Abstract Algebra - LPU Distance Education (LPUDE)ebooks.lpude.in/.../year_1/DMTH403_ABSTRACT_ALGEBRA.pdfB have no common elements, and so A B = , the empty set. When the intersection

Abstract�AlgebraDMTH403

ABSTRACT ALGEBRA

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SYLLABUS

Abstract Algebra

Objectives:

To learn about the structure as group, ring and field.

To gain knowledge about homomorphisms, isomorphisms, cosets, quotient groups, and the isomorphism theorems,rings, ideals, ring homeomorphisms, isomorphisms and its theorems.

To learn about fields, quotient fields and field extensions Galois Theory also.

Sr. No. Content

1 Groups : Definition and examples, Quotient groups, Cyclic groups, Permutation

groups and The alternating groups, Subgroups, normal subgroups and the

commutator subgroup, Generating sets, Lagrange's Theorem and Cayley's

theorem

2 Homomorphisms and Automorphisms, Direct products. External and internal

direct products,

3 Structure of finite abelian groups, Conjugate elements and class equations of

finite groups, Sylow's theorems and their simple applications.

4 Solvable groups,Jordan-Holder Theorem, Rings, Subrings, Ideals and their

operations

5 Factor rings and Homomorphisms, Integral domains

6 Polynomial rings,The field of quotients Euclidean domains, Principal Ideal

Domains, Unique factorization domain

7 Prime fields, finite and algebraic extensions, Roots of a polynomial

8 splitting fields; existence and uniqueness, Separable extensions, Finite fields; the

structure, the existence of GF (pn)

9 Galois theory :Normal extensions, Galois groups

10 Symmetric functions, fundamental theorem, Constructible polygons, Solvability

by radicals

CONTENTS

Unit 1: Generating Sets 1

Unit 2: Groups 20

Unit 3: Subgroups 37

Unit 4: Lagrange's Theorem 50

Unit 5: Normal Subgroups 60

Unit 6: Group Isomorphism 70

Unit 7: Homomorphism Theorem 82

Unit 8: Permutation Groups 90

Unit 9: Direct Products 101

Unit 10: Finite Abelian Groups 112

Unit 11: Conjugate Elements 123

Unit 12: Sylow�s Theorems 128

Unit 13: Solvable Groups 134

Unit 14: Rings 140

Unit 15: Subrings 153

Unit 16: Ideals 158

Unit 17: Ring Homomorphisms 168

Unit 18: Integral Domains 180

Unit 19: The Field of Quotient Euclidean Domains 188

Unit 20: Principal Ideal Domains 196

Unit 21: Unique Factorization Domains 207

Unit 22: Polynomial Rings 213

Unit 23: Division of Algorithm 222

Unit 24: Irreducibility and Field Extensions 227

Unit 25: Roots of a Polynomial 237

Unit 26: Splitting Fields, Existence and Uniqueness 244

Unit 27: Separable Extensions 249

Unit 28: Galois Theory 263

Unit 29: Computing Galois Groups 270

Unit 30: Invariant Subfield 274

Unit 31: The Galois Group of a Polynomial 279

Unit 32: Solvability by Radicals 290

6 LOVELY PROFESSIONAL UNIVERSITY

Corporate and Business Law

LOVELY PROFESSIONAL UNIVERSITY 1

Unit 1: Generating Sets

NotesUnit 1: Generating Sets

CONTENTS

Objectives

Introduction

1.1 Sets

1.2 Cartesian Product

1.3 Relations

1.4 Functions

1.4.1 Composition of Functions

1.5 Some Number Theory

1.5.1 Principle of Induction

1.5.2 Divisibility in Z

1.6 Summary

1.7 Keywords

1.8 Review Questions

1.9 Further Readings

Objectives

After studying this unit, you will be able to:

Explain the operations on sets

Define Cartesian products of sets

Describe the equivalence classes

Define different kinds of functions

Explain the division algorithm and unique prime factorisation theorem

Introduction

In this unit, we will discuss some basic ideas concerning sets and functions. These concepts areelementary to the study of any branch of mathematics, in particular of algebra. In the unit, wediscuss some basic number theory. The primary aim of this section is to assemble a few facts thatwill be required in the rest of the course. We also hope to give you a glimpse of the elegance ofnumber theory. It is this sophistication that led the mathematician Gauss to call number theorythe "queen of mathematics". Let us start explaining these concepts one-by-one.


Abstract Algebra

Notes 1.1 Sets

You must have used the word 'set' off and on in your conversations to describe any collection.In mathematics, the term set is used to describe any well defined collection of objects, that is,every set should be so described that given any object it should be clear whether the given objectbelongs to the set or not.

For instance, the collection N of all natural numbers is well defined, and hence is a set. But thecollection of all rich people is not a set, because there is no way of deciding whether a humanbeing is rich or not.

If S is a set, an object a in the collection S is called an element of S. This fact is expressed insymbols as a S (read as "a is in S" or "a belongs to S"). If a is not in S, we write a S. For example,3 R the set of real numbers. But 1 R .

A set with no element in it is called the empty set, and is denoted by the Greek letter (phi). Forexample, the set of all natural numbers less than 1 is .

There are usually two way of describing a non-empty set:

(1) Roster method, and (2) Set builder method.

Roster Method: In this method, we list all the elements of the set within braces. For instance, thecollection of all positive divisors of 48 contains 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48 as its elements. Sothis set may be written as (1, 2, 3,4, 6, 8, 12, 16, 24, 48).

In this description of a set, the following two conventions are followed:

Convention 1: The order in which the elements of the set are listed is not important.

Convention 2: No element is written more than once, that is, every element must be writtenexactly once.

For example, consider the set S of all integers between 1 1

1 and 4 .2 4

Obviously, these integers

are 2, 3 and 4. So we may write S = (2, 3, 4).

We may also write S = (3, 2, 4), but we must not write S = (2, 3, 2, 4). Why? Isn�t this whatConvention 2 says?

The roster method is sometimes used to list the elements of a large set also. In this case we maynot want to list all the elements of the set. We list a few, enough to give an indication of the restof the elements. For example, the set of integers lying between 0 and 100 is (0, 1, 2 ,........., 100),and the set of all integers is Z = (0, +1, !2, ........ }.

Another method that we can use for describing a set is the Set Builder Method.

Set Builder Method: In this method we first try to find a property, which characterises theelements of the set, that is, a property P which all the elements of the set possess, and which noother objects possess. Then we describe the set as

{x | x has property P), or as

{x : x has property P).

This is to be read as �the set of all x such that x has property P�. For example, the set of all integerscan also be written as

Z = {x | x is an integer}.



NotesSome other sets that you may be familiar with are

Q, the set of rational numbers = a

a,b Z, b 0b

R, the set of real numbers

C, the set of complex numbers = (a+ib | a, b R). (Here i 1. )

Let us now see what subsets are.

Subsets: Consider the sets A = (1, 3, 4) and B = (1, 4). Here every element of B is also an element

of A. In such a case, that is, when every element of a set B is an element of a set A, we say that Bis a subset of A, and we write this as B A.

It is obvious that if A is any set, then every element of A is certainly an element of A. So, for

every set A, A A.

Also, for any set A, A.

Now consider the set S = (1, 3, 5, 15) and T = (2, 3, 5, 7}. Is S T? No, because not every element

of S is in T; for example, 1 S but 1 T. In this case we say that S is not a subset of T, and denoteit by S T.

Note that if B is not a subset of A, there must be an element of B which is not an element of A. In

mathematical notation this can be written as � x B such that x� A�.

We can now say that two sets A and B are equal (i.e., have precisely the same elements) if andonly if A B and B A.

Let us now look at some operations on sets. We will briefly discuss the operations of union,intersection and complementation on sets.

Union: If A and B are subsets of a set S, we can collect the elements of both to get a new set. This

set is called their union. Formally, we define the union of A and B to be the set of all thoseelements of S which are in A or in B. We denote the union of A and B by A B. Thus,

A B=(X S | X A or x B).

For example, if A = {1, 2} and B = {4, 6, 7}, then A B = {1, 2, 4, 6, 7}.

Again, if A = {l, 2, 3, 4) and B = (2, 4, 6, 8), then A B = {l, 2, 3, 4, 6, 8). Observe that 2 and 4 are inboth A and B, but when we write A B, we write these elements only once, in accordance with

Convention 2 given earlier.

Can you see that, for any set A, A U A = A?

Now we will extend the definition of union to define the union of more than two sets.

If A1, A

2, A

3, ..........,A

k are k subsets of a set S, then their union A

1 A

2 . ...... A

k is the set of

elements which belong to at least one of these sets. That is,

A1 A

2 ........ A

k = {x S | x A

i for some i = 1, 2 ,......, k).

The expression A1 A

2 ........ A

k is often abbreviated to

k

ii 1

A .


Abstract Algebra

Notes If is a collection of subsets of a set S, then we can define the union of all members of by

A

A

= (x S | X " A for some A ).

Now let us look at another way of obtaining a new set from two or more given sets.

Intersection: If A and B are two subsets of a set S, we can collect the elements that are common to

both A and B. We call this set the intersection of A, and B (denoted by A B. So,

A B = { x S | X " A and x B } .

Thus, if P = {1, 2, 3, 4} and Q = {2, 4, 6, 8}, then PQ = {2, 4},

Can you see that, for any set A, A A = A?

Now suppose A = {1, 2) and B = (4, 6, 7). Then what is A B? We observe that, in this case, A and

B have no common elements, and so A B = , the empty set.

When the intersection of two sets is , we say that the two sets are disjoint (or mutually disjoint).For example, the sets (1, 4) and (0, 5, 7, 14) are disjoint.

The definition of intersection can be extended to any number of sets. Thus, the intersection of ksubsets A

1, A

2 ,....., A

k of a set S is

A, A2 ......... A, = { x E S | x E A

i for each i = 1, 2,........, k }.

We can shorten the expression A1 A

2 ......... A

k to

k

ii 1

A .

In general, if is a collection of subsets of a set S, then we can define the intersection of all themembers of by

A P

A

= { X S | X A V A } [ V denotes �forever�]

Apart from the operations of unions and intersections, there is another operation on sets, namely,the operation of taking differences.

Differences: Consider the sets A = { 1, 2, 3] and B = [2, 3, 4]. Now the set of all elements of A thatare not in B is {1}. We call this set the difference A \ B. Similarly, the difference B \A is the set ofelements of B that are not in A, that is, {4}.

Thus, for any two subsets A and B of a set S,

A\B = { x S | x A and x B }

When we are working with elements and subsets of a single set X, we say that the set X is theuniversal set. Suppose X is the universal set and A X. Then the set of all elements of X which arenot in A is called the complement of A and is denoted by A�, AC or X \ A.

Thus,

Ac = {x X | x A }.

For example, if X = [a, b, p, q , r) and A = {a, p, q], then Ac = (b, r).

1.2 Cartesian Product

An interesting set that can be formed from two given sets is their Cartesian product, named afterthe French philosopher and mathematician Rene Descartes (1596 - 1650). He also invented theCartesian co-ordinate system.



NotesLet A and B be two sets. Consider the pair (a, b), in which the first element is from A and thesecond from B. Then (a, b) is called an ordered pair. In an ordered pair the order in which the twoelements are written is important. Thus, (a, b) and (b,a) are different ordered pairs. Two orderedpairs (a, b) and (c, d) are called equal, or the same, if

a = c and b = d.

Definition: The Cartesian product A x B, of the sets A and B, is the set of all possible ordered pairs(a, b), where a A, b B.

For example, if A = {l , 2 , 3} and B = (4, 6), then

A × B = { (1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6) }.

Also note that

B × A = { (4, 1), (4, 2), (4, 3), (6, 1), (6, 2), (6, 3) } and A x B B x A.

Let us make some remarks about the Cartesian product here.

Remarks: (i) A x B = iff A = or B = .

(ii) If A has m elements and B has n elements, then A x B has mn elements. B x A also has mnelements. But the elements of B x A need not be the same as the elements of A x B, as you have justseen.

We can also define the Cartesian product of more than two sets in a similar way. Thus, if A1, A

2,

A3, .......... A

n, are n sets, we can define their Cartesian product as

A1 x A

2x .......x A

n = { (a

l, a

2,......., a

n) | a

1 A

l,.........., a

n A

n}.

For example, if R is the set of all real numbers, then

R x R = { (al, a

2) | a

1 R, a

2 R }

R x R x R = { (al, a

2, a

3) | a

i R for i = 1, 2, 3 ), and so on. It is customary to write R2 for R x R and

Rn for R x .......... x R (n times).

Now, you know that every point in a plane has two coordinates, x and y. Also, every orderedpair (x, y) of real numbers defines the coordinates of a point in the plane. So, we can say that R2

represents a plane. In fact, R2 is the Cartesian product of the x-axis and the y-axis. In the same wayR3 represents three-dimensional space, and Rn represents n-dimensional space, for any n 1.Note that R represents a line.

1.3 Relations

You are already familiar with the concept of a relationship between people. For example, aparent-child relationship exists between A and B if and only if A is a parent of B or B is a parentof A.

In mathematics, a relation R on a set S is a relationship between the elements of S. If a S isrelated to b S by means of this relation, we write a R b or (a, b) R. From the latter notation wesee that R S × S. And this is exactly how we define a relation on a set.

Definition: A relation R on a set S is a subset of S × S.

For example, if N is the set of natural numbers and R is the relation �is a multiple of�, then15 R 5, but not 5 R 15. That is, (15, 5) R but (5, 15) R. Here R N × N.

Again, if Q is the set of all rational numbers and R is the relation �is greater than�, then 3 R 2(because 3 > 2).


Abstract Algebra

Notes Definition: A relation R defined on a set S is said to be

(i) reflexive if we have aRa, V a S.

(ii) symmetric if aRb bRa V a, b S.

(iii) transitive if aRb and bRc aRc V a, b, c S.

To get used to these concepts, consider the following examples.

Example: Consider the relation R on Z given by �aRb if and only if a > b�. Determinewhether R is reflexive, symmetric and transitive.

Solution: Since a > a is not true, aRa is not true. Hence, R is not reflexive.

If a > b, then certainly b > a is not true. That is, aRb does not imply bRa. Hence, R is notsymmetric.

Since a > b and b > c implies a > c, we find that aRb, bRc implies aRc. Thus, R is transitive.

Example: Let S be a non-empty set. Let (S) denote the set of all subsets of S, i.e.,(S) = (A | A S}. We call p (S) the power set of S.

Define the relation R on (S) by

R= { (A,B) | A, B (S) and A B ].

Check whether R is reflexive, symmetric or transitive.

Solution: Since A A V A (S), R is reflexive.

If A B, B need not be contained in A. (In fact, A B and B A A = B.) Thus, R is not symmetric.

If A B and B C, then A C V A, B, C (S). Thus, R is transitive.

A very important property of an equivalence relation on a set S is that it divides S into a numberof mutually disjoint subsets, that is, it partitions S. Let us see how this happens.

Let R be an equivalence relation on the set S. Let a S. Then the set { b S | aRb } is called theequivalence class of a in S. It is just the set of elements in S which are related to a. We denote itby [a].

This is

[1] = { n | 1 R n ; n N }

= { n 1 n N and 5 divides 1-n)

= { n I n N and 5 divides n-1)

= { 1, 6, 11, 16, 21 ........ }.

Similarly,

[2] = { n | n N and 5 divides n�2 }

= { 2, 7, 12, 17, 22. ......... },

[3] = { 3, 8, 13, 18, 23 .......... },

[4] = { 4 , 9, 14 , 19, 24, . ........ },

[5] = { 5, 10, 15, 20, 25, ....... },



NotesNote that

(i) [1] and [6] are not disjoint. In fact, [1] = [6]. Similarly, [2] = [7], and so on.

(ii) N = [I] U [2] U [3] U [4] U [5], and the sets on the right hand side arc mutually disjoint.

We will prove these observations in general in the following theorem.

Theorem 1: Let R be an equivalence relation on a set S. For a S, let [a] denote the equivalenceclass of a. Then

(a) a [a],

(b) b [a] [a] = [b],

(c) S = a S

[a]

(d) if a, b S, then [a] [b] = or [a] = [b].

Proof: (a) Since R is an equivalence relation, it is reflexive.

aRa V a S. a [a].

(b) Firstly, assume that b " [a]. We will show that [a] [b] and [b] [a]. For this, let x [a].Then xRa.

We also know that aRb. Thus, by transitivity of R, we have xRb, i.e., x [b]. [a] [b]. We cansimilarly show that [b] [a].

[a] = [b].

Conversely, assume that [a] = [b]. Then b [b] = [a]. b [a].

(c) Since [a] S V a S, a S

[a] S

.

Conversely, let x S. Then x [x] by (a) above. [x] is one of the sets in the collection whose union

is a S

[a]

.

Hence, x a S

[a]

. So, S a S

[a]

.

Thus, S a S

[a]

and a S

[a] S

, proving (c).

(d) Suppose [a] [b] . Let x [a] [b].

Then x [a] and x [b]

[x] = [a] and [x] = [b], by (b) above.

[a] = [b].

Note that, in Theorem 1, distinct sets on the right hand side of (c) are mutually disjoint becauseof (d). Therefore, (c) expresses S as a union of mutually disjoint subsets of S; that is, we have apartition of S into equivalence classes.

Let us look at some more examples of partitioning a set into equivalence classes.


Abstract Algebra

Notes

Example: Let S be the set of straight lines in R × R. Consider the relation on S given by�L

1 R L

2 iff L

1 = L

2 or L

1 is parallel to L

2. Show that R is an equivalence relation, What are the

equivalence classes in S?

Solution: R is reflexive, symmetric and transitive. Thus, R is an equivalence relation. Now, takeany line L

1 (see Figure 1.1).

Figure 1.1: The Equivalence Class of L1

Let L be the line through (0, 0) and parallel to L1. Then L [L

1]. Thus, [L] = [L,]. In this way the

distinct lines through (0, 0) give distinct equivalence classes into which S is partitioned. Eachequivalence class [L] consists of all the lines in the plane that are parallel to L.

In the next section we will briefly discuss a concept that you may be familiar with, namely,functions.

1.4 Functions

Recall that a function f from a non-empty set A to a non-empty set B is a rule which associateswith every element of A exactly one element of B. This is written as f : A B. If f associates witha A, the element b of B, we write f(a) = b. A is called the domain of f, and the set f(A) = { f(a) |a A] is called the range of f. The range of f is a subset of B.

i.e., f(A) B. B is called the codomain of f.

Note that

(i) For each element of A, we associate some element of B.

(ii) For each element of A, we associate only one element of B.

(iii) Two or more elements of A could be associated with the same element of B.

For example, let A = { l, 2, 3 }, B = { l, 2, 3, 4, 5, 6, 7, 8, 9, l0 }. Define f : A B by f(1) = 1, f(2) = 4,f(3) = 9. Then f is a function with domain A and range {l, 4, 9}. In this case we can also writef(x) = x2 for each x A or f : A B : f(x) = x2. We will often use this notation for defining anyfunction.



NotesIf we define g : A B by g(1) = 1, g(2) = 1, g(3) = 4, then g is also a function. The domain of gremains the same, namely, A. But the range of g is {1, 4}.

Remark: We can also consider a function f : A B to be the subset { (a, f(a)) | a A } of A × B.

Now let us look at functions with special properties.

Definition: A function f : A B is called One-one (or injective) if f associates different elementsof A with different elements of B, i.e., if a

1, a

2 A and al a,, then f(a

l) f(a

2). In other words, f is

1-1 if f(a1) = f(a

2) a

1 = a

2.

In the examples given above, the function f is one-one. The function g is not one-one because 1and 2 are distinct elements of A, but g(1) = g(2).

Now consider another example of sets and functions.

Let A = (1, 2, 3), B = { p, q, r }. Let f : A B be defined by f(1) = q, f(2) = r, f(3) = p. Then f is a function.Here the range of f = B = codomain of f. This is an example of an onto function, as you shall see.

Definition: A function f : A B is called onto (or surjective) if the range of f is B, i.e., if, for eachb B, there is an a A such that f(a) = b. In other words, f is onto if f(A) = B.

For another important example of a surjective function, consider two non-empty sets A and B.We define the function

1 : A × B A :

1((a, b)) = a.

l is called the projection of A × B onto A. You

can see that the range of 1 is the whole of A. Therefore,

1 is onto. Similarly,

: A × B B :

2 ((a,

b)) = b, the projection of A × B onto B, is a surjective function.

If a function is both one-one and onto, it is called bijective, or a bijection.

Consider the following example that you will use again and again.

Example: Let A be any set. The function IA : A A : I

A(a) = a is called the identity function

on A. Show that I, is bijective.

Solution: For any a A, IA (a) = a. Thus, the range of I

A is the whole of A. That is, I

A is onto.

IA is also 1-1 because if a

1, a

2 A such that a

1 a

2, then I

A (a

1) I

A (a

2).

Thus, IA is bijective.

If f : A B is a bijection, then we also say that the sets A and B are equivalent. Any set which isequivalent to the set { 1, 2, 3 ,............, n}, for some n N, is called a finite set. A set that is not finiteis called an infinite set.

Convention: The empty set is assumed to be finite.

Let us now see what the inverse image of a function is.

Definition: Let A and B be two sets and f : A B be a function, Then, for any subset S of B, theinverse image of S under f is the set

f�1(S) = { a A | f(a) S }.

For example, 1AI (A) = { a A | I

A(a) A } = A.

f�1({ 1, 2, 3 }) = { n N | f(n) { 1 , 2 , 3 } }

= { n N | n + 5 { 1,2,3 }}

= , the empty set.

We now give some nice theorems involving the inverse image of a function.


Abstract Algebra

Notes Theorem 2: Let f : A B be a function. Then,

(a) for any subset S of B, f(f�1 (S) ) S.

(b) for any subset X of A, X f�1 (f(X)).

Proof: We will prove (a) and you can prove (b). Let b " f(f �1 (S) ). Then, by definition, 3 a f �1 (S)such that b = f(a). But a f �1 (S) f(a) S. That is, b S.

Thus, f(f�1 (S) ) S.

Now let us look at the most important way of producing new functions from given ones.

1.4.1 Composition of Functions

If f : A B and g : C D are functions and if the range of f is a subset of C, there is a natural wayof combining g and f to yield a new function h : A # D. Let us see how.

For each x A, h(x) is defined by the formula h(x) = g(f(x)).

Note that f(x) is in the range of f, so that f(x) C. Therefore, g(f(x)) is defined and is an elementof D. This function h is called the composition of g and f and is written as g o f. The domain ofg o f is A and its codomain is D. In most cases that we will be dealing with we will have B = C.Let us look at some examples.

Example: Let f : R R and g : R R be defined by f(x) = x2 and g(x) = x + 1. What isg o f ? What is f o g ?

Solution: We observe that the range of f is a subset of R, the domain of g. Therefore, g o f is

defined. By definition, V x R, g o f(x) = g(f(x)) = f(x) + 1 = x2 + 1.

Now, let us find f o g. Again, it is easy to see that f o g is defined. V x R,

f o g(x) = f(g(x)) = (g(x))2 = (x + 1)2.

So f o g and g o f are both defined. But g o f f o g . (For example, g o f(1) f o g(l).)

Example: Let A = {1, 2, 3}, B = {p, q, r} and C = {x, y}. Let f : A B be defined by f(1) = p,f(2) = p, f(3) = r. Let g : B C be defined by g(p) = x, g(q) = y, g(r) = y. Determine if f o g and g of can be defined.

Solution: For f o g to be defined, it is necessary that the range of g should be a subset of thedomain of f. In this case the range of g is C and the domain of f is A. As C is not a subset of A,f o g cannot be defined.

Since the range of f, which is (p, r), is a subset of B, the domain of g, we see that g o f is defined.Also g o f : A C is such that

g o f(l) = g(f(1)) = g(p) = x,

g o f(2) = g(f(2)) = g(p) = x,

g o f(3) = g(f(3)) = g(r) = y.

In this example note that g is surjective, and so is g o f.

We now come to a theorem which shows us that the identity function behaves like the number1 R does for multiplication. That is, if we take the composition of any function f with a suitableidentity function, we get the same function f.



NotesTheorem 3: Let A be a set. For every function f : A A, we have f o IA = I

A o f = f.

Proof: Since both f and IA are defined from A to A, both the compositions f o I

A and I

A o f are

defined. Moreover, V x A,

f o IA(x) = f(I

A(x)) = f (x), so f o I

A = f.

Also, V X A, IA o f(x) = I

A (f(x)) = f(x), so I

A o f = f.

In the case of real numbers, you know that given any real number x + 0, $ y % 0 such that xy = 1.y is called the inverse of x. Similarly, we can define an inverse function for a given function.

Definition: Let f : A B be a given function. If there exists a function g : B A such that f o g =I

B and g o f = I

A, then we say that g is the inverse of f, and we write g = f �1.

For example, consider f : R R defined by f(x) = x + 3. If we define g : R R by g(x) = x � 3, then

f o g(x) = f(g(x)) = g(x) + 3 = (x � 3) + 3 = x V x R. Hence, f o g = IR. You can also verify that

g o f = IR. So g = f�1.

Note that in this example f adds 3 to x and g does the opposite � it subtracts 3 from x. Thus, thekey to finding the inverse of a given function is : try to retrieve x from f(x).

For example, let f : R R be defined by f(x) = 3x + 5. How can we retrieve x from 3x + 5? The

answer is �first subtract 5 and then divide by 3�. So, we try g(x) = x 5

(x) .3

And we find

g o f(x) = g(f(x)) = f(x) 5 (3x 5) 5

x.3 3

Also, f o g(x) = 3(g(x)) + 5 = (x 5)

3 5 x V x R.3

Let�s see if you�ve understood the process of extracting the inverse of a function,

Do all functions have an inverse? No, as the following example shows.

Example: Let f : R R be the constant function given by f(x) = 1 V x " R. What is theinverse of f ?

Solution: If f has an inverse g : R R, we have f o g = IR, i.e., V x R, f o g(x) = x. Now take

x = 5. We should have f o g (5) = 5, i.e., f(g(5)) = 5. But f(g(5)) = 1, since f(x) = 1 V x. So we reach acontradiction. Therefore, f has no inverse.

In view of this example, we naturally ask for necessary and sufficient conditions for f to have aninverse. The answer is given by the following theorem.

Theorem 4: A function f ; A B has an inverse if and only if f is bijective.

Proof: Firstly, suppose f is bijective. We shall define a function g : B A and prove that g = f�1.

Let b B. Since f is onto, there is some a " A such that f(a) = b. Since f is one-one, there is only onesuch a A. We take this unique element a of A as g(b). That is, given b B, we define g(b) = a,where f(a) = b.

Note that, since f is onto, B = { f(a) | a A). Then, we are simply defining g : B A by g(f(a)) =a. This automatically ensures that g o f = I

A.

Now, let b B and g(b) = a. Then f(a) = b, by definition of g. Therefore, f o g(b) = f(g(b)) = f(a) =b. Hence, f o g = I

B.


Abstract Algebra

Notes So, f o g = IB and g o f = I

A. This proves that g = f �1.

Conversely, suppose f has an inverse and that g = f �1. We must prove that f is one-one and onto.

Note g o f is 1 � 1 f is 1 � 1

Suppose f(a1) = f(a

2). Then g(f(a

1)) = g(f(a

2)).

g o f(a1) = g o f(a

2)

a = a

2, because g o f = I

A.

So, f is one-one.

Note g o f is onto g is onto.

Next, given b B, we have f o g = IB, So that f o g(b) = I

B(b) = b,

i.e., f(g(b)) = b. That is, f is onto.

Hence, the theorem is proved.

1.5 Some Number Theory

In this section we will spell out certain factorisation properties of integers that we will usethroughout the course. For this we first need to present the principle of finite induction.

1.5.1 Principle of Induction

We will first state an axiom of the integers that we will often use implicitly, namely, thewell-ordering principle. We start with a definition.

Definition: Let S be a non-empty subset of Z. An element a S is called a least element (or a

minimum element) of S if a b V b S. For example, N has a least element, namely, 1. But Zhas no least element. In fact, many subsets of Z, like 2Z, (-1, -2, -3, ...... ), etc., don�t have leastelements.

The following axiom tells us of some sets that have a least element.

Well-ordering Principle: Every non-empty subset of N has a least element.

You may be surprised to know that this principle is actually equivalent to the principle of finiteinduction, which we now state.

Theorem 5: Let S N such that

(i) 1 S, and

(ii) whenever k S, then k + l S.

Then S = N.

This theorem is further equivalent to:



NotesTheorem 6: Let S N such that

(i) 1 S, and

(ii) if m S V m < k, then k S.

Then S = N

We will not prove the equivalence of the well-ordering principle and Theorems 5 and 6 in thiscourse, since the proof is slightly technical.

Let us rewrite Theorems 5 and 6 in the forms that we will normally use.

Theorem 5: Let P(n) be a statement about a positive integer n such that

(i) P(1) is true, and

(ii) if P(k) is true for some k N, then P(k + l) is true.

Then, P(n) is true for all n N.

Theorem 6: Let P(n) be a statement about a positive integer n such that

(i) P(1) is me, and

(ii) if P(m) is true for all positive integers m < k, then P(k) is true.

Then P(n) is true for all n N.

The equivalent statements given above are very useful for proving a lot of results in algebra.As we go along, we will often use the principle of induction in whichever form is convenient.Let us look at an example.

Example: Prove that 13 + 23 + ............... + n3 = 2 2n (n 1)

4

for every n N.

Solution: Let Sn = 13 + ............ + n3, and let P(n) be the statement that

2 2

n

n (n 1)S .

4

Since 2 2

1

1 2S ,

4

P(I) is true.

Now, suppose P(n � 1) is true, i.e., 2 2

n 1

(n 1) nS

4

Then, S, = 13 + ............ + (n � 1)3 + n3

= Sn�1

+ n3.

=2 2

3(n 1) nn ,

4

since P(n � 1) is true.

=2 2n (n 1) 4n

4

=2 2n (n 1)

4

Thus, P(n) is true.


Abstract Algebra

Notes Therefore, by the principle of induction, P(n) is true for all n in N.

Now, use the principle of induction to prove the following property of numbers that you musthave used time and again.

1.5.2 Divisibility in Z

One of the fundamental ideas of number theory is the divisibility of integers.

Definition: Let a, b " , a % 0. Then, we say that a divides b if there exists an integer c such thatb = ac. We write this as a | b and say that a is a divisor (or factor) of b, or b is divisible by a, orb is a multiple of a.

If a does not divide b we write a × b.

We give some properties of divisibility of integers in the following exercise. You can provethem very easily.

We will now give a result, to prove which we use Theorem 5.

Theorem 7 (Division Algorithm): Let a, b Z, b > 0. Then there exist unique integers q, r suchthat a = qb + r, .where 0 5 r < b.

Proof: We will first prove that q and r exist. Then we will show that they are unique. To provetheir existence, we will consider three different situations : a = 0, a > 0, a < 0.

Case 1 (a = 0): Take q = 0, r = 0. Then a = qb + r.

Case 2 (a > 0): Let P(n) be the statement that n = qb + r for some q, r Z, 0 r < b.

Now let us see if P(l) is true.

If b = l, we can take q = l, r = 0, and thus, 1 = 1.1 + 0.

If b l, then take q = 0, r = 1, i.e., 1 = 0.b + l.

So, P(1) is true.

Now suppose P(n � 1) is me, i.e., (n � 1) = qlb + r

1 for some q

l, r

1 Z, 0 r

1 < b. But then r

1 b � 1,

i.e., r1 + 1 b. Therefore,

1 1 1

1 1

q b (r 1), if (r 1) bn

(q 1)b 0,if r 1 b

This shows that P(n) is true. Hence, by Theorem 5, P(n) is true, for any n N. That is, for a > 0, a= qb + r, q, r Z, 0 r < b.

Case 3 (a < 0): Here (-a) > 0. Therefore, by Case 2, we can write

(�a) = qb + r�, 0 r� < b

i.e., ( q)b, if r 0

a(�q � 1)b + (b - r ), if 0 < r < b

This proves the existence of the integers q, r with the required properties.

Now let q�, r� be in Z such that a = qb + r and a = q�b + r�, where 0 5 r, r� < b. Then r � r� = b(q� � q).Thus, b | (r � r�). But | r � r� | < b. Hence, r � r� = 0, i.e., r = r� and q = q�. So we have proved theuniqueness of q and r.

In the expression, a = qb + r, 0 & r < b, r is called the remainder obtained when a is divided by b.



NotesDefinition: Let a, b Z. c Z is called a common divisor of a and b if c | a and c | b. For example,2 is a common divisor of 2 and 4. You know that 1 and �1 are common divisors of a and b, for anya, b Z. Thus, a pair of integers do have more than one common divisor. This fact leads us to thefollowing definition.

Definition: An integer d is said to be a greatest common divisor (g.c.d. in short) of two non-zerointegers a and b if

(i) d | a and d | b, and

(ii) if c | a and c | b, then c | d.

Note that if d and d� are two g.c.d s of a and b, then (ii) says that d | d� and d� | d. Thus, d = d�.But then only one of them is positive. This unique positive g.c.d. is denoted by (a, b).

We will now show that (a, b) exists for any non-zero integers a and b. You will also see howuseful the well-ordering principle is.

Theorem 8: Any two non-zero integers a and b have a g.c.d., and (a, b) = ma+nb, for some m,n Z.

Proof: Let S = {xa + yb | x, y Z, (xa + yb) > 0).

Since a2 + b2 > 0, a2 + b2 S, i.e., S 0. But then, by the well-ordering principle, S has a leastelement, say d = ma + nb for some m, n Z. We show that d = (a, b).

Now d S. Therefore, d > 0. So, by this division algorithm we can write

a = qd + r, 0 r < d. Thus,

r = a � qd = a � q(ma + nb) = (1 � qm)a + (�qn)b.

Now, if r 0, then r S, which contradicts the minimality of d in S. Thus, r = 0, i.e., a = qd, i.e.,d | a. We can similarly show that d | b. Thus, d is a common divisor of a and b.

Now, let c be an integer such that c | a and c | b.

Then a = a1c, b = b

1c for some a

1, b

1 Z.

But then d = ma + nb = malc + nb

1c = (ma

1 + nb

1)c. Thus, c | d, So we have shown that d is a g.c.d.

In fact, it is the unique positive g.c.d. (a,b).

For example, the g.c.d. of 2 and 10 is 2 = 1.2 + 0.10, and the g.c.d, of 2 and 3 is 1 = (�1)2 + l(3).

Pairs of integers whose g.c.d. is 1 have a special name.

Definition: If (a,b) = 1, then the two integers a and b are said to be relatively prime (or coprime)to each other.

Using Theorem 8, we can say that a and b are coprime to each other iff there, exist m, n Z suchthat 1 = ma + nb.

The next theorem shows us a nice property of relatively prime numbers.

Theorem 9: If a,b Z; such that (a, b) = 1 and b | ac, then b | d.

Proof: We know that 3 m, n Z such that 1 = ma + nb. Then c = c.1 = c(ma+nb) = mac + nbc.

Now, b | ac and b | bk. b | (mac + nbc). Thus, b | c.

Let us now discuss prime factorisation.

Definition: A natural number p ( 1) is called a prime if its only divisors are 1 and p. If a naturalnumber n (% 1) is not a prime, then it is called a composite number.


Abstract Algebra

Notes For example, 2 and 3 are prime numbers, while 4 is a composite number.

Note that, if p is a prime number and a Z such that p × A, then (p,a) = 1.

Now consider the number 50. We can write 50 = 2 × 5 × 5 as a ,product of primes. In fact we canalways express any natural number as a product of primes. This is what the unique primefactorisation theorem says.

Theorem 10 (Unique Prime Factorisation Theorem): Every integer n > 1 can be written asn = p

1.p

2 ........ p

n, where p

l, ........ p

n are prime numbers. This representation is unique, except for

the order in which the prime factors occur.

Proof: We will first prove the existence of such a factorisation. Let P(n) be the statement thatn + l is a product of primes. P(1) is true, because 2 is a prime number itself.

Now let us assume that P(m) is true for all positive integers m < k. We want to show that P(k) istrue. If (k+l) is a prime, P(k) is true. If k+l is not a prime, hen we can write k + l = m, m

2, where

1 < m, < k + l and 1 < m2 < k + l. But then P(m

l � 1) and P(m

2 � 1) are both true. Thus, m

l = p

1p

2. ....

pr, m

2 = q

1q

2 .........q

s, where p

l,p

2, .......P

r, q

l, q

2, .......q

s are primes. Thus,

k + 1 = pl p

2 ... p

r q

1 q

2 .... q

s, i .e., P(k) is true. Hence, by Theorem 6, P(n) is true for every n N.

Now let us show that the factorisation is unique.

Let n = pl p

2 ... p, = q

1 q

2 ... q

s, where

pl, p

2, ......p

1. q

l, q

2, ......, q

s are primes. We will use induction on t.

If t = 1, then p1 = q

1, q

2 ....... q

s. But p

1 is a prime. Thus, its only factors are 1 and itself.

Thus, s = 1 and p1 = q

1.

Now suppose t > 1 and the uniqueness holds for a product of t�1 primes. Now p1 | q

1q

2 ......q

s and

hence, p1 | q

i for some i, By re-ordering q

1 ....... q, we can assume that p

1 | q

1. But both

p1 and q

1 are primes. Therefore, p

1 = q

l. But then p

2 ..... p, = q

2 ....... q,. So, by induction, t�1 = s�1

and p2 ,........ p

t are the same as q

2. ......q,, in some order.

Hence, we have proved the uniqueness of the factorisation.

The primes that occur in the factorisation of a number may be repeated, just as 5 is repeated inthe factorisation 50 = 2 × 5 × 5. By collecting the same primes together we can give the followingcorollary to Theorem 10.

Corollary: Any natural number n can be uniquely written as n = plml p

2m2 .....P

rmr where for i = 1,

2, ....... r, each mi N and each p

i is a prime with 1 < p

1 < p

2 < .... < p,.

As an application of Theorem 10, we give the following important theorem, due to the ancientGreek mathematician Euclid.

Theorem 11: There are infinitely many primes.

Proof: Assume that the set P of prime numbers is finite, say

P = { p1, p

2, .... p

n}. Consider the natural number

n = (p1p

2 ......... p

n) + 1

Now, suppose some pi | n. Then p

i | (n � p

1p

2 ....... p,), i.e., p

i | 1, a contradiction.

Therefore, no pi divides n. But since n > 1, Theorem 10 says that n must have a prime factor. We

reach a contradiction. Therefore, the set of primes must be infinite.



NotesSelf Assessment

1. Let A = {2, {4, 5}, 4}. Which statement is correct?

(a) 5 is an element of A.

(b) {5} is an element of A.

(c) {4, 5} is an element of A.

(d) {5} is a subset of A.

2. Which of these sets is finite?

(a) {x | x is even}

(b) {x | x < 5}

(c) {1, 2, 3,...}

(d) {1, 2, 3,...,999, 1000}

3. Which of these sets is not a null set?

(a) A = {x | 6x = 24 and 3x = 1}

(b) B = {x | x + 10 = 10}

(c) C = {x | x is a man older than 200 years}

(d) D = {x | x < x}

4. Let D E. Suppose a D and b E. Which of the following statements must be true?

(a) c D

(b) b D

(c) a E

(d) a D

5. Let A = {x | x is even}, B = {1, 2, 3... 99, 100}, C = {3, 5, 7, 9}, D = {101, 102} and E = {101, 103,105}. Which of these sets can equal S if S A and S and B are disjoint?

(a) A (b) B (c) C

(d) D (e) E

6. Which statement best describes the Venn diagram below?

(a) A = B

(b) A and B are not comparable

(c) A B

(d) A B

7. Which set S does the power set 2S = {, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} come from?

(a) {{1},{2},{3}}

(b) {1, 2, 3}

(c) {{1, 2}, {2, 3}, {1, 3}}

(d) {{1, 2, 3}}


Abstract Algebra

Notes 1.6 Summary

In this unit we have covered the following points:

Some properties of sets and subsets.

The union, intersection, difference and complements of sets.

The Cartesian product of Sets.

Relations in general, and equivalence relations in particular.

The definition of a function, a 1-1 function, an unto function and a bijective function.

The composition of functions.

The well-ordering principle, which states that every subset of N has a least element.

The principle of finite induction, which states that : If P(n) is a statement about some n Nsuch that

(i) P(1) is true, and

(ii) if P(k) is true for some k N, then P(k + l) is true,

then P(n) is true for every n N.

The principle of finite induction can also be stated as:

If P(n) is a statement about some n N such that

(i) P(l) is true, and

(ii) if P(m) is true for every positive integer m < k, then P(k) is true,

then P(n) is true for every n N.

Note that the well-ordering principle is equivalent to the principle of finite induction.

Properties of divisibility in Z, like the division algorithm and unique prime factorisation.

1.7 Keywords

Empty Set: A set with no element in it is called the empty set, and is denoted by the Greek letter (phi). For example, the set of all natural numbers less than 1 is .

Roster Method: It is sometimes used to list the elements of a large set also. In this case we maynot want to list all the elements of the set.

Union: If A and B are subsets of a set S, we can collect the elements of both to get a new set. Thisset is called their union.


1. Let C = {1, 2, 3, 4} and D = {1, 3, 5, 7, 9}. How many elements does the set C D contain?How many elements does the set CD contain?

2. Let U = {1, 2, 3... 8, 9}, B = {1, 3, 5, and 7} and C = {2, 3, 4, 5, 6}. How many elements does theset (B C)� contain? How many elements does the set (C � B)� contain?

3. Let S = {a, b}. How many elements does the power set 2S contain?

4. Let S = {1, 2, 3}. How many subsets does S contain?



Notes5. Let a, b, c be non-zero integers. Then

(a) a|0, ±1|a, ±a|a.

(b) a|b ac|bc

(c) a|b and b|c a|c

(d) a|b and b|a a = ±b

(e) c|a and c|b c | (ax + by) x, y Z.

6. If p is a prime and p|ab, then show that p|a or p|b.

7. If p is a prime and p|a1a

2.....a

n, then show that p|a

i for some i = 1,....,n.

Answers: Self Assessment

1. A is consisted of elements: 3, {4, 5} and 8, so {4, 5} is an element of A

2. {1,2,3, ..., 1000} is finite, because it is consisted of final number of elements.

3. Set B is not an empty set because it contains one element. The only element of the set B iszero. B = {0}

4. a E is true, because a D and D E means that every element from D is contained in E.

5. The correct answer is E, because E consists of even numbers as elements and the intersectionof sets S and B is a null set.

6. A B is the correct answer because A is a superset of B.

7. The correct answer is {1, 2, 3} because all subsets of {1, 2, 3} are , {1}, {2}, {3}, {1, 2}, {1, 3},{2, 3}, {1, 2, 3}.


Books Dan Saracino: Abstract Algebra; A First Course.

Mitchell and Mitchell: An Introduction to Abstract Algebra.

John B. Fraleigh: An Introduction to Abstract Algebra (Relevant Portion).

Online links www.jmilne.org/math/CourseNotes/

www.math.niu.edu

www.maths.tcd.ie/

archives.math.utk.edu

http://www.jmilne.org/math/CourseNotes/

http://www.math.niu.edu

http://www.maths.tcd.ie/


Abstract Algebra

Notes Unit 2: Groups

CONTENTS

Objectives

Introduction

2.1 Binary Operations

2.1.1 Operation �.� Table

2.2 Group

2.2.1 Abelian Group

2.3 Properties of Groups

2.4 Different Types of Group

2.4.1 Integers Modulo n

2.4.2 The Symmetric Group

2.4.3 Complex Numbers

2.5 Summary

2.6 Keywords



Objectives


Discuss the binary operations

Explain the term abelian and non-abelian groups

Describe the cancellation laws and laws of indices for various groups

Discuss the properties of integers modulo n, permutations and complex numbers

Introduction

The theory of groups is one of the oldest branches of abstract algebra. It has many applicationsin mathematics and in the other sciences. Group theory has helped in developing physics,chemistry and computer science. Its own roots go back to the work of the eighteenth centurymathematicians Lagrange, Ruffini and Galois.

In this unit, we will study about the group theory in detail. We surge fine groups and give someexamples. After that we understand details of some properties of groups that the elements of agroup satisfy. Let us discuss all these one by one.


Unit 2: Groups

Notes2.1 Binary Operations

As you all know common operations of addition and multiplication in R, Q and C. All theseoperations are examples of binary operations. It can be defined as:

Definition: Let S be a non-empty set. Any function * : S × S S is called a binary operationon S.

So, a binary operation associates a unique element of S to every ordered pair of elements of S.

For a binary operation * on S and (a, b) S × S, we denote *(a, b) by a*b.

We will use symbols like +, �, ×, , o, * and A to denote binary operations.

Let us look at some examples.

(i) + and x are binary operations on Z. In fact, we have + (a, b)= a + b and × (a, b) = a × b a,b Z. We will normally denote a × b by ab.

(ii) Let (S) be the set of all subsets of S. Then the operations and are binary operations

on (S), since A B and A B are in (S) for all subsets A and B of S.

(iii) Let X be a non-empty set and F(X) be the family of all functions f : X X. Then the

composition of functions is a binary operation on F(X), since fog F(X) f, g F(X).

After defining a non-empty set lets define properties of binary operations.

Definition: Let * be a binary operation on a set S. We say that

(i) * is closed on a subset T of S, if a * b T a, b T

(ii) * is associative if, for all a, b, c S, (a * b) * c = a * (b * c).

(iii) * is commutative if, for all a, b S, a * b = b * a.

For example, the operations of addition and multiplication on R are commutative as well asassociative. But, subtraction is neither commutative nor associative on R. Why? Is a � b = b � a or(a � b) � c = a � (b � c) 4) a, b, c R ? No, for example, 1 � 2 ! 2 � 1, and (1 � 2) � 3 1 � (2 � 3). Alsosubtraction is not closed on N R. because 1 N. 2 N but 1 � 2 N.

Note A binary operation on S is always closed on S, but may not be closed on a subsetof S.

Task For the following binary operations defined on R, determine whether they arecommutative or associative. Are they closed on N?

(a) x y = x + y � 5

(b) x * y = 2(x + y)

(c) x y = x y

2

for all x, y R.


Abstract Algebra

Notes As you are familiar with the equation such as a(b + c) = ab + ac and (b + c)a = bc +ba a, b, c R.

As this equation explains that multiplication distributes over addition in R. In general we candefine this as.

Definition: If o and * are two binary operations on a set S, then we say that * is distributive over

o if a, b, c S, we have a * (b o c) = (a * b) o (a * c) and (b o c) * a = (b * a) o (c * a).

For example, let a * b = a b

2

a, b R. Then a(b a c) = b c ab ac

a2 2

= ab * ac, and (b * c)a

= b c ba ca

a ba * ca a,b,c R.2 2

Hence, multiplication is distributive over *.

Let us now look deeper at some binary operations. You know that, for any a " R, a + 0 =a, 0 + a= a and a + (�a) = (�a) + a = 0. We say that 0 is the identity element for addition and (�a) is thenegative or additive inverse of a.

Definition: Let *.be a binary operation on a set S. If there is an element e S such that a S,a * e = a and e * a = a, then e is called an identity element for *.

For a S, we say that b S is an inverse of a, if a * b = e and b * a = e. In this case we usually writeb = a�1.

Let us first discuss the uniqueness of identity element for *, and uniqueness of the inverse of anelement with respect to *, if it exists. After that we will discuss the examples related to identityelements.

Theorem 1: Let * be a binary operation on a set S. Then

(a) if * has an identity element, it must be unique.

(b) if * is associative and s S has an inverse with respect to *, it must be unique.

Proof: (a) Suppose e and e� are both identity elements for *.

Then e = e * e�, since e� is an identity element.

= e�, since e is an identity element.

That is, e = e�. Hence, the identity element is unique.

(b) Suppose there exist a, b S such that s * a = e = a * s and s * b = e = b * s, e being the identityelement for *, Then

a = a * e = a * ( s * b )

= (a * s) * b, since * is associative.

= e * b = b .

That is, a = b.

Hence, the inverse of s is unique.

This uniqueness theorem allows us to say the identity element and the inverse, henceforth.

A binary operation may or may not have an identity element. For example, the operation ofaddition on N has no identity element.


Unit 2: Groups

NotesSimilarly, an element may not have an inverse with respect to a binary operation. For example,2 " Z has no inverse with respect to multiplication on Z, does it?

Now let us consider the following examples.

Example: If the binary operation # : R × R R is defined by a b = a + b � 1, prove that has an identity. If x R, determine the inverse of x with respect to , if it exists.

Solution: We are looking for some e R such that a e = a = e a a R. So we want e R such

that a + e � 1 = a a R. Obviously, e = 1 will satisfy this. Also, 1 a= a a R. Hence, 1 is theidentity element of .

For x R, if b is the inverse of x, we should have b x = 1.

i.e., b + x � 1 = 1, i.e., b = 2 � x. Indeed, (2 � x) x = (2 � x) + x � 1 = l.

Also, x (2 � x) = x + 2 � x � 1 =l. So, x�1 = 2 � x.

Example: Let S be a non-empty set. Consider (S), the set of all subsets of S. Are and

commutative or associative operations on (S)? Do identity elements and inverses of elementsof (S) exist with respect to these operations?

Solution: Since A B = B A and A B = B A A, B (S), the operations of union andintersection are commutative. You can see that the empty set and the set S are the identities ofthe operations of union and intersection, respectively. Since S , there is no B (S) such thatS B = . In fact, for any A (S) such that A # IS, A does not have an inverse with respect tounion. Similarly, any proper subset of S does not have an inverse with respect to intersection.

When the set S under consideration is small, we can represent the way a binary operation on Sacts by a table.

2.1.1 Operation �.� Table

Let S be a finite set and * be a binary operation on S. We can represent the binary operation bya square table, called an operation table or a Cayley table. The Cayley table is named after thefamous mathematician Arthur Cayley (1821-1895).

To write this table, we first list the elements of S vertically as well as horizontally, in the sameorder. Then we write a * b in the table at the intersection of the row headed by a and the columnheaded by b.

For example, if S = (�1, 0, 1) and the binary operation is multiplication, denoted by., then it canbe represented by the following table.

. �1 0 1

�1 (�1) . (�1)

= 1

(�1) . 0

= 0

(�1) . 1

= �1

0 0 . (�1)

= 0

0.0

= 0

0.1

= �1

1 1 . (�1)

= �1

1.0

= 0

1.1

=1


Abstract Algebra

Notes Conversely, if we are given a table, we can define a binary operation on S. For example, we candefine the operation * on S = {1, 2, 3} by the following table.

* 1 2 3

1 1 2 3

2 3 1 2

3 2 3 1

From this table we see that, for instance, 1 * 2 = 2 and 2 * 3 = 2.

Now 2 * 1 = 3 and 1 * 2 = 2. 2 * 1 1 * 2. That is, * is not commutative.

Again, (2 * 1) * 3 = 3 * 3 = 1 and 2 * (1 * 3) = 2

(2 * 1) * 3 2 * (1 * 3). , * is not associative.

See how much information a mere table can give !

Now consider the following definition.

Definition: Let * be a binary operation on a non-empty set S and let a,, . . . . . .,ak+1

, S. We definethe product a, * . ... .. * a

k+1, as follows:

If k = 1, a, * a2 is a well defined element in S.

If al * ..... * a, is defined, then

a, * .... . . * ak+l

= (a, * . ..... * a,) * ak+1

We use this definition in the following result.

Theorem 2: Let a,, ...... ,am+n

be elements in a set S with an associative binary operation *. Then(a, * ..,...* a,) * (a,,, * ...... * a

m+n) = a

1 * ...... * a

m+n.

Proof: We use induction on n. That is, we will show that the statement is true for n = 1. Then,assuming that it is true for n � 1, we will prove it for n.

If n = 1, our definition above gives us

(a, .......* a,) * a,,, = a, *...... * am+1

.

Now, assume that

(a, *.......* a,) * (a,,, ........ am+n�1

) = a, * ....... a,,,,

Then

(a, * ......... * a) * (a,,, .......... a)

= (a1 *.........* a,) * ((a

m+1 * ..... * a

m+n�1) * a

m+n)

= ( (a, *......* a,,) * (a,,, *......* a,,,) ) * a, since * is associative

= (a1 ......... a

m+n�1) * a

m+n by induction

= a, * ....... a, by definition.

Hence, the result holds for all n.

We will use Theorem 2 quite often in this course, without explicitly referring to it.

Now that we have discussed binary operations let us talk about groups.


Unit 2: Groups

Notes2.2 Group

After understanding the concept of binary operations. Let us start defining group.

Definition: Let G be a non-empty set and * be a binary operation on G. We say that the pair(G, * ) is a group if

G 1) * is associative:�

G 2) G contains an identity element e for * , and

G 3) every element in G has an inverse in G with respect to *.

We will now give some examples of groups.

Example: Show that (Z, +) is a group, but (Z,.) is not.

Solution: + is an associative binary operation on Z. The identity element with respect to + is 0,and the inverse of any n Z is (�n). Thus, (Z, +) satisfies GI, G2 md G3. Therefore, it is a group.

Now, multiplication in Z is associative and 1 Z is the multiplicative identity. But does everyelement in Z have a multiplicative inverse? No. For instance, 0 and 2 have no inverses withrespect to �.� Therefore, (Z,.) is not a group.

Note that (Z,.) is a semigroup since it satisfies GI. So, there exist semigroups that aren�t groups!

Actually, to show that (G, *) is a group it is sufficient to show that * satisfies the followingaxioms.

G 1�) * is associative.

G 2�) e G such that a * e = a a G .

G 3�) Given a G, 3 b G such that a * b = e.

What we are saying is that the two sets of axioms are equivalent. The difference between themis the following:

In the first set we need to prove that e is a two-sided identity and that the inverse b of any a Gsatisfies a * b = e and b * a = e. In the second set we only need to prove that e is a one-sided identityand that the inverse b of any a G only satisfies a * b = e.

In fact, these axioms are also equivalent to

G 1�) * is associative.

G 2") 3 e G such that i * a = a a G.

G 3") Given a G, 3 b G such that b * a = e.

Clearly, if * satisfies GI, G2 and G3, then it also satisfies Gl�, G2' and G3'. The following theoremtells us that if * satisfies the second set of axioms, then it satisfies the first set too.

Theorem 3: Let (G, * ) satisfy Gl�, G2� and G3�. Then e * a = a a G. Also, given a G, if bGsuch that a * b = e, then b * a = e. Thus, (G, *) satisfies G1, G2 and G3.

To prove this theorem, we need the following result.

Lemma 1: Let (G, * ) satisfy Gl�, G2' and G3'. If a G such that a *a = a, then a = e.

Proof: By G3' we know that 3 b G such that a * b = e.


Abstract Algebra

Notes Now (a * a) * b = a * b = e.

Also a * (a * b) = a * e = a. Therefore, by Gl�, a = e.

Now we will use this lemma to prove Theorem 3.

Proof of Theorem 3: G1 holds since G1 and G1' are the same axiom. We will next prove that G3is true. Let a G. By G3' 3 b G such that a * b = e. We will show that

b * a = e. Now,

(b * a) * (b * a) = (b * (a * b)) * a = (b * e) * a = b * a .

Therefore, by Lemma 1, b * a = e. Therefore, G3 is true.

Now we will show that G2 holds. Let a G. Then by G2', for a G, a * e = a. Since G3 holds, b G such that a * b = b * a = e. Then

e * a = (a * b) * a = a * (b * a) = a * e = a .

That is, G2 also holds.

Thus, (G, *) satisfies G1, G2 and G3.

Example: Let G = { ±1, i }, i = 1. Let the binary operation be multiplication. Showthat (G) is a group.

Solution: The table of the operation is

1 �1 i �i

1 1 �1 i �i

�1 �1 1 �i i

i i �i �1 1

�i �i i 1 �1

This table shows us that a.l = a a G. Therefore, 1 is the identity element. It also shows us that(G) satisfies G3. Therefore, (G) is a group.

Note that G = {1, x, x2, x3}, where x = i.

2.2.1 Abelian Group

Definition: If (G, *) is a group, where G is a finite set consisting of n elements, then we say that(G, *) is a Finite group of order n. If G is an infinite set, then we say that (G,*) is an infinite group.

If * is a commutative binary operation we say that (G, *) is a commutative group, or an abeliangroup. Abelian groups are named after the gifted young Norwegian mathematician Niels HenrikAbel.

Now let us discuss an example of a non-commutative (or non-abelian) group. Before doing thisexample recall that an m x n matrix over a Set S is a rectangular arrangement of elements of S inm rows and n columns.

NoteIf A =

a b

c d

then ad-bc is called the determinant of A and is written as det

A or |A|


Unit 2: Groups

Notes

Example: Let G be the set of all 2 x 2 matrices with non-zero determinant. That is,

a bG a, b, c, d r, ad � bc 0

c d

Consider G with the usual matrix multiplication, i.e, for

a p qb ap br aq bsA and p in G, A.P

c r sd cp dr cq ds

Show that (G, �) is a group.

Solution: First we show that . is a binary operation, that is, A, P G A.P G.

Now,

det (A.P) = det A. det P # 0. since det A 0, det P 0.

Hence, A.P G for all A, P in G.

Note det (AB) = (det A) (det B)

We also know that matrix multiplication is associative and 1 0

0 1

is the multiplicative identity.

Now, for A = a b

c d

in G. the mamx

d b1 0*ad bc ad bcB is such that det B 0 and AB .

c a 0 1ad bcad bc ad bc

Thus, B = A�1. (Note that we have used the axiom G3' here, and not G3.) This shows that the set ofall 2 × 2 matrices over R with non-zero determinant forms a group under multiplication. Since

1 2 0 1 2 1and

3 4 1 0 4 3

0 1 1 2 3 4

1 0 3 4 1 2

we see that this group is not commutative.

And now another example of an abelian group.

Example: Consider the set of all translations of R2,

2 2a,b a,bT {f : R R |f (x, y) (x a, y b) for some fixed a, b R}

Note that each element fa,b

in T is represented by a point (a, b) in R2. Show that (T, o) is a group,where o denotes the composition of functions.


Abstract Algebra

Notes Solution: Let us see if o is a binary operation on T.

Now fa,b

o fc,d

(x , y) = fa,b

(x + c, y + d) = (x + c + a, y + d + b)

= fa+c, b+d

(x,y) for any (x, y) R2.

fa,b

ofc,d

= fa+c, b+d

T.

Thus, o is a binary operation on T.

Now, fa,b

o f0,0

= fa,b

fa,b

T.

Therefore, f0,0

is the identity element.

Also, fa,b

o f-a, -b

= f0,0

fa,b

T.

Therefore, f -a,�

is the inverse of fa,b

T.

Thus, (T, o) satisfies G1�, G2' and G3', and hence is a group.

Note that fa,b

o fc,d

= fc,d

o fa,b

fa,b

, fc,d

T. Therefore, (T, o) is abelian.

2.3 Properties of Groups

Before understanding the properties of group lets first give notational conventions.

Convention: Let us, we will denote a group (G, *) by G, if there is no danger of confusion. Wewill also denote a * b by ab, for a, b G, and say that we are multiplying a and b. The letter e willcontinue to denote the group identity.

Now let us discuss a simple theorem.

Theorem 4: Let G be a group. Then

(a) (a�1)�1 = a for every a G.

(b) (ab)�1 = b�1 a�1 for all a, b G.

Proof: (a) By the definition of inverse,

(a-1)-1 (a-1) = e = (a-1) (a-1)-1.

But, a a-1 = a-1 a = e also, Thus, by Theorem 1 (b), (a-1)-1 = a.

(b) For a, b G, ab G. Therefore, (ab)-1 G and is the unique element satisfying (ab) (ab)-1

= (ab)-l (ab) = e.

However, (ab) (b�1 a-1) = ((ab) b-1) a-1

= (a (b b-1) a-1)

= (a e) a-1

= aa-1

=e

Similarly, (b-1 a-1) (ab) = e.

Thus, by uniqueness of the inverse we get (ab)�1 = b�1 a-1.

Note that, for a group G, (ab)-1 = a-1 b-1 a, b G only if G is abelian.

You know that whenever ba = ca or ab = ac for a, b, c in R*, we can conclude that b = C. That is,we can cancel a. This fact is true for any group.


Unit 2: Groups

NotesTheorem 5: For a, b, c in a group G,

(a) ab = ac b = c. (This is known as the left cancellation law.)

(b) ba = ca b = c. (This is known as the right cancellation law.)

Proof: We will prove (a) and leave you to prove (b).

(a) Let ab = ac. Multiplying both sides on the left hand side by a�I " G, we get

a-1 (ab) = a�1 (ac)

(a�1 a) b = (a-1 a) c

eb = ec, e being the identity element.

b = c.

Now let us prove another property of groups.

Theorem 6: For elements a. b in a group G, the equations ax = b and ya = b have unique solutionsin G.

Proof: We will first show that these linear equations do have solulions in G, and then we willshow that the solutions are unique.

For a, b G, consider a-1 b G. We find that a(a-1 b) = (aa-1) b = eb = b. Thus, a-1 b satisfies theequation ax = b, i.e., ax = b has a solution in G.

But is this the only solution? Suppose x1, x

2 are two solutions of ax = b in G. Then ax, = b = ax

2. By

the left cancellation law, we get xl = x

2. Thus, a-1 b is the unique solution in G.

Similarly, using the right cancellation law, we can show that ba-1 is the unique solution ofya = b in G.

Now we will illustrate the property given in Theorem 6.

Example: Consider 2 3 1 5

A ,B1 2 0 4

in GL, (R)

Find the solution of AX = B.

Solution: From Theorem 6, we know that X = A-1 B. Now,

A�1 = 2 3

1 2

A�1B = 2 2

X.1 3

In the next example we consider an important group.

Example: Let S be a non-empty set. Consider (S) with the binary operation ofsymmetric difference A, given by

A B=(A\B) (B\A) A, B (S).

Show that ((S), A) is an abelian group. What is the unique solution for the equation Y A=B?

Solution: A is an associative binary operation. This can be seen by using the facts that

A\B=A BC, (A B)C = AC BC, (A B)C = AC BC


Abstract Algebra

Notes and that and are commutative and associative. A is also commutative since A A B

= B A A, B (S).

Also, is the identity element since A A = A A (S).

Further, any element is its own inverse, since A A A = A (S).

Thus, ((S). A) is an abelian group.

For A, B in ((S), A) we want to solve Y A A = B. But we know that A is its own inverse. So, byTheorem 6, Y = B A A-1 = B A A is the unique solution. What we have also proved is that (B A A)A A = B for any A, B in (S).

Definition: Let G be a group. For a G, we define

(i) a0 = e.

(ii) an = an-1 . a, if n > 0

(iii) a�n = (a-1)n, if n > 0.

n is called the exponent (or index) of the integral power an of a.

Thus, by definition a1 = a, a2 = a . a, a3 = a2 . a, and so on.

Notes When the notation used for the binary operation is addition, an becomesna. For example, f a any a Z,

na = 0 if a = 0,

na = a + a+ ... +a (n times) if n > 0,

na = (�a) + (�a) + .... + (�a) (�n times) if n < 0.

Let us now prove some laws of indices for group elements.

Theorem 7: Let G be a group. For a " G and m, n " Z,

(a) (an)-1 = a-n = (a-1)n, (b) am, an = am+n, (c) (am)n = amn

Proof: We prove (a) and (b), and leave the proof of (c) to you.

(a) If n = 0, clearly (an)-l = a-0 = (a-1)n.

Now suppose n > 0. Since aa-1 = e, We see that

e = en = (aa�1)n

= (aa-1) (aa-1) .... (aa-1) (n times)

= an (a-1)a, since a and a-1 compute.

(an)-1 = (a-1)n.

Also, (a-1)n = a-n, by definition.

(an)-1 = (a-1)a = a-a when n > 0.

If n < 0, then (�n) > 0 and

(an)-1 = [a-(n)]-1


Unit 2: Groups

Notes= [(a-a)-1}-1, by the case n > 0

= ad

Also, (a-1)n = (a-l)-(-n)

= [(a-1)-1]-n, by the case n > 0

= a*.

So, in this case too,

(an)-1 = a-n = (a-1)n.

(b) If m = 0 or n = 0, then am+n = am . an. Suppose m 0 and n 0.

Case 1 (m > 0 and n > 0): We prove the proposition by induction on n.

If n = 1, then am . a = am+1, by definition.

Now assume that am . an-1 = am+n-1

Then, am . an = am(an-1 . a) = (am . an-1) a = am+n-1 . a = am+n . Thus, by the principle of induction, (a) holdsfor all m > 0 and n > 0.

Case 2 (m < 0 and n < 0): Then (-m) 0 and (-n) > 0. Thus, by Case 1, a-n . a-m = a-(n+m) = adwn). Takinginverses of both the sides and using (a), we get,

gm+n = (a-n . a-m)-1 = (a-m)-1 . (a-n)-1 = am . an.

Case 3 (m > 0, n < 0 such that m + n 0): Then, by Case 1, am+n . a-n = am. Multiplying both sides onthe right by an = (a-n)-1, we get am+n = am. an.

Case 4 ( m > 0, n < 0 such that m+n < 0): By Case 2, a-m. am+n= an. Multiplying both sides on the leftby am = (a-m)-1, we get am+n = am . an.

The cases when m < 0 and n > 0 are similar to Cases 3 and 4. Hence, awn = am . an for all a G andm, n Z.

2.4 Different Types of Group

2.4.1 Integers Modulo n

Consider the set of integers, Z, and n N. Let us define the relation of congruence on Z by : a iscongruent to b modulo n if n divides a-b. We write this as a b (mod n). For example, 4 1(mod 3), since 3 | (4-1).

Similarly, (-5) 2(mod 7) and 30 0 (mod 6).

is an equivalence relation, and hence partitions Z into disjoint equivalence classes called

congruence classes modulo n. We denote the class containing r by r.

Thus, r ={ m Z | m r (modn) }.

So an integer m belongs to r for some r, 0 r < n, iff n | (r-m), i.e., iff r�m = kn, for some k Z.

r = { r + kn | k

Now, if m n, then the division algorithm says that m = nq + r for some q, r Z, 0 r < n. Thatis, m r (mod n), for some r = 0, .,..., n-1. Therefore, all the congruence classes modulo n are

n0,1, ....., n 1. Let Z {0,1,2, ....., n 1}. We define the operation + on Zn by a b a b.


Abstract Algebra

NotesIs this operation well defined? To check this, we have to see that if a b c d in Z

n, then

a b c d.

Now, a b (mod n) and c d (mod n). Hence, there exist integers kl and k

2 such that a � b = k

1n

and c � d = k2n. But then (a + c) � (b + d) = (a � b) + (c � d) = (k

1 + k

2)n.

a c b d .

Thus, + is a binary operation on Z.

For example, 2 2 0 in Z4 since 2 + 2 = 4 and 4 0(mod 4).

Now, let us show that (Zn, +) is a commutative group.

(i) a b a b b a b a a, b Zn, i.e.,

addition is commutative in Z

(ii) a b c b c a (b c)

na b c (a b) c (a b) c a, b, c Z ,

i.e., addition is associative in Z,.

(iii) a + 0 = a = 0 + a a Z,, i.e., 0 is the identity for addition,

(iv) ~ or ; Zn, nn Z such that a + n � a = n = 0 = n � a+ a.

Thus, every element a in Zn has an inverse with respect to addition.

The properties (i) to (iv) show that (Zn, +) is an abelian group.

Actually we can also define multiplication on Zn by a . b = ab. Then, nb = b a a, b Z . Also,

n(a b)c a(b c) a, b, c Z . Thus, multiplication in Zn is a commutative and associative binary

operation.

Z, also has a multiplicative identity, namely, 1.

But (Z,, .) is not a group. This is because every element of Zn, for example Q, does not have a

multiplicative inverse.

But, suppose we consider the non-zero elements of Zn, that is, *

nZ ,. . Is this a group? For example,

*4Z 1, 2, 3 is not a group because * is not even a binary operation on *

4Z , since

* *42 . 2 0 Z . But Z ,. is an abelian group for any prime p.

2.4.2 The Symmetric Group

We will now discuss the symmetric group briefly. In Next Unit we will discuss this group inmore detail.


Unit 2: Groups

NotesLet X be a non-empty set. We have seen that the composition of functions defines a binaryoperation on the set F(X) of all functions from X to X. This binary operation is associative.

IX, the identity map, is the identity in F(X).

Now consider the subset S(X) of F(X) given by

S(X) = {f F(X) | f is bijective }.

So f S(X) iff f-l : X X exists. Remember that f o f-1 = f-1 . o f = IX. This also shows that f-1 S(X).

Now, for all f, g in S(X),

(g o f) o (f-1 o g-1) = IX = (f-1 o g-1) o (g o f), i.e., g o f S(X).

Thus, o is a binary operation on S(X).

Let us check that (S(X), o) is a group.

(i) o is associative since (fog) o h = f o (g o h) f, g, h E S(X).

(ii) IX is the identity element because f o I

X = I

X o f f S(X).

(iii) f-1 is the inverse off, for any f S(X).

Thus, (S(X), o) is a group. It is called the symmetric group on X.

If the set X is finite, say X = (1, 2, 3 ...... n), then we denote S(X) by S,, and each f S, is called apermutation on n symbols.

Suppose we want to construct an element f in Sn. We can start by choosing f(1). Now, f(1) can

be any one of the n symbols 1,2, ..... n. Having chosen f(l), we can choose f(2) from the set{ l,2 ........ n } \ { f(l) }, i.e., in (n � 1) ways. This is because f is 1 � 1. Inductively, after choosing f(i),we can choose f(i + l) in (n � i) ways. Thus, f can be chosen in (1 × 2 × .... × n) = n ! ways, i.e.,S

n contains n ! elements.

For our convenience, we represent f S, by

1 2 .......... n

..........f(1) f(2) f(n)

For example, 1 2 3 4

2 4 3 1

represents the function f : (1, 2. 3. 4) {1. 2, 3. 4) : f (1) = 2, f(2) = 4,

f (3) = 3, f (4) = 1. The elements in the top row can be placed in any order as long as the order ofthe elements in the bottom row is changed accordingly.

Thus, 2 1 3 4

4 2 3 1

also represents the same function f.

Definition: We say that f Sn is a cycle of length r if there are x,, ...., x, in X = { 1, 2, ....., n) such that

f(xi) = x

i+1 for 1 i r � 1, f(x

r) = x

1 and f(t) = t for t # x, ..., x,. In this case f is written as (x

1 .... x,).

For example, by f = (2 4 5 10) S10

, we mean f (2) = 4, f (4) = 5, f (5) = 10, f (10) = 2 and f (j) = j forj # 2, 4, 5, 10.

i.e., 1 2 3 4 5 6 7 8 9 10

f1 4 3 5 10 6 7 8 9 2


Abstract Algebra

Notes

Note In the notation of a cycle, we don�t mention the elements that are left fixed

by the permutation. Similarly, the permutation 2 5

5 3

is the cycle (1 2 5 3 4 ) in S5.

Now let us see how we calculate the composition of two permutations. Consider the followingexample in S

5.

° =1 2 3 4 5 1 2 3 4 5

2 5 4 3 1 5 3 4 1 2

=

1 2 3 4 5

(1) (2) (3) (4) (5)

=

1 2 3 4 5

(5) (3) (4) (1) (2)

=1 2 3 4 5

(2 4).1 4 3 2 5

since 1, 3 and 4 are left fixed.

And now let us talk of a group that you may be familiar with, without knowing that it is agroup.

2.4.3 Complex Numbers

In this sub-section we will show that the set of complex numbers forms a group with respect toaddition. Some of you may not be acquainted with some basic properties or complex numbers.

Consider the set C of all ordered pairs (x, y) of real numbers. i.e.. we take C = R × R. Defineaddition (+) and multiplication (.) in C as follows:

(x1, y

1) + (x

2, y

2) = (x

1 + x

2, y

1 + y

2) and

(x1, Y

1) . (x

2, y

2) = (x

1 x

2 � y

1 y

2, x

1y

2 + x

2y

1)

for (x1 . y

1) and (x

2, y

2) in C.

This gives us an algebraic system (C, +, .) called the system of complex numbers. We mustremember that two complex numbers (x

1, y

1) and (x

2, y

2) are equal iff x

1 = x

2 and y

1 = y

2.

You can verify that + and . are commutative and associative.

Moreover,

(i) (0, 0) is the additive identity.

(ii) for (x, y) in C, (�x, �y) is its additive inverse.

(iii) (1, 0) is the multiplicative identity.

(iv) if (x, y) (0, 0) in C, then either x2 > 0 or y2 > 0.


Unit 2: Groups

NotesHence, x2 + y2 > 0. Then

2 2 2 2

yx(x,y). ,

x y x y

= 2 2 2 2 2 2 2 2

( y) yx xx . y . , x . y

x y x y x y x y

= (1, 0)

Thus, 2 2 2 2

yx,

x y x y

is the multiplicative inverse of (x, y) in C.

Thus, (C, +) is a group and (C*,.) is a group. (As usual, C* denotes the set of non-zero complexnumbers.)

Now let us see what we have covered in this unit.

Self Assessment

1. For a binary operation * on S and (a, b) S × S, we denote *(a, b) by ...............

(a) a * b (b) (a, b)*

(c) ab* (d) ba*

2. Binary operations associates a ............... of S to every ordered pair of elements of S.

(a) same element (b) different element

(c) unique element (d) single set element

3. Suppose their exist a, b S such that S * a = e = a * s and S * b = e = b * s, e being the identityelement for *, then

(a) a = b (b) b�1 = a

(c) a�1 = b (d) a2 = b

4. For x R, if b is inverse of x, the you should have b x = 1. Then x-1 is equal to

(a) 2 � x (b) x � 2

(c) 2�1 x (d) 2 + x�1

5. ............... are named after the gifted young Norwegian mathematician Niels Henrik Abel.

(a) Abelian group (b) Sub group

(c) Normal group (d) Cyclic group

2.5 Summary

Here we discussed various types of binary operations.

Also defined and given examples of groups.

We proved and used the cancellation laws and laws of indices for group elements.

In this unit we discussed the group of integers modulo n, the symmetric group and thegroup of complex numbers.


Abstract Algebra

Notes 2.6 Keywords

Binary Operation: A binary operation on S is always closed on S, but may not be closed on a subsetof S.

Abelian Group: If (G, *) is a group, where G is a finite set consisting of n elements, then we saythat (G, *) is a Finite group of order n. If G is an infinite set, then we say that (G,*) is an infinitegroup.


1. Obtain the identity element, if it exists, for the operations 31 2 4

32 4 1

.

2. For x R, obtain x-1 (if it exists) for each of the operations 31 2 4

32 4 1

.

3. Show that (Q, +) and (R, +) are groups.

4. Calculate (1 3) ° (1 2) in S3.

5. Write the inverse of the following in S3:

(a) (1 2)

(b) (1 3 2)

Show that (1 2) ° (1 3 2)�1 (1 2)-1 ° (1 3 2)-1. (This shows that in Theorem 4(b) we can�t write(ab)-1 = a-1b-1.)






www.math.niu.edu

www.maths.tcd.ie/






Unit 3: Subgroups

NotesUnit 3: Subgroups

CONTENTS

Objectives

Introduction

3.1 Subgroups

3.2 Properties of Subgroups

3.3 Cyclic Groups

3.4 Summary

3.5 Keywords



Objectives


Define subgroups

Explain the intersection, union and product of two subgroups

Describe structure and properties of cyclic groups

Introduction

In the last unit, you have studied about the algebraic structures of integers, rational numbers,real numbers and complex numbers. You have got an idea that, not only is Z Q R C, butthe operations of addition and multiplication coincide in these sets. In the present unit, you willgo through more examples of subsets of groups which are groups in their own right. Suchstructures are rightfully named subgroups. We will discuss some of their properties also. Wewill see some cases in which we obtain a group from a few elements of the group. In particular,we will study cases of groups that can be built up by a single element of the group.

3.1 Subgroups

In the previous unit, you have already read the concept of group. You also noted that group (Z+),(Q+) and (R+) are the member of a bigger group (C+) complex number. These all groups thatcontained in bigger group are not just subsets but groups.

All these are examples of subgroup. Lets define subgroup.

Definition: Let (G,*) be a group. A non-empty subset H of G is called a subgroup of G if

(i) a * b H a, b H, i.e., * is a binary operation on H,

(ii) (H,*) is itself a group.


Abstract Algebra

Notes So, by definition, (Z,+) is a subgroup of (Q,+), (R,+) and (C,+).

Now, if (H,*) is a subgroup of (G,*), can the identity element in (H,*) be different from theidentify element in (G,*)? Let us see. If h is the identity of (H,*), then, for any a H, h * a = a * h= a. However, a H G. Thus, a * e = e * a = a, where e is the identity in G. Therefore, h * a =e * a.

By right cancellation in (G,*)w,e get h = e.

Thus, whenever (H, *) is a subgroup of (G,*), e H.

Remark 1: (H,*) is a subgroup of (G, *) if and only if

(i) e H,

(ii) a, b H a * b H,

(iii) a H a-1 H.

We would also like to make an important remark about notation here.

Remark 2: If (H,*) is a subgroup of (G,*), we shall just say that H is a subgroup of G, provided thatthere is no confusion about the binary operations. We will also denote this fact by H G.

Now let us first discuss an important necessary and sufficient condition for a subset to be asubgroup.

Theorem 1: Let H be a non-empty subset of a group G. Then H is a subgroup of G iff a, b H ab-1 H.

Proof: Firstly, let us assume that H G. Then, by Remark l, a, b H a, b-1 H

ab-1 H.

Conversely, since H , a H. But then, aa-1 = e H.

Again, for any a H, ea-1 = a-1 H.

Finally, if a, b H, then a, b-1 H. Thus, a (b-1)-1 = ab H, i.e.,

H is closed under the binary operation of the group.

Therefore, by Remark 1, H is a subgroup.

Note A subgroup of an abelian group is abelian.

Example: Consider the group (C*.,). Show that

S = { z C | |z| = 1 } is a subgroup of C*.

Solution: S , since 1 S. Also, for any z1, z

2 S,

|z1 z

2�1| = |z

1| |z

2�1| = |z

1|

2

11.

|z |

Hence, zl z

2-1 S. Therefore, by Theorem 1, S C*.


Unit 3: Subgroups

Notes

Example: Consider G = M2×3

(C), the set of all 2 × 3 matrices over C. Check that (G,+) is anabelian group. Show that

0 a bS a, b, c C

0 0 c

is a subgroup of G.

Solution: We define addition on G By

a c p q r a p c rb b q

e s t u e td f d s f u

You can see that + is a binary operation on G. 0 = 0 0 0

0 0 0

is the additive identity and a c

d f

is the inverse of a cb

ed f

G.

Since, a + b = b + a a, b C, + is also abelian.

Therefore, (G,+) is an abelian group.

Now, since O S, S . Also for

0 a 0 eb d, S

0 0 c 0 0 f

, we see that

0 a 0 e 0b d a d b e,

0 0 c 0 0 0 0f c f

S.

= S I G.

Example: Consider the set of all invertible 3 × 3 matrices over R, GL3 (R). That is,

A GL3(R) iff det (A) 0. Show that SL

3 (R) = (A GL

3(R) | det(A) = I ) is a subgroup of (GL

3(R),.).

Solution: The 3 × 3 identity matrix is in SL3(R). Therefore, SL

3(R) .

Now, for A, B SL3(R),

det (AB-1) = det (A) det(B-1) = 1

1,det(B)

since det (A) = 1 and det (B) = I.

AB-1 SL3(R)

SL3(R) I GL

3(R).

Example: Any non-trivial subgroup of (Z, +) is of the form mZ, where m N andmZ = { mt | t Z) = { 0, m, ± 2m, ± 3m,....... ).

Solution: We will first show that mZ is a subgroup of Z. Then we will show that if H is asubgroup of Z, H # {0}, then H = mZ, for some m N.


Abstract Algebra

Notes Now, 0 mZ. Therefore, mZ . Also, for mr, ms mZ, mr-ms = m(r-s) mZ.

Therefore, mZ is a subgroup of Z.

Note that m is the least positive integer in mZ.

Now, let H (0) be a subgroup of Z and S = { i | i > 0, i H).

Since H # {0), there is a non-zero integer k in H. If k > 0, then k S. If k < 0, then (-k) S, since(�k) H and (�k) > 0.

Hence, S .

Clearly, S N. Thus, by the well-ordering principle S has a least element, say s. That is, s is theleast positive integer that belongs to H.

Now sZ H. Why? Well, consider any element st sZ.

If t = 0, then st = 0 H.

If t > 0, then st = s + s + ..... + s (t times) H.

If t < 0, then st = (�s) + (�s) + ....+ (�s) (�t times) H.

Therefore, st H t Z. That is, sZ H.

Now, let m H. By the division algorithm m = ns + r for some n, r Z, 0 r < s. Thus, r = m � ns.But H is a subgroup of Z and m, ns H. Thus, r H. By minimality of s in S, we must have r = 0,i.e., m = ns. Thus, H sZ.

So we have proved that H = sZ.

You know that the polar form of a non-zero complex number z C is z = r (cos + i sine), wherer = | z | and is an argument of z. Moreover, if

1, is an argument of z

1 and

2 that of z

2. then

l

+ 2 is an argument of z

l z

2. Using this we will try to find the nth roots of 1, where n N.

If z = r (cos + i sin !) is an nth root of 1, then zn = 1.

Thus, by De Moivre�s theorem,

1 = zn = rn (cos n + i sin n), that is,

cos (0) + i sin (0) = rn (cos n + i sin n). ................. (1)

Equating the modulus of both the sides of (1). we get rn = 1, i.e., r =l.

On comparing the arguments of both sides of (1), we see that 0 + 2nk (k Z) and n arearguments of the same complex number. Thus, n can take any one of the values 2k, k Z. Does

this mean that as k ranges over Z and ranges over 2nk

n we get distinct nth roots of 1? Let us

find out. Now, cos 2nk

n + i sin

2zkn

= 2 m 2 m

cos i sinn n

if and only if 2nk 2nm

2ntn n

for

some t Z. This will happen iff k = m + nt, i.e., k m (mod n). Thus, corresponding to every r

in Z, we get an nth root of unity, z = 2 r 2 r

cos i sin , 0 r n;n n

and these are all the nth roots

of unity.

For example, If n = 6, we get the 6th roots of 1 as z0, z

l, z

2, z

3, z

4 and z

5, where zj =

2 j 2 jcos i sin , j 0,1,2,3,4,5.

6 6

In Figure 3.1 you can see that all these lie on the unit circle (i.e.,

the circle of radius one with centre (0, 0)). They form the vertices of a regular hexagon.


Unit 3: Subgroups

NotesFigure 3.1: 6th Roots of Unity

Now, let = 2 2

cos i sinn n . Then all the nth roots of 1 are 1, , 2, .......,n-1, since

j 2 j 2 jcos i sin for 0 j n 1

n n

(using De Moivre�s theorem).

Note is the Greek letter omega.

Example: Show that Un (C�, .).

Solution: Clearly, U, # 0. Now, let 1, j U,.

Then, by the division algorithm, we can write i + j = qn + r for q, r Z, 0 r n � 1. But theni . j = i+j= qn+r = (n)q. r = r U,, since n = 1. Thus, U, is closed under multiplication.

Finally, if i U,, then 0 n � i n � 1 and i . n�i = n = 1; i.e., n-i is the inverse of oi for all1 i < n. Hence, U

n is a subgroup of C*.

Note that Un, is a finite group of order n and is a subgroup of an infinite group, C*. So, for every

natural number n we have a finite subgroup of order n of C*.

Before ending this we will introduce you to a subgroup that you will use off and on.

Definition: The centre of a group G, denoted by Z(G), is the set

z(G) = {G G | xg = gx x G}.

Thus, Z(G) is the set of those elements of G that commute with every element of G.


Abstract Algebra

Notes For example, if G is abelian, then Z(G) = G.

We will now show that Z(G) G.

Theorem 2: The centre of any group G is a subgroup of G.

Proof: Since e Z(G), Z(G) . Now,

a Z(G) ax =xa x G.

x = a-1 xa x G, pre-multiplying by a-1.

xa-1 = a-1 x x G, post-multiplying by a-1.

a-1 Z(G).

Also, for any a, b Z(G) and for any x G, (ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x (ab).

ab Z(G).

Thus, Z(G) is subgroup of G.

3.2 Properties of Subgroups

After discussing the term subgroup let us start understanding the important properties ofsubgroup.

Theorem 3: Let G be a group, H be a subgroup of G and K be a subgroup of H. Then K is asubgroup of G.

Proof: Since K H, K and ab-1 K a, b K. Therefore, K G.

Let us discuss at subgroups of Z, in the context of Theorem 3.

Example: In earlier example we have seen that my subgroup of Z is of the form mZ forsome m N. Let mZ and kZ be two subgroups of Z. Show that rnZ is a subgroup of kZ iff k | m.

Solution: We need to show that mZ kZ k | m. Now mZ kZ m mZ kZ m kZ m = kr for some r Z k | m.

Conversely, suppose k | m.

Then, m = kr for some r Z. Now consider any n mZ, and let t Z such that n = mt.

Then n = mt = (kr) t = k(rt) kZ.

Hence, mZ kZ.

Thus, mZ kZ iff k | m.

Theorem 4: If H and K are two subgroups of a group G, then H K is also a subgroup of G.

Proof: Since e H and e K, where e is the identity of G, e H K.

Thus, H K .

Now, let a, b H K. By Theorem 1 , it is enough to show that ab-1 H K. Now, since a, b

H, ab-1 H. Similarly, since a, b K, ab-1 K. Thus, ab-1 H K.

Hence, H K is a subgroup of G.


Unit 3: Subgroups

NotesThe whole argument of Theorem 4 remains valid if we take a family of subgroups instead of justtwo subgroups. Hence, we have the following result.

Theorem 4 (a): If i i 1H

is a family of subgroups of a group G, then i

i 1

H

Hi is also a subgroup

of G.

Now question arises that does the union of two or more subgroup is again a subgroup. Lets see

its true or not. Consider the, two subgroups 2Z and 3Z of Z. Let S = 2Z 32. Now, 3 32 S,2 22 S, but 1 = 3 � 2 is neither in 2Z nor in 3Z. Hence, S is not a subgroup of (Z, +). Thus, if A

and B are subgroups of G, A B need not be a subgroup of G. But, if A B, then A B = B is a

subgroup of G. The next exercise says that this is the only situation in which A B is a subgroupof G.

Let us now see what we mean by the product of two subsets of a group G.

Definition: Let G be a group and A, B be non-empty subsets of G.

The product of A and B is the set AB = { ab | a A, b B).

For example, (2Z) (3Z) = { (2m) (3m) | m, n Z)

= { 6mn | m, n Z }

= 62.

In this example we find that the product of two subgroups is a subgroup. But is that always so?Consider the group

S3 = {I, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}, and its subgroups H = { I, (1 2) } and K = { I, (1 3)).

(Remember, (1 2) is the permutation 1 2 3

2 1 3

and (1 2 3) is the permutation 1 2 3

2 3 1

.

Now HK = {I × I , I × ( 1 3 ), (1 2) × I,(1 2) × (1 3)}

= { I, (1 3), (1 2), (1 3 2) }

HK is not a subgroup of G, since it is not even closed under composition. (Note that (1 3) (1 2)= (1 2 3) HK.)

So, when will the product of two subgroups be a subgroup? The following result answers thisquestion.

Theorem 5: Let H and K be subgroups of a group G. Then HK is a subgroup of G if and only if HK= KH.

Proof: Firstly, assume that HK G. We will show that HK = KH; Let hk HK. Then

(hk)-1 = k-1 h-1 HK, since HK G.

Therefore, k-1 h-1 = h1 k

l for some h

i H, k

1 K. But then hk = (k-1 h-1)-1 = k

1-1 h

1-1 KH. Thus,

HK KH.

Now, we will show that KH HK. Let kh KH. Then (kh)-1 = h-1 k-1 HK. But HK G. Therefore,((kh)-1)-1 HK, that is, kh HK. Thus, KH HK.

Hence, we have shown that HK = KH.

Conversely, assume that HK = KH. We have to prove that HK G. Since e = e2 HK, HK .Now, let a, b HK. Then a = hk and b = h

1 k

1 for some h, h

l H and k, k

1 K.


Abstract Algebra

Notes Then ab-1 = (hk) (k1

-1 h1

-1) = h [ (kk1

-1) h1

-1].

Now, (kk1

-1) h1

-1 KH = HK, Therefore, 3 h2k

2 HK such that (kk

1-1)h

1-1 = h

2k

2.

Then, ab-1 = h(h2k

2) = (hh

2)k

2 HK.

Thus, by Theorem 1, HK G.

The following result is a nice corollary to Theorem 5.

Corollary: If H and K are subgroups of an abelian group G, then HK is a subgroup of G.

3.3 Cyclic Groups

Let us understand the meaning of cyclic group.

Let G be any group and S a subset of G. Consider the family F of all subgroups of G that containS, that is,

F = { H | H G and S H } .

We claim that F . Why? Doesn�t G F? Now, by Theorem 4'(a), H F

H

is a subgroup of G.

Note that

(i)H F

S H.

(ii)H F

H

is the smallest subgroup of G containing S. (Because if K is a subgroup of G

containing S, then K F. Therefore, H F

H

K.)

These observations lead us to the following definition.

Definition: If S is a subset of a group G, then the smallest subgroup of G containing S is called thesubgroup generated by the set S, and is written as <S>.

Thus, <S> = { H | H G, S H }.

If S = , then <S> = {e).

If <S> = G, then we say that G is generated by the set S, and that S is a set of generators of G.

If the set S is finite, we say that G is finitely generated.

We will give an alternative way of describing <S>. This definition is much easier to work withthan the previous one.

Theorem 6: If S is a non-empty subset of a group G, then

<S> = k1 2 nn n1 2 k i 1 ka a ..... a |a S for 1 i k, n , ..., n Z .

Proof: Let A = k1 2 nn n1 2 k i 1 ka a ..... a |a S for 1 i k, n , ..., n Z .

Since a1,.., a

k S <S> and <S> is a subgroup of G, 1n

1a <S>

i = 1, ...., k. Therefore, k1 2 nn n1 2 ka a ..... a <S>, i.e., A <S>.


Unit 3: Subgroups

NotesNow, let us see why <S> A. We will show that A is a subgroup containing S. Then, by thedefinition of <S>, it will follow that <S> A.

Since any a S can be written as a = a1, S A.

Since S , A .

Now let x, y A. Then x = k1 2 nn n1 2 ka a ..... a

y = 1 2 rm m m1 2 rb b .... b a

j , b

j S for 1 i k, 1 j r.

Then xy-1 = k1 2 1 2 r1nn n m m m

1 2 k 1 2 ra a ..... a b b .... b

= k1 2 1 1nn n m m1 2 k 1 1a a ..... a b .... b A.

Thus, by Theorem 1, A is a subgroup of G. Thus, A is a subgroup of G containing S. And hence,<S> A.

This shows that <S> = A.

Note that, if (G, +) is a group generated by S, then any element of G is of the form n1 a

l + n

2 a

2 +

..... + nr a

r, where a

1, a

2,....., a, S and n

1, n

2, ....., n

r, Z.

For example, Z is generated by the set of odd integers S = { ± 1, f3, ± 5, ......). Let us see why. Letm Z. Then m = 2r

s where r 0 and s S. Thus, m <S>. And hence, <S> = Z.

Definition: A group G is called a cyclic group if G = < {a) > for some a E G. We usually write< {a) > as < a >.

Note that < a > = { an | n " Z ).

A subgroup H of a group G is called a cyclic subgroup if it is a cyclic group. Thus, < (12) > is acyclic subgroup of S

3 and 22 = <2> is a cyclic subgroup of Z.

We would like to make the following remarks here.

Remark: (i) If K G and a K, then < a > K. This is because < a > is the smallest subgroup ofG containing a.

(ii) All the elements of < a > = { an | n Z) may or may not be distinct. For example, takea = (12) S

3.

Then < (1 2) > = { I, (1 2)), since (1 2)2 = I, (1 2)3 = (1 2), and so on.

We will now prove a nice property of cyclic groups.

Theorem 7: Every cyclic group is abelian.

Proof: Let G = < a > = { an | n Z). Then, for any x, y in G, there exist m, n Z such that x = am, y= an. But, then, xy = am. an am+n = an+m = an. am = yx. Thus, xy = yx for all x, y in G.

That is G is abelian.

Note Theorem 7 says that every cyclic group is abelian. But this does not mean thatevery abelian group is cyclic.


Abstract Algebra

Notes

Example: Consider the set K4 = {e, a, b, ab] and the binary operation on K

4 given by the

table.

× e a b ab

e e a b ab

a a e ab b

b b ab e a

ab ab b a e

The table shows that (K4, .) is a group.

This group is called the Klein 4-group, after the pioneering German group theorist Felix Klein.

Example: Show that K4 is an abelian but not cyclic.

Solution: From the table we can see that K4 is an abelian. If it were cyclic, it would have to be

generated by e, a, b or ab. Now, < e > = {e}. Also, a1 = a, a2 = e, a�= a, and so on.

Therefore, < a > = { e, a }. Similarly, = { e, b } and < ab > = { e, ab).

Therefore, K4 can�t be generated by e, a, b or ab.

Thus, K4 is not cyclic.

Theorem 8: Any subgroup of a cyclic group is cyclic.

Proof: Let G = < x > be a cyclic group and H be a subgroup.

If H = {e}, then H = < e >, and hence, H is cyclic.

Suppose H {e}. Then 3 n Z such that xn H, n 0. Since H is a subgroup, (xn)-1 = x-n H.Therefore, there exists a positive integer m (i.e., n or -n) such that x

m H. Thus, the set S = {t N

| xt H) is not empty. By the well-ordering principle S has a least element, say k. We will showthat H = < xk >.

Now, <xk > H, since xk H.

Conversely, let xn be an arbitrary element in H. By the division algorithm n = mk + r where m,r Z, 0 r k � 1. But then xr = xn-mk = xn . (xk)-m H, since xn, xk H. But k is the least positiveinteger such that xk H. Therefore, xr can be in H only if r = 0. And then, n = mk and xn = (xk)m < xk >. Thus, H < xk >. Hence, H = < xk >, that is, H is cyclic.

Now, Theorem 8 says that every subgroup of a cyclic group is cyclic. But the converse is not true.That is, we can have groups whose proper subgroups are all cyclic, without the group beingcyclic.

Consider the group S3, of all permutations on 3 symbols. Its proper subgroups are

A = 

B = <(1 2)>

C = <(1 3)>

D = <(2 3)>

E = <(1 2 3)>.

As you can see, all these are cyclic. But, you know that S3 itself is not cyclic.


Unit 3: Subgroups

NotesNow we state a corollary to Theorem 8, in which we write down the important point made in theproof of Theorem 8.

Corollary: Let H {e} be a subgroup of < a >. Then H = < an >, where n is the least positive integersuch that an H.

Self Assessment

1. Subgroup of an abelian group is ..................

(a) normal (b) abelian

(c) cyclic (d) homomorphism

2. If H be a non-empty subset of a group G. Then H is a subgroup of G if a, b H, then

(a) ab-1 H (b) a-1b H

(c) ab-1 H (d) a H, b�1 H

3. .................. is a Greek letter omega.

(a) (b)

(c) (d)

4. Let G be a group, H be subgroup of G and be subgroup of H then k is a .................. of G.

(a) normal group (b) cyclic group

(c) abelian group (d) subgroup

5. Let H and k be subgroups of a group G. Then KH = HK then (HK)�1 = ..................

(a) k-1, h-1 Hk (b) hk kh Hk

(c) x/h Hk (d) H/k Hk

3.4 Summary

In this unit we have covered the following points.

Here we discussed the definition and examples of subgroups.

The intersection of subgroups is a subgroup.

The union of two subgroups H and K is a subgroup if and only if H K or K H.

The product of two subgroups H and K is a subgroup if md only if HK = KH.

The definition of a generating set.

A cyclic group is abelian, but the converse need not be true.

Any subgroup of a cyclic group is cyclic, but the converse need not be true.

3.5 Keywords

Subgroup: Let (G,*) be a group. A non-empty subset H of G is called a subgroup of G if

(i) a * b H a, b H, i.e., * is a binary operation on H,

(ii) (H,*) is itself a group.


Abstract Algebra

Notes Cyclic Groups: Let G be any group and S a subset of G. Consider the family F of all subgroups ofG that contain S, that is, F = { H | H G and S H }.


1. Find all cyclic subgroups of Z24

×.

2. In Z20

×, find two subgroups of order 4, one that is cyclic and one that is not cyclic.

(a) Find the cyclic subgroup of S7 generated by the element (1, 2, 3)(5, 7). (b) Find a

subgroup of S7 that contains 12 elements. You do not have to list all of the elements if you

can explain why there must be 12, and why they must form a subgroup.

3. In G = Z21

×, show that

4. H = { [x]21

| x 1 (mod 3) } and K = { [x]21

| x 1 (mod 7) } are subgroups of G.

5. Let G be an abelian group, and let n be a fixed positive integer. Show that

N = { g in G | g = an for some a in G } is a subgroup of G.

6. Suppose that p is a prime number of the form p = 2n + 1.

(a) Show that in Zp

× the order of [2]p is 2n.

(b) Use part (a) to prove that n must be a power of 2.

7. In the multiplicative group C× of complex numbers, find the order of the elements

= 2 2

i2 2

and = 2 2

i2 2

.

8. Let K be the following subset of GL2 (R).

K = a b

d a,c 2b,ad bc 0c d

9. Show that K is a subgroup of GL2 (R).

10. Compute the centralizer in GL2 (R) of the matrix

1 0

1 0

.

11. Let G be the subgroup of GL2 (R) defined by G =

m bm 0 .

0 1

12. Let A = 1 1

0 1

and B = 1 0

.0 1

Find the centralizers C(A) and C(B), and show that C(A)

C(B) = Z(G), where Z(G) is the center of G.


1. (b) 2. (a) 3. (a) 4. (d) 5. (a)


Unit 3: Subgroups

Notes3.7 Further Readings





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Abstract Algebra

Notes Unit 4: Lagrange's Theorem

CONTENTS

Objectives

Introduction

4.1 Cosets

4.2 Lagrange's Theorem

4.3 Summary

4.4 Keywords



Objectives


Discuss the cosets of a subgroup

Explain the partition a group into disjoint cosets of a subgroup

Prove and explain Lagrange's theorem

Introduction

In the last unit, you have studied about the subgroup and different properties of subgroups. Inthis unit, you will learn the concept of cosets and also see how a subgroup can partition a groupinto equivalence classes. You can use cosets to prove a very useful result about the number ofelements in a subgroup. In the present era, this elementary theorem is known as Lagrange'stheorem, though Lagrange proved it for subgroups of S only. Let us understand these conceptswith the help of examples and theorem.

4.1 Cosets

First of all we will discuss cosets. Cosets means the product of two subset of a particular group.

In a case when one of the subsets consists of single element only, we will go through a situationi.e.,

H(x) = {hx | h H}.

where H is a subgroup of G and x G, we will denote H{x} by Hx.

Definition: Let H be a subgroup of a group G, and let x G. We call the set

Hx = {hx | h H}

a right coset of H in G. The element x is a representative of Hx.

We can similarly define the left coset


Unit 4: Lagrange's Theorem

NotesxH={xh | h H} .

Note that, if the group operation is +, then the right and left cosets of H in (G,+) represented byx G are

H + x = { h + x | h H} and x + H = { x + h | h H }, respectively.

Example: Show that H is a right as well as a left coset of a subgroup H in a group G.

Solution: Consider the right coset of H in G represented by e, the identity of G. Then

He = { he | h H } = ( h | h H } = H .

Similarly, eH = H.

Thus, H is a right as well as left coset of H in G.

Example: What are the right cosets of 4Z in Z?

Solution: Now H = 4Z = { ......, -8, - 4, 0, 4, 8, 12, ..... }

The right cosets of H are

H + 0 = H, using Example.

H+ 1 ={ ....., �11, �7, �3, 1, 5, 9, 13, .... )

H + 2 = { ....., �10, �6, �2, 2, 6, 10, 14, .... )

H + 3 = { ....., �9, �5, �1, 3, 7, 11, 15, .... )

H + 4 = { ....., �8, �4, 0, 4, 8, 12 ,.....) = H

Similarly, you can see that H + 5 = H + 1, H + 6 = H + 2, and so on.

You can also check that H � 1 = H + 3, H � 2 = H + 2, H � 3 = H + l, and so on.

Thus, the distinct right cosets are H, H + 1, H + 2 and H + 3.

In general, the distinct right cosets of H (= nZ) in Z are H, H + 1, ....., H + (n � 1). Similarly, thedistinct left cosets of H (= nZ) in Z are H, 1 + H, 2 + H, ....., (n � 1) + H.

After understanding the concept of cosets. Let us discuss some basic and important properties ofcosets.

Theorem 1: Let H be a subgroup of a group G and let x, y G.

Then

(a) X HX

(b) Hx = H x H.

(c) Hx = H xy�1 H.

Proof: (a) Since x = ex and e H, we find that x Hx.

(b) Firstly, let us assume that Hx = H. Then, since x Hx, x H.

Conversely, let us assume that x H. We will show that Hx H and H Hx. Now any elementof Hx is of the form hx, where h H. This is in H, since h H and x H. Thus, Hx H. Again, leth H. Then h = (hx-l) x Hx, since hx-1 H.

H HX.


Abstract Algebra

Notes H = Hx.

(c) Hx = Hy Hxy- l = Hyy-1 = He = H xy�1 H, by (b)

Conversely, xy-1 H Hxy-l = H = Hxy-1y = Hy Hx = Hy.

Thus, we have proved (c).

The properties listed in Theorem 1 are not only true for right cosets.

Note Along the lines of the proof of Theorem 1, we can prove that if H is a subgroup ofG and x, y G, then

(a) x xH.

(b) xH = H x H.

(c) x H = yH x-1y H.

Example: Let G = Sg = {I, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} and H be the cyclic subgroup of Ggenerated by (1 2 3). Obtain the left cosets of H in G.

Solution: Two cosets are

H = { I, (1 2 3). (1 3 2)) and

(1 2)H = { (1 2), (1 2) × (1 2 3), (1 2) × (1 3 2))

= { (1 2) , (2 3), (1 3)}

For the other cosets you can apply Theorem 1 to see that

(1 2)H = (2 3)H = (1 3)H and

(1 2 3)H = H = (1 3 2)H.

Example: Consider the following set of 8, 2 × 2 matrices over C, Q, = (± I, ± A , ± B, ± C},where

1 0 0 1 0 i i 0I , A , B , C and i 1.

0 1 1 0 i 0 0 i

You can check that the following relations hold between the elements of Q8:

I2 = I , A2 = B2 = C2 = �I,

AB = C = �BA, BC = A = �CB, CA = B = �AC.

Therefore, Q8 is a non-abelian group under matrix multiplication.

Show that the subgroup H = < A > has only two distinct right cosets in Q8.

Solution: H = < A > = { I, A, A2, A3} = {I, A, �I, �A},

since A4 = I, A5 = A, and so on.

Therefore, HB = {B, C, �B, �C} , using the relations given above.

Using Theorem 1 (b), we see that



NotesH = HI = HA = H(�I) = H(�A).

Using Theorem 1 (c), we see that

HB = HC = H(�B) = H(�C).

Therefore, H has only two distinct right cosets in Q8, H and HB.

We will now show that each group can be written as the union of disjoint cosets of any of itssubgroups. For this, first we define a relation on the elements of G.

Definition: Let H be a subgroup of a group G. We define a relation �~� on G by x � y iff x y-l H,where x, y G. Thus, from Theorem 1 we see that x � y iff Hx = Hy.

We will prove that this relation is an equivalence relation.

Theorem 2: Let H be a subgroup of a group G. Then the relation ~ defined by �x ~ y iff xy-1 H�is an equivalence relation. The equivalence classes are the right cosets of H in G.

Proof: We need to prove that ~ is reflexive, symmetric and transitive.

Firstly, for any x G, xx-1 = e H. x ~ x, that is, ~ is reflexive.

Secondly, if x ~ y for any x, y G, then xy-1 H.

(xy-1)-1 = yx-1 H. Thus, y ~ x. That is, ~ is symmetric.

Finally, if x, y, z G such that x ~ y and y ~ z, then xy-1 H and yz-1 H.

(xy-1) (yz-1) = x (y-1y)z-1 = xz-1 H. x ~ Z.

That is , ~ is transitive.

Thus, ~ is an equivalence relation.

The equivalence class determined by x G is

[ x l = { y G | y ~ x } = { y G | xy-1 H}.

Now, we will show that [x] = Hx. So, let y [X}. Then Hy = Hx, by Theorem 1. And since y Hy,y Hx.

Therefore, [x] Hx.

Now, consider any element hx of Hx. Then x(hx)-1 = xx-1h-1 = h-1 H.

Therefore, hx ~ x. That is, hx [x]. This is true for any hx Hx. Therefore, Hx [x].

Thus, we have shown that [x] = Hx.

Remark: If Hx and Hy are two right cosets of a subgroup H in G, then Wx = W

y or HXHY =

Note that what Theorem 2 and the remark above say is that any subgroup H of a group Gpartitions G into disjoint right cosets.

On exactly the same lines as above we can state that:

(i) any two left cosets of H in G are identical or disjoint, and

(ii) G is the disjoint union of the distinct left cosets of H in G.

4.2 Lagrange's Theorem

To understand this theorem first we have to define the order of a finite group, after that we willshow that the order of any subgroup divides the order of the group.


Abstract Algebra

Notes So let us start with a definition.

Definition: The order of a finite group G is the number of elements in G. It is denoted by o(G).

For example, o(S3) = 6 and o(A

3) = 3. Remember, A

3 = {I, (1 2 3), (1 3 2)}!

You can also see that o(Zn) = n. And, you know that o(S

n) = n!.

Now, let G be a finite group and H be a subgroup of G. We define a function f between the set ofright cosets of H in G and the set of left cosets of H in G by

f : { Hx | x G } { yH | y G } : f(Hx) = x�1H.

Definition: Let H be a subgroup of a finite group G. We call the number of distinct cosets of H inG the index of H in G, and denote it by | G : H |.

Thus, we see that | S3 : A

3 | = 2.

Note that, if we take H = {e}, then | G : {e} | = o(G), since {e}g = {g) g G and {e)g {e}g�if g g�.

Now let us look at the order of subgroups. In last unit you saw that the orders of the subgroupsof S

3 are 1, 2, 3 and 6. All these divide o(S

3) = 6. This fact is part of a fundamental theorem about

finite groups. Its beginnings appeared in a paper in 1770, written by the famous Frenchmathematician Lagrange. He proved the result for permutation groups only. The general resultwas probably proved by the famous mathematician Evariste Galois in 1830.

Theorem 3 (Lagrange): Let H be a subgroup of a finite group G. Then

o(G) = o(H) | G : H |. Thus, o(H) divides o(G) and | G : H | divides o(G).

Proof: You know that we can write G as a union of disjoint right cosets of H in G. So, if Hxl, Hx

2,

...., Hx, are all the distinct right cosets of H in G, we have

G = Hx1 Hx

2 ... Hx

r�( 1 )

and | G : H | = r.

We know that | Hx1 | = | Hx

2 | = ... = | Hx

r | = o(H).

Thus, the total number of elements in the union on the right hand side of ( I ) is

o(H) + o(H) + .....+ o(H) (r times) = r o(H).

Therefore, (1) says that o(G) = r o(H)

= o(H) | G : H |.

You will see the power of Lagrange�s theorem when we get down to obtaining all the subgroupsof a finite group.

For example, suppose we are asked to find all the subgroups of a group G of order 35. Then theonly possible subgroups are those of order 1, 5, 7 and 35. So, for example, we don�t need to wastetime looking for subgroups of order 2 or 4.

In fact, we can prove quite a few nice results by using Lagrange�s theorem. Let us prove someresults about the order of an element.

Definition: Let G be a group and g G. Then the order of g is the order of the cyclic subgroup< g >, if < g > is finite. We denote this finite number by o(g). If < g > is an infinite subgroup of G,we say that g is of infinite order.



NotesNow, let g G have finite order. Then the set {e, g, g2, ...} is finite, since G is finite. Therefore, allthe powers of g can�t be distinct. Therefore, gr = gs for some r > s. Then gr-s = e and r-s N. Thus,the set { t N | gt = e } is non-empty. So, by the well ordering principle it has a least element. Letn be the least positive integer such that gn= e.

Then

< g > = {e, g, g2, ...., gn-1}.

Therefore, o(g) = o(< g >) = n.

That is, o(g) is the least positive integer n such that gn = e.

Note If g (G, + ), then o(g) is the least positive integer n such that ng = e.

Now suppose g G is of infinite order. Then, for m 11. gm gn. (Because, if gm = gn, thengm-n = e, which shows that < g > is a finite group.) We will use this fact while provingTheorem 5.

Theorem 4: Let G be a group and g G be of order n. Then gm = e for some m N iff n | m.

Proof: We will first show that gm = e ! n (m.F or this consider the set S = { r Z | gr = e }.

Now, n S. Also, if a, b S, then ga = e = gb. Hence, ga-b = ga (gb)-1 = e. Therefore, a-b S. Thus,S Z.

Note So, from last unit, we see that S = nZ. Remember, n is the least positive integerin S!

Now if gm = e for some m N, then m S = nZ. Therefore, n / m.

Now let us show that n | m gm = e. Since n | m, m = nt for some t E Z. Then gm = gnt = (gn)t

= et = e. Hence, the theorem is proved.

We will now use Theorem 4 to prove a result about the orders of elements in a cyclic group.

Theorem 5: Let G = < g > be a cyclic group.

(a) If g is of infinite order then gm is also of infinite order for every m Z.

(b) If o(g) = n, then

o(gm) =

nm 1, ..., n 1.

n, m ((n,m) is the g.c.d. of n and m.)

Proof: (a) An element is of infinite order iff all its powers are distinct. We know that all thepowers of g are distinct. We have to show that all the powers of gm are distinct. If possible, let(gm)t = (gm)w, Then gmt = gmw, But then mt = mw, and hence, t = w. This shows that the powers of gm

are all distinct, and hence gm is of infinite order.

(b) Since o(g) = n, G = {e, g, ........ gn-1 ) . < gm >, being a subgroup of G, must be of finite order. Thus,

gm is of finite order. Let o(gm) = t. We will show that t =

n.

n, m


Abstract Algebra

Notes Now, gmt = (gm)t = e n | tm, by Theorem 4.

Let d = (n, m). We can then write n = n1d, m = m

1d, where (m

1, n

1,) = 1.

Then

1

n nn .

d n, m

Now, n | tm n | tm1d n

1d | tm

1d n | tm

1.

But (n, m1) = 1. Therefore, n

1 | t. ...... (1)

Also, (gm)n1 = gm1dn1 = gm1n = (gn)m1 = em1 = e.

Thus, by definition of o(gm) and Theorem 4, we have

t | n1. ...... (2)

(1) and (2) show that

1

m

nt n .

n, m

ni.e., o(g ) .

(n, m)

Using this result we know that o

12

12(4) in Z is 3.

12, 4

Theorem 6: Every group of prime order is cyclic.

Proof: Let G be a group of prime order p. Since p 1, a G such that a e. Theorem 4, o(a) | p.Therefore, o(a) = 1 or o(a) = p. Since a e, o(a) 2.

Thus, o(a) = p, i.e., o(< a >) = p. So, < a > G such that o(< a >) = b(G). Therefore, < a > = G, that is,G is cyclic.

Using Theorems 3 and 6, we can immediately say that all the proper subgroups of a group oforder 35 are cyclic.

Now let us look at groups of composite order.

Theorem 7: If G is a finite group such that o(G) is neither 1 nor a prime, then G has nontrivialproper subgroups.

Proof: If G is not cyclic, then any a G, a e, generates a proper non-trivial subgroup < a >.

Now, suppose G is cyclic, say C = < x >, where o(x) = mn (m, n 1).

Then, (xm)n = xmn = e. Thus, by Theorem 4, o(xm) n < o(G).

Thus <xm> is a proper non-trivial subgroup of G.

We first define the Euler phi-function, named after the Swiss mathematician Leonard Euler(1707-1783).

Definition: We define the Euler phi-function : N N as follows :

(1) = I, and

(n) = number of natural numbers < n and relatively prime to n, for n 2 2.



NotesFor example, (2) = I and (6) = 2 (since the only positive integers < 6 and relatively prime to 6 are1 and 5).

We will now prove a lemma, which will be needed to prove the theorem that follows it. Thislemma also gives us examples of subgroups of Z,, for every n 2 2.

Lemma: Let G = G {r Z,|(r,n) 1}, where n 2 2. Then (G,.) is a group, where r . s =

rs r, s Zn. Further, o(G) = (n).

Proof: We first check that G is closed under multiplication.

Now, r, s G (r, n) = 1 and (s, n) = 1 (rs, n) = 1.

rs G. Therefore, is a binary operation on G.

1 G, and is the identity.

Now, for any r G, (r, n) = 1.

ar + bn = 1 for some a, b Z

n | ar � 1

ar 1 (mod n),

a r 1.

1

a r

Further, a G, because if a and n have a common factor other than 1; then this factor will �divide

ar + bn = 1. But that is not possible.

Thus, every element in G has an inverse.

Therefore, (G, .) is a group.

In fact, it is the group of the elements of Z, that have multiplicative inverses.

Since G consists of all those r Z, such that r < n and (r, n) = I, o(G) = (n).

Lemma and Lagrange�s theorem immediately give us the following result due to themathematicians Euler and Pierre Fermat.

Theorem 8 (Euler-Fermat): Let a N and n 2 such that (a,n) = 1.

Then,, a(n) I (mod n).

Proof:

1. Leonhard Euler published a proof in 1789. Using modern terminology, one may prove thetheorem as follows: the numbers a which are relatively prime to n form a group undermultiplication mod n, the group G of (multiplicative) units of the ring Z/nZ. This grouphas (n) elements. The element a : = a (mod n) is a member of the group G, and the ordero(a) of a (the least k > 0 such that ak = 1) must have a multiple equal to the size of G.(The order of a is the size of the subgroup of G generated by a, and Lagrange�s theoremstates that the size of any subgroup of G divides the size of G.)

Thus for some integer M > 0, M· o(a) = (n). Therefore, a(n) = ao(a)·M = (ao(a))M = 1M = 1. Thismeans that a(n) = 1 (mod n).


Abstract Algebra

Notes 2. Another direct proof: If a is coprime to n, then multiplication by a permutes the residueclasses mod n that are co prime to n; in other words, (writing R for the set consisting of the(n) different such classes) the sets { x : x in R } and { ax : x in R } are equal; therefore, the twoproducts over all of the elements in each set are equal. Hence, P a(n)P (mod n) where P isthe product over all of the elements in the first set. Since P is coprime to n, it follows thata(n) 1 (mod n).

Self Assessment

1. If H is a subgroup of a group G, and let x G then Hx = {Hx | h H}. Thus Hx called as

(a) Left coset of H is G (b) Right coset of H is G

(c) Subgroup of G (d) Cyclic group of G

2. If H is .................., the alternating group on 3 symbols

(a) A1 (b) A3

(c) A4 (d) A5

3. Every group of prime order is ..................

(a) normal (b) cyclic

(c) subgroup (d) abelian

4. Two left cosets of a .................. are disjoint or identical

(a) normal (b) cyclic

(c) subgroup (d) abelian

5. af(n) = 1 (modn) where a, n N, (a, n) = 1 and

(a) n 2 (b) n 2

(c) n = 2 (d) n 2

4.3 Summary

The definition and examples of right and left cosets of a subgroup.

Two left (right) cosets of a subgroup are disjoint or identical.

Any subgroup partitions a group into disjoint left (or right) cosets of the subgroup.

The definition of the order of a group and the order of an element of a group.

The proof of Lagrange�s theorem, which slates that if H is a subgroup of a finite group G,then o(G) = o(H) | G : H |. But, if m | o(G), then G need not have a subgroup of order.

The following consequences of Lagrange�s theorem:

Every group of prime order is cyclic.

a(n) = 1 (mod n), where a, n N, (a,n) = 1 and n 2.



Notes4.4 Keywords

Coset: Let H be a subgroup of a group G, and let x G. We call the set Hx = {hx | h H} a rightcoset of H in G.

Lagrange: Let H be a subgroup of a finite group G. Then o(G) = o(H) | G : H |. Thus, o(H) divideso(G) and | G : H | divides o(G).


1. Obtain the left and right cosets of H = < (1 2) > in S3. Show that Hx xH for some x S

3.

2. Show that K = {I, �I} is a subgroup of Q8. Obtain all its right cosets in Q

8.

3. Let H be a subgroup of a group G. Show that there is a one-to-one correspondence betweenthe elements of H and those of any right or left coset of H.

(Hint: Show that the mapping f : H Hx : f(h) = hx is a bijection.)

4. Write Z as a union of disjoint cosets of 5Z.

5. Check that f is a bijection.

6. What are the orders of

(a) (1 2) S3, (b) I S

4,

(c)0 1

1 0

Q8, (d) 43 Z

(e) 1 R?


1. (b) 2. (b) 3. (d) 4. (c) 5. (a)






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Abstract Algebra

Notes Unit 5: Normal Subgroups

CONTENTS

Objectives

Introduction

5.1 Normal Subgroups

5.2 Quotient Groups

5.3 Summary

5.4 Keywords



Objectives


Discuss the concept of normal subgroups

Explain the Quotient group

Introduction

In earlier units, you have studied about the term subgroups and cosets. In this unit, we willdiscuss a special class of subgroups known as normal subgroups. You will also come to knowabout that the cosets of such a subgroup form a group with respect to a suitably defined operation.These groups are called quotient groups. After discussing these concepts, we will also discusssome examples related to this concept.

5.1 Normal Subgroups

In the last unit, you have studied about coset of a subgroup also introduced with a fact that leftcoset aH, not be same as the right coset Ha.

But this fact is true for certain subgroup for which Ha and aH represented by the same elementcoincide.

In group theory, these types of subgroup are very important and this type of a subgroup has aspecial name. This subgroup is referred to normal subgroup.

Definition: A subgroup N of a group G is called a normal subgroup of G if Nx = xN x G, and

we write this as N G.

For example, any group G has two normal subgroups, namely, {e} and G itself. Can you seewhy? Well, {e}x = {x} = x{e}, for any x G, and Gx = G = xG, for any x G.


Unit 5: Normal Subgroups

NotesLet us consider an example.

Example: Show that every �subgroup of Z is normal in Z.

Solution: As you know that if H is a subgroup of Z, then H = mZ, for some m Z. Now, for anyz Z,

H + z = { ..., - 3 m + z , - 2 m + z , - m + z , z , m + z , 2 m + z ,...}

= { ..., z - 3m, z � 2m, z � m, z, z + m, z + 2m,} {since + is commutative)

= z + H.

H Z.

Above example is a special case of the fact that every subgroup of a commutative group is anormal subgroup.

Let us now prove a result that gives equivalent conditions for a subgroup to be normal.

Theorem 1: Let H be a subgroup of a group G. The following statements are equivalent.

(a) H is normal in G.

(b) -lg Hg H g G.

(c) -lg Hg H g G.

Proof: We will show that (a) (b) (c) (a). This will show that the three statements areequivalent.

(a) (b) : Since (a) is true, Hg = gH g E G. We want to prove (b). For this, consider

g-1Hg for g E G. Let g-1hg g-1Hg.

Since hg Hg = gH, 3 h1 H such that hg = gh

1.

g-1hg = g-1gh1 = h

1 H

(b) holds.

Note g-1Hg = {g-1hg | h H }

(b) (c) : Now, we know that (b) holds, i.e., for g G, g-1Hg H. We want to show

that H g-1Hg. Let h H. Then

h = ehe = (g-1g) h (g-1g)

= g-1 (ghg-1) g

= g-1{ (g-1)-1 hg-1 } g g-1Hg, since (g-1)-1hg-1 (g-1)-1 H(g-1) H.

H g-1Hg.

g-1Hg = H g G.


Abstract Algebra

Notes (c) (a) : For any g G, we know that g-1Hg = H.

g(g-1Hg) = gH, that a, Hg = gH.

H G, that is, (a) holds.

Note Ha = Hb Hac = Hbc for any a, b, c G.

We would like to make the following remark about Theorem 1.

Remark: Theorem 1 says that 1H G g Hg H g e G. This does not mean that

g hg = h h H and g G .

For example, you have shown that A3 S

3. Therefore, by Theorem 1,

(1 2)-1 A3(1 2) = A

3. But, (1 2)-1(1 3 2) (1 2) (1 3 2). In fact, it is (1 2 3).

Theorem 2: Every subgroup of a commutative group is normal.

Proof: Let G be an abelian group, and H G. For any g G and h H, g-1hg = (g-1g)h = h E H.

g-1Hg H. Thus, H G.

Theorem 2 says that if G is abelian, then all its subgroups are normal. Unfortunately, the converseof this is not true. That is, there are non-commutative groups whose subgroups are all normal.We will give you an example after doing Theorem 3. Let us first look at another example of anormal subgroup.

Example: Consider the Klein 4-group, K4, given in table below. Show that both its

subgroups < a > and are normal.

× e a b ab

e e a b ab

a a e ab b

b b ab e a

ab ab b a e

Solution: Consider the table of the operation given in table. Note that a and b are of order 2.Therefore, a = a-1 and b = b-1. Also note that ba = ab.

Now, let H = < a > = {e, a}. We will check that H K4, that is, g-1hg H g K

4 and h E H.

Now, g-1eg = e E H g E K4.

Further, e-1ae = a H, a-1aa = a H, b-1ab = bab = a H and (ab)-1 a(ab)

= b-1 (a-1aa)b = bab = a E H.

H K4.

By a similar proof we can show that K4.



NotesIn above Example, both < a > and are of index 2 in K4. We have the following result about

such subgroups.

Theorem 3: Every subgroup of a group G of index 2 is normal in G.

Proof: Let N G such that | G : N | = 2. Let the two right cosets of N be N and Nx, and the twoleft cosets be N and yN.

Now, G = N yN, and x G. x N or x yN.

Since N Nx = , x N. x yN, xN = yN.

To show that N G, we need to show that Nx = xN.

Now, for any n N, nx G = N xN. Therefore, nx N or nx xN.

But nx N, since x N. nx xN.

Thus, Nx xN.

By a similar argument we can show that xN Nx.

Nx = xN, and N G.

We will use this theorem to show that, for any n 2, the alternating group A, is a normalsubgroup of S

n.

In fact, if you go back b, you can see that A4 S

4, since Lagrange�s theorem implies that

44 4

4

o(S ) 4!|S : A | 2.

o(A ) 12

Consider the quaternion group Q8, which we discussed earlier. It has the following 6 subgroups:

Ho = (I}, HI = {I, � I}, H

2 = (I, � I, A, � A), H

3 = {I, � I, B, � B},

H4 = {I, � I, C, � C), H

5 = Q

8.

You know that H0 and H

5 are normal in Q

8. Using Theorem 3, you can see that Hz, H

3 and H

4 are

normal in Qe.

By actual multiplication you can see that

11 1 8 1 8g H g H g Q . H Q .

Therefore, all the subgroups of Q8 are normal.

But, you know that Q8 is non-abelian (for instance, AB = � BA).

So far we have given examples of normal subgroups. Let us look at an example of a subgroupthat isn�t normal.

Example: Show that the subgroup < (1 2) > of S3 is not normal.

Solution: We have to find g S3 such that g-1(1 2)g < (1 2) >.

Let us try g = (1 2 3).

Then, g-1(l 2)g = (3 2 1) (1 2) (1 2 3)

= (3 2 1) (2 3) = (1 3) < (1 2)>


Abstract Algebra

Notes Therefore, < (1 2) > is not normal in S3.

In earlier unit we proved that if H I G and K H, then K I G. That is, �� is a transitive relation. But

� � is not a transitive relation. That is, if H N and N G, it is not necessary that H G.

Theorem 4: Let H and K be normal subgroups of a group G. Then H K G:

Proof: From Theorem 4 of Unit 3, you know that H K G. We have to show that

g-1xg H K x H K and g G.

Now, let x EH K and g G. Then x H and H G. g-1xg H.

Similarly, g-1xg K. g-1xg H K

Thus, H K G.

Example: Let G be the group generated by

{ x, y | x2 = e, y4 = e, xy = y-1x }

L e t H = < x > a n d K = < y > .

Then show that K G, H G and G = HK.

Solution: Note that the elements of G are, of the form xi $, where i = 0, 1 and j = 0, 1, 2, 3

G = {e, x, xy, xy2, xy3, y , y2, y3}

| G : K | = 2. Thus, by Theorem 3, K G.

Note that we can�t apply Theorem 2, since G is non-abelian (as xy = y-1x and y y-l).

Now let us see if H G.

Consider y-1xy. Now y-1xy = xy2, because y-1x = xy.

If xy2 H, then xy2 = e or xy2 = x. (Remember o(x) = 2, so that x-1 = x.)

Now, xy2 = e y2 = x-1 = x

y3 = xy = y-1x

y4 = x

e = x, a contradiction.

Again xy2 = x y2 = e, a contradiction..

Y-1xy = xy2 H, and hence, H G.

Finally, from the definition of G you see that G = HK.

The group G is of order 8 and is called the dihedral group, D8. It is the group of symmetries of a

square, that is, its elements represent the different ways in which two copies of a square can beplaced so that one covers the other. A geometric interpretation of its generators is shown infigure 5.1.

Take y to be a rotation of the Euclidean plane about the origin through ,2

and x the reflection

about the vertical axis.



NotesFigure 5.1: Geometric Representation of the Generators of D8

We can generalise D8 to the dihedral group

D2n

= < { x, y | x2 = e, yn = e, xy = y-1x} >, for n > 2.

5.2 Quotient Groups

Here we will use a property of normal subgroups to create a new group. This group is analogousto the concept of quotient spaces given in the Linear Algebra course.

Let H be a normal subgroup of a group G. Then gH = Hg for every g G. Consider the collectionof all cosets of H in G. (Note that since H G, we need not write �left coset� or �right coset; simply�coset� is enough.) We denote this set by G/H. Now, for x, y H, we have

(Hx) (Hy) = H(xH)y, using associativity,

= HHxy, using normality of H,

= Hxy, since HH = H because H is a subgroup.

Now, we define the product of two cosets Hx and Hy and G/H by (Hx)(Hy) = Hxy for all x, y in G.

As this definition seems to depend on the way in which we represent a coset. Let us discuss thisin detail. Suppose C

1 and C

2 are two cosets, say C

1 = Hx and C

2 = Hy. Then C

1C

2 = Hxy. But C

1 and

C2 can be written in the form Hx and Hy in several ways. So, you may ask : Does C

1C

2 depend on

the particular way of writing C1 and C

2?

In other words, if C1 = Hx = Hx

1 and C

2 = Hy = Hy

1, then is C

1C

2 = Hxy or is C

1C

2 = Hx

1y

1?

Actually, we will show you that Hxy = Hx1y

1, that is, the product of cosets is well-defined.

Since Hx = Hx1 and Hy = Hy

1, xx

1-1 H, yy

1-1 H.

(xy) (x1 y

1)-1 = (xy) (y

1-1 x

1-1) = x (yy

l-l) x

l-1

= x (yy1

-1)x-1 (xx1

-1) H, since xx1 H and H G

i.e:, (xy) (x1y

1)-1 H.

Hxy = Hx1y

1.

So, we have shown you that multiplication is a well-defined binary operation on G/H.

We will now show that (G/H,.) is a group.

Theorem 5: Let H be a normal subgroup of a group G and G/H denote the set of all cosets of Hin G. Then G/H becomes a group under multiplication defined by Hx . Hy = Hxy, x, y G. Thecoset H = He is the identity of G/H and the inverse of Hx is the coset Hx-1


Abstract Algebra

Notes Proof: We have already observed that the product of two cosets is a coset.

This multiplication is also associative, since

((Hx) (Hy)) (Hz) = (Hxy) (Hz)

= Hxyz, as the product in G is associative,

= Hx (yz)

= (Hx) ( Hyz)

= (Hx) ((Hy) (Hz)) for x, y, z G.

Now, if e is the identity of G, then Hx, He = Hxe = Hx and He. Hx = Hex = Hx for every x G.Thus, He = H is the identity element of G/H.

Also, for any x G, Hx Hx-1 = Hx x-1 = He = Hx-1x = Hx-1.Hx.

Thus, the inverse of Hx is Hx-1.

So, we have proved that G/H, the set of all cosets of a normal subgroup H in G, forms a groupwith respect to the multiplication defined by Hx.Hy = Hxy. This group is called the quotientgroup (or factor group) of G by H.

Note that the order of the quotient group G/H is the index of H in G. Thus, by Lagrange�stheorem you know that if G is a finite group, then

o(G)o(G/H)

o(H)

Also note that if (G, +) is an abelian group and H G, then H G. Further, the operation on

G/H is defined by (H + x) + (H + y) = H + (x + y).

Let us look at a few examples of quotient groups.

Example: Obtain the group G/H, where G = S3 and H = A

3 = {I, (1 2 3), (1 3 2)}.

Solution: Firstly, note bat A3 S

3, since |S

3 : A

3| = 2.

You know that G/H is a group of order 2 whose elements are H and (1 2) H.

Example: Show that the group Z/nZ is of order n.

Solution: The elements of Z/nZ are of the form a + nZ = {a + kn | k Z).

Thus, the elements of Z/nZ are precisely the congruence classes modulo n, that is, the elementsof Z

n.

Thus, Z/nZ = {0,1,2,....,n 1}.

o(Z/nZ) = n.

Note that addition in Z/nZ is given a + b = a + b

Definition: Let G be a group and x, y G. Then x-1y-1 xy is called the commutator of x and y. It isdenoted by [x, y].

The subgroup of G generated by the set of all commutators is called the commutator subgroupof G. It is denoted by [G, G].



NotesFor example, if G is a commutative group, then

x-1y-1xy = x-1xy-1y = e x, y G. [G, G] = {e}.

Theorem 6: Let G be a group. Then [G, G] is a normal subgroup of G. Further, G/[G, G] iscommutative.

Proof: We must show that, for any commutator x-1y-1xy and for any g G, g-1 (x-1y-1xy)g [G,G].

Now g-1(x-1y-1xy)g = (g-1xg)-1 (g-1yg)-1 (g-1xg) (g-1yg) [G, G].

[G, G] G.

For the rest of the proof let us denote [G, G] by H, for convenience.

Now, for x, y G,

HxHY = H y Hx Hxy = Hyx (xy) (yx)-1 H

Thus, since xy x-1 y-1 H x, y G, HxHy = HyHx x, y G. That is, G/H is abelian.

Note We have defined the quotient group G/H only if H G. But if H G we canstill define G/H to be the set of all left (or right) cosets of H in G. But, in this case G/H willnot be a group.

Remark: If H is a subgroup of G, then the product of cosets of H is defined only when H G.

This is because, if HxHy = Hxy x, y G, then, in particular,

Hx-1Hx = Hx-1x =He = H x G.

Therefore, any h H, x-1hx = ex-1hx Hx-1 Hx= H.

That is, X-1 Hx H for any x E G.

H G.

Self Assessment

1. Every subgroup of a ................... group is a normal subgroup.

(a) associative (b) large

(c) cyclic (d) commutative

2. g-1Hg = ................... if h H where H is a subgroup of G.

(a) g-1hg (b) gh-1g

(c) gh-1g-1 (d) ghg

3. The group of G is of order ................... is called dihedral group.

(a) 6 (b) 7

(c) 8 (d) 9

4. Every group of index ................... is normal.

(a) 4 (b) 5

(c) 3 (d) 2


Abstract Algebra

Notes 5. If H and k are normal subgroup of group G, then so is H ................... k.

(a) (b)

(c) = (d)

5.3 Summary

We discussed here:

The definition and examples of a normal subgroup.

Every subgroup of an abelian group is normal.

Every subgroup of index 2 is normal.

If H and K are normal subgroups of a group G, then so is H K.

The product of two normal subgroups is a normal subgroup.

If H N and N G, then H need not be normal in G.

The definition and examples of it quotient group.

If G is abelian, then every quotient group of G is abelian. The converse is not true.

The quotient group corresponding to the commutator subgroup is commutative.

The set of left (or right) cosets of H in G is a group if and only if H G.

5.4 Keywords

Normal Subgroup: A subgroup N of a group G is called a normal subgroup of 6 if Nx = xN x

G, and we write this as N 6.

Dihedral Group, D8: It is the group of symmetries of a square, that is, its elements represent the

different ways in which two copies of a square can be placed so that one covers the other.

Quotient Group: If C1 = Hx = Hx

1 and C

2 = Hy = Hy

1, then is C

1C

2 = Hxy or is C

1C

2 = Hx

1y

1?

Actually, we will show you that Hxy = Hx1y

1, that is, the product of cosets is well-defined.


1. Show that A3 S

3.

2. Consider the subgroup SL2(R) = {A E GL

2(R) | det(A) = 1} of GL

2(R). Using the facts that det

(AB) = det (A) det (B) and det (A�1) = 1

,det(A)

prove that SL2(R) GL

2(R).

3. Consider the group of all 2 × 2 diagonal matrices over R*, with respect to multiplication.How many of its subgroups are normal.

4. Show that Z(G), the centre of G, is normal in G. (Remember that Z(G) = {x G| xg = gxg G}).

5. Show that <(2 3)> is not normal in S3.



Notes6. Prove that if H G and K G, then HK G.

7. Prove that if H G , K G, then HK G.


1. (d) 2. (a) 3. (c) 4. (d) 5. (b)






www.math.niu.edu

www.maths.tcd.ie/






Abstract Algebra

Notes Unit 6: Group Isomorphism

CONTENTS

Objectives

Introduction

6.1 Homomorphisms

6.2 Isomorphisms

6.3 Group Isomorphism

6.4 Summary

6.5 Keywords



Objectives


Explain the concept of homomorphism

Describe Isomorphism

Introduction

In the last unit, you have studied about the normal groups and the concept of quotient group.In this unit, we will discuss various properties of those functions between groups which preservethe algebraic structure of their domain groups. These functions are called groupHomomorphisms. This term was introduced by the mathematician Klein in 1893. This conceptis analogous to the concept of a vector space homomorphism, as you studied in the earlier unit.In this unit, you will also get an idea about a very important mathematical idea�isomorphism.

6.1 Homomorphisms

Let us start our study of functions from one group to another with an example.

Consider the groups (Z, f) and ({1, - 1},). If we define

f : Z {1, �1} by f(n) = 1, if n is even

1, if n is odd,

then you can see that f(a + b) = f(a).f(b) a, b Z. What we have just seen is an example of a

homomorphism, a function that preserves the algebraic structure of its domain.

Definition: Let (G1, *

1) and (G

2,*

2) be two groups. A mapping f : G

1 � G

2 is said to be a group

homomorphism (or just a homomorphism), if

f(x *1 y) = f(x) *

2 f(y) x, y G

1.


Unit 6: Group Isomorphism

NotesNote that a homomorphism f from G1 to G

2 carries the product x *

1 y in G

1 to the product

f(x) *2 f(y) in G

2.

Note The word �homomorphism� is derived from two Greek words �homos�,meaning �link�, and �morphe�, meaning �form�.

Let us define two sets related to a given homomorphism.


1) and (G

2, *

2) be two groups and f : G

1 G

be a homomorphism. Then we

define

(i) the image of f to be the set

Im f = {f(x) 1 x G1}.

(ii) the kernel of f to be the set

Ker f = {x G1 | f(x) = e

2}, where e

2 is the identity of G

2.

Note that Im f G2, and Ker f = f-1 ({e

2} G

1.

Example: Consider the two groups (R, +) and (R*,.). Show that the map exp : (R, +) (R*,.): exp(r) = er is a group homomorphism. Also find Im exp and Ker exp.

Solution: For any r1, r

2 R, we know that 1 2 1 2r r r re e .e .

:. exp(rl + r

2) = exp(r

1).exp(r

2).

Hence, exp is a homomorphism from the additive group of real numbers to the multiplicativegroup of non-zero real numbers.

Now, Im exp = {exp(r) | r R} = {er | r R},

Also, Ker exp = {r R | er = l} = {0}.

Note that examples takes the identity 0 of R to the identity 1 of R*. example also carries theadditive inverse � r of r. to the multiplicative inverse of exp (r).

Example: Consider the groups (R, +) and (C, +) and define f : (C, +) (R, +) by f(x + iy) =x, the real part of x + iy. Show that f is a homomorphism. What are Im f and Ker f?

Solution: Take any two elements a + ib and c + id in C. Then,

f((a + ib) + (c + id)) = f((a + c) + i(b + d)) = a + c = f(a + ib) + f(c + id)

Therefore, f is a group homomorphism.

Imf = {f(x + iy) | x, y R } = { x | x R ) = R.

So, f is a surjective function

Ker f = { x + iy C | f ( x + iy ) = 0 } = { x + iy C | x = 0 }

= { iy | y E R }, the set of purely imaginary numbers.

Note that f carries the additive identity of C to the additive identity of R and ( � z) to � f(z), for anyz C.


Abstract Algebra

Notes In Examples 1 and 2 we observed that the homomorphisms carried the identity to the identityand the inverse to the inverse. In fact, these observations can be proved for any grouphomomorphism.

Theorem 1: Let f ; (G1, *

1) (G

2, *

2) be a group homomorphism.

Then

(a) f(el) = e

2, where e


1 and e


2.

(b) f(x-1) = [f(x)]-1 for all x in G1.

Proof: (a) Let x G1. Then we have e

1 * 1x = x. Hence,

f(x) = f(el *

1 x) = f(e

1) *

2 f(x), since f is a homomorphism. But

f(x) = e2 *

2 f(x) in G

2.

Thus, f(e1) *

2 f(x)= e

2 *

2 f(x).

So, by the right cancellation law in G2, f(e

1) =e

2.

(b) Now, for any x G1, f(x) *

2 f(x-1) = f(x *

1x-1) = f(e

1) = e

2.

Similarly, f(x-1) *2 f(x) = e

2.

Hence, f(x-1) = [f(x)]-1 x E G1.

Note that the converse of Theorem 1 is false. That is, if f : G1 G

2 is a function such that f(e

1) = e

2

and [f(x)]-1 = f(x-l) = f(x-1) x G1, then f need not be a homomorphism. For example, consider

f : Z Z : f(0) = 0 and ,

n 1 n 0f(n)

n 1 n 0

Since f(l + 1) f(1) + f(l), f is not a homomorphism. But f(e1) = e

2 and f(n) = - f(- n) n Z.

Let us look at a few more examples of homomorphisms now. We can get one important class ofhomomorphisms from quotient groups.

Example: Let H G. Consider the map p : G G/H : p(x) = Hx. Show that p is ahomomorphism. Also show that p is onto. What is Ker p?

Solution: For x, y G, p(xy) = Hxy = Hx Hy = p(x) p(y). Therefore, p is a homomorphism.

Now, Im p = { p(x) ( x G} = ( Hx | x G } = G/H. Therefore, p is onto.

Ker p = { x G | p(x) = H ). (Remember, H is the identity of G/H.)

= ( x G | H x = H }

= { x G | x H ), by Theorem 1 of Unit 4.

= H.

In this example you can see that Ker p G. You can also check that Theorem 1 is true here.

Example: Let H be a subgroup of a group G. Show that the map i : H G, i(h) = h is ahomomorphism. This function is called the inclusion map.



NotesSolution: Since i(h1 h

2) = h

1 h

2 = i(h

1)i(h

2) h

1, h

2 H. i is a group homomorphism.

Let us briefly look at the inclusion map in the context of symmetric groups. Consider twonatural numbers m and n, where m I n.

Then, we can consider Sm

Sn, where any Sm, written as

1 2 .... m,

....(1) (2) (m)

is considered to be the same as

1 2 .... m m 1 .... n

.... m 1 .... n(1) (2) (m)

S

n, i.e., (k) = k for m + 1 k n.

Then we can define an inclusion map i : Sm

Sr.

For example, under i : S3 S

4, (1 2) goes to

1 2 3 4.

2 1 4 4

We will now prove some results about homomorphisms. Henceforth, for convenience, we shalldrop the notation for the binary operation, and write a * b as ab.

Now let us look at the composition of two homomorphisms. Is it a homomorphism? Let us see.�

Theorem 2: If f : G1 G

2 and g : G

2 G

3 are two group homomorphisms, then the composite

map g . f : G1 G

3 is also a group homomorphism.

Proof: Let x. y G. Then

g o f(xy) = g(f(x : y))

= g(f(x)f(y), since f is a homomorphism.

= g(f(x)) g(f(g)), since g is a homomorphism.

= g o f(x).g o f(y).

Thus g, f is a homomorphism.

Theorem 3: Let f : G1 G

2 be a group homomorphism. Then

(a) Ker f is a normal subgroup of G1.

(b) Im f is a subgroup of G2.

Proof: (a) Since [(e1) = e

2, e

1 Ker f. Ker f .

Now, if x, y Ker f, then f(x) = e2 and f(y) = e

2.

f(xy-1) = f(x) f(y-1) = f(x) [f(y)]-1 = e2.

xy-1 Ker f.

Therefore, by Theorem 1 of Unit 3, Ker f G1. Now, for any y G

1 and x E Ker f,

f(y-1xy) = f(y-1) f(x)f(y)

= [f(y)]-1e2f(y), since f(x) = e

2 and by Theorem 1

= e2.

Ker f G1.


Abstract Algebra

Notes (b) Im f , since f(el) E Im f.

Now, let x2, y

2 Im f. Then x

1, y

1 G

1 such that f(x

1) = x

2 and f(y

1) = y

2.

x2y

2-1 = f(x

1) f(y

1-1) = f(x

1y

1-1) Imf.

Im f G2.

Using this result, we can immediately see that the set of purely imaginary numbers is a normalsubgroup of C.

Consider : (R, +) (C*,.) (x) = cos x + i sin x. We have seen that(x + y) = (x)(y), that is, is agroup homomorphism. Now (x) = 1 iff x = 2n for some n Z . Thus, by Theorem 3, Ker =(2 n | n Z) is a normal subgroup of (R. +). Note that this is cyclic, and 2n is a generator.

Similarly, Im is a subgroup of C*. This consists of all the complex numbers with absolute value1, i.e., the complex numbers on the circle with radius 1 unit and centre (0, 0).

You may have noticed that sometimes the kernel of a homomorphism is {e} and sometimes it isa large subgroup. Does the size of the kernel indicate anything? We will prove that ahomomorphism is 1 � 1 iff its kernel is {e}.

Theorem 4: Let f : G1 G

2 be a group homomorphism. Then f is injective iff Ker f = {e

1}, where

e1 is the identity element of the group G

1.

Proof: Firstly, assume that f is injective. Let x Ker f. Then f(x) = e2, i.e., f(x) = f(e

1). But f is 1 � 1.

x = e1.

Thus, Kerf = {e1}.

Conversely, suppose Ker f = {e1]. Let x, y G

1 such that

f(x) = f(y). Then f(xy-1) = f(x) f(y-1)

= f(x) [f(y)]-1 = e2.

xy-1 Ker f = {e1}. xy-1 = e1 and x = y.

This shows that f is injective.

So, by using Theorem 4, we can immediately say that any inclusion i : B G is 1-1, sinceKer i = {e}.

Let us consider another example.

Example: Consider the group T of translations of R2. We define a map : (R2 + ) (T, o) by4 (a, b) = f

a, b. Show that is an onto homomorphism, which is also 1-1.

Solution: For (a, b), (c, d) in R*, we have seen that

fa+c,h+d

= fa,b

o fc.d

((a, b) + (c, d)) = (a, b) o (c, d).

Thus, , is a homomorphism of groups.

Now, any element of T is -f(a, b). Therefore, is surjective. We now show that is also injective.

Let (R, b) Ker . Then $(a, b) = f0, 0

i.e., fa, b

= f0,0

fa,b

(0, 0) = f0, 0

(0, 0),

i.e., (a, b) = (0, 0)



Notes Ker = {1 (0, 0)}

is 1-1.

So we have proved that ! is a homomorphism, which is bijective.

And now let us look at a very useful property of a homomorphism that is surjective.

Theorem 5: Iff : G1 G

2 is an onto group homomorphism and S is a subset that generates G

1,

then f(S) generates G2.

Proof: We know that

G1 = < S > = { 1 2 mr r r

1 2 mx x ...x | m N, x1 S, r

1 Z for all i). We will show that

G2 = < f(S) >

Let x G2, Since f is surjective, there exists y G

1 such that f(y) = x. Since y G

1, y = 1 mr r

1 mx ...x ,

for some m N, where xi S and ri Z, 1 i m.

Thus, x = f (y) = 1 mr r1 mf x ...x

= 1 mr r1 m(f(x )) ... (f(x )) , since f is a homomorphism.

x < f(S) >. since f(x1) f(S) for every i = 1, 2, ..., r.

Thus G2 = < f(S) >.

So far you have seen examples of various kinds of homomorphisms-injective, surjective andbijective. Let us now look at bijective homomorphism in particular.

6.2 Isomorphisms

Definition: Let G1 and G

2 be two groups. A homomorphism f : G

1 G

2 is called an isomorphism

if f is 1-1 and onto.

In this case we say that the group G1 is isomorphic to the group G

2 or G

1 and G

2 are isomorphic.

We denote this fact by G1 G

2.

An isomorphism of a group G onto itself is called an automorphism of G. For example, theidentity� function IG : G G : I

G(x) = x is an automorphism.

Note The word �isomorphisms� is derived from the Greek word �ISOS� meaning�equal�.

Let us look at another example of an isomorphism.

Example: Consider the set G = a b

a, b R .ab

Then G is a group with respect to matrix addition.

Show that f : G C : f a b

ab

= a + ib is an isomorphism.


Abstract Algebra

Notes Solution: Let us first verify that f is a homomorphism. Now, for any two elements

a cb dand in G,

a cb d

a c a cb d b dr f (a c) i(b d)

a c a cb d (b d)

= (a + ib) + (c + id)

= a cb d

f fa cb d

Therefore, f is a homomorphism.

Now, Ker f = a a 0 0b b

f a ib 0 a 0,b 0a a 0 0b b

Therefore, by Theorem 4, f is 1-1.

Finally, since Im f = C. f is surjective

Therefore, f is an isomorphism.

We would like to make an important remark now.

Remark: If G1 and G

2 are isomorphic groups, they must have the same algebraic structure and

satisfy the same algebraic properties. For example, any group isomorphic to a finite group mustbe finite and of the same order. Thus, two isomorphic groups are algebraically indistinguishablesystems.

The following result is one of the consequences of isomorphic groups being algebraically alike

Theorem 6: If f : G H is a group isomorphism and Y G, then < x > < f (x)> ,

Therefore.

(i) if s is of finite order, then o(x) = o(f(s)).

(ii) if x is of infinite order, so is f(x).

Proof: If we restrict f to any subgroup K of G, we have the function f | K ; K f(K), Since f isbijective, sc is its restriction f | k ; k f(K) for any subgroup K of G. In particular, for any x G,< x > f(< x >) = < f(x) >,

Now if x has finite order, then o(x) = o(< x >) = o(< f(x) >) = o(f(x)), proving (i)

To prove (ii) assume hat x is of infinite order. Then < x > is an infinite group.

Therefore, < f(x) > is an infinite group, and hence, f(x) is of infinite order. So, we have proved (ii).

Example: Show that (R*,.) is not isomorphic to (C*,.).

Solution: Suppose they are isomorphic, and f : C* � R* is an isomorphism. Then

o(i) = o(f(i)), by Theorem 6, Now o(i) = 4. o(f(i)) = 4.



NotesHowever, the order of any real number different from ±1 is infinite: and o(1) = 1, o(�1) = 2.

So we reach a contradiction. Therefore, our supposition must be wrong. That is, R* and C* arenot isomorphic.

You must have noticed that the definition of an isomorphism just says that the map is bijective,i.e., the inverse map exists. It does not tell us any properties of the inverse. The next result doesso.

Theorem 7: If f : G1 G

2 is an isomorphism of groups, then f-1 : G

2 G

is also an isomorphism.

Proof: You know that f-1 is bijective. So, we only need to show that f-1 is a homomorphism. Let a�,b� G

2 and a = f-1 (a�), b = f-1 (b�). Then f(a)= a� and f(b)= b�.

Therefore, f(ab) = f(a) f(b) = a�b�. On applying f-1, we get

f-1 (a�b�) = ab = f-1 (a�) f-l (b�), Thus,

f-1 (a�b�) = f-1 (a�) f-1(b�) for all a�, b� G2.

Hence, f-1 is an isomorphism.

From Theorem 7 we can immediately say that

-1 : T R2 : -1(fa,b,

) = (a, b) is an isomorphism.

Theorem 7 says that if GI G

2, then G

2 G

1. We will be using this result quite often.

6.3 Group Isomorphism

In abstract algebra, a group isomorphism is a function between two groups that sets up a one-to-one correspondence between the elements of the groups in a way that respects the given groupoperations. If there exists an isomorphism between two groups, then the groups are calledisomorphic. From the standpoint of group theory, isomorphic groups have the same propertiesand need not be distinguished.

Definition and Notation

Given two groups (G, *) and (H, ), a group isomorphism from (G, *) to (H, ) is a bijectivegroup homomorphism from G to H. Spelled out, this means that a group isomorphism is abijective function f : G H such that for all u and v in G it holds that

f(u * v) = f(u) f(v).

The two groups (G, *) and (H, ) are isomorphic if an isomorphism exists. This is written:

(G, *) (H, )

Often shorter and more simple notations can be used. Often there is no ambiguity about thegroup operation, and it can be omitted:

G H

Sometimes one can even simply write G = H. Whether such a notation is possible withoutconfusion or ambiguity depends on context. For example, the equals sign is not very suitablewhen the groups are both subgroups of the same group.

Conversely, given a group (G, *), a set H, and a bijection f : G H, we can make H a group

(H, ) by defining

f(u) f(v) = f(u * v).

If H = G and = * then the bijection is an automorphism (q.v.)


Abstract Algebra

Notes Intuitively, group theorists view two isomorphic groups as follows: For every element g of agroup G, there exists an element h of H such that h �behaves in the same way� as g (operates withother elements of the group in the same way as g). For instance, if g generates G, then so does h.This implies in particular that G and H are in bijective correspondence. So the definition of anisomorphism is quite natural.

An isomorphism of groups may equivalently be defined as an invertible morphism in thecategory of groups, where invertible here means has a two-sided inverse.

Examples:

1. The group of all real numbers with addition, (, +), is isomorphic to the group of allpositive real numbers with multiplication (+, ×):

(, +) (+, ×)

via the isomorphism

f(x) = ex

(see exponential function).

2. The group of integers (with addition) is a subgroup of , and the factor group / isisomorphic to the group S1 of complex numbers of absolute value 1 (with multiplication):

/ S1

An isomorphism is given by

f(x + ) = e2x1

for every x in .

3. The Klein four-group is isomorphic to the direct product of two copies of 2 = /2

(see modular arithmetic), and can therefore be written 2 ×

2. Another notation is Dih

2,

because it is a dihedral group.

4. Generalizing this, for all odd n, Dih2n

is isomorphic with the direct product of Dihn and Z

2.

5. If (G, *) is an infinite cyclic group, then (G, *) is isomorphic to the integers (with theaddition operation). From an algebraic point of view, this means that the set of all integers(with the addition operation) is the �only� infinite cyclic group.

Some groups can be proven to be isomorphic, relying on the axiom of choice, but the proof doesnot indicate how to construct a concrete isomorphism.

1. The group (, +) is isomorphic to the group (, +) of all complex numbers with addition.

2. The group (*, ·) of non-zero complex numbers with multiplication as operation isisomorphic to the group S1 mentioned above.

Properties

The kernel of an isomorphism from (G, *) to (H, ), is always {eG} where e

G is the identity

of the group (G, *)

If (G, *) is isomorphic to (H, ), and if G is abelian then so is H.

If (G, *) is a group that is isomorphic to (H, ) [where f is the isomorphism], then if abelongs to G and has order n, then so does f(a).



Notes If (G, *) is a locally finite group that is isomorphic to (H, ), then (H, ) is also locally

finite.

We also go through that �group properties� are always preserved by isomorphisms.

Cyclic Groups

All cyclic groups of a given order are isomorphic to (n, +

n).

Let G be a cyclic group and n be the order of G. G is then the group generated by < x > = {e,x,...,xn � 1}. We will show that

G (n, +

n)

Define

: G n = {0, 1,..., n � 1}, so that (xa) = a. Clearly, is bijective.

Then

(xa . xb) = (xa+b) = a + b = (xa) + n(xb) which proves that G

n, +n.

Consequences

From the definition, it follows that any isomorphism f : G H will map the identity element ofG to the identity element of H,

f(eG) = e

H

that it will map inverses to inverses,

f(u�1) = [f(u)]�1

and more generally, nth powers to nth powers,

f(un) = [f(u)]n

for all u in G, and that the inverse map f�1 : H G is also a group isomorphism.

The relation �being isomorphic� satisfies all the axioms of an equivalence relation. If f is anisomorphism between two groups G and H, then everything that is true about G that is onlyrelated to the group structure can be translated via f into a true ditto statement about H, and viceversa.

Self Assessment

1. Let H be a subgroup of a Group G. Then H G, i(h) = h is a homomorphism. This functionis called the .................

(a) inclusion map (b) normal function

(c) cyclic (d) abelian

2. gof (x, y) is equal to:

(a) gof(x) . gof(y) (b) gof(x) + gof(y)

(c) gof(x-1) . gof(y-1) (d) gof(x) . gof(y-1)

3. Let f : G1 G

2 be a group homomorphism thus her f is a ................. of G.

(a) subgroup (b) normal

(c) cyclic (d) abelian


Abstract Algebra

Notes 6.4 Summary

We have discussed here:

The definition and example of a group homomorphism.

Let f : G1 G

2 be a group homomorphism. Then f(e

1) = e

2,

[f(x)]-1 = f(x-1), Im f G2, Ker f G

1.

A homomorphism is 1-1 iff its kernel is the trivial subgroup.

The definition and examples of a group isomorphism.

Two groups are isomorphic iff they have exactly the same algebraic structure.

The composition of group homomorphisms (isomorphisms) is a group homomorphism(isomorphism).

6.5 Keywords

Homomorphism: Homomorphism is derived from two Greek words �homos�, meaning �link�,and �morphe�, meaning �form�.

Inclusion Map: Let H be a subgroup of a group G. Show that the map i : H G, i(h) = h is ahomomorphism. This function is called the inclusion map.


1. Show that f : (R*,.) (R, 4) : f(x) = inx, the natural logarithm of x, is a group homomorphism.Find Ker f and Im f also.

2. Is f : (GL3(R)

3,) (w*,.) : f(A) = det(A) a homomorphism? If so, obtain Ker f and Im f.

3. Define the natural homomorphism p from S3 to S

3/A

3. Does (1 2) E Ker p? Does (1 2) E

Im p?

4. Let S = [z C| |z| = 1}.

Define f : (R, +) (S,.) L f(x) = eInx, where n is a fixed positive integer. Is f a homomorphism?If so find Ker f.


1. (a) 2. (a) 3. (b)







Notes


www.math.niu.edu

www.maths.tcd.ie/






Abstract Algebra

Notes Unit 7: Homomorphism Theorem

CONTENTS

Objectives

Introduction

7.1 Fundamental Theorem of Homomorphism

7.2 Automorphisms

7.3 Summary

7.4 Keywords



Objectives


Discuss fundamental theorem of homomorphism

Explain the concept of automorphism

Introduction

After understanding the concept of isomorphisms. Let us prove some result about the relationshipbetween homomorphisms and quotient groups. The first result is the Fundamental Theorem ofHomomorphism for groups. It is called �fundamental� because a lot of group theory dependsupon this result. This result is also called the first isomorphism theorem.

7.1 Fundamental Theorem of Homomorphism

Theorem 1 (Fundamental Theorem of Homomorphism): Let G1 and G

2 be two groups and f : G

1

G2 be a group homomorphism. Then

G1/Ker f Im f.

In particular, if f is onto, then G1/Ker f G

2.

Proof: Let Ker f = H. Note that H G1. Let us define the function

: G1/H Im f : (Hx) = f(x).

At first glance it seems that the definition of depends on the coset representative. But wewill show that if x, y G

1 such that Hx = Hy, then (Hx) = (Hy). This will prove that is a

well-defined function.

Now, Hx = Hy xy-1 H = Ker f f(xy-1) = e2, the identity of G

2.

f(x)[f(y)]-1 = e2 f(x) = f(y).

(Hx) = (HY).


Unit 7: Homomorphism Theorem

NotesTherefore, y is a well-defined function,

Now, let us check that is a homomorphism. For Hx, Hy G1/H,

(Hx)(Hy)) = (Hxy)

= f(xy)

= f(x) f(y), since f is a homomorphism.

= (H) (HY)

Therefore, is a group homomorphism.

Next, let us see whether is bijective or not.

Now, (Hx) = (Hy) for Hx, HY in G1/H

f(x) = f(y)

f(x) [f(y)l-1 = e2

f(xy1) = e2

xy-1 Ker f = H.

Hx = Hy

Thus, , is 1-1.

Also, any element of Im f is f(x) = (Hx), where x G1.

Im = Im f.

So, we have proved that is bijective, and hence, an isomorphism. Thus, G1/Ker f = Im f.

Now, if f is surjective, Im f = G2. Thus, in this case G

1/Ker f G

2.

The situation in Theorem 1 can be shown in the following diagram.

Here, p is the natural homomorphism.

The diagram says that if you first apply p, and then , to the elements of G1, it is the same as

applying f to them. That is,

p = f.

Also, note that Theorem 1 says that two elements of G1 have the same image under f iff they

belong to the same coset of Ker f.

Let us look at a few examples.

One of the simplest situations we can consider is IG : G G. On applying Theorem 1 here, we see

that G/{e} G. We will be using this identification of G/{e) and G quite often.

Example: Prove that C/R R.


Abstract Algebra

Notes Solution: Define f : C R : f(a + ib) = b. Then f is a homomorphism, Ker f = R and Im f = R.Therefore, on applying Theorem 1 we see that C/R R.

Example: Consider f : Z ({1, - 1),.) : 1, if n is even

f(n)1, if n is odd.

At the beginning, you saw that f is a homomorphism. Obtain Ker f and Im f. What does Theorem1 say in this case?

Solution: Let Ze and Z

o denote the set of even and odd integers, respectively. Then

Ker f = {n Z | f(n) = 1 } = Z,

Im f = {f(n) | n Z ) = { l , � 1}

Thus, by Theorem 1, Z/Z, { 1, - 1 }.

This also tells us that o(Z/Ze) = 2. The two cosets of Z

e in Z are Z

e and *.

{ Ze, Z

o } { 1, -1 }.

Example: Show that GL2(R)/SL

2(R) R*, where SL

2(R) = {A GL

2(R) | det (A) = 1 }.,

Solution: We know that the function

f : GL2(R) R* : f(A) = der(A) is a homomorphism. Now, Ker f = SL

2(R).

Also, Im f = R*, since any r R* can be written as det 1 0

.0 1

Thus, using Theorem 1, GL2(R)/SL

2(R) R*.

Now-we will use the Fundamental Theorem of Homomorphism to prove a very importantresult which classifies all cyclic groups.

Theorem 2: Any cyclic group is isomorphic to (Z, +) or (Zn, +).

Proof: Let G = < x > be a cyclic group. Define

f : Z G : f(n)=xn.

f is a homomorphism because

f(n + m) = xn+m = xn . xm = f(n) f(m).

Also note that Im f = G.

Now, we have two possibilities for Ker I Ker f = {0) or Ker f {0}.

Case 1 (Ker f = {0)): In this case f is 1-1. Therefore, f is an isomorphism. Therefore, by Theorem 7of unit 6, f-1 is an isomorphism. That is, G (Z, +).

Case 2 (Ker f # {0)): Since Ker f Z, we know that Ker f = nZ, for some n N. Therefore, by theFundamental Theorem of Homomorphism, Z/nZ G.

G = Z/nZ = (Zn, +).

Over here note that since < x > = Zn, o(x) = n. So, a finite cyclic group is isomorphic to Z

n, where

n is the order of the group.

Theorem 3: If H and K are subgroups of a group G, with K normal in G, then H/(H K) (HK)/K.



NotesProof: We must first verify that the quotient groups H/(H K) and (HK)/K are well defined.

You know that H K H. You know that HK G. Again, you know that K HK. Thus, the

given quotient groups are meaningful.

Now, what we want to do is to find an onto homomorphism f : H � (HK)/K with kernel H K.Then we can apply the Fundamental Theorem of Homomorphism and get the result. We definef : H (HK)/K : f(h) = hK.

Now, for x, y H,

f(xy) = xyK = (xK) (yK) = f(x) f(y).

Therefore, f is a homomorphism.

We will show that Im f = (HK)/K. Now, take any element hK Im f. Since h H, h HK

hK (HK)/K. Im f (HK)/K. On the other hand, any element of (HK)/K is

hkK = hK, since k K.

hkK Im f. (HK)/K Im f.

Im f = (HK)/K.

Finally, Ker f = { h H | f(h) = K } = { h H hK = K }

= { h H | h K }

= H K .

Thus, on applying the Fundamental Theorem, we get H / (H K) (HK) / K

We would like to make a remark here.

Remark: If H and K are subgroups of (G.+ ), then Theorem 3 says that

(H + K) / K H/H K.

Theorem 4: Let H and K be normal subgroups of a group G such that K H. Then (G/K)/(H/K) G/H.

Proof: We will define a homomorphism from G/K onto G/H, whose kernel will turn out to beH/K.

Consider f : G/K G/H : f(Kx) = Hx. f is well-defined because Kx = Ky tor x, y G

xy-1 K H xy-1 H Hx = Hy (Kx) = f(Ky)

7.2 Automorphisms

Let us start discussing the concept of automorphism

Let G be a group. Consider

Aut G = { f : G G | f is an isomorphism }.

You have already seen that the identity map IG Aut G. You know that Aut G is closed under the

binary operation of composition. Iff E Aut G, then f-1 Aut G. We summarise this discussion inthe following theorem.

An isomorphism from a group (G,*) to itself is called an Automorphisms of this group. Thus itis a bijection f : G G such that

f(u) * f(v) = f(u * v).


Abstract Algebra

Notes An automorphism always maps the identity to itself. The image under an automorphism of aconjugacy class is always a conjugacy class (the same or another). The image of an element hasthe same order as that element.

The composition of two automorphisms is again an automorphism, and with this operation theset of all automorphisms of a group G, denoted by Aut(G), forms itself a group, the automorphismgroup of G.

For all Abelian groups there is at least the automorphism that replaces the group elements bytheir inverses. However, in groups where all elements are equal to their inverse this is thetrivial automorphism, e.g. in the Klein four-group. For that group all permutations of the threenon-identity elements are automorphisms, so the automorphism group is isomorphic to S

3 and

Dih3.

In Zp for a prime number p, one non-identity element can be replaced by any other, with

corresponding changes in the other elements. The Automorphisms group is isomorphic toZ

p � 1. For example, for n = 7, multiplying all elements of Z

7 by 3, modulo 7, is an automorphism

of order 6 in the automorphism group, because 36 = 1 (modulo 7), while lower powers do notgive 1. Thus this automorphism generates Z

6. There is one more automorphism with this property:

multiplying all elements of Z7 by 5, modulo 7. Therefore, these two correspond to the elements

1 and 5 of Z6, in that order or conversely.

The automorphism group of Z6 is isomorphic to Z

2, because only each of the two elements 1 and

5 generate Z6, so apart from the identity we can only interchange these.

The automorphism group of Z2 × Z

2 × Z

2 = Dih

2 × Z

2 has order 168, as can be found as follows.

All 7 non-identity elements play the same role, so we can choose which plays the role of (1,0,0).Any of the remaining 6 can be chosen to play the role of (0, 1, 0). This determines whichcorresponds to (1, 1, 0). For (0, 0, 1) we can choose from 4, which determines the rest. Thus wehave 7 × 6 × 4 = 168 automorphisms. They correspond to those of the Fano plane, of which the 7points correspond to the 7 non-identity elements. The lines connecting three points correspondto the group operation: a, b, and c on one line means a + b = c, a + c = b, and b + c = a. See alsogeneral linear group over finite fields.

For Abelian groups all automorphisms except the trivial one are called outer automorphisms.

Non-Abelian groups have a non-trivial inner automorphism group, and possibly also outerAutomorphisms.

Theorem 5: Let G be a group. Then Aut G, the set of automorphisms of G, is a group.

Let us look at an example of Aut G.

Example: Show that Aut Z Z1.

Solution: Let f : Z Z be an automorphism. Let f(1) = n. We will show that n = 1

or � 1. Since f is onto and 1 Z , m Zsuch that f(m) = l, i.e., mf(l)=l, ie., m=l.

n = 1 or n = �1.

Thus, there are only two elements in Aut Z, I and �I.

So Aut Z = < � I > Z2.

Now given an element of a group G. We will define an automorphism of G corresponding to it.

Consider a fixed element g G. Define

fg : G G : f

g(x) = gxg-1.



NotesWe will show that fg is an automorphism of G.

(i) f, is a homomorphism : If x, y G, then

fg(xy) = g(xy) g-1

= gx(e) yg-1, where e is the identity of G.

= gx(g-1g) yg-1

= (gxg-1) (gyg-1)

= fg(x) f

g(dy).

(ii) fg is 1-1 : For x, y G, f

g(x) = f

g(y) gxg-1 = gyg-1 x = y, by the cancellation laws in G.

(iii) fg is onto : If y G, then

Y = (gg-1)y(gg-1)

= (g-1yg)g-1

= fg(g-1yg) lm f

g.

Thus, f, is an automorphism of G.

Definition: fg is called an inner automorphism of G induced by the element g in G. The subset of

Aut G consisting of all inner automorphism of G is denoted by Inn G.

For example, Let us compute fg(1). f

g(l 3) and f

g(1 2 3), where g = (1 2). Note that g-1 = (1 2) = g.

Now, fg (1) = g a I o g-1 = I,

fg(l 3) = (1 2) (1 3) (1 2) = (2 3).

fg(l 2 3) = (1 2)(1 2 3)(1 2) = (1 3 2).

Theorem 6: Let G be a group. Then Inn G is a normal subgroup of Aut G. i

Proof: Inn G is non-empty, because IG = f

e Inn G, where e is the identity in G.

Now, let us see if f, o fh Inn G for g, h G.

For any x G, fg . f

n(x) = fg(hxh-1)

= g(hxh-1) g-1

= (gh)x (gh)-1

= fgh(x)

Thus, fgh

= fg o f

h, i.e., Inn G is closed under composition. Also f

e = I

G belongs to Inn G.

Now, for fg Inn G, 3 fg-1 Inn G such that

fg o f

g-1 = f

gg-1 = f

e = I

G. Similarly, f

g-1o f

g = I

G.

Thus, fg

-1 = (fg)-1. That is every element of Inn G has an inverse in Inn G.

This proves that Inn G is a subgroup of Aut G.

Now, to prove that Inn Aut G, let Aut G and fg Inn G. Then, for any x G

-1 × fg × (x) = -1 × f

g( (x))

= -1 (g(x)g-1)

= -1 (g) -1((x)) -1(g-1)


Abstract Algebra

Notes = -1 (g) x[-1 (g)]-1

= 1 (g )f (x).

(Note that -1(g) G.)

-1 o fg o = 1 (g )f

Inn G Aut G and f

g Inn G.

Inn G Aut G.

Now we will prove an interesting result which relates the cosets of the centre of a group G to

lnn G. Recall that the centre of G, Z(G) = { x G | xg = gx g G }.

Theorem 7: Let G be a group. Then G/Z(G) Inn G.

Proof: As usual, we will use the powerful Fundamental Theorem of Homomorphism to provethis result.

We define f : G Aut G : f(g) = fg.

Firstly, f is a homomorphism because for g, h G,

f(gh) = fgh

= I, o fh (sec proof of Theorem 13)

= [(g) o f(h).

Next, Im F = ( fg, 1 g G ) = Inn G.

Finally, Ker f = ( g G | f, = IG }

= { g G [ fg(x) = x x G }

= { g G | gxg-1 = x x G }

= { g G | gx = xg x G }

= Z(G).

Therefore, by the Fundamental Theorem,

G/Z(G) Inn G.

Self Assessment

1. An isomorphism of a group G itself is called as an ................. of G.

(a) Homomorphism (b) automorphism

(c) Herf (d) one-to-one function

2. The word isomorphisms is derived from Greek word ISOS meaning .................

(a) equal (b) unequal

(c) bijective (d) subjective

7.3 Summary

The proof of the Fundamental Theorem of Homomorphism, which says that if f : G1 G

2

is a group homomorphism, then G1/Ker f Im f,



Notes Any infinite cyclic group is isomorphic to (Z, +). Any finite cyclic group of order n isisomorphic to ( Z , +).

Let G be a group, H G, K G. Then H/(H K) HK)/K.

Let G be a group, H G, K G, K H, Then (G/K)/(H/K) G/H.

The set of automorphism of a group G, Aut G, is a group with respect to the compositionof functions.

Inn G Aut G, for any group G.

G/Z(G) Inn G, for any group G.

7.4 Keywords

Group Homomorphism: Iff : G1 G

2 and g : G

2 G

3 are two group homomorphisms, then the

composite map g . f : G1 G

3 is also a group homomorphism.

Isomorphisms: Let G1 and G

2 be two groups. A homomorphism f : G

1 G

2 is called an

isomorphism if f is 1-1 and onto.


1. Let G be a group and H G. Show that there exists a group G1 and a homomorphism

f : G G1 such that Ker f = H.

2. Show that the homomorphic image of a cyclic group is cyclic i.e., if G is a cyclic group andf : G G� is a homomorphism, then f(G) is cyclic.

3. Show that Z = nZ, for a fixed integer n,

(Hint: Consider f : (Z, +) (nZ, +) : f(k) = nk)

4. Is f : Z Z : f(x) = 0 a homomorphism? An isomorphism?


1. (b) 2. (a)






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Abstract Algebra

Notes Unit 8: Permutation Groups

CONTENTS

Objectives

Introduction

8.1 Symmetric Group

8.2 Cyclic Decomposition

8.3 Alternating Group

8.4 Cayley�s Theorem

8.5 Summary

8.6 Keywords



Objectives


Discuss the concept of permutation group

Explain the symmetric group

Describe the cyclic decomposition

Prove and use Cayley�s Theorem

Introduction

In earlier classes, you have studied about the symmetric group. As you have often seen inprevious units, the symmetric groups S, as well as its subgroups, have provided us a lot ofexamples. The symmetric groups and their subgroups are called permutation groups. It was thestudy of permutation groups and groups of transformations that gave the foundation to grouptheory. In this unit, we will prove a result by the mathematician Cayley, which says that everygroup is isomorphic to permutations group. This result is what makes permutation groups soimportant.

8.1 Symmetric Group

In earlier units, you have studied that a permutation on n non-empty set X is a bijective functionfrom X onto X. We denote the set of all permutations on X by S(X).

Suppose X is a finite set having n elements. For simplicity, we take these elements to be1, 2, . . . , n. Then, we denote the set of all permutations on these n symbols by S

n.


Unit 8: Permutation Groups

Notes

We represent any f Sn in n 2-line form as

1 2 .... nf .

....f(1) f(2) f(n)

Now, there are n possibilities for f(l), namely, 1, 2, . . . , n. Once f(1) has been specified, there are(n � 1) possibilities for f(2), namely, {1, 2, . . . , n} \ {f(1)}. This is because f is 1-1. Thus, there aren(n � 1) choices for f(1) and f(2). Continuing in this manner, we see that there are n! differentways in which f can be defined. Therefore, S, has n! element.

Now, let us discuss at the algebraic structure of S(X), for any set X. The composition ofpermutations is a binary operation on S(X). To help you regain practice in computing thecomposition of permutations, consider an example.

Let f = 4

1 2 3 4 1 2 3 4and g be in S .

2 4 1 3 4 1 3 2

Then, to get fog we first apply g and then apply f.

f o g(1) = f(g(1)) = f(4) = 3.

f o g (2) = f(g(2)) f(1) = 2.

f o g (3) = f(g(3)) = f(3) = 1.

f o g (4) = f(g(4)) = f(2) = 4.

f o g = 1 2 3 4

3 2 1 4

We show this process diagrammatically in Figure 8.1.

Figure 8.1: (1 2 3 4) o (1 4 2) in S4

Now, let us go back to S(X), for any set k.

Theorem 1: Let X be a non-empty set. Then the system (S(X), 0 ) forms a group, called thesymmetric group of X.

Thus, Sn is a group of order n!. We call S

n, the symmetric group of degree n. Note that if f S

n, then

1....f(1) f(2) f(n)

f .1 2 .... n


Abstract Algebra

Notes Remark: From now we will refer to the composition of permutations as multiplication ofpermutations. We will also drop the composition sign. Thus, we will write f o g as fg.

The two-line notation that we have used for a permutation is rather cumbersome. In the nextsection we will see how to use a shorter notation.

8.2 Cyclic Decomposition

Let us first discuss what a cycle is.

Consider the permutation f = 2 4

1 2

. Choose any one of the symbols say 1.

Now, we write down a left hand bracket followed by I : (1

Since f maps 1 to 3, we write 3 after 1 : (1 3

Since f maps 3 to 4, we write 4 after 3 : (1 3 4

Since f maps 4 to 2, we write 2 after 4 : ( l 3 4 2

Since f maps 2 to 1 (the symbol we started with),we close the brackets after the symbol (1 3 4 2)

Thus, we write f = (1 3 4 2). This means that f maps each symbol to the symbol on its right, exceptfor the final symbol in the brackets, which is mapped to the first.

If we had chosen 3 as our starting symbol we would have obtained the expression (3 4 2 1) for f.However, this means exactly the same as (1 3 4 2), because both denote the permutation whichwe have represented diagrammatically in Figure 8.2.

Figure 8.2: (1 3 4 2)

1

2

3

4

Such a permutation is called a 4-cycle, or a cycle of length 4. Figure 8.2 can give you an indicationas to why we give this name.

Let us give a definition now.

Definition: A permutation f Sn, is called an r-cycle (or cycle of length r) if there are r distinct

integers i1, i

2,, i

3, . . . , i

r lying between 1 and n such that

f(i1) = i

2, f(i

2) = i

3, . . . . . , f(

i,-1) = i

r, f(i

r) = i

1.

and f(k) = k k {i1, i

2 , . . . , i

r).

Then, we write f = (i1 i

2 . . . . . i

r).



NotesIn particular, 2-cycles are called transpositions. For example, the permutation f = (2 3) S3 is a

transposition. Here f(1) = 1, f(2) = 3 and f(3) = 2.

Later you will see that transpositions play a very important role in the theory of permutations.

Now consider any 1-cycle (i) in S,. It is simply the identity permutation 1 2 .... n

I ,1 2 .... n

since

it maps i to i and the other (n - 1) symbols to themselves.

Let us see some examples of cycles in S3 (1 2 3) is the 3-cycle that takes 1 to 2, 2 to 3 and 3 to 1. There

are also 3 transpositions in S3, namely, (1 2), (1 3) and (2 3).

Now, can we express any permutation as a cycle? No. Consider the following example from S5.

Let g be the permutation defined by

g = 1 2 3 4 5

.3 5 4 1 2

If we start with the symbol 1 and apply the procedure for obtaining a cycle to g, we obtain(1 3 4) after three steps, Because, g maps 4 to 1, we close the brackets, even though we have notyet written down all the symbols. Now we simply choose another symbol that has not appearedso far, say 2, and start the procedure of writing a cycle again. Thus, we obtain another cycle (2 5).Now, all the symbols are exhausted.

g = (1 3 4) (2 5).

We call this expression for g a product of a 3-cycle and a transposition. In Figure 8.3 we representg by a diagram which shows the 3-cycle and the 2-cycle clearly.

Figure 8.3: (1 3 4) (2 5)

Because of the arbitrary choice of symbol at the beginning of each cycle, there are many ways ofexpressing g. For example,

g = (4 1 3) (2 5) = (2 5) (1 3 4) = (5 2) (3 4 1).

That is, we can write the product of the separate cycles in any order, and the choice of the startingelement within each cycle is arbitrary.

So, you see that g can�t be written as a cycle; it is a product of disjoint cycles.

Definition: We call two cycle disjoint if they have no symbol in common. Thus, disjoint cyclesmove disjoint sets of elements, (Note that f ! S,, moves a symbol i if f(i) i. We say that f fixesi if f(i) = i.)

So, for example, the cycles (1 2) and (3 4) in S4 are disjoint. But (1 2) and (1 4) are not disjoint, since

they both move 1.


Abstract Algebra

Notes Note that if f and g are disjoint, then fg=-gf, since f and g move disjoint sets of symbols.

Now let us examine one more example. Let h be the permutation in S5 defined by

h = 1 2 3 4 5

.4 2 3 5 1

Following our previous rules, we obtain

h = (1 4 5) (2) (3),

because each of the symbols 2 and 3 is left unchanged by h. By convention, we don�t include the1-cycles (2) and (3) in the expression for h unless we wish to emphasize them, since they justrepresent the identity permutation. Thus, we simply write h = (1 4 5).

The same process that we have just used is true for any cycle. That is, any r-cycle (i1 i

2 . . . . . i

r) can

be written as (i1 i

r) (i

1 i

1) . . . . . (i

1 i

2), a product of transpositions.

Now we will use Theorem 2 to state a result which shows why transpositions are so importantin the theory of permutations.

Theorem 2: Every permutation in Sn (n 2) can be written as a product of transpositions.

Proof: The proof is really very simple. By Theorem 1 every permutation, apart from I, is aproduct of disjoint cycles. Also, you have just seen that every cycle is a product of transpositions.Hence, every permutation, apart from I, is a product of transpositions.

Also, I = (1 2) (1 2). Thus, I is also a product of transpositions. So, the theorem is proved.

Let us see how Theorem 3 works in practice. This is the same as (1 4) (1 2) (1 3) (1 5).

Similarly, the permutation 1 2 3 4 5 6

3 6 4 1 2 5

= (1 3 4) (2 6 5) = (1 4) (1 3) (2 5) (2 6).

The decomposition given in Theorem 3 leads us to a subgroup of Sn that we will now discuss.

8.3 Alternating Group

You have seen that a permutation in Sn can be written as a product of transpositions. But all such

representations have one thing in common � if a permutation in Sn is the product of an odd

number of transpositions in one such representation, then it will be a product of an odd numberof transpositions in any such representation. Similarly, if f S

n is a product of an even number

of transpositions in one representation, then f is a product of an even number of transpositionsin any such representation. To see this fact we need the concept of the signature or sign function.

Definition: The signature of f Sn, (n 2) is defined to be

sign i , j 1

f(i) � f(i)f

j � i

For example, for f = (1 2 3) S3,

sign f = f(2) � f(1) f(3) � f(1) f(3) � f(2)

. .2 � 1 3 � 1 3 � 2



Notes

3 � 2 1 2 1 3= 1.

1 2 1

Similarly, iff = (1 2) S3, then

sign f = f(2) � f(1) f(3) � f(1) f(3) � f(2)

. .2 � 1 3 � 1 3 � 2

= 1 2 3 � 2 3 1

1.1 2 1

Henceforth, whenever we talk of sign f, we shall assume that f Sn for some n 2.

Theorem 3: Let f, g S,. Then sign (f o g) = (sign f) (sign g).

Proof: By definition,

sign fog n

i , j 1i j

f(g(j)) � f(g(i))j � i

i , j i , j

f(g( j)) f(g(i)) g( j) g(i).

g( j) g(i) j i

Now, as i and j take all possible pairs of distinct values from 1 to n, so do g(i) and g(j), since g isa bijection.

i j

f(g( j)) f(g(i))sign f.

g( j) g(i)

sign (fog) = (sign f) (sign g).

Now we will show that Im (sign) = (1, � 1}.

Theorem 4: (a) If t S, is a transposition, then sign t = � 1.

(b) sign f = 1 or - 1 f S,.

(c) Im (sign) = (1, �1).

Proof: (a) Let t = (p q), where p < q.

Now, only one factor of sign t involves both p and q, namely,

t (q) � t p � 1= 1.

q � p q - p

Every factor of sign t that doesn�t contain p or q equals 1, since

t(i) t( j) i j1, if i, j p, q.

i j i j

The remaining factors contain either p or q, but not both. These can be paired together to formone of the following products.

t(i) � t(p) t(i) � t(q) i � q i � p. = . l, if i > q,

i � p i � q i � p i � q


Abstract Algebra

Notes t(i) � t(p) t(q) � t(i) i � q p � i= . l, if q > i > p,

i � p q � i i � p q � i

t(p) � t(i) t(q) � t(i) q � i p � i= . l, if i > p,

p � i q � i p � i q � i

Taking the values of all the factors of sign t, we see that sign t = �1.

(b) Let f S,. By Theorem 3 we know that f = t1,t

2 .... t, for some transpositions t

1, ..... t

r in S

n.

sign f = sign (t1 t

2 . . . . t,)

= (sign t1) (sign t

2) . . . . . sign (t

r), by Theorem 3.

= (-1)r, by (a) above.

sign f = 1 or �1.

(c) We know that Im (sign) {1, � 1}.

We also know that sign t = �1, for any transposition t; and sign I = 1.

{1, � 1} Im {sign}

Im (sign) = {1, �1}.

Now, we are in a position to prove what we said at the beginning of this section.

Theorem 5: Let f S, and let

f = t1t

2 ..... t

r = t

1� t

2�..... t

4�

be two factorisations of f into a product of transpositions. Then either both r and s are evenintegers, or both are odd integers.

Proof: We apply the function sign: Sn {1, �1} to f = t

1t

2 . . . . t

r.

By Theorem 4 we see that

sign f = (sign t1) (sign t

2) ...... (sign t

r) = (- 1)�.

sign (t1� t

2� . . . t

s�) = (�1) substituting t

1� t

2�. . . ts� for f.

that is, ( -1)s = (�1)r.

This can only happen if both s and r are even, or both are odd.

So, we have shown that for f S, the number of factors occurring in any factorisation of f intotransposition is always even or always odd. Therefore, the following definition is meaningful.

Definition: A permutation f Sn, is called even if it can be written as a product of an even sign

number of transposition. f is called odd if it can be represented as a product of an odd number oftranspositions.

For example, (1 2) S3 is an odd permutation. In fact, any transposition is an odd permutation.

On the other hand, any 3-cycle is an even permutation, since

(i j k) = (i k) (i j)

Now, we define an important subset of Sn, namely,

A, = (f Sn, | f is even).

We�ll show that A,, Sn, and that o(A

n) =

n!,

2 for n 2.



NotesTheorem 6: The set A,, of even permutations in S,, forms a normal subgroup of S

n, of order

n!.

2

Proof: Consider the signature function,

sign : Sn (1, �1).

Note that (1, �1) is a group with respect to multiplication. Now, Im (sign) = (1, �1). Let us obtainKer (sign).

Ker (sign) = { f Sn | sign f = 1 }

= (f Sn | f is even)

= A.

A Sn.

Further, by the Fundamental Theorem of Homomorphism

Sn/A

n (1, �1).

o(Sn/A

n) = 2, that is, n

n

o(S )2.

o(A )

o(An) = no(S ) n!.

2 2

Note that this theorem says that the number of even permutations in S, equals the number ofodd permutations in S,.

Theorem 6 leads us to the following definition.

Definition: A,, the group of even permutations in Sn, is called the alternating group of

degree n.

Let us look at an example that you have already seen in previous units, A3. Now, Theorem 6 says

that o(A3) =

3!3.

2 Since (1 2 3) = (1 3) (1 2), (1 2 3) A

3. Similarly,

(1 3 2) A3. Of course, I A

3.

A3 = {I, (1 2 3), (1 3 2)).

A fact that we have used in the example above is that an r-cycle is odd if r is even, and even if ris odd. This is because (i

1i

2 .... i

r,) = (i

1 i,) (i

1 i

r-1) . . . . . . (i

1 i

2), a product of (r � 1) transpositions.

Now, for a moment, let us go back to Unit 4 and Lagrange�s theorem. This theorem says that theorder of the subgroup of a finite group divides the order of the group. We also said that ifn |o(G), then G need not have a subgroup of order n. Now that you know what A

4 looks like, we

are in a position to illustrate this statement.

We will show that A4 has no subgroup of order 6, even though 6 | o (A

4). Suppose such a

subgroup H exists. Then o(H) = 6, o (A4) = 12. (A

4 : H | = 2. H A4 (see Theorem 3, Unit 5).

Now, A4/H is a group of order 2.

(Hg)2 = H g A4. (Remember H is the identity of A

4/H.)

g2 H g A4.

Now, (1 2 3) E A4. (1 2 3)* = (1 3 2) H.


Abstract Algebra

Notes Similarly, (1 3 2)2 = (1 2 3) H. By the same reasoning (1 4 2), (1 2 4), (1 4 3), (1 3 4), (2 3 4), (2 4 3)are also distinct element of H. Of course, I H.

Thus, H contains at least 9 elements.

o(H) 9. This contradicts our assumption that o(H) = 6.

Therefore, A4 has no subgroup of order 6.

We use A4 to provide another example too. (See how useful A

4 is!) In earlier unit we�d said that

if H N and N G, then H need not be normal in 6. Well, here�s the example.�

Consider the subset V4 = {I, (1 2) (3 4), (1 4) (2 3), (1 3) (2 4)) of A

4.

Now, let H = {I, (1 2) (3 4)}. Then H is a subgroup of index 2 in V4. H A V

4.

So, H V4, V

4 A

4. But H A

4. Why? Well, (1 2 3) A

4 is such that

(1 2 3)-1 (1 2) (3 4) (1 2 3) = (1 3) (2 4) H.

And now let us see why permutation groups are so important in group theory.

8.4 Cayley�s Theorem

Most finite groups that first appeared in mathematics were groups of permutations. It was theEnglish mathematician Clayley who first realised that every group has the algebraic structureof a subgroup of S(X), for some set X. In this section we will discuss Cayley�s result and some ofits applications.

Theorem 7 (Cayley): Any group G is isomorphic to a subgroup of the symmetric group S(G).

Proof: For a G, we define the left multiplication function

fa : G G : f

4(x) = ax.

fa is 1-1, since

fa(x) = fa(y) ax = ay x = y x, y E G.

fa is onto, since any x E G is f, (a � �x).

:. fa S(G) a G.

(Note that S(G) is the symmetric group on the set G.)

Now we define a function f : G S(G) : f(a) = fa.

We will show that f is an injective homomorphism. For this we note that

(fao

,fb) (x) = f

a(bx) = abx = f

ab (x) a, b G.

f(ab) = fab

= fa o f

b = f(a) of (b) a, b G.

That is, f is a homomorphism.

Now, Ker f = (a G | fa = IG

)

= ( a G | fa(x)=x x G }

= ( a G |a x = x x G }

= {e}.



NotesThus, by the Fundamental Theorem of Homomorphism,

G/Ker f Im f S(G),

that is, G is isomorphic to a subgroup of S(G).

As an example of Cayley�s theorem, we will show you that the Klein 4-group K4 is isomorphic

to the subgroup V4 of S

4. The multiplication table for K

4 is

. e a b c

e e a b c

a a e c b

b b c e a

c c b a e

Self Assessment

1. If .................. is a group of order n!. Then we call S, the symmetric group of define n.

(a) Sn (b) Sn

(c) 1nS (d) Sn-1

2. Every permutation is Sn (n ..................) can be written as produce of transposition

(a) n 2 (b) n 3

(c) n 4 (d) n 5

3. If t S, is a transposition then sight = ..................

(a) �1 (b) 1

(c) 0 (d) 2

4. Any group G is .................. to a subgroup of the symmetric group S(G)

(a) isomorphic (b) homomorphic

(c) automorphic (d) surjective

5. Any group is isomorphic to a .................. group.

(a) normal group (b) subgroup

(c) cyclic group (d) permutation group

8.5 Summary

The symmetric group S(X), for any set X, and the group S,, in particular.

The definitions and some properties of cycles and transpositions.

Any non-identity permutation in Sn can be expressed as a disjoint product of cycles.

Any permutation in Sn (n 2) can be written as a product of transpositions.

The homomorphism sign : Sn � {1, � 1}, n 2.

Odd and even permutations.

A,, the set of even permutations in S,, is a normal subgroup of Sn of order

n!,

2 for

Any group is isomorphic to a permutation group.


Abstract Algebra

Notes 8.6 Keywords

Symmetric Group: Let X be a non-empty set. Then the system (S(Xj, 0) forms a group, called thesymmetric group of X.

Permutation: A permutation f S, is called an r-cycle (or cycle of length r) if there are r distinctintegers i

1, i

2, i

3, . . . , i

r lying between 1 and n.


1. Show that (S,, °) is a non-commutative group for n 3.

(Hint: Check the 1 2 3

2 3 1

and 1 2 3

3 2 1

don�t commute.)

2. Write down 2 transpositions, 2 3-cycles and a 5-cycle in S5.

3. Show that every permutation in Sn is a cyclic iff n < 4.

4. Iff = (i2 i

2 , ....i) S,, then show that f-1 = (i

r i

r-1 .... i

2i

1).

5. Iff is an r-cycle, then show that o(f) = r, i.e., fr = I and f5 I, if s < r.

(Hint: If f = (i1, i

2 ....i), then f(i

1) = i

2, r2(i

1) = i

3,....,fr-1(i

1) = i

m)

6. Express the following cycles as products of transpositions.

(a) (1 3 5) (b) (5 3 1)

(c) (2 4 5 3)

7. Write the permutation in E3(b) as a product of transpositions.

8. Show that (1 2 .... 10) = (1 2) (2 3) . . . (9 10).

9. Check that (V4, °) is a normal subgroup of A

4.


1. (b) 2. (a) 3. (a)

4. (a) 5. (d)






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Unit 9: Direct Products

NotesUnit 9: Direct Products

CONTENTS

Objectives

Introduction

9.1 Direct Product of Groups

9.1.1 External Direct Product

9.1.2 Internal Direct Product

9.2 Sylow Theorems

9.3 Groups of Order 1 to 10

9.4 Summary

9.5 Keywords



Objectives


Discuss direct product of groups

State Sylow theorem

Explain groups of order 1 to 10.

Introduction

In the last unit, we have studied about permutation group. This unit will provide you theinformation related to 15 finite groups and direct products. Let us understand all these one byone.

9.1 Direct Product of Groups

In this section, we will discuss a very important method of constructing new groups. We willfirst see how two groups can be combined to form a third group. Then we will see how twosubgroups of a group can be combined to form another subgroup.

9.1.1 External Direct Product

In this sub-section we will construct a new group from two or more groups that we alreadyhave.

Let (G1, *

1) and ((G

2, *

2) be two groups. Consider their Cartesian product G = G

1 × G

2 = {(x, y) |

x G1, y G

2}.


Abstract Algebra

Notes Can we define a binary operation on G by using the operations on G1 and G

2? Let us try the

method, namely, component-wise multiplication. That is, we define the operation * on G by

(a,b) * (c, d)=(a*1 c, b*

2d) a, c G

1, b, d G

2.

So, you have proved that G = G1 × G

2 is a group with respect to *. We call G the external direct

product of (G1, *

1) and (G

2, *

2).

For example, R2 is the external direct product of R with itself.

Another example is the direct product (Z, +) × (R*, .) in which the operation is given by (m, X) *(n, y) = (m + n, xy).

We can also define the external direct product of 3, 4 or more groups on the same lines.


1), (G

2, *

2), . . . . . , (G,, *

n) be n groups. Their external direct product is the

group (G, *), where

G = G1 × G

2 ..... × G

n and

Thus, Rn is the external direct product of n copies of R,

We would like to make a remark about notation now.

Remark: Henceforth, we will assume that all the operations *, *1,. . . , *

n are multiplication, unless

mentioned otherwise. Thus, the operation on

G = G1 × G

2 × ..... × G

n will be given by

(a1, ..... , a) . (b

1, ..... , b)

= (a1b

1, a

2b

2, .... , a

ab

a,) V a

i, b

i G

i.

Now, let G be the external direct product G1 × G

2. Consider the projection map

T1 : G

1 × G

2 G

1 : !

1 : (x, y) = x.

Then 1 is a group homomorphism, since

1 ((a, b) (c, d)) =

1 (ac, bd)

= ac

= 1 (a, b)

1 (c, d)

1 is also onto, because any x G

1 is !

1 (x, e

2)

Now, let us look at Ker 1.

Ker 1 = {(x, y) G

× G

2 |

1 (x, y) = e

1}

= {(e1, y) | y Gz} = {e

1} × G

2.

{e} × G2 G

1 × G

2.

Also, by the Fundamental Theorem of Homomorphism (G1 × G

2)/({e

1} × Gz) G

1.

We can similarly prove that G1 × {e

2} G

1 × G

2 and (G

1 × G

2)/(G

1 × {e

2}) G

2.

So, far we have seen the construction of G1 × G

2 from two groups G

1 and G

2. Now we will see

under what conditions we can express a group as a direct product of its subgroups.

9.1.2 Internal Direct Product

Let us begin by recalling from Unit 5 that if H and K are normal subgroups of a group G, then HKis a normal subgroup of G. We are interested in the case when HK is the whole of G. We have thefollowing definition.



NotesDefinition: Let H and K be normal subgroups of a group G. We call G the internal direct productof H and K if

G = HK and H K = {e}.

We write this fact as G = H × K.

For example, let us consider the familiar Klein 4-group

K4 = {e, a, b, ab}, where a2 = e, b2 = e and ab = ba.

Let H = <a> and K = . Then H K = {e). Also, K4 = HK.

K4 = H × K.

Note that H Z2 and K Z

2 K

4 Z

2 × Z

2.

For another example, consider Z10

. It is the internal direct product of its subgroups H = {0, 5}

and K = {0, 2, 4, 6, 8}. This is because

(i) Z10

= H + K, since any element of Z10

is the sum of an element of H and an element of K, and

(ii) H K= {0} .

Now, can an external direct product also be an internal direct product? What does it say? It saysthat the external product of G

1 × G

2 is the internal product (G

1 × {e

2}) × ({e

1} × G

2).

We would like to make a remark here.

Remark: Let H and K be normal subgroups of a group G. Then the internal direct product of Hand K is isomorphic to the external direct product of H and K. Therefore, when we talk of aninternal direct product of subgroups we can drop the word internal, and just say �direct productof subgroups�.

Let us now extend the definition of the internal direct product of two subgroups to that ofseveral subgroups.

Definition: A group G is the internal direct product of its normal subgroups H1, H

2, .... ,H

n if

(i) G = H1 H

2 .... H

n and

(ii) Hi H

1 .... H

i-1, H

i+1 .... H

n = {e} i = 1, .... , n.

For example, look at the group G generated by {a, b, c}, where a2 = e = b2 = c2 and ab = ba, ac = ca,bc = cb. This is the internal direct product of < a >, and < c >. That is G Z

2 × Z

2 × Z

2.

Now, can every group be written as an internal direct product of two or more of its propernormal subgroups? Consider Z. Suppose Z = H × K, where H, K are subgroups of Z.

You know that H = < m > and K = < n > for some m, n Z. Then mn H K. But if H × K is a directproduct, H K = {0}. So, we reach a contradiction. Therefore, Z can�t be written as an internaldirect product of two subgroups.

By the same reasoning we can say that Z can�t be expressed as H1 × H

2 × ..... × H

n, where Hi Z

i = 1, 2, .... , n.

When a group is an internal direct product of its subgroups, it satisfies the following theorem.

Theorem 1: Let a group G be the internal direct product of its subgroups H and K. Then

(a) each x G can be uniquely expressed as x = hk, where h H, k K; and

(b) hk = kh h H , k K .


Abstract Algebra

Notes Proof: (a) We know that G = HK. Therefore, if x G, then x = hk, for some h H, k K. Nowsuppose x = h

1k

1 also, where h

1 H and k

1 K. Then hk = h

1k

1.

h1

-1 h = k1k-1. Now h

1-1 h H.

Also, since h1

-1h = k1k-1 K, h

1-1 h K. h

1-1 h H K = {e}.

h1

-1h = e, which implies that h = h1.

Similarly, k1 k-1 = e, So that k

1 = k.

Thus, the representation of x as the product of an element of H and an element of K is unique.

(b) The best way to show that two elements x and y commute is to show that their commutatorx-1y-1 xy is identity. So, let h H and k K and consider h-�k-�hk. Since K G, h-1k-1 h K.

h-1k-1hk K.

By similar reasoning, h-1k-1hk H. h-1k-1hk H K = {e}.

h-1 k-1hk = e, that is, hk = kh.

Now let us look at the relationship between internal direct products and quotient groups.

Theorem 2: Let H and K be normal subgroups of a group G such that G = H × K. Then G/H Kand G/K H.

Proof: We will use Theorem 8 of Unit 6 to prove this result.

Now G = HK and H K = {e}. Therefore,

G/H = HK/H K/H K = K/{e) K.

We can similarly prove that G/K H.

Theorem 3: Let G be a finite group and H and K be its subgroups such that G = H X K.

Then o(G) = o(H) o(K).

9.2 Sylow Theorems

In Unit 4 we proved Lagrange�s theorem, which says that the order of a subgroup of a finitegroup divides the order of the group. We also said that if G is a finite cyclic group and m | o(G),then G has a subgroup of order. But if G is not cyclic, this statement need not be true, as you haveseen in the previous unit. In this context, in 1845 the mathematician Cauchy proved the followinguseful result.

Theorem 4: If a prime p divides the order of a finite group G, then G contains an element oforder p.

The proof of this result involves a knowledge of group theory that is beyond the scope of thiscourse. Therefore, we omit it.

Theorem 5: If a prime p divides the order of a finite group G, then G contains a subgroup oforder p.

Proof: Just take the cyclic subgroup generated by an element of order p. This element existsbecause of Theorem 4.

So, by Theorem 5 we know that any group of order 30 will have a subgroup of order 2, asubgroup of order 3 and a subgroup of order 5. In 1872 Ludwig Sylow, a Norwegian



Notesmathematician, proved a remarkable extension of Cauchy�s result. This result, called the firstSylow theorem, has turned out to be the basis of finite group theory. Using this result we cansay, for example, that any group of order 100 has subgroups of order 2, 4, 5 and 25.

Theorem 6 (First Sylow Theorem): Let G be a finite group such that o (G) = pnm, where p is a

prime, n 1 and (p, m) = 1. Then G contains a subgroup of order pk k = 1, . . . , n.

We shall not prove this result or the next two Sylow theorems either. But, after stating all theseresults we shall show how useful they are.

The next theorem involves the concepts of conjugacy and Sylow p-subgroups which we nowdefine.

Definition: Two subgroups H and K of a group G are conjugate in G if g G such thatK = g-1Hg and then K is called a conjugate of H in G.

Now we define Sylow p-subgroups.

Definition: Let G be a finite group and p be a prime such that pn | o(G) but pn+1 o(G), for some n 1. Then a subgroup of G of order pn is called a Sylow p-subgroup of G.

So, if o(G) = pnm, (p, m) = I, then a subgroup of G of order p� is a Sylow p-subgroup. Theorem 6says that this subgroup always exists. But, a group may have more than one Sylow p-subgroup.The next result tells us how two Sylow p-subgroups of a group are related.

Theorem 7 (Second Sylow Theorem): Let G be a group such that o(G) = pnm, (p, m) = 1, p a prime.Then any two Sylow p-subgroups of G are conjugate in G.

And now let us see how many Sylow p-subgroups a group can have.

Theorem 8 (Third Sylow Theorem): Let G be a group of order pnm, where (p, m) = 1 and p is aprime. Then n

p, the number of distinct Sylow p-subgroups of G, is given by n

p = 1 + kp for some

k 0. And further, np | o(G).

We would like to make a remark about the actual use of Theorem 8.

Remark: Theorem 8 says that np, 1

(mod p). (n

p, pn) = 1. Also, since np | o(G), using Theorem 9

of Unit 1 we find that np | m. This fact helps us to cut down the possibilities for n,, as you will see

in the following examples.

Example: Show that any group of order 15 is cyclic.

Solution: Let G be a group of order 15 = 3 × 5. Theorem 6 says that G has a Sylow 3-subgroup.Theorem 8 says that the number of such subgroups must divide 15 and must be congruent tol(mod 3). In fact, by Remark 3 the number of such subgroups must divide 5 and must be congruentto l(mod 3). Thus, the only possibility is 1. Therefore, G has a unique Sylow 3-subgroup, say H.

Hence, by Theorem 7 we know that H G. Since H is of prime order, it is cyclic.

Similarly, we know that G has a subgroup of order 5. The total number of such subgroups is 1,6or 11 and must divide 3. Thus, the only possibility is 1. So G has a unique subgroup of order 5, say

K. Then K G and K is cyclic.

Now, let us look at H K. Let x H K. Then x H and x K.

o(x) | o(H) and o(x) | o(K) i.e., o(x) | 3 and o(x) | 5.

o(x) = 1. x = e. That is, H K = {e}. Also,

G = HK.

So, G = H × K Z3 X Z

5 = Z

15,


Abstract Algebra

Notes

Example: Show that a group G of order 30 either has a normal subgroup of order 5 or anormal subgroup of order 3, i.e. G is not simple. A group G is called simple if its only normalsubgroups.

Solution: Since 30 = 2 × 3 × 5, G has a Sylow 2-subgroup, a Sylow 3-subgroup and a Sylow5-subgroup. The number of Sylow 5-subgroups is of the form 1 + 5k and divides 6. Therefore, itcan be 1 or 6. If it is 1, then the Sylow 5-subgroup is normal in G.

On the other hand, suppose the number of Sylow 5-subgroups is 6. Each of these subgroups aredistinct cyclic groups of order 5, the only common element being e. Thus, together they contain24 + 1 = 25 elements of the group. So, we are left with 5 elements of the group which are of order2 or 3. Now, the number of Sylow 3-subgroups can be 1 or 10. We can�t have 10 Sylow3-subgroups, because we only have at most 5 elements of the group which are of order 3. So, ifthe group has 6 Sylow 5-groups then it has only 1 Sylow 3-subgroup.

Now let us use the powerful Sylow theorems to classify groups of order 1 to 10. In the process wewill show you the algebraic structure of several types of finite groups.

9.3 Groups of Order 1 to 10

Here, we will apply the results of the above discussion to study some finite groups. In particular,we will list all the groups of order 1 to 10, up to isomorphism.

We start with proving a very useful result.

Theorem 9: Let G be a group such that o(G) = pq, where p, q are primes such that p > q and

q | p � 1. Then G is cyclic.

Proof: Let P be a Sylow p-subgroup and Q be a Sylow q-subgroup of G. Then o(P) = p ando(Q) = q. Now, any group of prime order is cyclic, so P = < x > and Q = < y > for some x, y G.By the third Sylow theorem, the number n

p of subgroups of order p can be 1 , 1+ p, 1 + 2p, . . . , and

it must divide q. But p > q. Therefore, the only possibility for np is 1. Thus, there exists only one

Sylow p-subgroup, i.e., P. Further, by Sylow�s second theorem P G.

Again, the number of distinct Sylow q-subgroups of G is nq, = 1 + kq for some k, and n, | p. Since

p is a prime, its only factors are 1 and p. n, = 1 or nq = p. Now if 1 + kq = p, then q | p � 1. But

we started by assuming that 9 | p - 1. So we reach a contradiction. Thus, nq = 1 is the only

possibility. Thus, the Sylow q-subgroup Q is normal in G.

Now we want to show that G = P × Q. For this, let us consider P Q. The order of any element ofP Q must divide p as well as q, and hence it must divide (p, q) = 1.

P Q = {e}. o(PQ) = o(P) o(Q) = pq = o(G). G = PQ.

So we find that G = P × Q Zp × Z

p Z

pq,

Therefore, G is cyclic of order pq.

Using Theorem 9, we can immediately say that any group of order 15 is cyclic. Similarly, ifo(G) = 35, then G�is cyclic.

Now if q | p � 1, then does o(G) = pq imply that G is cyclic? Well, consider S3. You know that o(S

3)

= 6 = 2.3, but S3 is not cyclic. In fact, we have the following result.

Theorem 10: Let G be a group such that o(G) = 2p, where p is an odd prime, Then either G is cyclicor G is isomorphic to the dihedral group D

2p of order 2p.



Notes(Recall that D2p

= < (x, y | xP = e = Y2 and yx = x-ly} > .)

Proof: As in the proof of Theorem 9, there exists a subgroup P = < x > of order p with

P G and a subgroup Q = < y > of order 2, since p > 2. Since (2, p) = 1,

P Q = {e}. o(PQ) = o(G).

G = PQ.

Now, two cases arise, namely, when Q G and when Q G.

If Q G, then G = P × Q. And then G = <xy>.

If Q is not normal in G, then G must be non-abelian.

(Remember that every subgroup of an abelian group is normal.)

xy yx. y-1xy x.

Now, since P = < x > G, y-1xy P. y-1xy = x�, for some r = 2 ,... . , p - 1.

Therefore, y-2xy2 = y-1(y-1xy) = y�1xry = (y-1xy)r = (xr)r = 2rx

x = 2rx , since o(y) = 2.

2 1rx 6

.

But o(x) = p. Therefore, by Theorem 4 of Unit 4, p | r2 � 1, i.e., p | (r � 1) (r + 1)

p | ( r � 1 ) or p | ( r + 1 ). But 2 r p � 1. p = r + l,

i.e., r = p � 1. So we see that

y-1 xy = xr =� xp-1 = x-1

So, G = PQ = < {x, y | xp = e, y2 = e, y-1xy = x-1 > , which is exactly the same algebraic structure asthat of D

2p.

G = D2p

= {e, x, x2, ... , xp-1, y, xy, x2 y, . . . . , xp-1y]

Example: What are the possible algebraic structures of a group of order 6?

Solution: Let G be a group of order 6. Then, by theorem 10, G Z6 or G Ds. You must have

already noted that S3 D

6. So, if G is not cyclic, then G S

3.

Now, from Theorem 6 of Unit 4, we know that if o(G) is a prime, then G is cyclic. Thus, groupsof orders 2, 3, 5 and 7 are cyclic. This fact allows us to classify all groups whose orders are 1, 2, 3,5, 6, 7 or 10. What about the structure of groups of order 4 = 22 and 9 = 32? Such groups are coveredby the following result.

Theorem 11: If G is a group of order p2, p a prime, then G is abelian.

We will not prove this result, since its proof is beyond the scope of this course. But, using thistheorem, we, can easily classify groups of order p2.

Theorem 12: Let G be a group such that o(G) = p2, where p is a prime. Then either G is cyclic orG = Z

p × Z

p, a direct product of two cyclic groups of order p.

Proof: Suppose G has an element a of order p2. Then G = < a > .

On the other hand, suppose G has no element of order. Then, for any x E G, o(x) = I or o(x) = p.


Abstract Algebra

Notes Let x G, x e and H = < x > . Since x e, o(H) 1

o(H) = p.

Therefore, y G such that y H. Then, by the same reasoning, K = < y > is of order p. Both Hand K are normal in G, since G is abelian.

We want to show that G = H × K. For this, consider H K. Now H K H.

o(H K) | o(H) = p. o(H K) = 1 or o (H K) = p.

If o(H K) = p, then H K = H, and by similar reasoning, H K = K. But then,

H = K. y H, a contradiction.

o(H K) = 1, i.e., H K = {e}.

So, H G, K G, H K = {e} and o(HK) = p2 = o(6).

G = H × K Zp × Z

p.

So far we have shown the algebraic structure of all groups of order 1 to 10, except groups of order8. Now we will list the classification of groups of order 8.

If G is an abelian group of order 8, then

(i) G Z8, the cyclic group or order 8, or

(ii) G Z4 × Z

2, or

(iii) G Z2 × Z

2 × Z

2.

If G is a non-abelian group of order 8, then

(i) G Q8, the quaternion group discussed in Unit 4, or

(ii) G D8, the dihedral group discussed in Unit 5.

So, we have seen what the algebraic structure of any group of order 1, 2, . . . . , 10 must be. Wehave said that this classification is up to isomorphism. So, for example, any group of order 10 isisomorphic to Z

10 or D

10. It need not be equal to either of them.

Self Assessment

1. Let a group G be the ................... product of its subgroups H and k. Then hk = kh h H,k K.

(a) external (b) internal

(c) finite (d) infinite

2. Let H and k be normal subgroups of a group G such that G = H × k. Then G/H ...................and G/k H

(a) k (b) H

(c) H-1 (d) k-1�

3. Let G ................... be and H and k be its subgroup such that G = H × k. Thus O(G) = O(H) o(k).

(a) external (b) internal




Notes4. If a prime p divides the order of a finite group G, then G contains an element of ...................

(a) P (b) G

(c) Q (d) R

5. If a prime P divides the order of a finite group G, then G contains a ................... of order P.

(a) subgroup (b) normal

(c) cycle (d) permutation

9.4 Summary

In this unit we have discussed the following points:

The definition and examples of external direct products of groups.

The definition and examples of internal direct products of normal subgroups.

If (m, n) = 1, then Zm

× Zn Z

mn.

o(H × K) = o(H) o(K).

The statement and application of Sylow�s theorems, which state that: Let G be a finite

group of order pnm, where p is a prime and p | m. Then

G contains a subgroup of order pk k = 1, ... , n;

any two Sylow p-subgroups are conjugate in G;

the number of distinct Sylow p-subgroups of G is congruent to 1 (mod p) and divideso(G) (in fact, it divides m).

Let o(G) = pq, p a prime, p > q, q | p � 1. Then G is cyclic.

Let o(G) = p2, p a prime. Then

G is abelian.

G is cyclic or G Zp × Z

p.

The classification of groups of order 1 to 10, which we give in the following table.

O(G) Algebraic Structure

1 {e}

2 Z2

3 Z3

4 Z4 or Z2 × Z2

5 Z5

6 Z6 or S3

7 Z7

8 Z8 or Z4 × Z2 or Z2 × Z2 × Z2 (if G is abelian)

Q8 or D8 (if G is non-abelian)

9 Z9 or Z3 × Z3

10 Z10 or D10


Abstract Algebra

Notes 9.5 Keywords

External Direct Product: Let (G1, *

1), (G

2, *

2), . . . . . , (G

n, *

n) be n groups. Their external direct

product is the group (G, *), where

G = G1 × G

2 ..... × G

n and

Thus, Rn is the external direct product of n copies of R.

Internal Direct Product: Let H and K be normal subgroups of a group G. We call G the internaldirect product of H and K if

G = HK and H K = {e}.

We write this fact as G = H × K.

Sylow p-subgroup: Let G be a finite group and p be a prime such that pn | o(G) but pn+1 o(G), forsome n 1. Then a subgroup of G of order pn is called a Sylow p-subgroup of G.


1. Show that the binary operation * on G is associative. Find its identity element and theinverse of any element (x, y) in G.

2. Show that G1 × G

2 = G

2 × G

1, for any two groups G

1 and G

2.

3. Show that G1 × G

2 is the product of its normal subgroup H = G

1 × {e

2} and K = {e

1} × G

2. Also

show that (G1 × {e

2}) ({e

1} × G

2) = {(e

1, e

2)}.

4. Prove that P(G1 × G

3) = Z(G

1) × Z(G

2), where Z(G

3) denotes the centre of G (see Theorem 2

of unit 3).

5. Let A and B be cyclic groups of order m and n, respectively, where (m, n) = 1. Prove thatA × B is cyclic of order mn.

(Hint: Define f : Z Zm

× Zn : f(r) = (r + mZ, r + nZ). Then apply the Fundamental theorem

of Homomorphism to show that Zm

× Zn Z

mn.

6. Let H and K be normal subgroups of G which satisfy (a) of Theorem 1. Then show thatG = H × K.

7. Use Theorem 2 to prove Theorem 3.


1. (b) 2. (a) 3. (c) 4. (a) 5. (a)







Notes


www.math.niu.edu

www.maths.tcd.ie/






Abstract Algebra

Notes Unit 10: Finite Abelian Groups

CONTENTS

Objectives

Introduction

10.1 Definition

10.2 Properties

10.3 Notation

10.4 Summary

10.5 Keywords



Objectives


Define finite abelian group

Explain the properties of finite abelian group

Discuss the notation of finite abelian group

Introduction

A group for which the elements commute (i.e., AB = BA for all elements A and B) is called a finiteabelian group. All cyclic groups are finite abelian, but a finite abelian group is not necessarilycyclic. All subgroups of a finite abelian group are normal. In a finite abelian group, each elementis in a conjugacy class by itself, and the character table involves powers of a single elementknown as a group generator. In Mathematica, the function finite abelian group[{n

1, n

2 ...}] represents the direct product of the cyclic groups of degrees n

1 n

2 ...

10.1 Definition

A finite abelian group is a set, A, together with an operation �� that combines any two elementsa and b to form another element denoted a � b. The symbol �� is a general placeholder for aconcretely given operation. To qualify as a finite abelian group, the set and operation,(A, �), must satisfy five requirements known as the finite abelian group axioms:

Closure

For all a, b in A, the result of the operation a � b is also in A.


Unit 10: Finite Abelian Groups

NotesAssociatively

For all a, b and c in A, the equation (a � b) � c = a � (b � c) holds.

Identity Element

There exists an element e in A, such that for all elements a in A, the equation e � a = a � e = a holds.

Inverse Element

For each a in A, there exists an element b in A such that a � b = b � a = e, where e is the identityelement.

Commutatively

For all a, b in A, a � b = b � a.

More compactly, a finite abelian group is a commutative group. A group in which the groupoperation is not commutative is called a �non-finite abelian group� or �non-commutative group�.

You should notice that any field is a finite abelian group under addition. Furthermore, undermultiplication, the set of non-zero elements of any field must also form a finite abelian group.Of course, in this case the two operations are not independent�they are connected by thedistributive laws.

The definition of a finite abelian group is also useful in discussing vector spaces and modules.In fact, we can define a vector space to be a finite abelian group together with a scalar multiplicationsatisfying the relevant axioms. Using this definition of a vector space as a model, we can state thedefinition of a module in the following way.

10.2 Properties

Let us assume that, If n is a natural number and x is an element of a finite abelian group G writtenadditively, then nx can be defined as x + x + ... + x (n summands) and (�n)x = �(nx). In this way, Gbecomes a module over the ring Z of integers. In fact, the modules over Z can be identified withthe finite abelian groups.

Theorems about finite abelian groups can often be generalized to theorems about modules overan arbitrary principal ideal domain. A typical example is the classification of finitely generatedfinite abelian groups which is a specialization of the structure theorem for finitely generatedmodules over a principal ideal domain. In the case of finitely generated finite abelian groups,this theorem guarantees that a finite abelian group splits as a direct sum of a torsion group anda free finite abelian group. The former may be written as a direct sum of finitely many groupsof the form Z/pkZ for p prime, and the latter is a direct sum of finitely many copies of Z.

If f, g : G H are two group homomorphisms between finite abelian groups, then their sumf + g, defined by (f + g)(x) = f(x) + g(x), is again a homomorphism. (This is not true if H is a non-finite abelian group.) The set Hom (G, H) of all group homomorphisms from G to H thus turnsinto a finite abelian group in its own right.

Somewhat kind to the dimension of vector spaces, every finite abelian group has a rank. It isdefined as the cardinality of the largest set of linearly independent elements of the group. Theintegers and the rational numbers have rank one, as well as every subgroup of the rationals.


Abstract Algebra

Notes 10.3 Notation

There are two main notational conventions for finite abelian groups: �+� additive and �.�

multiplicative.

Convention Operation Identity Powers Inverse

Addition x + y 0 nx �x

Multiplication x * y or xy e or 1 xn x�1

Generally, the multiplicative notation is the usual notation for groups, while the additive notationis the usual notation for modules. The additive notation may also be used to emphasize that aparticular group is abelian, whenever both abelian and non-finite abelian groups are considered.

Multiplication Table

To verify that a finite group is abelian, a table (matrix) - known as a Cayley table - can beconstructed in a similar fashion to a multiplication table. If the group is G = {g

1 = e, g

2, ..., g

n} under

the operation �, the (i, j)�th entry of this table contains the product gi . g

j. The group is abelian if

and only if this table is symmetric about the main diagonal.

This is true since if the group is abelian, then gi . g

j = g

j . g

i. This implies that the (i, j)�th entry of the

table equals the (j, i)�th entry, thus the table is symmetric about the main diagonal.

Examples:

1. For the integers and the operation addition �+�, denoted (Z,+), the operation + combinesany two integers to form a third integer, addition is associative, zero is the additiveidentity, every integer n has an additive inverse, �n, and the addition operation iscommutative since m + n = n + m for any two integers m and n.

2. Every cyclic group G is abelian, because if x, y are in G, then xy = aman = am + n = an + m = anam =yx. Thus the integers, Z, form a finite abelian group under addition, as do the integersmodulo n, Z/nZ.

3. Every ring is a finite abelian group with respect to its addition operation. In a commutativering the invertible elements, or units, form an abelian multiplicative group. In particular,the real numbers are a finite abelian group under addition, and the non-zero real numbersare a finite abelian group under multiplication.

4. Every subgroup of a finite abelian group is normal, so each subgroup gives rise to a quotientgroup. Subgroups, quotients, and direct sums of finite abelian groups are again abelian.

In general, matrices, even invertible matrices, do not form a finite abelian group undermultiplication because matrix multiplication is generally not commutative. However, somegroups of matrices are finite abelian groups under matrix multiplication - one example is thegroup of 2 x 2 rotation matrices.

Example: Find all finite abelian groups of order 108 (up to isomorphism).



NotesSolution: The prime factorization is 108 = 22 · 33. There are two possible groups of order 4: Z4 and

Z2 × Z

2 . There are three possible groups of order 27: Z

27 , Z

9 × Z

3 , and Z

3 × Z

3 × Z

3 . This gives us

the following possible groups:

Z4 × Z

27

Z2 × Z

2 × Z

27

Z4 × Z

9 × Z

3

Z2 × Z

2 × Z

9 × Z

3

Z4 × Z

3 × Z

3 × Z

3

Z2 × Z

2 × Z

3 × Z

3 × Z

3 .

Example: Let G and H be finite abelian groups, and assume that G × G is isomorphic toH × H. Prove that G is isomorphic to H.

Solution: Let p be a prime divisor of |G|, and let q = pm be the order of a cyclic component of G.If G has k such components, then G × G has 2k components of order q. An isomorphism betweenG × G and H × H must preserve these components, so it follows that H also has k cyclic componentsof order q. Since this is true for every such q, it follows that G H

Example: Let G be a finite abelian group which has 8 elements of order 3, 18 elements oforder 9, and no other elements besides the identity. Find (with proof) the decomposition of G asa direct product of cyclic groups.

Solution: We have |G| = 27. First, G is not cyclic since there is no element of order 27. Since thereare elements of order 9, G must have Z

9 as a factor. To give a total of 27 elements, the only

possibility is G Z9 × Z

3.

Check: The elements 3 and 6 have order 3 in Z9, while 1 and 2 have order 3 in Z

3. Thus, the

following 8 elements have order 3 in the direct product: (3, 0), (6, 0), (3, 1), (6, 1), (3, 2), (6, 2),(0, 1), and (0, 2).

Example: Let G be a finite abelian group such that |G| = 216. If | 6 G | = 6, determine G upto isomorphism.

Solution: We have 216 = 23 · 33, and 6G Z2 × Z

3 since it has order 6. Let H be the Sylow

2-subgroup of G, which must have 8 elements. Then multiplication by 3 defines an automorphismof H, so we only need to consider 2H. Since 2H Z

2, we know that there are elements not of order

2, and that H is not cyclic, since 2 Z8 Z

4. We conclude that H Z

4 × Z

2.

A similar argument shows that the Sylow 3-subgroup K of G, which has 27 elements, must beisomorphic to Z

9 × Z

3.

Using the decomposition, we see that

G Z4 × Z

2 × Z

9 × Z

3.

(If you prefer the form of the decomposition, you can also give the answer in the form G Z36

× Z6.)

Example: Apply both structure theorems to give the two decompositions of the finite

abelian group 216Z

Solution: 216Z 8Z × 27Z Z

2 × Z

2 × 27Z


Abstract Algebra

Notes Since 27 is a power of an odd prime, it follows that 27Z is cyclic. This can also be shown directly

by guessing that 2 is a generator.

Since 27Z has order 33 - 32 = 18, an element can only have order 1, 2, 3, 6, 9 or 18. We have

22 = 4,

23 = 8,

26 82 10, and

29 23 · 26 8 · 10 -1,

so it follows that 2 must be a generator.

We conclude that 216Z Z

2 × Z

2 × Z

18.

To give the first decomposition, states that any finite abelian group is isomorphic to a directproduct of cyclic groups of prime power order. In this decomposition we need to split Z

18 up into

cyclic subgrops of prime power order, so we finally get the decomposition

216Z Z2 × Z

2 × Z

2 × Z

9.

On the other hand, the second decomposition, where any finite finite abelian group is written asa direct product of cyclic groups in which the orders any component is a divisor of the previousone. To do this we need to group together the largest prime powers that we can. In the firstdecomposition, we can combine Z

2 and Z

9 to get Z

18 as the first component. We end up with

216Z Z18

× Z2 × Z

2

as the second way of breaking 216Z up into a direct product of cyclic subgroups.

Example: Let G and H be finite abelian groups, and assume that they have the followingproperty. For each positive integer m, G and H have the same number of elements of order m.Prove that G and H are isomorphic.

Solution: We give a proof by induction on the order of |G|. The statement is clearly true forgroups of order 2 and 3, so suppose that G and H are given, and the statement holds for allgroups of lower order. Let p be a prime divisor of |G|, and let G

p and H

p be the Sylow

p-subgroups of G and H, respectively. Since the Sylow subgroups contain all elements of ordera power of p, the induction hypothesis applies to G

p and H

p. If we can show that G

p H

p for all p,

then it will follow that G H, since G and H are direct products of their Sylow subgroups.

Let x be an element of Gp with maximal order q = pm. Then < x > is a direct factor of G

p, so there

is a subgroup G� with Gp = < x > × G�. By the same argument we can write H

p = < y > × H�, where

y has the same order as x.

Now consider < xp > × G� and < yp > × H�. To construct each of these subgroups we have removedelements of the form (xk, g�), where xk has order q and g� is any element of G�. Because x hasmaximal order in a p-group, in each case the order of g� is a divisor of q, and so (xk, g�) has orderq since the order of an element in a direct product is the least common multiple of the orders ofthe components. Thus to construct each of these subgroups we have removed (pm � pm-1) · |G�|elements, each having order q. It follows from the hypothesis that we are left with the samenumber of elements of each order, and so the induction hypothesis implies that < xp > × G� and< yp > × H� are isomorphic. But then G� H�, and so G

p H

p, completing the proof.



NotesProposition: Every finite abelian group has a natural structure as a module over the ring Z.

As with vector spaces, one goal is to be able to express a finite abelian group in terms of simplerbuilding blocks. For vector spaces we can use one-dimensional spaces as the building blocks; forfinite abelian groups, it seems natural to use the simple finite abelian groups.

Recall that in an arbitrary group G, a subgroup N G is called a normal subgroup if gxg�1 N,for all x N and all g G. Then G is said to be a simple group if its only normal subgroups are{1} and G. If the group A is abelian, then all subgroups are normal, and so A is simple iff its onlysubgroups are the trivial subgroup (0) and the improper subgroup A. The same definition isgiven for modules: a nonzero module M is a simple module if its only submodules are (0) andM. When you view a finite abelian group as a Z-module, then, of course, the two definitionscoincide.

Note Any cyclic finite abelian group is isomorphic to Z or Zn, for some n.

Outline of the Proof: Let A be a cyclic finite abelian group that is generated by the singleelement a. Define the group homomorphism f : Z A by setting f(n) = na, for all n Z. Note thatf maps Z onto A since f(Z) = Za = A. If f is one-to-one, then A is isomorphic to Z. If f is notone-to-one, we need to use the fundamental homomorphism theorem and the fact that everysubgroup of Z is cyclic to show that A is isomorphic to Z

n, where n is the smallest positive

integer such that na = 0.

Proposition: A finite abelian group is simple iff it is isomorphic to Zp, for some prime number p.

Proof: First, let A be a finite abelian group isomorphic to Zp, where p is a prime number. The

isomorphism preserves the subgroup structure, so we only need to know that Zp has no proper

nontrivial subgroups. This follows from the general correspondence between subgroups of Zn

and divisors of n, since p is prime precisely when its only divisors are ±1 and ±p, which correspondto the subgroups Z

p and (0), respectively.

Conversely, suppose that A is a simple finite abelian group. Since A is nonzero, pick anynonzero element a A. Then the set Za = {na | n Z} is a nonzero subgroup of A, so byassumption it must be equal to A. This shows that A is a cyclic group. Furthermore, A can�t beinfinite, since then it would be isomorphic to Z and would have infinitely many subgroups. Weconclude that A is finite, and hence isomorphic to Z

n, for some n. Once again, the correspondence

between subgroups of Zn and divisors of n shows that if Z

n is simple, then n must be a prime

number.

A module M is said to be semisimple if it can be expressed as a sum (possibly infinite) of simplesubmodules. Although the situation for finite abelian groups is more complicated than forvector spaces, it is natural to ask whether all finite abelian groups are semisimple.

Example: The group Z4 is not a semisimple Z-module. First, Z

4 is not a simple group.

Secondly, it cannot be written non-trivially as a direct sum of any subgroups, since its subgroupslie in a chain Z

4 2Z

4 (0), and no two proper nonzero subgroups intersect in (0).

Example: The group Z6 is a semisimple Z-module. To see this, define f : Z

6 Z

2 Z

3 by

setting f(0) = (0, 0), f(1) = (1, 1), f(2) = (0, 2), f(3) = (1, 0), f(4) = (0, 1), f(5) = (1, 2). You can check thatthis defines an isomorphism, showing that Z

6 is isomorphic to a direct sum of simple finite

abelian groups.


Abstract Algebra

Notes The function defined in the example is a special case of a more general result that is usuallyreferred to as the Chinese remainder theorem (this result is given more generally for rings. Theproof of the next proposition makes use of the same function.

Proposition: If k = mn, where m and n are relatively prime integers, then Zk is isomorphic to

Zm

Zn.

Outline of the Proof: Define f : Zk Z

m Z

n by f([x]

k) = ([x]

m, [x]

n), for all x Z. Here I have been

a bit more careful, by using [x]k to denote the congruence class of x, modulo k. It is not hard to

show that f preserves addition. The sets Zk and Z

m Z

n are finite and have the same number of

elements, so f is one-to-one iff it is onto, and therefore proving one of these conditions will givethe other. (Actually, it isn�t hard to see how to prove both conditions.) Showing that f is one-to-one depends on the fact that if x is an integer having both m and n as factors, then it must havemn as a factor since m and n are relatively prime. On the other hand, the usual statement of theChinese remainder theorem is precisely the condition that f is an onto function.

Corollary: Any finite cyclic group is isomorphic to a direct sum of cyclic groups of prime powerorder.

The corollary depends on an important result in Z: every positive integer can be factored into aproduct of prime numbers. Grouping the primes together, the proof of the corollary uses inductionon the number of distinct primes in the factorization.

This basic result has implications for all finite groups. The cyclic group Zn also has a ring

structure, and the isomorphism that proves the corollary is actually an isomorphism of rings,not just of finite abelian groups. To use this observation, suppose that A is a finite finite abeliangroup. Let n be the smallest positive integer such that na = 0 for all a A. (This number might befamiliar to you in reference to a multiplicative group G, where it is called the exponent of thegroup, and is the smallest positive integer n such that gn = 1 for all g G.)

You can check that because na = 0 for all a A, we can actually give A the structure of aZ

n-module.

Next we can apply a general result that if a ring R can be written as a direct sum R = I1 . . .I

n

of two-sided ideals, then each Ij is a ring in its own right, and every left R-module M splits up

into a direct sum M1 . . . M

n, where M

j is a module over I

j . Applying this to Z

n, we can write

Zn as a direct sum of rings of the form Z

pk , where p is a prime, and then the group A breaks up

into A1 . . . A

n, where each A

j is a p-group, for some prime p. (Recall that a group G is a

p-group if every element of G has order p.) This argument proves the next lemma. (You can alsoprove it using Sylow subgroups, if you know about them.)

Every finite abelian group can be written as a direct sum of p-groups.

The decomposition into p-groups occurs in one and only one way. Then it is possible to provethat each of the p-groups splits up into cyclic groups of prime power order, and so we have thefollowing fundamental structure theorem for finite abelian groups.

Theorem 1: Any finite abelian group is isomorphic to a direct sum of cyclic groups of primepower order.

A proof of the fundamental structure theorem, let us first discuss some of the directions itsuggests for module theory. First of all, the hope was to construct finite abelian groups out ofones of prime order, not prime power order. The only way to do this is to stack them on top ofeach other, instead of having a direct sum in which the simple groups are lined up one beside theother. To see what I mean by �stacking� the groups, think of Z

4 and its subgroups Z

4 2Z

4 (0).

It might be better to picture them vertically.



NotesZ4

|

2Z4

|

(0)

The subgroup 2Z4 = {0, 2} Z2 is simple, and so is the factor module Z

4/2Z

4 Z

2. This having Z

2

stacked on top of Z2, and the group is structured so tightly that you can�t even find an isomorphism

to rearrange the factors.

A module M is said to have a composition series of length n if there is a chain of submodules M= M

0 M

1 . . . M

n = (0) for which each factor module M

i�1/M

i is a simple module. Thus, we

would say that Z4 has a composition series of length 2. This gives a measurement that equals the

dimension, in the case of a vector space. It is also true that the length of a cyclic group of orderpn is precisely n. It can be shown that if M has a composition series of length n, then every othercomposition series also has length n, so this is an invariant of the module. Furthermore, thesame simple modules show up in both series, with the same multiplicity.

The idea of a composition series is related to two other conditions on modules. A module is saidto satisfy the ascending chain condition, or ACC, if it has no infinite chain of ascending submodules;it is said to satisfy the descending chain condition, or DCC, if it has no infinite chain of descendingsubmodules. Modules satisfying these conditions are called Noetherian or Artinian, respectively.A module has finite length iff it satisfies both the ACC and DCC. As an example to keep in mind,let�s look at the ring of integers, which has ACC but not DCC. Since mZ nZ iff n | m,generators get smaller as you go up in Z, and larger as you go down. Any set of positive integershas a smallest element, so we can�t have any infinite ascending chains, but, for example, we canconstruct the infinite descending chain 2Z 4Z 8Z ... .

The cyclic groups of prime power order play a crucial role in the structure of finite abeliangroups precisely because they cannot be split up any further. A module M can be expressed as adirect sum of two submodules M

1 and M

2 iff M

1 M

2 = (0) and M

1 + M

2 = M. In the case of a cyclic

group of prime power order, the subgroups form a descending chain, and so any two nonzerosubgroups have a nonzero intersection. A module is called indecomposable if it cannot bewritten as a direct sum of two nonzero submodules. With this terminology, the cyclic groups ofprime power order are precisely the indecomposable finite abelian groups. The major results inthis direction are (the Krull-Schmidt theorem), which show that any module with finite lengthcan be written as a direct sum of indecomposable submodules, and this decomposition is uniqueup to isomorphism and the order of the summands.

After this rather lengthy preview, or review, as the case may be, it is time to move on to studygeneral rings and modules. The next results present a proof of the structure theorem for finiteabelian groups, but you should feel free to skip them.

Lemma: Let A be a finite abelian p-group.

(a) Let a A be an element of maximal order, and let b + Za be any coset of A/Za. Then thereexists d A such that d + Za = b + Za and Zd Za = (0).

(b) Let a A be an element of maximal order. Then there exists a subgroup B with A Za B.

Proof: (a) The outline of part (a) is to let s be the smallest positive integer such that sb Za. Thenwe solve the equation sb = sx for elements x Za and let d = b � x.


Abstract Algebra

Notes Using o(x) for the order of an element x, let s be the order of b + Za in the factor group G/Za.Then sb Za, and we can write sb = (qt)a for some exponent qt such that t = p for some and

p | q. Then qa is a generator for Za, since q is relatively prime to o(a). Since s is a divisor of the

order of b, we have o(b)/s = o(sb) = o((qt)a) = o(a)/t, or simply, o(b) . t = o(a) . s. All of these arepowers of p, and so o(b) o(a) implies that s|t, say t = ms. Then x = (qm)a is a solution of theequation sb = sx. If d = b � x, then d + Za = b + Za and so sd = sb � sx = sb � sb = 0. Therefore,Zd Za = (0), since nd Za implies n(b � x) = nb � nx Za. Thus, nb Za implies n(b + Za) = Zain G/Za, so s|n and nd = 0.

(b) The outline of this part is to factor out Za and use induction to decompose A/Za into a directsum of cyclic groups. Then part (a) can be used to choose the right preimages of the generatorsof A/Za to generate the complement B of Za.

We use induction on the order of A. If |A| is prime, then A is cyclic and there is nothing to prove.Consequently, we may assume that the statement of the lemma holds for all groups of order lessthan |A| = p. If A is cyclic, then we are done. If not, let Za be a maximal cyclic subgroup, and usethe induction hypothesis repeatedly to write A/Za as a direct sum B

1 B

2 . . . B

n of cyclic

subgroups.

We next use part (a) to choose, for each i, a coset ai + Za that corresponds to a generator of A

i such

that Zai Za = (0). We claim that A Za B for the smallest subgroup B = Za

1 + Za

2 + · · · + Za

n

that contains a1, a

2, . . ., a

n.

First, if x Za (Za

1 +· · · + Za

n), then x = m

1a

1 +· · · + m

na

n Za for some coefficients m

1, . . . ,m

n.

Thus x + Za = (m1a

1 + · · · + m

na

n) + Za = Za, and since A/Za is a direct sum, this implies that

mia

i + Za = Za for each i. But then m

ia

i Za, and so m

ia

i = 0 since Za

i Za = (0). Thus x = 0.

Next, given x A, express the coset x + Za as (m1a

1 +· · · + m

na

n) + Za for coefficients m

1, . . ., m

n.

Then x xZa, and so x = ma + m1a

1 + · · · + m

na

n for some m.

Thus, we have shown that Za B = (0) and A = Za + B, so A Za B.

Theorem 2 (Fundamental Theorem of Finite Abelian Groups): Any finite abelian group isisomorphic to a direct sum of cyclic groups of prime power order. Any two such decompositionshave the same number of factors of each order.

Proof: We first decompose any finite abelian group A into a direct sum of p-groups, and then wecan use the previous lemma to write each of these groups as a direct sum of cyclic subgroups.

Uniqueness is shown by induction on |A|. It is enough to prove the uniqueness for a givenp-group. Suppose that

Zp

1 Z

p

2 · · · Z

p

n = Z

p

1 Z

p

2 · · · Z

p

m

where 1

2 . . .

n and

1

2 . . .

m. Consider the subgroups in which each element has

been multiplied by p. By induction, 1 � 1 =

1 � 1, . . ., which gives

1 =

1, . . ., with the possible

exception of the i�s and

j�s that equal 1. But the groups have the same order, and this determines

that each has the same number of factors isomorphic to Zp. This completes the proof.

Self Assessment

1. A .................. is a set, A together with an operations �.�. That combines any two elements aand b to form another element denoted a.b.

(a) cyclic (b) permutation

(c) abelian (d) normal



Notes2. In a finite abelian group, each element is in a conjugacy class by itself and the charactertable involve powers of a single element known as a ..................

(a) group generator (b) group connector

(c) group and subgroup (d) normal group element

3. In mathematica, the function finite abelian group {n1, n

2, .... } represents .................. product

of the cyclic group of degree n1n

2 ..................

(a) direct (b) indirect

(c) single (d) external

4. In commutative ring .................. the elements, or unit, from an abelian multiplicationgroups.

(a) inversible (b) vertible

(c) direct (d) finite

5. Every subgroup of a finite abelian group is normal, so each subgroup gives rest to a.................. group.

(a) cyclic (b) permutation

(c) quotient (d) multiplicative

10.4 Summary

A finite abelian group is a set, A, together with an operation �� that combines any twoelements a and b to form another element denoted a � b. The symbol �� is a generalplaceholder for a concretely given operation. To qualify as a finite abelian group, the setand operation, (A, �), must satisfy five requirements known as the finite Abelian groupaxioms.

Generally, the multiplicative notation is the usual notation for groups, while the additivenotation is the usual notation for modules. The additive notation may also be used toemphasize that a particular group is abelian, whenever both abelian and non-finite abeliangroups are considered.

For the integers and the operation addition �+�, denoted (Z,+), the operation + combinesany two integers to form a third integer, addition is associative, zero is the additiveidentity, every integer n has an additive inverse, �n, and the addition operation iscommutative since m + n = n + m for any two integers m and n.

Every cyclic group G is abelian, because if x, y are in G, then xy = aman = am + n = an + m = anam =yx. Thus the integers, Z, form a finite abelian group under addition, as do the integersmodulo n, Z/nZ.

10.5 Keywords

Finite Abelian Group: A finite abelian group is a set, A, together with an operation �� thatcombines any two elements a and b to form another element denoted a � b.

Multiplication: The multiplicative notation is the usual notation for groups, while the additivenotation is the usual notation for modules.

Cyclic Group: Every cyclic group G is abelian, because if x, y are in G, then xy = aman = am + n =an + m = anam = yx.


Abstract Algebra

Notes 10.6 Review Questions

1. Compute all possible finite abelian groups of order n. What is the largest n for which itwill work?

2. Find all finite abelian group of order less than or equal to 40 up to isomorphism.

3. Find all finite abelian groups of order 200 to 720 up to isomorphism.

4. Show that the infinite direct product G = Z2 × Z

2 × . . . is not finitely generated.

5. Let G be a finite abelian group of order m. If n divides m, prove that G has a subgroup oforder n.


1. (c) 2. (a) 3. (a) 4. (a) 5. (c)






www.math.niu.edu

www.maths.tcd.ie/






Unit 11: Conjugate Elements

NotesUnit 11: Conjugate Elements

CONTENTS

Objectives

Introduction

11.1 Conjugate Subgroup

11.2 Summary

11.3 Keywords



Objectives


Define conjugate subgroup

Discuss conjugacy class of an element

Introduction

In the last unit, you have studied about finite abelian group. If G is a group and X is an arbitraryset, a group action of an element g G and x X is a product, gx giving in x many problem inalgebra may best be attached in group actions. In this unit, you will get the information relatedto conjugate elements.

11.1 Conjugate Subgroup

Definition: Let G be a group, and let x, y be elements of G. Then y is said to be a conjugate of xif there exists an element a in G such that y = axa-1.

If H and K are subgroups of G, then K is said to be a conjugate subgroup of H if there exists anelement a in G such that K = aHa-1.

Proposition 1:

(a) Conjugacy of elements defines an equivalence relation on any group G.

(b) Conjugacy of subgroups defines an equivalence relation on the set of all subgroups of G.

Definition: Let G be a group. For any element x in G, the set

{ a in G | axa-1 = x }

is called the centralizer of x in G, denoted by C(x).

For any subgroup H of G, the set

{ a in G | aHa-1 = H }

is called the normalizer of H in G, denoted by N(H).


Abstract Algebra

Notes Proposition 2: Let G be a group and let x be an element of G. Then C(x) is a subgroup of G.

Proposition 3: Let x be an element of the group G. Then the elements of the conjugacy class of xare in one-to-one correspondence with the left cosets of the centralizer C(x) of x in G.

Example: Two permutations are conjugate in Sn if and only if they have the same shape

(i.e., the same number of disjoint cycles, of the same lengths). Thus, in particular, cycles of thesame length are always conjugate.

Theorem 1: [Conjugacy class Equation] Let G be a finite group. Then

| G | = | Z(G) | + [ g : C(x) ]

where the sum ranges over one element x from each nontrivial conjugacy class.

Definition: A group of order pn, with p a prime number and n 1, is called a p-group.

Theorem 2: [Burnside] Let p be a prime number. The center of any p-group is nontrivial.

Corollary 1: Any group of order p2 (where p is prime) is abelian.

Theorem 3: [Cauchy] If G is a finite group and p is a prime divisor of the order of G, then Gcontains an element of order p.

Example: Prove that if the center of the group G has index n, then every conjugacy class ofG has at most n elements.

Solution: The conjugacy class of an element a in G has [G : C(a)] elements. Since the center Z(G)is contained in C(a), we have [G : C(a)] [G : Z(G)] = n. (In fact, [G : C(a)] must be a divisor of n.)

Example: Find all finite groups that have exactly two conjugacy classes.

Solution: Suppose that |G| = n. The identity element forms one conjugacy class, so the secondconjugacy class must have n-1 elements. But the number of elements in any conjugacy class is adivisor of |G|, so the only way that n-1 is a divisor of n is if n = 2.

Example: Let G = D12

, given by generators a, b with |a|=6, |b|=2, and ba=a-1b. Let H ={ 1, a3, b, a3b }. Find the normalizer of H in G and find the subgroups of G that are conjugate to H.

Solution: The normalizer of H is a subgroup containing H, so since H has index 3, either NG (H)

= H or NG (H) = G. Choose any element not in H to do the first conjugation.

aHa-1 = { 1, a(a3)a5, aba5, a(a3b)a5 } = { 1, a3, a2b, a5b }

This computation shows that a is not in the normalizer, so NG (H) = H. Conjugating by any

element in the same left coset aH = { a, a4, ab, a4b } will give the same subgroup. Therefore, itmakes sense to choose a2 to do the next computation.

a2Ha-2 = { 1, a3, a2ba4, a2(a3b)a4 } = { 1, a3, a4b, ab }

Comment: It is interesting to note that an earlier problem shows that b, a2b, and a4b form oneconjugacy class, while ab, a3b, and a5 b form a second conjugacy class. In the above computations,notice how the orbits of individual elements combine to give the orbit of a subgroup.

Example: Write out the class equation for the dihedral group Dn. Note that you will need

two cases: when n is even, and when n is odd.



NotesSolution: When n is odd the center is trivial and elements of the form ai b are all conjugate.Elements of the form ai are conjugate in pairs; am a-m since a2m 1. We can write the classequation in the following form:

|G| = 1 + ((n-1)/2) · 2 + n

When n is even, the center has two elements. (The element an/2 is conjugate to itself since it isequal to a-n/2. This shows that Z(G) = { 1, an/2 }.) Therefore, elements of the form ai b split into twoconjugacy classes. In this case the class equation has the following form:

|G| = 2 + ((n-2)/2) · 2 + 2 · (n/2)

Example: Show that for all n 4, the centralizer of the element (1,2)(3,4) in Sn has order

8· (n-4)!. Determine the elements in the centralizer of ((1,2)(3,4)).

Solution: The conjugates of a = (1,2)(3,4) in Sn are the permutations of the form (a,b) (c,d). The

number of ways to construct such a permutation is

n(n-1)/2 · (n-2)(n-3)/2 · 1/2 ,

and dividing this into n! gives the order 8 · (n-4)! of the centralizer.

We first compute the centralizer of a in S4. The elements (1, 2) and (3, 4) clearly commute with

(1, 2) (3, 4). Note that a is the square of b = (1, 3, 2, 4); it follows that the centralizer contains, so b3 = (1, 4, 2, 3) also belongs. Computing products of these elements shows that we mustinclude (1, 3)(2, 4) and (1, 4)(2, 3), and this gives the required total of 8 elements.

To find the centralizer of a in Sn, any of the elements listed above can be multiplied by any

permutation disjoint from (1, 2)(3, 4). This produces the required total |C(a)| = 8 · (n-4)!.

Self Assessment

1. Let G be a group and let x be an elements of the G. Then L(x) is a ............... of G.

(a) Normal subgroup (b) Cyclic subgroup

(c) Subgroup (d) Permutation group

2. Any group of order p2 is ...............

(a) permutation (b) abelian

(c) cyclic (d) finite

3. If G is a ............... group and P is a prime divisor of the order of G, then G contains anelement of order P.

(a) direct (b) external

(c) internal (d) finite

4. Let P be a prime number. The center of any P-group is ...............

(a) trivial (b) non-trivial


5. A group of order pn, with P is a prime number and n ............... is called a p-group.

(a) a = 1 (b) b > 1

(c) c < 1 (d) d 1


Abstract Algebra

Notes 11.2 Summary

Let G be a group, and let x,y be elements of G. Then y is said to be a conjugate of x if thereexists an element a in G such that y = axa-1.

If H and K are subgroups of G, then K is said to be a conjugate subgroup of H if there existsan element a in G such that K = aHa-1.

Conjugacy of elements defines an equivalence relation on any group G.

Conjugacy of subgroups defines an equivalence relation on the set of all subgroups of G.

Let G be a group. For any element x in G, the set

{ a in G | axa-1 = x }


For any subgroup H of G, the set

{ a in G | aHa-1 = H }

is called the normalizer of H in G, denoted by N(H).

Let G be a group and let x be an element of G. Then C(x) is a subgroup of G.

Let x be an element of the group G. Then the elements of the conjugacy class of x are inone-to-one correspondence with the left cosets of the centralizer C(x) of x in G.

11.3 Keywords

Conjugate Element: If H and K are subgroups of G, then K is said to be a conjugate subgroup ofH if there exists an element a in G such that K = aHa-1.

Centralizer: Let G be a group. For any element x in G, the set

{ a in G | axa-1 = x }



1. Compute the G-equivalence classes of X for each of the G-sets X = {1, �2, 24, 5, 6} andG = {(1), (1, 2) (3, 4, 5) ; (1 2) (3 4 5), (1 2) (3 8 4)} for each x X verify |G| = |O

x| |G

x|.

2. Write the class equation for S5 and for |Gx|

3. Let P be prime. Show that the number of different abelian groups of order Pn is the sameas the number of conjugacy class in S

n.

4. Let a G, show that for any g G, gc(a)g-1 = c(gag-1).

5. Let |G| = pn and suppose that |Z(G)| = pn-1 for p prime. Prove that G is abelian.

6. Let G be a group with order pn, where p is prime and X a finite G-set. If XG = {x X : gx = x

for all g G} is the set of elements in X fixed by the group actions, then prove that|X| = |X

G| (mod

p).


1. (c) 2. (b) 3. (d) 4. (b) 5. (d)








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Abstract Algebra

Notes Unit 12: Sylow�s Theorems

CONTENTS

Objectives

Introduction

12.1 The Sylow Theorems

12.1.1 A Proof of Sylow�s Theorems

12.2 Summary

12.3 Keywords



Objectives


Discuss Sylow�s Theorem

Describe examples of Sylow�s Theorem

Introduction

We already know that the converse of Lagrange�s Theorem is false. If G is a group of order m andn divides m, then G does not necessarily possess a subgroup of order n. For example, A

4 has

order 12 but does not possess a subgroup of order 6. However, the Sylow Theorems do providea partial converse for Lagrange�s Theorem: in certain cases they guarantee us subgroups ofspecific orders. These theorems yield a powerful set of tools for the classification of all finitenon-abelian groups.

12.1 The Sylow Theorems

We will use the idea of group actions to prove the Sylow Theorems. Recall for a moment whatit means for G to act on itself by conjugation and how conjugacy classes are distributed in thegroup according to the class equation. A group G acts on itself by conjugation via the map (g, x) gxg-1. Let x

1,...,x

k be representatives from each of the distinct conjugacy classes of G that consist

of more than one element. Then the class equation can be written as

|G| = |Z(G)| + [G : C(x1)] + ... + [G : C(x

k)],

where Z(G) = {g G : gx = xg for all x G} is the center of G and C(xi) = {g G : gx

i = x

ig} is the

centralizer subgroup of xi.

We now begin our investigation of the Sylow Theorems by examining subgroups of order p,where p is prime. A group G is a p-group if every element in G has as its order a power of p,where p is a prime number. A subgroup of a group G is a p-subgroup if it is a p-group.


Unit 12: Sylow�s Theorems

Notes12.1.1 A Proof of Sylow�s Theorems

In this handout, we give proofs of the three Sylow theorems which are slightly different fromthe ones in the book. Recall the following lemma:

Lemma: Let p be a prime number, and let G be a p-group (a finite group of order pk for somek 1) acting on a finite set S. Then the number of fixed points of the action is congruent to |S|modulo p.

We make the following definition: if G has order pkm with p | m, a Sylow p-subgroup of G is

a subgroup of order pk.

Theorem (Sylow�s First Theorem): If G is a finite group of order n = pkm with p prime and

p | m, then G has a subgroup of order pk. In other words, if Sylp(G) denotes the set of Sylow

p-subgroups of G, then Sylp(G) 0 .

Proof. The proof is by induction on |G|, the base case |G| = 1 being trivial. If there exists a

proper subgroup H of G such that p | [G : H], then a Sylow p-subgroup of H is also a a Sylow

p-subgroup of G and we�re finished by induction. So without loss of generality, we may assumethat p | [G : H] whenever H < G. From the class equation, it follows that p | |Z

G|. By Cauchy�s

theorem, there exists a subgroup N ZG of order p, which is necessarily normal in G. Let G =

G/N, so | G | = pk�1m. By induction, G has a subgroup P of order pk�1. Let P be the subgroup

of G containing N which corresponds to P by the first isomorphism theorem. Then

|P| = |P/N| . |N| = pk�1 . p = pk,

so that P is a Sylow p-subgroup of G as desired.

Theorem (Sylow�s Second Theorem): If G is a finite group and p is a prime number, then allSylow p-subgroups of G are conjugate to one another.

Proof: We show more precisely that if H is any subgroup of G of p-power order and P is anySylow p-subgroup of G, then there exists x G such that H xPx�1. (This implies the theorem,since if H Syl

p(G) then |H| = |P| = |xPx�1|, which implies that H = xPx�1, so that H is conjugate

to P.) Note that H acts on G/P (the set of left cosets of P in G) by left multiplication. Let Fix denotethe elements of G/P fixed by this action. Then |Fix| |G/P| (mod p) by the Lemma. Since

p | m = |G/P|, |Fix| 0, and thus Fix 0 ;. Let xP be a left coset fixed by the action. Then

hxP = xP h H x�1 Hx P,

so that H xPx�1 as desired.

Theorem (Sylow�s Third Theorem): If G is a finite group and p is a prime number, letn

p = |Syl

p(G)|. Then n

p | |G| and n

p 1 (mod p).

Proof: We consider the action of G on Sylp(G) by conjugation. By the second Sylow theorem, thisaction is transitive, so there is just one orbit. Hence n

p, which is the size of this orbit, divides |G|.

To prove the congruence np 1 (mod p), we fix a Sylow p-subgroup P Syl

p(G) and consider the

action of P on Sylp(G) by conjugation. Let Fix denote the set of fixed points of this action. Note

that Q Fix P NG(Q), and in particular P Fix. If Q Fix, then P, Q N

G(Q) are both Sylow

p-subgroups of NG(Q), so they are conjugate in N

G(Q) (again by the second Sylow theorem). But

Q is a normal subgroup of NG(Q), so P = Q. Thus Fix = {P}, and in particular |Fix| = 1. By the

Lemma, np 1 (mod p) as desired.


Abstract Algebra

Notes The more precise fact established in our proof of Sylow�s Second Theorem yields the followinguseful result:

Corollary: If G is a finite group and p is a prime number, then any subgroup of G of p-powerorder is contained in some Sylow p-subgroup.

Since G acts transitively by conjugation on Sylp(G), and the stabilizer of P Syl

p(G) is N

G(P), we

deduce that np = [G : NG(P)] for any P Syl

p(G).

Therefore:

Corollary: If G is a finite group and p is a prime number, let np be the number of Sylow

p-subgroups of G. Then the following are equivalent:

1. np = 1.

2. Every Sylow p-subgroup of G is normal.

3. Some Sylow p-subgroup of G is normal.

Example: By direct computation, find the number of Sylow 3-subgroups and the numberof Sylow 5-subgroups of the symmetric group S

5. Check that your calculations are consistent

with the Sylow theorems.

Solution: In S5 there are ( 5 · 4 · 3 ) / 3 = 20 three cycles. These will split up into 10 subgroups of

order 3. This number is congruent to 1 mod 3, and is a divisor of 5 · 4 · 2.

There are ( 5! ) / 5 = 24 five cycles. These will split up into 6 subgroups of order 5. This numberis congruent to 1 mod 5, and is a divisor of 4 · 3 · 2.

Example: How many elements of order 7 are there in a simple group of order 168?

Solution: First, 168 = 23 . 3 . 7. The number of Sylow 7-subgroups must be congruent to 1 mod 7and must be a divisor of 24. The only possibilities are 1 and 8. If there is no proper normalsubgroup, then the number must be 8. The subgroups all have the identity in common, leaving8 · 6 = 48 elements of order 7.

Example: Prove that a group of order 48 must have a normal subgroup of order 8 or 16.

Solution: The number of Sylow 2-subgroups is 1 or 3. In the first case there is a normal subgroupof order 16. In the second case, let G act by conjugation on the Sylow 2-subgroups. This producesa homomorphism from G into S

3. Because of the action, the image cannot consist of just 2

elements. On the other hand, since no Sylow 2-subgroup is normal, the kernel cannot have 16elements. The only possibility is that the homomorphism maps G onto S

3, and so the kernel is a

normal subgroup of order 48 / 6 = 8.

Example: Let G be a group of order 340. Prove that G has a normal cyclic subgroup oforder 85 and an abelian subgroup of order 4.

Solution: First, 340 = 22 . 5 . 17. There exists a Sylow 2-subgroup of order 4, and it must be abelian.No divisor of 68 = 22 . 17 is congruent to 1 mod 5, so the Sylow 5-subgroup is normal. Similarly,then Sylow 17-subgroup is normal. These subgroups have trivial intersection, so their productis a direct product, and hence must be cyclic of order 85 = 5 . 17. The product of two normalsubgroups is again normal, so this produces the required normal subgroup of order 85.



Notes

Example: Show that there is no simple group of order 200.

Solution: Since 200 = 23 . 52, the number of Sylow 5-subgroups is congruent to 1 mod 5 and adivisor of 8. Thus there is only one Sylow 5-subgroup, and it is a proper nontrivial normalsubgroup.

Example: Show that a group of order 108 has a normal subgroup of order 9 or 27.

Solution: Let S be a Sylow 3-subgroup of G. Then [G:S] = 4, since |G| = 22 33, so we can let G actby multiplication on the cosets of S. This defines a homomorphism µ : G -> S

4, so it follows that

| µ(G) | is a divisor of 12, since it must be a common divisor of 108 and 24. Thus | ker(µ) | 9,and it follows that ker(µ) S, so | ker(µ) | must be a divisor of 27. It follows that | ker(µ) | = 9or | ker(µ) | = 27.

Example: If p is a prime number, find all Sylow p-subgroups of the symmetric group Sp.

Solution: Since |Sp| = p!, and p is a prime number, the highest power of p that divides |S

p| is p.

Therefore, the Sylow p-subgroups are precisely the cyclic subgroups of order p, each generatedby a p-cycle. There are (p-1)! = p! / p ways to construct a p-cycle (a

1, . . . , a

p). The subgroup

generated by a given p-cycle will contain the identity and the p-1 powers of the cycle. Twodifferent such subgroups intersect in the identity, since they are of prime order, so the totalnumber of subgroups of order p in S

p is (p-2)! = (p-1)! / (p-1).

Example: Prove that if G is a group of order 56, then G has a normal Sylow 2-subgroupor a normal Sylow 7-subgroup.

Solution: The number of Sylow 7-subgroups is either 1 or 8. Eight Sylow 7-subgroups wouldyield 48 elements of order 7, and so the remaining 8 elements would constitute the (unique)Sylow 2-subgroup.

Example: Prove that if N is a normal subgroup of G that contains a Sylow p-subgroup ofG, then the number of Sylow p-subgroups of N is the same as that of G.

Solution: Suppose that N contains the Sylow p-subgroup P. Then since N is normal it alsocontains all of the conjugates of P. But this means that N contains all of the Sylow p-subgroupsof G, since they are all conjugate. We conclude that N and G have the same number of Sylowp-subgroups.

Example: Prove that if G is a group of order 105, then G has a normal Sylow 5-subgroupand a normal Sylow 7-subgroup.

Solution: The notation np(G) will be used for the number of Sylow p-subgroups of G. Since 105

= 3 · 5 · 7, we have n3(G) = 1 or 7, n

5(G) = 1 or 21, and n

7(G) = 1 or 15 for the numbers of Sylow

subgroups. Let P be a Sylow 5-subgroup and let Q be a Sylow 7-subgroup. At least one of thesesubgroups must be normal, since otherwise we would have 21 · 4 elements of order 5 and 15 · 6elements of order 7. Therefore, PQ is a subgroup, and it must be normal since its index is thesmallest prime divisor of |G|, so we can apply the result in the previous problem. Since PQ isnormal and contains a Sylow 5-subgroup, we can reduce to the number 35 when considering thenumber of Sylow 5-subgroups, and thus n

5(G) = n

5(PQ) = 1. Similarly, since PQ is normal and

contains a Sylow 7-subgroup, we have n7(G) = n

7(PQ) = 1.


Abstract Algebra

Notes Self Assessment

1. A group G is a p-group of every element in G has its order a power of ................

(a) g (b) p

(c) g-1 (d) p-1

2. If G is a finite group. Then G is p-group of and only if |G| = ................

(a) p-p (b) pp

(c) pn (d) pn

3. Let P be a Sylow p-subgroups of a ................ G and let x have as its order a power of p. Ifx-1p(x) = p. Then x p.

(a) indirect (b) infinite

(c) finite (d) direct

4. A subgroup of a group G is a p- ................ if it is a p-group.

(a) subgroup (b) normal group

(c) infinite group (d) cyclic group

5. How many elements of order 7 are there is a simple group of order 168.

(a) 7 (b) 8

(c) 9 (d) 48

12.2 Summary

Let G be a finite group and p a prime such that p divides the order of G. Then G contains asubgroup of order p.

(First Sylow Theorem) Let G be a finite group and p a prime such that pr divides |G|.Then G contains a subgroup of order pr.

Let P be a Sylow p-subgroup of a finite group G and let x have as its order a power of p.If x-1Px = P. Then x P.

Let H and K be subgroups of G. The number of distinct H-conjugates of K is[H : N(K) H].

(Second Sylow Theorem) Let G be a finite group and p a prime dividing |G|. Then allSylow p-subgroups of G are conjugate. That is, if P

1 and P

2 are two Sylow p-subgroups,

there exists a g G such that gP1g-1 = P

2.

12.3 Keywords

Cauchy: Let G be a finite group and p a prime such that p divides the order of G. Then G containsa subgroup of order p.

First Sylow Theorem: Let G be a finite group and p a prime such that pr divides |G|. Then Gcontains a subgroup of order pr.



Notes12.4 Review Questions

1. What are the order of all Sylow p-subgroups where G has order 18, 24, 54 and 80?

2. Find all the Sylow 3-subgroups of S4 and show that they are all conjugate.

3. Show that every group of order 45 has a normal subgroup of order 9.

4. Let H be a Sylow p-subgroup of G. Prove that H ps the only Sylow p-subgroup of Gcontained in N(H).

5. Prove that no group of order 96 is simple.

6. If H is normal subgroup of a finite group G and |H| = pk for some prime p, show that H isa contained in every Sylow p-subgroup of G.


1. (b) 2. (c) 3. (c) 4. (a) 5. (d)






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Abstract Algebra

Notes Unit 13: Solvable Groups

CONTENTS

Objectives

Introduction

13.1 Solvable Group

13.2 Summary

13.3 Keywords



Objectives


Discuss the solvable groups

Describe examples of solvable group

Introduction

In the earlier unit, you have studied about the conjugate elements and Sylow�s Theorem. Thisunit will equip you with more information related to solvable group.

13.1 Solvable Group

Definition: The group G is said to be solvable if there exists a finite chain of subgroupsG = N

0 N

1 ··· N

n such that

(i) Ni is a normal subgroup in N

i-1 for i = 1, 2, ... ,n,

(ii) Ni-1

/ Ni is abelian for i = 1, 2, ..., n, and

(iii) Nn = {e}.

Proposition: A finite group G is solvable if and only if there exists a finite chain of subgroupsG = N

0 N

1 ... N

n such that


i-1 for i = 1, 2, . . ., n,

(ii) Ni-1

/ Ni is cyclic of prime order for i = 1, 2, . . ., n, and

(iii) Nn = {e}.

Theorem 1: Let p be a prime number. Any finite p-group is solvable.

Definition: Let G be a group. An element g in G is called a commutator if

g = aba-1b-1

for elements a, b in G.


Unit 13: Solvable Groups

NotesThe smallest subgroup that contains all commutators of G is called the commutator subgroup orderived subgroup of G, and is denoted by G�.

Proposition: Let G be a group with commutator subgroup G�.

(a) The subgroup G� is normal in G, and the factor group G/G� is abelian.

(b) If N is any normal subgroup of G, then the factor group G/N is abelian if and only ifG� N.

Definition: Let G be a group. The subgroup (G� )� is called the second derived subgroup of G. Wedefine G(k) inductively as (G(k-1))�, and call it the k th derived subgroup.

Theorem 2: A group G is solvable if and only if G(n) = {e} for some positive integer n.

Corollary: Let G be a group.

(a) If G is solvable, then so is any subgroup or homomorphic image of G.

(b) If N is a normal subgroup of G such that both N and G/N are solvable, then G is solvable.

Definition: Let G be a group. A chain of subgroups G = N0 N

1 ... N

n such that


i-1 for i = 1, 2, . . . ,n,

(ii) Ni-1

/ Ni is simple for i = 1, 2, . . . ,n, and

(iii) Nn = {e}

is called a composition series for G.

The factor groups Ni-1

/ Ni are called the composition factors determined by the series.

Theorem 3: [Jordan-Hölder] Any two composition series for a finite group have the samelength. Furthermore, there exists a one-to-one correspondence between composition factors ofthe two composition series under which corresponding composition factors are isomorphic.

Example: Let p be a prime and let G be a non-abelian group of order p3. Show that thecenter Z(G) of G equals the commutator subgroup G� of G.

Solution: Since G is non-abelian, we have |Z(G)| = p. (The center is nontrivial, and if |Z(G)| =p2, then G/Z(G) is cyclic, the text implies that G is abelian.) On the other hand, any group oforder p2 is abelian, so G/Z(G) is abelian, which implies that G� Z(G). Since G is nonabelian, G� {e},and therefore G� = Z(G).

Example: Prove that Dn is solvable for all n.

One approach is to compute the commutator subgroup of Dn, using the standard description

Dn = { ai bj | 0 i < n, 0 j < 2, o(a) = n, o(b) = 2, ba = a-1b }

We must find all elements of the form xyx-1y-1, for x,y in Dn. We consider the cases x = ai or x = aib

and y = aj or y = ajb.

Case 1: If x = ai and y = aj, the commutator is trivial.

Case 2: If x = ai and y = ajb, then xyx-1y-1 = aiajba-iajb = aiajaibajb = aiajaia-jb2 = a2i, and thus each evenpower of a is a commutator.

Case 3: If x = ajb and y = ai, we get the inverse of the element in Case 2.

Case 4: If x = aib and y = ajb, then


Abstract Algebra

Notes xyx-1y-1 = aibajbaibajb, and so we get

xyx-1y-1 = aia-jb2aia-jb2 = a2(i-j), and again we get even powers of a.

Thus the commutator subgroup Dn� is either < a > (if n is odd) or < a2 > (if n is even). In either case,

the commutator subgroup is abelian, so Dn�� = {e}.

Example: Prove that any group of order 588 is solvable, given that any group of order 12is solvable.

We have 588 = 22 · 3 · 72. Let S be the Sylow 7-subgroup. It must be normal, since 1 is the onlydivisor of 12 that is 1 mod 7. By assumption, G / S is solvable since | G / S | = 12. Furthermore,S is solvable since it is a p-group. Since both S and G / S are solvable, it follows from Corollarythat G is solvable.

Example: Let G be a group of order 780=22 · 3 · 5 · 13. Assume that G is not solvable. Whatare the composition factors of G? (Assume that the only nonabelian simple group of order 60 isA

5.)

The Sylow 13-subgroup N is normal, since 1 is the only divisor of 60 that is 1 mod 13. Using thefact that the smallest simple nonabelian group has order 60, we see that the factor G/N must besimple, since otherwise each composition factor would be abelian and G would be solvable.Thus the composition factors are Z

13 and A

5.

Theorem-[Jordan-Hölder] Any two composition series for a finite group have the same length.Furthermore, there exists a one-to-one correspondence between composition factors of the twocomposition series under which corresponding composition factors are isomorphic.

Let |G| = N. We first prove existence, using induction on N. If N = 1 (or, more generally, if G issimple) the result is clear. Now suppose G is not simple. Choose a maximal proper normalsubgroup G1 of G. Then G1 has a Jordan-Hölder decomposition by induction, which produces aJordan-Hölder decomposition for G.

To prove uniqueness, we use induction on the length n of the decomposition series. If n=1 thenG is simple and we are done. For n > 1, suppose that

G G1 G2 Gn = {1}

and

G G1 G2 Gm=1

are two decompositions of G . If G1 = G1 then we�re done (apply the induction hypothesis to G1),so assume G1/G1 . Set H : = G1 G1 and choose a decomposition series H H

1 Hk = {1} for H.

By the second isomorphism theorem, G1/H=G1G1/G1=G/G1 (the last equality is because G1G1is a normal subgroup of G properly containing G1). In particular, H is a normal subgroup of G1with simple quotient. But then

G1 G2 ... Gn

and

G1 H ... Hk

are two decomposition series for G1, and hence have the same simple quotients by the inductionhypothesis; likewise for the G1 series. Therefore, n=m. Moreover, since G/G1=G1/H andG/G1=G1/H (by the second isomorphism theorem), we have now accounted for all of thesimple quotients, and shown that they are the same.



NotesSelf Assessment

1. Let P be a prime number. Any ............... p-group is solvable

(a) infinite (b) direct

(c) finite (d) indirect

2. The smallest subgroup that contains all commutations of G is called as ...............

(a) commutator subgroup (b) normal subgroup

(c) generator subgroup (d) cyclic subgroup

3. If x = ai and y = ..............., the commutator is trivial

(a) aj (b) a-1

(c) a-j (d) -1y

4. Let G be a group the subgroup is called the ............... of G.

(a) normal subgroup (b) second derived subgroup

(c) composition series (d) cyclic series

13.2 Summary

The group G is said to be solvable if there exists a finite chain of subgroups G = N0 N

1

··· Nn such that


i-1 for i = 1, 2, ..., n,

(ii) Ni-1

/ Ni is abelian for i = 1, 2, ..., n, and

(iii) Nn = {e}.

A finite group G is solvable if and only if there exists a finite chain of subgroups G = N0

N1 ... N

n such that


i-1 for i = 1, 2, . . ., n,

(ii) Ni-1

/ Ni is cyclic of prime order for i = 1, 2, . . ., n, and

(iii) Nn = {e}.

Let p be a prime number. Any finite p-group is solvable.

Let G be a group. An element g in G is called a commutator if

g = aba-1b-1

for elements a,b in G.

The smallest subgroup that contains all commutators of G is called the commutatorsubgroup or derived subgroup of G, and is denoted by G�.

Let G be a group. A chain of subgroups G = N0 N

1 ... N

n such that


i-1 for i = 1, 2, . . ., n,

(ii) Ni-1

/ Ni is simple for i = 1, 2, . . ., n, and

(iii) Nn = {e}


The factor groups Ni-1

/ Ni are called the composition factors determined by the series.


Abstract Algebra

Notes 13.3 Keywords

Commutator Subgroup: The smallest subgroup that contains all commutators of G is called thecommutator subgroup or derived subgroup of G, and is denoted by G�.

Let G be a group. A chain of subgroups G = N0 N

1 ... N

n such that


i-1 for i = 1, 2, . . ., n,

(ii) Ni-1

/ Ni is simple for i = 1, 2, . . ., n, and

(iii) Nn = {e}


Jordan-Hölder: Any two composition series for a finite group have the same length. Furthermore,there exists a one-to-one correspondence between composition factors of the two compositionseries under


1. Prove the normal series

Z60

{ 3 } { 15 } { 0 }

Z60

{ 4 } { 20 } { 0 }

of the group Z60

are isomorphic.

2. Let G and H be solvable groups. Show G × H is also solvable.

3. If G has a composition series and if N is a proper normal subgroup of G, Show the n existsa composition series containing N.

4. Let N be a normal subgroup of G. If N and G/N have composition series, then G must alsohave a composition series.

5. Let N be a normal subgroup of G if N and G/N are solvable groups. Show that G is alsosolvable group.

6. Prove that G is a solvable group if and only if G has a series of subgroups G = Pn P

n-1

... P1 P

0 = { e }

where pi is normal in p

i+1 and the order p

i+1/p

i is prime.


1. (c) 2. (a) 3. (a) 4. (b)







Notes


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Abstract Algebra

Notes Unit 14: Rings

CONTENTS

Objectives

Introduction

14.1 What is a Ring?

14.2 Elementary Properties

14.3 Two Types of Rings

14.4 Summary

14.5 Keywords



Objectives


Define and give examples of rings

Discuss some elementary properties of rings from the defining axioms of a ring

Define and give examples of commutative rings, rings with identity and commutativerings with identity

Introduction

With this unit, we start the study of algebraic system with two binary operations satisfyingcertain properties. Z, Q and R are examples of such a system, which we shall call a ring.

Now, you know that both addition and multiplication are binary operations on Z. Further, Z isan abelian group under addition. Though it is not a group under multiplication, multiplicationis associative. Also, addition and multiplication are related by the distributive laws

a(b + c) = ab + nc, and (a + b)c = ac + bc

for all integers a, b and c. We generalise these very properties of the binary operations to definea ring in general. This definition is given by the famous algebraist Emmy Noether.

After defining rings we will provide several examples of rings. You will also learn aboutsome propertics of rings that follow from the definition itself. Finally, we shall discuss certaintypes of rings that are obtained when we impose more restrictions on the �multiplication� inthe ring.

As the contents suggest, this unit lays the foundation for the rest of this course. So make sure thatyou have attained the following objectives before going to the next unit.


Unit 14: Rings

Notes14.1 What is a Ring?

You are familiar with Z, the set of integers. You also know that it is a group with respect toaddition. Is it a group with respect to multiplication too? No. But multiplication is associativeand distributes over addition. These properties of addition and multiplication of integers allowus to say that the system (Z, +, .) is a ring. But, what do we mean by a ring?

Definition: A non-empty set R together with two binary operations, we mean usually calledaddition (denoted by f) and multiplication (denoted by .), is called a ring if the following axiomsare satisfied:

R 1) a + b = b + a for all a, b in R, i.e., addition is commutative.

R 2) (a + b) + c = a + (b + c) for all a, b, c in R, is., addition is associative.

R 3) There exists an element (denoted by 0) of R such that

a + 0 = a = 0 + a for all a in R, i.e., R has an additive identity.

R 4) For each a in R, there exists x in R such that a + x =: 0 = x + a, i.e., every elements of R hasan additive inverse.

R 5) (a . b).c = a.(b . c) for all a, b, c in R, i.e., multiplication is associative.

R 6) a.(b + c) = a . b + a . c, and

( a t b ) . i = a . d + b . c

for all a, b, c in R,

i.e., multiplication distributes over addition from the left as well as the right.

The axioms RI-R4 say that (R, +) is an abelian group. The axiom R5 says that multiplication isassociative. Hence, we can say that the system (R, +, .) is a ring if

(i) (R, +) is an abelian group,

(ii) (R, .) is a semigroup, and

(iii) for all a, b, c in R, a.(b + c) = a . b t a . c, and (a + b ) = a . c + b . c.

As you know that the addition identity 0 is unique, and each element a of R has a unique additiveinverse (denoted by - a). We call the element 0 the zero element of the ring.

By convention, we write a � b for a + ( �b).

Let us look at some examples of rings now. You have already seen that Z is a ring. What aboutthe sets Q and R? Do (Q, +, .) and (R, +, .) satisfy the axioms R1 � R6? They do.

Therefore, these systems are rings.

The following example provides us with another set of examples of rings

Example: Show that (nZ, +, .) is a ring, where n Z.

Solution: You know that nZ = { nm I m Z } is an abelian group with respect to addition. Youalso know that multiplication in nZ is associative and distributes over addition from the right aswell as the left. Thus, nZ is a ring under the usual addition and multiplication.

So far the examples that we have considered have been infinite rings, that is, their underlyingsets have been infinite sets. Now let us look at a finite ring, that is, a ring (R, +. .) where R is a


Abstract Algebra

Notes finite set. Our example is the set Z,,. Let us briefly recall the construction of Z,, the set of residueclasses modulo n.

If a and b are integers, we say that a is congruent to b modulo n if a � b is divisible by n; insymbols, a b (mod n) if n I (a � b). The relation �congruence modulo n� is an equivalencerelation in Z. The equivalence class containing the integer a is

a = { b Z ( a - b is divisible by n }

= { a + m p | m Z }.

It is called the congruence class of a modulo n or the residue class of a modulo n. The set of allequivalence classes is denoted by Z,,. So

Z , , = {0,1,2,...,n 1}.

We define addition and multiplication of classes in terms of their representatives by

a b a b and

a . b ab a, b Zn.

To help you regain some practice in adding and multiplying in Z,,, consider the followingCayley tables for Z

n.

Now let us go back to looking for a finite ring.

Example: Show that (Zn, +, .) is a ring.

Solution: You already know that (Zn, +) is an abelian group, and that multiplication is associative

in Z,. Now we need to see if the axiom R6 is satisfied.

For any na, b c Z ,

a.(b c) a.(b c) a.b a.c a.b a.c ab ac

Similarly, ( a + 6) . = L.; + b.c a, b c Z, .

So, (Zn, +,) satisfies the axioms R1-R6. Therefore, it is a ring.

Now let us look at a ring whose underlying set is a subset of C.

Example: Consider the set

Z + iZ = { m + in | m and n are integers }, where i2 = � I.


Unit 14: Rings

NotesWe define �+� and �.� in Z + iZ to be the usual addition and multiplication of complex numbers.Thus, foram + in and s + it in Z + iZ,

(m + in) + (s + it) = (m + s) + i(n + t), and

(m + in) . (s + it) = (p � nt) + i(mt + ns).

Verify that Z + iZ is a ring under this addition and multiplication. (This ring is called the ring ofGaussian integers, after the mathematician Carl Friedrich Gauss.)

Solution: Check that (Z + iZ, +) is a subgroup of (C, I�). Thus, the axioms RI-R4 are satisfied. Youcan also check that

((a + ib) . (c + id)) . (m + in) = (a + ib) . ((c + id) . (m + in))

a + ib, c + id, m + in Z + iZ.

This shows that R5 is also satisfied.

Finally, you can check that the right distributive law holds, i.e.,

((a + ib) + (c + id)) . (m + in) = (a + ib) . (m + in) + (c + id) . (m + in) for any a + ib, c + id, m +in Z + iZ.

Similarly, you can check that the left distributive law holds. Thus, (Z + iZ, + , .) is a ring. Theoperations that we consider in it are not the usual addition and multiplication.

Example: Let X be a non-empty set, (XI ) be the collection of all subsets of X and Adenote the symmetric difference operation. Show that ((X), A, n) is a ring.

Solution: For any two subsets A and B of X,

A B = (A\B) (B\A)

It is clear that ( (X), A) is an abelian group. You also know that is associative. Now let us seeif distributes over A.

Let A, B, C E (X). Then

A (B C) = A [(B\C) (C\B)]

= [A (B\C)][A (C\ B)], since n distributes over U.

= [(A B)\(A C)][(A C)\(A B)], since distributes over complementation.

= (A B) A (A C).

So, the left distributive law holds.

Also, (B C) A = A (B C), since is commutative.

= (A B) P (A C)

= (B A) A ( C A).

Therefore, the right distributive law holds also.

Therefore, ( (X), A, ) is a ring.

So� far you have seen examples of rings in which both the operations defined on the ring havebeen commutative. This is not so in the next example.

Example: Consider the set

M2(R) =

11 1212 21

21 22

a a a, a , a , and a; are real numbers

a a


Abstract Algebra

Notes Show that M2(R) is a ring with respect to addition and multiplication of matrices.

Solution: You can check that (M2(R), +) is an abelian group. You can also verify the associative

property for multiplication. We now show that A . (B + C) = A . B A- A . C for A, B, C in M2(R).

A . (B + C) = 11 12 11 1211 12

21 22 21 2221 22

a a c cb b

a a c cb b

= 11 12 11 11 12 12

21 22 21 21 22 22

a a b c b c

a a b c b c

=

11 11 11 12 21 21 11 12 12 12 22 22

21 11 11 22 21 21 21 12 12 22 22 22

a b c a b c a b c a b c

a b c a b c a b c a b c

=

11 11 12 21 11 11 12 21 11 12 12 22 11 12 12 22

21 11 22 21 21 11 22 21 21 12 22 22 21 12 22 22

a b a c a c a c a b a b a c a c

a b a c a c a c a b a b a c a c

= 11 11 12 21 11 12 12 2211 11 12 21 11 12 12 22

21 11 22 21 21 12 12 2221 11 22 21 21 12 22 22

a c a c a c a ca b a b a b a b

a c a c a c a ca b a b a b a b

= 11 12 11 12 11 1211 12

21 22 21 22 21 2221 22

a a a a c cb b. .

a a a a c cb b

= A.B + A.C

In the same way we can obtain the other distributive law, i.e., (A + B) . C = A. C + B . C A, B,C M

2(R).

Thus, M2(R) is a ring under matrix addition and multiplication.

Note Multiplication over M2(R) is not commutative. So, we can�t say that the left

distributive law implies the right distributive law in this case.

Example: Consider the class of all continuous real valued functions defined on the closedinterval [0, 1]. We denote this by C [0, 1]. If f and g are two continuous functions on [0, 1], wedefine f + g and fg as

(f + g) (x) = f(x) + g(x) (i.e., pointwise addition)

and (f. g) (x) = f(x). g(x) (i.e., pointwise multiplication)

for every x [0, 1]. From the Calculus course you know that the function f + g and fg are definedand continuous on [0, 1], i.e., if f and g C[0, 1], then both f + g and f .g are in C [0, 1]. Show thatC [0, 1] is a ring with respect to + and

Solution: Since addition in R is associative and commutative, so is addition in C [0, 1]. Theadditive identity of C [0, 1] is the zero function. The additive inverse off C [0, 1] is (�f), where

(�f)(x) = � f(A) x [0, 1]. See figure 14.1 for a visual interpretation of (- f). Thus, (C [0, 1], +) isan abelian group. Again, since multiplication in R is associative, so is multiplication in C [0, 1].


Unit 14: Rings

NotesFigure 14.1: The Graphs of f and (�f) over [0, 1]

Now let us see if the axiom R6 holds.

To prove f . (g + h) = f . g + f . h, we consider (f . (g + h) (x) for any x in [0, 1].

Now (f . (g + h)(x) = f(x) (g + h) (x)

= f(x) g(x) +h(x))

= f(x) g(x) + f(x)h(x), since, distributes over + in R.

= (f.g)(x) + (f.h)(x)

Since multiplication is commutative in C [0, 1], the other distributive law also holds. Thus, R6 istrue for C [0, 1]. Therefore, (C [0, 1], +, .) is a ring.

This ring is called the ring of continuous functions on [0, 1].

The next example also deals with functions.

Example: Let (A, +) be an abelian group. The set of all endomorphisms of A is

End A = ( f : A � A | f(a + b) = f(a) + f(b) a, b A )

For f, g End A, we define f + g and f . g as

(f + g) (a) = f(a) + g( a), and ...(1)

(f. g) (a) = fog(a) = f(g(a)) a A

Show that (End A, +,.) is a ring. (This ring is called the endomorphism ring of A.)

Solution: Let us first check that + and . defined by (1) are binary operations on End A.

For all a, b A,

(f + g) (a + b) = f(a + b) f g(a + b)

= (f(a) + f(b)) + (g(a) + g(b))

= (f(a) + g(a)) + (f(b) + g(b))

= (f + g) (a) + (f + g) (b), and


Abstract Algebra

Notes (f. g) (a + b) = f(g(a + b))

= f(g(a) + g(b))

= f(g(a)) + f(g(b))

= (f . g)(a) + (f . g) (b)

Thus, f + g and f . g End A.

Now let us see if (End A, +, .) satisfies Rl-R6.

Since + in the abelian group A is associative and commutative, so is + in End A. The zerohomomorphism on A is the zero element in End A. (� f) is the additive inverse of f E End A. Thus,(End A, +) is an abelian group.

You also know that the composition of functions is an associative operation in End A.

Finally, to check R6 we look at f . (g + h) for any f, g, h End A. Now for any a A,

[f . (g + h)l (a) = f((g + h) (8))

= f(g(a) + h(a))

= f(g(a)) + f(h(a))

= (f . g) (a) + (f . h) (a)

= (f . g + f . h) (a)

f.(g + h) = f . g + f . h.

We can similarly prove that (f + g) . h = f . h + g . h.

Thus, R1-R6 are true for End A.

Hence, (End A, +, .) is a ring.

Note It is not commutative since fog need not be equal to gof for f, g End A.

Now, let us look at the Cartesian product of rings.

Example: Let (A, +,.) and (B, + , ) be two rings. Show that their Cartesian product

A X B is a ring with respect to and * defined by

(a, b) (a�, b�) = (a + a�, b + b�), and

(a, b) * (a�, b�) = (a . a�, b b�)

for a11 (a, b), (a�, b�) in A X B.

Solution: We have defined the addition and multiplication in A X B componentwise. The zeroelement of A X B is (0, 0). The additive inverse of (a, b) is (�a, � b), where � b denotes the inverseof b with respect to � .

Since the multiplications in A and B are associative, * is associative in A × B. Again, using the factthat R6 holds for A and B, we can show that R6 holds for A × B. Thus, (A × B, 0, *) is a ring.


Unit 14: Rings

Notes14.2 Elementary Properties

In this section we will prove some simple but important properties of rings which are immediateconsequences of the definition of a ring. As we go along you must not forget that for any ring R,(R, +) is an abelian group. Hence, the results obtained for groups in the earlier units are applicableto the abelian group (R, +). In particular,

(i) the zero element, 0, and the additive inverse of any element is unique.

(ii) the cancellation law holds for addition; i.e., a, b, c R , a + c = b + c a = b.

As we have mentioned earlier, we will write a � b for a + (�b) and ab for a. b, where a, b R.

So let us state some properties which follow from the axiom R6, mainly.

Theorem 1: Let R be a ring. Then, for any a, b, c R,

(i) a0 = 0 = 0a,

(ii) a(�b) = (�a)b = �(ab),

(iii) (� a) (� b) = ab,

(iv) a(b � c) = ab � ac, and

Proof:

(i) Now, 0 + 0 = 0

a(0 + 0) = a0

a0 + a0 = a0, applying the distributive law.

= a0 + 0, since 0 is the additive identity.

a0 = 0, by the cancellation law for (R, +).

Using the other distributive law, we can similarly show that 0a = 0.

Thus, a0 = 0 = 0a for all a R.

(ii) From the definition of additive inverse, we know that b + (� b) = 0.

Now, 0 = a0, from (i) above.

= a(b + (� b)), as 0 = b + (� b).

= ab + a(� b), by distributivity.

Now, ab + [� (ab)] = 0 and ab + a(� b) = 0. But you know that the additive inverse of anelement is unique.

Hence, we get � (ab) = a(� b).

In the same manner, using the fact that a + (�a) = 0, we get � (ab) = (� a)b.

Thus, a(� b) = (� a)b = � (ab) for all a, b R.

(iii) For a, b R,

(� a) (� b) = � (a(� b)), from (ii) above.

= a(� (� b)), from (ii) above.

= ab, since b is the additive inverse of (� b).


Abstract Algebra

Notes (iv) For a, b, c R,

a(b - c) = a(b + (� c))

= ab + a(� c), by distributivity.

= ab + (� (ac)), from (ii) above.

= ab � ac.

If k is an integer (k 2) such that the sum of k elements in a ring R is defined, we define the sumof (k + 1) elements a

l, a

2 ..., a

k+l in R, taken in that order, as a

1 + .., + a

k+1 = (a

k + ..... + a

k) + a

k+1.

In the same way if k is a positive integer such that the product of k elements in R is defined, wedefine the product of (k + 1) elements a

1, a

2, ..., a

k+l (taken in that order) as

al.a

12 ... a

k+1 = (a

1.a

2 .... .a

k) . a

k+1.

As we did for groups, we can obtain laws of indices in the case of rings also with respect to both+ and ., in fact, we have the following results for any ring R.

(i) If m and n are positive integers and a R, then

am . an = am+n, and

(am)� = amn.

(ii) If m and n are arbitrary integers and a, b R, then

(n + m)a = na + ma,

(nm)a = n(ma) = m(na),

n(a + b) = na + nb,

m(ab) = (ma)b = a(mb), and

(ma) (nb) = mn (ab) = (mna)b.

(iii) If a1 + a

2, ..., a,, b

1, ..., b

n, R then

(a1, + ... + a

m) ( b

1 + ... + b

n)

= a l b

l + ... + a

1b

n + a

2a

1 + ... + a

2b

n + ... + a

mb

1 + ... + a

mb

n.

There are several other properties of rings that we will be discussing throughout this block. Fornow let us look closely at two types of rings, which are classified according to the behaviour ofthe multiplication defined on them.

14.3 Two Types of Rings

The definition of a ring guarantees that the binary operation multiplication is associative and,along with +, satisfies the distributive laws. Nothing more is said about the properties ofmultiplication. If we place restrictions on this operation we get several types of rings. Let usintroduce you to two of them now.

Definition: We say that a ring (R, +, .) is commutative if . is commutative, i.e., if ab = ba for all a,b R.

For example, Z, Q and R are commutative rings.

Definition: We say that a ring (R, +, .) is a ring with identity (or with unity) if R has an identityelement with respect to multiplication, i.e., if there exists an element e in R such that

ae = ea = a for all a R.


Unit 14: Rings

NotesCan you think of such a ring? Aren�t Z, Q and R examples of a ring with identity?

Definition: We say that a ring (R, +, .) is a commutative ring with unity, if it is a commutativering and has the multiplicative identity element 1.

Thus, the rings Z, Q, Rand C are all commutative rings with unity. The integer 1 is themultiplicative identity in all these rings.

We can also find commutative rings which are not rings with identity. For example, 2Z, the ringof all even integers is commutative. But it has no multiplicative identity.

Similarly, we can find rings with identity which are not commutative. For example, M2(R) has

the unit element 1 0

.0 1

But it is not commutative. For instance,

if A = 1 0 0 1

and B ,2 0 0 2

then

AB =1 0 0 1 0 1

and2 0 0 2 0 2

BA = 0 1 1 0 2 0

and0 2 2 0 4 0

Thus, AB BA.

Now, can the trivial ring be a ring with identity? Since 0 . 0 = 0, 0 is also the multiplicativeidentity for this ring. So (( 0 ), +, .) is a ring with identity in which the additive and identitiescoincide. But, if R is not the trivial ring we have the following result.

Theorem 2: Let R be a ring with identity 1. If R { 0 } then the elements 0 and 1 are distinct.

Proof: Since R { 0 } , a R, a 0. Now suppose 0 = 1. Then a = a . 1 = a . 0 = 0 (by Theorem 1).That is, a = 0, a contradiction. Thus, our supposition is wrong. That is, 0 1.

Now let us go back when will A × B be commutative? A × B is commutative if and only if boththe rings A and B are commutative. Let us see why. For convenience we will denote the operationsin all three rings A, B and A × B by + and . . Let (a, h) and (a�, b�) A × B.

Then (a, b) . (a�, b�) = (a�, b�) . (a, b)

( a.a �, b . b�) = (a�. a, b� . b)

a.a� = a�.a and b . b�= b�. b .

Thus, A × B is commutative iff both A and B are commutative rings.

We can similarly show that A × B is with unity iff A and B are with unity. If A and B haveidentities e

1 and e

2 respectively, then the identity of A × B is (e

1, e

2).

Now we will give an important example of a non-commutative ring with identity. This is thering of real quaternions. It was first described by the Irish mathematician William RowanHamilton (1805-1865). It plays an important role in geometry, number theory and the study ofmechanics.


Abstract Algebra

Notes

Example: Let H = ( a + bi + cj + dk | a, b, c, d R ), where i, j, k are symbols that satisfyi2 = � 1 = j2 = k2, ij = k = � ji, jk = i = � kj ki = j = � ik.

We define addition and multiplication in H by

(a + bi + cj + dk) + (ai + b

li + c

ij + d

1k )

= (a + a1) (b + b

1)

i + (c � c

1)

j t (d + d

1)k, and

(a + bi + cj + dk) (a1 + b

li + cj+ d

1k) = (aa

l � bb

1 � cc

1 � dd

1) + (ab

1+ ha

1 + cd

1 � dc

1)

i + (ac

1 � bd

l + ca

1

+ db1)j � (ad

1 + bc

1 � cb

1 + da

1)k

(This multiplication may seem complicated. But it is not so. It is simply performed as forpolynomials, keeping the relationships between i, j and k in mind.)

Show that H is a ring.

Solution: Note that ( ± 1, ± i, ± j, ± k ] is the group QH.

Now, you can verify that (H, +) is an abelian group in which the additive identity is 0 = 0 + 0i +0j + 0k, multiplication in H is associative, the distributive laws hold and

I = 1 + 0i + 0j + 0k is the unity in H.

Do you agree that H is not a commutative ring? You will if You remember that ij ji, forexample.

So far, in this unit we have discussed various types of rings. We have seen examples of commutativeand non-commutative rings. Though non-commutative rings are very important for the sake ofsimplicity we shall only deal with commutative rings henceforth. Thus, from now on, for us aring will always mean a commutative ring. We would like you to remember that both + and .are commutative in a commutative ring.

Now, let us summarise what we have done in this unit.

Self Assessment

1. For each a in R. There exists X in R such that a + X = :0 = ................ i.e. every elements of Rhas an additive inverse.

(a) a . x (b) x + a

(c) x-1 + a (d) a-1 + x

2. If a and b are integers, we say that a is congruent to b modulo n : f ................ is divisibleby n.

(a) a + b (b) a � b

(c) a . b (d) a/b

3. A × B is with unity if A and B are with unity. If A and B have identies e1 and e

2 respectively,

then the identity of A × B is ................

(a) e1 + e

2(b) e

2 + 1

1e

(c) (e, e2) (d) 1 1

1 2(e ,e )

4. The ................ for addition and multiplication and the generalised distributive law.

(a) law of indices (b) Ring

(c) Subring (d) ideal


Unit 14: Rings

Notes5. If m and n are arbitrary integers and a, b R then (n + m) a = na + ma and n(a + b) =

(a) na + nb (b) an + bn

(c) nab + nba (d) an + bn-1

14.4 Summary

In this unit we discussed the following points.

Definition and examples of a ring.

Some properties of a ring like

a . 0 = 0 = 0 . a,

a(� b) = � (ab) = (� a) b,

(� a) (� b) = ab,

a(b � c) = ab � ac,

(b � c)a = ba � ca

a, b, c in a ring R.

The laws of indices for addition and multiplication, and the generalised distributive law.

Commutative rings, rings with unity and commutative rings with unity.

Henceforth, we will always assume that a ring means a commutative ring, unless otherwisementioned.

14.5 Keywords

Ring: A non-empty set R together with two binary operations, usually called addition (denotedby f) and multiplication (denoted by .), is called a ring if the following axioms are satisfied.

Commutative Rings: We say that a ring (R, +, .) is commutative if . is commutative, i.e., if ab = bafor all a, b R. For example, Z, Q and R are commutative rings.


1. Write out the Cayley tables for addition and multiplication in *6Z , the set of non-zero

elements of Z6. Is *

6(Z , ,'.) a ring? Why?

2. Show that the set Q 2Q {p 2q|p,q Q} is a ring with respect to addition and

multiplication of real numbers.

3. Let R = a 0

a,b are real numbers .0 b

Show that R is a ring under matrix addition and

multiplication.

4. Let R = a 0

a,b are real numbers .0b

Prove that R is a ring under matrix addition and

multiplication.


Abstract Algebra

Notes 5. Why is ((X), , ) not a ring?

6. Show that { 0 } is a ring with respect to the usual addition and multiplication. (This is calledthe trivial ring.)

7. Prove that the only ring R in which the two operations are equal (i.e., a + b = ab a,b R) is the trivial ring.

8. Show that the set of matrices x x

x Rx x

is a commutative ring with unity.

9. Let R be a Boolean ring (i.e., a2 = a a R). Show that a = �a a R. Hence show that Rmust be commutative.


1. (b) 2. (a) 3. (c) 4. (a) 5. (a)






www.math.niu.edu

www.maths.tcd.ie/






Unit 15: Subrings

NotesUnit 15: Subrings

CONTENTS

Objectives

Introduction

15.1 Subrings

15.2 Summary

15.3 Keyword



Objectives


Discuss examples of subrings and ideals of some familiar rings

Explain whether a subset of a ring is a subring or not

Describe whether a subset of a ring is an ideal or not

Define and give examples of quotient rings

Introduction

In this unit, we will study various concepts in ring theory corresponding to some of those thatwe have discussed in group theory. We will start with the notion of a subring, which correspondsto that of a subgroup, as you may have guessed already.

Then we will take a close look at a special kind of subring, called an ideal. You will see that theideals in a ring play the role of normal subgroups in a group. That is, they help us to define anotion in ring theory corresponding to that of a quotient group, namely, a quotient ring.

After defining quotient rings, we will look at several examples of such rings. But you will onlybe able to realise the importance of quotient rings in the future units.

We hope that you will be able to meet the following objectives of this unit, because only thenyou will be comfortable in the future units of this course.

15.1 Subrings

In last unit we introduced you to the concept of subgroups of a group. In this unit we willintroduce you to an analogous notion for rings. Remember that for us a ring means acommutative ring.

In the previous unit you saw that, not only is Z Q, but Z and Q are rings with respect to thesame operations. This shows that Z is n subring of Q, as you will now realise.


Abstract Algebra

Notes Definition: Let (R, +, .) be a ring and S be a subset of R. We say that S is a subring of R, if(S, +, .) is itself a ring, i.e., S is a ring with respect to the operations on R.

For example, we can say that 2Z, the set of even integers, is a subring of Z.

Before giving more examples, let us analyse the definition of a subring. The definition says thata subring of a ring R is a ring with respect to the operations on R. Now, the distributive,commutative and associative laws hold good in R. Therefore, they hold good in any subset ofR also. So, to prove that a subset S of R is a ring we don�t need to check all the 6 axioms R1-R6for S. It is enough to check that

(i) S is closed under both + and . ,

(ii) 0 S, and

(iii) for each a S, � a S.

If S satisfies these three conditions, then S is a subring of R. So we have an alternative definitionfor a subring.

Definition: Let S be a subset of a ring (R, +, .). S is called a subring of R if

(i) S is closed under + and . , i.e., a + b, a. b S whenever a, b S,

(ii) 0 S, and

(iii) for each a S, - a S.

Even this definition can be improved upon. For this, recall from Unit 3 that (S, +) (R, +) if a � b S whenever a, b S. This observation allows us to give a set of conditions for a subset to be asubring, which are easy to verify.

Theorem 1: Let S be a non-empty subset of (R, +, .). Then S is a subring of R if and only if

(a) x � y S x, y S; and

(b) xy S x, y S.

Proof: We need to show that S is a subring of R according to our definition iff S satisfies (a) and(b). Now, S is a subring of R iff (S, f ) (R, f ) and S is closed under multiplication, i.e., iff (a)and (b) hold.

So, we have proved the theorem.

This theorem allows us a neat way of showing that a subset is a subring.


We have already noted that Z is a subring of Q. In fact, you can use Theorem 1 to check that Z issubring of R, C and Z + iZ too. You can also verify that Q is a subring of R, C and

Q 2Q {a 2 | , Q}.

Example: Consider Z6, the ring of integers modulo 6. Show, that 3Z

6 = (3.0, 3.1, ....., 3.5) is

a subring of Z6.

Solution: Firstly, do you agree that 3Z6 = (0, 3)? Remember that 6 = 0, 9 = 3, and so on. Also,

0 � 5 = � 3 = 5. Thus, x - y 3Z6 x, y 3Z

6. You can also verify that xy 3Z

6 x, y 3Z

6. Thus,

by Theorem 1, 3Z6 is a subring of Z

6.


Unit 15: Subrings

Notes

Example: Consider the ring (X). Show that S = { , X ) is a subring of (X).

Solution: Note that A A A = A (X). A = � A in (X).

Now, to apply Theorem 1 we first note that S is non-empty.

Next, = S, X X = S,

X = X S, = S, X X = X S, X = S.

Thus, by Theorem 1, S is a subring of (X).

For each proper subset of X we get a subring of (X). Thus, a ring can have, several subrings. Letus consider two subrings of the ring Z2.

Example: Show that S = ( (n, 0) } | n Z } is a subring of Z × Z. Also show that

D = ((n,n) | n Z } is a subring of Z × Z.

Solution: You can recall the ring structure of Z2. Both S and D are non-empty. Both of themsatisfy (a) and (b) of Theorem 1. Thus, S and D are both subrings of Z2.

We would like to make a remark here which is based on the examples of subrings that you haveseen so far.

Remark: (i) If R is a ring with identity, a subring of R may or may not be with identity. Forexample, the ring Z has identity 1, but its subring nZ (n 2) is without identity.

(ii) The identity of a subring, if it exists, may not coincide with the identity of the ring. Forexample, the identity of the ring Z × Z is (1, 1). But the identity of its subring Z × {0} is (1, 0).

Example: Let R be a ring and a R. Show that the set aR = ( ax | x R } is a subring of R.

Solution: Since R , aR . Now, for any two elements ax and ay of aR,

ax � ay = a(x � y) aR and (ax) (ay) = a(xay) aR.

Thus, by Theorem 1, aR is a subring of R.

Using Example we can immediately say that nmZ is a subring of Zn m E Z. This also shows

us a fact that we have already seen : nZ is a subring of Z n Z.

Now let us look at some properties of subrings. From Unit 3 you know that the intersection oftwo or more subgroups is a subgroup. The following result says that the same is true forsubrings.

Theorem 2: Let S1 and S

2 be subrings of a ring R. Then S

1 S

2 is also a subring of R.

Proof: Since 0 E S1 and 0 E S

2, 0 E S

1S

2. S

1S

2 .

Now, let x, y S1 S

2. Then x, y E S

1 and x, y S

2. Thus, by Theorem 1, x � y and xy are in S

1 as

well as in S2, i.e,, they lie in S

1 S

2.

Thus, S1 S

2 is a subring of R.

On the same lines as the proof above we can prove that the intersection of any family of subringsof a ring R is a subring of R.

Now let us look at the Cartesian product of subrings.


Abstract Algebra

Notes Theorem 3: Let S1 and S

2 be subrings of the rings R

1 and R

2, respectively. Then S

1 × S

2 is a subring

of R1 × R

2.

Proof: Since S1 and S

2 are subrings of R

1 and R

2, S

1 and S

2 . S

1 × S

2 .

Now, let (a, b) and (a�, b�) S1 × S

2. Then a, a� E S

1 and b, b� E S

2. As S

1 and S

2 are subrings, a � a�,

a. a� S1 and b � b�, b b� S

2.

(We are using + and . for both R1 and R

2 here, for convenience.) Hence,

(a, b) � (a�, b�) = (a � a�, b � b�) S1 × Sz, and

(a, b) . (a�, b�) = (aa�, bb�) S1 × S

2.

Thus, by Theorem 1, S1 × S

2 is a subring of R

1 × R

2.

Self Assessment

1. Let S be a subset of a ring (R, +,.). S is called ................ of R of O S

(a) ring (b) subring

(c) polynomial ring (d) ideals

2. S be a ................ subset of (R, +,.). Then S is a subring of R. If and only if x � y SY X, y S,

(a) empty (b) non-empty

(c) null set (d) real set

3. Let R be a ring and a R then the set aR = (ax | x R) is a ................ of R.

(a) subring (b) ring

(c) ideal (d) polynomial

15.2 Summary

Let (R, +, .) be a ring and S be a subset of R. We say that S is a subring of R, if (S, +, .) is itselfa ring, i.e., S is a ring with respect to the operations on R.

For example, we can say that 2Z, the set of even integers, is a subring of Z.

Before giving more examples, let us analyse the definition of a subring. The definitionsays that a subring of a ring R is a ring with respect to the operations on R. Now, thedistributive, commutative and associative laws hold good in R. Therefore, they hold goodin any subset of R also. So, to prove that a subset S of R is a ring we don�t need to check allthe 6 axioms R1-R6 for S. It is enough to check that

(i) S is closed under both + and . ,

(ii) 0 S, and

(iii) for each a S, � a S.

If S satisfies these three conditions, then S is a subring of R. So we have an alternativedefinition for a subring.

Let S be a subset of a ring (R, +, .). S is called a subring of R if

(i) S is closed under + and . , i.e., a + b, a. b E S whenever a, b E S,


Unit 15: Subrings

Notes(ii) 0 E S, and

(iii) for each a E S, - a E S.

Even this definition can be improved upon. For this, recall from Unit 3 that (S, f ) (R, +)ifa � b E S whenever a, b E S. This observation allows us to give a set of conditions for asubset to be a subring, which are easy to verify.

15.3 Keyword

Subring: Let (R, +, .) be a ring and S be a subset of R. We say that S is a subring of R, if (S, +, .) isitself a ring, i.e., S is a ring with respect to the operations on R.


1. Show that S = a 0

a,b Z .0 b

is a subring of R =

a 0a,b R .

0 b

Does S have a unit

element?

If yes, then is the unit element the same as that of R?

2. For any ring R, show that {0} and R are its subrings.

3. Show that if A is subring of B and B is a subring of C, then A is a subring of C.

4. Give an example of a subset of Z which is not a subring.

5. Show that a,3 and 0,2,4 are proper ideal of Z6.


1. (b) 2. (b) 3. (a)






www.math.niu.edu

www.maths.tcd.ie/






Abstract Algebra

Notes Unit 16: Ideals

CONTENTS

Objectives

Introduction

16.1 Quotient Rings

16.2 Summary

16.3 Keywords



Objectives


Discuss ideals of some familiar rings

Explain whether a subset of a ring is an ideal or not

Define and give examples of quotient rings

Introduction

In earlier unit, you have studied normal subgroups and the role that they play in group theory.You saw that the most important reason for the existence of normal subgroups is that they allowus to define quotient groups. In ring theory, we would like to define a similar concept, aquotient ring. In this unit, we will discuss a class of subrings. These subrings are called ideals.While exploring algebraic number theory, the 19th century mathematicians Dedekind, Kroneckerand others developed this concept. Let us see how we can use it to define a quotient ring.

Consider a ring (R, + , .) and a subring I of R. As (R, +) is an abelian group, the subgroup, I isnormal in (R, +), and hence the set R/I = ( a + 1 | a R }, of all cosets of I in R, is group under thebinary operation + given by

(a + I) + (b + I) = (a + b) + I ..... (1)

for all a + I, b + I R/I. We wish to define. on R/I so as to make R/I a ring. You may think thatthe most natural way to do so is to define

(a + I) . (b + I) = a b + I a + 1, b + I R ..... (2)

But, is this well defined? Not always. For instance, consider the subring Z of R and the set ofcosets of Z in R. Now, since 1 = 1 � 0 Z , 1 + Z = 0 + Z.

Therefore, we must have

( 2 + Z) . (1 + Z) = ( 2 + Z) . (0 + Z), i.e,, 2 + Z = O + Z, i.e., 2 Z.

But this is a contradiction. Thus, our definition of multiplication is not valid for the set R/Z.


Unit 16: Ideals

NotesBut, it is valid for R/I if we add some conditions on I: What should these conditions be? Toanswer this, assume that the multiplication in (2) is well defined.

Then, (r + I). (0 + I) = r . 0 + I = 0 + I = I for all r R.

Now, you know that if x I, then x + I = 0 + I = I.

As we have assumed that is well defined, we get

(r + I) (x + I) = (r + I) . (0 + I) = 0 + I whenever r R, x I.

i.e., rx + I = I whenever r R, x I

Thus, rx I, whenever r R, x I.

So, if � . � is well defined we see that the subring I must satisfy the additional condition thatrx I whenever r R and x I.

We will prove that this extra condition on I is enough to make the operation a well defined oneand (R/I, +, .) a ring. In this unit we will consider the subrings I of R on which we impose thecondition rx I whenever r R and x I.

Definition: We call a non-empty subset I of a ring (R, +, .) an ideal of R if

(i) a � b I for all a, b I, and

(ii) ra I for all r R and a I.

Over here we would like to remark that we are always assuming that our rings are commutative.In the case of non-commutative rings the definition of an ideal is partially modified as follows.

A non-empty subset I of a non-commutative ring R is an ideal if

(i) a � b I a, b I, and

(ii) ra I and ar I a I, r R.

Now let us go back to commutative rings. From the definition we see that a subring I of a ringR is an ideal of R iff ra I r R a and a I.

Let us consider some examples. You saw that for any ring R, the set (0) is a subring. In fact, it isan ideal of R called the trivial ideal of R. Other ideals, if they exist, are known as non-trivialideals of R.

You can also verify that every ring is an ideal of itself. If an ideal I of a ring R is such that I R,then I is called a proper ideal of R.

For example, if n 0,1, then the subring nZ = { nm | m Z ) is a proper non-trivial ideal of Z. Thisis because for any z Z and nm nZ, z(nm) = n(zm) nZ.

Example: Let X be an infinite set. Consider I, the class of all finite subsets of X. Show thatI is an ideal of (X).

Solution: I = { A | A is a finite subset of X }. Note that

(i) I, i.e., the zero element of (X) is in I,

(ii) A � B = A + (�B) = A + B, as B = �B in (X) = A B.

Thus, if A, B I, then A - B is again a finite subset of X, and hence A � B I.

(iii) AB = A B. Now, whenever A is a finite subset of X and B is any element of (X), ABis a finite subset of X. Thus, A I and B P (X) AB I .

Hence, I is an ideal of (X).


Abstract Algebra

Notes

Example: Let X be a set and Y be a non-empty subset of X. Show that

I = { A (x) | AY = } is an ideal of (X).

In particular, if we take Y = {x0}; where X

0 is a fixed element of X, then

I = { A (X) | x0 A } is an ideal of (X).

Solution: Firstly, 1,

Secondly, A, B E I,

(A � B ) Y = (A B ) Y = (A Y) (B Y ) = = , so that A � B I.

Finally, for A I and B E (X),

(AB) Y = (A B) Y = (A Y) B = B = , So that AB I

Thus, I is an ideal of (X).

Example: Consider the ring C[0, 1]

Let M = ( f C[0, 1] | f(1/2) = 0 ). Show that M is an ideal of C[0, 1].

Solution: The zero element 0 is defined by 0(x) = 0 for all x [0, 1]. Since 0(1/2) = 0, O E M.

Also, if f, g M, , then (f � g) (l /2 ) = f (1/2 ) � g (1/2 ) = 0 � 0 = 0.

So, f � g M .

Next, iff M and g C [0, 1] then (fg) (1/2) = f(1/2) g (1/2) = 0 g(1/2) = 0, so fg M.

Thus, M is an ideal of C[0, 1].

When you study Unit 17, you will see that M is the kernel of the homomorphism

: C[0, 1] R : (f) = f(1/2).

Example: For any ring R and al, a

2 R, show that Ra

1 + Ra

2 = { x

1a

1 + x

2a

2 | x

1, x

2 R )

is an ideal of R.

Solution: Firstly, 0 = 0a1 t 0a

2. 0 Ra

1 + Ra

2.

Next, (x1a

1 + x

2a

2) � (y

1a

1 + y

2a

2)

= ( x1 � y

1)a

1+ (x

2 � y

2)a

2 Ra

1 + Ra

2 x

1, x

2, y

1, y

2 R.

Finally, for r R and x1a

1 + x

2a

2 Ra

1 + Ra

2,

r(x1a

1 x

2a

2) = rx

1a

1 + rx

2a

2 Ra

1 + Ra

2.

Thus, Ra1 + Ra

2 is an ideal of R.

This method of obtaining ideals can be extended to give ideals of the form { x1a

1 + x

2a

2 + ... + x

na

n

| xi R } for fixed elements a

1..,... , a, of R. Such ideals crop up again and again in ring theory. We

give them a special name.

Definition: Let a1, ....., a, be given elements of a ring R. Then the ideal generated by a

1, ....., a,, is

Ra1 + Ra

2 + ... + Ra

n = (x

1a

l + x

2a

2 + ... + x

na

n | x, E R ). a

1, ....., a,, are called the generators of this ideal.

We also denote this ideal by < a1, a

2, ....., an >


Unit 16: Ideals

NotesWhen n = 1, the ideal we get is called a principal ideal. Thus, if a R, then Ra = < a > is a principalideal of R. In the next unit you will be using principal ideals quite a lot.

Tasks 1. Let R be a ring with identity. Show that < 1 > = R.

2. Find the principal ideals of Z10

generated by 3 and 5 .

Definition: An element a of a ring R is called nilpotent if there exists a positive integer n suchthat a� = 0.

For example, 3 and 6 are nilpotent elements of Z9, since

23 9 0 and

26 36 0. Also, in any

ring R, 0 is a nilpotent element.

Now consider the following example.

Example: Let R be a ring. Show that the set of nilpotent elements of R is an ideal of R.

This ideal is called the nil radical of R.

Solution: Let N = { a R | an = 0 for some positive integer n }. Then 0 EN.

Also, if a, b N, then an = 0 and bm = 0 for some positive integers m and n.

Now, m n

m n m n r m n rr

r 0

(a b) C a ( b)

For each r = 0, 1, ....., m + n, either r n or m + n � r m, and hence, either ar = 0 or bm+n-r = 0. Thus,the term ar bm+n-r = 0. S0 (a � b)m+n = 0.

Thus, a � b N whenever a, b N.

Finally, if a N, a� = 0 for some positive integer n, and hence, for any

r R, (ar)n = anrn = 0, i.e., ar N.

So, N is an ideal of R.

Let us see what the nil radicals of some familiar rings are. For the rings Z, Q, R or C, N = {0}, sincethe power of any non-zero element of these rings is non-zero.

For Z4, N = {0,2}.

Theorem 1: Let R be a ring with identity 1. If I is an ideal of R and I E I, then I = R.

Proof: We know that I R. We want to prove that R I. Let r E R. Since 1 E I and I is an ideal ofR, r = r . l I. So, R I. Hence I = R.

Using this result we can immediately say that Z is not an ideal of Q. Does this also tell us whetherQ is an ideal of R or not�? Certainly Since 1 Q and Q R, Q can�t be an ideal of R.

Now let us shift our attention to the algebra of ideals. In the previous section we proved that theintersection of subrings is a subring. We will now show that the intersection of ideals is an ideal.We will also show that the sum of ideals is an ideal and a suitably defined product of ideals is anideal.


Abstract Algebra

Notes Theorem 2: If I and J are ideals of a ring R, then

(a) IJ,

(b) I + J = { a + b | a I and b J }, and

(c) IJ = { x R | x is a finite sum a1b

1 + ... + a

mb

m, where ai 1 and bi J } are ideals of R.

Proof: (a) From Theorem 2 you know that I J is a subring of R. Now, if a l J, then a I anda J. Therefore, ax I and a J for all x in R. So ax I J for all a I J and x R. Thus, I Jis an ideal of R.

(b) Firstly, 0 = 0 + 0 l + J I + J = f.

Secondly, if x, y 1 + J, then x = a1 + b

1 and y = a

2 + b

2 for some a

1, a

2, I and b

1, b

2 J.

So x � y = (a1 + b

1) � (a

2 + b

2) = (a

1 � a

2) (b

1 � b

2) I + J.

Finally, let x I + J and r R. Then x = a + b for some a I and b J. Now

xr = (a + b)r = ar +br I + J, as a I implies ar I and b J implies br J for all r R.

Thus, I + J is an ideal of R.

(c) Firstly, IJ , since I and J .

Next, let x, y IJ. Then x = a1b

1 + ... + a

mb

m and

y = a�1b�

1 + ... + a�

nb�

n for some a

l, ..., a, a�

1,.. ., a�, I and b

i,.., b

m, b�

1,...., , b�

n J.

x � y = (a1b

1 + ... + a

mb

m) � (a�

1b�

1 + ... + a�

nb�

n)

= a1b

1 + ... + a

mb

m + (� a�

1)b�

1 + ... + (� a�

n)b�

n

which is a finite sum of elements of the form ab with a I and b J.

So, x � y IJ.

Finally, let x IJ say x = a1b

1 + ... + a

nb

n with a, 1 and b, E J. Then, for any r E R

xr = (a1b

1 + ... + a

nb

n)r = a

1(b

1r ) ... + a

n(b

nr),

which is a finite sum of elements of the form ab with a 1 and b J.

(Note that bi J b

ir J for all r in R.)

Thus, IJ is an ideal of R.

Over here, we would like to remark that if we define IJ = { ab | a E I, b J }, then IJ need not evenbe a subring, leave alone being an ideal. This is because if x, y E IJ, then with this definition of IJit is not necessary that x � y IJ.

Let us now look at the relationship between the ideals obtained. Let us first look at the followingparticular situation:

R = Z, I = 22 and J = 10Z. Then I J = J, since J I. Also, any element of I + J is ot the form x = 2n+ 10m, where n, m 2. Thus, x = 2(n + 5m) 2Z. On the other hand, 2Z = f I + J. Thus , I + J =< 2 , 1 0 > = < 2 >.

Similarly, you can see that IJ = < 20 >.

Note that IJ IJ I I + J.

In fact, these inclusions are true for any I and J. We show the relationship in figure 16.1.


Unit 16: Ideals

NotesFigure 16.1: The Ideal Hierarchy

16.1 Quotient Rings

You have studied quotient groups. You know that given a normal subgroup N of a group G, theset of all cosets of N is a group and is called the quotient group associated with the normalsubgroup N. Using ideals, we will now define a similar concept for rings. At the beginning wesaid that if (R, +, .) is a ring and I is a subring of R such that

(R/I, +, .) is a ring, where + and . are defined by

(X + I) + (y + I) = (x + y) + I and

( x + I ) . ( y + I ) = x y + I x + I , y + I R / I ,

then the subring I should satisfy the extra condition that rx I whenever r R and x I, i.e.,I should be an ideal. We now show that if I satisfies this extra condition then the operations thatwe have defined on R/I are well defined.

From group theory we know that (R/I, +) is an abelian group. So we only need to check that iswell defined, i.e., if

a + I = a� + I, b + I = b� + I, then ab + I = a�b� + I.

Now, since a + I = a� + I, a � a� I.

Let a � a� = x. Similarly, b � b� 1, say b � b� = y.

Then ab = (a� + x ) (b� + y ) = a�b� + (xb�+ a�y + xy).

ab � a�b� I, since x I. y I and I is an ideal of R.

ab + I = a�b� + I.


Abstract Algebra

Notes Thus, is well defined on R/I.

Now our aim is to prove the following result.

Theorem 3: Let R be a ring and I be an ideal in R. Then R/I is a ring with respect to addition andmultiplication defined by

(X + I) + (y + I) = (x + y) + I, and

( x + I ) . ( y + I ) = x y + I x,y R.

Proof: As we have noted earlier, (R/I, +) is an abelian group. So, to prove that R/I is a ring weonly need to check that . is commutative, associative and distributive over +.

Now,

(i) . is commutative : (a + I). (b + I) = ab + I = ba + I = (b + I), (a 4- I) for all a + I,b + I R/I.

(ii) . is associative : � � a, b, c R

((a + I). (b + I)). (c + I) = (ab + I). (C + I)

= (ab)c + I

= a(bc) + I

= (a + I) . ((b + I) . (c + I))

(iii) Distributive law : Let a + I, b + I, c + I RA. Then

(a + I). ((b + I) + (c + 1)) = (a + I) [(b + 6) + I]

= a(b + c) + I

= (ab + ac) + I

= (ab + I) + (ac + I).

= (a + I). (b + I) + (a + I).(c -1 I)

Thus, R/I is a ring.

This ring is called the quotient ring of R by the ideal I.

Let us look at some examples. We start with the example that �gave rise to the terminology�R mod I�.

Example: Let R = Z and I = nZ. What is R/I?

Solution: You have seen that nZ is, an ideal of Z. From Unit 2 you know that

Z/nZ = { nZ., I + nZ, ..., (n � 1) + nZ }.

= {0,1,....,n 1}, the same as the set of equivalence classes modulo n.

So, R/I is the ring Zn.

Now let us look at an ideal of Zn, where n = 8.

Example: Let R = Z8. Show that I = {0,4} is an ideal of R. Construct the Cayley tables for

+ and, in R/I.


Unit 16: Ideals

NotesSolution: I = 4 R, and hence is an ideal of R. From group theory you know that the number 8 of

elements in R/I = o(R/I) = o(R) 8

4.o(I) 2

You can see that these elements are

0 + 1 = {0, 4}, 1 + 1 = {1, 5} , 2 + 1 = {2, 6}, 3 + I = (3, 7}.

The Cayley tables for + and . in R/I are

Self Assessment

1. A non-empty subset I of a ring (R+,.) an ................. of R of a � b I for all a, b I.

(a) ring (b) subring

(c) polynomial (d) ideal

2. If n 0, 1. Then the subring nZ = {nm | m Z} is a proper ................. ideal of Z.

(a) non-trivial (b) trivial

(c) direct (d) indirect

3. X be a set and Y be a non-empty subset of X. Then I = {A (x) | A ................. y = } is an idealof (x).

(a) (b)

(c) (d)

4. If I and J are ideals of a ring R, then I ................. J are ideals ring R.

(a) (b)

(c) (d)

5. A normal subgroup N of a group G, the set of all cosets of N is a group and is called................. associated with the normal subgroup N.

(a) quotient group (b) ring

(c) subring (d) ideal

16.2 Summary

We call a non-empty subset I of a ring (R, +, .) an ideal of R if


(ii) ra I for all r R and a I.


Abstract Algebra

Notes Over here we would like to remark that we are always assuming that our rings arecommutative. In the case of non-commutative rings the definition of an ideal is partiallymodified as follows.

A non-empty subset I of a non-commutative ring R is an ideal if

(i) a � b I a, b I, and

(ii) ra I and ar I a I, r R.

Now let us go back to commutative rings. From the definition we see that a subring I of a

ring R is an ideal of R iff ra I r R a and a I.

You can also verify that every ring is an ideal of itself. If an ideal I of a ring R is such thatI R, then I is called a proper ideal of R.

For example, if n 0,1, then the subring nZ = { nm | m Z ) is a proper non-trivial idealof Z. This is because for any z Z and nrn nZ, z(nm) = n(zm) nZ.

An element a of a ring R is called nilpotent if there exists a positive integer n such thata� = 0.

For example, 3 and 6 are nilpotent elements of Z9, since

23 9 0 and

26 36 0. Also,

in any ring R, 0 is a nilpotent element.

16.3 Keywords

Proper Ideal: We call a non-empty subset I of a ring (R, +, .) an ideal of R if


every ring is an ideal of itself. If an ideal I of a ring R is such that I R, then I is called a properideal of R.

Nilpotent: An element a of a ring R is called nilpotent if there exists a positive integer n such thata� = 0.

Quotient Group: A normal subgroup N of a group G, the set of all cosets of N is a group and iscalled the quotient group associated with the normal subgroup N.

This ring is called the quotient ring of R by the ideal I.


1. Let S be a subring of a ring R. Can we always define a ring homomorphism whose domainis R and kernel is S? Why?

2. Prove Theorem 8.

3. In the situation of Theorem 8 prove that

(a) if g o f is 1 � 1, then so is f.

(b) if g o f is onto, then so is g.

4. Use Theorem 8 to show that the function h : Z × Z Z2 defined by h((n, m)) = m is a

homomorphism.

5. Which of the following functions are ring isomorphisms?

(a) f : Z : f(n) = n


Unit 16: Ideals

Notes(b) f : Z 5Z : f(n) = 5n

(c) f : C C : f(z) = z , the complex conjugate of z.

6. Show that the composition of isomorphisms is an isomorphism.


1. (d) 2. (a) 3. (a) 4. (c) 5. (a)






www.math.niu.edu

www.maths.tcd.ie/






Abstract Algebra

Notes Unit 17: Ring Homomorphisms

CONTENTS

Objectives

Introduction

17.1 Homomorphisms

17.2 Properties of Homomorphisms

17.3 The Isomorphism Theorems

17.4 Summary

17.5 Keyword



Objectives


Discuss whether a function is a ring homomorphism or not

Explain the kernel arid image of any homomorphism

Explain examples of ring homomorphisms and isomorphisms

Prove and use some properties of a ring homomorphism; state, prove and apply theFundamental Theorem of Homomorphisms for rings

Introduction

You have studied about the functions between groups that preserve the binary operation. Youalso saw how useful they were for studying the structure of a group. In this unit, we will discussfunctions between rings which preserve the two binary operations. Such functions are calledring homomorphisms. You will see how homomorphisms allow us to investigate the algebraicnature of a ring.

If a homomorphism is a bijection, it is called an isomorphism. The role of isomorphisms in ringtheory, as in group theory, is to identify algebraically identical systems. That is why they areimportant. We will discuss them also.

Finally, we will show you the interrelationship between ring homomorphism, ideals andquotient rings.

17.1 Homomorphisms

Analogous to the notion of a group homomorphism, we have the concept of a ringhomomorphism. Recall that a group homomorphism preserves the group operation of its domain.So it is natural to expect a ring homomorphism to preserve the ring structure of its domain.Consider the following definition.


Unit 17: Ring Homomorphisms

NotesDefinition: Let (R1, +, . ) and (R

1,+ . . ) be two rings and f : R

1 � R

2 be a map. We say that f is a ring

homomorphisms if

f(a + b) = f(a) 4 � f(b), and

f(a . b) = f(a) . f(b) for all a, b in R1.

Note that the + and . occurring on the left hand sides of the equations in the definition above aredefined on R

1, while the + and . occurring on the right hand sides are defined on R

2.

So, we can say that f : R1 � R

2 is a homomorphism if

(i) the image of a sum is the sum of the images, and

(ii) the image of a product is the product of the images.

Thus, the ring homomorphism f is also a group homomorphisms from (R1,+ ) into (R

2, +).

Just as we did in Unit 6, before giving some examples of homomorphisms let us define thekernel and image of a homomorphism. As is to be expected, these definitions are analogous tothe corresponding ones in Unit 6.

Definition: Let R1 and R

2 be two rings and f : R

1 � R

2 be a ring homomorphism. Then we define

(i) the image of f to be the set lm f = {f(x) | x R1},

(ii) the kernel off to be the set Ker f = {x R1 | f(x) = 0).

Note that lm f R2 and Ker f R

1,

If Im f = R2, f is called an epimorphism or an onto homomorphism, and then R

2 is called the

homomorphic image of R1.

Now let us look at some examples.

Example: Let R be a ring. Show that the identity map IK is a ring homomorphism. What are

Ker IR and Im I

R?

Solution: Let x, y R. Then

IR(x + y) = x + y = I

R(x) + I

R(y), and

IR(xy) = xy = I

R{(X) I

R(y).

Thus, IR(xy) = xy = I

R(x) . I

R(y).

Thus, IR is a ring homomorphism.

Ker IR = { x R | I

R(x) = 0 }

=.{x R | x = 0)

= {0}

Im

IR = {(I

R(x) [ x R ]}

={x | x R ]

= R.

Thus, IR, is a surjection, and hence an epimorphism.

Example: Let s N. Show that the map f : Z � Z, given by f(m) = m for all m Z is ahomomorphism. Obtain Ker f and Im f also.


Abstract Algebra

Notes Solution: For any m, n Z,

f(m + n) = m t n = m n = f(m) + f(n), and

f(mn) = mn = m n = f(m) f(n).

Therefore, f is a ring homomorphism.

Now, Kerf = (m Z | f(m) = 0 |

= {m Z | m= 01

= (m Z | m 0 (mod s))

= sz.

Im f = (f(m) | m Z)

= ( m | m Z)

= Zs,

showing that f is an epimorphism.

Example: Consider the map f : Z6 � Z

3 : f(n (mod 6)) = n(mod 3). Show that f is a ring

homomorphism. What is Ker f?

Solution: Firstly, for any n, m Z,

f(n(mod 6) + m(mod 6)) = f((n + m) (mod 6)) = (n + m) (mod 3)

= n (mod 3) + m(mod 3)

= f(n (mod 6)) + f(m(mod 6))

You can similarly show that

f(n(mod 6) . m(mod 6)) = f(n(mod 6)) . f(m(mod 6)).

Thus, f is a ring homomorphism.

Ker f = {n(mod.6) | n 0(mod 3)) = {n(mod 6) | n 3Z)

= {0,3}, bar denoting �mod 6�.

Before discussing any more examples, we would like to make a remark about terminology. Infuture we will use the term �homomorphism� for �ring homomorphism�. You may rememberthat we also did this in the case of group homomorphisms.

Now let us look at some more examples.

Example: Consider the ring C[0, 1] of all real valued continuous functions defined on theclosed interval [0, 1].

Define : C[0, 1] + R : (f) = f(1/2). Show that is a homomorphism.

Solution: Let f and g C [0, 1]

Then (f + g) (x) = f(x) f g(x) and

(fg) (x) = f(x) g(x) for all x C[0, 1].



NotesNow, (f + g) = (f + g) (1/2) = f(1/2) + g(1/2) = (f) + (g), and

f(fg) = (fg) (1/2) = 1 1

f g (f) (g).2 2

Thus, is a homomorphism.

Example: Consider the ring R = a 0

a, b R0 b

under matrix addition and multiplication.

Show that the map I : Z 13 : f(n) = n 0

0 n

is a homomorphism.

Solution: Note that f(n) = nI, where I is the identity matrix of order 2. Now you can check that f(n

+ m) = f(n) + f(m) and f(nm) = f(n) f(m) n, m Z. Thus, f is a homomorphism.

Example: Consider the ring (X) of Unit 14.

Let Y be a non-empty subset of X .

Define f : (X) (Y) by f(A) I= AY for all A in ( X ) . Show that f is a homomorphism.Does Y� Ker f ? What is Im f ?

Solution: For any A and B in (X),

f(A B) = f((A\ B) (B\ A))

= ((A\B) (B\AY))

= ((A\B)Y) ((B\A)Y)

= ((AY)\(BY))((BY)\(AY))

= (f(A)\ f(B)(f(B) \f(A))

= f(A) A f(B), and

f(AB) = (AB)Y

= (AB)(YY)

= (AY)(BY), sinceis associative and commutative.

I = f(A)f(B).

So, f is a ring homomorphism from (X) into (Y).

Now, the zero element of (Y) is . Therefore,

Ker f = { A (X) | AY = ) . Yc Ker f.

We will show that f is surjective.

Now, Im f = {AY | A (x)]

Thus, Im f (Y). To show that (Y) Im f, take any B (Y).

Then B (X) and f(B) � BY = B. Thus, B Im f.

Therefore, Im f = (Y).

Thus, f is an onto homomorphism.


Abstract Algebra

Notes

Tasks 1. Let A and B be two rings. Show that the projection map P : A × B A : p(x, y)= x is a homomorphism. What are Ker p and Im p?

2. Is f : Z + 2Z Z + 2Z : f(a + 2b ) = a � 2b a homomorphism?

3. Show that the map : C[0, 1] R × R : (f = (f(0), f(1)) is a homomorphism.

Having discussed many examples, let us obtain some basic results about ring homomorphisms.

17.2 Properties of Homomorphisms

Let us start by listing some properties that show how a homomorphism preserves the structureof its domain. The following result is only a restatement of Theorem 1 of Unit 6.

Theorem 1: Let f : R1 + R

2 be a homomorphism from a ring R

1 into a ring R

2. Then

(a) f(0) = 0,

(b) f(� x) = � f(x) x R1, and

(c) f (x � y) = f(x) � f(y) x, y R1.

Proof: Since f is a group homomorphism from (R1, + ) to (R

2, + ), we can apply Theorem 1 of

Unit 6 to get the result.

Theorem 2: Let f : R1 � R

2 be a ring homomorphism. Then

(a) if S is a subring of R1, f(S) is a subring of R

2;

(b) if T is a subring of R2, f -1 (T) is a subring of R

1.

Proof: We will prove (b) and leave the proof of (a) to you. Let us use Theorem 1 of Unit 16.

Firstly, since T , f-1 (T) . Next, let a, b f -1(T). Then f(a), f(b) T

f(a) � f(b) T and f(a) f(b) T

f(a � b) T and f(ab) T

a � b f�1 (T) and ab f -1(T)

f-1(T) is a subring.

Now, it is natural to expect an analogue of Theorem 2 for ideals. But consider the inclusion i : Z� R : i(x) = x. You know that 22 is an ideal of Z. But is i(2Z) (i.e., 22) an ideal of R? No. For example,

2 22, 14

R, but 1 1

2. 2Z.4 2 Thus, the homomorphic image of an ideal need not be an ideal.

But, all is not lost. We have the following result.


2 be a ring homomorphism.

(a) Iff is surjective and I is an ideal of R1, then f (I) is an ideal of R,.

(b) If I is an ideal of R2, then f-1(1) is an ideal of R

1 and Ker f f-1(J).

Proof: Here we will prove (a)

Firstly, since I is a subring of R1, f(1) is a subring of R

2.



NotesSecondly, take any f(x) f(1) and r R2. Since f is surjective, s R

1 such that f(s) = r.

Then

rf(x) = f(s) f(x) = f(sx) f(I), since sx I.

Thus, f(1) is an ideal of R2.

Now, consider an epimorphism f : R S and an ideal I in R. By Theorem 3 you know that f(1) isan ideal of S and f-1(f(I)) is an ideal of R. How are I and f-1(f(I)) related? Clearly, I f-1(f(1)).Can f-1(f(T)) contain elements of R\I? Remember that Ker f f-1(f(1)) also. Thus,

I + Ker f f-1(f(1)). In fact, I + Ker f = f-1(f(1)). Let us see why.

Let x f-1(f(l)). Then f(x) f(1). Therefore, f(x) = f(y) for sdme y I. Then

f(x � y) = 0.

x � y Ker f, i.e., x y + Ker f I + Ker f.

f-1(f(I)) I + Ker f.

Thus, f-1(f(I)) = I + Ker f.

This tells us that if Ker f I, then

f-1(f(I)) = I (since Kerf I + Ker f = I).

Theorem 4: Let f : R S be an onto ring homomorphism. Then

(a) if I is an ideal in R containing Ker f, I = f-1(f(I))

(b) the mapping 1 � f(I) defines a one-to-one correspondence between the set of ideals of Rcontaining Ker f and the set of ideals of S.

Proof: We have proved (a) in the discussion above. Let us prove (b) now.

Let A be the set of ideals of R containing Ker f, and B be the set of ideals of S.

Define : A B : 4(I) = f(I).

We want to show that is one-one and onto.

is onto : If J B then f-1 (J) A and Ker f f-1 (J) by Theorem 3.

Now (f-1(J)) = f(f-1(J)) = J,

is one-one : If I1 and I

2 are ideals in R containing Ker f, then

(I1) = (I

2) f(I

1) = f(I

2)

f-1(f(I1)) = f-1(f(I

2))

I1 = I

2, by (a).

Thus, is bijective.

And now let us look closely at the sets Ker f and Im f, where f is a ring homomorphism. In Unit6 we proved that iff : G

1 � G

2 is a group homomorphism then Ker f is a normal subgroup of G

1

and Im f is a subgroup of G2. We have an analogous result for ring homomorphisms, which you

may have already realised from the examples you have studied so far.


2 be a ring homomorphism. Then

(a) Ker f is an ideal of R1.

(b) Im f is a subring of R2.


Abstract Algebra

Notes Proof: (a) Since (0) is an ideal of R2, by Theorem 3(b) we know that f-1({o}) is an ideal of R

1. But

f-1({o}) = Ker f.

Thus, we have shown that Ker f is an ideal of R1.

(b) Since R1 is a subring of R

1, f(R

1) is a subring of R

2, by Theorem 2(a). Thus, Im f is a subring of

R2.

This result is very useful for showing that certain sets are ideals. For example, from Theorem 5you can immediately say that {0,3} is an ideal of Z

6. As we go along you will see more examples

of this use of Theorem 5.

Let us look a little more closely at the kernel of a homomorphism. In fact, let us prove a resultanalogous to Theorem 4 of Unit 6.


2 be a homomorphism. Then f is injective iff Ker f = {0}

Proof: f is injective iff f is an injective group homomorphism from (R1, +) into (R

2, + ). This is true

iff Ker f = {0}, by Theorem 4 of Unit 6. So, our result is proved.

So far we have seen that given a ring homomorphism f : R � S, we can obtain an ideal of R,namely, Ker f. Now, given an ideal I of a ring R can we define a homomorphism f so that

Ker f = I?

The following theorem answers this question. Before going to the theorem recall the definitionof quotient rings.

Theorem 7: If I is an ideal of a ring R, then there exists a ring homomorphism f : R R/I whosekernel is I.

Proof: Let us define f : R R/I by f(a) = a + I for all a R. Let us see iff is a homomorphism. Forthis take any a, b R. Then

f(a + b) = (a + b) + I = (a + I) + (b + I) = f(a) + f(b), and

f(ab) = ab + I = (a + I) (b + I) = f(a) f(b).

Thus, f is a homomorphism.

Further, Kerf = {a R | f(a) = 0 + I} = { a R | a + I = I }

= {a R | a I} = I .

Thus, the theorem is proved.

Also note that the homomorphism f is onto.

We call the homomorphism defined in the proof above the canonical (or natural) homomorphismfrom R onto R/I.

Now let us look at the behaviour of the composition of homomorphisms. We are sure you findthe following result quite unsurprising.

Theorem 8: Let R1, R

2 and R

3 be rings and f : R

1 � R

2, and g : R

2 R

3 be ring homomorphisms.

Then their composition gof : R1 R

3 given by (gof (x) = g(f(x)) for all x R

1 is a ring

homomorphism.

The proof of this result is on the same lines as the proof of the corresponding result inUnit 6.



Notes17.3 The Isomorphism Theorems

We discussed group isomorphisms and various results involving them. In this section we willdo the same thing for rings. So, let us start by defining a ring isomorphism.

Definition: Let R1 and R

2 be two rings. A function f : R

1 R

2 is called a ring isomorphism (or

simply an isomorphism) if

(i) f is a ring homomorphism,

(ii) f is 1 � 1, and

(iii) f is onto.

Thus, a homomorphism that is bijective is an isomorphism.

An isomorphism of a ring R onto itself is called an automorphism of R.

Iff : R1 R

2 is an isomorphism, we say that R

1 is isomorphic to R

2, and denote it by R

1 R

2.

Remark: Two rings are isomorphic if and only if they are algebraically identical. That is,isomorphic rings must have exactly the same algebraic properties. Thus, if R

1 is a ring with

identity then it cannot be isomorphic to a ring without identity. Similarly, if the only ideals ofR

1 are {0} and itself, then any ring isomorphic to R

1 must have this property too.

And now, let us go back to Unit 6 for a moment. Over there we proved the Fundamental Theoremof Homomorphism for groups, according to which the homomorphic image of a group G isisomorphic to a quotient group of G, Now we will prove a similar result for rings, namely, the firstisomorphism theorem or the Fundamental Theorem of Homomorphism for rings.

Theorem 9 (The Fundamental Theorem of Homomorphism): Let f : R S be a ringhomomorphism. Then R/Ker f Im f. In particular, iff is surjective, then R/Ker f S.

Proof: Firstly, note that K/Ker f is a well defined quotient ring since Ker f is an ideal of R. Forconvenience, let us put Ker f = I. Let us define

: R/I � S by (x t I) = f(x).

As in the case of Theorem 8 of Unit 6, we need to check that , is well defined, i.e., if

x + I = y + I then (x + I) = (y + I).

Now, x t I = y + Ix � y I = Ker f f(x � y) = 0 f(x) = f(y)

(x + I) = (y + I).

Thus, is well defined.

Now let us see whether , is an isomorphism or not.

(i) , is a homomorphism : Let x, y R. Then

((x + I) + (y + I)) = (x + y + I) = f(x + y) = f(x) + f(y)

= (x + I) + (y + I), and

((x + 1) (y + 1)) = (xy + 1) = f(xy) = f(x) f(y)

= (x + 1)(y + 1)

Thus, is a ring homomorphism.

(ii) Im = Im f : Since (x + I) = f(x) Im f x R, Im y Im f. Also, any element of Im f isof the form f(x) = (x + I) for some x R. Thus, Im f Im . So, Im = Im f.


Abstract Algebra

Notes (iii) is 1 � 1. To show this let x, y R such that

(x + I) = (y + I). Then f(x) = f(y),

so that f(x � y) = 0, i.e., x � y Ker f = I.

i.e., x + I = y + I.

Thus, is I - I

So, we have shown that R/Ker f 1m f.

Thus, iff is onto, then Im f = S and R/Ker f S.

Note that this result says that f is the composition o , where is the canonical homomorphism:R � R/I : (a) = a + I. This can be diagrammatically shown as

Let us look at some examples of the use of the Fundamental Theorem.

Consider p : Z � Zm

: p(n) = n.p is an epimorphism and Ker p = {n|n = 0 in Z,,,} = mZ.

Therefore, Z/mZ Z,

(Note that we have often used the fact that Z/mZ and Zm

are the same.)

As another example, consider the projection map

p : R1 × R

2 R

1 : p(a, b) = a, where R

1 and R

2 are rings. Then p is onto and its kernel is ((0, b) |

b R2}, which is isomorphic to R

2.

Therefore, (R1 × R

2)/R

2 R

1.

Let us now apply Theorem 9 to prove that any ring homomorphism from a ring R onto Z isuniquely determined by its kernel. That is, we can�t have two different ring homomorphismsfrom R onto Z with the same kernel. (Note that this is not true for group homomorphisms. Infact, you know that I

z and � I

z are distinct homomorphisms from Z onto itself with the same

kernel, {0}. To prove this statement we need the following result.

Theorem 10: The only non-trivial ring homomorphism from Z into itself is Iz.

Proof: Let f : Z - Z be a non-trivial homomorphism. Let n be a positive integer.

Then n = 1 + 1 + ..... + 1 (n times). Therefore, . ,

f(n) = f(1) + f(1) + ..... -1 f(1) (n times) = n f(1).

On the other hand, if n is a negative integer, then �n is a positive integer. Therefore, f(�n) = (�n)f(l), i.e., �f(n) = � nf(1), since f is a homomorphism. Thus, f(n) = n f(1) in this case too.

Also f(0) = 0 = of(1).



NotesThus, f(n) = nf(1) n Z ..... (1)

Now, since f is a non-trivial homomorphism, f(m) # 0 for some m Z.

Then, f(m) = f(m . 1) = f(m) f(1).

Cancelling f(m) on both sides we get f(1) = 1.

Therefore, from (1) we see that

f(n) = n n Z, i.e., f = Iz.

This theorem has an important corollary.

Corollary: Let R be a ring isomorphic. to Z. If f and g are two isomorphisms from R onto Z, thenf = g.

Proof: The composition f.g-� is an isomorphism from Z. onto itself. Therefore, by Theorem 10,fog-1 = Iz, i.e., f = g.

We are now in a position to prove the following result.

Theorem 11: Let R be a ring and f and g be homomorphisms from R onto Z such that Ker f =Ker g. Then f = g.

Proof: By Theorem 9 we have isomorphisms

r : R/Ker f Z and

g : R/Ker g Z.

Since Ker f = Ker g, r and

g are isomorphisms of the same ring onto Z. Thus, by the corollary

above, r =

g.

Also, the canonical maps r : R R/Ker f and g : R R/Ker g are the same since Ker f = Ker g.

f = r o f =

g o

g = g.

Let us halt our discussion of homomorphisms here and briefly recall what we have done in thisunit. Of course, we have not finished with these functions. We will be going back to them againand again in the future units.

Self Assessment

1. If R1 + R

2 be two rings and f : R

1 R

2 be a ring ................. then we define imf = {f(x) | x R

1}.

(a) isomorphisms (b) automorphism

(c) homomorphism (d) polynomial

2. If im f = R2, f is called an ................. or onto homomorphism, then R

2 is called the

homomorphic image of RZ.

(a) epimorphism (b) hemomorphism

(c) isomorphism (d) analogous

3. Two rings are isomorphic if and only if they are algebraically .................

(a) designed (b) identical

(c) onto (d) isomorphic

4. A homomorphism that is ................. is an isomorphism

(a) subjective (b) bijective

(c) onto (d) injective


Abstract Algebra

Notes 5. The only ................. ring homomorphism from Z into itself is Z2.



17.4 Summary

1. The definition of a ring homomorphism, its kernel and its image, along with severalexamples.

2. The direct or inverse image of a subring under a homomorphism is a subring.

3. Iff : R - S is a ring homomorphism, then

(i) Im f is a subring of S,

(ii) Ker f is an ideal of R,

(iii) f-1(1) is an ideal of R for every ideal I of S.

(iv) iff is surjective, then f(I) is an ideal of S.

4. A homomorphism is injective iff its kernel is {0}.

5. The composition of homomorphisms is a homomorphism.

6. The definition and examples of a ring isomorphism.

7. The proof and applications of the Fundamental Theorem of Homomorphism which saysthat iff : R S is a ring homomorphism, then R/Ker f Im f.

17.5 Keyword

Isomorphism: If a homomorphism is a bijection, it is called an isomorphism.


1. Which of the following rings are not fields?

2Z, Z5, Z

6, Q × Q

2. Will a subring of a field be a field? Why?

3. Show that char (X) = 2, where X is a non-empty set.

4. Let R be a ring and char R = m. What is char (R × R)?


1. (c) 2. (a) 3. (b) 4. (b) 5. (b)







Notes


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Abstract Algebra

Notes Unit 18: Integral Domains

CONTENTS

Objectives

Introduction

18.1 Integral Domains

18.2 Field

18.3 Summary

18.4 Keywords



Objectives


Discuss whether an algebraic system is an integral domain or not

Explain the characteristic of any ring

Describe whether an algebraic system is a field or not

Introduction

In the earlier units, we have introduced you to rings, and then to special rings whose specialitylay in the properties of their multiplication. In this unit, we will introduce you to yet anothertype of ring, namely, an integral domain. You will see that an integral domain is a ring withidentity in which the product of two non-zero elements is again a non-zero element. We willdiscuss various properties of such rings.

Next, we will look at rings like Q, R, C, and Z,, (where p is a prime number). In these rings, thenon-zero elements form an abelian group under multiplication. Such rings are called fields.These structures are very useful, one reason being that we can �divide� in them.

Related to integral domains and fields are certain special ideals called prime ideals and maximalideals. In this unit, we will also discuss them and their corresponding quotient rings.

18.1 Integral Domains

You know that the product of two non-zero integers is a non-zero integer, i.e., if m, n Z such.

that m 0, n 0, then mn 0. Now consider the ring Z6. We find that 2 0 and 3 0 , yet

2 . 3 0. So, we find that the product of the non-zero elements 2 and 3 in Z6 is zero.

As you will soon realise, this shows that 2 (and 3) is a zero divisor, i.e., 0 is divisible by 2

(and 3 ).

So, let us see what a zero divisor is.


Unit 18: Integral Domains

NotesDefinition: A non-zero element a in a ring R is called a zero divisor in R if there exists: anon-zero element b in R such that ab = 0.

Now do you agree that 2 is a zero divisor in Z,? What about 3 in Z4? Since 3 x 0 for every

non-zero x in Z4, 3 is not a zero divisor in Z

4.

Now let us look at an example of a zero divisor in C[0, l]. Consider the function

f C[0, 1] given by

1x , 0 x 1/2

2f(x)0,1/2 x 1

Let us define g : [0, 1] + R by

0, 0 x 1/2g(x)

x 1/2,1/2 x 1

Then g C[0, 1], g 0 and (fg) (x) = 0 x [0,1]. Thus, fg is the zero function. Hence, f is a zerodivisor in C[0, 1].

For another example, consider the Cartesian product of two non-trivial rings A and B. For everya 0 in A, (a, 0) is a zero divisor in A × B. This is because, for any b 0 in B. (a . 0) (0.b) = (0.0).

Now let us look at the ring (X), where X is a set with at least two elements, Each non-emptyproper subset A of X is a zero divisor because A.XC = AAC = , the zero element of (X).

Let us now talk of a type of ring that is without zero divisors.

Definition: We call a non-zero ring R an integral domain if

(i) R is with identity, and

(ii) R has no zero divisors.

Thus, an integral domain is a non-zero ring wilh identity in which the product of two non-zeroelements is a non-zero element.

This kind of ring gets its name from the set of integers, one of its best known examples. Otherexamples of domains that immediately come to mind are Q, R and C. What about C[0,1]? Youhave already seen that it has zero divisors. Thus C[0,l] is not a domain.

Note Several authors often shorten the term �integral domain� to �domain�. We willdo so too.

The next result gives us an important class of examples of integral domains.

Theorem 1: Zp is an integral domain iff p is a prime number.

Proof: Firstly, let let us assume that p is a prime number. Then you know that Zp is a non-zero

ring with identity. Let us see if it has zero divisors. For this, suppose pa,b Z satisfy a,b 0.

Then ab 0, i.e., p | ab. Since p is a prime number, we see that p | a or p | b. Thus, a 0 or b 0.


Abstract Algebra

NotesWhat we have shown is that if a 0 and b 6, then ab 6. Thus, Z

p is without zero divisors, and

hence, is domain.

Conversely, we will show that if p is not a prime, then Zp is not a domain, So, suppose p is not

a prime. If p = 1, then Z , is the trivial ring, which is not a domain.

If p is composite number and m | p, you know that m Zp is a zero divisor. Thus, Z

p has zero

divisors. Hence, it is not a domain.

Task Which of the following rings are not domains? Why?

Z4, Z

5, 2Z, Z + iZ, R × R, {0}.

Now consider a ring R. We know that the cancellation law for addition holds in R, i.e., wheneveracb = acc in R, then b = c. But, does ab = ac imply b = c? It need not. For example, 0.1 = 0.2 in Z but1 # 2. So, if a = 0, ab = ac need not imply b = c. But, if a # 0 and ab = ac, is it true that b = c�? We willprove that this is true for integral domains.

Theorem 2: A ring R has no zero divisors if and only if the cancellation law for multiplicationholds in R (i.e., if a, b, c R such that a 0 and ab = ac, then b = c.)

Proof: Let us first assume that R contains no zero divisors. Assume that a, b, c R such thata 0 and ab = ac. Then a(b � c) = ab � ac = 0. As a 0, and R has no zero divisors, we get b � c = 0,i.e., b = c.

Thus if ab = ad and a 0, then b = c.

Conversely, assume that the cancellation law for multiplication holds in R. Let a R such thata 0. Suppose ab = 0 for some b R. Then ab = 0 = a0. Using the cancellation law for multiplication,we get b = 0. So, a is not a zero divisor, i.e., R has no zero divisors.

Using this theorem we can immediately say that the cancellation law holds for multiplication inan integral domain.

Now let us introduce a number associated with an integral domain in fact, with any ring.

For this let us look at Z4 first. We know that 44x 0 x Z . In fact, 8x = 0 and 12 x = 0 also for

any x Z4.

But 4 is the least element of the set { n N | nx = 0 x Z4 ). This shows that 4 is the

characteristic of Z4, as you will see now.

Definition: Let R be a ring. The least positive integer n such that nx = 0 x R is called the

characteristic of R. If there is no positive integer n such that nx = 0 x R, then we say that thecharacteristic of R is zero.

We denote the characteristic of the ring R by char R.

You can see that char Zn = n and char Z = 0.

Now let us look at a nice result for integral domains. It helps in considerably reducing ourlabour when we want to obtain the characteristic of a domain.



NotesTheorem 3: Let m be a positive integer and R be an integral domain. Then the followingconditions are equivalent.

(a) m1 = 0.

(b) ma = 0 for all a R.

(c) ma = 0 for some a 0 in R.

Proof: We will prove (a) (b) (c) (a).

(a) (b) : We know that m 1 = 0.

Thus, for any a R, ma = m (la) = (ml) (a) = 0a = 0, i.e., (b) holds.

(b) (c) : If ma = 0 a R, then if is certainly true for some a 0 in R.

(c) (a) : Let ma = 0 for some a 0 in R. Then 0 = ma = m (la) = (ml) a. As a 0 and R is withoutzero divisors, we get m

1 = 0.

What Theorem 3 tells us is that to find the characteristic of a domain we only need to look at theset in {n.1 | n N}.


(i) char Q = 0, since n.1 0 for any n N.

(ii) Similarly, char R = 0 and char C = 0.

(iii) You have already seen that chat Z, = n. Thus, for any positive integer n, there exists a ringwith characteristic n.

Now let us look at a peculiarity of the characteristic of a domain.

Theorem 4: The characteristic of an integral domain is either zero or a prime number.?

Proof: Let R be a domain. We will prove that if the characteristic of K is not zero, then it is aprime number. So suppose char R = m, where m 0. So m is the least positive integer such thatm.1 = 0. We will show that m is a prime number by supposing that it is not, and then proving thatour supposition is wrong.

So suppose m = st, where s, t N, 1 < s < m and 1 < t < m. Then m.1 = 0 (st).l = 0 * (s.1) (t.1) = 0.As R is without zero divisors, we get s.1 = 0 or t.1 = 0. But, s and t are less than m. So, we reach acontradiction to the fact that m = char R. Therefore, our assumption that m = st, where 1 < s < m,1 < t < m is wrong. Thus, the only factors of m are 1 and itself. That is, m is a prime number.

We will now see what algebraic structure we get after we impose certain restrictions on themultiplication of a domain.

18.2 Field

Let (R, +, .) be a ring. We know that (R, +) is an abelian group. We also know that the operationis commutative and associative. But (R,.) is not an abelian group. Actually, even if R has identity,(R,.) will never be a group since there is no element a R such that a.0 = 1. But can (R\{0}) be agroup? It can, in some cases. For example, from Unit 2 you know that Q* and R* are groups withrespect to multiplication. This allows us to say that Q and R are fields, a term we will now define.

Definition: A ring (R, +,.) is called a field if (R\{0},.) is an abelian group.


Abstract Algebra

Notes Thus, for a, system (R,+,.) to be a field it must satisfy the ring axioms R1 to R

6 as well as the

following axioms.

(i) is commutative,

(ii) R has identity (which we denote by 1) and 1 0, and

(iii) every non-zero element x in R has a multiplicative inverse, which we denote by x�1.

Just as a matter of information we would like to tell you that a ring that satisfies only (ii) and (iii)above, is called a division ring or a skew field or a non-commutative field. Such rings are veryimportant in the study of algebra, but we will not be discussing them in this course.

Let us go back to fields now. The notion of a field evolved daring the 19th century through theresearch of the German mathematicians Richard Dedekind and Leopold Kronecker in algebraicnumber theory. Dedekind used the German word Körper, which means field, for this concept.This is why you will often find that a field is denoted by K.

As you may have realised, two of the best known examples of fields are R and C. These were thefields that Dedekind considered. Yet another example of a field is the following ring.

Example: Show that Q 2Q {a 2b|a,b Q} is field.

Solution: From Unit 14 you know that F Q 2Q is a commutative ring with identity 1 + 2.

Now, let a 2b be a non-zero element of F. Then either a 0 or b 0. Now, using therationalisation process, we see that

1

a 2b

=2 2

1 1 2b a fiba 2ba 2b (a 2b)(a 2b)

= 2 2 2 2

1 ( b)2 F

a 2b a 2b

(Note that a2 � 2b2 0, since 2 is not rational and either a 0 or b 0.)

Thus, every non-zero element has a multiplicative inverse. Therefore, Q 2Q is a field.

Can you think of an example of a ring that is not a field? Does every non-zero integer have amultiplicative inverse in Z? No. Thus, Z is not a field.

By now you have seen several examples of fields. Have you observed that all of them happen tobe integral domains also? This is not a coincidence. In fact, we have the following result.

Theorem 5: Every field is an integral domain.

Proof: Let F be a field. Then F {0} and 1 B. We need to see if F has zero divisors. So let a andb be elements of F such that ab = 0 and a 0. As a 0 and P is a field, a-1 exists.

Hence, b = I . b = (a � la) b = ad-1 (ab) = a-1 0. Hence, if a 0 and ab = 0, we get b = 0, i.e., F has nozero divisors. Thus, F is a domain.

Theorem 5 may immediately prompt you to ask if every domain is a field. You have alreadyseen that Z is a domain but not a field. But if we restrict ourselves to finite domains, we find thatthey are fields.



NotesTheorem 6: Every finite integral domain is a field.

Proof: Let R = {a, = 0, a1 = 1, a

2,....., a,] be a finite domain. Then R is commutative also. To show

that R is a field we must show that every non-zero element of R has a multiplicative inverse.

So, let a = aj be a non-zero element of R (i.e., i 0). Consider the elements aa

1, ..., aa

n. For every

j 0, aj 0; and since a 0, we get aa

j 0.

Hence, the set { aa1, ..., aa

n } G (a,, ..., a,}.

Also, aa, , aa ,..., aa, are all distinct elements of the set {a,, ...., a,}, since aaj = na

k a

j = a

j, using the

cancellation law for multiplication.

Thus, {aa1, ...., aa

n} = [a; ,...., a

n}.

In particular, a, = aaj, i.e., 1 = aa

j for some j. Thus, a is invertible in R. Hence every non-zero

element of R has a multiplicative inverse. Thus, M is a field.

Using this result we can now prove a theorem which generates several examples of finite fields.

Theorem 7: Zn is a field if and only if n is a prime number.

Proof: From theorem 1 you know that Zn is a domain if and only if n is a prime number. You also

know that Zn has only n elements. Now we can apply Theorem 6 to obtain the result.

Theorem 7 unleashes a load of examples of fields : Z2, Z

3, Z

5, Z

7,, and so on. Looking at these

examples, and other examples of fields, can you say anything about the characteristic of a field?In fact, using Theorems 4 and 5 we can say that.

Theorem 8: The characteristic of a field is either zero or n prime number.

So far the examples of finite fields that you have seen have consisted of p elements, for someprime p. In the following exercise we give you an example of a finite field for which this is notso.

Theorem 9: Let R be a ring with identity. Then R is a field if and only if R and {0} are the onlyideals of R.

Proof: Let us first assume that R is a field. Let I be an ideal of R. If I {0), there exists, a non-zeroelement x I. As x 0 and R is a field, xy = 1 for some y R. Since x I and I is an ideal, xy I,i.e., 1 I.

Conversely, assume that R and { 0 } are the only ideals of R. Now, let a 0 be an element of R.Then you know that the set Ra = [ra | r R] is a non-zero ideal of R. Therefore, Ra = R.

Now, 1 R = Ra. Therefore, 1 = ba for some b R, i.e., a-1 exists. Thus, every non-zero elementof R has a multiplicative inverse. Therefore, R is a field.

Using Theorem 9, we can obtain some interesting facts about field homomorphisms (i.e., ringhomomorphisms from one field to another). We give them to you in the form of an exercise.

Now that we have discussed domains and fields, let us look at certain ideals of a ring, withrespect to which the quotient rings are domains or fields.

Self Assessment

1. Several authors often shorten the term .................... to domain.

(a) integral domain (b) abstract domain

(c) different domain (d) prime domain


Abstract Algebra

Notes 2. Zp is an integral domain if p is a .................... number.

(a) even (b) odd

(c) prime (d) integer

3. A ring R has .................... zero divisor if and only if the cancellation law for multiplicationholds in R.

(a) 1 (b) 2

(c) 0 (d) 3

4. A ring (R, +,.) is called a .................... if (R | { 0 }.) is an abelian group.

(a) field (b) domain

(c) range (d) ideal

5. Every .................... integral domain is a field.

(a) infinite (b) finite


18.3 Summary

The definition and examples of an integral domain.

The definition and examples of a field.

Every field is a domain.

A finite domain is a field.

The characteristic of any domain or field is either zero or a prime number.

18.4 Keywords

Zero Divisor: A non-zero element a in a ring R is called a zero divisor in R if there exists: a non-zero element b in R such that ab = 0.

Prime Number: Zp is an integral domain iff p is a prime number.

Abelian Group: A ring (R, +,.) is called a field if (R\{0},.) is an abelian group.


1. Let n N and m | n | < m < n. Then show that m is a zero divisor in Zn.

2. List all the zero divisors in Z.

3. For which rings with unity will 1 be a zero divisor?

4. Let R be a ring and a R be a zero divisor. Then show that every element of the principalideal Ra is a zero divisor.

5. In a domain, show that the only solutions of the equation x2 = x are x = 0 and x = 1.

6. Prove that 0 is the only nilpotent element in a domain.



NotesAnswers: Self Assessment

1. (a) 2. (c) 3. (b) 4. (c) 5. (a)






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Abstract Algebra

Notes Unit 19: The Field of Quotient Euclidean Domains

CONTENTS

Objectives

Introduction

19.1 Prime and Maximal Ideals

19.2 Field of Quotients

19.3 Summary

19.4 Keywords



Objectives


Discuss whether an algebraic system is an integral domain or not

Define and identify prime ideals and maximal ideals

Prove and use simple properties of integral domains and fields

Construct or identify the field of quotients of an integral domain

Introduction

Finally, we shall see how to construct the smallest field that contains a given integral domain.This is essentially the way that Q is constructed from Z. We call such a field the field of quotientsof the corresponding integral domain.

In this unit, we have tried to introduce you to a lot of new concepts. You may need some time tograsp them. Take as much time as you need. But by the time you finish it, make sure that youhave attained the knowledge of following topics.

19.1 Prime and Maximal Ideals

In �Z� we know that if p is a prime number and p divides the product of two integers a and b, theneither p divides a or p divides b. In other words, if ab pZ, then either a pZ or b pZ. Becauseof this property we say that pZ is a prime ideal, a term we will define now.

Definition: A proper ideal P of a ring R is called a prime ideal of R if whenever ab P for a, b R, then either a P or b P.

You can see that {0} is a prime ideal of Z because ab {0} a {0} or b {0}, where a,b Z.


Unit 19: The Field of Quotient Euclidean Domains

NotesAnother example of a prime ideal is

Example: Let R be an integral domain. Show that I = ((0, x) | x R) is a prime ideal of R x R.

Solution: Firstly, you know that I is an ideal of R x R. Next, it is a proper ideal since I R x R.Now, let us check if I is a prime ideal or not. For this let (a

1, b

1), (a

2, b

2) R x R such that (a

1, b

1)

(a2, b

2) I. Then (a

1a

2, b

1b

2) = (0, x) for some x R

a1a

2 = 0, i.e., a, = 0 or a

2 = 0, since R is a domain. Therefore, (a, b

1) I or (a

2, b

2) I. Thus, I is a

prime ideal.

Now we will, prove the relationship between integral domains and prime ideals.

Theorem 1: An ideal P of a ring R with identity is a prime ideal of R if and only if the quotientring R/P is an integral domain.

Proof: Let us first assume that P is a prime ideal of R. Since R has identity, so has R/P. Now, leta + P and b + P be in R/P such that (ai � P) (b + P) = P, the zero element of R/P. Then ab+P = P, i.e.,ab P. As P is a prime ideal of R either a P or b P. So either a + P = P or b+P = P.

Thus, R/P has no zero divisors.

Hence, R/P is an integral domain.

Conversely, assume that R/P is an integral domain. Let a, b R such that ab P. Then ab + P =P in R/P, i.e., (a + P) (b + P) = P in R/P. As R/P is an integral domain, either a + P = P or b + P =P, i.e., either a E P or b P. This shows that P is a prime ideal of R.

An ideal mZ of Z is prime iff m is a prime number. Can we generalise this relationship betweenprime numbers and prime ideals in Z to any integral domain? To answer this let us first try andsuitably generalise the concepts of divisibility and prime elements.

Definition: In a ring R, we say that an elements divides an element b if b = ra for some r R. Inthis case we also say that a is a factor of b, or a is a divisor of b.

Thus, 3 divides 6 in Z7, since 3.2 6.

Now let us see what a prime element is.

Definition: A non-zero element p of an integral domain R is called n prime element if

(i) p does not have a multiplicative inverse, and

(ii) whenever a, b R and p | ab, then p | a or p | b.

Can you say what the prime elements of Z are? They are precisely the prime numbers and theirnegatives.

Now that we know what a prime element is, let us see if we can relate prime ideals and primeelements in an integral domain.

Theorem 2: Let R be an integral domain. A non-zero element p R is n prime element if andonly if Rp is a prime ideal of R.

Proof: Let us first assume that p is a prime element in R. Since p does not have a multiplicativeinverse, 1 Rp. Thus, Rp is a proper ideal of R. Now let a, b R such that ab Rp. Then ab = rpfor some r R

p | a or p | b, since p is a prime element.

a = xp or b = xp for some x R.

a p or b Rp .

Thus ab Rp either a Rp or b G Rp, i.e., Rp is a prime ideal of R.


Abstract Algebra

Notes

Note x R has a multiplicative inverse iff Rx = R.

Conversely, assume that Rp is a prime ideal. Then Rp R, Thus, 1 Rp, and hence, p does nothave a multiplicative inverse. Now suppose p divides ab, where a, b R. Then ab = rp far somer R, i.e., ab Rp.

As Rp is a prime ideal, either a Rp or b Rp. Hence, either p | a or p | b. Thus, p is a primeelement in R.

Theorem 2 is very useful for checking whether an element is a prime element or not, or forfinding out when a principal ideal is a prime ideal.

Prime ideals have several useful properties.

Now consider the ideal 22 in Z. Suppose the ideal nZ in Z is such that 2Z nZ Z. Then n | 2. n= 1or n = 2. nZ = Z or nZ = 2Z.

This shows that no ideal can lie between 2Z and Z. That is, 22 is maximal among the properideals of Z that contain it. So we say that it is a �maximal ideal�. Let us define this expression.

Definition: A proper ideal M of a ring R is called a maximal ideal if whenever I is an ideal of Rsuch that M I R, then either I = M or I = R.

Thus, a proper ideal M is a maximal ideal if there is no proper ideal of R which contains it. Anexample that comes to mind immediately is the zero ideal in any field F. This is maximalbecause you know that the only other ideal of F is F itself.

To generate more examples of maximal ideals, we can use the following characterisation of suchideals.

Theorem 3: Let R be a ring with identity. An ideal M in R is maximal if and only if R/M is a field.

Proof: Let us first assume that M is a maximal ideal of R. We want to prove that R/M is a field.For this, it is enough to prove that R/M has no non-zero proper ideals. So, let I be an ideal ofR/M. Consider the canonical homomorphism : R R/M : (r) = r + M. Then, you know that-1 (I) is an ideal of R containing M, the kernel of . Since M is a maximal ideal of R. 1(I) = M or

-1(I) = R. Therefore, I = (-1 (I)) is either (M) or (R). That is, I = {0} or I = R/M, where; = O +M

= M. Thus, R/M is a field.

Conversely, let M be an ideal of R such that R/M is a field. Then the only ideals of R/M are

{0} and R/M. Let I be an ideal of R containing M. Then, as above (1) = {0} or, (I) = R/M.

I = -1((1)) is M or R. Therefore, M is a maximal ideal of R.

Corollary: Every maximal ideal of a ring with identity is a prime ideal.

Now, the corollary is a one-way statement. What about the converse? That is, is every primeideal maximal? What about the zero ideal in Z? Since Z is a domain but not a field and Z Z/{0},Z/{0} is a domain but not a field. Thus. (0) is a prime ideal but not a maximal ideal of Z.

Example: Show that an ideal mZ of Z is maximal iff m is a prime number.

Solution: You know that Z,,, is a field iff m is a prime number. You also know that Z/mZ = Zm

.Z | mZ is a field iff m is prime. Hence, mZ is maximal in Z iff m is a prime number.



Notes

Example: Show that 122Z is a maximal ideal of Z12

, whereas (0, 4,8) is not.

Solution: You know that Z12

= Z/12Z and 122Z = 2Z/12Z. We see that Z12

/ 122Z = (Z/12Z)/

(2Z/12Z) = Z2, which is a field. Therefore, 122Z {0,2,4,6,8,10} is maximal in Z

12.

Now 12 12 12{0, 4, 8} = 4Z 2Z Z .

Therefore, {0, 4, 8} is not maximal in Z12

.

We first introduced you to a special ideal of a ring, called a prime ideal. Its speciality lies in thefact that the quotient ring corresponding to it is an integral domain.

Then we discussed a special kind of prime ideal, i.e., a maximal ideal.

19.2 Field of Quotients

Consider Z and Q. You know that every element of Q is of the form a

,b

where a Z and b Z*.

Actually, we can also denote ab

by the ordered pair (a, b) Z × Z*. Now, in Q we know that

a cb d = - iff ad = bc. Let us put a similar relation on the elements of Z × Z*

Now, we also know that the operations on Q are given by

a c ad bc a c a c a cand . , Q.

b d bd b d b d b d

Keeping these in mind we can define operations on Z × Z*. Then we can suitably define anequivalence relation on Z × Z* to get a field isomorphic to Q.

We can generalise this procedure to obtain a field from any integral domain. So, take an integraldomain R. Let K be the following set of ordered pairs:

K= {(a,b) ) a , b R and b 0)

We define a relation ~ in K by

(a, b) ~ (c, d) if ad = bc.

We claim that ~ is an equivalence relation. Let us see if this is so.

(i) (a, b) ~ (a, b) (a, b) K, since R is commutative. Thus, ~ is reflexive.

(ii) Let (a, b), (c, d) K such that (a, b) ~ (c, d). Then ad = bc, i.e., cb = da. Therefore, (c, d) ~(a, b). Thus, ~ is symmetric.

(iii) Finally, let (a,b), (c,d), ( u, v) K such that (a,b) � (c,d) and (c,d) ~ (u,v ). Then ad = bc andcv = du. Therefore, (ad) v = (bc)v = bdu, i.e., avd =bud. Thus, by the cancellation law formultiplication (which is valid for a domain), we get av = bu, i.e., (a,b) � (u,v). Thus, � istransitive.

Hence, ~ is an equivalence relation.


Abstract Algebra

Notes Let us denote the equivalence class that contains (a,b) by [a,b]. Thus,

[a,b] = {(c,d) | c,d R, d 0 and ad = bc }.

Let F be the set of all equivalence classes of K with respect to ~.

Let us define + and in F as follows. (It might help you to keep in mind the rules for adding andmultiplying rational numbers.)

[a,b] + [c,d] = [ad+bc,bd] and

[a,b].[c,d] = [ac,bd].

Do you think + and . are binary operations on F?

Note b # 0 and d # 0 in the integral domain R imply bd # 0. So, the right-handsides of the equations given above are well defined equivalence classes. Thus, the sum andproduct of two elements in F is again an element in F.

We must make sure that these operations are well defined.

So, let [a,b] = [a�,b�] and [c,d] = [c�,d�]. We have to show that [a,b] + [c,d] = [a�,b�] + [c�,d�],i.e., [ad+bc,bd] = [a�d�+b�c�,b�d�].

Now, (ad+bc) b�d� � (a�d� + b�c�) bd

= ab�dd�, + cd�bb� � a�bdd� � cdbb�

= (ab� � a�b)dd� + (cd� - c�d) bb�

= (0) dd� + (0)bb�, since (a,b) - (a�, b�) and (c,d) ~ (c�,d�).

= 0

Hence, [ad+bc,bd] = [a� d� + b�c�,brd�], i.e., + is well defined.

Now, let us show that [a,b] . [c,d] = [a�,br] . [c�,d�],

i.e., [ac,bd] = [a�c� b�d�].

Consider (ac) (b�d�) - (bd) (a�c�)

= ab�cd� � ba�dc� = ba�cd� � ba� cd�, since ab� = bar and cd� = dc�

= 0

Therefore, [ac,bd] = [a�c�,b�d�]. Hence,. is well defined.

We will now prove that F is a field.

(i) + is associative : For [a,b], [c,d], [u,v] E F,

([a,b] + [c,d]) + [u,v] = [ad+bc,bd] + [u,v]

= [(ad+bc)v + ubd, Wv]

= [adv + b(cv+ud), bdv]

= [a,b] + [cv+ud,dv]

= [a,b] + ([c,d]+ [u,v])



Notes(ii) + is commutative :.For [a,b], [c,d] F,

[a,b] + [c,d] = [ad+bc,bd] = [cb+da,db] = [c,d] + [a,b]

(iii) [0,1] is the additive identity for F : For [a,b] F,

[0,1] + [a,b] = [0.b+l.a, l.b] = [a,b]

(iv) The additive inverse of [a,b] F is [�a,b] :

[a,b] + [�a,b] = [ab-ab,b2] = [0,b2] = [0,1], since 0.1 = 0.b2.

We would like you to prove the rest of the requirements for F to be a field.

So we have put our heads together and proved that F is a field.

Now, let us define f : R F : f(.a) = [a,1]. We want to show that f is a monomorphism.

Firstly, for a, b R,

f(a+b) = [a+b,1] = [a,]] + [b,l].

= f(a) + f(b), and

Thus, f is a ring homomorphism.

Next, let a,b R such that f(a) = f(b). Then [a,1] = [b,l], i.e., a = b. Therefore, f is 1�1.

Thus, f is a monomorphism.

So, Im f = (R) is a subring of F which is isomorphic to R.

As you know, isomorphic structures are algebraically identical.

So, we can identify R with f(R), and think of R as a subring of F. Now, any element of F is of theform [a,b] = [a,1] [l,b] = [a,l] [b,l]-1 = f(a) f(b)-1, where b 0. Thus, identifying x R with f(x) f(R),we can say that any element of F is of the form ab-1, where a,b R, b 0.

All that we have discussed adds up to the proof of the following theorem.

Theorem 4: Let R be an integral domain. Then R can be embedded in a field F such that everyelement of F has the form ab-1 for a, b R, b 0.

The field F whose existence we have just proved is called the field of quotients (or the field offraction) of R.

Thus, Q is the field of quotient of Z. What is the field of quotients of R? The following theoremanswers this question.

Theorem 5: Iff : R K is a monomorphism of an integral domain R into a field K, then thereexists a monomorphism g : F K : g([a,1]) = f(a), where F is the field of quotients of R.

It says that the-field of quotients of an integral domain is the smallest field containing it. Thus,the field of quotients of any field is the field itself.

So, the field of quotients of R is R and of Zp is Z

p, where p is a prime number.

Self Assessment

1. An ideal P of a ring R is called a/an .................. ideal of R. If whenever ab P for a, b R,then either a P or b P.

(a) prime ideal (b) odd ideal

(c) even ideal (d) integer ideal


Abstract Algebra

Notes 2. An ideal P of a ring R with identity is a prime ideal of R. If and only if the .................. R/Pis an integral domain.

(a) polynomial ring (b) subring

(c) quotient ring (d) ideal ring

3. If x R, it has multiplicative inverse iff RX = ..................

(a) R (b) RX-1

(c) XR (d) X

4. A proper ideal m of a ring R is called maximal ideal of whenever I is an ideal of R such thatm .................. I .................. R then either I = m or I = R.

(a) , (b) ,

(c) , (d) ,

5. If R be a ring with identity. An ideal M in R is maximal if and only if .................. is a field.

(a) R.M (b) R/M

(c) M/R (d) R+M

19.3 Summary

The characteristic of any domain or field is either zero or a prime number.

The definition and examples of prime and maximal ideals.

The proof and use of the fact that a proper ideal I of a ring R with identity is prime(or maximal) iff R/I is an integral domain (or a field).

Every maximal ideal is a prime ideal.

An element p of an integral domain R is prime iff the principal ideal pR is a prime ideal ofR.

Z, is a field iff n is a prime number.

The construction of the field of quotients of an integral domain.

19.4 Keywords

Prime Ideal: A ideal P of a ring R is called a prime ideal of R if whenever ab P for a, b R, theneither a P or b P.

Proper Ideal: A proper ideal M of a ring R is called a maximal ideal if whenever I is an ideal ofR such that M I R, then either I = M or I = R.

Maximal Ideal: Every maximal ideal of a ring with identity is a prime ideal.


1. Let F be a field. Show that F, with the Euclidean valuation d defined by d(a) = 1 a F/{0}, is a Euclidean domain.

2. Let F be a field. Define the function

d : F(x)\{0} N {0} : d(f(x)) = deg f(x).

Show that d is a Euclidean valuation on F[x], and hence, F[x] is a Euclidean domain.



Notes3. Find all the units in

(a) Z (b) Z6 (c) Z/5Z (d) Z + IZ

4. Let R be an integral domain. Show that

(a) u is a unit in R iff u | 1.

(b) for a, b R, a | b and b | a iff a and b are associates in R.

5. Which of the following polynomials is irreducible? Give reasons for your choice.

(a) x2 � 2x + 1 R[x] (b) x2 + x + 1 C[x]

(c) x � i C[x] (d) x3 � 3x2 + 2x + 5 R[x].


1. (a) 2. (c) 3. (a) 4. (c) 5. (b)






www.math.niu.edu

www.maths.tcd.ie/






Abstract Algebra

Notes Unit 20: Principal Ideal Domains

CONTENTS

Objectives

Introduction

20.1 Euclidean Domain

20.2 Principal Ideal Domain (PID)

20.3 Summary

20.4 Keywords



Objectives


Discuss the Principal Ideal Domains

Describe theorem related to Principal Ideal Domains

Introduction

In the last unit, you have studied about field and integer domain. In this unit, you will studyabout Principal Ideal Domains.

20.1 Euclidean Domain

In earlier classes you have seen that Z and F[x] satisfy a division algorithm. There are manyother domains that have this property. Here we will introduce you to them and discuss some oftheir properties. Let us start with a definition.

Definition: Let R be an integral domain. We say that a function d : R \ (0) NU (0) is a Euclideanvaluation on R if the following conditions are satisfied:

(i) d(a) d (ab) a, b R \ {0}, and

(ii) for any a, b k, b 0 3 q, r R such that

a = bq+r, where r = 0 or d(r) < d(b).

And then R is called a Euclidean domain.

Thus, a domain on which we can define a Euclidean valuation is a Euclidean domain.

Let us consider an example.

Example: Show that Z is a Euclidean domain.

Solution: Define, d : Z N{0} : d(n) = |n|.


Unit 20: Principal Ideal Domains

NotesThen, for any a,b Z \ {O],

d(ab)= |ab| = |a| |b| |a| (since | b| for b 0)

= d(a).

i.e., d(a) d(ab).

Further, the division algorithm in Z says that if a, b Z, b 0, then 3 q, r Z such that

a = bq + r, where r = 0 or 0 < |r| < |b|.

i.e:, a = bq+r, where r = 0 or d(r) < d(b).

Hence, d is a Euclidean valuation and Z is a Euclidean domain.

Let us now discuss some properties of Euclidean domains. The first property involves theconcept of units. So let us define this concept. Note that this definition is valid for any integraldomain.

Definition: Let R be an integral domain. An element a R is called a unit (or an invertibleelement) in R, if we can find an element b R, such that ab = 1, i.e., if a has a multiplicativeinverse.

For example, both 1 and -1 are units in Z since 1.1 = 1 and (-1).(-1) = 1.

!Caution The difference between a unit in R and the unity in R. The unity is the identitywith respect to multiplication and is certainly a unit. But a ring can have other units ton, asyou have just seen in the case of Z.

Now, can we obtain all the units in a domain? You know that every non-zero element in a fieldF is invertible. Thus, the set of units of F� is F \ {0}. Let us look at some examples.

Example: Obtain all the units in F[x], where F is a field.

Solution: Let f (x) P[ x] be a unit, Then g(x) F[x] such that f(x) g(x) = 1. Therefore,

deg f(x) g(x)) = deg(1) = 0, i.e.,

deg f(x) + deg g(x) = 0

Since deg f(x) and deg g(x) are non-negative integers, this equation can hold only if deg f(x) = 0= deg g(x). Thus, f(x) must be a non-zero constant, i.e., an element of F\ {0}. Thus, the units of F[x]are the non-zero elements of F. That is, the units of F and F[x] coincide.

Example: Find all the units in R = a b 5|a,b Z .

Solution: Let a b 5 be a unit in R. Then there exists

c d 5 R such that

a b 5 c d 5 = 1

(ac � 5bd) + (bc + ad) 5 = 1

ac � 5bd = 1 and bc+ad = 0

abc � 5b2d = b and bc+ad = 0

a(�ad) � 5b2d = b, substituting bc = �ad.


Abstract Algebra

Notes So, if b 0, then (a2 + 5b2) | b, which is not possible.

b = 0.

Thus, the only units of R are the invertible elements of Z.

Theorem 1: Let R be a Euclidean domain with Euclidean valuation d. Then, for any a R \ {0},d(a) = d(l) iff a is a unit in K.

Proof: Let us first assume that a R\ {0] with d(a) = d(1).

By the division algorithm in R, q, r R such that 1 = aq+r,

where r = 0 or d(r) < d(a) = d(1).

Now, if r 0, d(r) = d(r.1) d(1). Thus, d(r) < d(1) can�t happen.

Thus, the only possibility for r is r = 0,

Therefore, 1 = aq, so that a is a unit.

Conversely, assume that a is a unit in R. Let b R such that ab = 1. Then d(a) d(ab) = d(1). Butwe know that d(a) = d(a.1) d(1). So, we must have d(a) = d(1).

Using this theorem, we can immediately solve Example, since f(x) is a unit in F[x] iff deg f(x) =deg (1 ) = 0.

Now let us look at the ideals of a Euclidean domain.

Theorem 2: Let R be a Euclidean domain with Euclidean valuation d. Then every ideal I % of Ris of the form I = Ra for some a R.

Proof: If I = (01, then I = Ka, where a = 0. So let us assume that I {0}. Then I\ {0} is non-empty.Consider the set {d(a) | a I \{0}). The well ordering principle this set has a minimal element.Let this be d(b), where b e I \ {0}. We will show that I = Rb.

Since b 1 and I is an ideal of R,

Rb I. ...(1)

Now take any a I. Since I R and R is a Euclidean domain, we can find q, r R such that

a = bq + r, where r = 0 or d(r) < d(b).

Now, b I bq I. Also, a I. Therefore, r = a � bq I.

But r = 0 or d(r) < d(b), The way we have chosen d(b), d(r) < d(b) is not possible.

Therefore, r = 0, and hence, a = bq Rb.

Thus, I Rb. ...(2)

From (1) and (2) we get

I = Rb.

Thus, every ideal I of a Euclidean domain R with Euclidean valuation d is principal, and isgenerated by a I, where d(a) is a minimal element of the set {d(x) | x I \ (0) }.

Tasks 1. Show that every ideal of F[x] is principal, where F is a field.

2. Using Z as an example, show that the set

3. S = (a R\ (0) | d(a) > d(1) } (0) is not an ideal of the Euclidean domain Rwith Euclidean valuation d.



Notes20.2 Principal Ideal Domain (PID)

In the previous section you have proved that every ideal of F[x] is principal, where F is a field.There are several other integral domains, apart from Euclidean domains, which have this property.We give such rings a very appropriate name.

Definition: We call an integral domain R a principal ideal domain (PID, in short) if every idealin R is a principal ideal.

Note Every Euclidean domain is a PID

Thus, Z is a PID. Can you think of another example of a PID? What about Q and Q[x]? In fact, byTheorem 2 all Euclidean domains are PIDs. But, the converse is not true. That is, every principalideal domain is not a Euclidean domain.

For example, the ring of all complex numbers of the form b

a 1 i 19 ,2

where a, b Z, is a

principal ideal domain, but not a Euclidean domain.

Now let us look at an example of an integral domain that is not a PID.

Example: Show that Z[x] is not a PID,

Solution: You know that Z[x] is a domain, since Z is one. We will show that all its ideals are notprincipal. Consider the ideal of Z[x] generated by 2 and x, i.e., < 2,x>. We want to show that< 2, x > <f(x)> for any f(x) Z[X].

On the contrary, suppose that 3 f(x) Z[x] such that < 2, x > = < f(x) >. Clearly, f(x) 0.

Also, 3 g(x), h(x) Z[x] such that

2 = f(x) g(x) and x = f(x) h(x).

Thus, deg f(x) + deg g(x) = deg 2 = 0 ...(1)

and deg f(x)+deg h(x) = deg x = 1 ...(2)

(I) shows that deg f(x) = 0, i.e., f(x) Z, say f(x) = n.

Then (2) shows that deg h(x) = 1. Let h(x) = ax+b with a,b Z.

Then x =f(x) h(x) = n(ax+b).

Comparing the coefficients on either side of this equation, we see that na = 1 and nb = 0. Thus, nis a unit in Z, that is, n = If I.

Therefore, 1 < f(x) > = < x,2 >. Thus, we can write

I = x (a0 +a

1x+ ...+a

rxr ) + 2(b

0+b

1x+ .... +b

sxs), where a

i,b

j Z i = 0, l,.. ...., r and j = 0, 1,...,s.

Now, on comparing the constant term on either side we see that 1 = 2b0. This can�t be true, since

2 is not invertible in Z. So we reach a contradiction.

Thus, < x,2 > is not a principal ideal.

Thus, Z[x] is not a P.I.D.


Abstract Algebra

Notes We will now discuss some properties of divisibility in PIDs. If R is a ring and a,b R, with a 0,then a divides b if there exists c R such that b = ac.

Definition: Given two elements a and b in a ring. R, we say that c R is a common divisor of aand b if c | a and c | b.

An element d R is a greatest common divisor (g.c.d, in short) of a, b R if

(i) d | a and d | b, and

(ii) for any common divisor c of a and b, c | d.

For example, in Z a g.c.d of 5 and 15 is 5 , and a g.c.d of 5 and 7 is 1.

We will show you that if the g.c.d of two elements exists, it is unique up to units, i.e., if d and dare two g.c.ds of a and b, then d=ud� , for some unit u.

So now let us prove the following result.

Theorem 3: Let R be an integral domain and a, b R. If a g.c.d of a and b exists, then it is uniqueup to units.

Proof: So, let d and d� be two g.c.ds of a and b. Since d is a common divisor and d� is a g.c.d, weget d | d� . Similarly, we get d�|d. Thus, we see that d and d� are associates in R. Thus, the g.c.d ofa and b is unique up to units.

Theorem 3 allows us to say the g.c.d instead of a g.c.d. We denote the g.c.d of a and b by (a,b).(This notation is also used for elements of R × R. But there should be no cause for confusion. Thecontext will clarify what we are using the notation for.

How do we obtain the g.c.d of two elements in practice? How did we do it in Z? We looked at thecommon factors of the two elements and their product turned out to be the required g.c.d.We will use the same method in the following example.

Example: In Q[x] find the g.c.d of

p(x) = x2 + 3x � 10 and

q(x) = 6x2 � 10x � 4

Solution: By the quadratic formula, we know that the roots of p(x) are 2 and �5, and the roots ofq(x) are 2 and �1/3.

Therefore, p(x) = (x � 2) (x + 5) and q(x) = 2(x � 2) (3x + 1).

The g.c.d of p(x) and q(x) is the product of the common factors of p(x) and q(x), which is (x � 2).

Let us consider the g.c.d of elements in a PD.

Theorem 4: Let R be a PID and a, b R. Then (a, b) exists and is of the form ax + by for some x,y R.

Proof: Consider the ideal <a, b>. Since R is a PID, this ideal must be principal also. Let d R suchthat <a, b> = <d>. We will show that the g.c,d of a and b is d.

Since a <d>, d | a, Similarly, d | b.

Now suppose c R such that c | a and c | b.

Since d E <a,b>, 3 x, y R such that d = ax+by.

Since c | a and c | b, c | (ax+by), i.e., c | d.



NotesThus, we have shown that d = (a,b), and d= ax+by for some x.y R.

The fact that F[x] is a PID gives-us the following corollary to Theorem 4.

Corollary: Let F be a field. Then any two polynomials f(x) and g(x) in F[x] have a g.c.d which isof the form a(x)f(x)+b(x)g(x) for some a(x), b(x) F[x].

For example, (c), (x�1) = 15

(x3 � 2x2 + 6x � 5) + ( x)

5

(x2 � 2x + 1).

Now you can use Theorem 4 to prove the following exercise about relatively prime elements ina PID, i.e., pairs of elements whose g.c.d is 1.

Let us now discuss a concept related to that of a prime element of a domain.

Definition: Let R be an integral domain. We say that an element x R is irreducible if

(i) x is not a unit, and

(ii) if x = ab with a,b R, then a is a unit or b is a unit.

Thus, an element is irreducible if it cannot be factored in a non-trivial way, i.e., its only factorsare its associates and the units in the ring.

So, for example, the irreducible elements of Z are the prime numbers and their associates. Thismeans that an element in Z is prime iff it is irreducible.

Another domain in which we can find several examples is F[x], where F is a field. Let us look atthe irreducible elements in R[x] and C[x], i.e., the irreducible polynomials over R and C. Considerthe following important theorem about polynomials in C[x]. You have already come across thisin the Linear Algebra course.

Theorem 5 (Fundamental Theorem of Algebra): Any non-constant polynomial in C[x] has aroot in C.

Does this tell us anything about the irreducible polynomials over C? Yes. In fact, we can alsowrite it as:

Theorem 5: A polynomial is irreducible in C[x] iff it is linear.

A corollary to this result is:

Theorem 6: Any irreducible polynomial in R[x] has degree 1 or degree 2.

We will not prove these results here but we will use them often when discussing polynomialsover R or C. You can use them to solve the following exercise.

Let us now discuss the relationship between prime and irreducible elements in a PID.

Theorem 7: In a PID an element is prime iff it is irreducible.

Proof: Let R be a PID and x R be irreducible. Let x | ab, where a, b R. Suppose x I a.

Then (x,a) = 1, since the only factor of x is itself, up to units. Thus, xb, Thus, x is prime.

Task Let R be a domain and p R be a prime element. Show that p is irreducible.(Hint: Suppose p = ab. Then p | ab. If p | a, then show that b must be a unit.)


Abstract Algebra

Notes Now, why do you think we have said that Theorem, 7 is true for a PID only? You can see that oneway is true for any domain. Is the other way true for any domain? That is, is every irreducibleelement of a domain prime? You will get an answer to this question.

Example: Just now we will look at some uses of Theorem 7.

Theorem 7 allows us to give a lot of examples of prime elements of F[x]. For example, any linearpolynomial over F is irreducible, and hence prime. In the next unit we will particularly considerirreducibility (and hence primeness) over Q[x].

Now we would like to prove a further analogy between prime elements in a PID and primenumbers, namely, a result analogous. For this we will first show g very interesting property ofthe ideals of a PID. This property called the ascending chain condition, says that any increasingchain of ideals in a PID must stop after a finite number of steps.

Theorem 8: Let R be a PID and I1,I

2,.. .. .. be an infinite sequence of ideals of R satisfying.

11 1

2 ....

Then 3 m N such that I, = Im+1

= Im+2

= ....

Proof: Consider the set I = I, I2 ... = n

m 1

I

. We will prove that I is an ideal of R.

Firstly, I , since I1 and I

1 I.

Secondly, if a,b I, then a I, and b Is for some r,s N.

Assume r s. Then Is I,. Therefore, a, b I,. Since I, is an ideal of R, a�b I, I. Thus,

a�b I a, b I.

Finally, let x R and a I. Then a I, for some r N.

xa I, I. Thus, whenever x R and a I, xa I.

Thus, I is an ideal of R. Since R is a PID, I = <a> for some a R. Since a I, a I, for some m N.

Then I I,. But I, I. So we. see that I = Im

.

Now, I, Im+1

mTherefore, I, = I

m+1

Similarly, I, = Im+2

and so on. Thus, Im = Im+1

= Im+2

= ...

Now, for a moment let us go back, where we discussed prime ideals. Over there we said that anelement p R is prime iff is a prime ideal of R. If R is a PID, we shall use Theorem 7 to makea stronger statement.

Theorem 9: Let R be a PID. An ideal < a > is a maximal ideal of R iff a is a prime element of R.

Proof: If < a > is a maximal ideal of R, then it is a prime ideal of R. Therefore, a is a prime elementof R.

Conversely, let a be prime and let I be an ideal of R such that < a > I. Since R is a PID, I = for some b R. We will show that b is a unit in R.

 = R, i.e., I = R.

Now, < a > a = bc for some c R. Since a is irreducible, either b is an associate of a orb is a unit in R. But if b is an associate of a, then = <a>, a contradiction. Therefore, b is a unitin R. Therefore, I = R.



NotesThus, <a> is a maximal ideal of R.

What Theorem 9 says is that the prime ideals and maximal Ideals coincide in a PID.

Now, take any integer n. Then we can have n = 0, or n = ± 1, or n has a prime factor. This propertyof integers is true for the elements of any PID, as you will see now.

Theorem 10: Let R be a PID and a be a non-zero non-invertible element of R. Then there is someprime element p in R such that a.

Proof: If a is prime, take p = a. Otherwise, we can write a =albl, where neither a, nor b1 is an

associate of a. Then < a > < a1 >. If a

1 is prime, take p = a

1. Otherwise, we can write a

1 = a

2b

2,

where neither a2 nor b

2 is an associate of a,. Then <a

1> < a

2 >. Continuing in this way we get

an increasing chain

<a> <a1> <a

2> ...

By Theorem 8, this chain stops with some < a, >. Then a, will be prime, since it doesn�t have anynon-trivial factors. Take p = a,, and the theorem is proved.

And now we are in a position to prove that any non-zero non-invertible element of a PID can beuniquely written as a finite product of prime elements (i.e., irreducible elements).

Theorem 11: Let Rt be a PID. Let a R such that a 0 and a is not a unit. Then a = p1, p

2....p

r, where

p1,p

2.... p

r, are prime elements of R.

Proof: If a is a prime element, there is nothing to prove. If not, then P1 | a, for some prime p

1 in

R, by Theorem 10. Let a = pla

l. If p

1a

1. If a

1 is a prime, we are through. Otherwise P

2 | a, for some

prime p2 in R. Let a

1, = p

2a

2. Then a = p

1p

2a

2. If a

2 is a prime, we are through. Otherwise we

continue the process. Note that since al is a non-trivial factor of a, <a> <a1>. Similarly, <a

1>

< a2 >. So, as the process continues we get an increasing chain of ideals,

<a> <a1> <a

2> ...

in the PID R. Just as in the proof of Theorem 10, this chain ends at < a, > for some m N, and a,is irreducible.

Hence, the process stops after m steps, i.e., we can write a = p1p

2 ... p

m, where p

i is a prime element

of R i = 1, .... m.

Thus, any non-zero non-invertible element in a PID can be factorised into a product of primes.What is interesting about this factorisation is the following result that you have already provedfor Z in Unit 1.

Theorem 12: Let R be a PID and a 0 be non-invertible in R. Let a = p1p

2....p

n = q

1q

2....q

m, where

pi and q

j are prime elements of R. Then n = m and each p

i is an associate of some q

j for 1 i | n,

1 j | m.

Before going into the proof of this result, we ask you to prove a property of prime elements thatyou will need in the proof.

Task Use induction on n to prove that if p is a prime element in an integral domainR and if p | a

1a

2 ... a, (where a

l, a

2,.. .., a, R), then p | a

i, for some i = 1, 2. .... n.

Now let us start the proof of Theorem 12.


Abstract Algebra

Notes Proof: Since p1p

2, ...p, = q

1q

2 ... ,.q

m, p

1 | q

1q

2. ... q

m,.

Thus, p1 | q

j for some j = 1. .... ..,m. By changing the order of the q

i, if necessary, we can assume

that j = 1, i.e., p1 | q. Let q

1 = p

lu

l. Since q

1 is irreducible, u

1 must be a unit in R. So p

1 and q

1 are

associates. Now we have

p1p

2 = P

n (p

1u

1)q

2....q

m.

Cancelling p1 from both sides, we get

p2p

3...p

n = u

1q

2...,q

m.

Now, if m > n, we can apply the same process to p2, p

3, and so on.

Then we will get

1 = u1u

2 .... u

n q

n+1 .... q

m.

This shows that qn+1

is a unit. But this contradicts the fact that qn+1

is irreducible.

Thus, m n.

Interchanging the roles of the ps and qs and by using a similar argument, we get n m.

Thus, n = m.

During the proof we have also shown that each pi is an associate of some qj, and vice versa.

What Theorem 12 says is that any two prime factorisations of an element in a PID are identical,apart from the order in which the factors appear and apart from replacement of the factors bytheir associates.

Thus, Theorems 11 and 12 say that every non-zero element in a PID R, which is not a unit, can beexpressed uniquely (up to associates) as a product of a finite number of prime elements.

For example, x2 � 1 R[x] can be written as (x-1)(x+1) or (x-1) (x-1) or [2(x-tl)] [2(x-1)] in R[x].

The property that we have shown for a PID in Theorems 11 and 12 is true for several otherdomains also. Let us discuss such rings now.

Self Assessment

1. An integral domains R a .................. if every ideal in R is a principle ideal.

(a) principle ideal domain (b) unique ideal domain

(c) special ideal domain (d) range ideal domain

2. g.c.d. represent ..................

(a) greatest common divisor (b) greatest common dividend

(c) greatest common domain (d) greatest commutated domain

3. Let R be .................. and a, b R, if a g.c.d. of a and b exists, then it is unique up to unit.

(a) domain and range (b) integral domain

(c) UID (d) SID

4. PID stands for ..................

(a) principal integral domain (b) pair ideal domain

(c) principle ideal domain (d) principle ideal divisor



Notes5. Any irreducible polynomials R[x] has degree 1 or define ..................

(a) n (b) 2

(c) 4 (d) 5

20.3 Summary

The definition and examples of a Euclidean domain.

Z any field and any polynomial ring over a field are Euclidean domains.

Units, associates, factors, the g.c.d. of two elements, prime elements and irreducible elementsin an integral domain.

The definition and examples of a principal ideal domain (PID).

Every Euclidean domain is a PID, but the converse is not true.

Thus, Z, F and F[x] are PIDs, for any field F.

The g.c.d. of any two elements a and b in a PID R exists and is of the form ax + by for somex, y R.

The Fundamental Theorem of Algebra: Any non-constant polynomial over C has all itsroots in C.

20.4 Keywords

Euclidean Domain: An integral domain D is called a Euclidean domain if for each non-zeroelement x in D there is assigned a non-negative integer (x) such that

(i) (ab) (b) for all non-zero a,b in D, and

(ii) for any non-zero elements a,b in D there exist q,r in D such that a = bq + r, where eitherr = 0 or (r) < (b).

UID: Let R be a commutative ring with identity. A non-zero element p of R is said to beirreducible if

(i) p is not a unit of R, and

(ii) if p = ab for a,b in R, then either a or b is a unit of R.

Any principal ideal domain is a unique factorization domain.


1. Show that a subring of a PID need not be PID.

2. Will any quotient ring of a PID be a PID? Why? Remember that a PID must be an integraldomain.

3. Let R be an integral domain. Show that

(a) u is a unit in R iff u | 1.

(b) for a, b R, a | b and b | a iff a and b are associates in R.


Abstract Algebra

Notes 4. Find the g.c.d. of

(a) 2 and 6 in Z/ 8 , (b) x2 + 8x + 15 and x2 + 12x + 35 in Z[x],

(c) x3 � 2x2 + 6x � 5 and x2 � 2x + 1 in Q[x].

5. Let R be a PID and a, b, c R such that a (bc. Show that if (a, b) = 1, then a | c.

(Hint: By Theorem 4, x, y R such that ax + b = 1.)


1. (a) 2. (a) 3. (b) 4. (c) 5. (b)






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Unit 21: Unique Factorization Domains

NotesUnit 21: Unique Factorization Domains

CONTENTS

Objectives

Introduction

21.1 Unique Factorisation Domain (UFD)

21.2 Summary

21.3 Keyword



Objectives


Discuss unique factorization domains

Explain theorems of UID

Introduction

In this unit, we shall look at special kinds of integral domains. These domains were mainlystudied with a view to develop number theory. Let us say a few introductory sentences aboutthem.

You saw that the division algorithm holds for F[x], where F is a field. You saw that it holdsfor Z. Such integral domains are called Euclidean domains.

We shall look at some domains which are algebraically very similar to Z. These are the principalideal domains, so called because every ideal in them is principal.

Finally, we shall discuss domains in which every non-zero non-invertible element can be uniquelyfactorised in a particular way. Such domains are very appropriately called unique factorisationdomains. While discussing them, we shall introduce you to irreducible elements of a domain.

While going through the unit, you will also see the relationship between Euclidean domains,principal ideal domains and unique factorisation domains.

21.1 Unique Factorisation Domain (UFD)

Here we shall look at some details of a class of domains that include PDs.

Definition: We call an integral domain R a unique factorisation domain (UFD, in short) if everynon-zero element of R which is not a unit in R can be uniquely expressed as a product of a finitenumber of irreducible elements of R.


Abstract Algebra

Notes Thus, if R is a UFD and a R, with a 0 and a being non-invertible, then

(i) a can be written as a product of a finite number of irreducible elements, and

(ii) if a = p1p

2.. ..p

n = q

1q

2 ..... q

m, be two factorisations into irreducibles, then n = m and each

pi is an associate of some q

j, where 1 | i | n, 1 j m.

Thus, F[x] is a UFD for any field F.

Also, since any Euclidean domain is a PID, it is also a UFD. You directly proved that Z is a UFD.Why don�t you go through that proof and then try and solve the exercises.

Now we give you an example of a domain which is not a UFD (and hence, neither a PID nor aEuclidean domain).

ED implies PID implies UFD

Theorem 1: Every Euclidean domain is a principal ideal domain.

Proof: For any ideal l, take a nonzero element of minimal norm b. Then l must be generated byb, because for any a l we have a = bq + r for some q, r with N(r) < N(b), and we must haver = 0 otherwise r would be a nonzero element of smaller norm than b, which is a contradiction.

Fact: If R is a UFD then R[x] is also a UFD.

Theorem 2: Every principal ideal domain is a unique factorization domain.

Proof: We show it is impossible to find an infinite sequence a1, a

2,... such that a

1 is divisible by a

i+1

but is not an associate. Once done we can iteratively factor an element as we are guaranteed thisprocess terminates.

Suppose such a sequence exists. Then the ai generate the sequence of distinct principal ideals (a

1)

(a2) ... The union of these ideals is some principal ideal (a). So a (a

n) for some n, which

implies (ai) = (a

n) for all i n, a contradiction.

Uniqueness: Each irreducible p generates a maximal ideal (p) because if (p) (a) R then p = abfor some b R implying that a or b is a unit, thus (a) = (p) or (a) = R. Thus R/(p) is a field. Nextsuppose a member of R has two factorizations

p1 ... p

r = q

1 ... q

s

Consider the ideals (pi) . (q

i). Relabel so that p

1 generates a minimal ideal amongst these

(in other words, (p1) does not strictly contain another one of the ideals). Now we show (p

1) = (q

i)

for some i. Suppose not. Then (p1) does not contain any q

i, thus q

i is nonzero modulo (p

1) for

all i, which is a contradiction because the left-hand side of the above equation is zero modulo(p

1).

Relabel so that (p1) = (q

1). Then p

1 = uq

1 for some unit u. Cancelling gives up

2 ... p

r = q

2 ... q

z. The

element up2 is also irreducible, so by induction we have that factorization is unique.

The converse of the above theorem is not always true. Consider the ring [x]. The ideal (2, x) isnot principal: suppose (2, x) = (a) for some a. Since this ideal contains the even integers, a mustbe some integer (multiplication never reduces the degree of an element), and in fact it must be(an associate of) 2. But (2) does not contain polynomials with odd coefficients, so (2, x) = (2).

Example: Show that Z[ 5] {a b 5|a,b Z} is not a UFD.

Solution: Let us define a function

f : Z [ 5 ] N U {0} by f(a+b 5 ) = a2+ 5b2.



NotesThis function is the norm function, and is usually denoted by N.

You can check that this function has the property that

f() = f() f() , , Z [ 5 ].

Now, 9 has two factorisations in Z[ 5], namely,

9 = 3.3 = ( 2 + �5 ) ( 2 � �5 ).

You have already shown that the only units of Z [ 5 ] are 1 and �1. Thus, no two of 3, 2+ 5

and 2 � 5 are associates of each other.

Also, each of them is irreducible. For suppose any one of them, say 2 + 5 , is reducible. Then

2+ 5 = for some non-invertible a, Z[ 5 ].

Applying the function f we see that

f ( 2 + �5 ) = f() f(),

i.e., 9 = f() f().

Since f(), f() N and a, are not units, the only possibilities are f() = 3 = f().

So, if a = a + b 5 , then a2 + 5b2 = 3.

But, if b 0, then a2 + 5b2 5; and if b = 0, then a2 = 3 is not possible in Z. So we reach a

contradiction. Therefore, our assumption that 2 + 5 is reducible is wrong. That is, 2 + 5 isirreducible.

Similarly, we can show that 3 and 2� 5 are irreducible. Thus, the factorisation of 9 as a product

of irreducible elements is not unique. Therefore, Z[ 5 ] is not a UFD.

From this example you can also see that an irreducible element need not be a prime element.

For example, 2 + 5 is irreducible and 2+ 5 |3.3, but 2+ 5 | 3 . Thus, 2 + 5 is not aprime element.

Now let us discuss some properties of a UFD. The first property says that any two elements of aUFD have a g.c.d. and their g.c.d. is the product of all their common factors. Here we will use thefact that any element a in a UFD R can be written as

1 2 nr r r1 2 na p p ...p

where the pis are distinct irreducible elements of R. For example, in Z[x] we have

x3 � x2 � x + 1 = (x � 1) (x + l) (x � 1 ) = (x � 1)2 (x + 1).

So, let us prove the following result.

Theorem 3: Any two elements of a UFD have a g.c.d.

Proof: Let R be a UFD and a.b R.

Let 1 2 n 1 2 nr r r s s s1 2 n 1 2 na p p ...p and b p p ...p


Abstract Algebra

Notes where p1, p

2, ..., p

n are distinct irreducible elements of R and r

i and s

i are non-negative integers

i = 12, ..., n.

(If some pi does not occur in the factorisation of a, then the corresponding r

i = 0. Similarly, if

some pi is not a factor of b, then the corresponding s

i = 0. For example, take 20 and 15 in Z. Then

20 = 22 × 30 × 5 and 15 = 20 × 31 × 51.)

Now, let t, = min (ri, s

i) i = 1, 2 ,....,n .

Then 1 2 nt t t1 2 nd p p ...p divides a as well as b, since t

i r

i and t

i s

i i = 1, 2, ...., n.

Now, let c | a and c | b. Then every irreducible factor of c must be an irreducible factor of a andof b, because of the unique factorisation property.

Thus, 1 2 nm m m1 2 nc p p ...p , where m

i r

i and m

i s

i i = 1,2, ...,n . Thus, m

i t

i i = 1,2 ,..., n.

Therefore, c | d.

Hence, d = (a, b).

Over there we found a non-UFD in which an irreducible element need not be a prime element.The following result says that this distinction between irreducible and prime elements can onlyoccur in a domain that is not a UFD.

Theorem 4: Let R be a UFD. An element of R is prime iff it is irreducible.

Proof: We know that every prime in R is irreducible. So let us prove the converse.

Let a R be irreducible and let a | bc, where b, c R.

Consider (a,b). Since a is irreducible, (a,b) = 1 or (a,b) = a.

If (a,b) = a, a| b.

If (a,b) = I, then d| b . Let bc = ad, where d R.

Let 1 2 m 1 2 nr r r s s s1 2 m 1 2 nb p p ...p , and c q q ...q , be irreducible factorisations of b and c. Since bc = ad

and a is irreducible, a must be one of the pis or one of the qjs. Since a | b, a p

i for any i.

Therefore, a = qj for some j. That is, a | c.

Thus, if (a,b) = 1 then alc.

So, we have shown that a/ bc a | b or a | c.

Hence, a is prime.

Theorem 5: Let R be a UFD. Then R[x] is a UFD.

We will not prove this result here, even though it is very useful to mathematicians. But let usapply it. You can use it to solve the following exercises.

Lemma: Let D be a unique factorization domain, and let p be an irreducible element of D. If a,bare in D and p|ab, then p|a or p|b.

Definition: Let D be a unique factorization domain. A non-constant polynomial

f(x) = an xn + a

n-1 xn-1 + · · · + a

1 x + a

0 in D[x]

is called primitive if there is no irreducible element p in D such that p | ai for all i.



NotesLemma: The product of two primitive polynomials is primitive.

Lemma: Let Q be the quotient field of D, and let f(x) be a polynomial in Q[x]. Then f(x) can bewritten in the form f(x) = (a/b)f*(x), where f*(x) is a primitive element of D[x], a,b are in D, anda and b have no common irreducible divisors. This expression is unique, up to units of D.

Lemma: Let D be a unique factorization domain, let Q be the quotient field of D, and let f(x) bea primitive polynomial in D[x]. Then f(x) is irreducible in D[x] if and only if f(x) is irreducible inQ[x].

Theorem 6: If D is a unique factorization domain, then so is the ring D[x] of polynomials withcoefficients in D.

Corollary: For any field F, the ring of polynomials

F[x1 , x

2 , ... , x

n]

in n indeterminates is a unique factorization domain.

For example, the ring Z [ 5 ] is not a unique factorization domain.

Self Assessment

1. If R is a UFD and a R, with a 0 and being a .................., then a can be written as a productof finite number of irreducible elements.

(a) invertible (b) non-invertible

(c) external (d) infinite

2. Any euclidean domains is a PID, it is also a ..................

(a) integral domain (b) UFD

(c) SFD (d) Ideal

3. Let R be a UFD. Then R(x) is a ..................

(a) UFD (b) SFD

(c) PID (d) Special range domain

4. In a UFD an element is prime iff it is ..................

(a) reducible (b) finite

(c) irreducible (d) infinite

5. Any two elements in a .................. have g.c.d.

(a) SFD (b) PID

(c) UFD (d) Domain

21.2 Summary

In a PID every prime ideal is a maximal ideal.

The definition and examples of a unique factorisation domain (UFD).

Every PID is a UFD, but the converse is not true. Thus, Z, F� and F[x] are UFDs, for anyfield F.


Abstract Algebra

Notes In a UFD (and hence, in a PID) an element is prime iff it is irreducible.

Any two elements in a UFD have a g.c.d.

If R is a UFD. then so is K[x].

21.3 Keyword

Unique Factorisation Domain: We call an integral domain R a unique factorisation domain(UFD, in short) if every non-zero element of R which is not a unit in R can be uniquely expressedas a product of a finite number of irreducible elements of R.


1. Directly prove that F[x] is a UFD, for any field F.

(Hint: Suppose you want to factorise f(x). Then use induction on deg f(x)).

2. Give two different prime factorisations of 10 in Z.

3. Give two different factorisations of 6 as a product of irreducible elements in Z[ 5].

4. Give an example of a UFD which is not a PID.

5. If p is an irreducible element of a UFD R, then is it irreducible in every quotient ring of R?

6. Is the quotient ring of a UFD a UFD? Why?

7. Is a subring of a UFD a UFD? Why?


1. (b) 2. (b) 3. (a) 4. (c) 5. (c)






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Unit 22: Polynomial Rings

NotesUnit 22: Polynomial Rings

CONTENTS

Objectives

Introduction

22.1 Ring of Polynomials

22.2 Some Properties of R[x]

22.3 Summary

22.4 Keywords



Objectives


Identify polynomials over a given ring

Prove and use the fact that R [x], the set of polynomials over a ring R, is a ring

Relate certain properties of R[x] to those of R

Introduction

In the earlier units, you must have come across expressions of the form x+1, x2+2x+1, and so on.These are examples of polynomial. You have also dealt with polynomials in the course of LinearAlgebra. In this unit, we will discuss sets whose elements are polynomials of the type a

0 + a, x +

... + anxn, where a

0, a

1,......, a,, are elements of a ring R. You will see that this set, denoted by R [x],

is a ring also.

You may wonder why we are talking of polynomial rings in a block on domains and fields. Thereason for this is that we want to focus on a particular case, namely, R [x], where R is a domain.This will turn out to be a domain also, with a lot of useful properties. In particular, the ring ofpolynomials over a field satisfies a division algorithm, which is similar to the one satisfied byZ. We will prove this property and use it to show how many roots any polynomial over a fieldcan have.

22.1 Ring of Polynomials

As we have said above, you may already be familiar with expressions of the type 1 + x, 2 + 3x +4x2, x5-1, and so on. These are examples of polynomials over the ring Z. Do these examplessuggest to you what a polynomial over any ring R is ? Let�s hope that your definition agreeswith the following one.

Definition: A polynomial over a ring R in the indeterminate x is an expression of the form

a0x0 + a

1x1 + a

2x2 + ... + a

nxn,


Abstract Algebra

Notes where n is a non-negative integer and a0, a,, ..., a

n R.

While discussing polynomials we will observe the following conventions. We will

(i) write x0 as 1, so that we will write a0 for a

0x0,

(ii) write x1 as x,

(iii) write xm instead of 1 .xm (i.e., when am

= l),

(iv) omit terms of the type O.xm.

Thus, the polynomial 2 + 3x2 � 1.x3 is 2x0 + 0.x1 + 3x2 + (�1)x3.

Henceforth, whenever we use the word polynomial, we will mean a polynomial in the

indeterminate x. We will also be using the shorter notation n

ii

i 0

a x

for the polynomial

a0 + a

1x+ ... + a

nxn.

Let us consider a few mox basic definitions related to a polynomial.

Definition: Let an + a, x + ... + a, xn be a polynomial over a ring R. Each of a

0 ,a

l, . . ., a, is a coefficient

of this polynomial. If a, 0, we call a, the leading coefficient of this polynomial.

If a1 = 0 = a

2 = ... = a

n , we get the constant polynomial, a

0. Thus, every element of R is a constant

polynomial.

In particular, the constant polynomial 0 is the zero polynomial.

It has no leading coefficient.

Now, there is a natural way of associating a non-negative integer with any non-zero polynomial.

Definition: Let a,, + a, x + . . . + a,, xn be a polynomial over a ring R, where a, 0. Then we call theinteger n the degree of this polynomial, and we write

ni

ii 0

deg a x n,of a, , 0.

We define the degree of the zero polynomial to be � . Thus, deg 0 = �.

Let us consider some examples.

(i) 3x2 + 4x + 5 is a polynomial of degree 2, whose coefficients belong to the ring of integersZ. Its leading coefficient is 3.

(ii) x2 + 2x4 + 6x + 8 is a polynomial of degree 4, with coefficients in Z and leading coefficient2. (Note that this polynomial can be rewritten as 8 + 6x + x2 + 2x4.)

(iii) Let R be a ring and r R, r 0. Then r is a polynomial of degree 0, with leading coefficientr.

Before giving more examples we would like to set up some notation.

Notation: We will denote the set of all polynomials over a ring R by R[x]. (Please note the use ofthe square brackets [ ]. Do not use any other kind of brackets because R [x] and R (x) denotedifferent sets.)

Thus, R[x] = n

i1 i

i 0

a x a R i 0, 1,...n, where n 0, n Z .

We will also often denote a polynomial a0 + a

1 x + . . . + a

n xn by f(x), p (x), q(x), etc.



NotesThus, an example of an element from Z

4 [x] is f(x) = 2 x2 + 3 x + i.

Here deg f(x) = 2, and the leading coefficient of f(x) is 2 .

Now, for ring R, we would like to see if you can define operations on the set R [x] so that itbecomes a ring. For this purpose we define the operations of addition and multiplication ofpolynomials.

Definition: Let f(x) = a,, + a1x + .. + a, xn and g (x) = b0 + b, x + .. + b

mxm be two polynomials in

R[X]. Let us assume that m 2 n. Then their sum f(x) + g(x) is given by

f(x) + g(x) = (a,, + b0) + (a, + b

1)x + .. + (a

n+ b,) x

n + b

n+1 xn+l .. + b

mxm.

For example, consider the two polynomials p(x), q(x).in Z[x] given by

p(x) = 1 + 2x + 3x2, q(x) = 4 + 5x + 7x3

Then

p(x) + q(x) = (1+4) + (2+5)x + (3+0) x2 + 7x3 = 5 + 7x + 3x2 + 7x3.

Note that p (x) + q (x) Z [X] and that

From the definition given above, it seems that deg (f(x)+g(x)) = max (deg f (x), deg g (x)). But thisis not always the case. For example, consider p(x) = 1 + x2 and q (x) = 2 + 3x � x2 in Z [X].

Then p(x) + q(x) = (1+2) + (0+3)x + (1-1)x2 = 3 a 3x.

Here deg (p(x) + q (x)) = 1 < max (deg p(x), deg q(x)).

So, what we can say is that

deg (f(x) + g(x)) max (deg f(x), deg g(x))

f(x), g(x) R[x].

Now let us define the product of polynomials.

Definition: If f(x) = a,, + a1x + .. + a, xn and g(x) = b

0 + b, x + .. + b

mxm are two polynomials in R [x],

we define their product f(x). g(x) by

f(x) . g(x) = c0 a c

1x +.. + c

m+nxm+n

where c1 = a

1b

0, + a

i-1 b

1 + .... a

0b

i i = 0,l ,... ; m + n.

Note that ai = 0 for i > n and b

i = 0 for i > m,

As an illustration, let us multiply the following polynomials in Z[x] :

p(x) = 1 � x + 2x3, q(x) = 2 + 5x + 7x2.

Here a, = 1, a, = �1, a2 = 0, a

3 = 2, b

0 = 2, b, = 5, b

2 = 7.

Thus, p(x) q(x) = 5

ii

i=0

c x , where

c0 = a

0b

0 = 2,

c1 = a

1b

0 + a

0b

1 = 3,

c2 = a

2b

0 + a

l b

1 + a

0b

2 = 2,

c3 = a

3b

0 + a

2b

1 + a

1b

2 + a

0b

3 = �3 (since b

3 = 0).


Abstract Algebra

Notes c4 = a

4b

0 + a

3b

1 + a

2b

2 +a

1b

3 + a

0b

4 = 10 (since a

4 = 0 = b

4).

c5 = a

5b

0 +a

4b

1 + a

3b

2 + a

2b

3 +a

1b

4 + a

0b, = 14 (since a

5 = 0 = b

5).

So p(x). q(x) = 2 + 3x +2x2 - 3x3 + 10x4 + 14x5.

Note that p(x). q(x) Z[X], and deg (p(x) q(x)) = 5 = deg p (x) + deg q (x).

As another example, consider

2 6p(x) 1 2x,q(x) 2 3x Z [x].

Then, p(x). q(x) = 2 3 22 + 4x + 3x + 6x = 2 + 4x + 3x .

Here, deg (p(x). q(x)) = 2 < deg p (x) + deg q (x) (since deg p (x) = 1, deg q (x) = 2).

In the next section we will show you that

deg (f(x) g(x)) deg f(x) + deg g(x)

By now you must have got used to addition and multiplication of polynomials. We would liketo prove that for any ring R, R[x] is a ring with respect to these operations. For this we must notethat by definition, + and . are binary operations over R [x].

Now let us prove the following theorem. It is true for any ring, commutative or not,

Theorem 1: If R is a ring, then so is R[x], where x is an indeterminate.

Proof: We need to establish the axioms R1 � R6 of Unit 14 for (R[x], + , .).

(i) Addition is Commutative: We need to show that

p(x) + q(x) = q(x) + p(x) for any p(x) , q(x) R [x].

Let p (x) = a0 + a

1x + ... + a,xn, and

q(x) = b0 + b

1x + ... + b

mxm be in R[x].

Then, p (x) + q(x) = c0 + c

1x + ... + c

1xt,

where ci = a

i + b

i and t = max (m,n).

Similarly,

q(x) + p(x) = d0 + d

1x + ... + d

sxs,

where di = b

i + a

i, s = max (n, m) = t.

Since addition is commutative in R, c, =di i 0.

So we have

p(x) + q(x) = q(x) + p(x).

(ii) Addition is Associative: Again, by using the associativity of addition in R, we can showthat if p(x), q(x), s(x) R[x], then

{p(x) + q (x)} + s(x) = p(x) + {q(x) + s(x)}.

(iii) Additive Identity: The zero polynomial is the additive identity in R [x]. This is because,for any p(x) = a

0 + a, x+ ... + a

nxn

R[x],

0 + p(x) = (0 + a,) + (0 +a1)x + ... +(0 + a

n)xn

= a0 + a

l x + ... +a

nxn

= p(x)



Notes(iv) Additive Inverse: For p (x) = a, + a1x +... + a

nxn R[x], consider the polynomial �p(x) =

�a, �a1x � ... �a

nxn, � a

i being the additive inverse of ai in R. Then

p(x) + (�p(x)) = (a,, �a,,) + (a1 � a

1) x + ... + (a

1 � a

n)xn

= 0 + 0.x + 0.x2+ ... + 0.xn

= 0.

Therefore, � p(x) is the additive inverse of p(x).

(v) Multiplication is Associative:

Let p(x) =a, +a1x + ... +a

nxn,

q(x) = b0 + b

1x+ ... +b

mxm,

and t (x) = d0 +d

1x + ... + d

rxr, be in R [x]

Then

p(x) . q(x) = c0 + c

1 x + .. . + c

sxs

, where s = m + n and

ck = akb

0 + a

k�1b

1 + ... + a

0b

k k = 0,l, ...,s .

Therefore,

{p(x) . q(x)} t (x) = e0 + e

1 x + ... +e

1xt,

where t = s + r = m+n+r and

ek = c

kd

0 + c

k-1d

1 + ... + c

0d

k

= (akb

0 + ... + a

0b

k)d

0 + (a

k-1,b

0 + ... + a

0b

k-1) d

1 + ... + a

0b

0d

k.

Similarly, we can show that the coefficient of xk (for any k 0) in p(x) (q (x) t(x))

is akb

0d

0 + a

k-1, (b

1d

0 + b

0d

1) + ... + a

0(b

kd

0 + b

k-1, d

1 + ... + b

0d

k)

= ek, by using the properties of + and . in R.

Hence, {p(x).q(x)} . t(x) = p(x) . {q (x). t (x)}

(vi) Multiplication Distributes over Addition:

Let p(x) = a0 +a

1x + ... + a

nx

n,

q(x) = b0 + b

lx + ...+ b

mxm

and t(x) = d0 + d, x + . . . + d

r xr be in R[x],

The coefficient of xk in p (x). (q(x) + t (x)) is

ck = a

k (b

0 + d

0) + a(b

1 + d

1) + (b

1 + d

1) + ... + a, (b

k + d

k).

And the coefficient of xk in p (x) q (x) + p (x) t(x) is

(akb

0 + a

k-1b

1 + ... + a

0b

k) + (a

kd

0 + a

k-1d

1 + ... +a

0d

k),

= ak(b

0 + d

0) +a

k-1 (b + d

1) + ... +a

0 (b

k + d

k) = c

k

This is true k 0.

Hence, p (x) . (q(x) + t (x) ] = p (x) . q(x) + p (x). t.(x).

Similarly, we can prove that

{q(x)+t(x).p(x)=q(x).p(x)+t(x).p(x)

Thus, R [x] is a ring.


Abstract Algebra

Notes Note that the definitions and theorem in this section are true for any ring. We have not restrictedourselves to commutative rings. But, the case that we are really interested in is when R is adomain. In the next section we will progress towards this case.

22.2 Some Properties of R[x]

In the previous section you must have realised the intimate relationship between the operationson a ring R and the operations on R [x]. The next theorem reinforces this fact.

Theorem 2: Let R be a ring.

(a) If R is commutative, so is R [x].

(b) If R has identity, so does R [x].

Proof: (a) Let p (x) = a0 + a, x + .. . + a,xn and

q(x) = b0 + b

1x + ... + b

mx

m be in R[x].

Then (x) . q(x) = c0 + c

1 x + .. . + c

sxs, where s = m + n and

ck = a

kb

0 + a

k-1 b

1 +. . .+ a

0b

k

= bka

0 + b

k-1a

14 ... + b

1a

k-1 + b

0a

k, since both addition and multiplication are commutative in R.

= coefficient of xk in q (x) p(x).

Thus, for every i 0 the coefficients of x1 in p(x) q(x) and q(x) p(x) are equal

Hence, P(x) q(x) = q(x) p(x).

(b) We know that R has identity 1. We will prove that the constant polynomial 1 is the identityof R [X]. Take any

p(x) = a0 + a

1x + ... + a

nxn R[x].

Then 1.p(x) = c0 + c

1x + ... + c

nxn (since deg 1 = 0),

where ck = a

k , 1 + ak

-1 . 0 + a

k-2 . 0 + ... + a

0.0 = a

k

Thus, 1, p(x) = p (x).

Similarly, p(x). 1 = p(x).

This shows that 1 is the identity of R[x]. ,

In the following exercise we ask you to check if the converse of Theorem 2 is true.

Now let us explicitly state a result which will help in showing us that R is a domain iff R [x} is adomain, This result follows just from the definition of multiplication of polynomials.

Theorem 3: Let R be a ring and f (x) and g (x) be two non-zero elements of R [x]. Then deg (f(x) g(x)) deg f(x) + deg g (x), with equality if R is an integral domain.

Proof: Let f (x) = a0 + a

1x + ... + a

nxn, a

n 0,

and g (x) = b0 + b

1x + ... + b

mxm, b

m 0.

Then deg f (x) = n, deg g (x) = m. We know that

f (x). g (x) = c0 +c

1x + ... + c

m+n × m+n,

where ck = a

kb

0 + a, b

1 + ... + a

0b

k,.



NotesSince an+1

, an+2

, ... and bm+1

, bm+2

. .. . are all zero,

cm+n

= anb

m

Now, if R is without zero divisors, then anb

m 0, since a, 0

and b 0. Thus, in this case,

deg (f(x) g (x)) = deg f(x) + deg g (x).

On the other hand, if R has zero divisors, it can happen that a,b, = 0. In this case,

deg (f (x) g (x)) < m+n = deg f(x) + deg g(x).

Thus, our theorem is proved.

The following result follows immediately from Theorem 3.

Theorem 4: R [x] is an integral domain <=>. R is an integral domain.

Proof: From Theorem 2, we know that R is a commutative ring with identity iff R[x] is acommutative ring with identity. Thus, to prove this theorem we need to prove that. R is withoutzero divisors iff R [x] is without zero divisors.

So let us first assume that R is without zero divisors.

Let p(x) = a0 + a

lx+ ... + a

nx

n, and q(x) = b

0 + b

1 x +... +b

mx

m

be in R [x], where a, 0 and b, 0.

Then, in Theorem 3 we have seen that deg .(p (x) q (x)) = m + n 0.

Thus, p (x) q (x) 0

Thus, R [x] is without zero divisors.

Conversely, let us assume that R [x] is without zero divisors. Let a and b be non-zero elementsof R. Then they are non-zero elements of R [x] also. Therefore, ab 0. Thus, R is without zerodivisors. So, we have proved the theorem.

Now, you have seen that many properties of the ring R carry over to R�[x]. Thus, if F is a field, weshould expect F[x] to be a field also, But this is not so. F[x] can never be a field.

This is because any polynomial of positive degree in F|x| does not have a multiplicativeinverse. Let us see why.

Let f (x) F [x] and deg f (x) = n > 0. Suppose g (x) F [x] such that f (x) g (x) = 1. Then

0 = deg 1 = deg (f(x) g (x)) = deg f(x) + deg g (x), since F [x] is a domain.

= n + deg g (x) n > 0.

We reach a contradiction.

Thus, F [x] cannot be a field.

But there are several very interesting properties of F [x], which are similar to those of Z, the setof integers. In the next section we shall discuss the properties of division in F [x].

Self Assessment

1. A polynomial over a ring R in determinate X is an expression of the form .................

(a) a0x0 + a

1x1 + a

2x2 + ...... a

nxn (b) a0x

1 + a2x

2 + a3x

3 + ...... anx

n

(c) a-1x + a-1x2 + a-1x

3 ...... a-1x

n(d) a

0x-1 + a

1x-1 + a

2x-3 ...... a

nx-n


Abstract Algebra

Notes 2. The degree of the zero polynomial to be ................. thus degree 0 = .................

(a) � , � (b) ,

(c) , �1 (d) �1,

3. 3x2 + 4x + 5 is a polynomial of degree ................., whose coefficients belong to the ring ofintegers Z its leading coefficients is .................

(a) 4, 5 (b) 2, 3

(c) 2, 4 (d) 2, 5

4. x2 + 2x4 + 6x + 8 is a polynomial of degree ................. with coefficient 2.

(a) 4 (b) 5

(c) 6 (d) 8

5. Let R be a ring and r R, R ................. 0. Then r is polynomial degree of 0 with leadingcoefficient r.

(a) = (b)

(c) (d)

22.3 Summary

The definition and examples of polynomials over a ring.

The ring structure df R[x], where R is a ring.

R is a commutative ring with identity iff R[x] is a commutative ring with identity.

R is an integral domain iff R[x] is an integral domain.

22.4 Keywords

Polynomial: A polynomial over a ring R in the indeterminate x is an expression of the form

a0x0 + a

1x1 + a

2x2 + ... + a

nxn,

where n is a non-negative integer and a0, a,, ..., a

n R.

Coefficient of Polynomial: Let an + a, x + ... + a, xn be a polynomial over a ring R. Each of a

0 ,a

l, .

. ., a, is a coefficient of this polynomial. If a, 0, we call a, the leading coefficient of thispolynomial.


1. Identify the polynomials from the following expressions. Which of these are elements ofZ[x]?

(a) x6 + x5 + x4 + x2 + x + 1 (b) 22

2 1x x

x x

(c) 23x 2x 5 (d) 2 31 1 11 x x x

2 3 4

(e) x1/2 + 2x3/2 + 3x5/2 (f) �5



Notes2. Calculate

(a) (2 + 3x2 + 4x3) + (5x + x3) in Z[x]. (b) 2 37(6 2x ) (1 2x 5x ) in Z [x].

(c) (1 + x) (1 + 2x + x2) in Z[x]. (d) 23(1 x) (1 2x x ) in Z [x]

(e) (2 + x + x2) (5x + x3) in Z[x]

3. If R is a ring such that R[x] is commutative and has identity, then

(a) is R commutative?

(b) does R have identity?


1. (a) 2. (a) 3. (b) 4. (a) 5. (b)






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Abstract Algebra

Notes Unit 23: Division of Algorithm

CONTENTS

Objectives

Introduction

23.1 The Division Algorithm

23.2 Summary

23.3 Keywords



Objectives


Prove and use the division algorithm for F[X], where F is a field

Discuss examples related to algorithms

Introduction

In the last unit, you have studied about polynomials rings. In this unit, we will discuss thedivision of algorithm.

23.1 The Division Algorithm

We have discussed various properties of divisibility in Z. In particular, we proved the divisionalgorithm for integers. We will now do the same for polynomials over a field F.

Theorem 1 (Division Algorithm): Let F be a field. Let f(x) and g(x) be two polynomials in F[x],with g(x) 0. Then

(a) there exist two polynomials q(x) and r (x) in F [X] such that

f (x) = q (x) g (x) + r (x), where deg r(x) < deg g (x).

(b) the polynomials q (x) and r (x) are unique.

Proof: (a) If deg f (x) < deg g (x), we can choose q (x) = 0.

Then f(x) = 0. g(x) + f (x), where deg f(x) < deg g (x).

Now, let us assume that deg f(x) deg g (x).

Let f(x) = a0 + a

1x + . . . +a

nxn, a, 0, and

g(x) = b0 + b

1x + ... + b

mxm, b

m 0, with n m.


Unit 23: Division of Algorithm

NotesWe shall apply the principle of induction on deg f(x), i.e., n.

If n = 0, then rn = 0, since g(x) 0. Now

f(x) = a,,, g(x) = b,, and hence

f(x) = (a,, b0

-l) b0 + 0 = q (x) g (x) + r (x), where q(x) = a

0b

0-1 and r(x) = 0.

Thus,

f(x) = q(x) g(x) + r (x), where deg r(x) < deg g(x).

So the algorithm is true when n = 0. Let us assume that the algorithm is valid for all polynomialsof degree n � 1 and how to establish that it is true for f(x). Consider the polynomial

f1(x) = f(x) � a

nb

m-1xn-m g(x)

= (a,, + a, x +. . .+anxn) � (a

nb

m-1 b

0xn-m+a

nb

m-1 b

1xn-m+1 +...+ a

nb

m-1b

mxn)

Thus, the coefficient of xn in f, (x) is zero; and hence,

deg f, (x) n-1.

By the induction hypothesis, there exist q, (x) and r (x) in

F[x] such that f, (x) = q, (x) g(x) + r(x), where deg r(x) < deg g(x).

Substituting the value of f,(x), we get

f(x)-anb

m-1g(x) = q

1(x) g(x) + r(x),

i.e., f(x) = {anb

m-1 xn-m +q

1(x)} g(x) + (x)

= q(x) g(x)+r(x), where q(x) = anb

m-1 xn-m +q

1(x)

and deg r(x) < deg g(x).

Therefore, the algorithm is true for f(x), and hence for all polynomials in F[x].

(b) Now let us show that q(x) and r(x) are uniquely determined.

If possible, let

f(x) = q1(x) g(x) + r

1(x), where deg r

1(x) < deg g(x).

and

f(x) =q2(x) g(x)+r

2(x), where deg r

2(x) < deg g(x).

Then

q1(x) g(x)+r

1(x) = q

2(x) g(x)+r

2(x), so that

{q1(x) � q

2(x)} g(x) = r

2(x) � r

1(x) ...(1)

Now if q1(x) q

2(x), then deg {q

1(x) � q

2(x)} 0, so that

deg [{q1(x) � q

2(x) g(x)] deg g(x).

On the other hand, deg {r2(x) - r

1(x)} < deg g(x), since

deg r2(x) < deg g(x) and deg r

1(x) < deg g(x).

But this contradicts Equation (1). Hence. Equation (1) will remain valid only if

q1(x) � q

2(x) = 0. And then r

2(x) � r

1(x) = 0,

i.e., q1(x) = q

2(x) and r

1(x) = r

2(x).


Abstract Algebra

Notes Thus, we have proved the uniqueness of q(x) and r(x) in the expression f(x) = q(x) g(x)+r(x). Hereq(x) is called the quotient and r(x) is called the remainder obtained on dividing f(x) by g(x).

Now, what happens if we take g(x) of Theorem 1 to be a linear polynomial? We get the remaindertheorem. Before proving it let us set up some notation.

Notation: Let R be a ring and f(x) R[x]. Let

f(x) = a0 + a

lx + ... +a

nxn.

Then, for any r R, we define

that is, f(r) is the value of f(x) obtained by substituting r for x.

Thus, if f(x) = 1 + x + x2 Z[x], then

f(2) = 1 + 2 + 4 = 7 and f(0) = 1 + 0 + 0 = 1.

Let us now prove the remainder theorem, which is a corollary to the division algorithm.

Theorem 2 (Remainder Theorem): Let F be a field. If f(x) P[x] and b F, then there exists aunique polynomial q(x) F[x] such that f(x) = (i-b) q(x)+f(b).

Proof: Let g(x) = x-b. Then, applying the division algorithm to f(x) and g(x), we can find uniqueq(x) and r(x) in F[x], such that

f(x) = q(x)g(x) + r(x)

= q(x) (x � b) + r(x), where deg r(x) < deg g(x) = 1.

Since deg r(x) < 1, r (x) is an element of F, say a.

So, f(x) = (x - b)q(x) + a,

Substituting b for x, we get

f(b) = (b � b) q(b) + a

= 0.q(b) + a= a

Thus, a = f(b).

Therefore, f(x) = (x-b) q(x)+f(b).

Note that deg f(x) = deg(x-b)+deg q(x) = l+deg q(x).

Therefore, deg q(x) = deg f(x)-1.

Let us apply the division algorithm in a few situations now.

Example: Express x4 + x3 + 5x2 � x as

(x2 + x + 1) q(x) + r(x) in Q[x].

Solution: We will apply long division of polynomials to solve this problem.

2 4 3 2

4 3 2

2

2

x + x + 1) x + x + 5x x

x x x

4x x

4x 4x 4

5x 4


Unit 23: Division of Algorithm

NotesNow, since the degree of the remainder -5x- 4 is less than deg .(x2+x+1), we stop the process. Weget

x4 + x3 + 5x2 � x = (x2 + x + 1) (x2 + 4) � (5x + 4)

Here the quotient is x2 + 4 and the remainder is � (5x+4).

Now, let us see what happens when the remainder in the expression f = qg + r is zero.

Self Assessment

1. Let F be a field. Let f(x) and g(x) be two polynomials is f[x], with g(x) 0, then thepolynomial q(x) and r(x) an ...................

(a) unique (b) deficient


2. If deg f(x) < deg g(x) we can chosen q(x) = 0. Then f(x) = 0.g(x) + f(x) where degf(x) ...................deg g(x).

(a) < (b) >

(c) (d)

3. x4 + x3 + 5x2 � x is equal to ...................

(a) (x2 + x + 1) (q(x) + r(x) is Q[x])

(b) (x + x2 + 1) (q-1(x) + r-1(x) in Q[x])

(c) (x + x2 + 1)-1 (q(x) + r(x) in Q[x])

(d) q(x)-1 + q(x)2 + (x + x2 + 1) in Q[x]

4. ................... theorem said that let F be a field, if F[x] P[x] and b F, then there exists aunique polynomial q(x) F[x] such that f(x) = (i - b) q(x) + F(b)

(a) remainder theorem (b) division algorithm

(c) contradiction theorem (d) division matrix

23.2 Summary

The division algorithm in F[x], where F is a field, which states that if f(x), g(x) F(x),g(x) 0, then there exist unique q(x), r(x) F[x] with f(x) = q(x) g(x)+r(x) and deg r(x)< deg g(x).

a F is a root of f(x) F[x] iff (x�a) | f(x).

A non-zero polynomial of degree n over a field F can have at the most n roots.

23.3 Keywords

Division Algorithm: Let F be a field. Let f(x) and g(x) be two polynomials in F[x], with g(x) 0.

Remainder Theorem: Let F be a field. If f(x) P[x] and b F, then there exists a unique polynomialq(x) F[x] such that f(x) = (i-b) q(x)+f(b).


Abstract Algebra

Notes 23.4 Review Questions

1. Express f as gq + r, where deg r < degg, in each of the following cases.

(a) f = x4 + 1, g = x3 in Q[x]

(b) 3 23f = x + 2x � x + 1, g = x + 1 in Z [x].

(c) f = x3 � 1, g = x � 1 in R[x].

2. You know that if p, q Z, q 0, then pq

can be written as the sum of an integer and a

fraction * with | m | < | q |. What is the analogous property for elements of F[x]?


1. (a) 2. (a) 3. (a) 4. (a)






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Unit 24: Irreducibility and Field Extensions

NotesUnit 24: Irreducibility and Field Extensions

CONTENTS

Objectives

Introduction

24.1 Irreducibility in Q[x]

24.2 Field Extensions

24.2.1 Prime Fields

24.2.2 Finite Fields

24.3 Summary

24.4 Keywords



Objectives


Prove and use Eisenstein�s criterion for irreducibility in Z[x] and Q[x]

Obtain field extensions of a field F from F[x]

Obtain the prime field of any field

Use the fact that any finite field F has pn elements, where char F = p and dim zp F = n

Introduction

We have discussed various kinds of integral domains, including unique factorisation domains.Over there you saw that Z[x] and Q[x] are UFDs. Thus, the prime and irreducible elementscoincide in these rings. In this unit, we will give you a method for obtaining the prime(or irreducible) elements of Z[x] and Q[x]. This is the Eisenstein criterion, which can also be usedfor obtaining the irreducible elements of any polynomial ring over a UFD.

After this, we will introduce you to the field extensions and subfields. We will use irreduciblepolynomials for obtaining field extensions of a field F from F[x]. We will also show you thatevery field is a field extension of Q or Z, for some prime p. Because of this, we call Q and the Z

ps

prime fields. We will discuss these fields briefly.

Finally, we will look at finite fields. These fields were introduced by the young Frenchmathematician Evariste Galois while he was exploring number theory. We will discuss someproperties of finite fields which will show us how to classify them.

Before reading this unit ,we suggest that you go through the definitions of irreducibility.


Abstract Algebra

Notes 24.1 Irreducibility in Q[x]

We introduced you to irreducible polynomials in F[x], where F is a field. We also stated theFundamental Theorem of Algebra, which said that a polynomial over C is irreducible iff it islinear. You also learnt that if a polynomial over R is irreducible, it must have degree 1 ordegree 2. Thus, any polynomial over R of degree more than 2 is reducible. And, using thequadratic formula, we know which quadratic polynomials over R are irreducible.

Now let us look at polynomials over Q. Again, as for any field F, a linear polynomial over Q isirreducible. Also, by using the quadratic formula we can explicitly obtain the roots of anyquadratic polynomial over Q, and hence figure out whether it is irreducible or not. But, can youtell whether 2x7 + 3x5 � 6x4 + 3x3 + 12 is irreducible over Q. This criterion was discovered by thenineteenth century mathematician Ferdinand Eisenstein. In this section we will build up thetheory for proving this useful criterion.

Let us start with a definition.

Definition: Let f(x) = a, + a1x + . .. + a

nxn Z[x]. We define the content of f[x] to be the g.c.d. of the

integers a0, a

1,,..., a,.

We say that f(x) is primitive if the content of f(x) is 1.

For example, the content of 3x2 + 6x + 12 is the g.c.d. of 3, 6 and 12, i.e., 3. Thus, this polynomialis not primitive. But x5 + 3x2 + 4x � 5 is primitive, since the g.c.d of 1, 0, 0, 3, 4, �5 is 1.

We will now prove that the product of primitive polynomials is a primitive polynomial. Thisresult is well known as Gauss� lemma.

Theorem 1: Let f(x) and g(x) be primitive polynomials. Then so is f(x) g(x).

Proof: Let f(x) = a0 + a

1x + ... + a

nxn Z[x] and

g(x) = b0 + b

1x + ... + b

mxm Z[x], where the

g.c.d. of a0, a

1, ..., a, is 1 and the g.c.d. of b

0, b

1..., b

m is 1. Now

f(x) g(x) = c0 + c

1x + ... + c

m+nxm+n

where c, = a0b

k + a

1b

k-1 + ... + a

kb

0.

To prove the result we shall assume that it is false, and then reach a contradiction. So, supposethat f(x) g(x) is not primitive. Then the g.c.d. of c

0, c

1...., c

m+n is greater than 1, and hence some

prime p must divide it. Thus, p | ci i = 0, 1, ..., m+n. Since f(x) is primitive, p does not divide

some ai. Let r be the least integer such that p| a

r. Similarly, let s be the least integer such that

p| bs.

Now consider

cr+s

= a0b

r+s + a

Ib

r+s-1 + ... + a

rb

s + ... + a

r+s b

0

= arb

s + (a

0b

r+s + a

Ib

r+s-1 + ... + a

r-1 b

s+1 + a

r+1b

s-1 + ... + a

r+s b

0)

By our choice of r and s, p | a0, p | a

1, ..., p | a

r-1, and p | b

0, p | b

1, ..., p | b

s-1. Also p | c

r+s,

Therefore, p | cr+s

� (a0b

r+s +... + a

r-1 b

s+1 + a

r+1 b

s-1 + ... + a

r+s b

0)

i.e., p | ar b,.

p ( a, or p | bs, since p is a prime.



NotesBut p | a

r and p | b

s. So we reach a contradiction. Therefore, our supposition is false. That is, our

theorem is true.

Let us shift our attention to polynomials over Q now.

Consider any polynomial over Q, say f(x) = 3 23 1 1x + x + 3x + .

2 5 3 If we take the l.c.m of the

denominators, is., of 2, 5, 1 and 3, i.e., 30 and multiply f(x) by it, what do we get? We get

30f(x) = 45x3 + 6x2 + 90x + 10 Z[x]

Using the same process, we can multiply any f(x) Q[x] by a suitable integer d so that df(x), Z[X]. We will use this fact while relating irreducibility in Q[x] with irreducibility in Z[x].

Theorem 2: If f(x) Z[x] is irreducible in Z[x], then it is irreducible in Q[x].

Proof: Let us suppose that f(x) is not irreducible over Q[x]. Then we should reach a contradiction.So let f(x) = g(x) h(x) in Q[x], where neither g(x) nor h(x) is unit, i.e., deg g(x) > 0, deg h(x) > 0. Sinceg(x) Q[x]. m Z such that mg(x) Z[x]. Similarly, n Z such that nh(x) Z[x]. Then,

mnf(x) = mg(x) nh(x) ... (1)

Thus, (1) gives us

mnrf1(x) = stg

1(x)h

1(x) ...(2)

Since g1(x) and h

1(x) are primitive, Theorem 1 says that g

1(x) h

1(x) is primitive. Thus, the content

of the right hand side polynomial in (2) is st. But the content of the left hand side polynomial in(2) is mnr. Thus, (2) says that mnr = st.

Hence, using the cancellation law in (2), we get f,(x) = g,(x) h1(x).

Therefore, f(x) = rf1(x) = (rg

1(x)) h

1(x) in Z[x], where neither rp

1(x) nor h

1(x) is a unit. This

contradicts the fact that f(x) is irreducible in Z[x].

Thus, our supposition is false. Hence, f(x) must be irreducible in Q[x].

What this result says is that to check irreducibility of ii polynomial in Q[x], it is enough to checkit in Z[x]. And, for checking it in Z[x] we have the terrific Eisenstein�s criterion, that we mentionedat the beginning.

Theorem 3 (Eisenstein�s Criterion): Let f(x) = a0 + a

lx + ... + a,,xn Z[x]. Suppose that for some

prime number p;

(i) P | an,

(ii) p | a0, p | a

1, ..., p | a

n-1, and

(iii) p2 | a0.

Then f(x) is irreducible in Z[x] (and hence Q[x]).

Proof: Suppose f(x) is reducible in Z[x].

Let f(x) = g(x) h(x),

where g(x) = b0 + b

1x + ... + b,, xm, m > 0 and

h(x) = c0 + c

1 x + ... + c

rxr, r > 0.


Abstract Algebra

Notes Then n = deg f = deg g + deg h = m + r, and

ak = b

0 c

k + b

1 c

k-l ... + b

k c

0 k = 0, 1 ..., n.

Now a0 = b

0c

0. We know that p | a

0. Thus, p | b

0c

0, p | b

0 or p | c

0. Since p2 | a

0, p cannot divide

both b0 and c

0. Let us suppose that p | b

0 and p k CJ

Now let us look at a,, = b, c,. Since p | a, we see that p | bm

and p | cr. Thus, we see that for some

i, p | bi. Let k be the least integer such that p | b

k. Note that 0 < k m < n.

Therefore, p|ak.

Since p|ak and p|b

0, p | b

1, ..., p | b

k�1, we see that p(a

k � (b

0c

k + .... + b

k�1c

1), i. e.,

p (bkc

0. But p | b

k and p | c

0. So we reach a contradiction.

r Thus, f(x) must be irreducible in Z[x].

Let us illustrate the use of this criterion.

Example: Is 2x7 + 3x5 � 6x4 + 3x3 + 12 irreducible in Q[x]?

Solution: By looking at the coefficients we see that the prime number 3 satisfies the conditionsgiven in Eisenstein�s criterion. Therefore, the given polynomial is irreducible in Q[x].

Example: Let p be a prime number. Is Q[x]/<x3 � p > a field?

Solution: You know that for any field F, if f(x) is irreducible in F[x], then <f(x)> is a maximalideal of F[x].

Now, by Eisenstein�s criterion, x3-p is irreducible since p satisfies the conditions given inTheorem 3. Therefore, <x3 � p> is a maximal ideal of Q[x].

You also know that if R is a ring, and M is a maximal ideal of R, then R/M is a field.

Thus, Q[x] /<x3 � p> is a field.

Example: Let p be a prime number. Show that

f(x) = xp-1 + xp-2 + .... + x + 1 is irreducible in Z[x], f(x) is called the pth cyclotornic polynomial.

Solution: To start with, we would like you to note that f(x) = g(x) h(x) in Z[x] iff f(x + 1) = g(x + 1)h(x + 1) in Z[x]. Thus, f(x) is irreducible in Z[x] iff f(x + l) is irreducible in Z[x].

Now, f(x) = px 1

x 1

f(x + 1) =

px 1 1

x

= 1x

(xp + pC1 xp-1 + ... + pC

p-1 x + 1 � 1), (by the binomial theorem)

= xp-1 + pxp-2 + pC2xp-3 + ... + pC

p-2 x + p.



NotesNow apply Eisenstein�s criterion taking p as the prime. We find that f(x+l) is irreducible.

Therefore, f(x) is irreducible.

So far we have used the fact that if f(x) Z[x] is irreducible over Z, then it is also irreducible overQ. Do you think we can have a similar relationship between irreducibility in Q[x] and R[xl? To

answer this, consider f(x) = x2 - 2. This is irreducible in Q[x], but f(x) = (x 2 )(x 2 ) in R[x].

Thus, we cannot extend irreducibility over Q to irreducibility over W.

But, we can generalise the fact that irreducibility in Z[x] implies irreducibility in Q[x]. This is notonly true for Z and Q; it is true for any UFD R and its field of quotients F. Let us state thisrelationship explicitly.

Theorem 4: Let R be a UFD with field of quotients F.

(i) If f(x) R[x] is an irreducible primitive polynomial, then it is also irreducible in F[x].

(ii) (Eisenstein�s Criterion) Let f(x) = a0 + a

1x + ... + a, xn R[x] and p R be a prime element

such that p | a,, p2 | a0 and p | a

i for 0 i < n. Then f(x) is irreducible in F[x].

The proof of this result is on the same lines as that of Theorems 2 and 3. We will not be doing ithere. But if you are interested, you should try and prove the result yourself.

Now, we have already pointed out that if F is a field and f(x) is irreducible over F, then F[x]/<f(x)> is field. How is this field related to F? That is part of what we will discuss in the nextsection.

24.2 Field Extensions

We shall discuss subfields and field extensions. To start with let us define these terms. By nowthe definition may be quite obvious to you.

Definition: A non-empty subset S of a field F is called a subfield of F if it is a field with respectto the operations on F. If S$F, then S is galled a proper subfield of F.

A field K is called a field extension of F if F is a subfield of K. Thus, Q is a subfield of R and R isa field extension of Q. Similarly, C is a field extension of Q as well as of R.

Note that a non-empty subset S of a field F is a subfield of F if

(i) S is a subgroup of (F,+), and

(ii) the set of all non-zero elements of S forms a subgroup of the group of non-zero elementsof F under multiplication.

Theorem 5: A non-empty subset S of a field F is a subfield of F if and only if

(i) a S, b S a � b S, and

(ii) a S , b S , b 0 ab-1 S.

Now, let us look at a particular field extension of a field F. Since F[x] is an integral domain, wecan obtain its field of quotients. We denote this field by F(x). Then F is a subfield of F(x). Thus,F(x) is a field extension of F. Its elements are expressions of the form f,( x) where f(x), g(x) F[x]and g(x) # 0.

g(x)

There is another way of obtaining a field extension of a field F from F[x]. We can look at quotientrings of F[x] by its maximal ideals. You know that an ideal is maximal in F[x] iff it is generatedby an irreducible polynomial over F. So, F[x]/<f(x)> is a field iff f(x) is irreducible over F.


Abstract Algebra

Notes Now, given any f(x) F[x], such that deg f(x) > 0, we will show that there is a field monomorphismfrom F into F[x]/d(x)>. This will show that F[x)/<f(x)> contains an isomorphic copy of F; andhence, we can say, that it contains F. So, let us define 0 : F F[x]/d(x)>: (a) = a + <f(x)>.

Then (a+b) = (a) + (b), and

(ab) = (a) (b).

Thus, is a ring homomorphism.

What is Ker ?

Ker = {a F] a + <f(x)> = <f(x)>}

= {a F | a <f(x)>}

= {a F | f(x) | a}

= {0}, since deg f > 0 and deg a 0.

Thus, is 1-1, and hence an inclusion.

Hence, F is embedded in F[x]/<f(x)>.

Thus, if f(x) is irreducible in F[x], then F[x]/<f(x)> is a field extension of F.

Well, we have looked at field extensions of any field F. Now let us look at certain fields, one ofwhich F will be an extension of.

24.2.1 Prime Fields

Let us consider any field F. Can we say anything about what its subfields look like? Yes, we cansay something about one of its subfields. Let us prove this very startling and useful fact.

Theorem 6: Every field contains a subfield isomorphic to Q or to Zp, for some prime number p.

Proof: Let F be a field. Define a function f : Z F : f(n) = n.1 = 1 + 1 + .... + 1 (n times).

f is a ring homomorphism and Ker f = pZ, where p is the characteristic of F.

You know that char F = 0 or char F = p, a prime. So let us look at these two cases separately.

Case 1 (char F = 0): In this case f is one-one. Z = f(Z). Thus, f(Z) is an integral domain containedin the field F. Since F is a field, it will also contain the field of quotients of f(Z). This will beisomorphic� to the field of quotients of Z, i.e., Q. Thus, F has a subfield which is isomorphic to Q.

Case 2 (char F = p, for some prime p):

Since p is a prime number, Z/pZ is a field.

Also, by applying the Fundamental Theorem of Homomorphism to f, we get Z/pZ f(Z). Thus,f(Z) is isomorphic to Z

p and is contained in F. Hence, F has a subfield isomorphic to Z

p.

Let us Theorem 6 slightly. What it says is that:

Let F be a field.

(i) If char F = 0, then F has a subfield isomorphic to Q.

(ii) If char F = p, then P has a subfield isomorphic to Z.

Because of this property of Q arid Zp (where p is a prime number) we call these fields primefields.



NotesThus, the prime fields are Q, Z2, Z

3, Z

5, etc.

We call the subfield isomorphic to a prime field (obtained in Theorem 6), the prime subfield ofthe given field.

Let us again reword Theorem 6 in terms of field extensions. What it says is that every field is aWeld extension of a prime field.

Now, suppose a field F is an extension of a field K. Are the prime subfields of K and F isomorphicor not? To answer this let us look at char K and char F. We want to know if char K = char F or not.Since F is a field extension of K, the unity of F and K is the same, namely, 1. Therefore, the leastpositive integer n such that n.1 = 0 is the same for F as well as K. Thus, char K = char F. Therefore,the prime subfields of K and F are isomorphic.

A very important fact that a field is a prime field iff it has no proper subfields.

Now let us look at certain field extensions of the fields Zp.

24.2.2 Finite Fields

You have dealt a lot with the finite fields Zp. Now we will look at field extensions of these fields.

You know that any finite field F has characteristic p, for some prime p. And then F is an extensionof Z. Suppose P contains q elements. Then q must be a power of p. That is what we will provenow.

Theorem 7: Let F be a finite field having q elements and characteristic p. Then q = pn, somepositive integer n.

The proof of this result uses the concepts of a vector space and its basis.

Proof: Since char F = p, F has a prime subfield which is isomorphic to Zp. We lose nothing if we

assume that the prime subfield is Zp. We first show that F is a vector space over Z

p with finite

dimension.

Recall that a set V is a vector space over a field K if:

(i) we can define a binary operation + on V such that (V, +) is an abelian group,

(ii) we can define a �scalar multiplication. : K × V V such that a, b K and v, w V,

a(a + w) = a.v + a.w

(a + b).v = a.v + b.w

(ab). v = a. (b.v)

1.v = v.

Now, we know that (F, +) is an abelian group. We also know that the multiplication in F willsatisfy all the conditions that the scalar multiplication should satisfy. Thus, F is a vector spaceover 2,. Since F is a finite field, it has a finite dimension over Z

p. Let dim Z

p F = n. Then we can

find a,. .., an, a F such that

F = Zpa

1 + Z

pa

2 + .. + Z

pa

n.

We will show that F has pn elements.

Now, any element of F is of the form

b1a

1, + b

2a

2 + ... +, b

na

n, where b,, . .., b

n Zp.

Now, since o(Zp) = p, b1 can be any one of its p elements.


Abstract Algebra

Notes Similarly, each of b2, b

3, ...., b

n has p choices. And, corresponding to each of these choices we get

a distinct element of F. Thus, the number of elements in F is p × p × ... × p (n times) = pn.

The utility of this result is something similar to that of Lagrange�s theorem. Using this result weknow that, for instance, no field of order 26 exists. But does a field of order 25 exist? DoesTheorem 7 answer this question? It only says that a field of order 25 can exist. But it does not saythat it does exist. The following exciting result, the proof of which is beyond the scope of thiscourse, gives us the required answer. This result was obtained by the American mathematicianE.H. Moore in 1893.

Theorem 8: For any prime number p and n N, there exists a field with pn elements. Moreover,any two finite fields having the same number of elements are isomorphic.

Self Assessment

1. If f(x) Z(x) is irreducible over Q[x]. Then it is .............. in Q[x].

(a) reducible (b) irreducible

(c) direct (d) finite

2. A non-empty subsets of a field F is called a .............. of F it is a field with respect to theoperation on F.

(a) subfield (b) field domain

(c) range field (d) extension

3. Every field contains a subfield is o morphic to Q or to Zp for r some .............. P.

(a) prime (b) finite

(c) infinite (d) external

4. Let F be a finite having of elements and characteristics P, then q = .............., some positiveinteger n.

(a) p-1 (b) pn

(c) xpn (d) p.xp

5. For any prime number P and n N, then exists a field with Pn elements. Move over, andtwo .............. fields having the same number of elements are isomorphic.

(a) infinite (b) finite

(c) direct (d) extension

24.3 Summary

Gauss lemma, i.e., the product of primitive polynomials is primitive.

For any n N, we can obtain an irreducible polynomial over Q of degree n.

Definitions and examples of subfields and field extensions.

Different ways of obtaining field extensions of a field F from F[x].



Notes Eisenstein�s irreducibility criterion for polynomials over Z and Q. This states that iff(x) = a

0 + a, x + . . . + a

nxn Z[x] and there is a prime p Z such that

p | ai i = 0 , 1 . ..., n � 1.

p | a, and

p | a0,

then f(x) is irreducible over Z� (and hence over Q).

Every field contains a subfield isomorphic to a prime field.

The prime fields are Q or Zp, for some prime p.

The number of elements in a finite field F is pn�, where char F = p and dim zpF = n.

Given a prime number p�and n N, there exists a field containing pn elements. Any twofinite fields with the same number of elements are isomorphic.

If F is a finite field with pn elements, then npx x is a product of pn linear polynomials

over F.

24.4 Keywords

Eisenstein�s Criterion: Let f(x) = a0 + a

lx + ... + a,,xn Z[x]. Suppose that for some prime number

p; (i) P | an, (ii) p | a

0, p | a

1,...p|a

n�1

Subfield of F: A non-empty subset S of a field F is called a subfield of F if it is a field with respectto the operations on F. If S$F, then S is galled a proper subfield of F.


1. What are the contents of the following polynomials over Z?

(a) 1 + x + x2 + x3 + x4 (b) 7x4 � 7

(c) 5(2x2 � 1) (x + 2)

2. Prove that any polynomial f(x) Z[x] can be written as dg(x), where d is the content of f(x)and g(x) is a primitive polynomial.

3. For any n N and prime number p, show that xn � p is irreducible over Q[x]. Note that thisshows us that we can obtain irreducible polynomials of any degree over Q[x].

4. If a0 + a

1x + ... + a, xn Z[x] is irreducible in Q[x], can you always find a prime p that satisfied

the conditions (i), (ii) and (iii) of Theorem 3?

5. Which of the following elements of Z[x] are irreducible over Q?

(a) x2 � 12 (b) 8x3 + 6x2 � 9x + 24

(c) 5x + 1

6. Let p be a prime: integer. Let a be a non-zero non-unit square-free integer, i.e., 2b | a for

any b Z. Show that Z[x]/<xp + a> is an integral domain.

7. Show that ppx a Z [x] is not irreducible for any a Z.


Abstract Algebra

Notes Answers: Self Assessment

1. (b) 2. (a) 3. (a) 4. (b) 5. (b)






www.math.niu.edu

www.maths.tcd.ie/






Unit 25: Roots of a Polynomial

NotesUnit 25: Roots of a Polynomial

CONTENTS

Objectives

Introduction

25.1 Roots of Polynomials

25.2 Summary

25.3 Keywords



Objectives


Define roots of polynomials

Discuss examples of roots of polynomial

Introduction

You have seen when we can say that an element in a ring divides another element. Let us recallthe definition in the context of F[x], where F is a field.

25.1 Roots of Polynomials

Definition: Let f(x) and g(x) be in F[x], where F is a field and g(x) 0. We say that g(x) dividesf(x)(or g(x) is a factor of f(x), or f(x) is divisible by gi(x)) if there-exists q(x) F[x] such that

f(x) = q(x) g(x).

We write g(x) | f(x) for �g(x) divides f(x)�, and g(x) | f(x) for �g(x) does not divide f(x)�.

Now, if f(x) F[x] and g(x) F[x], where g(x) 0, when g(x) | f(x)? We find that g(x) | f(x) ifr(x) = 0.

Definition: Let F be a field and f(x) F[x]. We say that an element a F is a root (or zero) of f(x)if f(n) = 0.

For example, 1 is a root of x2 � 1 R[x], since 12 � 1 = 0.

Similarly, �1 is a root of f(x) = 3 2 1 1x x x Q[x],

2 2 since

f(�1) = �1 +11 1

0.2 2


Abstract Algebra

Notes Let F be a field and f (x) F[x]. Then a P is a root of f(x) if and only if (x�a) | f(x)).

We can generalise this criterion to define a root of multiplicity m of a polynomial in F[x].

Definition: Let F be a field and f(x) F[x]. We say that a F is a root of multiplicity m (where mis a positive integer) of

f(x) if (x � a)m | f(x), but (x�a)m+1 | f(x).

For example, 3 is a root of multiplicity 2 of the polynomial (x�3)2 (x+2) Q[x]; and (�2) is a rootof multiplicity 1 of this polynomial.

Now, is it easy to obtain all the roots of a given polynomial? Any linear polynomial ax+b F[x]will have only one root, namely, -a-1b. This is because ax+b = 0 iff x = -a-1b.

In the case of a quadratic polynomial ax2 + bx + c F[x], you know that its two roots are obtainedby applying the quadratic formula

2b b 4ac2a

For polynomials of higher degree we may be able to obtain some roots by trial and error. Forexample, consider f(x) = x5 � 2x + 1 R[x]. Then, we try out x = 1 and find f(1) = 0. So, we find that1 is a zero of f(x). But this method doesn�t give us all the roots of f(x).

As we have just seen, it is not easy to find all the roots of a given polynomial. But, we can givea definite result about the number of roots of a polynomial.

Theorem 1: Let f(x) be a non-zero polynomial of degree n over a field F:Then f(x) has at most nroots in F.

Proof: If n = 0, then f(x) is a non-zero constant polynomial.

Thus, it has no roots, and hence, it has at most 0 ( = n) roots in F.

So, let, us assume that n 1. We will use the principle of induction on n. If deg f(x) = 1,

then

f(x) = a0 + a

1x, where a

0, a, F and a, 0.

So f(x) has only one root, namely, (�a1

-1 a0).

Now assume that the theorem is true for all polynomials in F[x] of degree n. We will show thatthe number of roots of f(x) n.

If f(x) has no root in F, then the number of roots of f(x) in F is 0 S n. So, suppose f(x) has a roota F.

Then f(x) = (x � a) g(x), where deg g(x) = n�1.

Hence, by the induction hypothesis g(x) has at most n�1 roots in F, say a1,....,a

n-1. Now,

ai is a root of g(x) g(a

i) = 0 f(a

i) = (a,�a) g(a

i) = 0

a ai is a root of f(x) i = 1, ..., n � 1.

Thus, each root of g(x) is a root of f(x).

Now, b F is a root of f(x) iff f(b) = 0, i.e., iff (b � a) g(b) = 0, i.e., iff b � a = 0 or g(b) = 0, since F isan integral domain. Thus, b is a root of f(x) iff b = a or b is a root of g(x). So, the only roots off(x) are a and a

1, ..., a

n-1. Thus, f(x) has at the most n roots, and so, the theorem is true for n.



NotesHence, the theorem is true for all n 1.

Using this result we know that, for example, x3�1 Q[x] can�t have more than 3 roots in Q.

In Theorem 1 we have not spoken about the roots being distinct. But an obvious corollary ofTheorem 1 is that

if f(x) F[x] is of degree n, then f(x) has st most n distinct roots in F.

We will use this result to prove the following useful theorem.

Theorem 2: Let f(x) and g(x) be two non-zero polynomials of degree n over the field F. If there

exist n+l distinct elements a,,.. .,an+1

, in F such that f(ai) = g(a

i) i = I , ..., n+l, then f(x) = g(x).

Proof: Consider the polynomial h(x) = f(x) = g(x)

Then deg h(x) n, but it has n + l distinct roots a,, ..., an+1

.

This is impossible, unless h(x) = 0, i.e., f(x) = g(x).

Example: Prove that 36x 5x Z [x] has more roots than its degree. (Note that Z

6 is not

a field.)

Solution: Since the ring is finite, it is easy for us to run through all its elements and check whichof them, are roots of

3f(x) x 5x.

So, by substitution we find that

f(0) = 0 = f(1) = f(2) = f(3) = f(4) = f(5).

In fact, every element of Z6 is a zero of f(x). Thus, f(x) has 6 zeros, while deg f(x) = 3.

So far, we have been saying that a polynomial of degree n over F has at most n roots in Fa. It canhappen that the polynomial has no root in F. For example, consider the polynomial x2 + 1 R[x].You know that it can have 2 roots in R, at the most. But as you know, this has no roots in R (it hastwo roots, i and �i, in C).

We can find many other examples of such polynomials in R[x]. We call such polynomialsirreducible over R. We shall discuss them in detail in the next units.

Now let us end this unit by seeing what we have covered in it.

Definition: Let F be a set on which two binary operations are defined, called addition andmultiplication, and denoted by + and · respectively. Then F is called a field with respect to theseoperations if the following properties hold:

(i) Closure: For all a,b in F the sum a + b and the product a . b are uniquely defined and belongto F.

(ii) Associative Laws: For all a,b,c in F,

a + (b + c) = (a + b) + c and a· (b· c) = (a· b)· c.

(iii) Commutative Laws: For all a,b in F,

a + b = b + a and a · b = b· a.

(iv) Distributive Laws: For all a, b, c in F,

a· (b + c) = (a· b) + (a· c) and (a + b)· c = (a· c) + (b· c).


Abstract Algebra

Notes (v) Identity Elements: The set F contains an additive identity element, denoted by 0, such thatfor all a in F,

a + 0 = a and 0 + a = a.

The set F also contains a multiplicative identity element, denoted by 1 (and assumed to bedifferent from 0) such that for all a in F,

a· 1 = a and 1· a = a.

(vi) Inverse Elements: For each a in F, the equations

a + x = 0 and x + a = 0

have a solution x in F, called an additive inverse of a, and denoted by -a. For each nonzeroelement a in F, the equations

a· x = 1 and x· a = 1

have a solution x in F, called a multiplicative inverse of a, and denoted by a-1.

Definition: Let F be a field. For am

, am-1

, . . . , a1, a

0 in F, an expression of the form

am

xm + am-1

xm-1 + · · · + a1x + a

0

is called a polynomial over F in the indeterminate x with coefficients am

, am-1

, . . . , a0. The set of

all polynomials with coefficients in F is denoted by F[x]. If n is the largest nonnegative integersuch that a

n 0, then we say that the polynomial

f(x) = anxn + · · · + a

0

has degree n, written deg(f(x)) = n, and an is called the leading coefficient of f(x). If the leading

coefficient is 1, then f(x) is said to be monic.

Two polynomials are equal by definition if they have the same degree and all correspondingcoefficients are equal. It is important to distinguish between the polynomial f(x) as an elementof F[x] and the corresponding polynomial function from F into F defined by substituting elementsof F in place of x. If f(x) = a

mxm + · · · + a

0 and c is an element of F, then f(c) = a

mcm + · · · + a

0. In fact,

if F is a finite field, it is possible to have two different polynomials that define the same polynomialfunction. For example, let F be the field Z

5 and consider the polynomials x5 - 2x + 1 and 4x + 1. For

any c in Z5, by Fermat�s theorem we have c5 c (mod 5), and so

c5 - 2c + 1 - c + 1 4c + 1 (mod 5),

which shows that x5 - 2x + 1 and 4x + 1 are identical, as functions.

For the polynomials

f(x) = am

xm + am-1

xm-1 + · · · + a1x + a

0

and

g(x) = bnxn + b

n-1xn-1 + · · · + b

1x + b

0,

the sum of f(x) and g(x) is defined by just adding corresponding coefficients. The product f(x)g(x)is defined to be

am

bnxn+m + · · · + (a

2b

0 + a

1b

1 + a

0b

2)x2 + (a

1b

0 + a

0b

1)x + a

0b

0.

The coefficient ck of xk in f(x)g(x) can be described by the formula

ck =

i

k

k ii 0ab .



NotesThis definition of the product is consistent with what we would expect to obtain using a naiveapproach: Expand the product using the distributive law repeatedly (this amounts to multiplyingeach term be every other) and then collect similar terms.

Proposition: If f(x) and g(x) are non-zero polynomials in F[x], then f(x)g(x) is non-zero anddeg(f(x)g(x)) = deg(f(x)) + deg(g(x)).

Corollary: If f(x),g(x),h(x) are polynomials in F[x], and f(x) is not the zero polynomial, thenf(x)g(x) = f(x)h(x) implies g(x) = h(x).

Definition: Let f(x),g(x) be polynomials in F[x]. If f(x) = q(x)g(x) for some q(x) in F[x], then we saythat g(x) is a factor or divisor of f(x), and we write g(x) | f(x). The set of all polynomials divisibleby g(x) will be denoted by < g(x) >.

Lemma: For any element c in F, and any positive integer k,

(x - c) | (xk - ck).

Theorem 3: Let f(x) be a non-zero polynomial in F[x], and let c be an element of F. Then thereexists a polynomial q(x) in F[x] such that

f(x) = q(x)(x - c) + f(c).

Moreover, if f(x) = q1(x)(x - c) + k, where q

1(x) is in F[x] and k is in F, then q

1(x) = q(x) and k = f(c).

Definition: Let f(x) = am

xm + · · + a0 belong to F[x]. An element c in F is called a root of the

polynomial f(x) if f(c) = 0, that is, if c is a solution of the polynomial equation f(x) = 0 .

Corollary: Let f(x) be a non-zero polynomial in F[x], and let c be an element of F. Then c is a rootof f(x) if and only if x-c is a factor of f(x). That is,

f(c) = 0 if and only if (x-c) | f(x).

Corollary: A polynomial of degree n with coefficients in the field F has at most n distinct rootsin F.

Self Assessment

1. Let F be a field and f(x) F[x] then we say that an element a F is a root of f(x) off(n) = ...............

(a) 1 (b) 2

(c) 0 (d) 2-1

2. 1 is a root of x2 � 1 R[x], since 12 � 1 = ...............

(a) 2 (b) 1

(c) 0 (d) �1

3. ............... is a root of f(x) = x3 + x2 + 1 1

x2 2

Q[x]

(a) 1 (b) 2

(c) �1 (d) �2

4. If n = 0 then f(x) is a non-zero ............... polynomial

(a) constant (b) degree

(c) range (d) power


Abstract Algebra

Notes 5. x3 + 5x z6[x] has ............... roots than its degree.

(a) 2 (b) 3

(c) 1 (d) more

25.2 Summary

If f(x) and g(x) are non-zero polynomials in F[x], then f(x)g(x) is non-zero and deg(f(x)g(x))= deg(f(x)) + deg(g(x)).

If f(x),g(x),h(x) are polynomials in F[x], and f(x) is not the zero polynomial, then f(x)g(x) =f(x)h(x) implies g(x) = h(x).

Let f(x),g(x) be polynomials in F[x]. If f(x) = q(x)g(x) for some q(x) in F[x], then we say thatg(x) is a factor or divisor of f(x), and we write g(x) | f(x). The set of all polynomialsdivisible by g(x) will be denoted by < g(x) >.

For any element c in F, and any positive integer k,

(x - c) | (xk - ck).

Let f(x) be a non-zero polynomial in F[x], and let c be an element of F. Then there exists apolynomial q(x) in F[x] such that

f(x) = q(x)(x - c) + f(c).

Moreover, if f(x) = q1(x)(x - c) + k, where q

1(x) is in F[x] and k is in F, then q

1(x) = q(x) and

k = f(c).

Let f(x) = am

xm + · · + a0 belong to F[x]. An element c in F is called a root of the polynomial

f(x) if f(c) = 0, that is, if c is a solution of the polynomial equation f(x) = 0 .

Let f(x) be a non-zero polynomial in F[x], and let c be an element of F. Then c is a root of f(x)if and only if x-c is a factor of f(x). That is,

f(c) = 0 if and only if (x-c) | f(x).

A polynomial of degree n with coefficients in the field F has at most n distinct roots in F.

25.3 Keywords

Field: Let F be a set on which two binary operations are defined, called addition and multiplication,and denoted by + and · respectively. Then F is called a field with respect to these operations.

Identity Elements: The set F contains an additive identity element, denoted by 0, such that for alla in F,

a + 0 = a and 0 + a = a.

Inverse Elements: For each a in F, the equations

a + x = 0 and x + a = 0

have a solution x in F, called an additive inverse of a, and denoted by -a. For each non-zeroelement a in F, the equations

a· x = 1 and x· a = 1




1. Let F be a field and f(x) F[x] with deg f(x) 1. Let a F. Show that f(x) is divisible byx � a iff f(a) = 0.

2. Find the roots of the following polynomials, along with their multiplicity.

(a) 21 5f(x) x x 3 Q[x]

2 2 (b) 2

3f(x) x x 1 Z [x]

(c) 4 35f(x) x 2x 2x 1 Z [x]

3. Let F be a field and a F. Define a function

: F[x] F : f(f(x)) = f(x)

This function is the evaluation at a.

Show that

(a) f is an onto ring homomorphism.

(b) f (b) = b b F.

(c) Ker f = <x � a>

So, what does the Fundamental Theorem of Homomorphism say in this case?

4. Let p be a prime number. Consider p 1px 1 Z [x].

Use the fact that Zp is a group of order

p to show that every non-zero element of Zp is a root of xp-1 � 1. Thus, show that xp-1 � 1 =

(x 1)(x 2)...(x p 1).

5. The polynomial x4 + 4 can be factored into linear factors in Z5[x]. Find this factorisation.


1. (c) 2. (c) 3. (c) 4. (a) 5. (d)






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Abstract Algebra

Notes Unit 26: Splitting Fields, Existence and Uniqueness

CONTENTS

Objectives

Introduction

26.1 Extension Field

26.2 Summary

26.3 Keywords



Objectives


Discuss splitting field

Describe extension field and theorem related to extension

Introduction

Beginning with a field K, and a polynomial f(x) K, we need to construct the smallest possibleextension field F of K that contains all of the roots of f(x). This will be called a splitting field forf(x) over K. The word �the� is justified by proving that any two splitting fields are isomorphic.

Let F be an extension field of K and let u F. If there exists a non-zero polynomial f(x) K[x]such that f(u) = 0, then u is said to be algebraic over K. If not, then u is said to be transcendentalover K.

26.1 Extension Field

If F is an extension field of K, and u F is algebraic over K, then there exists a unique monicirreducible polynomial p(x) K[x] such that p(u) = 0. It is the monic polynomial of minimaldegree that has u as a root, and if f(x) is any polynomial in K[x] with f(u) = 0, thenp(x) | f(x).

Alternate proof: The proof in the text uses some elementary ring theory. Then decided to includea proof that depends only on basic facts about polynomials.

Assume that u F is algebraic over K, and let I be the set of all polynomials f(x) K[x] such thatf(u) = 0. The division algorithm for polynomials can be used to show that if p(x) is a non-zeromonic polynomial in I of minimal degree, then p(x) is a generator for I, and thus p(x) | f(x)whenever f(u) = 0.

Furthermore, p(x) must be an irreducible polynomial, since if p(x) = g(x)h(x) for g(x); h(x) K[x],then g(u)h(u) = p(u) = 0, and so either g(u) = 0 or h(u) = 0 since F is a field. From the choice of p(x)as a polynomial of minimal degree that has u as a root, we see that either g(x) or h(x) has thesame degree as p(x), and so p(x) must be irreducible.


Unit 26: Splitting Fields, Existence and Uniqueness

NotesIn the above proof, the monic polynomial p(x) of minimal degree in K[x] such that p(u) = 0 iscalled the minimal polynomial of u over K, and its degree is called the degree of u over K.

Let F be an extension field of K, and let u1, u

2, ..., u

n F. The smallest subfield of F that contains

K and u1, u

2,..., u

n will be denoted by K(u

1, u

2,..., u

n). It is called the extension field of K generated

by u1, u2,...., u

n. If F = K(u) for a single element u F, then F is said to be a simple extension of K.

Let F be an extension field of K, and let u F. Since K(u) is a field, it must contain all elements ofthe form

2 m0 1 2 m

2 n0 1 2 n

a + a u + a u + ... + a u,

b + b u + b u + ... + b u

where ai, b

j K for i = 1,..., m and j = 1,... n. In fact, this set describes K(u), and if u is transcendental

over K, this description cannot be simplified. On the other hand, if u is algebraic over K, then thedenominator in the above expression is unnecessary, and the degree of the numerator can beassumed to be less than the degree of the minimal polynomial of u over K.

If F is an extension field of K, then the multiplication of F defines a scalar multiplication,considering the elements of K as scalars and the elements of F as vectors. This gives F thestructure of a vector space over K, and allows us to make use of the concept of the dimension ofa vector space. The next result describes the structure of the extension field obtained by adjoiningan algebraic element.

Definition: Let F be an extension field of K and let u F be an element algebraic over K.

(a) K(u) K[x]= hp(x)i, where p(x) is the minimal polynomial of u over K.

(b) If the minimal polynomial of u over K has degree n, then K(u) is an n-dimensional vectorspace over K.

Alternate proof: The standard proof uses the ring homomorphism : K[x] F defined byevaluation at u. Then the image of is K(u), and the kernel is the ideal of K[x] generated by theminimal polynomial p(x) of u over K. Since p(x) is irreducible, ker() is a prime ideal, and soK[x] = ker() is a field because every nonzero prime ideal of a principal ideal domain is maximal.Thus K(u) is a field since K(u) 245= K[x]= ker().

The usual proof involves some ring theory, but the actual ideas of the proof are much simpler.To give an elementary proof, define : K[x]= {p(x)} K(u) by ([f(x)]) = f(u), for all congruenceclasses [f(x)] of polynomials (modulo p(x)). This mapping makes sense because K(u) contains u,together with all of the elements of K, and so it must contain any expression of the form a

0 +a

1u+

... + am

um, where ai K, for each subscript i. The function is well-defined, since it is also

independent of the choice of a representative of [f(x)]. In fact, if g(x) K[x] and f(x) is equivalentto g(x), then f(x) � g(x) = q(x)p(x) for some q(x) K[x], and so f(u) � g(u) = q(u)p(u) = 0, showingthat ([f(x)]) = ([g(x)]).

Since the function simply substitutes u into the polynomial f(x), and it is not difficult to showthat it preserves addition and multiplication. It follows from the definition of p(x) that is one-to-one. Suppose that f(x) represents a nonzero congruence class in K[x]= {p(x)}. Then

p(x) | f(x), and so f(x) is relatively prime to p(x) since it is irreducible. Therefore, there exist

polynomials a(x) and b(x) in K[x] such that a(x)f(x) + b(x)p(x) = 1. It follows that [a(x)][f(x)] = [1]for the corresponding equivalence classes, and this shows that K[x] /{p(x)} is a field. Thus theimage E of in F must be subfield of F. On the one hand, E contains u and K, and on the otherhand, we have already shown that E must contain any expression of the form a

0 + a

1u + ... + a

mum,

where ai K. It follows that E = K(u), and we have the desired isomorphism.

(b) It follows from the description of K(u) in part (a) that if p(x) has degree n, then the setB = {1, u, u2,..., un-1} is a basis for K(u) over K.


Abstract Algebra

Notes Let F be an extension field of K. The dimension of F as a vector space over K is called the degreeof F over K, denoted by [F : K]. If the dimension of F over K is finite, then F is said to be a finiteextension of K. Let F be an extension field of K and let u F. The following conditions areequivalent: (1) u is algebraic over K; (2) K(u) is a finite extension of K; (3) u belongs to a finiteextension of K.

Never underestimate the power of counting: the next result is crucial. If we have a tower ofextensions K E F, where E is finite over K and F is finite over E, then F is finite over K, and[F : K] = [F : E][E : K]. This has a useful corollary, which states that the degree of any element ofF is a divisor of [F : K].

Let K be a field and let f(x) = a0 + a

1x + ... + a

nxn be a polynomial in K[x] of degree n > 0. An

extension field F of K is called a splitting field for f(x) over K if there exist elements r1, r

2,..., r

n

F such that

(i) f(x) = an(x � r

1)(x � r

2) ... (x � rn) and

(ii) F = K(r1, r

2,..., r

n).

In this situation we usually say that f(x) splits over the field F. The elements r1, r

2,..., r

n are roots

of f(x), and so F is obtained by adjoining to K a complete set of roots of f(x). An inductionargument (on the degree of f(x)) can be given to show that splitting fields always exist. Theoremstates that if f(x) K[x] is a polynomial of degree n > 0, then there exists a splitting field F for f(x)over K, with [F : K] n!.

The uniqueness of splitting fields follows from two lemmas. Let : K L be an isomorphism offields. Let F be an extension field of K such that F = K(u) for an algebraic element u F. Let p(x)be the minimal polynomial of u over K. If v is any root of the image q(x) of p(x) under , andE = L(v), then there is a unique way to extend to an isomorphism : F E such that (u) = v and(a) = (a) for all a K. The required isomorphism : K(u) L(v) must have the form

(a0 + a

1u + ... + a

n-1un-1) = (a

0) + (a

1)v + ... + (a

n-1)vn-1

The second lemma is stated as follows. Let F be a splitting field for the polynomial f(x) K[x]. If : K L is a field isomorphism that maps f(x) to g(x) L[x] and E is a splitting field for g(x) overL, then there exists an isomorphism : F E such that (a) = (a) for all a K. The proof usesinduction on the degree of f(x), together with the previous lemma.

Theorem states that the splitting field over the field K of a polynomial f(x) K[x] is unique upto isomorphism. Among other things, this result has important consequences for finite fields.

Self Assessment

1. If F is an extension field k and u F is algebraic over K, then their exists a ...............

(a) different (b) finite

(c) infinite (d) unique

2. The monic polynomial P(x) of minimal degree in K[x] such that P(u) = 0 is called is............... of r over K and its degree is called the degree of u over K.

(a) maximal polynomial (b) minimal polynomial

(c) finite polynomial (d) infinite polynomial

3. The dimension of F as a vector space K is called the ............... of F over K, denoted by [f : k]

(a) range (b) domain

(c) degree (d) field


Unit 26: Splitting Fields, Existence and Uniqueness

Notes4. The splitting field over the field K of a polynomial f(x) K[x] is unique up to ...............

(a) homomorphism (b) isomorphism

(c) automorphism (d) finite extension

26.2 Summary

If F is an extension field of K, and u F is algebraic over K, then there exists a unique monicirreducible polynomial p(x) K[x] such that p(u) = 0. It is the monic polynomial of minimaldegree that has u as a root, and if f(x) is any polynomial in K[x] with f(u) = 0, thenp(x) | f(x).

Alternate Proof: The proof in the text uses some elementary ring theory. I�ve decided to includea proof that depends only on basic facts about polynomials.

Let F be an extension field of K, and let u1, u

2, ..., u

n F. The smallest subfield of F that contains

K and u1, u

2,..., u

n will be denoted by K(u

1, u

2,..., u

n). It is called the extension field of K generated

by u1, u

2,...., u

n. If F = K(u) for a single element u F, then F is said to be a simple extension of K.

Let F be an extension field of K, and let u F. Since K(u) is a field, it must contain all elements ofthe form

2 m0 1 2 m

2 n0 1 2 n

a + a u + a u + ... + a u,

b + b u + b u + ... + b u

where ai, b

j K for i = 1,..., m and j = 1,... n. In fact, this set describes K(u), and if u is transcendental

over K, this description cannot be simplified. On the other hand, if u is algebraic over K, then thedenominator in the above expression is unnecessary, and the degree of the numerator can beassumed to be less than the degree of the minimal polynomial of u over K.

If F is an extension field of K, then the multiplication of F defines a scalar multiplication,considering the elements of K as scalars and the elements of F as vectors. This gives F thestructure of a vector space over K, and allows us to make use of the concept of the dimension ofa vector space. The next result describes the structure of the extension field obtained by adjoiningan algebraic element.

The uniqueness of splitting fields follows from two lemmas. Let : K L be an isomorphism offields. Let F be an extension field of K such that F = K(u) for an algebraic element u F. Let p(x)be the minimal polynomial of u over K. If v is any root of the image q(x) of p(x) under , andE = L(v), then there is a unique way to extend to an isomorphism : F E such that (u) = v and(a) = (a) for all a K. The required isomorphism : K(u) L(v) must have the form

(a0 + a

1u + ... + a

n-1un-1) = (a

0) + (a

1)v + ... + (a

n-1)vn-1

The second lemma is stated as follows. Let F be a splitting field for the polynomial f(x) K[x].If : K L is a field isomorphism that maps f(x) to g(x) L[x] and E is a splitting field for g(x)over L, then there exists an isomorphism : F E such that (a) = (a) for all a K. The proof usesinduction on the degree of f(x), together with the previous lemma.

The splitting field over the field K of a polynomial f(x) K[x] is unique up to isomorphism.

26.3 Keywords

Splitting Field: Beginning with a field K, and a polynomial f(x) K, we need to construct thesmallest possible extension field F of K that contains all of the roots of f(x). This will be called asplitting field for f(x) over K.


Abstract Algebra

Notes Extension Field: Let F be an extension field of K and let u F. If there exists a nonzero polynomialf(x) K[x] such that f(u) = 0, then u is said to be algebraic over K. If not, then u is said to betranscendental over K.

In the above proof, the monic polynomial p(x) of minimal degree in K[x] such that p(u) = 0 iscalled the minimal polynomial of u over K, and its degree is called the degree of u over K.


1. Find the splitting field over Q for the polynomial x4 + 4.

2. Let p be a prime number. Find the splitting fields for xp � 1 over Q and over R.

3. Find the splitting field for x3 + x + 1 over Z2.

4. Find the degree of the splitting field over Z2 for the polynomial (x3 + x + 1)(x2 + x + 1).

5. Let F be an extension field of K. Show that the set of all elements of F that are algebraicover K is a subfield of F.

6. Let F be a field generated over the field K by u and v of relatively prime degrees m and n,respectively, over K. Prove that [F : K] = mn.

7. Let F E K be extension fields. Show that if F is algebraic over E and E is algebraic overK, then F is algebraic over K.

8. Let F K be an extension field, with u F. Show that if [K(u) : K] is an odd number, thenK(u2) = K(u).

9. Find the degree [F : Q], where F is the splitting field of the polynomial x3 � 11 over the fieldQ of rational numbers.

10. Determine the splitting field over Q for x4 + 2.

11. Determine the splitting field over Q for x4 + x2 + 1.

12. Factor x6 � 1 over Z7; factor x5 � 1 over Z

11.


1. (d) 2. (b) 3. (c) 4. (b)






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Unit 27: Separable Extensions

NotesUnit 27: Separable Extensions

CONTENTS

Objectives

Introduction

27.1 Separability

27.2 Summary

27.3 Keywords



Objectives


Define separability

Discuss examples related to separable extension

Introduction

In the last unit, you have studied about the splitting field and extension field. This unit willprovide you information related to separable extension.

27.1 Separability

Separability of a finite field extension L/K can be described in several different ways. Theoriginal definition is that every element of L is separable over K (that is, has a separableminimal polynomial in K[X]). We will give here three descriptions of separability for a finiteextension and use each of them to prove two theorems about separable extensions.

Theorem 1: Let L/K be a finite extension. Then L/K is separable if and only if the trace functionTr

L/K : L K is not identically 0.

The trace function is discussed in Appendix A.

Theorem 2: Let L/K be a finite extension. Then L/K is separable if and only if the ring KK L

has no non-zero nilpotent elements. When L/K is separable, the ring KK L is isomorphic to

[L:K ]K .

Example: Consider the extension Q( 2 )=Q. Since Q( 2 ) Q[X]/(X2 � 2), tensoring with

Q gives 2QQ Q 2 ; Q[X]/(X 2) Q[X]/((X 2 )(X 2 ) Q Q,

which is a product of 2 copies of Q (associated to the 2 roots of X2 2) and has no nilpotent

elements besides 0.


Abstract Algebra

Notes

Example: Consider the extension F2( u )/F

2(u). Since F

2( u ) F

2[X]/(X2 � u),

2

2 2 22 F (u) 2 2 2F (u) F ( u) F ( u) F (u)[X]/(X � u) = F (u)[X]/(X � u ) ,

which has the nonzero nilpotent element X � u .

Theorem 3: Let L/K be a finite extension. Then L is separable over K if and only if any derivationof K has a unique extension to a derivation of L.

For above two proofs, the reader should be comfortable with the fact that injectivity andsurjectivity of a linear map of vector spaces can be detected after a base extension: a linearmap is injective or surjective if and only if its base extension to a larger field is injective orsurjective.

Each of the three theorems above will be proved and then lead in its own way to proofs of thefollowing two theorems.

Theorem 4: If L = K(a1,....., a

r) and each a

i is separable over K then every element of L is separable

over K (so L/K is separable).

Theorem 5: Let L/K be a finite extension and F be an intermediate field. If L/F and F/K areseparable then L/K is separable.

We will use our new viewpoints to define separability for arbitrary (possibly non-algebraic)field extensions.

We want to show L/K is separable if and only if TrL/K

: L K is not identically 0. The trace mapis either identically 0 or it is onto, since it is K-linear with target K, so another way of puttingTheorem 1 is that we want to show L/K is separable if and only if the trace from L to K is onto.

Proof: We might as well take K to have positive characteristic p, since in characteristic 0 allfinite field extensions are separable and the trace is not identically 0 : TrL

/K(1) = [L : K] 0 in

characteristic 0.

If L/K is separable, by the primitive element theorem we can write L = K() where is separableover K. To show the trace is surjective for finite separable extensions, it suffices to prove surjectivityof the trace map on K()/K when K is any base field and is separable over K.

If L/K is inseparable, then there must be some a L which is inseparable over K. SinceTr

L/K = Tr

K()/K o Tr

L/K(), it success to prove the trace map on K()=K vanishes when is inseparable

over K.

For both cases of the field extension K()/K ( separable or inseparable over K), let have

minimal polynomial (X) in K[X]. Write (X) = (Xpm) where m is as large as possible, so (X)is separable. Thus (X) is separable if and only if m = 0.

Let n = deg = pmd, with d = deg . In K[X],

(X) = (X � 1) ... (X �

d)

for some i�s, which are all distinct since (X) is separable. Write

mpi ig , so the

i�s are distinct.

Then

m m m m mp p p p p1 d 1 d(X) (X ) (X ) ... (X ) (X ) ... (X ) .



NotesConsider now the extension of scalars up to K of the trace map TrK(a)/K

: K(a) K:

k(a)/K K KKTr id Tr : K K(a) K K K.

Tr is the trace map on KK K( ) as a K -vector space.

Since tensoring with a field extension preserves injectivity and surjectivity of a linear map,

TrK(a)/K

is onto Tr is onto; TrK(a)

=K Tr = 0

Since K() K[X] /((X)) as K-algebras, K( ) K[X]/( (X)) as K -algebras, and thus is isomorphic

to the direct product of the rings K [X]/mp

i(X � ). The trace is the sum of the traces to K on each

mpiK[X]/(X � ). Let�s look at the trace from

mpiK[X]/(X � ). to K .

In K [X], mp

iX � = mp

i(X � ) . Then m mp p

iK[X]/(X � ) = K[Y]/(Y ), where Y = X i. If m = 0, then

K [Y ] /mp(Y ) = K , so the trace to K is the identity. If m > 0, any element of K [Y ]/

mp(Y ) is the

sum of a constant plus a multiple of Y , which is a constant plus a nilpotent element (since Y

mod rpY is nilpotent). Any constant in K [Y ]/

mp(Y ) has trace 0 since pm = 0 in K (because

m > 0). A nilpotent element has trace 0. Thus the trace to K of any element of K [Y ] /mp(Y ) is 0.

To summarize, when is separable over K (i.e., m = 0), the trace map from K() to K is onto since

it is onto after extending scalars to K . When a is inseparable over K (i.e., m > 0), the trace mapis identically 0 since it vanishes after extending scalars.

Theorem 1 implies Theorem 4.

Proof. Set L0 = K, L

1 = K(

1) = L

0(

1), and more generally L

i = K(

1,....

i) = L

i-1(

i) for i 1. So we

have the tower of field extensions

K = L0 L

1 L

2 ... L

r-1 L

r = L.

By transitivity of the trace,

1 0 2 1 r r�1L/K L /L L /L L /LTr = Tr o Tr o o Tr

Since i is separable over K and the minimal polynomial of

i over L

i-1 divides its minimal

polynomial over K, i is separable over L

i-1. Therefore Tr

Li-1(

i)/L

i-1 : L

i L

i-1 is onto from the

proof of Theorem 1, so the composite map TrL/K : L K is onto. Therefore L/K is separable byTheorem 1.

Corollary: Theorem 1 implies Theorem 5.

Proof: By Theorem 1 and the hypothesis of Theorem 5, both TrL/F

and TrF/K

are onto. Therefore,their composite Tr

L/K is onto, so L/K is separable by Theorem 1.

Proof: We will begin with the case of a simple extension L = K(). Let (X) be the minimalpolynomial of over K, so L K[X]/((X)) as K-algebras and

KK L K[X]/ (X)

as K -algebras. This ring was considered in the proof of Theorem 1, where we saw its structure

is different when (X) is separable or inseparable. If (X) is separable in K[X], then K[X]/( K (X))


Abstract Algebra

Notes is a product of copies of the field K , so it has no non-zero nilpotent elements. If (X) is inseparable,

then K [X]/((X)) is a product of copies of rings K [Y ]/(mpY ) with m > 0, which all have

nonzero nilpotents.

Now we consider the structure of K K L when L/K is any finite extension.

First assume L/K is separable. By the primitive element theorem, we can write L = K() and

is separable over K. By the first paragraph of the proof, [L:K]KK L K since (X) has distinct

linear factors in K .

If L/K is inseparable, then some a L is inseparable over K. Tensoring the inclusion map

K() L up to K , we have an inclusion

K K

K() K K L.

The ring K K

K() has a non-zero nilpotent element by the first paragraph of the proof, so

K K

L does as well.

Corollary: The proof of Theorem 2 implies Theorem 4.

Proof: Make a tower of intermediate extensions in L/K as in (2.2). Note K is an algebraicclosure of every field L

i in the tower. Since

K K 1 L1K L (K L ) L

and L1 = K(

1) with

1 separable over K, the proof of Theorem 2 implies

1[L :K]

K 1K L K

as K -algebras. Therefore

11

1 1

[L :K ] [L :K ]K L LK L K L (K L)

Since L = L1(

2,...

r) with each

i separable over L

1, we can run through the same computation

for 2LK L as we did for

KK L, and we get 2 1

1 2

[L :L ]L LK L (K L) , so

2 1 1 2

2

[L :L ][L :K] [L :K]K L 2K L (K L) = (K L L) .

Repeating this enough, in the end we get

r

r

[L:K ][L :K]K LK L (K L) K .


Proof: The field K is an algebraic closure of F and L. Using Theorem 2,

K KL ( K

KF)

F L

K [F:K]F

L since F/K is separable

( K F L)[F:K]



Notes K [L:F][F:K] since L=F is separable

K [L:K]

Thus L/K is separable by Theorem 1.2.

Theorem 6: Let L/K be an extension of fields, and L be algebraic over K. Then is separableover K if and only if any derivation on K has a unique extension to a derivation on K().

Proof: When L is separable over K, Corollary B.10 shows any derivation on K extendsuniquely to a derivation on K().

Now suppose L is inseparable over K. Then �(X) = 0, where (X) is the minimal polynomialof over K. In particular �() = 0. We are going to use this vanishing of �() to construct anonzero derivation on K() which extends the zero derivation on K.

Then the zero derivation on K has two lifts to K(): the zero derivation on K() and this otherderivation we will construct.

Define Z : K() K() by Z(f()) = f�(), where f(X) K[X]. Is this well-defined?

Well, if f1() = f

2(), then f

1(X) f

2(X) mod (X), say

f1(X) = f

2(X) + (X)k(X).

Differentiating both sides with respect to X,

f�1(X) = f�

2(X) + (X)k�(X) + �(X)k(X):

Evaluating both sides at yields f�1 () = f�

2() since �() = 0. So Z : K() K() is well-defined.

It is left to the reader to check Z is a derivation on K(). This derivation kills K, but Z() = 1, soZ extends the zero derivation on K while not being the zero derivation itself.

The reader can check more generally that when is inseparable over K and K() is arbitrarythe map f() f�() is a derivation on K() that extends the zero derivation on K and sends to . So there are many extensions of the zero derivation on K to K(): one for each element ofK().

We need a lemma to put inseparable extensions into a convenient form for our derivationconstructions later.

Lemma: Let L/K be a finite inseparable field extension. Then there is an L and intermediatefield F such that L = F() and is inseparable over F.

Proof: Inseparable field extensions only occur in positive characteristic. Let p be the characteristicof K. Necessarily [L : K] > 1. Since L/K is inseparable, there is some L that is inseparable overK.

Write L = K(1,....

r). We will show by contradiction that some

i has to be inseparable over K.

Assume every i is separable over K. Then we can treat L/K as a succession of simple field

extensions as in (2.2), where Li = L

i-1(

i) with

i separable over L

i-1. By Theorem, any derivation

on Li-1

extends to a derivation on Li, so any derivation on K extends to a derivation on L.

Moreover, this extended derivation on L is unique. To show that, consider two derivations D andD� on L that are equal on K. Since L

1 = K(

1) and

1 is separable over K, the proof of Corollary

B.10 tells us that D and D� both send L1 to L

1 and are equal on L

1. Now using L

1 in place of K, D and

D� being equal on L1 implies they are equal on L

2 since L

2 = L

1(

2) and

2 is separable over L

1. We

can keep going like this until we get D = D� on Lr = L. As a special case of this uniqueness, the only

derivation on L which vanishes on K is the zero derivation on L.


Abstract Algebra

Notes Now replace K as base field with K(), over which the i�s are of course still separable. Then any

derivation on K() extends uniquely to a derivation on L. But in the proof of Theorem we sawthere is a non-zero derivation Z on K() that vanishes on K, and an extension of that to aderivation on L is non-zero on L and is zero on K. We have a contradiction of the uniqueness ofextensions, so in any set of field generators {

1,....,

r}, some

i must be inseparable in K.

Choose a generating set {1,....,

r} with as few inseparable elements as possible. At least one

i

is inseparable over K and we may assume that r is one of them. Set =

r and F = K(

1,....

r-1)

(so F = K if r = 1). Then L = F(). We will show by contradiction that must be inseparableover F, which is the point of the lemma.

Suppose is separable over F. Then is separable over the larger field F(p) since its minimalpolynomial over F(p) divides its minimal polynomial over F. Since is a root of Xp � p

F(p)[X], its (separable) minimal polynomial in F(p)[X] is a factor of this, so that polynomial

must be X � . Therefore, F(p). Taking pk-th powers for any k 0, k k+1p pa F(a ), so

k k+1p pF(a ) F(a ).

The reverse inclusion is obvious, so kpF(a ) =

k+1pF(a ) for all k 0. Therefore,

kp pk1 r-1 rL = F( ) = F( ) = K( ,..., , )

for any k 0. We can pick k so that kp

is separable over K (why?). Then the generating setkp

1 r 1 r{ ,..., ,a } has with one less inseparable element among the field generators. This

contradicts the choice of generators to have as few members in the list as possible that areinseparable over K, so has to be inseparable over F.

Proof: Assume L/K is separable, so by the primitive element theorem L = K() where isseparable over K. Any derivation on K can be extended (using Theorem) uniquely to a derivationon L.

If L/K is inseparable, then Lemma lets us write L = F() with inseparable over F, andF K. The, by a construction used in the proof of Theorem, f() f�() with f(X) F[X] is anonzero derivation on L which is zero on F, and thus also zero on the smaller field K. This showsthe zero derivation on K has a non-zero extension (and thus two extensions) to a derivation on L.


Proof: Again we consider the tower of field extensions (2.2). Since Li = L

i-1(

i) and

i is separable

over Li-1

, the proof shows any derivation on Li-1

extends uniquely to a derivation on Li. Therefore,

any derivation on K = L0 can be extended step-by-step through the tower (2.2) to a derivation on

Lr = L. By the argument in the proof of Lemma, this derivation on L is unique.

Lemma: Let L/K be a finite extension and F be an intermediate extension such that F/K isseparable. Then any derivation F L which sends K to K has values in F.

Proof: Pick F, so is separable over K. Now use Corollary B.10 to see the derivation F Lsends to an element of K() F.

Corollary: Theorem 3 implies Theorem 5.

Proof: To prove L/K is separable, we want to show any derivation on K has a unique extensionto a derivation on L. Since F/K is separable, a derivation on K extends to a derivation on F. SinceL/F is separable, a derivation on F extends to a derivation on L.



NotesFor uniqueness, let D1 and D

2 be derivations on L which extend the same derivation on K. Since

D1(K) K and D

2(K) K, we have D

1(F) F and D

2(F) F by Lemma. Then D

1 = D

2 on F since

F/K is separable, and D1 = D

2 on L since L/F is separable.

When L/K is an algebraic extension of possibly infinite degree, here is the way separability isdefined.

Definition: An algebraic extension L/K is called separable if every finite subextension of L=K isseparable. Equivalently, L=K is separable when every element of L is separable over K.

This definition makes no sense if L/K is not an algebraic extension since a non-algebraic extensionis not the union of its finite subextensions.

Theorem 1 has a problem in the infinite-degree case: there is no natural trace map. However, theconditions in Theorems 2 and 3 both make sense for a general L/K. (In the case of Theorem 2,

we have to drop the specification of KK L as a product of copies of K , and just leave the

statement about the tensor product having no non-zero nilpotent elements.) It is left to thereader to check for an infinite algebraic extension L/K that the conditions of Theorems 2 and 3match Definition.

The conditions in Theorems 2 and 3 both make sense if L/K is not algebraic, so they could eachpotentially be used to define separability of a completely arbitrary field extension. But there isa problem: for transcendental (that is, non-algebraic) extensions the conditions in Theorems 2and 3 are no longer equivalent. Indeed, take L = K(u), with u transcendental over K. Then

KK L = K(u) is a field, so the condition in Theorem 2 is satisfied. However, the zero derivation

on K has more than one extension to K(u): the zero derivation on K(u) and differentiation withrespect to u on K(u).

Definition: A commutative ring with no nonzero nilpotent elements is called reduced.

A domain is reduced, but a more worthwhile example is a product of domains, like F3 × Q[X],which is not a domain but is reduced.

Definition: An arbitrary field extension L/K is called separable when the ring KK L is reduced.

Using this definition, in characteristic 0 all field extensions are separable. In characteristic p, anypurely transcendental extension is separable. The condition in Theorem 3, that derivations onthe base field admit unique extensions to a larger field, characterizes not separable field extensionsin general, but separable algebraic field extensions.

A condition equivalent to that in Definition is that KF L is reduced as F runs over the finite

extensions of K.

The condition that KK L is reduced makes sense not just for field extensions L/K, but for any

commutative K-algebra. Define an arbitrary commutative K-algebra A to be separable when

the ring KK A is reduced. This condition is equivalent to A F being reduced for every finite

extension field F/K.

Example: Let A = K[X]/(f(X)) for any non-constant f(X) K[X]. The polynomial f(X) neednot be irreducible, so A might not be a field. It is a separable K-algebra precisely when f(X) is aseparable polynomial in K[X].

When [A : K] is finite, an analogue of Theorem 1 can be proved: A is a separable K-algebra if andonly if the trace pairing hx; yi = Tr

A/K(xy) from A × A to K is non-degenerate.


Abstract Algebra

Notes Traces

Let A be a finite-dimensional commutative K-algebra (with identity), such as a finite extensionfield of K or the product ring Kn or even a mixture of the two: a product of finite extensionsof K. To any a 2 A we associate the K-linear map m

: A A which is left multiplication by a:

x x:

For a; b A and K, ma+b

= ma + m

b and m

aa = am

a, so m

a is a K-linear map.

Definition: For a finite-dimensional K-algebra A, the trace of a A is the trace of ma.

That is, the trace of a is tr(ma) K, usually written as Tr

A/K(a), so Tr

A/K : A K. The trace from A

to K is K-linear, hence identically zero or surjective since K is a one-dimensional K-vector space.

Example: Since m1 is the identity function, Tr

A/K(1) = [A : K].

Example: Suppose a A is nilpotent: ar = 0 for some r 1. Then ram 0, so m

a is a nilpotent linear

transformation. Thus its eigenvalues are all 0, so TrA/K

(a) = 0.

We now consider a finite-dimensional L-algebra A with K a subfield of L such that [L : K] < . Wehave finite-dimensional algebras A/L, A/K, and L/K. The next theorem is called the transitivityof the trace.

Theorem 7: In the above notation, TrA/K

= TrL/K Tr

A/L. In particular, if a L, then Tr

A/K(a) =

[A : L]TrL/K

(a).

Proof: Let (e1; : : : ; em) be an ordered L-basis of A and (f1; : : : ; fn) be an ordered K-basis

of L. Thus as an ordered K-basis of A we can use

(e1f

1,....., e

1f

n,....., e

mf

1,....., e

mf

n):

For a A, let

m n

j ij i ij s ijrs ri 1 r 1

ae c e , c f b f ,

for cij L and b

ijrs K. Thus a(e

jf

s) = ijrs i ri r

b e f . So

[ma]

A/L = (c

ij), [m

cij ]

L/K = (b

ijrs), [m

a]

A/K = ([m

cij ]

L/K):

Thus

TrL/K

(TrA/L

(a)) = L/K iii

Tr ( c )

= L /K iii

Tr (c )

= iirri r

b

= TrA/K

(a).

Theorem 8: Let A and B be finite-dimensional K-algebras. For (a; b) in the product ring A × B,Tr

(A×B)/K(a, b) = Tr

A/K(a) + Tr

B/K(b).

Proof: Let e1,....., e

m be a K-basis of A and f

1,....., f

n be a K-basis of B. In A × B, e

i . f

j = 0. Therefore,

the matrix for multiplication by (a, b), with respect to the K-basis { eif

j }, is a block-diagonal

matrix a

b

0[m ],

0 [m ]

whose trace is TrA/K

(a) + TrB/K

(b).



NotesTheorem 9: Let A be a finite-dimensional K-algebra, L/K be a field extension, and B = LKA be

the base extension of A to an L-algebra. For a A, TrB/L

(1a) = TrA/K

(a).

Proof: Let e1,....., e

n be a K-basis of A. Write ae

j =

n

ij ii 1c e

, so the matrix for ma in this basis is (c

ij).

The tensors 1 e1,....., 1 en are an L-basis of B, and we have

(1 a) (1 ej) = 1 ae

j =

n

ij ii 1

c (1 e ),

so the matrix for m1a

on B is the same as the matrix for ma on A. Thus Tr

A/K(a) = Tr

B/L(1 a).

Remark: Because m1a

and ma have the same matrix representation, not only are their traces the

same but their characteristic polynomials are the same.

Theorem 10: Let A be a finite-dimensional K-algebra. For any field extension L/K, the baseextension by K of the trace map A K is the trace map L

K A L. That is, the function

idTrA/K

: LK A L which sends an elementary tensor x a to xTrA/K

(a) is the trace mapTr

(LKA)/L.

Proof: We want to show Tr(LKA)/L

(t) = (idTrA/K

)(t) for all t LKA. The elementary tensors

additively span LK A so it succes to check equality when t = x a for x K and a A. This means

we need to check Tr(LKA)/K

(x a) = xTrA/K

(a).

Pick a K-basis e1,..., e

n for A and write ae

j =

n

ij ii 1c e

with cij K. The elementary tensors

1 e1,..., 1 e

n are an L-basis of LK A and

n n

j j ij i ij ii 1 i 1

(x a)(1 e ) x ae c (x e ) c x(1 e )

by the definition of the L-vector space structure on LKA. So the matrix for multiplication by

x a in the basis {1 ei} is (c

ijx), which implies

n n

(L KA )/L ii A /Ki 1 i 1

Tr (x a) c x x cii xTr (a).

Derivations

A derivation is an abstraction of differentiation on polynomials. We want to work withderivations on fields, but polynomial rings will intervene, so we need to understand derivationson rings before we focus on fields.

Let R be a commutative ring and M be an R-module (e.g., M = R). A derivation on R with valuesin M is a map D : R M such that D(a + b) = D(a) + D(b) and D(ab) = aD(b) + bD(a). Easily, byinduction D(an) = nan-1D(a) for any n 1. When M = R, we will speak of a derivation on R.

Example: For any commutative ring A, differentiation with respect to X on A[X] is a derivationon A[X] (R = M = A[X]).

Example: Let R = A[X] and M = A as an R-module by f(X)a := f(0)a. Then D: R M by D(f)= f�(0) is a derivation.

Example: Let D : R R be a derivation. For iii

f(X) a X in R[X], set fD(X) = iii

D a X .

This is the application of D coefficentwise to f(X). The operation f fD is a derivation on R[X] (tocheck the product rule, it suffices to look at monomials).


Abstract Algebra

Notes If R = F2[u] and D is the usual u-derivative on F

2[u], then the polynomial f(X) = (u3 + u)X4 + uX3

+ u2X + 1 in R[X] has fD(X) = (u2 + 1)X4 + X3.

Any element of R satisfying D(a) = 0 is called a D-constant, or just a constant if the derivation isunderstood. The constants for a derivation form a subring. For instance, from the product rule,taking a = b = 1, we obtain D(1) = 0.

Example: The set of all constants for X-differentiation on K[X] is K when K has characteristic 0 andK[Xp] when K has characteristic p.

Example: If D: R R is a derivation and f fD is the corresponding derivation on R[X], its ring ofconstants is C[X], where C is the constants for D.

We will generally focus on derivations from R to R, although it will be convenient to allowR-modules as the target space for derivations in Corollary, which is used in the main text in theproofs of Theorem 3 and Lemma.

Example: Let�s check that any derivation on K[X] which has the elements of K among its constantshas the form D(f) = hf� for some h K[X]. (When h = 1, this is the usual X-derivative.)

When K is among the constants of D, D is K-linear: D(cf) = cD(f) + fD(c) = cD(f). Therefore, D isdetermined by what it does to a K-basis of K[X], such as the power functions Xn. By induction,D(Xn) = nXn-1D(X) for all n 1. Therefore, by linearity, D(f) = f�(X)D(X) for every f K[X]. Seth = D(X).

Theorem 11: Let R be a domain with fraction field K. Any derivation D: R K uniquely extends

to D : K K, given by the quotient rule: D (a/b) = (bD(a) � aD(b))/b2.

Proof: Suppose there is an extension of D to a derivation on K. Then, if x = a/b is in K (with a, b A), a = bx, so

D(a) = bD(x) + xD(b):

Therefore in K,

2

D(a) � xD(b) bD(a) � aD(b)D(x) =

b b

To see, conversely, that this formula does give a derivation D on K, first we check it is well-defined: if a/b = c/d (with b and d nonzero), then ad = bc, so

aD(d) + dD(a) = bD(c) + cD(b).

Therefore,

2 2

bD(a) � aD(b) dD(a) � cD(d)

b d =

2 2

2 2

d (bD(a) � aD(b)) � b (dD(c) � cD(d))

b d

=2 2

2 2

bd(dD(a) � bD(c)) � d aD(b) + b cD(d)

b d

=2 2

2 2

bd(cD(b) � aD(d)) � d aD(b) + b cD(d)

b d

= 2 2

(bc � ad)dD(b) � (ad � bc)bD(d)

b d

= 0 since ad = bc.

That D satisfies the sum and product rules is left to the reader to check.



NotesTheorem 12: Let L/K be a finite extension of fields, and D: K K be a derivation. Supposea L is separable over K, with minimal polynomial (X) K[X]. That is, (X) is irreducible inK[X], () = 0, and �() 0. Then D has a unique extension from K to a derivation on the fieldK(), and it is given by the rule

D(f()) = D D(a)f ( ) � f'( )

'(a)

for any f(X) K[X].

Proof: The rule (B.1) looks bizarre at first. To make it less so, we start by assuming D has anextension to K(), and prove by a direct computation that it must be given by the indicated

formula. For any K(), write = f(), where f(X) = r

i ii=0c X and c

i K. Then

D() = D(f()) = r

i i-1 Di i

i=0

(D(c ) + c (i D( ))) = f ( ) + f'( )D( ).

Taking f(X) = (X) to be the minimal polynomial of over K, f() = 0, so if D has an extension toK() then (B.2) becomes

0 = D() + �()D(),

which proves (since �() 0) that D() must be given by the formula �D()/�(). Plugging thisformula for D(), shows D() must be given by the formula. Since was a general element ofK(), this proves D has at most one extension to a derivation on K().

Now, to show the formula works, we start over and define

D(f()) : = D

D ( )f (a) � f'( ) .

'( )

We need to show this formula is well-defined.

Suppose f1() = f

2() for f

1; f

2 K[X]. Then f

1(X) f

2(X) mod (X), say

f1(X) = f

2(X) + (X)k(X)

for some k(X) K[X]. Differentiating both sides with respect to X in the usual way,

f�1(X) = f�

2(X) + (X)k�(X) + �(X)k(X).

Evaluating at X = ,

f�1() = f�

2() + �()k().

Since �() 0, we divide by �() and multiply through by �D() to get

D DD

1 2

(a) ( )�f' ( ) f ' (a) ( )k( ).

' a '( )

We want to add D1f ( ) to both sides. First, apply D to the coefficients in (B.3), which is a derivation

on K[X], to get

D D D D1 2 kf (X) f (X) (X) (X) (X)k(X).

Therefore,

D D D1 2f (a) f (a) ( )k(a).


Abstract Algebra

Notes Add this to both sides to get

D DD ' D D ' D1 2 2 2

DD '2 2

(a) (a)f ( ) f ( ) f ( ) ( )k( ) f ( ) ( )k( )

'(a) '(a)

( )f ( ) f (a) .

'( )

This proves the formula for a derivation on K() is well-defined. It is left to the reader to checkthis really is a derivation.

Example: In contrast with Theorem 12, consider K = Fp(u) and L = K() where is a root

of Xp � u K[X]. This is an inseparable irreducible polynomial over K. The u-derivative on Kdoes not have any extension to a derivation on L. Indeed, suppose the u-derivative on K has anextension to L, and call it D. Applying D to the equation ap = u gives

pp-1D() = D(u).

The left side is 0 since we�re in characteristic p. The right side is 1 since D is the u-derivative onF

p(u). This is a contradiction, so D does not exist.

Corollary: Let L/K be a finite extension of fields. For any derivation D: K L and Lwhich is separable over K, D has a unique extension to a derivation K() L. If D(K) K thenD(K()) K().

Proof: Follow the argument in the proof of Theorem 12, allowing derivations to have values inL rather than in K(). The formula for D(f()) still turns out to be the same as in (B.1). Inparticular, if D(K) K then the extension of D to a derivation on K() actually takes values inK().

Self Assessment

1. A ................... ring with no non-zero nilpotent element is called reduced.

(a) associative ring (b) commutative ring

(c) multiplicative ring (d) addition ring

2. An arbitrary field extension ................... is called separable when the ring uK L is reduced.

(a) L-1k (b) L/K

(c) K/L-1 (d) (L + K)-1

3. If L/K be a ................... extensions. Then L is separable over K. If and only if any derivationof K has a unique extension to a derivative of L.

(a) finite (b) infinite

(c) domain (d) split

4. The extension 2 2F ( u )/F /4). Since 22 2F ( u ) F x /x u, which then the non-zero

nilpotent element ...................

(a) X u (b) u X

(c) 1X u (d) 1u u x



Notes5. If D : R R is a derivatives and f F0 is the corresponding derivation on R[x] from its ringof constants in C[x], where ................... is the constant for ..................., f(x) =

11 i i

u

a x in R[x] and fd(x) D(a )X

(a) D, C (b) C, D

(c) X, C (d) C, D

27.2 Summary

Let L/K be a finite extension. Then L is separable over K if and only if any derivation of Khas a unique extension to a derivation of L.

If L = K(a1,....., a

r) and each a

i is separable over K then every element of L is separable over

K (so L/K is separable).

Let L/K be a finite extension and F be an intermediate field. If L/F and F/K are separablethen L/K is separable.

The proof of Theorem 2 implies Theorem 4.

Let L/K be an extension of fields, and L be algebraic over K. Then is separable over Kif and only if any derivation on K has a unique extension to a derivation on K().

A commutative ring with no nonzero nilpotent elements is called reduced.

A domain is reduced, but a more worthwhile example is a product of domains, likeF3 × Q[X], which is not a domain but is reduced.

An arbitrary field extension L/K is called separable when the ring KK L is reduced.

A derivation is an abstraction of differentiation on polynomials. We want to work withderivations on fields, but polynomial rings will intervene, so we need to understandderivations on rings before we focus on fields.

27.3 Keywords

Separability: Separability of a finite field extension L/K can be described in several differentways.

Commutative Ring: A commutative ring with no nonzero nilpotent elements is called reduced.

Domain: A domain is reduced, but a more worthwhile example is a product of domains, likeF3 × Q[X], which is not a domain but is reduced.

Derivation: A derivation is an abstraction of differentiation on polynomials.


1. Let R be a domain with fraction field K. Any derivation D: R K uniquely extends to

D : K K, given by the quotient rule: D (a/b) = (bD(a) � aD(b))/b2. Prove it.

2. Let L/K be a finite extension of fields, and D: K K be a derivation. Supposea L is separable over K, with minimal polynomial (X) K[X]. That is, (X) is irreduciblein K[X], () = 0, and �() 0. Then D has a unique extension from K to a derivation on thefield K(), and it is given by the rule.


Abstract Algebra

Notes Answers: Self Assessment

1. (b) 2. (b) 3. (a) 4. (a) 5. (b)






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Unit 28: Galois Theory

NotesUnit 28: Galois Theory

CONTENTS

Objectives

Introduction

28.1 Galois Theory

28.2 Repeated Roots

28.3 The Fundamental Theorem

28.4 Summary

28.5 Keywords



Objectives


Discuss Galois theory

Describe repeated roots

Introduction

In the last unit, you have studied about extension field. This unit will provide informationrelated to Galois theory.

28.1 Galois Theory

This gives the definition of the Galois group and some results that follow immediately from thedefinition. We can give the full story for Galois groups of finite fields.

We use the notation Aut(F) for the group of all automorphisms of F, that is, all one-to-onefunctions from F onto F that preserve addition and multiplication. The smallest subfield containingthe identity element 1 is called the prime subfield of F. If F has characteristic zero, then its primesubfield is isomorphic to Q, and if F has characteristic p, for some prime number p, then itsprime subfield is isomorphic to Z

p. In either case, for any automorphisms of F we must have

(x) = x for all elements in the prime subfield of F.

To study solvability by radicals of a polynomial equation f(x) = 0, we let K be the field generatedby the coefficients of f(x), and let F be a splitting field for f(x) over K. Galois consideredpermutations of the roots that leave the coefficient field fixed. The modern approach is toconsider the automorphism determined by these permutations. The first result is that if F is anextension field of K, then the set of all automorphism : F F such that (a) = a for all a K isa group under composition of functions. This justifies the following definitions.


Abstract Algebra

Notes Definition: Let F be an extension field of K. The set

{ Aut(F) | (a) = a for all a K }

is called the Galois group of F over K, denoted by Gal(F/K).

Definition: Let K be a field, let f(x) K[x], and let F be a splitting field for f(x) over K. ThenGal(F/K) is called the Galois group of f(x) over K, or the Galois group of the equation f(x) = 0over K.

Proposition states that if F is an extension field of K, and f(x) K[x], then any element ofGal(F/K) defines a permutation of the roots of f(x) that lie in F. The next theorem is extremelyimportant.

Theorem 1: Let K be a field, let f(x) K[x] have positive degree, and let F be a splitting field forf(x) over K. If no irreducible factor of f(x) has repeated roots, then j Gal(F=K)j = [F : K].

This result can be used to compute the Galois group of any finite extension of any finite field, butfirst we need to review the structure of finite fields. If F is a finite field of characteristic p, then itis a vector space over its prime subfield Z

p, and so it has pn elements, where [F : Z

p] = n. The

structure of F is determined by the following theorem.

Theorem 2: If F is a finite field with pn elements, then F is the splitting field of the polynomialnpx � x over the prime subfield of F.

The description of the splitting field of npx � x over Z

p shows that for each prime p and each

positive integer n, there exists a field with pn elements. The uniqueness of splitting fields showsthat two finite fields are isomorphic iff they have the same number of elements. The field withpn elements is called the Galois field of order pn, denoted by GF(pn). Every finite field is a simpleextension of its prime subfield, since the multiplicative group of nonzero elements is cyclic, andthis implies that for each positive integer n there exists an irreducible polynomial of degree n inZ

p[x].

If F is a field of characteristic p, and n Z+, then {a F | npa = a} is a subfield of F, and this

observation actually produces all subfields. In fact, Proposition 6.5.5 has the following statement:Let F be a field with pn elements. Each subfield of F has pm elements for some divisor m of n.

Conversely, for each positive divisor m of n there exists a unique subfield of F with pm elements.If F is a field of characteristic p, consider the function : F F defined by (x) = xp. Since F hascharacteristic p, we have (a + b) = (a + b)p = ap + bp = (a) + (b), because in the binomial expansionof (a + b)p each coefficient except those of ap and bp is zero. (The coefficient (p!)/(k!(p � k)!)contains p in the numerator but not the denominator since p is prime, and so it must be equal tozero in a field of characteristic p.) It is clear that preserves products, and so is a ringhomomorphism. Furthermore, since it is not the zero mapping, it must be one-to-one. If F isfinite, then must also be onto, and so in this case is called the Frobenius automorphism of F.

Note that n(x) = npx (Inductively, n(x) = (n-1(x))p =

n-1p(x ) p = npx .) Using an appropriate power

of the Frobenius automorphism, we can prove that the Galois group of any finite field must becyclic.

Theorem 3: Let K be a finite field and let F be an extension of K with [F : K] = m. ThenGal(F/K) is a cyclic group of order m.

Outline of the proof: We start with the observation that F has pn elements, for some positive

integer n. Then K has pr elements, for r = n/m, and F is the splitting field of npx x over its

prime subfield, and hence over K. Since f(x) has no repeated roots, to conclude that |Gal(F/K)|= m. Now define : F F to be the rth power of the Frobenius automorphism. That is, define



Notesrp(x) = x . To compute the order of in Gal(F/K), first note that m is the identity since m(x) =rm np px = x x for all x F. But cannot have lower degree, since this would give a polynomial

with too many roots. It follows that is a generator for Gal(F/K).

28.2 Repeated Roots

In computing the Galois group of a polynomial, it is important to know whether or not it hasrepeated roots. A field F is called perfect if no irreducible polynomial over F has repeated roots.This section includes the results that any field of characteristic zero is perfect, and that any finitefield is perfect.

In the previous section, we showed that the order of the Galois group of a polynomial with norepeated roots is equal to the degree of its splitting field over the base field. The first thing in thissection is to develop methods to determine whether or not a polynomial has repeated roots.

Let f(x) be a polynomial in K[x], and let F be a splitting field for f(x) over K. If f(x) has the

factorization f(x) = t1 mm1 t(x � r ) ... (x � r ) over F, then we say that the root r

i has multiplicity m

i.

If mi = 1, then r

i is called a simple root.

Let f(x) K[x], with t k

kk=0f(x) = a x . The formal derivative f�(x) of f(x) is defined by the formula

t k-1kk=0

f'(x) = ka x , where kak denotes the sum of a

k added to itself k times. It is not difficult to

show from this definition that the standard differentiation formulas hold. Proposition showsthat the polynomial f(x) K[x] has no multiple roots iff it is relatively prime to its formalderivative f�(x). Proposition shows that f(x) has no multiple roots unless char(K) = p 0 and f(x)has the form f(x) = a

0 + a

1xp + a

2x2p + ... + a

nxnp.

A polynomial f(x) over the field K is called separable if its irreducible factors have only simpleroots. An algebraic extension field F of K is called separable over K if the minimal polynomialof each element of F is separable. The field F is called perfect if every polynomial over F isseparable.

Theorem states that any field of characteristic zero is perfect, and a field of characteristicp > 0 is perfect if and only if each of its elements has a pth root in the field. It follows immediatelyfrom the theorem that any finite field is perfect.

To give an example of a field that is not perfect, let p be a prime number, and let K = Zp. Then in

the field K(x) of rational functions over K, the element x has no pth root. Therefore, this rationalfunction field is not perfect.

The extension field F of K is called a simple extension if there exists an element u F such thatF = K(u). In this case, u is called a primitive element. Note that if F is a finite field, then Theoremshows that the multiplicative group Fx is cyclic. If the generator of this group is a, then it is easyto see that F = K(a) for any subfield K. Theorem shows that any finite separable extension is asimple extension.

28.3 The Fundamental Theorem

Here we study the connection between subgroups of Gal(F/K) and fields between K and F. Thisis a critical step in proving that a polynomial is solvable by radicals if and only if its Galoisgroup is solvable.


Abstract Algebra

Notes Let F be a field, and let G be a subgroup of Aut(F). Then

{a F | (a) = a for all G}

is called the G-fixed subfield of F, or the G-invariant subfield of F, and is denoted by FG.(Proposition shows that FG is actually a subfield of F.) If F is the splitting field over K of aseparable polynomial and G = Gal(F/K), then Proposition shows that FG = K. Artin�s lemmaprovides the first really significant result of the section. It states that if G is a finite group ofautomorphism of the field F, and K = FG, then [F : K] |G|.

Let F be an algebraic extension of the field K. Then F is said to be a normal extension of K if everyirreducible polynomial in K[x] that contains a root in F is a product of linear factors in F[x]. Withthis definition, the following theorem and its corollary can be proved from previous results.

Theorem 4: The following are equivalent for an extension field F of K:

(1) F is the splitting field over K of a separable polynomial;

(2) K = FG for some finite group G of automorphism of F;

(3) F is a finite, normal, separable extension of K.

As a corollary, we obtain the fact that if F is an extension field of K such that K = FG for some finitegroup G of automorphisms of F, then G = Gal(F/K).

The next theorem is the centerpiece of Galois theory. In the context of the fundamental theorem,we say that two intermediate subfields E

1 and E

2 are conjugate if there exists Gal(F/K) such

that (E1) = E

2. Proposition states that if F is the splitting field of a separable polynomial over K,

and K E F, with H = Gal(F/E), then Gal(F/(E)) = H-1, for any Gal(F/K).

Theorem 5 (The Fundamental Theorem of Galois Theory): Let F be the splitting field of aseparable polynomial over the field K, and let G = Gal(F/K).

(a) There is a one-to-one order-reversing correspondence between subgroups of G and subfieldsof F that contain K:

(i) The subfield FH corresponds to the subgroup H, and H = Gal(F/FH).

(ii) If K E F, then the corresponding subgroup is Gal(F/E), and E = FGal(F/E).

(b) [F : FH] = |H| and [FH : K] = [G : H], for any subgroup H of G.

(c) Under the above correspondence, the subgroup H is normal iff FH is a normal extensionof K. In this case, Gal(E/K) Gal(F/K) / Gal(F/E).

For example, suppose that F is a finite field of characteristic p, and has pm elements. Then[F : GF(p)] = m, and so G = Gal(F= GF(p)) is a cyclic group of degree m by Corollary. Since G iscyclic, the subgroups of G are in one-to-one correspondence with the positive divisors of m.Proposition shows that the subfields of F are also in one-to-one correspondence withthe positive divisors of m. Remember that the smaller the subfield, the more automorphismswill leave it invariant. By the Fundamental Theorem of Galois Theory, a subfield E with[E : GF(p)] = k corresponds to the cyclic subgroup with index k, not to the cyclic subgroup oforder k.

Self Assessment

1. If F has characteristics zero, then its prime subfield is isomorphic to Q and if F hascharacteristics P, for some prime number P, then its prime subfield is ................ to Zp.

(a) homomorphic (b) isomorphic

(c) automorphism (d) polynomial



Notes2. Let F be extension field of K. The set { Q Aut(F) | Q(a) = a for all a K } is Galois groupis denoted by ................

(a) Gal(F/K) (b) Gal(u/F)

(c) Gal-1(K/F) (d) Gal(k × F)

3. Let K be a finite field and let F be an extension of K with [F : k] = m. Then Gal(F/k) is a................ group of order m.

(a) cyclic (b) polynomial

(c) permutation (d) finite

4. A polynomial f(x) over the field k is called ................ if its irreducible factors have onlysimple roots.

(a) spittery field (b) extension field

(c) separable (d) finite field

5. The ................ F of K is called simple extensions. If then exist an element u F. Such thatF = K(u).

(a) finite field (b) extension field

(c) separable field (d) spliting field

28.4 Summary

Let F be an extension field of K. The set

{ Aut(F) | (a) = a for all a K }


Let K be a field, let f(x) K[x], and let F be a splitting field for f(x) over K. Then Gal(F/K)is called the Galois group of f(x) over K, or the Galois group of the equation f(x) = 0 overK.

It states that if F is an extension field of K, and f(x) K[x], then any element of Gal(F/K)defines a permutation of the roots of f(x) that lie in F. The next theorem is extremelyimportant.

Let K be a field, let f(x) K[x] have positive degree, and let F be a splitting field for f(x)over K. If no irreducible factor of f(x) has repeated roots, then j Gal(F=K)j = [F : K].

This result can be used to compute the Galois group of any finite extension of any finitefield, but first we need to review the structure of finite fields. If F is a finite field ofcharacteristic p, then it is a vector space over its prime subfield Z

p, and so it has pn elements,

where [F : Zp] = n. The structure of F is determined by the following theorem.

If F is a finite field with pn elements, then F is the splitting field of the polynomial npx � x

over the prime subfield of F.

Let K be a finite field and let F be an extension of K with [F : K] = m. Then Gal(F/K) is acyclic group of order m.

(The fundamental theorem of Galois theory) Let F be the splitting field of a separablepolynomial over the field K, and let G = Gal(F/K).


Abstract Algebra

Notes 28.5 Keywords

Prime Subfield: If F is a finite field with pn elements, then F is the splitting field of the polynomialnpx � x over the prime subfield of F.

The Fundamental Theorem of Galois Theory: Let F be the splitting field of a separable polynomialover the field K, and let G = Gal(F/K).


1. Determine the group of all automorphisms of a field with 4 elements.

2. Let F be the splitting field in C of x4 + 1.

(a) Show that [F : Q] = 4.

(b) Find automorphisms of F that have fixed fields Q( 2 ), Q(i), and Q( 2 i),respectively.

3. Find the Galois group over Q of the polynomial x4 + 4.

4. Find the Galois groups of x3 � 2 over the fields Z5 and Z

11.

5. Find the Galois group of x4 � 1 over the field Z7.

6. Find the Galois group of x3 � 2 over the field Z7.

7. Let f(x) 2 Q[x] be irreducible over Q, and let F be the splitting field for f(x) over Q. If [F : Q]is odd, prove that all of the roots of f(x) are real.

8. Find an element with Q( 2 , i) = Q().

9. Find the Galois group of x6 � 1 over Z7.

10. Prove that if F is a field and K = FG for a finite group G of automorphisms of F, then thereare only finitely many subfields between F and K.

11. Let F be the splitting field over K of a separable polynomial. Prove that if Gal(F/K) iscyclic, then for each divisor d of [F : K] there is exactly one field E with K E F and[E : K] = d.

12. Let F be a finite, normal extension of Q for which |Gal(F=Q)| = 8 and each element ofGal(F/Q) has order 2. Find the number of subfields of F that have degree 4 over Q.

13. Let F be a finite, normal, separable extension of the field K. Suppose that the Galois groupGal(F/K) is isomorphic to D

7. Find the number of distinct subfields between F and K. How

many of these are normal extensions of K?

14. Show that F = Q(i, 2 ) is normal over Q; find its Galois group over Q, and find allintermediate fields between Q and F.

15. Let F = Q( 2 , 3 2 ). Find [F : Q] and prove that F is not normal over Q.

16. Find the order of the Galois group of x5 � 2 over Q.


1. (b) 2. (a) 3. (a) 4. (c) 5. (b)








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Abstract Algebra

Notes Unit 29: Computing Galois Groups

CONTENTS

Objectives

Introduction

29.1 Transitively Group

29.2 Separable Polynomial

29.3 Summary

29.4 Keywords



Objectives


Discuss transitively and transitive group

Explain computing Galois group

Introduction

In the last unit, you have studied about Galois theory. In this unit, you will get informationrelated to computing the Galois groups.

29.1 Transitively Group

Definition: Let G be a group acting on a set S. We say that G acts transitively on S if for each pairof elements x,y in S there exist an element g in G such that y = gx.

If G is a subgroup of the symmetric group Sn, then G is called a transitive group if it acts

transitively on the set { 1, 2, ... , n }.

29.2 Separable Polynomial

Proposition: Let f(x) be a separable polynomial over the field K, with roots r1 , ... , r

n in its

splitting field F. Then f(x) is irreducible over K if and only if Gal(F/K) acts transitively on theroots of f(x).

Lemma: Let p be a prime number, and let G be a transitive subgroup of Sp. Then any nontrivial

normal subgroup of G is also transitive.

Lemma: Let p be a prime number, and let G be a solvable, transitive subgroup of Sp. Then G

contains a cycle of length p.

Proposition: Let p be a prime number, and let G be a solvable, transitive subgroup of Sp. Then G

is a subgroup of the normalizer in Sp of a cyclic subgroup of order p.


Unit 29: Computing Galois Groups

NotesLet f(x) be a polynomial of degree n over the field K, and assume that f(x) has roots r1, r

2, ... , r

n in

its splitting field F. The element of F defined by

= (ri - r

j)2,

where the product is taken over all i, j with 1 i < j n, is called the discriminant of f(x).

It can be shown that the discriminant of any polynomial f(x) can be expressed as a polynomial inthe coefficients of f(x), with integer coefficients. This requires use of elementary symmetricfunctions, and lies beyond the scope of what we have chosen to cover in the book.

We have the following properties of the discriminant:

(i) 0 if and only if f(x) has distinct roots;

(ii) belongs to K;

(iii) If 0, then a permutation in Sn is even if and only if it leaves unchanged the sign of

i j1 i < j n(r - r )

Proposition: Let f(x) be a separable polynomial over the field K, with discriminant , and let F beits splitting field over K. Then every permutation in Gal(F/K) is even if and only if is the squareof some element in K.

We now restrict our attention to polynomials with rational coefficients. The next lemma showsthat in computing Galois groups it is enough to consider polynomials with integer coefficients.Then a powerful technique is to reduce the integer coefficients modulo a prime and consider theGalois group of the reduced equation over the field GF(p).

Lemma: Let f(x) = xn + an-1

xn-1 + · · · + a1 x + a

0 be a polynomial in Q[x], and assume that

ai = b

i / d for d, b

0, b

1, ... , b

n-1 in Z.

Then dn f(x/d) is monic with integer coefficients, and has the same splitting field over Q as f(x).

If p is a prime number, we have the natural mapping : Z[x] > Zp[x] which reduces each coefficient

modulo p. We will use the notation p(f(x)) = fp(x).

Theorem [Dedekind]: Let f(x) be a monic polynomial of degree n, with integer coefficients andGalois group G over Q, and let p be a prime such that f

p(x) has distinct roots. If f

p(x) factors in Z

p[x]

as a product of irreducible factors of degrees n1, n

2, ... , n

k, then G contains a permutation with the

cycle decomposition

(1,2, ... ,n1) (n

1+1, n

1+2, ... , n

1+n

2) · · · (n-n

k+1, ... ,n),

relative to a suitable ordering of the roots.

Self Assessment

1. IF G is a .................. of symmetric group sn then G is called transitive group. If it actstransitively on Set {1, 2, 3, n}

(a) sub group (b) cyclic group

(c) permutation group (d) finite group

2. Let P be a prime number and let G be a transitive subgroup of Sp. Then any ..................normal subgroup of G is also transitive.




Abstract Algebra

Notes 3. Let P be a prime number and G be a solvable, transitive subgroup of Sp. Then G is a

subgroup of the normalizer in Sp of a cyclic subgroup of order ..................

(a) P (b) G

(c) Sp

(d) S

4. If f(x) be a polynomial of degree n over the field k and assume that f(x) has roots r1, r

2,...r

n

in its splitting field F. Then element of F defined by

(a) = (r1 � r

3)2 (b) = 2(r

1 � r

j)2

(c) = (ri � r

j)-2 (d) = 3(r

1 � r

j)3

29.3 Summary

Let f(x) be a separable polynomial over the field K, with roots r1 , ... , r

n in its splitting field

F. Then f(x) is irreducible over K if and only if Gal(F/K) acts transitively on the roots off(x).

Let p be a prime number, and let G be a transitive subgroup of Sp. Then any non-trivial

normal subgroup of G is also transitive.

Let p be a prime number, and let G be a solvable, transitive subgroup of Sp. Then G

contains a cycle of length p.

Let p be a prime number, and let G be a solvable, transitive subgroup of Sp. Then G is a

subgroup of the normalizer in Sp of a cyclic subgroup of order p.

Let f(x) = xn + an-1

xn-1 + · · · + a1 x + a

0 be a polynomial in Q[x], and assume that

ai = b

i / d for d, b

0, b

1, ... , b

n-1 in Z.

Then dn f(x/d) is monic with integer coefficients, and has the same splitting field over Q asf(x).

If p is a prime number, we have the natural mapping : Z[x] > Zp[x] which reduces each

coefficient modulo p. We will use the notation p(f(x)) = fp(x).

Let f(x) be a monic polynomial of degree n, with integer coefficients and Galois group Gover Q, and let p be a prime such that f

p(x) has distinct roots. If f

p(x) factors in Z

p[x] as a

product of irreducible factors of degrees n1, n

2, ... , n

k, then G contains a permutation with

the cycle decomposition

(1,2, ... ,n1) (n

1+1, n

1+2, ... , n

1+n

2) · · · (n-n

k+1, ... ,n),

relative to a suitable ordering of the roots.

29.4 Keywords

Transitive Group: If G is a subgroup of the symmetric group Sn, then G is called a transitive

group if it acts transitively on the set { 1, 2, ... , n }.

Separable Polynomial: Let f(x) be a separable polynomial over the field K, with roots r1 , ... , r

n

in its splitting field F. Then f(x) is irreducible over K if and only if Gal(F/K) acts transitively onthe roots of f(x).


Unit 29: Computing Galois Groups


1. Give the order and describe a generator of the Galois group of GF (729) over GF(9).

2. Determine the Galois group of each of the following polynomials in Q[x]; hence, determinethe solvability of each of the polynomials

(a) x5 � 12x2 + 2 (b) x5 � 4x4 + 2x + 2

(c) x3 � 5 (d) x4 � x2 � 6

(e) x5 + 1 (f) (x2 � 2) (x2 + 2)

(g) x8 � 1 (h) x8 + 1

(i) x4 � 3x2 � 10

3. Find a primitive element in the splitting field of each of the following polynomials inQ[x].

(a) x4 � 1 (b) x4 � 2x2 � 15

(c) x4 � 8x2 + 15 (d) x3 � 2

4. Prove that the Galois group of an irreducible quadratic polynomial is isomorphic to Z2.

5. Prove that the Galois group of an irreducible cubic polynomial is isomorphic to S3 or Z

3.


1. (a) 2. (b) 3. (a) 4. (a)






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Abstract Algebra

Notes Unit 30: Invariant Subfield

CONTENTS

Objectives

Introduction

30.1 G-invariant Subfield

30.2 Summary

30.3 Keywords



Objectives


Define G-invariant subfield

Discuss examples related to subfield

Introduction

In the last unit, you have studied about computing Galois theory and groups. In this unit, youwill get information related to fundamental theorem.

30.1 G-invariant Subfield

Proposition: Let F be a field, and let G be a subgroup of Aut(F). Then

{ a in F | (a) = a for all in G }

is a subfield of F.

Definition: Let F be a field, and let G be a subgroup of Aut (F). Then

{ a in F | (a) = a for all in G }

is called the G-fixed subfield of F, or the G-invariant subfield of F, and is denoted by FG.

Proposition: If F is the splitting field over K of a separable polynomial and G = Gal(F/K), thenFG = K.

Lemma [Artin]: Let G be a finite group of automorphisms of the field F, and let K = FG. Then

[F : K] | G |.

Let F be an algebraic extension of the field K. Then F is said to be a normal extension of K if everyirreducible polynomial in K[x] that contains a root in F is a product of linear factors in F[x].


Unit 30: Invariant Subfield

NotesThe following conditions are equivalent for an extension field F of K:


(2) K = FG for some finite group G of automorphisms of F;


If F is an extension field of K such that K = FG for some finite group G of automorphisms of F,then G = Gal(F/K).

Example: The Galois group of GF(pn) over GF(p) is cyclic of order n, generated by theautomorphism defined by (x) = xp, for all x in GF(pn). This automorphism is usually known asthe Frobenius automorphism of GF(pn).

Let F be the splitting field of a separable polynomial over the field K, and let G = Gal(F/K).

(a) There is a one-to-one order-reversing correspondence between subgroups of G and subfieldsof F that contain K:

(i) If H is a subgroup of G, then the corresponding subfield is FH, and

H = Gal(F/FH).

(ii) If E is a subfield of F that contains K, then the corresponding subgroup of G isH = Gal(F/E), and

E = FH.

(b) For any subgroup H of G, we have

[F : FH] = | H| and [FH : K] = [G : H].

(c) Under the above correspondence, the subgroup H is normal if and only if the subfieldE = FH is a normal extension of K. In this case,

Gal(E/K) Gal(F/K)/Gal(F/E).

In the statement of the fundamental theorem we could have simply said that normal subgroupscorrespond to normal extensions. In the proof we noted that if E is a normal extension of K, then(E) E for all in Gal(F/K). In the context of the fundamental theorem, we say that twointermediate subfields E

1 and E

2 are conjugate if there exists in Gal(F/K) such that (E

1) = E

2.

The next result shows that the subfields conjugate to an intermediate subfield E correspond tothe subgroups conjugate to Gal(F/E). Thus E is a normal extension if and only if it is conjugateonly to itself.

Let F be the splitting field of a separable polynomial over the field K, and let E be a subfield suchthat K E F, with H = Gal(F/E). If is in Gal(F/K), then

Gal(F/(E)) = H -1.

[Fundamental Theorem of Algebra]: Any polynomial in C[x] has a root in C.

Example: Prove that if F is a field extension of K and K = FG for a finite group G ofautomorphisms of F, then there are only finitely many subfields between F and K.

Solution: The given condition is equivalent to the condition that F is the splitting field over K ofa separable polynomial. Since we must have G = Gal (F/K), the fundamental theorem of Galoistheory implies that the subfields between F and K are in one-to-one correspondence with thesubgroups of F. Because G is a finite group, it has only finitely many subgroups.


Abstract Algebra

Notes

Example: Let F be the splitting field over K of a separable polynomial. Prove that ifGal (F/K) is cyclic, then for each divisor d of [F:K] there is exactly one field E with K E F and[E:K] = d.

Solution: By assumption we are in the situation of the fundamental theorem of Galois theory, sothat there is a one-to-one order-reversing correspondence between subfields of F that contain Kand subgroups of G = Gal (F/K). Because G is cyclic of order [F:K], there is a one-to-onecorrespondence between subgroups of G and divisors of [F:K]. Thus for each divisor d of [F:K]there is a unique subgroup H of index d. By the fundamental theorem, [FH: K] = [G:H], and so E= F^H is the unique subfield with [E:K] = d.

Comment: Pay careful attention to the fact that the correspondence between subfields andsubgroups reverses the order

Example: Let F be a finite, normal extension of Q for which | Gal (F/Q) | = 8 and eachelement of Gal (F/Q) has order 2. Find the number of subfields of F that have degree 4 over Q.

Solution: Since F has characteristic zero, the extension is automatically separable, and so thefundamental theorem of Galois theory can be applied. Any subfield E of F must contain Q, itsprime subfield, and then [E:Q] = 4 iff [F:E] = 2, since [F:Q] = 8. Thus the subfields of F that havedegree 4 over Q correspond to the subgroups of Gal (F/Q) that have order 2. Because eachnontrivial element has order 2 there are precisely 7 such subgroups.

Example: Let F be a finite, normal, separable extension of the field K. Suppose that theGalois group Gal (F/K) is isomorphic to D

7. Find the number of distinct subfields between F and

K. How many of these are normal extensions of K?

Solution: The fundamental theorem of Galois theory converts this question into the question ofenumerating the subgroups of D

7, and determining which are normal. If we use the usual

description of D7 via generators a of order 7 and b of order 2, with ba = a -1 b, then a generates a

subgroup of order 7, while each element of the form ai b generates a subgroup of order 2, for0 i < 7. Thus there are 8 proper nontrivial subgroups of D

7, and the only one that is normal is

< a >, since it has |D7| / 2 elements. As you should recall from the description of the conjugacy

classes of D7 conjugating one of the 2-element subgroups by a produces a different subgroup,

showing that none of them are normal.

Example: Show that F = Q ( 2 ,i) is normal over Q; find its Galois group over Q, and findall intermediate fields between Q and F.

Solution: It is clear that F is the splitting field over Q of the polynomial (x2 + 1)(x2 � 2), and thispolynomial is certainly separable. Thus, F is a normal extension of Q.

It follows that the Galois group is isomorphic to Z2 × Z

2. Since the Galois group has 3 proper

nontrivial subgroups, there will be 3 intermediate subfields E with Q E F.

The existence of 3 nontrivial elements begins with the splitting field of x4+1 over Q.

Comment: Recall that Z7 is the splitting field of x7 � x = x(x6 � 1).

Self Assessment

1. If F is field and G be a subgroup of Aut(F). Then {a in F | Q(a) = aQ in G} is called ...............of F.


Unit 30: Invariant Subfield

Notes(a) G-invariant subfield (b) variant subfield

(c) finite field (d) domain subfield

2. G-invariant subfield is denoted by ...............

(a) GF (b) FG

(c) F.G (d) GF-1

3. G be a finite group of automorphisms of the field F and let K = FG then [F : K] ...............|G|.

(a) (b)

(c) = (d)

4. For any subgroup H of G, we have ............... and [FH : K] = [G : H]

(a) [F : FH] = |H| (b) [HF : F] |H|

(c) [F�1 : H] = |H| (d) [F�1 : H�1] |H|

5. Let F be an algebraic extensions of the field K. Then F is said to be a ............... of K. If everyirreducible polynomial in K[x] that contains a root in F is a product of linear factors in F[x]

(a) normal extension (b) finite extension

(c) infinite extension (d) subgroup extension

30.2 Summary

The following conditions are equivalent for an extension field F of K:


(2) K = FG for some finite group G of automorphisms of F;


If F is an extension field of K such that K = FG for some finite group G of automorphismsof F, then G = Gal(F/K).

Let F be the splitting field of a separable polynomial over the field K, and let G = Gal(F/K).

(a) There is a one-to-one order-reversing correspondence between subgroups of G andsubfields of F that contain K:

(i) If H is a subgroup of G, then the corresponding subfield is FH, and

H = Gal(F/FH).

(ii) If E is a subfield of F that contains K, then the corresponding subgroup of G isH = Gal(F/E), and

E = FH.

(b) For any subgroup H of G, we have

[F : FH] = | H| and [FH : K] = [G : H].

(c) Under the above correspondence, the subgroup H is normal if and only if the subfieldE = FH is a normal extension of K. In this case,

Gal(E/K) Gal(F/K) / Gal(F/E).


Abstract Algebra

Notes Let F be the splitting field of a separable polynomial over the field K, and let E be asubfield such that K E F, with H = Gal(F/E). If is in Gal(F/K), then

Gal(F/(E)) = H -1.

[Fundamental Theorem of Algebra] Any polynomial in C[x] has a root in C.

30.3 Keywords

Normal Extension: Let F be an algebraic extension of the field K. Then F is said to be a normalextension of K if every irreducible polynomial in K[x] that contains a root in F is a product oflinear factors in F[x].

Frobenius Automorphism: The Galois group of GF(pn) over GF(p) is cyclic of order n, generatedby the automorphism defined by (x) = xp, for all x in GF(pn). This automorphism is usuallyknown as the Frobenius automorphism of GF(pn).


1. Compute each of the following Galois groups. Which of these field extensions are normalfield extensions? If the extension is not normal, find a normal extension of Q in which theextension field is contained.

(a) G(Q( 30)/Q) (b) 4G(Q( 5)/Q)

(c) G(Q( 2 , 3 , 5)/Q) (d) 3G(Q( 2 , 2 ,i)/Q)

(e) G(Q( 6 ,i)/Q)

2. Let F K E be field. If E is a normal extension of F, show that E must also be a normalextension of K.

3. Let G be the Galois group of a polynomial of degree n. Prove that |G| divides n!.

4. Let F E. If f(x) is solvable over F, show that f(x) is also solvable over E.


1. (a) 2. (b) 3. (b) 4. (a) 5. (a)






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Unit 31: The Galois Group of a Polynomial

NotesUnit 31: The Galois Group of a Polynomial

CONTENTS

Objectives

Introduction

31.1 Galois Group of Polynomial

31.2 Fundamental Theorem of Symmetric Functions

31.3 Symmetric Polynomial

31.4 Constructible Polygon

31.5 Connection to Pascal�s Triangle

31.6 Summary

31.7 Keywords



Objectives


Discuss the Galois group of polynomial

Explain the theorem of Galois theory

Introduction

To study solvability by radicals of a polynomial equation f(x) = 0, we let K be the field generatedby the coefficients of f(x), and let F be a splitting field for f(x) over K. Galois consideredpermutations of the roots that leave the coefficient field fixed. The modern approach is toconsider the automorphisms determined by these permutations. We note that any automorphismof a field F must leave its prime subfield fixed.

31.1 Galois Group of Polynomial

Proposition: Let F be an extension field of K. The set of all automorphisms : F > F such that (a)= a for all a in K is a group under composition of functions.

Definition: Let F be an extension field of K. The set

{ in Aut(F) | (a) = a for all a in K }


Definition: Let K be a field, let f(x) be a polynomial in K[x], and let F be a splitting field for f(x)over K. Then Gal(F/K) is called the Galois group of f(x) over K, or the Galois group of theequation f(x) = 0 over K.


Abstract Algebra

Notes Proposition: Let F be an extension field of K, and let f(x) be a polynomial in K[x]. Then anyelement of Gal(F/K) defines a permutation of the roots of f(x) that lie in F.

Let f(x) be a polynomial in K[x] with no repeated roots and let F be a splitting field for f(x) overK. If : K > L is a field isomorphism that maps f(x) to g(x) in L[x] and E is a splitting field for g(x)over L, then there exist exactly [F:K] isomorphisms : F -> E such that (a) = (a) for all a in K.

Theorem: Let K be a field, let f(x) be a polynomial in K[x], and let F be a splitting field for f(x)over K. If f(x) has no repeated roots, then |Gal(F/K)| = [F:K].

Corollary: Let K be a finite field and let F be an extension of K with [F:K] = m. Then Gal(F/K) is a cyclic group of order m.

If we take K = Zp, where p is a prime number, and F is an extension of degree m, then the

generator of the cyclic group Gal(F/K) is the automorphism : F -> F defined by (x) = xp, for allx in F. This automorphism is called the Frobenius automorphism of F.

A symmetric function on n variables x1,..., x

n is a function that is unchanged by any permutation

of its variables. In most contexts, the term �symmetric function� refers to a polynomial on nvariables with this feature (more properly called a �symmetric polynomial�). Another type ofsymmetric functions is symmetric rational functions, which are the rational functions that areunchanged by permutation of variables.

The symmetric polynomials (respectively, symmetric rational functions) can be expressed aspolynomials (respectively, rational functions) in the elementary symmetric polynomials. Thisis called the fundamental theorem of symmetric functions.

A function f(x) is sometimes said to be symmetric about the y-axis if f(�x) = f(x). Examples of suchfunctions include |x| (the absolute value) and x2 (the parabola).

31.2 Fundamental Theorem of Symmetric Functions

Any symmetric polynomial (respectively, symmetric rational function) can be expressed as apolynomial (respectively, rational function) in the elementary symmetric polynomials on thosevariables.

There is a generalization of this theorem to polynomial invariants of permutation groups G,which states that any polynomial invariant f R [X

1,... X

n] can be represented as a finite linear

combination of special G-invariant orbit polynomials with symmetric functions as coefficients,i.e.,

1 1 n Gr special

f p ( ,..., ) orbit (t),

where p1 R [X

1, ..., X

n],

and 1, ...,

n are elementary symmetric functions, and t = 1e

1X , ..., nenX are special terms.

Furthermore, any special term t has a total degree n(n � 1)/2, and a maximal variable degree n � 1.

31.3 Symmetric Polynomial

A symmetric polynomial on n variables x1,..., x

n (also called a totally symmetric polynomial) is

a function that is unchanged by any permutation of its variables. In other words, the symmetricpolynomials satisfy

f(y1, y

2, ..., y

n) = f(x

1, x

2,..., x

n), ...(1)

where yi = x

(i) and being an arbitrary permutation of the indices 1, 2, ..., n.



NotesFor fixed n, the set of all symmetric polynomials in n variables forms an algebra ofdimension n. The coefficients of a univariate polynomial f(x) of degree n are algebraicallyindependent symmetric polynomials in the roots of f, and thus form a basis for the set of all suchsymmetric polynomials.

There are four common homogeneous bases for the symmetric polynomials, each of which isindexed by a partition (Dumitriu et al., 2004). Letting l be the length of , the elementaryfunctions e

, complete homogeneous functions h

, and power-sum functions p

are defined for

l = 1 by

e1

=1 1

1 2 1

j jj j ... j

x ...x

...(2)

h1

=1

njm

m ... mn l1 j 1

x

...(3)

p1

=n

j 1

x .

...(4)

and for l > 1 by

s =

i

l

i 1

s

...(5)

where s is one of e, h or p. In addition, the monomial functions m are defined as

m = 1 2 m

(1) s (2 ) (m )S

x x ... x ,

...(6)

where S is the set of permutations giving distinct terms in the sum and is considered to be

infinite.

As several different abbreviations and conventions are in common use, care must be taken whendetermining which symmetric polynomial is in use.

The elementary symmetric polynomials k (x

1, ..., x

n) (sometimes denoted

k or e

) on n variables

{x1, ..., x

n} are defined by

1(x

1, ..., x

n) = i

1 i n

x

...(7)

2(x

1, ..., x

n) = i j

1 i j n

x x

...(8)

3(x

1, ..., x

n) = i j k

1 i j k n

x x x

...(9)

4(x

1, ..., x

n) = i j k l

1 i j k l n

x x x x

...(10)

5(x

1, ..., x

n) = i

1 i n

x

...(12)

The kth elementary symmetric polynomial is implemented in Mathematica as SymmetricPolynomial [k, {x

1, ..., x

n}]. Symmetric Reduction [f, {x

1, ..., x

n}] gives a pair of polynomials

{p, q} in x1, ..., x

n where is the symmetric part and is the remainder.


Abstract Algebra

Notes Alternatively, j(x

1,..., x

n) can be defined as the coefficient of xn-j in the generating function

i1 i n

(x x ).

...(13)

For example, on four variables x1, ..., x

4, the elementary symmetric polynomials are

1(x

1, x

2, x

3, x

4) = x

1 + x

2 + x

3 + x

4...(14)

2(x

1, x

2, x

3, x

4) = x

1 x

2 + x

1 x

3 + x

1 x

4 + x

2 x

3 + x

2 x

4 + x

3 x

4...(15)

3(x

1, x

2, x

3, x

4) = x

1 x

2 x

3 + x

1 x

2 x

4 + x

1 x

3 x

4 + x

2 x

3 x

4...(16)

4(x

1, x

2, x

3, x

4) = x

1 x

2 x

3 x

4...(17)

The power sum Sp (x

1,..., x

n) is defined by

Sp(x

1, ..., x

n) =

npk

k 1

x .

...(18)

The relationship between * and 1,...,

p is given by the so-called Newton-Girard formulas. The

related function sp(

1, ...,

n) with arguments given by the elementary symmetric polynomials

(not xn) is defined by

sp(

1,...,

n) = (�1)p�1 S

p (x

1,...,x

n) ...(19)

=n

p 1 pk

k 1

( 1) x .

...(20)

It turns out that sp (

1, ...,

n) is given by the coefficients of the generating function

ln (1 + 1t +

2 t2 +

3 t3 + ...) = kk

k 1

st

k

...(21)

= 2 2 3 31 1 2 1 1 2 3

1 1t ( 2 )t ( 3 3 )t ...

2 3

so the first few values are

s1 =

1...(22)

s2 = 2

1 22 ...(23)

s3 = 3

1 1 2 3� 3 3 ...(24)

s4 = 4 2 2

1 1 2 2 1 3 44 2 4 4 . ...(25)

In general, sp can be computed from the determinant

sp =

1

2 1

p 1 3 2 1

34 2 1

p p 1 p 2 p 3 1

0 0 012 0 013 01( 1)4 0

1p

...(26)



Notes(Littlewood 1958, Cadogan 1971). In particular,

S1(x

1,..., x

n) =

n

k 1k 1

x

...(27)

S2(x

1,..., x

n) = 2

1 22 ...(28)

S3(x

1,..., x

n) = 3

1 1 2 33 3 ...(29)

S4(x

1,..., x

n) = 4 2 2

1 1 2 2 1 3 44 2 4 4 ...(30)

(Schroeppel 1972), as can be verified by plugging in and multiplying through.

31.4 Constructible Polygon

In mathematics, a constructible polygon is a regular polygon that can be constructed withcompass and straightedge. For example, a regular pentagon is constructible with compass andstraightedge while a regular heptagon is not.

Conditions for Constructibility

Some regular polygons are easy to construct with compass and straightedge; others are not. Thisled to the question being posed: is it possible to construct all regular n-gons with compass andstraightedge? If not, which n-gons are constructible and which are not?

Carl Friedrich Gauss proved the constructability of the regular 17-gon in 1796. Five years later,he developed the theory of Gaussian periods in his Disquisitiones Arithmeticae. This theoryallowed him to formulate a sufficient condition for the constructability of regular polygons.

A regular n-gon can be constructed with compass and straight edge if n is the product of a powerof 2 and any number of distinct Fermat primes.

Gauss stated without proof that this condition was also necessary, but never published his proof.A full proof of necessity was given by Pierre Wantzel in 1837. The result is known as the Gauss�Wantzel theorem.

Figure 3.1


Abstract Algebra

Notes Construction of the regular 257-gon

Detailed results by Gauss� theory

Only five Fermat primes are known:

F0 = 3, F

1 = 5, F

2 = 17, F

3 = 257, and F

4 = 65537 (sequence A019434 in OEIS)

The next twenty-eight Fermat numbers, F5 through F

32, are known to be composite.

Thus an n-gon is constructible if

n = 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, � (sequence A003401 in OEIS),

while an n-gon is not constructible with compass and straightedge if

n = 7, 9, 11, 13, 14, 18, 19, 21, 22, 23, 25, � (sequence A004169 in OEIS).

31.5 Connection to Pascal�s Triangle

There are 31 known numbers that are multiples of distinct Fermat primes, which correspond tothe 31 odd-sided regular polygons that are known to be constructible. These are 3, 5, 15, 17, 51,85, 255, 257, �, 4294967295. As John Conway commented in The Book of Numbers, these numbers,when written in binary, are equal to the first 32 rows of the modulo-2 Pascal�s triangle, minusthe top row. This pattern breaks down after there, as the 6th Fermat number is composite, so thefollowing rows do not correspond to constructible polygons. It is unknown whether any moreFermat primes exist, and is therefore unknown how many odd-sided constructible polygonsexist. In general, if there are x Fermat primes, then there are 2x�1 odd-sided constructible polygons.

General Theory

In the light of later work on Galois Theory, the principles of these proofs have been clarified.It is straightforward to show from analytic geometry that constructible lengths must come frombase lengths by the solution of some sequence of quadratic equations. In terms of field theory,such lengths must be contained in a field extension generated by a tower of quadratic extensions.It follows that a field generated by constructions will always have degree over the base field thatis a power of two.

In the specific case of a regular n-gon, the question reduces to the question of constructing alength

cos(2/n).

This number lies in the n-th cyclotomic field � and in fact in its real subfield, which is a totallyreal field and a rational vector space of dimension

½(n),

where (n) is Euler�s quotient function. Wantzel�s result comes down to a calculation showingthat (n) is a power of 2 precisely in the cases specified.

As for the construction of Gauss, when the Galois group is 2-group it follows that it has asequence of subgroups of orders

1, 2, 4, 8, ...

that are nested, each in the next something simple to prove by induction in this case of an abeliangroup. Therefore, there are subfields nested inside the cyclotomic field, each of degree 2 over theone before. Generators for each such field can be written down by Gaussian period theory.



NotesFor example for n = 17 there is a period that is a sum of eight roots of unity, one that is a sum offour roots of unity, and one that is the sum of two, which is cos(2/17).

Each of those is a root of a quadratic equation in terms of the one before. Moreover, theseequations have real rather than imaginary roots, so in principle can be solved by geometricconstruction: this because the work all goes on inside a totally real field.

In this way the result of Gauss can be understood in current terms; for actual calculation of theequations to be solved, the periods can be squared and compared with the �lower� periods, in aquite feasible algorithm.

Compass and Straightedge Constructions

Compass and straightedge constructions are known for all constructible polygons. If n = p· qwith p = 2 or p and q co-prime, an n-gon can be constructed from a p-gon and a q-gon.

If p = 2, draw a q-gon and bisect one of its central angles. From this, a 2q-gon can beconstructed.

If p > 2, inscribe a p-gon and a q-gon in the same circle in such a way that they share avertex. Because p and q are relatively prime, there exists integers a,b such that ap + bq = 1.Then 2a/q + 2b/p = 2/pq. From this, a p·q-gon can be constructed.

Thus one only has to find a compass and straightedge construction for n-gons where n is aFermat prime.

The construction for an equilateral triangle is simple and has been known since Antiquity.Constructions for the regular pentagon were described both by Euclid and by Ptolemy.

Although Gauss proved that the regular 17-gon is constructible, he didn�t actually showhow to do it. The first construction is due to Erchinger, a few years after Gauss� work.

The first explicit construction of a regular 257-gon was given by Friedrich Julius Richelot(1832).

A construction for a regular 65537-gon was first given by Johann Gustav Hermes (1894).The construction is very complex; Hermes spent 10 years completing the 200-pagemanuscript. (Conway has cast doubt on the validity of Hermes� construction, however.

Figure 31.2


Abstract Algebra

Notes Figure 31.3

From left to right, constructions of a 17-gon, 257-gon and 65537-gon.

Other Constructions

It should be stressed that the concept of constructible as discussed in this article applies specificallyto compass and straightedge construction. More constructions become possible if other tools areallowed. The so-called neusis constructions, for example, make use of a marked rulers.Theconstructions are a mathematical idealization and are assumed to be done exactly.

Example: Determine the group of all automorphisms of a field with 4 elements.

Solution: The automorphism group consists of two elements: the identity mapping and theFrobenius automorphism.

As you know this field with 4 elements can be constructed as F = Z2[x] / < x2+x+1 >. Letting a be

the coset of x, we have F = {0, 1, a, 1+a}. Any automorphism of F must leave 0 and 1 fixed, so theonly possibility for an automorphism other than the identity is to interchange a and 1+a. Is thisan automorphism? Since x2+x+1 0, we have x2 -x-1 x+1, so a2 = 1+a and (1+a)2 = 1+2a+a2 = a. Thusthe function that fixes 0 and 1 while interchanging a and 1+a is in fact the Frobenius automorphismof F.

Example: Let F be the splitting field in C of x4+1.

(i) Show that [F:Q] = 4.

Solution: The polynomial x8-1 factors over Q as x8-1 = (x4-1)(x4+1) = (x-1)(x+1)(x2+1)(x4 +1). Thefactor x4 +1 is irreducible over Q by Eisenstein�s criterion. The roots of x4+1 are thus the primitive

8th roots of unity, ± 2 / 2 ± 2 / 2i, and adjoining one of these roots also gives the others,together with i. Thus, the splitting field is obtained in one step, by adjoining one root of x4+1, soits degree over Q is 4.

It is clear that the splitting field can also be obtained by adjoining first 2 and then i, so it can

also be expressed as Q( 2 , i).



Notes(ii) Find automorphisms of F that have fixed fields Q( 2 ), Q(i), and Q( 2 i), respectively.

Solution: These subfields of Q( 2 , i) are the splitting fields of x2-2, x2+1, and x2+2, respectively.

Any automorphism must take roots to roots, so if is an automorphism of Q( 2 , i), we must

have ( 2 ) = ± 2 , and (i) = ± i. These possibilities must in fact define 4 automorphisms of thesplitting field.

If we define 1 ( 2 ) = 2 and

1 (i) = -i, then the subfield fixed by

1 is Q( 2 ). If we define

2( 2 ) = � 2 and

2 (i) = i, then the subfield fixed by

2 is Q(i). Finally, for

3 =

2

1 we have

3( 2 ) = � 2 and (i) = �i, and thus

3 ( 2 i) = 2 i, so

3 has Q( 2 i) as its fixed subfield.

Example: Find the Galois groups of x3 � 2 over the fields Z5 and Z

11.

Solution: The polynomial is not irreducible over Z5, since it factors as x3-2 = (x+2)(x2-2x-1). The

quadratic factor will have a splitting field of degree 2 over Z5, so the Galois group is cyclic of

order 2.

A search in Z11

for roots of x3-2 yields one and only one: x = 7. Then x3-2 can be factored as x3-2 =(x-7)(x2+7x+5), and the second factor must be irreducible. The splitting field has degree 2 overZ

11, and can be described as Z

11[x] / < x2+7x+5 >. Thus the Galois group is cyclic of order 2.

Example: Find the Galois group of x4-1 over the field Z7.

Solution: We first need to find the splitting field of x4-1 over Z7. We have x4-1 = (x-1)(x+1)(x2+1).

A quick check of ±2 and ±3 shows that they are not roots of x2+1 over Z7, so x2+1 is irreducible

over Z7. To obtain the splitting field we must adjoin a root of x2+1, so we get a splitting field

Z7[x] / < x2+1 > of degree 2 over Z

7.

The Galois group of x4-1 over Z7 is cyclic of order 2.

Example: Find the Galois group of x3-2 over the field Z7.

Solution: In this case, x3-2 has no roots in Z7, so it is irreducible. We first adjoin a root a of x3-2 to

Z7. The resulting extension Z

7(a) has degree 3 over Z

7, so it has 73 = 343 elements, and each

element is a root of the polynomial x343-x. Let b> be a generator of the multiplicative group ofthe extension. Then (b114)3 = b342 = 1, showing that Z

7(a) contains a non-trivial cube root of 1. It

follows that x3-2 has three distinct roots in Z7(a): a, ab114, and ab228, so therefore Z

7(a) is a splitting

field for x3-2 over Z7. Since the splitting field has degree 3 over Z

7, it follows the Galois group of

the polynomial is cyclic of order 3.

Self Assessment

1. Galois considered ................... of the roots that leave the coefficient field fixed.

(a) polynomial (b) permutation

(c) combination (d) range

2. The modern approach is to consider ................... determined by permutation.

(a) homomorphism (b) automorphism

(c) isomorphism (d) ideal and subfield


Abstract Algebra

Notes 3. Any automorphism of a field f must leave its prime ................... fixed.

(a) sub group (b) sub domain

(c) sub field (d) sub range

4. Given [F : Q] is equal to ...................

(a) 7 (b) 5

(c) 6 (d) 4

5. Automorphism of F that have fixed fields Q( 2 ),Q(i) and Q 2 ,i respectively.

(a) 1 2 3Q Q 2 ,Q Q(i),Q 2i 2i

(b) 1 2 3Q 2 ,Q i,Q 2i 0

(c) Q = Q2 = Q

3

(d) Q1 = Q

2�1 = Q

3�1

31.6 Summary

Let F be an extension field of K. The set of all automorphisms : F > F such that (a) = a forall a in K is a group under composition of functions.

Let F be an extension field of K. The set



Let K be a field, let f(x) be a polynomial in K[x], and let F be a splitting field for f(x) overK. Then Gal(F/K) is called the Galois group of f(x) over K, or the Galois group of theequation f(x) = 0 over K.

Let F be an extension field of K, and let f(x) be a polynomial in K[x]. Then any element ofGal(F/K) defines a permutation of the roots of f(x) that lie in F.

Let f(x) be a polynomial in K[x] with no repeated roots and let F be a splitting field for f(x)over K. If : K > L is a field isomorphism that maps f(x) to g(x) in L[x] and E is a splittingfield for g(x) over L, then there exist exactly [F:K] isomorphisms : F -> E such that (a) = (a)for all a in K.

Let K be a field, let f(x) be a polynomial in K[x], and let F be a splitting field for f(x) overK. If f(x) has no repeated roots, then |Gal(F/K)| = [F:K].

Let K be a finite field and let F be an extension of K with [F:K] = m. Then Gal(F/K) is acyclic group of order m.

If we take K = Zp, where p is a prime number, and F is an extension of degree m, then the

generator of the cyclic group Gal(F/K) is the automorphism : F -> F defined by(x) = xp, for all x in F. This automorphism is called the Frobenius automorphism of F.

31.7 Keywords

Galois Group: Let F be an extension field of K. The set





NotesGalois Group of the Equation: Let K be a field, let f(x) be a polynomial in K[x], and let F be asplitting field for f(x) over K. Then Gal(F/K) is called the Galois group of f(x) over K, or theGalois group of the equation f(x) = 0 over K.


1. Let p be prime. Prove that there exists a polynomial f(x) Q[x] of degree p with Galois

group isomorphic to Sp. Conclude that for each prime p with p 5 there exists a polynomial

of degree p that is not solvable by radicals.

2. Let p be a prime and Zp(t) be the field of rational functions over Z

p. Prove that f(x) = xp � t

is an irreducible polynomial in Zp(t)[x]. Show that f(x) is not separable.

3. Let E be an extension field of F. Suppose that K and L are two intermediate fields. If thereexists an element G(E/F) such that (K) = L, then K and L are said to be conjugate fields.Prove that K and L are conjugate if and only if G(E/K) and G(E/L) are conjugate subgroupsof G(E/F).

4. Let Aut(). If a is a positive real number, show that (a) > 0.

5. Let K be the splitting field of x3 + x2 + 1 2[x]. Prove or disprove that K is an extension by

radicals.

6. Let F be a field such that char F 2. Prove that the splitting field of f(x) = ax2 + bx + c is

F( ), where a = b2 � 4ac.

7. Prove or disprove: Two different subgroups of a Galois group will have different fixedfields.

8. Let K be the splitting field of a polynomial over F. If E is a field extension of F containedin K and [E : F] = 2, then E is the splitting field of some polynomial in F[x].


1. (b) 2. (b) 3. (c) 4. (d) 5. (a)






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Abstract Algebra

Notes Unit 32: Solvability by Radicals

CONTENTS

Objectives

Introduction

32.1 Radical Extension

32.2 Solvable by Radicals

32.3 Summary

32.4 Keywords



Objectives


Discuss the radical extension

Explain that a polynomial equation is solvable by radical

Introduction

In most results, in this section we will assume that the fields have characteristic zero, in order toguarantee that no irreducible polynomial has multiple roots. When we say that a polynomialequation is solvable by radicals, we mean that the solutions can be obtained from the coefficientsin a finite sequence of steps, each of which may involve addition, subtraction, multiplication,division, or taking nth roots. Only the extraction of an nth root leads to a larger field, and so ourformal definition is phrased in terms of subfields and adjunction of roots of xn-a for suitableelements a.

32.1 Radical Extension

Definition: An extension field F of K is called a radical extension of K if there exist elementsu

1, u

2, ... , u

m in F and positive integers n

1, n

2, ... , n

m such that

(i) F = K (u1, u

2, ... , u

m), and

(ii) u1

n1 is in K and u

in

i is in K ( u

1, ... , u

i-1 ) for i = 2, ... , m .

32.2 Solvable by Radicals

For a polynomial f(x) in K[x], the polynomial equation f(x) = 0 is said to be solvable by radicalsif there exists a radical extension F of K that contains all roots of f(x).

Proposition: Let F be the splitting field of xn - 1 over a field K of characteristic zero. ThenGal(F/K) is an abelian group.


Unit 32: Solvability by Radicals

NotesTheorem 1: Let K be a field of characteristic zero that contains all nth roots of unity, let a be anelement of K, and let F be the splitting field of xn-a over K. Then Gal(F/K) is a cyclic group whoseorder is a divisor of n.

Theorem 2: Let p be a prime number, let K be a field that contains all pth roots of unity, and letF be an extension of K. If [F:K] = |Gal(F/K)| = p, then F = K(u) for some u in F such that up is inK.

Lemma: Let K be a field of characteristic zero, and let E be a radical extension of K. Then thereexists an extension F of E that is a normal radical extension of K.

Theorem 3: Let f(x) be a polynomial over a field K of characteristic zero. The equationf(x) = 0 is solvable by radicals if and only if the Galois group of f(x) over K is solvable.

Sn is not solvable for n 5, and so to give an example of a polynomial equation of degree n that

is not solvable by radicals, we only need to find a polynomial of degree n whose Galois groupover Q is S

n.

Lemma: Any subgroup of S5 that contains both a transposition and a cycle of length 5 must be

equal to S5 itself.

Theorem 4: There exists a polynomial of degree 5 with rational coefficients that is not solvableby radicals

Example: Let f(x) be irreducible over Q, and let F be its splitting field over Q. Show that ifGal (F/Q) is abelian, then F = Q(u) for all roots u of f(x).

Solution: Since F has characteristic zero, we are in the situation of the fundamental theorem ofGalois theory. Because Gal (F/Q) is abelian, every intermediate extension between Q and F mustbe normal. Therefore, if we adjoin any root u of f(x), the extension Q(u) must contain all otherroots of f(x), since it is irreducible over Q. Thus Q(u) is a splitting field for f(x), so Q(u) = F.

Example: Find the Galois group of x9-1 over Q.

Solution: We can construct the splitting field F of x9-1 over Q by adjoining a primitive 9th rootof unity to Q. We have the factorization

x9-1 = (x3-1)(x6+x3+1)

= (x-1)(x2+x+1)(x6+x3+1).

Substituting x+1 in the last factor yields

(x+1)6+(x+1)3+1 = x6+6x5+15x4+ 21x3+18x2+9x+3.

This polynomial satisfies Eisenstein�s criterion for the prime 3, which implies that the factorx6+x3+1 is irreducible over Q. The roots of this factor are the primitive 9th roots of unity, so itfollows that [F:Q] = 6. Gal (F/Q) is isomorphic to a subgroup of Z

9× Since Z

9× is abelian of order

6, it is isomorphic to Z6. It follows that Gal (F/Q) Z

6.

Comment: The Galois group of xn-1 over Q is isomorphic to Zn

× and so the Galois group is cyclicof order (n) iff n = 2, 4, pk, or 2pk, for an odd prime p.

Example: Show that x4-x3+x2-x+1 is irreducible over Q, and use it to find the Galois groupof x10-1 over Q.


Abstract Algebra

Notes Solution: We can construct the splitting field F of x10-1 over Q by adjoining a primitive 10th rootof unity to Q. We have the factorization

x10-1 = (x5-1)(x5+1)

= (x-1)(x4+x3+x2+x+1) (x+1)(x4-x3+x2-x+1).

Substituting x-1 in the last factor yields

(x-1)4-(x-1)3+(x-1)2-(x-1)+1

= (x4-4x3+6x2-4x+1) - (x3-3x2+3x-1) + (x2-2x+1) - (x-1) + 1

= x4-5x3+10x2-10x+5.

This polynomial satisfies Eisenstein�s criterion for the prime 5, which implies that the factorx4-x3+x2-x+1 is irreducible over Q.

The roots of this factor are the primitive 10th roots of unity, so it follows that [F:Q] = 4. The proofof Theorem 1 shows that Gal (F/Q) Z

10× and so the Galois group is cyclic of order 4.

Example: Show that p(x) = x5-4x+2 is irreducible over Q, and find the number of realroots. Find the Galois group of p(x) over Q, and explain why the group is not solvable.

Solution: The polynomial p(x) is irreducible over Q since it satisfies Eisenstein�s criterion forp = 2. Since p(-2) = -22, p(-1) = 5, p(0) = 2, p(1) = �1, and p(2) = 26, we see that p(x) has a real rootbetween -2 and -1, another between 0 and 1, and a third between 1 and 2. The derivativep�(x) = 5x4-4 has two real roots, so p(x) has one relative maximum and one relative minimum,and thus it must have exactly three real roots. It follows as in the proof of Theorem 2 that theGalois group of p(x) over Q is S

5, and so it is not solvable.

Self Assessment

1. Let F be splitting field of xn � 1 over a field K of characteristic .................. then G(F/K) is anabelian group.

(a) 1 (b) 0

(c) �1 (d) �2

2. Let K be a field of characteristic zero and let be a .................. of K. Thus there exists anextension of F of that is normal radical extension.

(a) radical extension (b) solvable group

(c) Galois group (d) finite element

3. There exists a polynomial of degree .................. with rational co-efficients that is not solvableby radical.

(a) 4 (b) 5

(c) 6 (d) 7

4. Any subgroup of S5 that contains both a transposition and cycle of length .................. must

be equal to S5 itself.

(a) 4 (b) 5

(c) 3 (d) 6


Unit 32: Solvability by Radicals

Notes32.3 Summary

An extension field F of K is called a radical extension of K if there exist elementsu

1, u

2, ... , u


1, n

2, ... , n

m such that

(i) F = K (u1, u

2, ... , u

m), and

(ii) u1

n1 is in K and u

in

i is in K ( u

1, ... , u

i-1 ) for i = 2, ... , m .

For a polynomial f(x) in K[x], the polynomial equation f(x) = 0 is said to be solvable byradicals if there exists a radical extension F of K that contains all roots of f(x).

Let F be the splitting field of xn - 1 over a field K of characteristic zero. Then Gal(F/K) is anabelian group.

Let K be a field of characteristic zero that contains all nth roots of unity, let a be an elementof K, and let F be the splitting field of xn-a over K. Then Gal(F/K) is a cyclic group whoseorder is a divisor of n.

Let p be a prime number, let K be a field that contains all pth roots of unity, and let F be anextension of K. If [F:K] = |Gal(F/K)| = p, then F = K(u) for some u in F such that up is in K.

Let K be a field of characteristic zero, and let E be a radical extension of K. Then there existsan extension F of E that is a normal radical extension of K.

Let f(x) be a polynomial over a field K of characteristic zero. The equationf(x) = 0 is solvable by radicals if and only if the Galois group of f(x) over K is solvable.

Sn is not solvable for n 5, and so to give an example of a polynomial equation of degree

n that is not solvable by radicals, we only need to find a polynomial of degree n whoseGalois group over Q is S

n.

Any subgroup of S5 that contains both a transposition and a cycle of length 5 must be equal

to S5 itself.

There exists a polynomial of degree 5 with rational coefficients that is not solvable byradicals

32.4 Keywords

Radical Extension: An extension field F of K is called a radical extension of K if there existelements u

1, u

2, ... , u


1, n

2, ... , n

m such that

(i) F = K (u1, u

2, ... , u

m)

Solvable by Radicals: For a polynomial f(x) in K[x], the polynomial equation f(x) = 0 is said to besolvable by radicals if there exists a radical extension F of K that contains all roots of f(x).


1. We know that the cyclotomic polynomial

pp 1 p 2

p

x 1(x) x x ... x 1

x 1

is irreducible over for every prime p. Let w be a zero p(x), and consider the field ().

(a) Show that , 2,...,p-1 are distinct zeros of p(x), and conclude that they are all the

zeros of p(x).


Abstract Algebra

Notes (b) Show that G((w)/) is abelian of order p � 1.

(c) Show that the fixed field of G(()/) is .

2. Let F be a finite field of characteristic zero. Let E be a finite normal extension of F withGalois group G(E/F): Prove that F K L E if and only if {id} G(E/L) G(E/K) G(E/F).

3. Let F be a field of characteristic zero and let f(x) F[x] be a separable polynomial of degree

n. If E is the splitting field of f(x), let 1,...,

n be the roots of f(x) in E. Let i j i j( ).

We define the discriminant of f(x) to be 2.

(a) If f(x) = ax2 + bx + c, show that 2 = b2 � 4ac.

(b) If f(x) = x3 + px + q, show that 2 = �4p3 � 27q2.

(c) Prove that 2 is in F.

(d) If G(E/F) is a transposition of two roots of f(x), show that () = �.

(e) If G(E/F) is an even permutation of the roots of f(x), show that () = .

(f) Prove that G(E/F) is isomorphic to a subgroup of An if and only if F.

(g) Determine the Galois groups of x3 + 2x � 4 and x3 + x � 3.


1. (b) 2. (a) 3. (b) 4. (b)






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