Ker(φ). Then g|E is the identity map

from E to E, which implies that g(α) = α. It follows from EK = K[α]

that g : EK → EK is the identity map. Hence φ is injective. Since

[EK : K] = [E : E ∩K], φ is a bijection.

Exercises

1. Show that the condition that the characteristic of the field

is not equal to 2 is indispensable in Example 8.1.7 by the example

F2(x)/F2(x2).


2. Let ζ = (−1 +√−3)/2 and let F = Q[ζ]. Show that the cubic

extension F [ 3√

2]/F is Galois.

3. Let L/F be a finite Galois extension and let H1, H2 be two

subgroups of G(L/F) with E1, E2 as their fixed fields respectively.

1) Show that the fixed field of H1 ∩H2 is E1E2.

2) Show that E1 ∩ E2 is the fixed field of H1H2 where H1H2 is the

subgroup of G(L/F) generated by H1 ∪H2.

4. Let K/E,E/F be finite Galois extensions. Show by example

that K/F is not necessarily a Galois extension.

5. Let E/F be a finite extension and let L be the algebraic closure of

E. If there exist [E : F ] distinct embeddings from E/F into L/F, then

E/F is called a separable extension; if every embedding from E/F

into L/F carries E into E then E/F is called a normal extension.

Prove that E/F is a Galois extension if and only if E/F is separable

and normal.

6. Let F be a field and let f(x) ∈ F [x]. Let α1, . . . , αn be all zeros

of f(x) in the algebraic closure F of F. Then the field F [α1, . . . , αn] is

referred to as the splitting field of f(x).

Show that a finite extension E/F is a Galois extension if and only if

it is the splitting field of some f(x) ∈ F [x] that does not have multiple

zeros in the algebraic closure of F.

7. Let p be an odd prime number and let E/F be a Galois extension

whose Galois group G(E/F) is a non-abelian group of order 2p. Let K

be an intermediate field of E/F such that [E : K] = 2. Show that K/F

is not a Galois extension.

8. Let E/F be a finite Galois extension and let α ∈ E. Assume

that g(α) 6= α for every g ∈ G(E/F)\1. Show that E = F [α].

9. Let E/F be a finite extension. Show that the integer |Aut(E/F)|divides [E : F ].


8.2 * Solvable Extension and Solvability of Alge-

braic Equations by Radicals

Definition 8.2.1. A finite extension E/F if a solvable extension if

there is a finite chain of intermediate fields

F = E0 ⊂ E1 ⊂ · · · ⊂ En = E

such that Ei/Ei−1 is Galois and G(Ei/Ei−1) is abelian for 1 ≤ i ≤ n.

Let g be an automorphism of a field K. For any f(x) = anxn +

an−1xn−1 + · · ·+ a0 ∈ K[x] denote by f g(x) the polynomial g(an)xn +

g(an−1)xn−1 + · · ·+ g(a0) ∈ K[x].

Lemma 8.2.2. For a fixed automorphism g of a field K, the map

K[x] → K[x], f(x) 7→ f g(x) is a ring isomomorphism. A polynomial

f(x) ∈ K[x] is irreducible if and only if f g(x) is irreducible over K.

Proof. Let f(x) = anxn + an−1x

n−1 + · · ·+ a0 and h(x) = bnxn +

bn−1xn−1 + · · ·+ b0 be two polynomials in K[x] (in which an or bn are

not necessarily nonzero). Then

f g(x)+hg(x) = g(an+bn)xn+g(an−1+bn−1)xn−1+· · ·+g(a0+b0) = (f+h)g(x)

and the coefficient of xr in (fh)g(x) is∑

i+j=r g(aibj) =∑

i+j=r g(ai)g(bj),

which is exact the coefficient of xr in f g(x)hg(x). Hence (fh)g(x) =

f g(x)hg(x). This shows that the map Φg : f(x) 7→ f g(x) is a homomor-

phism.

Since Φg−1 is the inverse of Φg, the homomorphism Φg is an isomor-

phism.

The second statement is clear.

Lemma 8.2.3. Let E/K,K/F be finite Galois extensions and let E

be an algebraically closed field containing E. There exists a subfield L

of E such that

1) E ⊆ L;

2) L/F is a finite Galois extension;

8.2. * SOLVABLE EXTENSION AND SOLVABILITY OF ALGEBRAIC EQUATIONS BY RADICALS163

3) there is a chain of intermediate fields

K = L0 ⊆ L1 ⊆ · · · ⊆ Ln = L

such that Li/Li−1 is a Galois extension and G(Li/Li−1) is isomorphic

to a subgroup of G(E/K) for 1 ≤ i ≤ n.

Proof. Let σ1, . . . , σr be the set of all embeddings of E/F into

E/F in which σ1 is the identity map. Denote Ei = σi(E) for 1 ≤ i ≤ r.

Let L = E1 · · ·Er.

Since K/F is Galois, σi(K) = K. The group Aut(Ei/K) is isomor-

phic to G(E/K) via the correspondence τ 7→ σ−1i τσi. Hence

|Aut(Ei/K)| = |G(E/K)| = [E : K] = [Ei : K],

which implies that Ei/K is a Galois extension for each i.

Let Li = E1 · · ·Ei for 0 ≤ i ≤ n. Since Li = Li−1Ei, Proposition

8.1.16 implies that the extension Li/Li−1 is Galois and G(Li/Li−1) ∼=G(Ei/Ei ∩Li−1), which is isomorphic to a subgroup of G(E/K). So 3)

is proved.

Let σ be an embedding of L/F into E/F. Then σ(Ei) = σσi(E) ⊆L, since σσi is an embedding from E/F into E/F. This shows that

σ ∈ Aut(L/F). Hence every embedding of L/F into E/F is an auto-

morphism of L over F.

It remains to show that L/F is a Galois extension. It suffices to

show that for any a ∈ L\F there is some σ ∈ Aut(L/F) such that

σ(a) 6= a due to Corollary 8.1.12. This is proved in two separate cases.

Case 1) a ∈ K\F.Since K/F is Galois, there is some g ∈ G(K/F ) such that g(a) 6= a.

There exists an embedding σ : L → E from L/F into E/F such that

σ|K = g by Theorem 6.4.8. Then σ ∈ Aut(L/F) and σ(a) 6= a.

Case 2) a ∈ L\K.Then a ∈ Li\Li−1 for some 1 ≤ i ≤ n. Since Li/Li−1 is Galois, there

is some g′ ∈ G(Li/Li−1) such that g′(a) 6= a. Then the same argument

as in Case 1) shows that there is σ ∈ Aut(L/F) such that σ|Li= g′.


Lemma 8.2.4. For any finite solvable extension E/F there is a finite

extension L/E such that L/F is a finite solvable Galois extension.

Proof. Let

F = E0 ⊂ E1 ⊂ · · · ⊂ En = E

be a finite chain of intermediate fields such that Ei/Ei−1 is Galois and

G(Ei/Ei−1) is abelian for 1 ≤ i ≤ n.

The lemma is proved by induction on n. When n = 1 the lemma is

obviously true. Assume that n > 1. Since En−1/F is solvable, by induc-

tion hypothesis there is a finite extension K ′/En−1 such that K ′/F is a

finite solvable Galois extension. Let E be the algebraic closure of E and

let σ be an embedding of K ′/En−1 into E/En−1. Denote K = σ(K ′).

Then K is a finite extension of En−1 such that K/F is a finite solvable

Galois extension. It is obvious that En−1 ⊆ E ∩K.Since E/En−1 is a Galois extension, so is KE/K and G(KE/K) ∼=

G(E/E∩K) by Proposition 8.1.16. Since E∩K is an intermediate field

of E/En−1, G(E/E∩K) is a subgroup of the abelian group G(E/En−1).

Hence G(KE/K) is an abelian group.

By Lemma 8.2.3 there is a finite Galois extension L of F in E

containing KE and a chain of intermediate fields

KE = L0 ⊆ L1 ⊆ · · · ⊆ Lm = L

such that Li/Li−1 is Galois with an abelian Galois group for each 1 ≤i ≤ m.

Put all these extensions together and we obtain a chain of extensions

F ⊆ K ⊆ KE = L0 ⊆ L1 ⊆ · · · ⊆ Lm = L

in which K/F is a solvable extension, G(KE/K) is abelian and each

G(Li/Li−1) is abelian for 1 ≤ i ≤ n. Hence L/F is solvable.

Proposition 8.2.5. A finite Galois extension E/F is solvable if

and only if G(E/F) is a solvable group.


Proof. ⇒: If there are intermediate subfields

F = K0 ⊂ K1 ⊂ K2 · · · ⊂ Kn = E

such that Ki/Ki−1 is Galois with an abelian Galois group for 1 ≤ i ≤ n.

By the fundamental theorem of Galois theory, there is a subnormal

series

1 = G(E/Kn) / G(E/Kn−1) / · · · / G(E/K1) / G(E/K0) = G(E/F).

Since G(E/Ki−1)/G(E/Ki) ∼= G(Ki/Ki−1) is an abelian group for each

i, G(E/F) is a solvable group.

⇐: Assume that G(E/F) is a solvable group. Then there is a sub-

normal series

1 = H0 / H1 / · · · / Hn−1 / Hn = G(E/F)

such that Hi/Hi−1 is abelian for each 1 ≤ i ≤ n. It corresponds to a

sequence of intermediate fields

F = EHn ⊆ EHn−1 ⊆ · · · ⊆ EH1 ⊆ EH0 = E.

Since Hi /Hi+1, EHi/EHi+1 is a Galois extension with G(EHi/EHi+1) ∼=

Hi+1/Hi by the fundamental theorem of Galois theory. Therefore E/F

is a solvable extension.

Let us investigate a special type of Galois extensions, which is

closely related to the solvability of polynomial equations by radicals.

Proposition 8.2.6. Let p be a prime number and let F be a field of

characteristic zero. Assume that F contains all p-th roots of unity. Let

E/F be an extension and let α be an element of E such that αp ∈ F.

Then F [α]/F is a Galois extension with a cyclic Galois group.

Proof. If α ∈ F then the proposition becomes trivial. Hence we

assume that α /∈ F.Since the characteristic of F is zero, gcd(xp − 1, pxp−1) = 1, the

polynomial xp− 1 has p distinct zeros 1, ζ, ζ2, . . . , ζp−1 in the algebraic


closure of E by Lemma 3.4.11. Since 1, ζ, ζ2, . . . , ζp−1 ∈ F and αp =

a ∈ F by the given conditions, there is the following decomposition

xp − a = (x− α)(x− ζα)(x− ζ2α) · · · (x− ζp−1α).

Let f(x) be the minimal polynomial of α over F. Then deg(f) > 1 by

the assumption α /∈ F. Since f(x) is a factor of xp−a, at least one ζdα

is a zero of f(x) in which 1 ≤ d ≤ p− 1. Hence there is an embedding

g from F [α] to E such that g(α) = ζdα. Since g2, g3, . . . , gp−1 are also

embeddings of F [α]/F into E/F and gi(α) = ζdiα for 1 ≤ i ≤ p − 1,

there are p distinct embeddings from F [α]/F into E/F. If follows from

Lemma 8.1.4 that p ≤ [F [α] : F ]. Since [F [α] : F ] = deg(f) ≤ deg(xp−a) = p, we have p = [F [α] : F ] and f(x) = xp − a. By Theorem

8.1.15 F [α]/F is a Galois extension and its Galois group consists of the

elements 1, g, g2, g3, . . . , gp−1, which is a cyclic group of order p.

Let C be the field of complex numbers and let x1, x2, . . . , xn be

indeterminates. Let L = C(x1, . . . , xn) be the field of rational functions

over C with variables x1, x2, . . . , xn.

Every element σ in the symmetric group Sn determines a permuta-

tion of variables x1, . . . , xn, which induces an automorphism of the field

L. All such automorphisms form a subgroup G of Aut(L) isomorphic

to Sn. Let

F = a ∈ L|f(a) = a ∀f ∈ G.

It follows from Lemma 8.1.11 that L/F is a Galois extension and

G(L/F) = G ∼= Sn.

Proposition 8.2.7. If n > 4 then the Galois extension L/F is not

solvable.

Proof. By Theorem 1.6.5 Sn is not solvable for n > 4, L/F is not

solvable by Proposition 8.2.6.

Let

σ1 = x1 + · · ·+ xn,

σ2 = x1x2 + x1x3 + · · ·+ xn−1xn,


· · ·

σn = x1x2 · · ·xn

be the elementary symmetric polynomials.

Proposition 8.2.8. Under the above notation,

F = K(σ1, σ2, . . . , σn).

Proof. Since F consists of all fractions u(x1, . . . , xn)/v(x1, . . . , xn)

such that u and v are symmetric polynomials. It is well-known that

every symmetric polynomial can be written as h(σ1, . . . , σn) for some

polynomial h(y1, . . . , yn). Hence the field F is generated by elementary

polynomials.

Corollary 8.2.9. A polynomial equation in one variable with de-

gree greater or equal to 5 and with general coefficients is not solvable

by radicals.

Proof. A monic polynomial of general coefficients can be written

as

f(x) = xn − σ1xn−1 + · · ·+ (−1)iσix

n−i + (−1)nσn.

Starting from E0 = C(σ1, . . . , σn) construct a chain of extensions

E0 ⊂ E1 ⊂ E2 · · · ⊂ EN ,

where Ei = Ei−1[αi], αpi

i ∈ Ei−1.We may assume that each pi is a prime

number, since any radical is the composite of prime order radicals. For

example 6√a = 3

√√a. By Proposition 8.2.6 each extension Ei/Ei−1 is a

Galois extension with cyclic Galois group. Hence EN/E0 is a solvable

extension. By Lemma 8.2.4 there is a finite solvable Galois extension

L/E0 containing EN as an intermediate field.

Suppose thatEN contains all zeros x1, . . . , xn of f(x). Then C(x1, . . . , xn)

is an intermediate field of the solvable Galois extension L/C(σ1, . . . , σn).

Hence G(C(x1, . . . , xn)/C(σ1, . . . , σn)) is a quotient group of the solv-

able group G(L/E0). This contradicts Proposition 8.2.7.

To conclude this course we introduce the inverse Galois problem.


Proposition 8.2.10. Every finite group G is isomorphic to the

Galois group of some finite Galois extension.

Proof. By Cayley’s theorem, G is isomorphic to a subgroup of

some symmetric group Sn.

Since the Galois group of C(x1, . . . , xn)/C(σ1, . . . , σn) is Sn, there is

an intermediate subfield E such that G(C(x1, . . . , xn)/E) is isomorphic

to G by the fundamental theorem of Galois theory.

This proof is easy. But if the base field is required to be the field of

rational numbers, the problem becomes a long standing open problem

in algebra. The precise formulation is as follows:

Is any finite group isomorphic to the Galois group of some finite

Galois extension of Q?

Exercises

1. Let E/F be a finite Galois extension and let H be a subgroup of

G(E/F). Show that there is β ∈ E such that H = g ∈ G(E/F)|g(β) =

β.

2. Construct two finite Galois extensions of Q whose Galois groups

are cyclic groups of orders 3 and 4 respectively.

3. Let p be a prime number and let F = Fp(x) be the field of

rational functions in one variable over the finite field Fp. Let E =

F [y]/(yp − y + x). Show that E/F is a Galois extension of degree p.

Appendix A

Quadratic residues

Let p be a prime number. The map sq : F∗p → F∗p, t → t2 is a group

homomorphism.

Definition A.0.11. An integer n not divisible by p is called a

quadratic residue modulo p if n ∈ Im(sq), otherwise it is a quadratic

nonresidue modulo p.

For any prime number p, the Legendre symbol of an integer n

module p is defined as

(n

p

)=

0, if p|n1, if p does not divide n and n is a quadratic residue modulo p

−1, if p does not divide n and n is a quadratic nonresidue modulo p

Example A.0.12. 1, 2, 4 are quadratic residues modulo 7 while

3, 5, 6 are quadratic nonresidues modulo 7.

Lemma A.0.13. Let p be an odd prime and let 0, a1, . . . , ap−1 be a

complete residue system modulo p. Then there are exactly (p − 1)/2

quadratic residues among a1, . . . , ap−1.

Proof. This follows from(

ai

p

)= 1 if and only if ai ∈ Im(sq) and

|Im(sq)| =|F∗p|

|Ker(sq)|=p− 1

2.

169

170 APPENDIX A. QUADRATIC RESIDUES

Lemma A.0.14. The equality(mn

p

)=

(m

p

)(n

p

)holds for any integer m,n and any prime number p.

Proof. If p|mn then both sides are equal to zero.

Assume that none of m,n is divisible by p. By the definition of

Legendre symbol,

n 7→(n

p

)is a group homomorphism from F∗p to 1,−1. Hence the equlity holds.

Lemma A.0.15. Let n be an integer not divisible by an odd prime

p. Then (np) = 1 if and only if

np−12 ≡ 1 (mod p).

Proof. Since F∗p is a cyclic group of order p− 1,(n

p

)= 1 ⇔ n ∈ Im(sq) ⇔ n|Im(sq)| = 1.

Corollary A.0.16. Let p be an odd prime. Then(−1

p

)= (−1)

p−12 .

Lemma A.0.17. Let p be an odd prime. If p ≡ ±1 (mod 8), then(2p

)= 1, otherwise

(2p

)= −1.

Proof. Since 8 is coprime with p, Theorem 7.2.2 implies that there

exists an element α of order 8 in F∗p, i.e., α8 = 1 and αd 6= 1 for any

1 ≤ d < 8. Hence α4 = −1 and thus α2 + α−2 = 0. So 2 = (α+ α−1)2.

If p ≡ ±1 (mod 8), then

(α+ α−1)p = αp + α−p = α+ α−1.

171

Hence (α+ α−1)p−1 = 1. It follows that

2p−12 = (α+ α−1)p−1 = 1.

Lemma A.0.15 implies that (2

p

)= 1.

If p ≡ ±5 (mod 8) then

(α+ α−1)p = α5 + α−5 = −(α+ α−1).

So (α+ α−1)p−1 = −1. Hence

2p−12 = (α+ α−1)p−1 = −1.

It follows from Lemma A.0.15 that(2

p

)= −1.

The following theorem is one of the fundamental theorem in number

theory.

Theorem A.0.18 ( Gauss reciprocity law ). Let p, p′ be two distinct

prime numbers. Then(p

p′

)= (−1)

(p−1)(p′−1)4

(p′

p

).

Choose an element w of order p in F∗p′ . For any integers n,m such

that n ≡ m (mod p), the equality wn = wm holds. Hence f(x) = wx

is a function on Fp.

The element

y =∑x∈Fp

(x

p

)wx

is in Fp′ .

172 APPENDIX A. QUADRATIC RESIDUES

Lemma A.0.19.

y2 = (−1)(p−1)/2p.

Proof.

y2 =∑

x,z∈Fp

(xz

p

)wx+z

=∑u∈Fp

wu

∑x∈Fp

(x(u− x)

p

) .

When u = 0,

∑x∈Fp

(x(u− x)

p

)= (p− 1)

(−1

p

).

When u 6= 0,

∑x∈Fp

(x(u− x)

p

)=∑x∈F∗p

(−x2

p

)(1− x−1u

p

)

=

(−1

p

)∑x∈F∗p

(1− x−1u

p

)= −

(−1

p

).

It follows from ∑u∈F∗p

wu =

p−1∑i=1

wi = −1

that

y2 = p

(−1

p

)= (−1)(p−1)/2p.

Lemma A.0.20.

yp′−1 =

(p′

p

).

173

Proof. Since y ∈ Fp′ ,

yp′ =∑x∈Fp

(x

p

)wxp′

=

(p′

p

)∑x∈Fp

(xp′

p

)wxp′

=

(p′

p

)y.

Theorem A.0.18 follows from

yp′−1 = (y2)(p′−1)/2 = (−1)(p−1)(p′−1)

4 p(p′−1)/2 = (−1)(p−1)(p′−1)

4

(p

p′

)and Lemma A.0.20.

Example A.0.21.(29

43

)=

(43

29

)=

(14

29

)=

(2

29

)(7

29

)= −

(7

29

)= −

(29

7

)= −

(1

7

)= −1.

Appendix B

Every finite skew field is a

field

Let n be a natural number greater than 1 and let ζ = e2mπi/n =

cos(2mπ/n) + i sin(2mπ/n), in which m is a natural number such that

m < n and gcd(m,n) = 1. Then ζ is is called an n-th primitive root of

unity. Let C denote the set of all n-th primitive roots of unity.

The equation xn − 1 = 0 has n zeros 1, e2πi/n, e4πi/n, . . . , e2(n−1)πi/n

in the complex number field. They form a cyclic group of order n under

the multiplication. Then C is exactly the set of all elements of order n

in this cyclic group. The number of elements in C is the Euler function

φ(n).

If ζ, η ∈ C, then there is a natural number d such that gcd(d, n) = 1

and ζd = η.

Theorem B.0.22. Let f(x) =∏

ζ∈C(x− ζ). Then f(x) is a monic

irreducible polynomial in Z[x] such that f |xn − 1 and deg(f) = φ(n).

Proof. Choose any ζ ∈ C. Since ζn − 1 = 0, the complex number

ζ is algebraic over Q. Let g(x) be the minimal polynomial of ζ over Q.Then xn − 1 = g(x)h(x), for some monic polynomial h(x) over Q. It

follows from Gauss’s lemma that g(x), h(x) ∈ Z[x].

Let p be a prime that does not divide n. Suppose that g(ζp) 6= 0.

Then h(ζp) = 0, since g(ζp)h(ζp) = ζpn − 1 = 0. For any

u(x) = amxm + am−1x

m−1 + · · ·+ a1x+ a0 ∈ Z[x]

174

175

let

u(x) = amxm + am−1x

m−1 + · · ·+ a1x+ a0 ∈ Fp[x],

in which aj is the residue class aj modulo p. Then

xn − 1 = g(x)h(x)

holds in Fp[x]. Hence h(x)p = h(xp), since F (xp) = F (x)p for any

F (x) ∈ Fp[x]. Since p does not divide n, the polynomial xn − 1 has

no multiple zeros in Fp[x]. Hence g(x) is coprime with h(x), so g(x)

is coprime with h(xp). By Lemma 3.2.12 there exist u(x), v(x) ∈ Fp[x]

such that

u(x)g(x) + v(x)h(xp) = 1,

i.e., there exist u(x), v(x), w(x) ∈ Z[x] such that

u(x)g(x) + v(x)h(xp) = 1 + pw(x).

Substitute ζ for x and we obtain pw(ζ) + 1 = 0. Hence g(x)|pw(x) + 1

and so g(x)|1, which leads to contradiction. Hence g(ζp) = 0 for any

prime number p that does not divide n.

Every η ∈ C can be written as η = ζm for some integer m that is

coprime with n. Let m = p1p2 . . . ps be the prime decomposition of m.

Then every pj does not divide n. Hence g(η) = 0, which implies that

f(x) =∏

ζ∈C(x− ζ) divides g(x). In particular, deg(f) ≤ deg(g).

On the other hand, since g(x) is irreducible, Q[α] ∼= Q[ζ] holds for

every zero α of g(x). Hence every zero of g(x) is in C. So deg(g) ≤deg(f), which implies deg(f) = deg(g). Since both f(x) and g(x) are

monic and f(x)|g(x), it follows that f(x) = g(x).

The polynomial f(x) in Theorem B.0.22 is called the n-th cyclo-

tomic polynomial, denoted by Φn(x).

Lemma B.0.23. Let a be a real number greater than or equal to 2.

Then |Φn(a)| > a− 1.

Proof. Since

Φn(a) =∏ζ∈C

(a− ζ)

176 APPENDIX B. EVERY FINITE SKEW FIELD IS A FIELD

by the definition of Φn(x), so |a− ζ| > a− 1 ≥ 1 for each ζ ∈ C.

Theorem B.0.24 (MacLagen-Wedderburn). Every finite skew field

is a field.

Proof. Suppose that E is a finite non-commutative skew field. Let

F = a ∈ E|ab = ba for any b ∈ E.

We claim that F is a field. For any a1, a2 ∈ F and any b ∈ E, the

equalities

(a1 + a2)b = a1b+ a2b = ba1 + ba2 = b(a1 + a2)

and

(a1a2)b = a1(a2b) = a1(ba2) = (a1b)a2 = (ba1)a2 = b(a1a2)

hold. Hence a1 + a2, a1a2 ∈ F. For any a ∈ F\0 the equality ab = ba

implies ba−1 = a−1b and so a−1 ∈ F. It is evident that the multiplication

in F is commutative. Hence F is a field. Let q = |F |. Then q is a power

of a prime number.

The skew field E has a natural structure of vector space over F. So

|E| = qn, where n is the dimension of E over F.

The set E∗ of all nonzero elements of E is a finite group of order

qn − 1 under multiplication. Consider the conjugation action g(a) =

gag−1 of E∗ on itself. Let x1, . . . , xr be a complete set of representatives

of the orbits. Since E is not commutative by hypothesis, there exists

at least one orbit of length greater than one. We may assume that the

length of the orbit containing xi for 1 ≤ i ≤ m is equal to one and the

remaining orbits has length greater than one. Let Hi be the stabilizer

of xi for i > m. Then F ∗ ⊆ Hi. It is easy to verify that Hi ∪ 0 is

a subspace of the vector space E over F. Hence |Hi| = qni − 1 with

ni < n.

Since x1, . . . , xm are exactly all the elements of F ∗, we have

q − 1 +s∑

j=1

qn − 1

qnj − 1= qn − 1. (B.1)

177

It follows from nj < n that Φn(x)(xnj − 1)|xn − 1. Hence

xn − 1−s∑

j=1

xn − 1

xnj − 1= Φn(x)u(x).

Since Φn(x) is a monic polynomial with integral coefficients by Theorem

B.0.22, we have u(x) ∈ Z[x]. The equality

q − 1 = Φn(q)u(q)

holds by (B.1). Therefore q−1 = |Φn(q)||u(q)| ≥ |Φn(q)|, contradicting

Lemma B.0.23.

Appendix C

Solutions of a cubic equation

and Hilbert theorem 90

By a standard change of variables x = y + b1/3 a cubic equation y3 +

b1y2 + b2y + b3 = 0 is transformed into the form

x3 + a2x− a3 = 0. (C.1)

Let x1, x2, x3 be three roots of (C.1). Then

x1 + x2 + x3 = 0,

x1x2 + x1x3 + x2x3 = a2,

x1x2x3 = a3.

Let F = C(a2, a3), E = C(x1, x2, x3). Then E/F is a Galois exten-

sion with the symmetric group S3 as its Galois group. Denote by σ the

element in G(E/F ) carrying x1, x2, x3 to x2, x3, x1 respectively. Then σ

generates a cyclic normal subgroup H of G(E/F ). Let K be the fixed

field of H. By the fundamental theorem of Galois theory, K/F is a

quadratic extension and E/K is a cubic Galois extension. Let τ be the

element of G(E/F ) carrying x1, x2 into x2, x1 respectively and keeping

x3 unchanged. Then the image of τ in G(K/F ) ∼= G(E/F )/G(E/K)

is a generator of G(K/F ).

As a first step, we try to find a generator of the extension K/F. A

178

179

standard choice is

δ = (x1 − x2)(x2 − x3)(x3 − x1) ∈ K\F,

since σ(δ) = δ and τ(δ) = −δ. Hence K = F [δ]. In fact, δ2 = −4a32 −

27a23 is the discriminant of the equation (C.1).

The next step is to find the minimal equation of x1 in the Galois

cubic extension K[x1]/K. We introduce a useful technique.

Let ζ = e2πi/3 = −1/2 + i√

3/2 be a cubic root of unity. Then

1 + ζ + ζ2 = 0. (C.2)

Let

u = x1 + ζx2 + ζ2x3, v = x1 + ζ2x2 + ζx3.

Then

σ(u) = x2 + ζx3 + ζ2x1 = ζ2u, σ(v) = x2 + ζ2x3 + ζx1 = ζv.

Hence

σ(u3) = u3, σ(v3) = v3.

This means that u3, v3 ∈ F [δ]. Hence u, v are cubic roots of some

elements in F [δ].

By linear algebra, x1, x2, x3 are linear combinations of u, v. By the

definition of u, v, there is a system of linear equations1 1 1

1 ζ ζ2

1 ζ2 ζ

x1

x2

x3

=

0

u

v

.Multiplying

1

3

1 1 1

1 ζ2 ζ

1 ζ ζ2

to the both sides yields

x1 =u

3+v

3, x2 = ζ2u

3+ ζ

v

3, x3 = ζ

u

3+ ζ2v

3. (C.3)

180APPENDIX C. SOLUTIONS OF A CUBIC EQUATION AND HILBERT THEOREM 90

It remains to compute u, v. A simple manipulation shows that uv =

−3a2, so

u3v3 = −27a32. (C.4)

In terms x1, x2, x3 express u3 and v3 as

u3 = 9a3 + 3ζx21x2 + 3ζ2x1x

22 + 3ζx2

2x3 + 3ζ2x2x23 + 3ζx2

3x1 + 3ζ2x3x21,

v3 = 9a3 + 3ζ2x21x2 + 3ζx1x

22 + 3ζ2x2

2x3 + 3ζx2x23 + 3ζ2x2

3x1 + 3ζx3x21.

Then

u3 +v3 = 18a3−3(x21x2 +x1x

32 +x2

2x3 +x2x23 +x2

3x1 +x3x21) = 27a3 ∈ F.

Since u3, v3 ∈ F [δ], they can be written as

u3 = α+ βδ, v3 = α′ + β′δ,

with α, β, α′, β′ ∈ F and β, β′ 6= 0, for u3, v3 /∈ F. So β′ = −β since

u3 + v3 ∈ F. Then u3v3 ∈ F implies α′ = α. Hence

2α = u3 + v3 = 27a3,

which implies α = 27a3/2. It follows from (C.4) that

α2 − β2(−4a32 − 27a2

3) = −27a32.

So

(4a32 + 27a2

3)β2 = −27

4(4a3

2 + 27a23).

Thus β = ±√−27/2. By the symmetry of u3 and v3 we may assume

that β =√−27/2. Hence we have

(u/3)3 =a3

2+

√(a2

3

)3

+(a3

2

)2

,

(v/3)3 =a3

2−√(a2

3

)3

+(a3

2

)2

.

Since every nonzero complex number has three distinct cubic roots, we

181

are facing the problem of the choices. We can choose any cubic root

for u/3, which is always a generator of E/K. So u/3 can be written as

u/3 =3

√a3

2+

√(a2

3

)3

+(a3

2

)2

.

Once u/3 is chosen, v/3 is determined. It must satisfy the condition

u

3

v

3= −a2

3.

By using (C.3) we obtain

Theorem C.0.25 ( Cardano’s formula). The three roots of the cubic

equation x3 + a2x− a3 = 0 are

x1 =3

√a3

2+

√(a2

3

)3

+(a3

2

)2

+3

√a3

2−√(a2

3

)3

+(a3

2

)2

,

x2 = ζ2 3

√a3

2+

√(a2

3

)3

+(a3

2

)2

+ ζ3

√a3

2−√(a2

3

)3

+(a3

2

)2

,

x3 = ζ3

√a3

2+

√(a2

3

)3

+(a3

2

)2

+ ζ2 3

√a3

2−√(a2

3

)3

+(a3

2

)2

,

in which the two cubic root satisfy

3

√a3

2+

√(a2

3

)3

+(a3

2

)2 3

√a3

2−√(a2

3

)3

+(a3

2

)2

= −a2

3.

A key point in the proof is the existence of u such that σ(u) = ζ2u,

i.e. ζ2 = σ(u)/u. This can be regarded as a special case of the following

celebrated theorem.

Theorem C.0.26 (Hilbert theorem 90). Let E/F be a Galois ex-

tension of degree n such that G(E/F ) is a cyclic group generated by σ.

For any η ∈ E∗ the following two statements are equivalent

1) ησ(η)σ2(η) · · ·σn−1(η) = 1;

2) there exists u ∈ E∗ such that η = u/σ(u).

182APPENDIX C. SOLUTIONS OF A CUBIC EQUATION AND HILBERT THEOREM 90

Proof. 2) ⇒ 1): Assume that η = u/σ(u). Then

ησ(η)σ2(η) · · ·σn−1(η) =u

σ(u)

σ(u)

σ2(u)· · · σ

n−2(u)

σn−1(u)

σn−1(u)

u= 1.

1) ⇒ 2): Assume that ησ(η)σ2(η) · · ·σn−1(η) = 1.

Let f : E → E be a map carrying ξ ∈ E into

ξ + ησ(ξ) + ησ(η)σ2(ξ) + · · ·+ ησ(η) · · ·σn−2(η)σn−1(ξ).

It is evident that ησ(f(ξ)) = f(ξ) for arbitrary ξ ∈ E. It suffices to

show the existence of ξ ∈ E such that f(ξ) 6= 0. But this is an easy

consequence of Theorem 2.4.10.

When η ∈ F ∗ and ηn = 1, the condition 1) of Hilbert theorem 90

always holds. So u = ξ + ησ(ξ) + η2σ2(ξ) + · · · + ηn−1σn−1(ξ) 6= 0,

and ησ(u) = u. For this reason, the choice of u = x1 + ζx2 + ζ2x3 and

v = x1 + ζ2x2 + ζx3 in solving the cubic equation is closely related to

Hilbert theorem 90.

Appendix D

Solutions of quartic

equations

The formula for the solutions of a quartic equation can be obtained in

a similar way as for the cubic equation by using Galois theory.

The coefficient of x3 can be eliminated by a suitable change of vari-

able. So a quartic equation can be written as

x4 + a2x2 − a3x+ a4 = 0. (D.1)

Let x1, x2, x3, x4 be the four roots of (D.1). Then

x1 + x2 + x3 + x4 = 0,

x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4 = a2,

x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 = a3,

x1x2x3x4 = a4.

Let F = C(a2, a3, a4), E = C(x1, x2, x3, x4). Then E/F is a Ga-

lois extension with G(E/F ) ∼= S4. Every element in S4 determines a

permutation of x1, x2, x3, x4.

Start as a first step with a composition series of the symmetric

group S4 :

e ⊂ K ⊂ H ⊂ A4 ⊂ S4,

183

184 APPENDIX D. SOLUTIONS OF QUARTIC EQUATIONS

in which K is the cyclic subgroup generated by (12)(34) and H =

e, (12)(34), (13)(24), (14)(23). The factor group A4/H is of order 3

and all other factors have order 2.

According to the fundamental theorem of Galois theory, this com-

position series determines a chain of intermediate fields

E ⊃ E1 ⊃ E2 ⊃ E3 ⊃ F.

Since [E : E1] = 2 and G(E/E1) = e, (12)(34), the minimal

polynomial of x1 over E1 is

(x− x1)(x− x2) = x2 − (x1 + x2) + x1x2. (D.2)

whose coefficients x1 + x2 and x1x2 belong to E1 but not to E2, for

they are changed into x3 + x4 and x3x4 respectively under the action

of (13)(24).

Since (13)(24) is a generator of G(E1/E2), the minimal polynomials

of x1 + x2 and x1x2 over E2 are

(x− x1 − x2)(x− x3 − x4) = x2 + (x1 + x2)(x3 + x4) (D.3)

and

(x− x1x2)(x− x3x4) = x2 − (x1x2 + x3x4)x+ a4 (D.4)

respectively.

We need to find (x1 +x2)(x3 +x4), x1x2 +x3x4 and their conjugates

(x1 + x3)(x2 + x4), (x1 + x4)(x2 + x3), x1x3 + x2x4, x1x4 + x2x3 over

F. In fact, it is enough to know x1x2 + x3x4, x1x3 + x2x4, x1x4 + x2x3,

because

(x1 + x2)(x3 + x4) + x1x2 + x3x4 = a2,

(x1 + x3)(x2 + x4) + x1x3 + x2x4 = a2,

(x1 + x4)(x2 + x3) + x1x4 + x2x3 = a2.

Since H S4 and S4/H ∼= S3, E2/F is a Galois extension of degree

6 with the Galois group S3. This coincides with the Galois extension of

185

a cubic equation. The cubic equation satisfied by the elements x1x2 +

x3x4, x1x3 + x2x4, x1x4 + x2x3 is

(x− (x1x2 + x3x4))(x− (x1x3 + x2x4))(x− (x1x4 + x2x3))

= x3 − a2x2 − 4a4x+ 4a2a4 − a2

3

= 0.

This cubic equation is called the resolvent cubic of the quartic

equation (D.1). Let λ1, λ2, λ3 be its three roots. By the symmetry of

the three elements x1x2+x3x4, x1x3+x2x4, x1x4+x2x3, we may assume

that

x1x2 + x3x4 = λ1,

x1x3 + x2x4 = λ2,

x1x4 + x2x3 = λ3.

(x1 + x2)(x3 + x4) = a2 − λ1,

(x1 + x3)(x2 + x4) = a2 − λ2,

(x1 + x4)(x2 + x3) = a2 − λ3.

By (D.3) the equation satisfied by x1 + x2 and x3 + x4 is x2 = λ1 − a2.

Hence

x1 + x2 =√λ1 − a2, x3 + x4 = −

√λ1 − a2.

For the same reason

x1 + x3 =√λ2 − a2, x2 + x4 = −

√λ2 − a2.

It is not hard to see that the 4 different combinations of the choices

of two square roots only affect the order of x1, x2, x3, x4. So these

two square roots can be chosen arbitrarily. But once√λ1 − a2 and

186 APPENDIX D. SOLUTIONS OF QUARTIC EQUATIONS

√λ2 − a2 are chosen, x1 + x4 and x2 + x3 are determined, because

(x1 + x2)(x1 + x3)(x1 + x4)

=1

8(x1 + x2 − x3 − x4)(x1 + x3 − x2 − x4)(x1 + x4 − x2 − x3)

= a3.

So the element x1 + x4 =√λ3 − a2 satisfies√

λ1 − a2

√λ2 − a2

√λ3 − a2 = a3.

It follows from

2x1 = (x1+x2)+(x1+x3)+(x1+x4) =√λ1 − a2+

√λ2 − a2+

√λ3 − a2

that

x1 =1

2(√λ1 − a2 +

√λ2 − a2 +

√λ3 − a2),

x2 =1

2(√λ1 − a2 −

√λ2 − a2 −

√λ3 − a2),

x3 =1

2(−√λ1 − a2 +

√λ2 − a2 −

√λ3 − a2),

x4 =1

2(−√λ1 − a2 −

√λ2 − a2 +

√λ3 − a2).

Appendix E

Hints or solutions for

exercises

Exercises 1.1

1. Check that the binary operation in G satisfies the three condition

for a group.

(1) Law of associativity: (a ∗ b) ∗ c = [aln(b)]ln(c) = aln(b)ln(c) and

a ∗ (b ∗ c) = aln[bln(c)] = aln(b)ln(c) imply (a ∗ b) ∗ c = a ∗ (b ∗ c).(2) Existence of the identity element: Let e be the base of the

natural logarithm. Then e ∗ a = eln(a) = a, a ∗ e = aln(e) = a for every

a ∈ G. Hence e is the identity element.

(3) Existence of the inverse: For any a ∈ G, let b = e1/ln(a). Then

a ∗ b = aln(b) = (eln(a))ln(b) = eln(a)ln(b) = e,

b ∗ a = bln(a) = (eln(b))ln(a) = eln(b)ln(a) = e.

Hence b is the inverse of a. 2

2. The composition of two strictly increasing continuous functions

f(x), g(x) with f(0) = g(0) = 0, f(1) = g(1) = 1 is still a function sat-

isfying all conditions required by the set A. Hence the binary operation

is well-defined.

The associativity holds in A, since the composition of functions

satisfies the associative law. Define the function e(x) = x, (0 ≤ x ≤ 1).

Then e ∈ A, and fe = ef for any f ∈ A. Hence e is the identity element

187

188 APPENDIX E. HINTS OR SOLUTIONS FOR EXERCISES

of A. Finally, the inverse function f−1(x) of any strictly increasing

function f(x) exists and f−1(x) is also strictly increasing. Hence f−1 ∈A. It is obvious that ff−1(x) = x = f−1f(x) for any x ∈ [0, 1]. Hence

ff−1 = f−1f = e. 2

3. It suffices to show that

(a) ae = a for any a ∈ G;

(b) If a, b ∈ G satisfy ba = e, then ab = e.

Assume that a, b ∈ G satisfy ba = e. The first condition implies

bab = eb = b. (E.1)

By the second condition there exists c ∈ G such that cb = e. Multiply-

ing both sides of (E.1) by c from the left yields cbab = cb. It follows

that ab = e, which proves (b).

For any a ∈ G, the second condition and (b) implies the existence

of b ∈ G such that ab = ba = e. Hence ae = a(ba) = (ab)a = ea = a,

which proves (a). 2

4. Assume that n = |G| <∞. For any a ∈ G, there are two elements

among a, a2, a3, . . . , an+1 that are equal by the pigeon hole principle.

Assume that ai = aj(i < j). Then aj−i = e. 2

5. Let n be the order of ab. Then (ab)n = 1. Multiplying both

sides of (ab)n = 1 by a−1 from the left and by a from the right yields

(ba)n = 1. Hence the order of ba does not exceed that of ab. For the

same reason the order of ab does not exceed that of ba. Hence they are

equal. 2

6. The given condition implies ab = (ab)−1 = b−1a−1 for any a, b ∈G. It follows from a−1 = a, b−1 = b that ab = ba. 2

Exercises 1.2

1. Let A,B ∈ G. Then

A =

1 a b

0 1 c

0 0 1

, B =

1 a′ b′

0 1 c′

0 0 1

.

189

So

AB =

1 a+ a′ b+ ac′ + b′

0 1 c+ c′

0 0 1

∈ G,

A−1 =

1 −a ac− b

0 1 −c0 0 1

∈ G.Hence G is a subgroup of GL2(R).

The center of G consists of all

C =

1 x y

0 1 z

0 0 1

such that 1 a b

0 1 c

0 0 1

1 x y

0 1 z

0 0 1

=

1 x y

0 1 z

0 0 1

1 a b

0 1 c

0 0 1

for any a, b, c ∈ R. This means that1 a+ x b+ az + y

0 1 c+ z

0 0 1

=

1 x+ a y + xc+ b

0 1 z + c

0 0 1

.Hence az = xc for any a, c ∈ R, which holds if and only if x = z = 0.

Hence the center of G consists of all real matrices in the form1 0 y

0 1 0

0 0 1

2. 1) Let a ∈ C(Y ). Then ab = ba for any b ∈ X, since X ⊆ Y. This

means a ∈ C(X). Hence C(X) ⊇ C(Y ).

2) Let a ∈ X. Since ab = ba for any b ∈ C(X) by the definition of

C(X), we have a ∈ C(C(X)). Hence X ⊆ C(C(X)).


3) It follows by 2) that C(X) ⊆ C(C(C(X))). Let Y = C(C(X)).

Then X ⊆ Y by 2). Thus C(X) ⊇ C(Y ) = C(C(C(X))) by 1). Hence

C(X) = C(C(C(X))). 2

3. Assume that a ∈ H, b ∈ G, b 6= 1. Than a ∈ 〈b〉. Hence ab = ba,

which implies that every element H is in the center of G. 2

4. Let t be a generator of the cyclic group G. Then a = tn, b = tm,

in which n,m are non-negative integers. Suppose that neither a nor b

is a perfect square. Then n,m are odd numbers. So n+m is even, i.e.,

n+m = 2r for some integer r. Hence ab is the square of tr.

Let Q∗ be the group of all nonzero rational numbers under the

multiplication. Then 2, 3, 2 · 3 = 6 are not perfect squares. 2

Remark: This example shows that Q∗ is not a cyclic group.

5. First proof: Let a be an arbitrary element in H. Since H is closed

under multiplication, every power of a is in H. Since G is a finite group,

some power of a is equal to the identity element 1. Hence 1 ∈ H.Assume that an = 1. Let b = an−1. Then ab = ba = 1. Hence H is

a subgroup of G.

Second proof: For any a ∈ H, construct a map f : H → H, b 7→ ab.

The cancelation law implies that f is injective. Since H is a finite

set, f is surjective too. Hence there exists c ∈ H such that f(c) = a,

i.e., ac = a, which implies that c = 1. There exists d ∈ H such that

f(d) = 1 for the same reason. So ad = 1, and d = a−1. Therefore H is

a subgroup of G. 2

6. For any P,Q ∈ G the equality (PQ)TA(PQ) = QT (P TAP )Q =

QTAQ = A holds. Hence PQ ∈ G. Since P TAP = A for any P ∈ G,

we have A = (P T )−1AP−1 = (P−1)TAP−1. Hence P−1 ∈ G. 2

Exercises 1.3

1. Hint: Every element of S7 which is not the identity element

can be expressed as the product of disjoint cycles. It is of one of the

following patterns:

(*******),(******),(*****),(****),(***),(**),

(*****)(**),(****)(**),(***)(**),(**)(**),

191

(****)(***),(***)(***),

(***)(**)(**), (**)(**)(**).

Answer1,2,3,4,5,6,7,10,12. 2

2. Hint: 20 = 4× 5, 18 = 3× 6 = 2× 9 = 2× 2× 3× 3.

Solution: The element (1234)(56789) is an element of order 20.

Write an element σ of order 18 as the product of disjoint cycles

σ = σ1 · · ·σd such that the length of σi is not smaller than that of σi+1

for i = 1, . . . , d− 1. The sum of the lengths of all factors is less than or

equal to 9. The length of each σi is a factor of 18. So the length m1 of

σ1 is one of 9, 6, 3, 2. If m1 = 9, then σ is a 9-cycle, whose order is not

equal to 18. If m1 = 6, then σ has at most one more factor, which is

either a 3-cycle or a 2-cycle. In this case the order of σ is 6, not equal

to 18. If m1 ≤ 3, then the order of σ does not exceed 6. Hence there is

no element of order 18 in S9. 2

3. Let A,B,C,D be the upper-left, upper-right, lower-left and

lower-right corners of the rectangle respectively. The permutations

id, (AB)(CD), (AC)(BD), (AD)(BC)

represent the identity map, left-right symmetry, top-bottom symmetry

and rotation by 180 degrees respectively. These elements are contained

in the symmetric group of the rectangle.

Since every isometry carries the long side to long side, it must carry

the ordered pair (A,B) to (A,B), (B,A), (C,D) or (D,C). It must be

one of the transformation as listed above. 2

4. Hint: Connect the centers of the six faces of the cube properly to

form an octagon located inside the cube. So the cubic and the regular

octagon have the same group of symmetry. 2

Exercises 1.4

1. Left cosets: H, (13), (123), (23), (132)Right cosets: H, (13), (132), (23), (123) 2

2. The statement 1) does not hold. A counter-example is G = S3

and H is the subgroup of the previous exercise with a = (23). Then

(G : H) = 3 but a3 /∈ H.


The key of the proof of 2) is to use pigeon hold principle. If H has

n left cosets and none of a, a2, . . . , an is in H, then at least two of them

are in the same left coset. Assume that as ∈ arH, where 1 ≤ r < s ≤ n.

Then as−r = (ar)−1as ∈ H, contradiction. 2

3. Let A be the set of all left cosets of H ∩K in K and let B be

the set of all left cosets of H in G. Construct a map f : A→ B which

carries a(H ∩ K) to aH. Need to show that this map is well-defined.

Assume that a(H∩K) = a′(H∩K). Then a−1a′ ∈ H∩K. So a−1a′ ∈ H,and thus aH = a′H. Hence f is well-defined.

Check that f is injective. Assume that f(a(H∩K)) = f(a′(H∩K)),

i.e., aH = a′H. Then a−1a′ ∈ H. Since a, a′ ∈ K, so a−1a′ ∈ K. Hence

a−1a′ ∈ H ∩K, which implies that a(H ∩K) = a′(H ∩K). Hence f is

injective. Since B is a finite set, so is A. 2

4. Choose an arbitrary a ∈ K. Let H = a−1K. Then H is a subset

of G containing the identity element. It suffices to show that H is a

subgroup of G.

For any h ∈ H define a map f : H → G, g 7→ hg. The image of f

is ha−1K. Since h = ha−1a ∈ ha−1K, so h ∈ a−1K ∩ ha−1K. By the

hypothesis of the problem ha−1K = a−1K = H. Hence f is a bijection

from H to H. This implies that H is closed under multiplication. Since

1 = a−1a ∈ H, the element 1 is in the image of f. Thus there exists

b ∈ H such that hb = 1. 2

5. Assume that a1H1 = a2H2, then H1 = a−11 a2H2. Hence H1 is

a left coset of H2 containing the identity element, which implies that

H1 = H2.

In S3 let H1 = id, (12), H2 = id, (13), a1 = (132), a2 = (132).

Then

a1H1 = (132), (23) = H2a2.

2

Exercises 1.5

1. Let Hii∈Λ be a collection of normal subgroups of a group G.

Let H = ∩i∈ΛHi. For any g ∈ G and any h ∈ H, since h ∈ Hi for every

i ∈ Λ, g−1hg ∈ Hi for every i ∈ Λ. Hence g−1hg ∈ H. This shows that

H G. 2

193

2. Hint: the conjugate of the product of two transpositions is still

the product of two transpositions. 2

3. Let g−1Hg be a conjugate subgroup of H. Then |g−1Hg| = r.

Since G has only one subgroup of order r, g−1Hg = H. Hence H G.

2

4. The key point is to use the normalizerNG(H) = g ∈ G|g−1Hg =

H. It contains H. Since (G : H) = 5 is a prime number, Lagrange’s

theorem implies NG(H) is either H or G. It suffices to exclude the

possibility NG(H) = H.

It follows from aH = Hb that b−1aH = b−1Hb. Hence b−1aH con-

tains the identity element. So b−1Hb = H, which implies b ∈ NG(H).

The condition Hb 6= H implies b /∈ H. 2

5. It is known in Exercise 2) thatH2 = e, (12)(34), (13)(24), (14)(23)is a normal subgroup of S4. Let H1 = e, (12)(34). Then H1 H2 but

H1 is not a normal subgroup of S4. 2

6. Let a, a′ ∈ H, b, b′ ∈ K. Then

aba′b′ = aba′b−1bb′.

Since HG, so ba′b−1 ∈ H. Hence aba′b−1 ∈ H, bb′ ∈ K, which implies

aba′b′ ∈ HK. Hence HK is closed under multiplication.

Let a ∈ H, b ∈ K. Then (ab)−1 = b−1a−1 = (b−1a−1b)b−1 ∈ HK.

Hence HK is a subgroup of G and 1) is proved.

For 2), since HK = e, (12), (13), (132) contains 4 elements, HK

is not a subgroup of S3 by Lagrange’s theorem.

Assume that H G,K G. Let a ∈ H, b ∈ K, g ∈ G. Then

g−1(ab)g = (g−1ag)(g−1bg) ∈ HK. Hence HK G. 2

7. Divide the proof into two steps: 1) amn = e; 2) If k > 0 and

ak = e, then mn|k.1) Since m is the order of an in G, so (an)m = e. Hence anm = e.

2) Since ak = ak = e, so n|k. Assume that k = nq. Then (an)q = e.

Since m is the order of an in G, so m|q. Therefore mn|k. 2

8. Hint: Let g be a generator ofN. It suffices to show that g(a−1b−1ab) =

(a−1b−1ab)g for any a, b ∈ G.


Since NG, there exist integers m,n such that aga−1 = gn, bgb−1 =

gm. 2

9. Hint: Otherwise |G/Z| < 4, which would imply that G/Z is a

cyclic group. This would imply that G is an abelian group. 2

10. Hint: Any element g ∈ G can be written as bib1 · · · bi−1bi+1 · · · bn.Hence gaig

−1 = biaib−1i for any ai ∈ Hi. 2

11. For any a ∈ K, since |aHa−1| = n, either aHa−1 = H or

aHa−1 = K. If aHa−1 = K, then H = a−1Ka = K, leading to contra-

diction. Hence aHa−1 = H. Since G is generated byH∪K, g−1Hg = H

for every g ∈ G. 2

Exercises 1.6

1. Every element of order 6 in S6 is either a 6-cycle or the product

of a 3-cycle and a transposition. So it is an odd permutation. Thus

G 6⊆ A6. Hence (G : G ∩ A6) ≥ 2. Since (G : G ∩ A6) ≤ (S6 : A6) by

Ex.1.4.3, it follows that (G : G ∩ A6) = 2.

2. By Exercise 1.5.2, G = e, (12)(34), (13)(24), (14)(23) is a nor-

mal subgroup of S4. Since every element of G is an even permutation,

G is also a normal subgroup of A4. 2

3. For any natural number n let

Hn = σ ∈ G|σ(i) = i for all i > n.

ThenH1 ⊆ H2 ⊆ H3 · · · , andG =⋃∞

n=1Hn. It is evident thatHn = An

is simple when n ≥ 5.

Assume that N G and N 6= G. There exists g ∈ Hm\N for some

m. Hence g ∈ Hn for every n > m. Since N∩HnHn, so N∩Hn = efor every n ≥ max5,m. Therefore

N =∞⋃

n=1

(N ∩Hn) = e.

This shows that G is a simple group. 2

4. Hint: |G/(G∩An)| ≤ 2 implies |G∩An| > 1. Then use G∩AnG.

2

195

5. Hint: The order of each 3-cycle has order 3 or 1 in the quotient

group Sn/G. 2

6. When n < 3, Sn is abelian, so [Sn, Sn] and its subgroup of

commutators are trivial.

For n ≥ 3, it is obvious that [Sn, Sn] is a normal subgroup of G

contained in An. Since An is a non-abelian simple group when n ≥ 5,

[Sn, Sn] and its subgroup of commutators are equal to An.

[S3, S3] = A3, [[S3, S3], [S3, S3]] = 1.

It remains to treat the case n = 4. Since (12)(13)(12)(13) = (123),

so [S4, S4] = A4. Since A4 has only one non-trivial normal subgroup

N = id, (12)(34), (13)(24), (14)(23), [A4, A4] = A4 or N.

Let σ, τ ∈ A4. Claim that σ−1τ−1στ ∈ N. Since A4 is generated by

3-cycles, we may assume that σ, τ are 3-cycles such that στ 6= τσ. We

may assume that σ = (ijk) and τ = (ijm) or (jim), in which i, j, k,m

are distinct. It follows from

(ijk)−1(ijm)−1(ijk)(ijm) = (ij)(km) ∈ N

and

(ijk)−1(jim)−1(ijk)(jim) = (ik)(jm) ∈ N

that [A4, A4] = N. 2

7. It is routine to check that conjugacy is an equivalence relation.

Express an element σ ∈ Sn as a product of disjoint cycles τ1, . . . , τr

whose lengths are m1, . . . ,mr respectively. Then σ−1 is still a product

of r disjoint cycles of lengths m1, . . . ,mr. Hence σ is conjugate to σ−1.

If (123) ∈ A4 is conjugate to (132) = (123)−1, then there exists

τ ∈ A4 such that τ(123)τ−1 = (132). However such a τ can only be

(12), (13) or (23). None of them belongs to A4. 2

8. Without loss of generality may assume γ = (123 · · ·n). Then

βγβ−1 = (β(1)β(2) · · · β(n)). Since βγ = γβ, the equality βγβ−1 = γ

holds. So (β(1)β(2) · · · β(n)) = (123 · · ·n), which implies that β(1) =

k, β(2) = k + 1, . . . , β(n− k + 1) = n, β(n− k) = 1, . . . , β(n) = k − 1.

Hence β = γk. 2


Exercises 1.7

1. [f(e1)]2 = f(e21) = f(e1) implies f(e1) = e2.

e2 = f(e1) = f(a−1a) = f(a−1)f(a) implies f(a)−1 = f(a−1).

Assume that h ∈ Ker(f), g ∈ G1. Then f(g−1hg) = f(g)−1f(h)f(g) =

f(g)−1f(g) = e2. Hence Ker(f) G1. 2

2. For any a ∈ G define the map

sa : 1, 2, . . . , n → 1, 2, . . . , n

by abi ∈ bsa(i)H. The key point is to show that sa is bijective. Assume

that sa(i) = sa(j). Then abi = bsa(i)hi, abj = bsa(j)hj for some hi, hj ∈H. Hence

bih−1i = a−1bsa(i) = a−1bsa(j) = bjh

−1j .

This means that bi, bj are in the same left coset. So i = j. Therefore sa

is bijective.

It remains to verify that τ(a′a) = τ(a′)τ(a) for any a′, a ∈ G. As-

sume that a′bi = bsa′ (i)h′i, with h′i ∈ H. Then

τ(a′)τ(a) =n∏

i=1

h′ihi.

It follows from

(a′a)bi = a′bsa(i)hi = bsa′ (sa(i))h′sa(i)hi

that

τ(a′a) =n∏

i=1

h′sa(i)hi.

Since sa is a bijection from the set 1, 2, . . . , n to itself and H is an

abelian group,n∏

i=1

h′sa(i)hi =n∏

i=1

h′ihi

holds, which implies τ(a′a) = τ(a′)τ(a). 2

3. The additive group C is abelian while GL2(R) is not. 2

197

4. The second isomorphism theorem implies H/(N ∩H) ∼= HN/N.

So (H : N ∩H) = (HN : N). Since H and N are subgroups between

N ∩H and HN, as illustrated in the following diagram,

HN

vvvvvvvvv

HHHHHHHHH

H

HHHHHHHHH N

vvvvvvvvv

H ∩N

we have

(HN : N ∩H) = (HN : H)(H : N ∩H) = (HN : N)(N : N ∩H).

Hence (HN : H) = (N : N ∩ H). Lagrange’s theorem implies (HN :

H)|(G : H), (N : N ∩ H)||N |. Since (G : H) is coprime with |N |, the

equality (HN : H) = (N : N ∩H) = 1 holds. Hence N ⊆ H. 2

5. Hint: It is routine to check that G is an abelian group.

Then check that the map f : C∗ → G, z 7→ 1− z is an isomorphism.

2

6. Since the composite of two maps satisfies the associative law,

so does the binary operation in Aut(G). Let σ ∈ Aut(G). Since σ is a

bijection, its inverse map τ exists, i.e., τ(σ(g)) = g and σ(τ(g)) = g

for any g ∈ G. In order to show that τ is the inverse of σ in Aut(G),

we need to verify that τ is a homomorphism. Let g, g′ ∈ G. Then

σ(τ(g)τ(g′)) = σ(τ(g))σ(τ(g′)) = gg′.

So τ(gg′) = τ(g)τ(g′), which implies that τ is a homomorphism. 2

7. Let σ ∈ Aut(G) be an inner automorphism, i.e., there exists

a ∈ G such that σ(g) = a−1ga for any g ∈ G. Let τ ∈ Aut(G). Then

τ−1στ(g) = τ−1(a−1τ(g)a) = τ−1(a)−1gτ−1(a)

for any g ∈ G. So τ−1στ ∈ Inn(G). This means that Inn(G) Aut(G).


2

8. Construct a map f : G → Inn(G) in the following way: Every

a ∈ G determines an inner automorphism g 7→ aga−1. Define this inner

automorphism to be f(a). It is obvious that f is surjective.

f(aa′)(g) = (aa′)g(aa′)−1 = a(a′ga′−1)a−1 = f(a)[f(a′)(g)]

holds for any a, a′ ∈ G. Hence f(aa′) = f(a)f(a′). So f is a surjective

homomorphism. Since

a ∈ Ker(f) ⇔ f(a) is the identity map ⇔ aga−1 = g∀g ∈ G⇔ a ∈ C(G),

Inn(G) ∼= G/C(G) holds by the fundamental theorem of homomor-

phism. 2

9. May assume that G = Z, which has exact two generators 1 and

−1. Since an automorphism of Z is totally determined by the image of

the generator 1, there are at most two automorphisms. Since n 7→ −nis a nontrivial automorphism, we have |Aut(G)| = 2 2

10. For every nonzero rational number r, define

f : Q → Q, a 7→ ra.

It is easy to see that f ∈ Aut(Q).

Assume that g ∈ Aut(Q). Then r = g(1) is a nonzero rational

number. Since g is a group homomorphism, the equality

g(n) = g(1 + 1 + · · ·+ 1) = g(1) + g(1) + · · ·+ g(1) = ng(1) = nr

holds for any natural number n. Moreover, g(n) = nr holds for any

n ∈ Z. Let n/m ∈ Q, in which m,n ∈ Z,m > 0. Then

rn = g(n) = g(mn

m

)= mg

( nm

).

Hence g( nm

) = r nm, which implies that g(a) = ra for any a ∈ Q. It

follows that there is one to one correspondence between Aut(Q) and

199

the set Q∗ of nonzero rational numbers. Furthermore it is easy to show

that Aut(Q) is isomorphic to the multiplicative group Q∗. 2

11. N ∩ H = e implies H ∼= H/(H ∩ N). The second isomor-

phism theorem implies H/(H ∩ N) ∼= NH/N = G/N. Hence every

complement subgroup of N is isomorphic to G/N. 2

12. Hint: First show that S = r ∈ Q|nr ∈ Z is a cyclic subgroup

of G of order n.

Next show that S is the unique subgroup of order n by using Propo-

sition 1.7.12.

13. Suppose that HG. Then G/H contains a nontrivial subgroup

which corresponds with a nontrivial subgroup of G containing H but

not equal to H, contradicting to the hypothesis that H is a maximal

subgroup of G. 2

Exercises 1.8

1. Hint: Write out four normal subgroups of G1 and show that G

is a simple group.

If G is abelian, then every subgroup of G1 is normal. In particular

the subgroup generated by any element (a, a) with a ∈ G is a normal

subgroup.

2. Assume that f is surjective. Then there exists a map (not

required to be a homomorphism) h : G→ G such that f h = 1. Here

1 stands for the identity map. Since

u f = u− u v u = g u,

the equality u = g u h holds. Then 1 = g + u v implies

1 = g + g u h v = g (1 + u h v),

Hence g is surjective.

By symmetry, the surjectivity of g implies that of f. 2

3. Let G = Z and H = 2Z. The G is not isomorphic to the direct

product of H and Z/2Z.


Assume that G is an infinite cyclic group with generator a with

some a ∈ G. Define a map f : H × G → G by f(h, an) = han. Let

(h′, an′) be another element of H× G. Then f(hh′, anan′) = hh′an+n′ =

f(h, an)f(h′, an′), for G is an infinite cyclic group. Hence f is a ho-

momorphism. Let (h, an) ∈ Ker(f). Then han = 1. Hence an ∈ H, so

an = 1, which implies n = 0, h = 1. Hence f is a monomorphism. Since

every element of G belongs to a coset of H f is surjective. 2

4. Hint: Every element of Ni commutes with every element of Nj for

i 6= j. In order to show that G is abelian, it suffices to show that every

Ni is abelian. Assume that a, b ∈ N1. Then b = uv, u ∈ N2, v ∈ N3.

Hence ab = auv = uva = ba.

G = N1×N2 implies N2∼= G/N1. For the same reason, N3

∼= G/N1.

5. i) Let g = (a, e) ∈ S ∩ H. Express any k ∈ G as k = th, (t ∈T, h ∈ H. Then t = (e, b). Hence k−1gk = h−1gh ∈ S ∩ H. Hence

S ∩H G. By symmetry, T ∩H G too.

ii) Construct a homomorphism f : H 7→ T, (a, b) 7→ b. Since G =

SH, f is surjective. The condition S ∩ H = e implies that f is

injective. Hence H ∼= T. For the same reason H ∼= S.

iii) Let s1, s2 ∈ S. Then there exists t1 ∈ T such that (s1, t1) ∈ H.

It follows from HG that (s−12 s1s2, t1) = (s2, e)

−1(s1, t1)(s2, e)−1 ∈ H.

Hence (s−11 s−1

2 s1s2, e) = (s1, t1)−1(s−1

2 s1s2, t1) ∈ S∩H. So s−11 s−1

2 s1s2 =

e and S is abelian. For the same reason , T is abelian too. 2

Exercises 1.10

1. ex = x ⇒ x ∼ x.

x ∼ y ⇒ gx = y ⇒ g−1y = x ⇒ y ∼ x.

x ∼ y, y ∼ z ⇒ gx = y, g′y = z ⇒ (g′g)x = z ⇒ x ∼ z

Let [x] be the equivalence class represented by x ∈ S. For any y ∈ [x]

there exists g ∈ G such that gx = y, so y ∈ Gx. Hence [x] ⊆ Gx.

Conversely z ∈ Gx implies z = gx for some g ∈ G. Hence Gx ⊆ [x].

Therefore [x] = Gx. 2

2. Since every conjugacy class C(a) is an orbit under the conjugacy

action of G on itself, the order of G is the product of the length of C(a)

and the order of the stabilizer of a. This proves 1).

201

2) Since C(1G) contains only one element 1, so all conjugacy classes

contain same number of elements if and only if G is abelian.

3) Since 1G is a conjugacy class, G − 1G must be the other

conjugacy class. Since (|G| − 1)||G| by the result of 1), |G| is equal to

2. 2

3. Two square matrices are conjugate if and only if they are similar.

If P−1AP = B for some P ∈ GL2(C) then there exists Q ∈ SL2(C)

such that Q−1AQ = B. In fact, Q can be taken to be

Q =

[√|P | 0

0√|P |

]P

where |P | stands for the determinant of P. Hence it suffices to enumer-

ate all Jordan canonical forms of determinant one. They are[a 0

0 a−1

],

where a is a nonzero complex number, and[1 1

0 1

]or [

−1 1

0 −1

].

4. Let T = (a1, a2)|a1, a2 ∈ S, a1 6= a2. Then |T | = n(n− 1). The

group acts on T by g(a1, a2) = (ga1, ga2).

Let (x1, y1), (x2, y2) be two distinct elements of T. Then there exists

g ∈ G such that g(x1, y1) = (x2, y2). Hence the action of G on T is

transitive. Therefore |T | divides |G|. 2

5. Suppose that G = Stab(x)Stab(y). Choose g ∈ G such that

g(y) = x. Assume that g = hk, h ∈ Stab(x), k ∈ Stab(y). Then h(y) =

hk(y) = g(y) = x, contradicting h(x) = x. 2

6. It is easy to see that S4 is an orbit.

Let (5i1 · · · in) be an n+1-cycle. If n > 1 then (i1 · · · in)−1(5i1 · · · in) =


(5in). In this case (5i1 · · · in) and the transposition (5in) are in the same

orbit.

Assume that 1 ≤ i < j ≤ 4. Then (ij)(5i)(ij) = (5j). Hence all

transpositions in the form (5i) are in the same orbit.

Let g, h ∈ S5\S4. Then g = (5i1 · · · in)g1, h = (5j1 · · · jm)h1, in

which g1, h1 ∈ S4. There exist p, q ∈ S4 such that p(5i1 · · · in)q−1 =

(5j1 . . . jm). Let r = g−11 q−1h1. Then pgr = h.

In summary, there are only two orbits S4 and S5\S4 with lengths

24 and 96 respectively. 2

Exercises 2.1

1. If e, e′ are unities, then e = ee′ = e′.

2) 0 ·a+0 ·a = (0+0) ·a = 0 ·a implies 0 ·a = 0. a · 0 = 0 is proved

in a similar way.

3) (−a) · b + ab = (−a + a)b = 0 · b = 0 implies (−a) · b = −(ab).

a · (−b) = −(ab) is proved in a similar way. 2

2. Let b, b′ be two inverses of an invertible element a. Then b =

b · 1 = b(ab′) = (ba)b′ = 1 · b′ = b′. 2

3. (u + a)∑n−1

i=0 ui(−a)n−1−i = un − (−a)n = un implies that

u−n∑n−1

i=1 ui(−a)n−1−i is the inverse of u+ a. 2

4. It is obvious that A is an abelian group under addition. The

equality

[(f ∗ g) ∗ h](n) =∑d|n

(f ∗ g)(d)h(nd

)=

∑d|n

∑e|d

f(e)g

(d

e

)h(nd

)=

∑rst=n

f(r)g(s)h(t)

= [f ∗ (g ∗ h)](n).

holds for any f, g, h ∈ A. Hence the multiplication of A is associative.

203

The equalities

[(f + g) ∗ h](n) =∑d|n

(f(d) + g(d))h(nd

)=

∑d|n

f(d)h(nd

)+∑d|n

g(d)h(nd

)= (f ∗ h)(n) + (g ∗ h)(n).

and f ∗ g = g ∗ f for all f, g ∈ A imply that A satisfies the law of

commutativity for multiplication and the law of distributivity.

Let

u(n) =

1, n = 1

0, n > 1.

Then (u∗f)(n) = f(n) for any f ∈ A, which implies that u is the unity

of A. Hence A is a commutative ring. 2

5. Let F,E be two fields. Then (1F , 0E)(0F , 1E) = (0F , 0E). Hence

F × E is not an integral domain, far from being a field. 2

6. 2x = (2x)2 = 4x2 = 4x for any x ∈ A. Hence 2x = 0. 2

Exercises 2.2

1. It is known from Chapter 1 that⋂

λ∈Λ Iλ is a subgroup of R

as an additive group. Let a ∈⋂

λ∈Λ Iλ, b ∈ R. Since Iλ is an ideal

for every λ ∈ Λ and a ∈ Iλ, so ab, ba ∈ Iλ for every λ ∈ Λ. Hence

ab, ba ∈⋂

λ∈Λ Iλ. This shows that⋂

λ∈Λ Iλ is an ideal.

Let I and J be ideals of R. Then I + J is an additive subgroup of

R. Since c(a+ b) = ca+ cb ∈ I + J, (a+ b)c = ac+ bc ∈ I + J for any

c ∈ R, a ∈ I, b ∈ J, I + J is an ideal of R.

Assume that∑n

i aibi ∈ IJ(ai ∈ I, bi ∈ J) and c ∈ R. Then cai ∈I, bic ∈ J for every i. Hence

c

(n∑i

aibi

)=

n∑i

(cai)bi ∈ IJ,

(n∑i

aibi

)c =

n∑i

ai(bic) ∈ IJ.


Hence IJ is an ideal of R.

2. Hint: The three elements[0 1

0 0

],

[1 0

0 0

],

[0 0

0 1

]

generate three nontrivial principal ideals

I1 =

[0 b

0 0

]|b ∈ R

,

I2 =

[a b

0 0

]|a, b ∈ R

,

I3 =

[0 b

0 c

]|b, c ∈ R

respectively.

Let I be a nonzero ideal. Then I contains a nonzero element[a b

0 c

].

By discussing the cases a 6= 0, c 6= 0 and a = c = 0 one can see that I

is one of I1, I2, I3 and A. Hence there are five ideals.

2

3. Hint: Let I be a nonzero ideal. Need to show I = Mn×n(F ). Let

A = (aij) be a nonzero element in I. Then aij 6= 0 for some i, j. By

multiplication of suitable elementary matrices from left and(or) from

the right one can show that I contains all Ers, where Ers is the matrix

whose (r, s) entry is 1 and all other entries are zero. 2

4. Assume that k = pns, in which n > 1 and (p, s) = 1. Let a = ps.

Then a 6= 0 and an = 0.

Assume that k = p1 · · · pm where p1, . . . , pm are distinct prime num-

bers. Then an = 0 (with a ∈ Z) implies pi|an for all i. Then pi|a for all

i and so a = 0. 2

5. Since Ii + Ij = A for any i 6= j, there exist aj ∈ Ii, cj ∈ Ij such

205

that aj + cj = 1. Let bi =∏

j 6=i cj. Then

bi ≡ 0 (mod Ij)

for all j 6= i and

bi =∏j 6=i

(1− aj) ≡ 1 (mod Ii).

Let b1, . . . , bn be the elements satisfying the condition in 1). Then

b = a1b1 + · · ·+ anbn satisfies the required conditions. 2

6. Let a, b ∈ ∪∞i=0Ii. There exists n > 0 such that a, b ∈ In. Hence

a+ b ∈ In ⊆ ∪∞i=0Ii. Moreover ac, ca ∈ In ⊆ ∪∞i=0Ii for any c ∈ In. 2

7. The greatest common divisor 6 of 12, 48, 30. 2

8. Since I is a left ideal, a(bc− cb) ∈ I for any a, b, c ∈ R. It follows

from

(ab− ba)c = a(bc− cb) + (ac)b− b(ac) ∈ I

that I is an ideal. 2

Exercises 2.3

1. Let b ∈ Ker(f). Then b ∈ I, i.e., b = (ax− 1)g(x) for some

g(x) = cmxm + cm−1x

m−1 + · · ·+ c1x+ c0

with ci ∈ R. Hence

acmxm+1 + (acm−1 − cm)xm + · · ·+ (ac0 − c1)− c0 = b

holds. By compare the coefficients of x of both sides we obtain

acm = 0,

cm = acm−1,

· · ·

c1 = ac0,


b = −c0.

Therefore

am+1b = −am+1c0 = −amc1 = · · · = −a2cm−1 = −acm = 0.

Conversely assume that anb = 0. Then

−(ax− 1)(an−1xn−1 + an−2xn−2 + · · ·+ ax+ 1)b = b.

Hence b ∈ Ker(f). 2

2. Let a, b ∈ R. Then

f(a+b) = (a+b+I, a+b+J) = (a+I, a+J)+(b+I, b+J) = f(a)+f(b),

f(ab) = (ab+ I, ab+ J) = (a+ I, a+ J)(b+ I, b+ J) = f(a)f(b).

Hence f is a homomorphism.

Assume that a ∈ I ∩ J. Then a+ I = I, a+ J = J, so a ∈ Ker(f).

Conversely a ∈ Ker(f) implies a ∈ I, a ∈ J, which implies a ∈ I ∩ J.Hence Ker(f) = I ∩ J.

Assume that I+J = R. There exist u ∈ I, v ∈ J such that u+v = 1.

The equality

f(bu+av) = (bu+av+I, bu+av+J) = (av+I, bu+J) = (a+I, b+J)

holds for any a, b ∈ R. Hence f is an epimorphism. Conversely assume

that f is surjective. Then there exists a ∈ R such that f(a) = (1+I, J),

i.e., a ∈ J, 1 − a ∈ I. Hence 1 = (1 − a) + a ∈ I + J, which implies

I + J = R. 2

3. Assume that f(x, y, z) ∈ Ker(φ). There exists q(x, y, z) ∈F [x, y, z] and r(x, y) ∈ F [x, y] such that f(x, y, z) = q(x, y, z)(z −x2) + r(x, y). There also exist q1(x, y) ∈ F [x, y] and u(x), v(x) ∈ F [x]

such that r(x, y) = q1(x, y)(y2 − x3) + u(x)y + v(x). It follows from

f(t2, t3, t4) = 0 that u(t2)t3 + v(t2) = 0. Since u(t2)t3 has only terms

of odd degrees and v(t2) has terms of even degrees, u(x) = v(x) = 0.

Hence f is contained in the ideal generated by y2 − x3 and z − x2. 2

207

Exercises 2.4

1. 0 is a maximal ideal of the simple ring M2(R) but M2(R) is not

a field. 2

2. Let I = R × 0, J = 0 × R. Then I, J are ideals of R × R. Since

R× R/I ∼= R× R/J ∼= R, both ideals I and J are maximal ideals.

Let K be a maximal ideal of R×R. Since 0 is not a maximal ideal,

K contains a nonzero element (a, b). If a 6= 0, b 6= 0, then K = R× R,contradicting the assumption that K is a maximal ideal. Hence either

a = 0 or b = 0. It follows that K is either I or J. 2

3. Let p(x, y) = x + y2, q(x, y) = y + x2 + 2xy2 + y4 and let I

be the ideal generated by p(x, y) and q(x, y). Every element f(x, y) in

I can be written as f(x, y) = u(x, y)p(x, y) + v(x, y)q(x, y) for some

u(x, y), v(x, y). Hence f(0, 0) = 0 for every f(x, y) ∈ I. Hence 1 /∈ I

and I is a proper ideal.

Since

y = q(x, y)− p(x, y)2 ∈ I, x = p(x, y)− y2 ∈ I

and x, y generate a maximal ideal, so I is a maximal ideal. 2

4. The characteristic of k cannot be zero, otherwise (1+1)n = 2n 6=2 = 1n + 1n. So the characteristic of k is a prime number p. Write n as

n = pem, in which m is not divisible by p. Suppose that m > 1. Let

f(x) = (x+ 1)n − xn − 1 = (xpe

+ 1)m − xn − 1 = mxpe(m−1) + g(x),

in which deg(g) < pe(m − 1). Since the degree of f(x) is pe(m − 1)

and k is an infinite field, there exists a ∈ k such that f(a) 6= 0, which

implies (a+1)n 6= an +1n, contradicting the hypothesis of the problem.

2

5. 1) [a −bb a

]7→ a+ bi.

2) Any P ∈ GL2(R) induces an automorphism M 7→ P−1MP of the

ring M2(R). It suffices to find some P ∈ GL2(R) such that P−1AP ∈ S.


We may try

P =

[1 y

0 x

].

Then

P−1AP =

[1 − y

x

0 1x

][0 3

−4 1

][1 y

0 x

]=

[4yx

3x− y + 4y2

x

− 4x

1− 4yx

].

The condition P−1AP ∈ S yields the system of equations

4y

x= 1− 4y

x,

3x− y +4y2

x=

4

x.

One solution of this system is x = 8/√

47, y = 1/√

47.3) is a consequence of the fundamental theorem of algebra. 2

6. 1) Let P be a prime ideal. Assume that (a + P )(b + P ) = P.

Then ab ∈ P, so a ∈ P or b ∈ P, which implies that either a + P or

b+ P is zero in A/P. Hence A/P is an integral domain.

Conversely assume that A/P is an integral domain. Assume that

a /∈ P, b /∈ P. Then a + P, b + P are nonzero elements in the quotient

ring A/P. Hence (a+ P )(b+ P ) 6= P. This implies that ab /∈ P. Hence

P is a prime ideal.

2) Let I be a maximal ideal of A. Then A/I is a field, which is an

integral domain. It follows from 1) that I is a prime ideal. 2

Exercises 3.1

1. 1) existence: If deg(f) < deg(g), then q = 0, r = f will work.

Assume that deg(f) ≥ deg(g). Apply induction on deg(f). Let axn

and bxm be the leading terms of f and g respectively. Let h = f −ab−1xn−mg. Then deg(h) < deg(f). By induction hypothesis there exist

q1, r ∈ A[x] such that h = q1g + r, deg(r) < deg(g). Let q = q1 +

ab−1xn−m. Then f = qg + r.

2) uniqueness: Assume that q1, q2, r1, r2 ∈ A[x] satisfy f = q1g +

r1 = q2g + r2, where deg(r1) < deg(g), deg(r2) < deg(g). Then (q1 −

209

q2)g = r2 − r1. Hence deg(r2 − r1) = deg(q1 − q2) + deg(g). It follows

from deg(r2 − r1) < deg(g) that r2 − r1 = q1 − q2 = 0. 2

2.

f(x) = 2x2g(x) + x3 + 2x2 + 2x+ 1

= 2x2g(x) + xg(x) + x2 + 1

= 2x2g(x) + xg(x) + g(x) + 2x+ 2.

Hence f(x) = (2x2 + x+ 1)g(x) + 2x+ 2. 2

3. The necessity is the consequence of Theorem 3.2.10. Assume

that f(x)u(x) + v(x)g(x) = 1. If h(x) is a common divisor of f(x) and

g(x), then h(x) divides f(x)u(x)+v(x)g(x), i.e., h(x)|1. So deg(h) = 0.

Hence f(x) and g(x) are coprime. 2

4. 1) Assume that b/a = s + ti, in which s, t ∈ Q. There exist

m,n ∈ Z such that |s−m| ≤ 1/2, |t− n| ≤ 1/2. From

|b/a− (m+ in)| = |(s−m) + i(t− n)| =√

(s−m)2 + (t− n)2 < 1

it follows that N(b− a(m+ in)) < N(a).

2) Let I be an ideal of Z[i]. Since the ideal 0 is a principal ideal, we

need to consider nonzero ideals only, i.e., we may assume that I 6= 0.Choose a nonzero element g in I\0 such that N(g) reaches the

minimum value. Then N(g) > 0. For any f ∈ I, there exist q, r ∈ Z[i]

such that f = qg + r and N(r) < N(g) by the result of 1). Since

r = f − qg ∈ I, r can only be 0 by the minimality of N(g). Hence

f ∈ qg, which implies that f is in the ideal generated by g. Therefore

I is the principal ideal generated by g.

3)

18 + i = (11 + 7i) + (7− 6i),

11 + 7i = i(7− 6i) + 5,

7− 6i = (1− i)5 + (2− i),

5 = (2− i)(2 + i).

Hence gcd(11 + 7i, 18 + i) = 2− i.


Another method: N(18 + i) = 325 and N(11 + 7i) = 170 imply

that gcd(N(18 + i), N(11 + 7i)) = 5. If h = gcd(11 + 7i, 18 + i), then

N(h) = 1 or 5. If 18 + i and 11 + 7i are not coprime, then h is one of

2 + i, 2− i,−2 + i,−2− i, 1 + 2i, 1− 2i,−1 + 2i,−1− 2i. By checking

these eight possible solution one obtains h = 2 − i,−2 + i, 1 + 2i or

1− 2i. 2

5. Let I be the ideal generated by 2 and x. It consists of all elements

in Z[x] whose constant term is even. So I is a nontrivial ideal. Suppose

that I is a principal ideal generated by f(x) ∈ Z[x]. Then there are

g(x), h(x) ∈ Z[x] such that 2 = f(x)g(x), x = f(x)h(x). The equality

2 = f(x)g(x) implies that f(x) is equal to 2 or −2, contradicting the

equality x = f(x)h(x). 2

6. Construct an epimorphism φ : Q[x] → Q×Q, f 7→ (f(2), f(−2)).

Then Ker(φ) = (x2 − 4). Hence Q[x]/(x2 − 4) ∼= Q×Q. 2

Exercises 3.4

1.

x9 − x = x(x− 1)(x+ 1)(x6 + x4 + x2 + 1)

= x(x− 1)(x+ 1)(x2 + 1)(x4 + 1).

Let g(x) = x2 + 1. Since g(1) = g(2) = 2 6= 0, g(x) is irreducible.

Let f(x) = x4 + 1. Since f(1) = f(2) 6= 0, f(x) has not linear factors.

Suppose that f(x) = (x2 + ax+ b)(x2 + cx+ d). Then

bd = 1,

a+ c = 0,

b+ d+ ac = 0,

ad+ bc = 0.

The equalities a + c = 0 and ad + bc = 0 imply c(b− d) = 0. If c = 0,

then b = −d, and b2 = −1 due to bd = 1. But there is no element b ∈ F3

such that b2 = −1. Hence c 6= 0, b = d, a = −c. Since b+ d+ ac = 0, so

211

2b− 1 = 0, which implies b = 2. It follows that

f(x) = (x2 + x+ 2)(x2 + 2x+ 2).

Therefore

x9 − x = x(x− 1)(x+ 1)(x2 + 1)(x2 + x+ 2)(x2 + 2x+ 2)

is the irreducible decomposition of x9 − x. 2

2. First proof: let x3 + αx+ 1 = (x+ a)(x2 + bx+ c). Then

a+ b = 0, c+ ab = α, ac = 1.

Hence

1− a3 = aα.

It is obvious that a 6= 0. Since F ∗ = F − 0 is a group of order 3

under the multiplication, a3 = 1 for any a ∈ F ∗ by Lagrange’s theorem.

Hence 1− a3 = aα does not hold.

Second proof: since the characteristic of F is equal to 2, the equality

a+a = 0 holds for any a ∈ F. Denote f(x) = x3 +αx+1. Then f(0) =

1 6= 0. Assume that b ∈ F\0. Then b3 = 1. Hence f(b) = αb 6= 0. So

f(a) 6= 0 for any a ∈ F. 2

3. 1) Let m+ n√

5i,m′ + n′√

5i ∈ A. Then

(m+ n√

5i) + (m′ + n′√

5i) = (m+m′) + (n+ n′)√

5i ∈ A,

(m+ n√

5i)(m′ + n′√

5i) = (mm′ − 5nn′) + (mn′ + nm′)√

5i ∈ A.

Hence A is a subring of C.2) Assume that 3 = (m+ n

√5i)(m′ + n′

√5i). Then

9 = (m2 + 5n2)(m′2 + 5n′2).

It follows that m2+5n2 = 3,m′2+5n′2 = 3 or m2+5n2 = 1,m′2+5n′2 =

9 or m2 + 5n2 = 9,m′2 + 5n′2 = 1. It is easy to see that m2 + 5n2 =

3 has no integral solution. The integral solutions of m2 + 5n2 = 9


are m = ±3, n = 0 and the integral solutions of m′2 + 5n′2 = 1 are

m′ = ±1, n′ = 0. Hence one of m + n√

5i and m′ + n′√

5i is equal to

±1. Hence 3 is irreducible. It can be shown by the same method that

2 +√

5i and 2−√

5i are irreducible too. 2

3) This is because (2 +√

5i)/3 /∈ A.4) Because 9 = 3 · 3 = (2 +

√5i)(2−

√5i). 2

Exercises 3.5

1. Let h(x) = f(x)−g(x). If f(a) = g(a) for all a ∈ F, the h(a) = 0

for all a ∈ F. If h(x) 6= 0, then it has only finitely many zeros in F.

Hence h(x) = 0. 2

2. First proof: direct verification.

Let f(x), g(x) ∈ I. Then f(2) = f ′(2) = f ′′(2) = g(2) = g′(2) =

g′′(2) = 0, Hence f(2) + g(2) = (f + g)′(2) = (f + g)′′(2) = 0. So

f(x) + g(x) ∈ I. For any h(x) ∈ R[x], Since

(h(x)f(x))′ = h′(x)f(x) + h(x)f ′(x),

(h(x)f(x))′′ = h′′(x)f(x) + 2h′(x)f ′(x) + h(x)f ′′(x),

the values of the polynomials h(x)f(x), (h(x)f(x))′, (h(x)f(x))′′ at 2

are all equal to zero. Hence h(x)f(x) ∈ I. Therefore I is an ideal.

Second proof: Since the characteristic of R is zero, f(x) ∈ I if and

only if (x−2)3|f(x). Hence I is the principal ideal generated by (x−2)3.

J is not an ideal, since f(x) = (x− 3)2− 1 ∈ I, while xf(x) /∈ J. 2

3. Let f(x) ∈ Fq[x]. Then f(x) = q(x)(xq − x) + r(x), in which

deg(r) < r. Since aq − a = 0 for all a ∈ Fq, f(x) and r(x) represent

the same function on Fq. Hence every polynomial function on Fq is

represented by a polynomial of degree less than q. By Corollary 3.5.4

any two distinct polynomials of degrees less than q represent distinct

functions. It is obvious that qq is the total number of the maps from

Fq to Fq, as well as the number of polynomials of degree less than q. 2

Exercises 4.1

1. Assume that a, b, c ∈ R such that ax3 + b sin(x) + c cos(x) = 0.

213

By substitution x = 0 we obtain c = 0. Hence ax3 + b sin(x) = 0. Let

x = π and a = 0 is obtained. Hence b = 0. 2

2. Let n = dim(V ). Then f is represented by an n × n matrix A

over F under a basis of V. Then f is injective ⇔ the rank of A is equal

to n and f is surjective ⇔ the rank of A is equal to n. Hence f is

injective ⇔ f is surjective. 2

3. For any nonzero element a of R, define a map f : R→ R, x 7→ ax.

Since f is a linear transform and since R is an integral domain, f is

injective. Hence f is surjctive by the previous exercise. Hence there

exists b ∈ R such that f(b) = 1, which means that ab = 1. 2

Exercises 4.2

1. First proof: Every element f ∈ Aut(G), can be represented by

an n × n invertible matrix A over Fp. If Ap2= I, then (A − I)p2

= 0,

so all eigenvalues of A− I are equal to zero, which implies that A− I

is similar to an upper triangular matrix with zeros on the diagonal.

Hence (A− I)p = 0, i.e., Ap = I. Therefore there is no element of order

p2 in Aut(G).

Second proof: Since

|Aut(G)| = |GL2(Fp)| = (p2 − 1)(p2 − p)

is not divisible by p2, there is no element in Aut(G) of degree p2 by

Lagrange’s theorem. 2

Exercises 5.1

1. Let Γ be the collection of all subsets of 1, 2, . . . , n containing

two elements. Then G acts on Γ by gi, j = g(i), g(j). Since m is

odd, the number of elements(2m2

)= m(2m− 1) of Γ is also odd.

Since |G| = 2r, the length of every orbit is a power of 2. Since |Γ|is odd, there is at least one orbit whose length is equal to 1. 2

2. 1) is obvious.

2) Let G be an arbitrary finite group. For any a ∈ G, let f(a) ∈S(G) be the map carrying g ∈ G to ag. Then f determines a map

from G to S(G). It is easy to verify the f is a monomorphism. Hence


G is a subgroup of S(G). It follows from 1) that G is isomorphic to a

subgroup of Sn. 2

3. The group G acts on the set S = 1, 2, . . . , n in a natural way.

Let i1, . . . , ir be the complete system of representatives of the orbits.

Then

n =r∑

j=1

|G||Stab(ij)|

.

Since |G| is a power of 2, as long as Stab(ij) 6= G, the number |G||Stab(ij)|

is even. Since n is odd, Stab(ij) = G for some j. 2

Exercises 5.2

1. Since

|GLn(Fp)| = (pn − 1)(pn − p)(pn − p2) · · · (pn − pn−1),

|U | = pn(n−1)/2,

the index (GLn(Fp) : U) = (pn− 1)(pn−1− 1) · · · (p− 1) is not divisible

by p. Hence U is a Sylow p-subgroup of GLn(Fp).

Let Γ be the collection of all Sylow p-subgroups of GLn(Fp). By

Sylow’s theorem the action of GLn(Fp) on Γ is transitive. Hence |Γ| =(GLn(Fp) : Stab(U)). Since Stab(U) consists of all nonsingular upper

triangular matrices (see the proof below), so

|Stab(U)| = (p− 1)n|U |.

Therefore

|Γ| = (p+ 1)(p2 + p+ 1) · · · (pn−1 + pn−2 + · · ·+ p+ 1).

Finally show that Stab(U) is the set of all nonsingular upper trian-

gular matrices. It is obvious that every nonsingular upper triangular

matrix is in Stab(U). Assume that P = (pij)1≤i,j≤n ∈ Stab(U) is not

upper triangular, then there exist 1 ≤ s < r ≤ n such that prs 6= 0. Let

s be the minimal index such that prs 6= 0 for some s < r. Then pij = 0

for all 1 ≤ j < s, j < i ≤ n. Let M be a matrix whose diagonal entries

are equal to 1 and the (s, r)-entry is equal to 1 and the remaining en-

215

tries are zero. Then M ∈ U. Let N = P−1MP. Suppose that N ∈ U,

then the (s, s) entries of PN and MP are pss and pss +prs respectively.

This would contradict PN = MP. Hence N = P−1MP /∈ U, which

means that P /∈ Stab(U). 2

2. Suppose that H 6= G. Let n = |G|,m = |H|. Let S be the

collection of all subgroups ofG conjugate toH. ThenG acts transitively

on S by conjugation. Since H is contained in the stabilizer of H, the

inequality |Stab(H)| ≥ m holds. hence |S| ≤ n/m. Since H ∩gHg−1 6=∅ for every g ∈ G, the inequality |∪K∈SK| < m(n/m) = n holds. Hence

there exists g ∈ G− ∪K∈SK, which implies that g is not conjugate to

any element in H. 2

3. 1225 = 52 ·72. Let N be the number of Sylow 7-subgroups. Then

N ≡ 1 (mod 7) and N |52. Since the remainder of 5 and 25 divided

by 7 are not equal to 1, the equality N = 1 holds. Hence the Sylow

7-subgroup is normal.

The number r of Sylow 5-subgroups divides 72 and r ≡ 1 (mod 5).

Hence r = 1. Hence the Sylow 5-subgroup is normal.

Since the intersection of the Sylow 7-subgroup and the Sylow 5-

subgroup is 1, a group of order 1225 is isomorphic to the direct

product of its Sylow 7-subgroup and its Sylow 5-subgroup. Hence this

group is abelian, because every group whose order is the square of a

prime number is abelian by Example 5.1.5. 2

4. Let G be a group of order 640 = 275. Let n be the number of

its Sylow 2-subgroups. Then n|5. Hence n = 1 or 5. If n = 1 then the

Sylow 2-subgroup is nontrivial normal subgroup of G.

Assume that n = 5. Let P be a Sylow 2-subgroup. Let M be

the collection of all left cosets of P in G. Then |M | = (G : P ) = 5.

The group G acts on M by left multiplication. This action gives a

homomorphism φ : G → S5. It is obvious that |Im(φ)| > 1. Since

(G : Ker(φ)) = |Im(φ)| divides |S5| = 5! = 23 · 3 · 5, so |Ker(φ)| > 1.

Hence G contains a nontrivial normal subgroup Ker(φ). 2

5. 168 = 23 · 3 · 7. Let N be the number of Sylow 7-subgroups of G.

Since G is simple, N > 1. The condition N ≡ 1 (mod 7), N |48 implies

that N = 8. Hence |S| = 8.


Let H ′ be a Sylow 7-subgroup different from H. Then H 6⊆ NG(H ′),

because H ′ is the unique Sylow 7-subgroup of NG(H ′). Hence there

exists a ∈ H, such that aH ′a−1 6= H ′. It follows that H is the only

orbit of length one of S under the action of H. Since the length of every

other orbit is divisible by 7 and |S| = 8, there are only two orbits with

lengths 1 and 7 respectively. 2

6. |S4| = 24 = 233. S4 has three Sylow 2-subgroups. They are

P1 = id, (12)(34), (13)(24), (14)(23), (12), (34), (1423), (1324),

P2 = id, (12)(34), (13)(24), (14)(23), (13), (24), (1432), (1234),

P3 = id, (12)(34), (13)(24), (14)(23), (14), (23), (1342), (1243).

7. 1) Sylow’s theorem implies (G : N) = 50. Let H be a subgroup

between N and G. Then (H : N)|50. Assume that H 6= G. Then

(H : N) < 50. So (H : N) is one of 1, 2, 5, 10, 25. Since P is also

a Sylow 7-subgroup of H with NH(P ) = N, we have (H : N) ≡ 1

(mod 7), which implies (H : N) = 1. Hence N is a maximal subgroup

of G.

2) Let K = NG(Q). Then N ⊆ K. It follows from i) that K = N

or G. It suffices to show that K 6= N. Let R be a Sylow 5-subgroup of

G containing Q. Then (R : Q) > 1. By Proposition 5.1.4 Q is a proper

subgroup of NR(Q). Assume that |Q| = 5r, |NR(Q)| = 5r+s, s > 0.

Since Q is a sylow 5-subgroup of N, |N | = 5rt for some t not divisible

by 5. Hence |NR(Q)| does not divide |N |. So NR(Q) 6⊆ N, which implies

K 6= N. 2

8. Denote by A and B the sets of Sylow p-subgroups of G and G/H

respectively. Then G acts on A by conjugation. On the other hand,

for any g ∈ G,Q ∈ B, let gQ = gQg−1, in which g is the element in

G/H represented by g. This defines an action of G on B. By Sylow’s

theorem both actions are transitive.

Let π : G→ G/H denote the natural homomorphism. Pick Q ∈ Band a Sylow p-subgroup P of π−1(Q). Since (G : π−1(Q)) = (G/H :

π−1(Q)/H) = (G/H : Q) is not divisible by p, P is a Sylow p-subgroup

217

of G, i.e., P ∈ A. Since H G, PH is a subgroup of π−1(Q). Since P

is a Sylow p-subgroup of both π−1(Q) and PH, (π−1(Q) : PH) is not

divisible by p. On the other hand, (π−1(Q) : PH)|(π−1(Q) : H) = |Q|implies that (π−1(Q) : PH) is a power of p. Hence PH = π−1(Q), which

implies π(P ) = Q. So StabG(P ) ⊆ StabG(Q) and (G : StabG(Q))| (G :

StabG(P )). Hence the number of Sylow p-subgroups of G/H divides

that of G. 2

Exercises 5.3

1. Suppose that Q is generated by m1/n1, . . . ,mr/nr in which

m1, . . . ,mr, n1, . . . , nr ∈ Z. Then every natural number is equal to

a1m1/n1 + a2m2/n2 + · · ·+ armr/nr

for some a1, . . . , ar ∈ Z. This would imply that tn1 · · ·nr is an integer

for any rational number t, which is absurd. Hence Q is not finitely

generated. 2

2. An element [(n,m)] in A has finite order if and only if there exist

r ∈ N, t ∈ Z such tht rn = 6t, rm = 30t. This is equivalent to the

condition m = 5n. Hence the torsion subgroup of A is a cyclic group

of order 6 generated by [(1, 5)]. Since (1, 5) and (0, 1) generate Z2 and

[(0, 1)] is not an element of finite order, we have A ∼= Z⊕ (Z/6Z). 2

3. 80 = 245. Hence a abelian group of order 80 is the direct sum of

a cyclic group of order 5 and an abelian group of order 24. The answer

is

Z/24Z⊕ Z/5Z,

Z/23Z⊕ Z/2Z⊕ Z/5Z,

Z/22Z⊕ Z/22Z⊕ Z/5Z,

Z/22Z⊕ Z/2Z⊕ Z/2Z⊕ Z/5Z,

Z/2Z⊕ Z/2Z⊕ Z/2Z⊕ Z/2Z⊕ Z/5Z.

4. The Sylow 2-subgroup of A is the direct sum of a cyclic group of

order 8 and two cyclic groups of order 2. The Sylow 3-subgroup of A


is the direct sum of two cyclic groups of order 3. 2

5. Let A be the set of all (a1, a2) ∈ Q2 such that

[a1, a2]

[4 2

2 4

][b1

b2

]∈ Z ∀b1, b2 ∈ Z.

Since [4 2

2 4

]−1

=

[1/3 −1/6

−1/6 1/3

],

A is generated by u = (1/3,−1/6) and v = (−1/6, 1/3). The order of

the cosets A/Z2 represented by u and v are equal to 6. Since |A/Z2| =42 − 22 = 12, the quotient group A/Z2 is the direct sum of a cyclic

group of order six and a cyclic group of order 2. 2

6. Suppose that |G| has two distinct prime factors p and q. Let P be

a Sylow p-subgroup of G and let Q be a Sylow q-subgroup of G. Then

P 6⊆ Q and Q 6⊆ P hold. This would contradict the hypothesis of the

problem. Hence |G| = pn for some prime number p. Suppose that G is

not abelian. Then there exist x, y ∈ G, xy 6= yx. So x /∈ 〈y〉, y /∈ 〈x〉,contradiction. Hence G is abelian. If G were not cyclic, then G =

G1⊕G2 for some nontrivial abelian groups G1 and G2. Since inside G,

G1 does not contain G2 and G2 does not contain G1, a contradiction is

reached. Hence G is cyclic. 2

7. All finitely generated abelian groups of rank less than or equal

to 1. 2

Exercises 5.4

1. Apply induction on |G|. When |G| = 1 the proposition is obvi-

ously true. Assume that |G| = n > 1. Choose any maximal normal

subgroup G1 of G. That means that G1 G, G1 6= G, and there is no

other normal subgroups between G1 and G. This implies that G/G1 is

a simple group. Since |G1| < |G|, there exists a composition series

1 = Gr Gr−1 · · ·G2 G1

219

by induction hypothesis. Hence

1 = Gr Gr−1 · · ·G2 G1 G

is a composition series of G. 2

2. Let

H =

[1 a

0 1

] ∣∣∣a ∈ F .Then H G. Since G/H ∼= F ∗ × F ∗, both G/H and H are abelian

groups. Hence G is solvable. 2

Exercises 6.2

1. Since [Q(√

2) : Q] = 2, [Q(√

2,√

3) : Q(√

2)] = 2, the exten-

sion Q(√

2,√

3)/Q is finite. Hence every element in Q(√

2,√

3) is an

algebraic number. So√

2 +√

3 ∈ Q(√

2,√

3) is an algebraic number.

To find I it suffices to find the minimal polynomials of√

2 +√

3.

First contruct

g(x) = (x−√

2−√

3)(x−√

2 +√

3)

= (x−√

2)2 − 3

= x2 − 2√

2x− 1.

Next construct

f(x) = (x2 − 1− 2√

2)(x2 − 1 + 2√

2)

= (x2 − 1)2 − 8

= x4 − 2x2 − 7.

So I is the principal ideal generated by x4 − 2x2 − 7 2

2. Since x3−3x+4 is irreducible, it is coprime with x2 +x+1. Use

Euclidean algorithm to find a(x), b(x) ∈ Q[x] such that

(x3 − 3x+ 4)a(x) + (x2 + x+ 1)b(x) = 1.


We obtain

a(x) =3x

49+

8

49, b(x) = −3x2

49− 5x

49+

17

49.

Hence

(α2 + α+ 1)−1 = −3α2

49− 5α

49+

17

49.

2

3. Since β ∈ F (α), there exist a(x), b(x) ∈ F [x] such that

β =a(α)

b(α).

Let f(x) = a(x) − βb(x). Since β /∈ F, f(x) is a nonzero element in

F (β)[x]. It follows from f(α) = 0 that α is algebraic over F (β). 2

4. Since α2 ∈ F (α), the field F (α2) is an intermediate field of the

extension F (α)/F. For the polynomial f(x) = x2 − α2 ∈ F (α2)[x] the

equality f(α) = 0 holds. Hence [F (α) : F (α2)] ≤ 2. Since

[F (α) : F (α2)][F (α2) : F ] = [F (α) : F ] = 5,

so [F (α) : F (α2)] = 1. 2

5. Since both K and F [α] are intermediate fields of K[α]/F,

[K[α] : K][K : F ] = [K[α] : F [α]][F [α] : F ].

Since [F [α] : F ] = deg(p) and [K : F ] are coprime, so deg(p) divides

[K[α] : K]. Since p(α) = 0 implies [K[α] : K] ≤ deg(p), so deg(p) =

[K[α] : K]. Hence p(x) is irreducible over K. 2

Exercises 6.3

1. The kernel of the homomorphism λ : F [x] → E, g(x) 7→ g(α)

is a principal ideal generated by an irreducible polynomial. Since

f(x) ∈ Ker(λ) is irreducible, so f(x) generates Ker(λ). By the fun-

damental theorem of homomorphism λ induces an isomorphism λ :

F [x]/(f(x)) ∼= F (α).

For the same reason the homomorphism µ : F [x] → E, g(x) 7→ g(β)

221

induces an isomorphism µ : F [x]/(f(x)) ∼= F (β). The composite µλ−1

is an isomorphism from F [α] to F [β] keeping every element in F fixed.

2

2. Since

02 + 0 + 2 = 2 6= 0, 12 + 1 + 2 = 1 6= 0, 22 + 2 + 2 = 2 6= 0,

the polynomial x2 + x + 2 has no zeros in F3. Hence x2 + x + 2 is an

irreducible polynomial over F3. Note that this method works only for

polynomials of degree not exceeding 3.

There are nine elements in E :

0, 1, 2, α, α + 1, α+ 2, 2α, 2α+ 1, 2α+ 2.

Here α is a root of x2 + x+ 2 = 0.

The table of addition is

0 1 2 α α+1 α+2 2α 2α+1 2α+2

0 0 1 2 α α+1 α+2 2α 2α+1 2α+2

1 1 2 0 α+1 α+2 α 2α+1 2α+2 2α

2 2 0 1 α +2 α α+1 2α+2 2α 2α+1

α α α+1 α+2 2α 2α+1 2α+2 0 1 2

α+1 α+1 α+2 α 2α+1 2α+2 2α 1 2 0

α+2 α+2 α α+1 2α+2 2α 2α+1 2 0 1

2α 2α 2α+1 2α+2 0 1 2 α α+1 α+2

2α+1 2α+1 2α+2 2α 1 2 0 α+1 α+2 α

2α+2 2α+2 2α 2α+1 2 0 1 α+2 α α+1

The multiplication table (by using α2 = 2α+ 1) is


0 1 2 α α+1 α+2 2α 2α+1 2α+2

0 0 0 0 0 0 0 0 0 0

1 0 1 2 α α+1 α+2 2α 2α+1 2α+2

2 0 2 1 2α 2α+2 2α+1 α α+2 α+1

α 0 α 2α 2α+1 1 α+1 α+2 2α+2 2

α+1 0 α+1 2α+2 1 α+2 2α 2 α 2α+1

α+2 0 α+2 2α+1 α+1 2α 2 2α+2 1 α

2α 0 2α α α+2 2 2α+2 2α+1 α+1 1

2α+1 0 2α+1 α+2 2α+2 α 1 α+1 2 2α

2α+2 0 2α+2 α+1 2 2α+1 α 1 2α α+2

Exercises 6.4

1. Let Γ be the set of all ideals of A disjoint from S. Then Γ is a

partially ordered set under the inclusion relation. Since 0 /∈ S, the zero

ideal is a member of Γ, so Γ 6= ∅. Let ∆ be a totally ordered subset of

Γ. Let I = ∪J∈∆J. Then I ∩ S = ∅. Thus I is an upper bound of ∆.

By Zorn’s lemma, there exists a maximal element P in Γ.

Assume that a ∈ A\P, b ∈ A and ab ∈ P. Let J be the ideal

generated by P and a. Since P ⊆ J and P 6= J, the maximality of

P implies that J = A. So 1 ∈ J, which implies that 1 = ca + p with

c ∈ A and p ∈ P. Multiplying both sides by b yields b = cab + pb ∈ P.Therefore P is a prime ideal. 2

2. Suppose there is an isomorphism f : Q(x) → Q. Then f(a) = a

for all a ∈ Q. So f(x) is transcendental over Q, leading to contradiction.

2

3. Since x2 + 1 has no zero in F, the field F is not algebraically

closed. Hence F is not isomorphic to Q. 2

4. Since E/E and E/F are algebraic extensions, the extension E/F

is also algebraic. Since E is algebraically closed, E is the algebraic

closure of F by definition. 2

Exercises 6.5

1. 1) Let (x, y) ∈ C ∩D. Then x, y satisfy the system

x2 + y2 = 1,

223

(x− n)2 + y2 = n2.

Subtracting both sides yields x = 1/2n.

2) Double the length of the given line segment and then subdivide

the new line segment into six equal segments.

3) same method.

2. 1) Since (ζ + ζ)2 = ζ2 + 2 + ζ2, the equality

(ζ + ζ)2 + (ζ + ζ)− 1 = 0

holds. Hence

ζ + ζ =−1±

√5

2.

It follows from ζ + ζ > 0 that

ζ + ζ =−1 +

√5

2.

2) Let two points O,A be given. Constructing the regular pentagon

following the following steps:

• Draw a line passing O and A by a ruler.

• Construct a circle C with O as its center and OA as its radius.

• Use standard method to construct a line passing O and perpen-

dicular to OA. It meets the circle C at two points P, P ′.

• Construct a circle with A as it center and OA as its radius. This

circle meets the extension of the line segment OA at a new point Q.

• Connect the points P,Q by the ruler. The length of the line

segment PQ is equal to√

5.

• Construct a circle with Q as its center and OA as its radius. It

intersects the line segment PQ at a point R.

• Use standard method to construct the middle point S of the line

segment PR. Then the length of PS is equal to (√

5− 1)/2.

• Construct a circle with O as its center and PS as its radius. It

intersects the line segment OA at a point T.

• Use standard method to construct a line passing T and perpen-

dicular to OA. It intersects the circle C at two points U,U ′.


• The points A,U, U ′ are three adjacent vertices of a regular pen-

tagon. It is easy to construct the remaining two vertices. 2

Exercises 7.1

1. The number of monic polynomials of degrees 1, 2, 3 are 5, 25, 125

respectively. The number of reducible monic polynomials of degree two

is 5 + 10 = 15. Hence the number of irreducible monic polynomials of

degree two is 25− 15 = 10.

A reducible monic polynomial of degree three is either the product

of three linear monic polynomials or the product of a linear monic

polynomial and an irreducible quadratic monic polynomial. They can

be easily counted. So the number of irreducible monic polynomials of

degree three is 40. 2

2. Since the degree of an irreducible real polynomial is at most two,

the polynomial x4 + 1 is reducible over R. We have

x4 + 1 = (x− a)(x− a3)(x− a5)(x− a7)

in which

a =

√2

2(1 + i).

Hence x4 + 1 has two real factors (x− a)(x− a7) = x2 −√

2x+ 1 and

(x − a3)(x − a5) = x2 +√

2x + 1. Both factors are not in Q[x]. Hence

x4 + 1 is irreducible over Q.Since the characteristic of F16 is 2, the polynomial

x4 + 1 = (x+ 1)4

is reducible over F16. 2

3.

x4 − 4 = (x−√

2)(x+√

2)(x− i√

2)(x+ i√

2)

and

x3 − 2 = (x− 3√

2)(x− ζ3√

2)(x− ζ2 3√

2)

over C, in which

ζ = −1

2+

√3i

2.

225

x4 − 4 = (x−√

2)(x+√

2)(x2 + 2)

and

x3 − 2 = (x− 3√

2)(x2 +3√

2x+3√

4)

over R.

x4 − 4 = x4 − 1 = (x− 1)(x+ 1)(x2 + 1)

and

x3 − 2 = x3 + 1 = (x+ 1)3

over F3. 2

4. Let A be a finite integral domain. Every nonzero element a ∈ Adetermines an injective map

f : A→ A, x 7→ ax.

Since A is a finite set, the map f is surjective. Hence there exists

b ∈ A such that ab = 1. This means that every nonzero element of A

is invertible.

The commutative ring F2 × F2 is not a field, but it contains 4 ele-

ments.

2

Exercises 7.2

1. Let f(x) = xp − x− 1 and let α be a zero of f(x) in Fp. Then

αp = α+ 1.

Take p-th power of the both sides repeatedly and we obtain

αp2

= α+ 2,

αp3

= α+ 3,

· · ·

αpp−1

= α+ p− 1,


αpp

= α.

So α is a zero of the polynomial xpp − x. But α is not a zero of xpi − x

for any 1 ≤ i ≤ p− 1. This implies that [Fp[α] : Fp] = p. Hence f(x) is

an irreducible polynomial over Fp. 2

2. Let S = x2|x ∈ F. Since the equation x2 = t has at most two

solutions in F for any t ∈ F and it has a unique solution when t = 0,

so |S| > |F |/2. Let T = a−S = a−b|b ∈ S. Then |T | = |S| > |F |/2.By the pigeon hole principle, S ∩ T 6= ∅. This implies that there are

x, y ∈ F such that x2 = a− y2. 2

3. This is because

xp − x =∏a∈Fp

(x− a).

4. To decide whether there is a linear factor of f(x) = x4+x3+x+3

we need to check whether f(x) has a zero in F5. Since

f(0) = 3, f(1) = 1, f(2) = 4, f(3) = 4, f(4) = 2,

f(x) has no linear factor.

Suppose that x4 + x3 + x + 3 = (x2 + ax + b)(x2 + cx + d). By

comparing coefficients we obtain the relations

a+ c = 1,

b+ d+ ac = 0,

ad+ bc = 1,

bd = 3.

The equality c = 1 − a and d = 3/b can be used to eliminate c, d to

obtain

b+ 3/b = a2 − a,

3a/b+ b(1− a) = 1.

227

Let u(a) = a2 − a, v(b) = b+ 3/b. Then

u(0) = 0, u(1) = 0, u(2) = 2, u(3) = 1, u(4) = 2,

v(1) = 4, v(2) = 1, u(3) = 4, u(4) = 1.

Hence only a = 3, b = 2 or 4 satisfy the condition b+ 3/b = a2 − a.

Let g(a, b) = 3a/b+ b(1− a). Then

g(3, 2) = 2 + 1 = 3 6= 1, g(3, 4) = 0 6= 1.

Hence x4 + x3 + x + 3 is not a product of two quadratic polynomials.

Hence it is irreducible. 2

5. Assume that z 6= 0. Let x′ = x/z, y′ = y/z. Then y′q +y′ = x′q+1.

For any β ∈ Fq2 , let α1, . . . , αn be all zeros of the equation y′q+y′ = βq+1

in Fq2 . It is easy to check that this equation has no multiple zeros.

Hence n = q.

The equality

αq2

i + αqi = (αq

i + αi)q = βq2+q = βq+1 = αq

i + αi

implies αq2

i −αi = 0. So αi ∈ Fq2 . Hence the number of solutions of the

equation y′q + y′ = x′q+1 is equal to q3.

When z = 0 the solution is x = 0, z = 0 and y arbitrary. Therefore

the original equation has q3 + q2 solutions. 2

Exercises 8.1

1. Let σ ∈ Aut(F2(x)/F2(x2)). Then σ(x)2 = σ(x2) = x2. Hence

(σ(x)− x)2 = σ(x)2 − x2 = 0.

It follows that σ(x) = x, which implies that σ is an identity map. Hence

Aut(F2(x)/F2(x2)) contains only one element. However F2(x)/F2(x

2)

is a quadratic extension. Therefore, F2(x)/F2(x2) is not a Galois ex-

tension. 2

2. Since [F : Q] = 2 and [Q( 3√

2) : Q] = 3, the polynomial x3 − 2 is

irreducible over F. So the minimal polynomial of 3√

2 over F is x3 − 2.


It can be decomposed into

x3 − 2 = (x− 3√

2)(x− ζ3√

2)(x− ζ2 3√

2)

in F [ 3√

2]. Hence it is a Galois extension by Theorem 8.1.15. 2

3. 1) Let K be the fixed subfield of H1 ∩H2. Since (H1 ∩H2) ⊆ Hi

for i = 1, 2, so Ei ⊆ K for i = 1, 2. Hence E1E2 ⊆ K.

Since Ei ⊆ E1E2 for i = 1, 2, Aut(L/E1E2) ⊆ Aut(L/Ei) = Hi for

i = 1, 2. Hence Aut(L/E1E2) ⊆ H1 ∩H2.

It follows from Proposition 8.1.10 that L/E1E2 is a Galois exten-

sion. The relation Aut(L/E1E2) ⊆ H1 ∩H2 implies K ⊆ E1E2 by the

fundamental theorem of Galois theory.

2) Let K be the fixed field of H1H2. Since Hi ⊆ H1H2 for i = 1, 2,

so K ⊆ Ei for i = 1, 2. Hence K ⊆ E1 ∩ E2. Since E1 ∩ E2 ⊆ Ei for

i = 1, 2, so Hi = Aut(L/Ei) ⊆ Aut(L/E1 ∩ E2) for i = 1, 2. Hence

H1H2 ⊆ Aut(L/E1 ∩ E2). Therefore E1 ∩ E2 ⊆ K, for E1 ∩ E2 is the

fixed field of Aut(L/E1 ∩ E2). 2

4. Let F = Q, E = Q[√

2], K = Q[21/4]. Since [E : F ] = [K : E] =

2, the extensions K/E,E/F are Galois extensions. However K/F is

not a Galois extension. 2

5. ⇒: Assume that E/F is a Galois extension, then |Aut(E/F )| =[E : F ] by definition. By Proposition 8.1.4 the number of embed-

dings from E/F into L does not exceed [E : F ]. Since every element

in Aut(E/F ) is an embedding from E/F into L, the number of em-

beddings of E/F into L is exactly equal to [E : F ]. Hence E/F is a

separable extension. Since the number of embeddings from E/F into

L is equal to Aut(E/F ) and every such embedding carries E into E,

the extension E/F is normal.

⇐: Let σ1, . . . , σn be the set of all embeddings from E/F into L,

in which n = [E : F ]. Since E/F is normal, σ1, . . . , σn ∈ Aut(E/F ).

Hence [E : F ] ≤ |Aut(E/F )|. Therefore E/F is a Galois extension. 2

6. The necessity follows from Theorem 8.1.15. The sufficiency is

proved by induction on [E : F ].

If [E : F ] = 1 then there is nothing to be proved. Assume that

229

[E : F ] > 1. Then at least one αi is not in F. Let p(x) ∈ F [x] be its

minimal polynomial. Since p(x)|f(x), p(x) is decomposed into linear

factors in E[x]. Assume that p(x) = (x−α1) · · · (x−αr) without loss of

generality. Let K = F [α1, . . . , αr]. Theorem 8.1.15 implies that K/F is

a Galois extension and [E : K] < [E : F ]. By the induction hypothesis

E/K is a Galois extension.

Let f ∈ G(K/F ). Theorem 6.4.8 implies that there exists an em-

bedding g : E → E from E to the algebraic closure E of E such

that g|K = f. For any αj, the image σ(αj) is still a zero of f(x), so

σ(αj) ∈ F [α1, . . . , αn]. This implies g ∈ Aut(E/F ). Hence the homo-

morphism

φ : Aut(E/F ) → G(K/F ), g 7→ g|K

is surjective. Since Ker(φ) = G(E/K), we have

|Aut(E/F )| = |G(E/K)| · |G(K/F )| = [E : K][K : F ] = [E : F ].

Hence E/F is a Galois extension. 2

7. Since G(E/K) is a subgroup of G(E/F ) of order 2, it suffices

to show that G(E/K) is not a normal subgroup of G(E/F ). This is a

consequence of the following proposition.

If a finite group of order 2p, with p being prime, has a normal

subgroup of order 2, then G is abelian.

In fact, the both the Sylow p-subgroup and the Sylow 2-subgroup

are normal subgroups. Hence G is their direct product. Hence G is

abelian.

8. Since E/F [α] is a Galois extension, the equality [E : F [α]] =

|G(E/F [α])| holds. The given condition means that G(E/F [α]) is triv-

ial. Hence [E : F [α]] = 1. 2

9. Let K be the fixed field of Aut(E/F ). Lemma 8.1.11 implies

that E/K is a Galois Galois extension. Hence |Aut(E/F )| = [E : K]

divides [E : F ]. 2

Exercises 8.2

1. If F is a finite field, then every finite extension of F is a simple


extension, which implies that EH = F [β] for some β ∈ EH . It follows

from the fundamental theorem of Galois theory thatH = G(E/F [β]) =

g ∈ G(E/F )|g(β) = β.Assume that F is an infinite field. For any α ∈ EH , let Gα = g ∈

G(E/F )|g(α) = α. Then Gα is a subgroup of G(E/F ) containing H.

Choose β ∈ EH such that |Gβ| reaches the minimum. It suffices to

show that Gβ = H.

Assume that Gβ 6= H. Choose g ∈ Gβ\H. Then there exists α ∈ EH

such that g(α) 6= α.

Let h1, . . . , hr be all elements of G(E/F )\Gβ. Since F is an infinite

field, there exists a ∈ F such that

hi(aβ − α) 6= aβ − α

for all 1 ≤ i ≤ r. This means that Gaβ−α ⊆ Gβ. It follows from

g(aβ − α) 6= aβ − α

that Gaβ−α is a proper subgroup of Gβ, leading to contradiction. 2

2. Let ζ = e2πi/5. Then

x4 + x3 + x2 + x+ 1 = (x− ζ)(x− ζ2)(x− ζ3)(x− ζ4).

Hence Q[x]/(x4 + x3 + x2 + x+ 1) ∼= Q[ζ] is a quartic Galois extension

of Q whose Galois group is cyclic.

The cubic Galois extension is a little more complicated. The poly-

nomial f(x) = x3−12x+8 is a primitive polynomial (all coefficients are

integers with 1 as their greatest common divisor). If f(x) is reducible

in Q[x] then it can be decomposed into the product of two primitive

polynomials of lower degrees and so it has a zero in Z. Since every zero

of f(x) in Z divides 8, it can only be one of ±1,±2,±4,±8. It is easy

to check that none of these integers is a zero of f(x). Hence f(x) is

irreducible over Q. So E = Q[x]/(x3 − 12x + 8) is a cubic extension

of Q. In order to show that it is a Galois extension, we need to check

that f(x) has three distinct zeros in E. In fact, it is easy to check that

α, α2−82

and α2+82

are zeros of f(x).

231

3. Define the homomorphism

φj : F [y] → F [y], f(y) 7→ f(y + j).

for j = 0, 1, . . . , p− 1. They keep yp − y + x unchanged. So φj induces

an automorphism of E. It is obvious that φ0, . . . , φp−1 are p distinct

elements in Aut(E/F ).

It follows from [E : F ] ≤ p that |Aut(E/F )| = p = [E : F ].

Therefore E/F is a Galois extension of degree p. 2

Bibliography

[1] Apostol,T., Introduction to analytic number theory, Springer

Verlag, New York, Berlin, Heidelberg, 1976, ISBN 0-387-90163-9.

[2] Artin,M., Algebra, Prentice Hall, Englewood Cliffs, New Jersey,

1991, ISBN 0-13-004763-5.

[3] Bourbaki,N., Algebra, Chapters 1-3 (English translation),

Addison-Wesley, Reading, Massachusetts, 1974, ISBN 0-201-

00639-1.

[4] Bourbaki,N., Algebra, Chapters 4-7 (English translation),

Springer Verlag, New York, Berlin, Heidelberg, 1988, ISBN 0-387-

19375-8.

[5] Feng,K., Li,S., Za, J. and Zhang, P., Introduction to modern alge-

bra (in Chinese), Press of the University of Science and Technology

of China, Hofei, 2002ISBN 7-312-00041-X/O.47

[6] Lang,S., Algebra, Addison-Wesley, Reading, Massachusetts, 1971,

ISBN 0-201-04177-4.

[7] Li, K., Basic alstract algebra (in Chinese), Qinghua University

Press, Beijing, 2007ISBN 978-7-302-14407-6

[8] Moh. Z., Lan, Y. and Zhao, C., Algebra I and II (in Chinese),

Beijing University Press, Beijing, 1986ISBN 7-301-01371-x/O.222

[9] Serre, J-P., A course in arithmetic, Springer Verlag, Berlin, New

York, Heidelberg, 1973, ISBN 0-387-90041-1.

[10] Shafarevich,I., Algebra I, Springer Verlag, Berlin New York Hei-

derberg.

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BIBLIOGRAPHY 233

[11] de Sousa,P., Silver,J., Berkeley problems in mathematics, Springer

Verlag, 2004, ISBN 10: 0387008926.

[12] Vinberg,E., A course in algebra (translated from Russian) , Amer-

ican Mathematical Society, 2003, ISBN 0-8218-3318-9.

[13] Wan,Z., Algebra and Coding ,3rd ed.(in Chinese), Higher Educa-

tion Press, Beijing, 2007ISBN 978-7-04-021717-9

[14] Weil, A. Basic number theory, 3rd edition, Springer Verlag, New

York, Berlin, Heidelberg, 1988, ISBN 0-387-19375-8.

[15] Yao, M., Abstract algebra (in Chinese), Fudan University Press ,

Shanghai, 1998, ISBN 7-309-02096-0/O.183

Index

abelian group, 4, 7

algebraic closure, 131

algebraic element, 123

algebraic extension, 123, 125

algebraic number, 132

algebraic structure, 1

algebraically closed field, 131

alternating group, 29

associated elements, 78

automorphism, 33

binary operation, 1

Cardano’ formula , 181

Cayley’s theorem, 101

center, 8, 10, 26

centralizer, 8

characteristic, 66

Chinese Remainder Theorem, 60

commutative ring, 51

commutator subgroup, 26

composition series, 116

congruence, 57

conjugate subgroup, 25

coprime ideals, 59

coset, 18, 24

cubic equation, 181

cycle, 14

cyclic group, 9, 11, 21, 35

cyclotomic polynomial, 175

degree of a polynomial, 71

dihedral group, 16

direct product, 40

direct sum, 40

division algorithm, 72

division ring, 53

embedding, 151

endomorphism, 61

epimorphism, 33

equivalence class, 22

Euler function, 44, 174

Euler’s theorem, 44

even permutation, 14

factor group, 116

Fermat’s little theorem, 45

field, 53

field extension, 120

field of fractions, 65

field of rational functions, 77

finite fields, 142

finite group, 4

finitely generated extension, 122

finitely generated group, 9

finitely generated ideal, 59

First Isomorphism Theorem, 37

fixed subfield, 150

fundamental theorem of homomor-

phisms, 35

234

INDEX 235

Galois extension, 152

Galois group, 152

Galois theory, 148

Gaussian integer, 75

general linear group, 5

greatest common divisor, 74

group, 4

group action, 45

group of automorphisms, 39

group of symmetries, 16

Hilbert theorem 90, 181

homomorphism, 32, 60

ideal, 56

identity element, 2

image, 33

indeterminate, 72

index, 20

infinite group, 4

inner automorphism, 34

integral domain, 53

intermediate field, 120, 150

invariant subspace, 96

inverse, 3, 53

invertible element, 3

irreducible decomposition, 80

irreducible element, 79

isomorphism, 33, 61

kernel, 33, 61

Lagrange’s theorem, 21

law of associativity, 2

law of cancelation, 5

left coset, 18

left ideal, 57

left inverse, 53

left translation, 46

left zero-divisor, 53

maximal ideal, 66

minimal polynomial, 124

monic polynomial, 71

monoid, 3

monomial, 76

monomorphism, 33

multiple factor, 82

multiple zero, 88

multiplicity, 82

non-commutative ring, 51

normal extension, 161

normal subgroup, 23, 26

normalizer, 27

odd permutation, 14

orbit, 47

orbit formula, 99

order of a group, 4

partition, 20

permutation group, 12

polynomial, 71

polynomial function, 87

prime field, 67

prime ideal, 70

primitive element, 145

primitive polynomial, 83

principal ideal, 59

principal ideal domain, 74

projective general linear group, 25

quadratic nonresidue, 169

236 INDEX

quadratic residue, 169

quartic equation, 183

quaternion, 54

quotient, 72

quotient group, 23

quotient ring, 56

quotient space, 95

rank, 115

reciprocity law, 171

regular polyhedron, 18

remainder, 72

resolvent cubic, 185

right coset, 18

right ideal, 57

right inverse, 53

right zero-divisor, 53

ring, 51

Second Isomorphism Theorem, 37

semigroup, 3, 7

separable extension, 161

simple extension, 122

simple group, 27, 30

simple ring, 58

skew field, 53

solvable extension, 162

solvable group, 118

special linear group, 7

splitting field, 161

stabilizer, 47

subfield, 56

subgroup, 7

subnormal series, 116

subring, 55

subspace, 91

Sylow subgroup, 101

Sylow’s theorem, 102

symmetric group, 12

symmetric polynomial, 77

tetrahedron, 17

torsion, 115

transcendental element, 123

transcendental number, 132

transitive action, 48

transposition, 14

trivial ideal, 58

trivial subgroups, 7

unit, 53

unitary group, 49

unitary operation, 2

unity, 51

unque factorization domain, 79

vector space, 90

Wilson’s theorem, 146

zero-divisor, 53

Zorn’s lemma, 133

INDEX 237

Documents

Lectures on Modern Algebra - homepage.fudan.edu.cnhomepage.fudan.edu.cn/jgyang/files/2012/02/algebra.pdf · iii is indispensable. Due to its importance, modern algebra (or abstract