A Semester Course in Basic Abstract Algebra

Marcel B. FinanArkansas Tech University

c©All Rights Reserved

December 29, 2011

1

PREFACE

This book is an introduction to abstract algebra course for undergraduateseither at the junior or senior level. The course is usually taken by Mathemat-ics, Physics, Chemistry, and Engineering majors. The content of the bookcan be covered in a one semester time period. The chapters presented hererepresent the core of the subject, the basic idea of groups, rings, and fields.This book has the additional goal of introducing the axiomatic method andthe construction of proofs.This book has been designed for use either as a supplement of standardtextbooks or as a textbook for a formal course in an introductory abstractalgebra.

Marcel B. Finan,Arkansas Tech UniversityJanuary, 2004.

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Contents

0 Preliminary Notions 4

1 The Concept of a Mapping 32

2 Composition. Invertible Mappings 43

3 Binary Operations 53

4 Composition of Mappings as a Binary Operation 62

5 Definition and Examples of Groups 70

6 Permutation Groups 76

7 Subgroups 867.1 Definition and Examples of Subgroups . . . . . . . . . . . . . 867.2 The Alternating Group . . . . . . . . . . . . . . . . . . . . . . 90

8 Symmetry Groups 97

9 Equivalence Relations 101

10 The Division Algorithm. Congruence Modulo n 10810.1 Divisibility. The Division Algorithm . . . . . . . . . . . . . . . 10810.2 Congruence Modulo n. . . . . . . . . . . . . . . . . . . . . . . 111

11 Arithmetic Modulo n 116

12 Greatest Common Divisors. The Euclidean Algorithm 122

13 Least Common Multiple. The Fundamental Theorem of Arith-metic 129

14 Elementary Properties of Groups 136

15 Generated Groups. Direct Product 14615.1 Finitely and Infinitely Generated Groups . . . . . . . . . . . . 14615.2 Direct Product of Groups. . . . . . . . . . . . . . . . . . . . . 148

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16 Cosets 153

17 Lagrange’s Theorem 159

18 Group Isomorphisms 168

19 More Properties of Isomorphims 174

20 Cayley’s Theorem 181

21 Homomorphisms and Normal Subgroups 185

22 Quotient Groups 192

23 Isomorphism Theorems 198

24 Rings: Definition and Basic Results 204

25 Integral Domains. Subrings 211

26 Ideals and Quotient Rings 217

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0 Preliminary Notions

Throughout this book, we assume that the reader is familiar with the follow-ing number systems:

• The set of all positive integers

N = {1, 2, 3, · · ·}.

• The set of all integers

Z = {· · · ,−3,−2,−1, 0, 1, 2, 3, · · ·}.

• The set of all rational numbers

Q = {ab

: a, b ∈ Z with b 6= 0}.

• The set R of all real numbers.• The set of all complex numbers

C = {a+ bi : a, b ∈ R}

where i =√−1.

We start with this introductory section to present the fundamentals of math-ematical logic, mathematical proofs, and set theory.

Fundamentals of Mathematical LogicLogic is commonly known as the science of reasoning. We will develop someof the symbolic techniques required later in the book.

Definition 0.1A proposition is any meaningful statement that is either true or false, butnot both.

We will use lowercase letters, such as p, q, r, · · · , to represent propositions.We will also use the notation

p : 1 + 1 = 3

to define p to be the proposition 1 + 1 = 3.

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Definition 0.2The truth value of a proposition is true, denoted by T, if it is a truestatement and false, denoted by F, if it is a false statement.

Statements that are not propositions include questions and commands.

Example 0.1Which of the following are propositions? Give the truth value of the propo-sitions.a. 2 + 3 = 7.b. Julius Ceasar was president of the United States.c. What time is it?d. Be quiet !

Solution.a. A proposition with truth value (F).b. A proposition wiht truth value (F).c. Not a proposition since no truth value can be assigned to this statement.d. Not a proposition.

Definition 0.3Let p and q be propositions. The conjunction of p and q, denoted p ∧ q, isthe proposition: p and q. This proposition is defined to be true only whenboth p and q are true and it is false otherwise. The disjunction of p andq, denoted p ∨ q, is the proposition: p or q. The ’or’ is used in an inclusiveway. This proposition is false only when both p and q are false, otherwise itis true.

Example 0.2Let

p : 5 < 9

q : 9 < 7.

Construct the propositions p ∧ q and p ∨ q.

Solution.The conjunction of the propositions p and q is the proposition

p ∧ q : 5 < 9 and 9 < 7.

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The disjunction of the propositions p and q is the proposition

p ∨ q : 5 < 9 or 9 < 7.

Definition 0.4A truth table displays the relationships between the truth values of propo-sitions.

Below, we display the truth tables of p ∧ q and p ∨ q.

p q p ∧ qT T TT F FF T FF F F

p q p ∨ qT T TT F TF T TF F F

Definition 0.5The negation of p, denoted ∼ p, is the proposition not p.

The truth table of ∼ p is displayed below

p ∼ pT FF T

Example 0.3Find the negation of the proposition p : −5 < x ≤ 0.

Solution.The negation of p is the proposition ∼ p : x > 0 or x ≤ −5

Definition 0.6Two propositions are equivalent if they have exactly the same truth valuesunder all circumstances. We write p ≡ q.

Example 0.4a. Show that ∼ (p ∨ q) ≡∼ p∧ ∼ q.b. Show that ∼ (p ∧ q) ≡∼ p∨ ∼ q.c. Show that ∼ (∼ p) ≡ p.

a. and b. are known as DeMorgan’s laws.

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Solution.a.

p q ∼ p ∼ q p ∨ q ∼ (p ∨ q) ∼ p∧ ∼ qT T F F T F FT F F T T F FF T T F T F FF F T T F T T

b.p q ∼ p ∼ q p ∧ q ∼ (p ∧ q) ∼ p∨ ∼ qT T F F T F FT F F T F T TF T T F F T TF F T T F T T

c.p ∼ p ∼ (∼ p)T F TF T F

Definition 0.7Let p and q be propositions. The implication p =⇒ q is the the propositionthat is false only when p is true and q is false; otherwise it is true. p is calledthe hypothesis and q is called the conclusion. The connective =⇒ is calledthe conditional connective.

Example 0.5Construct the truth table of the implication p =⇒ q.

Solution.The truth table is

p q p =⇒ qT T TT F FF T TF F T

Example 0.6Show that p =⇒ q ≡∼ p ∨ q.

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Solution.

p q ∼ p p =⇒ q ∼ p ∨ qT T F T TT F F F FF T T T TF F T T T

It follows from the previous exercise that the proposition p =⇒ q is alwaystrue if the hypothesis p is false, regardless of the truth value of q. We saythat p =⇒ q is true by default or vacuously true.In terms of words the proposition p =⇒ q also reads:(a) if p then q.(b) p implies q.(c) p is a sufficient condition for q.(d) q is a necessary condition for p.(e) p only if q.

Definition 0.8The converse of p =⇒ q is the proposition q =⇒ p. The opposite or inverseof p =⇒ q is the proposition ∼ p =⇒∼ q. The contrapositive of p =⇒ q isthe proposition ∼ q =⇒∼ p.

Example 0.7Show that p =⇒ q ≡∼ q =⇒∼ p.

Solution.We use De Morgan’s laws as follows.

p =⇒ q ≡ ∼ p ∨ q≡ ∼ (p∧ ∼ q)≡ ∼ (∼ q ∧ p)≡ ∼∼ q∨ ∼ p≡ q∨ ∼ p≡ ∼ q =⇒∼ p

Example 0.8Using truth tables show the following:a. p =⇒ q 6≡ q =⇒ pb. p =⇒ q 6≡∼ p =⇒∼ q

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Solution.a. It suffices to show that ∼ p ∨ q 6≡∼ q ∨ p.

p q ∼ p ∼ q ∼ p ∨ q ∼ q ∨ pT T F F T TT F F T F 6= TF T T F T 6= FF F T T T T

b. We will show that ∼ p ∨ q 6≡ p∨ ∼ q.

p q ∼ p ∼ q ∼ p ∨ q p∨ ∼ qT T F F T TT F F T F 6= TF T T F T 6= FF F T T T T

Definition 0.9The biconditional proposition of p and q, denoted by p ⇐⇒ q, is thepropositional function that is true when both p and q have the same truthvalues and false if p and q have opposite truth values. Also reads, ”p if andonly if q” or ”p is a necessary and sufficient condition for q.”

Example 0.9Construct the truth table for p⇐⇒ q.

Solution.

p q p⇐⇒ qT T TT F FF T FF F T

Example 0.10Show that the biconditional proposition of p and q is logically equivalent tothe conjunction of the conditional propositions p =⇒ q and q =⇒ p.

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Solution.

p q p =⇒ q q =⇒ p p⇐⇒ q (p =⇒ q) ∧ (q =⇒ p)T T T T T TT F F T F FF T T F F FF F T T T T

Propositions and QuantifiersAnother way to generate propositions is by means of quantifiers. Let P (x)be a statement that depends on a variable x and which is defined on someset D. ”If P (x) is true for all values of x ∈ D” then such a statement definesa proposition. For example, suppose that P (n) : 2n is even where n ∈ N.Then the statement ”For all n ∈ N, P (n)” is always true and thus defines aproposition. Such a statement can be written as

∀n ∈ N, {2n is even}.

The symbol ∀ is called the universal quantifier.

Example 0.11Write in the form ∀x ∈ D,P (x) the proposition :” every real number is eitherpositive, negative or 0.”

Solution.∀x ∈ R, x > 0, x < 0, or x = 0.

The proposition ∀x ∈ D,P (x) is false if P (x) is false for at least one valueof x. In this case x is called a counterexample.

Example 0.12Show that the proposition ∀x ∈ R, x > 1

xis false.

Solution.A counterexample is x = 1

2. Clearly, 1

2< 2 = 1

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.

The notation ∃x ∈ D,P (x) is a proposition that is true if there is at leastone value of x ∈ D where P (x) is true; otherwise it is false. The symbol ∃ iscalled the existential quantifier.

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Example 0.13Let P (x) denote the statement ”x > 3.” What is the truth value of theproposition ∃x ∈ R, P (x).

Solution.Since 4 ∈ R and 4 > 3 then the given proposition is true.

Example 0.14Write the sets ∩i∈ISi and ∪i∈ISi using quantifiers.

Solution.Recall that ∩i∈ISi = {x|x is an element of Si for all i in I}. Using the uni-versal quantifier we obtain ∩i∈ISi = {x|∀i ∈ I, x ∈ Si}. Similarly, since∪i∈ISi = {x|x is in Si for some i ∈ I} then using the existential quantifier wecan write ∪i∈ISi = {x|∃i ∈ I, x ∈ Si}.

Example 0.15a. What is the negation of the proposition ∀x ∈ D,P (x)?b. What is the negation of the proposition ∃x ∈ D,P (x)?

Solution.a. ∃x ∈ D,∼ P (x).b. ∀x ∈ D,∼ P (x).

Example 0.16Write the negation of each of the following propositions:

a. Every polynomial function is continuous.b. There exists a triangle with the property that the sum of angles is greaterthan 180◦.

Solution.a. There exists a polynomial that is not continuous everywhere.b. For any triangle, the sum of the angles is less than or equal to 180◦.

Next we discuss statements that contain multiple quantifiers. A typical ex-ample is the definition of a limit. We say that L = limx→a f(x) if and only if∀ε > 0,∃ a positive number δ such that if |x− a| ≤ δ then |f(x)− L| < ε.

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Example 0.17a. Let P (x, y) denote the statement ”x+y = y+x.” What is the truth valueof the proposition (∀x ∈ R)(∀y ∈ R), P (x, y)?b. Let Q(x, y) denote the statement ”x+ y = 0.” What is the truth value ofthe proposition (∃y ∈ R)(∀x ∈ R), Q(x, y)?

Solution.a. The given proposition is always true.b. The proposition is false. For otherwise, one can choose x 6= −y to obtain0 6= x+ y = 0 which is impossible

Example 0.18Find the negation of the following propositions:a. ∀x∃y, P (x, y).b. ∃x∀y, P (x, y).

Solution.a. ∃x∀y,∼ P (x, y).b. ∀x∃y,∼ P (x, y)

Example 0.19The symbol ∃ ! stands for the phrase ”there exists a unique”. Which of thefollowing statements are true and which are false.

a. ∃ ! x ∈ R,∀y ∈ R, xy = y.b. ∃ ! integer x such that 1

xis an integer.

Solution.a. True. Let x = 1.b. False since 1 and −1 are both integers with integer reciprocals

As a final application of quantifiers we prove the following result knownas DeMorgan’s laws for sets.

Theorem 0.1Let {Ai}i∈I be a collection of sets. Then

(a) (∪i∈IAi)c = ∩i∈IAci .(b) (∩i∈IAi)c = ∪i∈IAci .

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Proof.(a) Let x ∈ (∪i∈IAi)c . Then by the definition of complement we have x ∈ Uand x /∈ ∪i∈IAi. Thus, x /∈ Ai, ∀i ∈ I. Hence, we have x ∈ U andx /∈ Ai,∀i ∈ I and this means that x ∈ Aci ,∀i ∈ I. Therefore, x ∈ ∩i∈IAci .This proves that (∪i∈IAi)c ⊆ ∩i∈IAci .Now, let x ∈ ∩i∈IAci . Then x /∈ Ai,∀i ∈ I. Hence, x /∈ ∪i∈IAi. Since x ∈ Uand x /∈ ∪i∈IAi then x ∈ (∪i∈IAi)c . This shows that ∩i∈IAci ⊆ (∪i∈IAi)c .(b) Similar to (a) and is left for the reader.

Methods of Mathematical ProofsMathematical proofs can be classified as either a direct proof or an indirectproof. In a direct proof one starts with the hypothesis of an implicationp =⇒ q and then prove that the conclusion is true. Any other method ofproof will be referred to as an indirect proof. In this section we study twomethods of indirect proofs, namely, the proof by contradiction and the proofby contrapositive.

• Proof by contradiction: With this method, a conditional statement”p =⇒ q” is proved by showing that if p were true and q were not true, thensome contradiction (absurdity) would result. We give a couple of problemswhere we use this method.

Example 0.20Show that if n2 is an even integer then so is n.

Solution.Suppose the contrary. That is suppose that n is odd. Then there is an integerk such that n = 2k + 1. In this case, n2 = 2(2k2 + 2k) + 1 is odd and thiscontradicts the assumption that n2 is even. Hence, n must be even

Example 0.21Show tthat the number

√2 is irrational.

Solution.Suppose not. That is, suppose that

√2 is rational. Then there exist two in-

tegers m and n with no common divisors such that√

2 = mn. Squaring both

sides of this equality we find that 2n2 = m2. Thus, m2 is even. By Problem0.20, m is even. That is, 2 divides m. But then m = 2k for some integer k.

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Taking the square we find that 2n2 = m2 = 4k2, that is n2 = 2k2. This saysthat n2 is even and by Example 0.20, n is even. We conclude that 2 dividesboth m and n and this contradcits our assumption that m and n have nocommon divisors. Hence,

√2 must be irrational

• Proof by contrapositive: We already know that p =⇒ q ≡∼ q =⇒∼ p.So to prove p =⇒ q we sometimes instead prove ∼ q =⇒∼ p.

Example 0.22Show that if n is an integer such that n2 is odd then n is also odd.

Proof.Suppose that n is an integer that is even. Then there exists an integer k suchthat n = 2k. But then n2 = 2(2k2) which is even.

Method of Proof by InductionWe want to prove that some predicate P (n) is true for any nonnegative inte-ger n ≥ n0. This is acheived by using the method of mathematical induction.The steps of this method are as follows:

(i) (Basis of induction) Show that P (n0) is true.(ii) (Induction hypothesis) Assume P (n) is true.(iii) (Induction step) Show that P (n+ 1) is true.

Example 0.23Use the technique of mathematical induction to show that

1 + 2 + 3 + · · ·+ n =n(n+ 1)

2, n ≥ 1.

Solution.Let S(n) = 1 + 2 + · · ·+ n. Then

(i) (Basis of induction) S(1) = 1 = 1(1+1)2

. That is, S(1)is true.

(ii) (Induction hypothesis) Assume S(n) is true. That is, S(n) = n(n+1)2

.

(iii) (Induction step) We must show that S(n+ 1) = (n+1)(n+2)2

. Indeed,

S(n+ 1) = 1 + 2 + · · ·+ n+ (n+ 1)= S(n) + (n+ 1)

= n(n+1)2

+ (n+ 1)

= (n+1)(n+2)2

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Example 0.24a. Use induction to prove that n < 2n for all non-negative integers n.b. Use induction to prove that 2n < n! for all non-negative integers n ≥ 4.

Solution.a. Let S(n) = 2n − n, n ≥ 0. We want to show that S(n) > 0 is valid for alln ≥ 0. By the method of mathematical induction we have(i) (Basis of induction) S(0) = 20 − 0 = 1 > 0. That is, S(0)is true.(ii) (Induction hypothesis) Assume S(n) is true. That is, S(n) > 0.(iii) (Induction step) We must show that S(n+ 1) > 0. Indeed,

S(n+ 1) = 2n+1 − (n+ 1)= 2n.2− n− 1= 2n(1 + 1)− n− 1= 2n − n+ 2n − 1= (2n − 1) + S(n)> 2n − 1 > 0

since the smallest value of n is 0 and in this case 20 − 1 = 0.b. Let S(n) = n!− 2n, n ≥ 4. We want to show that S(n) > 0 for all n ≥ 4.By the method of mathematical induction we have(i) (Basis of induction) S(4) = 4!− 24 = 8 > 0. That is, S(4)is true.(ii) (Induction hypothesis) Assume S(n) is true. That is, S(n) > 0, n ≥ 4.(iii) (Induction step) We must show that S(n+ 1) > 0. Indeed,

S(n+ 1) = (n+ 1)!− 2n+1

= (n+ 1)n!− 2n(1 + 1)= n!− 2n + nn!− 2n

> 2(n!− 2n) = 2S(n) > 0.

where we have used the fact that if n ≥ 1 then nn! ≥ n!

Example 0.25 (Bernoulli’s inequality)Let h > −1. Use induction to show that

(1 + nh) ≤ (1 + h)n, n ≥ 0.

Solution.Let S(n) = (1+h)n− (1+nh). We want to show that S(n) ≥ 0 for all n ≥ 0.

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We use mathematical induction as follows.(i) (Basis of induction) S(0) = (1 + h)0 − (1 + 0h) = 0. That is, S(0)is true.(ii) (Induction hypothesis) Assume S(n) is true. That is, S(n) ≥ 0, n ≥ 0.(iii) (Induction step) We must show that S(n+ 1) ≥ 0. Indeed,

S(n+ 1) = (1 + h)n+1 − (1 + (n+ 1)h)= (1 + h)(1 + h)n − nh− 1− h≥ (1 + h)(1 + nh)− nh− 1− h= nh2 ≥ 0.

Fundamentals of Set TheorySet is the most basic term in mathematics. In this section we introduce theconcept of sets and its various operations and then study the properties ofthese operations.

Definition 0.10We define a set A as a collection of well-defined objects (called elements ormembers of A) such that for any given object x either one (but not both)of the following holds:

• x belongs to A and we write x ∈ A.

• x does not belong to A, and in this case we write x 6∈ A.

We denote sets by capital letters A,B,C, · · · and elements by lowercase lettersa, b, c, · · · . Sets consisting of sets will be denoted by script letters.There are two different ways to represent a set. The first one is to list,without repetition, the elements of the set. The other way is to describe aproperty that characterizes the elements of the set. This is also known asthe set-builder notation.We define the empty set, denoted by ∅, to be the set with no elements.

Example 0.26List the elements of the following sets.a. {x|x is a real number such that x2 = 1}.b. {x|x is an integer such that x2 − 3 = 0}.

Solution.a. {−1, 1}.b. ∅

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Example 0.27Use a property to give a description of each of the following sets.a. {a, e, i, o, u}.b. {1, 3, 5, 7, 9}.

Solution.a. {x|x is a vowel}.b. {n ∈ N|n is odd and less than 10}

Definition 0.11Let A and B be two sets. We say that A is a subset of B, denoted byA ⊆ B, if and only if every element of A is also an element of B.

If there exists an element of A which is not in B then we write A 6⊆ B.

Example 0.28Suppose that A = {2, 4, 6}, B = {2, 6}, and C = {4, 6}. Determine which ofthese sets are subsets of which other of these sets.

Solution.B ⊆ A and C ⊆ A

If sets A and B are represented as regions in the plane, relationships be-tween A and B can be represented by pictures, called Venn diagrams.

Example 0.29Represent A ⊆ B using Venn diagram.

Solution.The Venn diagram is given in Figure 0.1

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Figure 0.1

Definition 0.12Two sets A and B are said to be equal if and only if A ⊆ B and B ⊆ A. Wewrite A = B.

It follows from the above definition that in order to show A = B it suffices toshow the double inclusions mentioned in the definition. For non-equal setswe write A 6= B.

Example 0.30Determine whether each of the following pairs of sets are equal.(a) {1, 3, 5} and {5, 3, 1}.(b) {{1}} and {1, {1}}.

Solution.(a) {1, 3, 5} = {5, 3, 1}.(b) {{1}} 6= {1, {1}} since 1 6∈ {{1}}

Definition 0.13Let A and B be two sets. We say that A is a proper subset of B, denotedby A ⊂ B, if A ⊆ B and A 6= B.

Thus, to show that A is a proper subset of B we must show that everyelement of A is an element of B and there is an element of B which is not inA.

Example 0.31Order the sets of numbers: C,Z,R,Q,N using ⊂ .

Solution.N ⊂ Z ⊂ Q ⊂ R ⊂ C.

Example 0.32Determine whether each of the following statements is true or false.(a) x ∈ {x} (b) {x} ⊆ {x} (c) {x} ∈ {x}(d) {x} ∈ {{x}} (e) ∅ ⊆ {x} (f) ∅ ∈ {x}

Solution.(a) True (b) True (c) False, since only x ∈ {x} (d) True (e) True (f) False,since ∅ is a set and {x} has an element which is not a set.

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Definition 0.14If U is a given set whose subsets are under consideration, then we call U auniversal set. Let U be a universal set and A,B be two subsets of U. Theabsolute complement of A (See Figure 0.2) is the set

Ac = {x ∈ U |x 6∈ A}.

The relative complement of A with respect to B (See Figure 0.3) is theset

B − A = {x ∈ U |x ∈ B and x 6∈ A}.

Example 0.33Let U = R. Consider the sets A = {x ∈ R|x < −1 or x > 1} andB = {x ∈ R|x ≤ 0}. Find

a. Ac.b. B − A.

Solution.a. Ac = [−1, 1].b. B − A = [−1, 0]

Definition 0.15Let A and B be two sets. The union of A and B is the set

A ∪B = {x|x ∈ A or x ∈ B}

where the ’or’ is inclusive.(See Figure 0.4)

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Figure 0.4

The above definition can be extended to more than two sets. More precisely,if A1, A2, · · · , are sets then

∪∞n=1An = {x|x ∈ Ai for some i ∈ N}.

Definition 0.16Let A and B be two sets. The intersection of A and B is the set

A ∩B = {x|x ∈ A and x ∈ B}.

(See Figure 0.5).

If A ∩B = ∅ we say that A and B are disjoint sets.

Figure 0.5

Given the sets A1, A2, · · · , we define

∩∞n=1An = {x|x ∈ Ai for all i ∈ N}.

Example 0.34Let A = {a, b, c}, B = {b, c, d}, and C = {b, c, e}.

a. Find A ∪ (B ∩ C), (A ∪ B) ∩ C, and (A ∪ B) ∩ (A ∪ C). Which ofthese sets are equal?b. Find A ∩ (B ∪ C), (A ∩ B) ∪ C, and (A ∩ B) ∪ (A ∩ C). Which of thesesets are equal?c. Find A− (B − C) and (A−B)− C. Are these sets equal?

Solution.a. A ∪ (B ∩ C) = A, (A ∪ B) ∩ C = {b, c}, (A ∪ B) ∩ (A ∪ C) = {a, b, c} =A ∪ (B ∩ C).b. A ∩ (B ∪ C) = {b, c}, (A ∩ B) ∪ C = C, (A ∩ B) ∪ (A ∩ C) = {b, c}. =A ∩ (B ∪ C).c. A− (B − C) = A and (A−B)− C = {a} 6= A− (B − C).

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Example 0.35For each n ≥ 1, let An = {x ∈ R : x < 1 + 1

n}. Show that

∩∞n=1An = {x ∈ R : x ≤ 1}.

Solution.The proof is by double inclusions method. Let y ∈ {x ∈ R : x ≤ 1}. Thenfor all positive integer n we have y ≤ 1 < 1 + 1

n. That is, y ∈ ∩∞n=1An. This

shows that {x ∈ R : x ≤ 1} ⊆ ∩∞n=1An.Conversely, let y ∈ ∩∞n=1An. Then y < 1 + 1

nfor all n ≥ 1. Now take the limit

of both sides as n→∞ to obtain y ≤ 1. That is, y ∈ {x ∈ R : x ≤ 1}. Thisshows that ∩∞n=1An ⊆ {x ∈ R : x ≤ 1}.

Definition 0.17The notation (a1, a2, · · · , an) is called an ordered n-tuples. We say thattwo n-tuples (a1, a2, · · · , an) and (b1, b2, · · · , bn) are equal if and only if a1 =b1, a2 = b2, · · · , an = bn.Given n sets A1, A2, · · · , An the Cartesian product of these sets is the set

A1 × A2 × · · · × An = {(a1, a2, · · · , an) : a1 ∈ A1, a2 ∈ A2, · · · , an ∈ An}

Example 0.36Let A = {x, y}, B = {1, 2, 3}, and C = {a, b}. Find A×B × C.

Solution.

A×B × C = {(x, 1, a), (x, 2, a), (x, 3, a), (y, 1, a), (y, 2, a),(y, 3, a), (x, 1, b), (x, 2, b), (x, 3, b), (y, 1, b)

(y, 2, b), (y, 3, b)}We end this section by discussing some basic properties of sets.The following problem shows that the operation ⊆ is reflexive and transitive,concepts that will be discussed in Section 9.

Example 0.37a. Show that A ⊆ A.b. Suppose that A,B,C are sets such that A ⊆ B and B ⊆ C. Show thatA ⊆ C.c. Find two sets A and B such that A ∈ B and A ⊆ B.

22

Solution.a. The proposition if x ∈ A then x ∈ A is always true. Thus, A ⊆ A.b. We need to show that every element of A is an element of C. Let x ∈ A.Since A ⊆ B then x ∈ B. But B ⊆ C so x ∈ C.c. A = {x} and B = {x, {x}}.

Theorem 0.2Let A and B be two sets. Thena. A ∩B ⊆ A and A ∩B ⊆ B.b. A ⊆ A ∪B and B ⊆ A ∪B.

Proof.a. If x ∈ A ∩ B then x ∈ A and x ∈ B. This still imply that x ∈ A. Hence,A ∩B ⊆ A. A similar argument holds for A ∩B ⊆ B.b. The proposition ”if x ∈ A then x ∈ A ∪ B” is always true. Hence,A ⊆ A ∪B. A similar argument holds for B ⊆ A ∪B

Theorem 0.3If A and B are subsets of U thena. A ∪ U = U.b. A ∪ A = A.c. A ∪ ∅ = A.d. A ∪B = B ∪ A.e. (A ∪B) ∪ C = A ∪ (B ∪ C).

Proof.a. Clearly, A ∪ U ⊆ U. Conversely, let x ∈ U . Then definitely, x ∈ A ∪ U.That is, U ⊆ A ∪ U.b. If x ∈ A then x ∈ A or x ∈ A. That is, x ∈ A ∪ A and consequentlyA ⊆ A ∪ A. Conversely, if x ∈ A ∪ A then x ∈ A. Hence, A ∪ A ⊆ A.c. If x ∈ A∪∅ then x ∈ A since x 6∈ ∅. Thus, A∪∅ ⊆ A. Conversely, if x ∈ Athen x ∈ A or x ∈ ∅. Hence, A ⊆ A ∪ ∅.d. If x ∈ A ∪ B then x ∈ A or x ∈ B. But this is the same thing as sayingx ∈ B or x ∈ A. That is, x ∈ B ∪ A. Now interchange the roles of A and Bto show that B ∪ A ⊆ A ∪B.e. Let x ∈ (A∪B)∪C. Then x ∈ (A∪B) or x ∈ C. Thus, (x ∈ A or x ∈ B)or x ∈ C. This implies x ∈ A or (x ∈ B or x ∈ C). Hence, x ∈ A ∪ (B ∪ C).The converse is similar

23

Theorem 0.4Let A and B be subsets of U . Thena. A ∩ U = A.b. A ∩ A = A.c. A ∩ ∅ = ∅.d. A ∩B = B ∩ A.e. (A ∩B) ∩ C = A ∩ (B ∩ C).

Proof.a. If x ∈ A ∩ U then x ∈ A. That is , A ∩ U ⊆ A. Conversely, let x ∈ A.Then definitely, x ∈ A and x ∈ U. That is, x ∈ A ∩ U. Hence, A ⊆ A ∩ U.b. If x ∈ A then x ∈ A and x ∈ A. That is, A ⊆ A ∩ A. Conversely, ifx ∈ A ∩ A then x ∈ A. Hence, A ∩ A ⊆ A.c. Clearly ∅ ⊆ A ∩ ∅. Conversely, if x ∈ A ∩ ∅ then x ∈ ∅. Hence, A ∩ ∅ ⊆ ∅.d. If x ∈ A ∩B then x ∈ A and x ∈ B. But this is the same thing as sayingx ∈ B and x ∈ A. That is, x ∈ B ∩A. Now interchange the roles of A and Bto show that B ∩ A ⊆ A ∩B.e. Let x ∈ (A ∩ B) ∩ C. Then x ∈ (A ∩ B) and x ∈ C. Thus, (x ∈ A andx ∈ B) and x ∈ C. This implies x ∈ A and (x ∈ B and x ∈ C). Hence,x ∈ A ∩ (B ∩ C). The converse is similar

Theorem 0.5If A,B, and C are subsets of U thena. A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C).b. A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C).

Proof.a. Let x ∈ A∩ (B∪C). Then x ∈ A and x ∈ B∪C. Thus, x ∈ A and (x ∈ Bor x ∈ C). This implies that (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C).Hence, x ∈ A ∩B or x ∈ A ∩ C, i.e. x ∈ (A ∩B) ∪ (A ∩ C). The converse issimilar.b. Let x ∈ A ∪ (B ∩ C). Then x ∈ A or x ∈ B ∩ C. Thus, x ∈ A or (x ∈ Band x ∈ C). This implies that (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C).Hence, x ∈ A ∪ B and x ∈ A ∪ C, i.e. x ∈ (A ∪ B) ∩ (A ∪ C). The converseis similar

Theorem 0.6 (De Morgan’s Laws)Let A and B be subsets of U then

24

a. (A ∪B)c = Ac ∩Bc.b. (A ∩B)c = Ac ∪Bc.

Proof.a. Let x ∈ (A ∪ B)c. Then x ∈ U and x 6∈ A ∪ B. Hence, x ∈ U and (x 6∈ Aand x 6∈ B). This implies that (x ∈ U and x 6∈ A) and (x ∈ U and x 6∈ B).It follows that x ∈ Ac ∩Bc. Now, go backward for the converse.b. Let x ∈ (A ∩ B)c. Then x ∈ U and x 6∈ A ∩ B. Hence, x ∈ U and (x 6∈ Aor x 6∈ B). This implies that (x ∈ U and x 6∈ B) or (x ∈ U and x 6∈ A). Itfollows that x ∈ Ac ∪Bc. The converse is similar

Definition 0.18A collection of nonempty subsets {A1, A2, · · · , An} of A is said to be a par-tition of A if and only if

(i) A = ∪nk=1Ak.(ii) Ai ∩ Aj = ∅ for all i 6= j, 1 ≤ i, j ≤ n.

Example 0.38Let A = {1, 2, 3, 4, 5, 6}, A1 = {1, 2}, A2 = {3, 4}, A3 = {5, 6}. Show that{A1, A2, A3} is a partition of A.

Solution.(i) A1 ∪ A2 ∪ A3 = A.(ii) A1 ∩ A2 = A1 ∩ A3 = A2 ∩ A3 = ∅.

Definition 0.19The number of elements of a set is called the cardinality of the set. Wewrite |A| to denote the cardinality of the set A. If A has a finite cardinalitywe say that A is a finite set. Otherwise, it is called infinite.

Example 0.39What is the cardinality of each of the following sets.(a) ∅.(b) {∅}.(c) {a, {a}, {a, {a}}}.

Solution.(a) |∅| = 0(b) |{∅}| = 1(c) |{a, {a}, {a, {a}}}| = 3

25

Definition 0.20Let A be a set. The power set of A, denoted by P(A), is the set of allpossible subsets of A.

Example 0.40Find the power set of A = {a, b, c}.

Solution.

P(A) = {∅, {a}, {b}, {c}, {a, b}, {a, c},{b, c}, {a, b, c}}

Example 0.41If A ⊆ B then P(A) ⊆ P(B).

Solution.Let X ∈ P(A). Then X ⊆ A. Since A ⊆ B then X ⊆ B. Hence, X ∈ P(B)

Example 0.42Let A be a non-empty set with n elements. Prove that |P(A)| = 2n.

Solution.If n = 0 then A = ∅ and in this case P(A) = {∅}. Thus |P(A)| = 1. Asinduction hypothesis, suppose that if |A| = n then |P(A)| = 2n. Let B =A∪ {an+1}. Then P(B) consists of all subsets of A and all subsets of A withthe element an+1 added to them. Hence, |P(B)| = 2n + 2n = 2 · 2n = 2n+1.

26

Review Problems

Exercise 0.1a. Show that p ∧ q ≡ q ∧ p and p ∨ q ≡ q ∨ p.b. Show that (p ∨ q) ∨ r ≡ p ∨ (q ∨ r) and (p ∧ q) ∧ r ≡ p ∧ (q ∧ r).c. Show that (p∧ q)∨ r ≡ (p∨ r)∧ (q ∨ r) and (p∨ q)∧ r ≡ (p∧ r)∨ (q ∧ r).

Exercise 0.2Construct the truth table for the proposition: ∼ p ∨ q =⇒ r.

Exercise 0.3Construct the truth table for the proposition: (p =⇒ r)⇐⇒ (q =⇒ r).

Exercise 0.4Write negations for each of the following propositions. (Assume that all vari-ables represent fixed quantities or entities, as appropriate.)

a. If P is a square, then P is a rectangle.b. If today is Thanksgiving, then tomorrow is Friday.c. If r is rational, then the decimal expansion of r is repeating.d. If n is prime, then n is odd or n is 2.e. If Tom is Ann’s father, then Jim is her uncle and Sue is her aunt.f. If n is divisible by 6, then n is divisible by 2 and n is divisible by 3.

Exercise 0.5Write the contrapositives for the propositions of Exercise 0.4.

Exercise 0.6Write the converse and inverse for the propositions of Exercise 0.4.

Exercise 0.7Use the proof by contradiction to prove the proposition ”There is no greatesteven integer.”

Exercise 0.8Prove by contradiction that the difference of any rational number and anyirrational number is irrational.

27

Exercise 0.9Use the proof by contraposition to show that if a product of two positive realnumbers is greater than 100, then at least one of the numbers is greater than10.

Exercise 0.10Use the proof by contradiction to show that the product of any nonzerorational number and any irrational number is irrational.

Exercise 0.11Use the method of induction to show that

2 + 4 + 6 + · · ·+ 2n = n2 + n

for all integers n ≥ 1.

Exercise 0.12Use mathematical induction to prove that

1 + 2 + 22 + · · ·+ 2n = 2n+1 − 1

for all integers n ≥ 0.

Exercise 0.13Use mathematical induction to show that

12 + 22 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6

for all integers n ≥ 1.

Exercise 0.14Use mathematical induction to show that

13 + 23 + · · ·+ n3 =

(n(n+ 1)

2

)2

for all integers n ≥ 1.

Exercise 0.15Use mathematical induction to show that

1

1 · 2+

1

2 · 3+ · · ·+ 1

n(n+ 1)=

n

n+ 1

for all integers n ≥ 1.

28

Exercise 0.16Let S(n) =

∑nk=1

k(k+1)!

. Evaluate S(1), S(2), S(3), S(4), and S(5). Make aconjecture about a formula for this sum for general n, and prove your con-jecture by mathematical induction.

Exercise 0.17Show that 2n < (n+ 2)! for all integers n ≥ 0.

Exercise 0.18a. Use mathematical induction to show that n3 > 2n + 1 for all integersn ≥ 2.b. Use mathematical induction to show that n! > n2 for all integers n ≥ 4.

Exercise 0.19Which of the following sets are equal?a. {a, b, c, d}b. {d, e, a, c}c. {d, b, a, c}d. {a, a, d, e, c, e}

Exercise 0.20Let A = {c, d, f, g}, B = {f, j}, and C = {d, g}. Answer each of the followingquestions. Give reasons for your answers.a. Is B ⊆ A?b. Is C ⊆ A?c. Is C ⊆ C?d. Is C a proper subset of A?

Exercise 0.21a. Is 3 ∈ {1, 2, 3}?b. Is 1 ⊆ {1}?c. Is {2} ∈ {1, 2}?d. Is {3} ∈ {1, {2}, {3}}?e. Is 1 ∈ {1}?f. Is {2} ⊆ {1, {2}, {3}}?g. Is {1} ⊆ {1, 2}?h. Is 1 ∈ {{1}, 2}?i. Is {1} ⊆ {1, {2}}?j. Is {1} ⊆ {1}?

29

Exercise 0.22Let A = {b, c, d, f, g} and B = {a, b, c}. Find each of the following:a. A ∪B.b. A ∩B.c. A−B.d. B − A.

Exercise 0.23Indicate which of the following relationships are true and which are false:a. Z+ ⊆ Q.b. R− ⊂ Q.c. Q ⊂ Z.d. Z+ ∪ Z− = Z.e. Q ∩ R = Q.f. Q ∪ Z = Z.g. Z+ ∩ R = Z+

h. Z ∪Q = Q.

Exercise 0.24Let A = {x, y, z, w} and B = {a, b}. List the elements of each of the followingsets:a. A×Bb. B × Ac. A× Ad. B ×B.

Exercise 0.25Let A,B, and C be sets. Prove that if A ⊆ B then A ∩ C ⊆ B ∩ C.

Exercise 0.26Find sets A,B, and C such that A ∩ C = B ∩ C but A 6= B.

Exercise 0.27Find sets A,B, and C such that A ∩ C ⊆ B ∩ C and A ∪ C ⊆ B ∪ C butA 6= B.

Exercise 0.28Let A and B be two sets. Prove that if A ⊆ B then Bc ⊆ Ac.

30

Exercise 0.29Let A,B, and C be sets. Prove that if A ⊆ C and B ⊆ C then A ∪B ⊆ C.

Exercise 0.30Let A,B, and C be sets. Show that A× (B ∪ C) = (A×B) ∪ (A× C).

Exercise 0.31Let A,B, and C be sets. Show that A× (B ∩ C) = (A×B) ∩ (A× C).

Exercise 0.32a. Is the number 0 in ∅? Why?b. Is ∅ = {∅}? Why?c. Is ∅ ∈ {∅}? Why?

Exercise 0.33Let A and B be two sets. Prove that (A−B) ∩ (A ∩B) = ∅.

Exercise 0.34Let A and B be two sets. Show that if A ⊆ B then A ∩Bc = ∅.

Exercise 0.35Let A,B and C be three sets. Prove that if A ⊆ B and B ∩ C = ∅ thenA ∩ C = ∅.

Exercise 0.36Find two sets A and B such that A ∩B = ∅ but A×B 6= ∅.

Exercise 0.37Suppose that A = {1, 2} and B = {2, 3}. Find each of the following:a. P(A ∩ B).b. P(A).c. P(A ∪ B).d. P(A× B).

Exercise 0.38a. Find P(∅).b. Find P(P(∅)).c. Find P(P(P(∅))).

31

Exercise 0.39Determine which of the following statements are true and which are false.Prove each statement that is true and give a counterexample for each state-ment that is false.a. P(A ∪ B) = P(A) ∪ P(B).b. P(A ∩ B) = P(A) ∩ P(B).c. P(A) ∪ P(B) ⊆ P(A ∪ B).d. P(A× B) = P(A)× P(B).

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1 The Concept of a Mapping

The concept of a mapping (aka function) is important throughout mathe-matics. We have been dealing with functions for a long time. You recallfrom calculus that a function is a rule which assigns with each real numberin the domain a unique real number in the codomain. So both the domainand codomain are subsets of the real numbers system. In this section, wewould like to define functions on domains and codomains other than the setof real numbers.

Definition 1.1If S and T are nonempty sets, then a mapping (or a function) from S intoT is a rule which assigns to each member of S a unique member in T. Wecall S the domain and T the codomain of the mapping.

In what follows, we will use the terms function and mapping interchangeably.If α is a function from S to T we shall adopt the notation:

α : S −→ T.

Note that every member x in the domain S is associated to a unique membery of the codomain T . In function notation, we write y = α(x). However,not every element in the codomain need be associated to an element in thedomain.

Example 1.1If α is a mapping from S to S and A is a subset of S then the rule ιA(x) = xdefines a mapping from A into A. We call ιA the identity mapping on A.

Example 1.2Assume that S and T are finite sets containing m and n elements, respec-tively. How many mappings are there from S to T?

Solution.The problem of finding the number of mappings from S to T is the sameas that of computing the number of different ways each element of S can beassigned an image in T. For the first element, there are m possibilities, for thesecond element there are also m possibilitites, etc, for the nth element thereare m possibilities. By the Principle of Counting, there are mn mappingsfrom S to T.

33

Remark 1.1In the notation y = α(x), x is sometimes referred to as the preimage of ywith respect to α.

Sometimes it is necessary to identify the elements of T which can be asso-ciated to some elements in the domain S. This subset of the codomain iscalled the image or range of α (denoted α(S)).

Definition 1.2If α : S −→ T is a mapping and A is a subset of S then the set of all imagesof the members of A will be denoted by α(A).(See Figure 1.1). In set-buildernotation

α(A) = {α(x) : x ∈ A}.

Figure 1.1

Example 1.3The rules α and β defined from the set S = {x, y, z} to T = {1, 2, 3} representmappings with α(S) = T and β(S) = {1, 3}, respectively.(See Figure 1.2) Onthe other hand, the rule γ does not define a mapping for two reasons: first,the member y has no image, and second, the member x is associated to twomembers 1 and 3 of T.

34

Figure 1.2

Example 1.4With α and β as in Example 1.3

α({x, z}) = {2, 3} and β({x, z}) = {1}.

The first operation of mappings that we consider is the equality of two map-pings.

Example 1.5 (Equality of two mappings)We say that two mappings α and β from S into T are equal if and only ifα(x) = β(x) for all x ∈ S, i.e. the range of α is equal to the range of β. Wewrite α = β. When two functions are not equal we write α 6= β. This occurs,when there is a member in the common domain such that α(x) 6= β(x).For example, the mappings α and β of Example 1.3 are different sinceα(y) 6= β(y). On the other hand, the mappings α : R −→ R and β : R −→ Rdefined by α(x) = (x+ 1)2 and β(x) = x2 + 2x+ 1 are equal.

Next, we introduce a family of mappings with the property that the range isthe whole codomain.

Definition 1.3A mapping α : S −→ T is called onto (or surjective) if and only if α(S) = T.That is, if and only if for every member in the codomain there is a memberin the domain associated to it. Using quantifiers, α is onto if and only if∀y ∈ T , ∃x ∈ S such that α(x) = y.

35

Example 1.6In terms of a Venn diagram, a mapping is onto if and only if each element inthe codomain has at least one arrow pointed to it. Thus, since α(S) = T inExample 1.3 then α is onto whereas β is not since β(S) 6= T.

Example 1.7The functions f(x) = x2 and g(x) = sin x are not onto as functions from Rto R. However, if the codomain is restricted to R+ then f is onto. Also, ifthe codomain of g is restricted to [−1, 1] then g is onto.

Example 1.8Let α : S → T be a mapping between finite sets such that the number ofelements of S is less than that of T. Can α be onto? Explain.

Solution.If T has more elements than S and since each element of S is associated toexactly one element in T then some elements in T has no preimages in S.Thus, α can not be onto.

Next, you recall from calculus that a function α from R to R is one-to-oneif and only if its graph satisfies the horizontal line test, i.e. every horizontalline crosses the graph of α at most once. That is, no two different membersof the domain of α share the same member in the range. This concept canbe generalized to any sets.

Definition 1.4A mapping α : S −→ T is called one-to-one (or injective) if and only iffor any x1, x2 ∈ S

x1 6= x2 implies α(x1) 6= α(x2)

that is, unequal elements in the domain of α have unequal images in therange.

Example 1.9In terms of a Venn diagram, a mapping is one-to-one if and only if no twoarrows point to a same member in the codomain. The function α in Exam-ple 1.3 is one-to-one whereas β is not since x 6= z but β(x) = β(z).

36

Example 1.10Show that the functions α(x) = x2 and β(x) = sinx defined on the set R arenot one-to-one functions. Modify the domain of each so that they becomeone-to-one functions.

Solution.The domain of α is the set of all real numbers. Since α(−1) = α(1) = 1 thenα is not one-to-one on its domain. Similarly, β is defined for all real numbers.Since β(π

2) = β(5π

2) = 1 then β is not one-to-one. These functions become

one-to-one if α is restricted to either the interval [0,∞) or the interval (−∞, 0]whereas β can be restricted to intervals of the form

[(2k − 1)π

2, (2k + 1)π

2

],

where k is an integer.

Example 1.11Let α : S → T be a mapping between two finite sets such that S has moreelements than its range. Can α be one-to-one? Explain.

Solution.If S has more elements than α(S) then there must exist at least two distinctmembers of S with the same image in α(S). Thus, by Definition 1.4, α cannot be one-to-one.

An equivalent statement to

x1 6= x2 implies α(x1) 6= α(x2) (x1, x2 ∈ S)

is its contrapositive, i.e. the statement

α(x1) = α(x2) implies x1 = x2.

This latter condition is usually much easier to work with than the one givenin the definition of one-to-one as shown in the next example.

Example 1.12The mapping α(x) = 2x − 1 defined on R is a one-to-one function. To seethis, suppose that x1 = x2. Then multiplying both sides of this equalityby 2 to obtain 2x1 = 2x2. Finally, subtract 1 from both sides to obtain2x1 − 1 = 2x2 − 1. That is, α(x1) = α(x2). Hence, α is one-to-one.

37

A mapping can be one-to-one but not onto, onto but not one-one, neitherone-to-one nor onto, or both one-to-one and onto. See Example 1.13. Wesingle out the last case in the next definition.

Definition 1.5A mapping α : S −→ T is called a one-to-one correspondence (or bijec-tive) if and only if α is both one-to-one and onto.

Example 1.13(a) The mapping α : R −→ R defined by α(x) = x3 is a one-to-one cor-respondence. To see this, for any y ∈ R we can find an x ∈ R such thatα(x) = y. Indeed, let x = 3

√y. Hence, α is onto. Now, if x3

1 = x32 then taking

the cube root of both sides to obtain x1 = x2. That is, α is one-to-one.(b) The function β : N −→ N defined by β(n) = 2n is one-to-one but notonto. Indeed, β(n1) = β(n2) implies n1 = n2 so that β is one-to-one. Thefact that β is not onto follows from the fact that no arrow is pointed to thenumbers 1, 3, 5, etc.(c) The function γ : N −→ N defined by

β(n) =

{n+1

2if n is odd

n2

if n is even

is onto. To see this, let n be a positive integer. If n is odd thenm = 2n−1 ∈ Nis also odd. Moreover, β(m) = m+1

2= n. Now, if n is even then m = 2n ∈ N

is also even and βm = m2

= n. β is not one-to-one since β3 = β4 = 2.(d) The Ceiling function α(x) = dxe is the piecewise defined function givenby

dxe = smallest integer greater than or equal to x.

α is neither one-to-one nor onto as seen in Figure 1.3.

38

Figure 1.3

39

Review Problems

Exercise 1.1A Sequence of real numbers {an}∞n=1 can be viewed as a function α. Find thedomain and the codomain of α.

Exercise 1.2Determine whether α : Q×Q→ Q given by

α(ab,c

d

)=a+ c

b+ d

is a mapping.

Exercise 1.3Find an example of a function familiar to you from calculus that satisfies theeqution α(x+ y) = α(x) · α(y). Such a function will be called a homomor-phism.

Exercise 1.4Define a mapping α : N → N such that the equation α(x) = n has twosolutions for each n ∈ N.

Exercise 1.5Consider the functions α : R→ R defined by α(x) = x+ 1, β : R−{1} → Rdefined by β(x) = x2−1

x−1, and γ : R → R defined by γ(x) = 2x+2

2. Which are

equal?

Exercise 1.6A function can be defined in terms of a subset of a Cartesian product of twosets. More precisely, if S and T are two nonempty sets and G is a subset ofthe Cartesian product S×T = {(a, b) : a ∈ S and b ∈ T} such that each firstcomponent of an element of G is associated to exactly one second componentthen G defines a function from S to T.Let S = {1, 2, 3} and T = {u, v, w}. Consider the subsets α = {(1, u), (2, v), (3, w)},β = {(1, u), (2, u), (3, u)}, γ = {(1, u), (3, w)}, and δ = {(1, u), (2, u), (2, v), (3, w)}of S × T. Determine those who are functions and those who are not.

40

Exercise 1.7Let S = {w, x, y, z} and T = {1, 2, 3, 4}, and define α : S → T by α(w) =2, α(x) = 4, α(y) = 2, α(z) = 2 and β(w) = 4, β(x) = 2, β(y) = 3, β(z) = 1.

(a) Is α one-to-one? Is β one-to-one?Is α onto? Is β onto?(b) Let A = {w, y} and B = {x, y, z}. Determine each of the following subsetsof T : α(A), β(B), α(A ∩B), β(A ∪B).

Exercise 1.8Assume that S and T are finite sets containing m and n elements, respec-tively.

(a) How many mappings are there from S to T?(b) How many one-to-one mappings are there from S to T?

Exercise 1.9A mapping f : R → R is one-to-one iff each horizontal line intersects thegraph of f at most once.

(a) Formulate a similar condition for f : R→ R to be onto.(b) Formulate a similar condition for f : R→ R to be one-to-one and onto.

Exercise 1.10Suppose X is a set with 6 elements and Y is a finite set with n elements.Complete the following.

1) There exists an injective α : X → Y if and only if n .2) There exists surjective α : X → Y iff n .3) There exists a bijective α : X → Y iff n .

Exercise 1.11(a) What does it mean for a function not to be one-to-one?(b) What does it mean for a function not to be onto?

Exercise 1.12Let α, β, γ be mappings from Z to Z defined by α(n) = 2n, β(n) = n+ 1 andγ(n) = n3 for each n ∈ Z.

(a) Which of the three mappings are onto?

41

(b) Which of the three mappings are one-to-one?(c) Determine α(N), β(N), and γ(N).

Exercise 1.13For each ordered pairs (a, b) of integers define a mapping αa,b : Z → Z byαa,b(n) = an+ b.

1. For which pairs (a, b) is αa,b onto?2. For which pairs (a, b) is αa,b one-to-one?

Exercise 1.14For each n ∈ Z, define the mapping fn : Z→ Z by fn(x) = nx.

(a) For which values of n is fn onto ?(b) For which values of n is fn one-to-one ?

Exercise 1.15Prove that if A and B are finite sets and f : A → B is a one-to-one corre-spondence then A and B have the same number of elements.

Exercise 1.16Let α : S → T be a function. Suppose that S and T are finite sets with thesame number of elements. Show that α is one-to-one if and only if it is onto.

Exercise 1.17Let S and T be two nonempty sets. Prove that there is a one-to-one corre-spondence between S × T and T × S.

Exercise 1.18Let S be a nonempty set. Show that the identity mapping ιS is one-to-oneand onto.

Exercise 1.19(a) Show that α : R → [−1, 1] defined by α(x) = sinx is surjective but notinjective.(b) Show that β : [0, π

2] → R defined by β(x) = sin x is injective but not

surjective.(c) Show that γ : [0, π

2]→ [0, 1] defined by γ(x) = sinx is bijective.

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Exercise 1.20A small town has only 500 residents. Must there be 2 residents with thesame birthday (month and day)? Why?

Exercise 1.21Let f : A → B and let {Eα}α∈Λ be a collection of subsets of A. The unionof Eαs is defined by ∪αEα = {x ∈ Eα : for some α ∈ Λ}. Similarly, theintersection is defined by ∩αEα = {x ∈ Eα : ∀α ∈ Λ}. Prove that

1. f [∪αEα] = ∪αf [Eα],2. f [∩αEα] ⊆ ∩αf [Eα].

Exercise 1.22Construct an example of a mapping α : S → T such that

(1) α[E ∩ F ] 6= α[E] ∩ α[F ], where E,F ⊆ S.(2) Show that equality in (1) holds only if α is one-to-one.

Exercise 1.23Let A1, A2, B1, and B2 be nonempty sets such that A1 ∩ A2 = ∅ and B1 ∩B2 = ∅. Suppose that fi : Ai → Bi, (i = 1, 2), are given functions. Definef : A1 ∪ A2 → B1 ∪B2 by

f(x) =

{f1(x) if x ∈ A1

f2(x) if x ∈ A2

Prove:(a) f is one-to-one if and only if f1 and f2 are one-to-one.(b) f is onto if and only if f1 and f2 are onto.

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2 Composition. Invertible Mappings

In this section we discuss two procedures for creating new mappings fromold ones, namely, the composition of mappings and invertible mappings.

Composition of Two MappingsComposition is the combination of two or more mappings to form a singlenew mapping.

Definition 2.1Let α : S −→ T and β : T −→ U be two mappings. We define the compo-sition of α followed by β, denoted by β ◦ α, to be the mapping

(β ◦ α)(x) = β(α(x))

for all x ∈ α.

Note carefully that in the notation β ◦α the mapping on the right is appliedfirst. See Figure 2.1

Figure 2.1

Example 2.1Let α and β be given by the Venn diagram of Figure 2.2.

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Figure 2.2

Then(β ◦ α)(x) = β(α(x)) = β(2) = c(β ◦ α)(y) = β(α(y)) = β(1) = b(β ◦ α)(z) = β(α(z)) = β(3) = a

Example 2.2Let f and g be two functions given as sets of (x, y) points:

f = {(−2, 3), (−1, 1), (0, 0), (1,−1), (2,−3)}, g = {(−3, 1), (−1,−2), (0, 2), (2, 2), (3, 1)}.

Find (f ◦ g)(0).

Solution.Since g(0) = 2 and f(2) = −3 then (f ◦ g)(0) = f(g(0)) = f(2) = −3.

Example 2.3The following example shows how to find the formula of the compositionmapping given the formulas for both α and β. Let α : R −→ R and β :R −→ R be defined by α(x) = x2 + 2 and β(x) = x − 1 respectively. Findβ ◦ α and α ◦ β.

Solution.Using the definition of composition we have

(β ◦ α)(x) = β(α(x))= β(x2 + 2)= (x2 + 2)− 1 = x2 + 1

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while(α ◦ β)(x) = α(β(x))

= α(x− 1)= (x− 1)2 + 2= x2 − 2x+ 3

This example, shows that, in general, α ◦ β and β ◦ α need not be equal.

In the following two theorems, we discuss the question of either composingtwo onto mappings or two one-to-one mappings.

Theorem 2.1Assume that α : S −→ T and β : T −→ U are two mappings.

(a) If α and β are onto then the composition β ◦ α is also onto.(b) If β ◦ α is onto then β is onto.

Proof.(a) The mapping β ◦ α is a mapping from S to U. So, let u ∈ U. Since β isonto then there is a t ∈ T such that β(t) = u. Now, since α is onto then thereis an s ∈ S such that α(s) = t. Thus, given u ∈ U we can find an s ∈ S suchthat (β ◦ α)(s) = β(α(s)) = β(t) = u. This says that β ◦ α is onto.

(b) Suppose now that β ◦ α is onto. Pick an arbitrary element u ∈ U. Sinceβ ◦α is onto then there is an s ∈ S such that β(α(s)) = u. Let t = α(s) ∈ T.Then β(t) = u. This shows that β is onto.

Example 2.4Consider the two mappings α : N −→ N defined by α(n) = 2n and β : N −→N defined by

β(n) =

{n+1

2if n is odd

n2

if n is even

Show that β and β ◦ α are onto but α is not.

Solution.First, we show that β is onto. Let n ∈ N. If n is even then 2n is even andβ(2n) = n. If n is odd then 2n − 1 is odd and β(2n − 1) = n. Thus, β isonto. One can easily check that β ◦ α = ιN. Since the identity map is ontothen β ◦α is onto. The mapping α is not onto since odd positive integers donot have preimages.

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Theorem 2.2Assume that α : S −→ T and β : T −→ U are two mappings.

(a) If α and β are one-to-one then β ◦ α is also one-to-one.(b) If β ◦ α is one-to-one then α is one-to-one.

Proof.(a) Suppose that α and β are one-to-one. Suppose that (β ◦ α)(s1) =(β ◦ α)(s2) for some s1, s2 ∈ S. This implies that β(α(s1)) = β(α(s2)). Sinceβ is one-to-one then α(s1) = α(s2). Now since α is one-to-one then s1 = s2.Thus, β ◦ α is one-to-one.

(b) Assume that β ◦ α is one-to-one. Suppose that α(s1) = α(s2). Sinceβ is a well-defined mapping then β(α(s1)) = β(α(s2)). Since β ◦ α is one-to-one then s1 = s2. This shows that α is one-to-one.

Example 2.5Consider the two mappings α : N −→ N defined by α(n) = 2n and β : N −→N defined by

β(n) =

{n+1

2if n is odd

n2

if n is even

Show that α and β ◦ α are one-to-one but β is not.

Solution.Since α(n1) = α(n2) implies 2n1 = 2n2 and this in turns implies that n1 = n2

then α is one-to-one. Since β ◦ α = ιN and ιN is one-to-one then β ◦ α isone-to-one. The mapping β is not one-to-one since β(1) = β(2) with 1 6= 2.

Example 2.6Let α : S −→ T, β : T −→ U, and γ : U −→ V be three mappings such thatγ ◦ (β ◦ α) and (γ ◦ β) ◦ α are well defined. Show that

γ ◦ (β ◦ α) = (γ ◦ β) ◦ α

Solution.Note first that the mappings γ◦(β◦α) and (γ◦β)◦α have the same codomainV. The following argument shows that γ ◦ (β ◦ α) and (γ ◦ β) ◦ α have the

47

same range.Let s be in S then

[γ ◦ (β ◦ α)](s) = γ((β ◦ α)(s))= γ(β(α(s)))= (γ ◦ β)(α(s))= [(γ ◦ β) ◦ α](s)

Invertible MappingsIn this section we consider special kind of mappings which have the propertythat for each output value we can work our way backwards to find the uniqueinput that produced it.

Let α : S −→ T and β : T −→ S be two given mappings.

Definition 2.2We say that β is an inverse of α if and only if β ◦ α = ιS and α ◦ β = ιT . Inthis case we say that α is invertible.

An invertible mapping has a unique inverse as shown in the next theorem.

Theorem 2.3If α : S → T is invertible then its inverse is unique.

Proof.Suppose that α1 : T −→ S and α2 : T −→ S are two inverses of α. Thenfrom Definition 2.2 we have α1 ◦ α = α2 ◦ α = ιS and α ◦ α1 = α ◦ α2 = ιT .We want to show that the mappings α1 and α2 are equal. That is, we mustshow that α1(t) = α2(t) for each t ∈ T . Indeed,

α1(t) = ιS(α1(t))= (α2 ◦ α)((α1(t))= α2((α ◦ α1)(t))= α2(ιT (t))= α2(t)

Thus, α1 = α2.

Definition 2.3We denote the unique inverse of a mapping α by α−1.

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Example 2.7Show that the mapping α in Example 2.1 is invertible and find its inverse.

SolutionThe inverse of α is defined by

α−1(1) = y, α−1(2) = x, α−1(3) = z.

One can easily check that α ◦α−1 = ιT and α−1 ◦α = ιS where S = {x, y, z}and T = {1, 2, 3}. Looking closely at the Venn diagram we see that α−1 isgotten by reversing the direction of the arrows under α.

The following theorem characterizes those mappings that are invertible.

Theorem 2.4A mapping α : S −→ T is invertible if and only if α is one-to-one and onto.

Proof.Suppose first that α is invertible with inverse α−1 : T −→ S. We will showthat α is both one-to-one and onto. To see that α is one-to-one, we assumethat α(s1) = α(s2), where s1, s2 ∈ S, and show that s1 = s2. Indeed,

s1 = ιS(s1) = (α−1 ◦ α)(s1)= α−1(α(s1)) = α−1(α(s2))= (α−1 ◦ α)(s2) = ιS(s2) = s2

Next, to show that α is onto we pick an arbitrary member t in T and showthat there is an s in S such that α(s) = t. Indeed, since t is in T thent = ιT (t) = (α ◦ α−1)(t) = α(α−1(t)) = α(s) where s = α−1(t) ∈ S. Thisshows that α is onto.Conversely, suppose that α is one-to-one and onto. We will find a mappingβ : T −→ S such that α ◦ β = ιT and β ◦ α = ιS. Let t ∈ T. Since α is ontothen there is an element s ∈ S such that α(s) = t. s is unique, for if s′ ∈ Sis such that α(s′) = t then α(s) = α(s′). But α is one-to-one so that s = s′.Hence, for each t ∈ T there is a unique s ∈ S such that α(s) = t. Defineβ : T −→ S by β(t) = s. Then β satisfies the following properties:

(α ◦ β)(t) = α(β(t)) = α(s) = t, t ∈ T

and(β ◦ α)(s) = β(α(s)) = β(t) = s, s ∈ S.

That is, α◦β = ιT and β◦α = ιS. According to Definition 2.2, α is invertible

49

Example 2.8Show that the function f : R −→ R defined by f(x) = x2 is not invertible.

Solution.Indeed, since f(−1) = f(1) and 1 6= −1 then f is not one-to-one. ByTheorem 2.4, f can’t be invertible.

Example 2.9Show that the function f(x) = x3 is invertible and find a formula for theinverse.

Solution.We have seen (See Example 1.13(a), Sec. 1.1) that f(x) = x3 is one-to-oneand onto so that it has an inverse. To find the inverse we proceed as follows:

1. Replace f(x) by y to obtain y = x3.2. Switch the letters x and y to obtain x = y3.3. Solve for y in terms of x to obtain y = 3

√x.

4. Replace y by f−1(x) to obtain f−1(x) = 3√x.

Theorem 2.5Let α : S −→ T and β : T −→ U be two given mappings.

(a) If α is invertible then α−1 is also invertible with (α−1)−1 = α.(b) If α and β are both invertible then β ◦ α is invertible with inverse(β ◦ α)−1 = α−1 ◦ β−1.

Proof.(a) Suppose α is invertible with inverse α−1 : T −→ S. Since α−1 ◦ α = ιSand α ◦ α−1 = ιT then by Definition 2.2, α−1 is invertible with inverse α.

(b) Suppose α and β are invertible. Then α−1◦α = ιS, β−1◦β = α◦α−1 = ιT ,

and β ◦ β−1 = ιU . Thus, for any u ∈ U we have

[(β ◦ α) ◦ (α−1 ◦ β−1)](u) = [β ◦ (α ◦ α−1) ◦ β−1](u)= β ◦ (ιT (β−1(u))= (β ◦ β−1)(u)= ιU(u) = u

It follows that (β ◦ α) ◦ (α−1 ◦ β−1) = ιU . Similarly, one can show that(α−1 ◦ β−1) ◦ (β ◦ α) = ιS. By Definition 2.2, β ◦ α is invertible with inverseα−1 ◦ β−1.

50

Review Problems

Exercise 2.1Let f and g be two functions given as sets of (x, y) points:

f = {(−2, 3), (−1, 1), (0, 0), (1,−1), (2,−3)}, g = {(−3, 1), (−1,−2), (0, 2), (2, 2), (3, 1)}.

Find (g ◦ f)(1).

Exercise 2.2Let f(x) = log5 (x+ 1) and g(x) = x2 + 1.

(a) Find (g ◦ f)(x).(b) Find the domain and range of g ◦ f.(c) Evaluate (g ◦ f)(4).

Exercise 2.3Given h(x) = (x+ 1)2 + 2(x+ 1)− 3. Determine two functions f(x) and g(x)such that h(x) = f(g(x)).

Exercise 2.4You work 40 hours a week at a furniture store. You receive a $220 weeklysalary, plus a 3% commision on sales over $5000. Assume that you sellenough this week to get the commission. Given the functions f(x) = 0.03xand g(x) = x − 5000, which of (f ◦ g)(x) and (g ◦ f)(x) represents yourcommission?

Exercise 2.5Let α, β, and γ be mappings from Z to Z defined by α(n) = 2n, β(n) = n+1,and γ(n) = n2. Write a formula for each of the composition below. Also, de-termine the range in each case.

(i) α ◦ α(ii) α ◦ β(iii) β ◦ γ.

Exercise 2.6Prove that if α : S → T then α ◦ ιS = α and ιT ◦ α = α.

51

Exercise 2.7Consider f and g, mappings R to R, defined by f(x) = sin x and g(x) = 2x.Is f ◦ g equal to g ◦ f?

Exercise 2.8(a) Prove that if α : S → T, β : T → U, γ : T → U, α is onto, and β◦α = γ◦α,then β = γ.(b) Give an example to show that the condition ”α is onto” cannot be omittedfrom Part (a).

Exercise 2.9(a) Prove that if β : S → T, γ : S → T, α : T → U, α is one-to-one, andα ◦ β = α ◦ γ, then β = γ.(b) Give an example to show that the condition ”α is one-to-one” cannot beomitted from Part (a).

Exercise 2.10Let f : R→ R be the map given by f(x) = x2. Let

A = [1, 2] = {x ∈ R| 1 ≤ x ≤ 2},

B = (−1, 1) = {x ∈ R| − 1 < x < 1},

C = (4, 9) = {x ∈ R| 4 < x < 9},

D = [0, 9] = {x ∈ R| 0 ≤ x ≤ 9}.

Find

(a) f [A] (b) f [B] (c) f−1[C] (d) f−1[D].

Exercise 2.11Suppose that α : S −→ T is a given mapping and A ⊆ S. We know thatα(A) is empty only when A is empty. If B ⊆ T then we define

α−1(B) = {x ∈ S : α(x) ∈ B}.

Give an example of a mapping α such that B 6= ∅ but α−1(B) = ∅.(Hint:Use the mapping in Exercise 2.10)

52

Exercise 2.12Let f : A→ B and let {Fα}α∈Λ be a collection of subsets of B. Prove that

1. f−1[∪αFα] = ∪αf−1[Fα],2. f−1[∩αFα] = ∩αf−1[Fα].

Exercise 2.13Let S = {−2, 1, 2} and T = {1, 4, 9}. Consider the mapping α : S → Tdefined by

α(−2) = 4, α(1) = 1, α(2) = 4.

(a) Show that α is neither one-to-one nor onto.(b) Show that α−1(α(A)) 6= A, where A = {−2, 1}.(c) Show that α(α−1(B)) 6= B, where B = {4, 9}.

Exercise 2.14(a) Let α : S → T, where S and T are nonempty. Prove that α has theproperty α−1(α(A)) = A for every subset of S if and only if α is one-to-one.(b) Prove that α has the property that α(α−1(B)) = B for every subset Bof T if and only if α in onto.

Exercise 2.15Let α : S → T and β : T → U. Show that if β ◦ α is invertible then β is ontoand α is one-to-one.

53

3 Binary Operations

We are used to addition and multiplication of real numbers. These operationscombine two real numbers to generate a unique single real number. So wecan look at these operations as functions on the set

R× R = {(a, b) : a ∈ R and b ∈ R}

defined by+ : R× R −→ R

(a, b) −→ a+ b

and· : R× R −→ R

(a, b) −→ a · bThese operations are examples of a binary operation. The general definitionof a binary operation is as follows.

Definition 3.1A binary operation on a set S is a mapping ∗ that assigns to each orderedpair of elements of S a uniquely determined element of S. That is, ∗ : S ×S −→ S is a mapping. The set S is said to be closed under the operation ∗.

The image ∗(a, b) will be denoted by a ∗ b.

Example 3.1Addition and multiplication are binary operations on the set Z of integersso that this set is closed under these operations. However, Z is not closedunder the operation of division since 1÷ 2 is not an integer.

Example 3.2The ”ordered pair” statement in Definition 3.1 is critical. For example,consider the binary operation ∗ defined on the set N by a ∗ b = ab. Then2 ∗ 3 = 23 = 8 and 3 ∗ 2 = 32 = 9. That is, 2 ∗ 3 6= 3 ∗ 2.

Example 3.3 (Cayley’s Tables)The idea of a binary operation is just a way to produce an element of a setfrom a given pair of ordered elements of the same set. In the case of a finiteset we could list the rule in a table which we’ll call a multiplication table orCayley’s table. For example, the following is the multiplication table of abinary operation ∗ : {a, b} −→ {a, b}.

54

* a ba a bb b a

In studying binary operations on sets, we tend to be interested in thoseoperations that have certain properties which we discuss next.

Definition 3.2A binary operation ∗ on a set S is said to be associative if it satisfies theassociative law:

a ∗ (b ∗ c) = (a ∗ b) ∗ c

for all a, b, c ∈ S.

The associative property allows us to speak of a ∗ b ∗ c without having toworry about whether we should find the answer to a ∗ b first and then thatanswer ”multiplied” by c rather than evaluate b ∗ c first and then ”multiply”a with that answer. Which ever way we process the expression we end upwith the same element of the set. Note though that it does not say we cando the product in any order (i.e. a∗b and b∗a may not have the same value).

Example 3.41. The operations ” + ” and · on R are associative.2. The operation ”− ” on R is not associative since 2− (3− 4) 6= (2− 3)− 4.(Notice that if the associative law fails for just one triple (a, b, c) then theoperation is not associative).3. The operation ∗ defined by a ∗ b = ab on the set N is not associative since2 ∗ (3 ∗ 2) = 512 and (2 ∗ 3) ∗ 2 = 64.

Definition 3.3A binary operation ∗ on a set S is said to be commutative if it satisfies thecondition:

a ∗ b = b ∗ a

for all a, b,∈ S. In this case, the order in which elements are combined doesnot matter.

Remark 3.1When a set with a binary operation is given by a Cayley’s table then the

55

operation is commutative if and only if equal elements appear in all posi-tions that are symmetrically placed relative to the diagonal from upper leftto lower right. That is, to check whether an operation defined by a Cayley’stable is commutative, simply draw a diagonal line from upper left to lowerright, and see if the table is symmetric about this line. For example, theoperation ∗ defined by the table below is commutative.

* a b c da a b c db b c d ac c d a bd d a b c

Example 3.5The binary operations of addition and multiplication on R are both commu-tative. However, the binary operation of subtraction on R does not satisfythe commutative law since 5− 7 6= 7− 5.

Example 3.6The binary operation on R defined by a ∗ b = a+ b− 1 is commutative since

a ∗ b = a+ b− 1 = b+ a− 1 = b ∗ a.

Example 3.7Show that the binary operation on R defined by a∗b = 1+ab is commutativebut not associative.

Solution.For any real numbers a and b we have a ∗ b = 1 + ab = 1 + ba = b ∗ a wherewe used the fact that multiplication in R is commutative. Now, by lettinga = 0, b = 1, and c = −1 then a∗ (b∗ c) = a∗ 0 = 1 and (a∗ b)∗ c = 1∗ c = 0.Thus, ∗ is not associative.

Definition 3.4Let S be a set on which there is a binary operation ∗. An element e of thisset is called a left identity if for all a ∈ S, we have e ∗ a = a. Similarly, anelement e is a right identity if a ∗ e = a for each a ∈ S.

56

Example 3.8Given a binary operation on a set.

1. There might be left identities which are not right identities and vice-versa. For example, the operation a ∗ b = a on the set R has 2 as a rightidentity which is not a left identity. The set R with the operation a ∗ b = bhas 2 as a left identity which is not a right identity.

2. There might be many left or right identity elements. The set R withthe operation a ∗ b = a, every number is a right identity. With the operationa ∗ b = b, every number is a left identity.

3. There might be no left or right identity elements. For example, theset {2, 3, 4, · · ·} has no left or right identity elements under the operationa ∗ b = a · b

We tend to be familiar with the situation in which there is a unique iden-tity. As soon as an operation has both a left and a right identity, they arenecessarily unique and equal as shown in the next theorem.

Theorem 3.1If S is a set with a binary operation ∗ that has a left identity element e1 anda right identity element e2 then e1 = e2 = e.

Proof.Let e1 ∈ S be a left identity element and e2 ∈ S be a right identity element.Then

e1 = e1 ∗ e2(since e2 is a right identity)= e2(since e1 is a left identity)

Definition 3.5An element which is both a right and left identity is called the identityelement(Some authors use the term two sided identity.) Thus, an elementis an identity if it leaves every element unchanged.

Remark 3.2Note that an identity (left or right or both) for one operation does not haveto be an identity for another operation. Think of addition and multiplicationon the reals where the identities are 0 and 1 respectively.

57

Example 3.9The operation a ∗ b = a + b − 1 on the set of integers has 1 as an identityelement since 1 ∗ a = 1 + a− 1 = a and a ∗ 1 = a+ 1− 1 = a for all integera.

Example 3.10Show that the operation a∗b = 1+ab on the set of integers Z has no identityelement.

—noindent Solution.If e is an identity element then we must have a ∗ e = a for all a ∈ Z. Inparticular, 1 ∗ e = e. But this imply that 1 + e = 1 or e = 0. Since 2 ∗ 0 6= 0then e does not exist.

Whenever a set has an identity element with respect to a binary operationon the set, it is then in order to raise the question of inverses.

Definition 3.6Suppose that an operation ∗ on a set S has an identity element e. Let a ∈ S.If there is an element b ∈ S such that a ∗ b = e then b is called a rightinverse of a. Similarly, if b ∗ a = e then b is called a left inverse.

Example 3.111. An element can have no left or right inverses. For example, the number 2has no left or right inverse with respect to multiplication on the set of integers.

2. There might be a left inverse which is not a right inverse and vice versa.For example, consider the set M(Z) of all functions from the set of integersinto itself. Then the operation of composition is a binary operation on M(Z).Consider the two functions f(n) = 2n and

g(n) =

{n2

if n is even4 if n is odd

Then (g ◦ f)(n) = n for all n ∈ Z. That is, g is a left inverse of f. However,since

(f ◦ g)(n) =

{n if n is even8 if n is odd

then g is not a right inverse since f ◦ g 6= ιZ

58

Suppose that an element a ∈ S has both a left inverse and a right inversewith respect to a binary operation ∗ on S. Under what condition are the twoinverses equal?

Theorem 3.2Let S be a set with an associative binary operation ∗ and identity elemente. Let a, b, c ∈ S be such that a ∗ b = e and c ∗ a = e. Then b = c.

Proof.Indeed,

b = e ∗ b= (c ∗ a) ∗ b= c ∗ (a ∗ b)= c ∗ e= c

Definition 3.7If a has both a left and right inverse then we say that a has two-sidedinverse or simply an inverse element.

Example 3.12Consider the operation ∗ on the set of integers defined by a ∗ b = a + b− 1.We will show that each integer has an inverse under this operation. Indeed,let x be an integer. Let y be a right inverse of x. Then x ∗ y = 1. That is,x+ y − 1 = 1. Solving for y we find y = −x+ 2. This is also a left inverse ofx since (−x+ 2) ∗ x = −x+ 2 + x− 1 = 1.

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Review Problems

Exercise 3.1Which of the following equations define operations on the set of integers? Ofthose that do, which are associative?Which are commutative? Which haveidentity elements?

1. a ∗ b = ab+ 1.2. a ∗ b = a.3. a ∗ b = a2 + b2.4. a ∗ b = 3.

Exercise 3.2Does (a, b)→ ab define an operation on the set of all integers?

Exercise 3.3If ∗ is an operation on S and T is a subset of S that is closed with respectto ∗ then two of the following three statements are necessarily true, but onemay be false. Which two are true?(a) If ∗ is associative on S, then ∗ is associative on T.(b) If there is an identity element for ∗ on S, then there is an identity elementfor ∗ on T.(c) If ∗ is commutative on S, then ∗ is commutative on T.

Exercise 3.4Complete the following table in such a way that makes ∗ commutative.

* a b c da a b db cc c d a bd a c

Exercise 3.5Determine the smallest subset A of Z such that 2 ∈ A and A is closed withrespect to addition.

Exercise 3.6Determine the smallest subset A of Q such that 2 ∈ A and A is closed withrespect to addition and division.

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Exercise 3.7How many different operations are there on a 1-element set? 2-element set?3- element set? n-element set?

Exercise 3.8Assume that ∗ is an associative operation on S and that a ∈ S. Let

C(a) = {x ∈ S : a ∗ x = x ∗ a}.

Prove that C(a) is closed with respect to ∗.

Exercise 3.9Assume that ∗ is an operation on S with identity element e and that

x ∗ (y ∗ z) = (x ∗ z) ∗ y

for all x, y, z ∈ S. Prove that ∗ is commutative and associative.

Exercise 3.10List all the binary operations defined on set with two elements. (See Exercise3.7)

Exercise 3.11Given the Cayley’s table for a binary operation ∗ defined on the set S ={a, b, c, d}

* a b c da b c a bb c d b ac a b c dd a b d d

(a) Is ∗ commutative? Why?(b) Determine whether there is an identity element in S for ∗.(c) If there is an identity element, which elements have inverses?

Exercise 3.12(a) Prove that the set of all injective mappings from S to S is closed undermapping composition.(b) Prove that the set of all surjective mappings from S to S is closed undermapping composition.

61

Exercise 3.13Let M be the set of all rectangular arrays of two rows and two columns[

a bc d

]Such an array is called a 2-by-2 matrix. Let addition be defined for

elements of M by[a bc d

]+

[x yz w

]=

[a+ x b+ yc+ z d+ w

](a) Prove that + is commutative.(b) Prove that + is associative.(c) Find the identity element for addition.(d) What is the additive inverse of[

a bc d

]in M?

Exercise 3.14LetM be the collection defined in the previous exercise. Define the operationof multiplication by[

a bc d

] [x yz w

]=

[ax+ bz ay + bwcx+ dz cy + dw

](a) Prove that multiplication is associative.(b) Prove that multiplication is not commutative.(c) Find the identity element of M with respect to multiplication.(d) Show that if ad− bc 6= 0 then[

a bc d

]−1

=

[d

ad−bc−b

ad−bc−c

ad−bca

ad−bc

]Exercise 3.15Suppose that there are two binary operations ∗ and # defined on a set S.We say that ∗ is distributive with respect to # if for all a, b, c ∈ S, we have

a ∗ (b#c) = (a ∗ b)#(a ∗ c).

Prove that: A ∩ (B ∪ C) = (A ∩B) ∪ (A ∩ C), for any sets A,B,C.

62

4 Composition of Mappings as a Binary Op-

eration

For a nonempty set S, letM(S) be the set of all mappings from S to S. Wehave seen that composition of mappings defines a binary operation inM(S).In this section, we want to study the properties of this operation.First, we discuss the question of commutativity and inverse elements. Ingeneral, composition is not commutative in M(S). Also, not every elementof M(S) is invertible.

Example 4.1Let S be a nonempty set.

(a) Show that composition in M(S) is not , in general, commutative.(b) Show that not every element of M(S) is invertible.

Solution.(a) Consider the setM(Z) of all functions from the set of integers into itself.Then the operation of composition is a binary operation onM(Z). Considerthe two functions α(n) = 2n and

β(n) =

{n2

if n is even4 if n is odd

Then (β ◦ α)(n) = n for all n ∈ Z. That is, β ◦ α = ιZ. However, since

(α ◦ β)(n) =

{n if n is even8 if n is odd

then β ◦ α 6= α ◦ β. Hence, composition is not commutative.

(b) From part (a), we see that β ◦ α = ιZ. Thus, β is a left inverse of α.However, α ◦ β 6= ιZ so that β is not a right inverse of α. Hence, α is notinvertible.

Under what conditions on S, composition in M(S) is commutative?

Theorem 4.1Let S be a nonempty set such that |S| < 2. Then composition of mappingsin M(S) is commutative.

63

Proof.We will show that if |S| ≥ 2 then composition of mappings is not commuta-tive. Indeed, if |S| ≥ 2 then there exist two distinct elements a and b of S.Define α, β : S −→ S as follows:

α(a) = a α(b) = aβ(a) = b β(b) = b.

Then (α ◦ β)(a) = α(β(b)) = α(b) = a and (β ◦ α)(a) = β(α(a)) = β(b) = b.Since a 6= b then α ◦ β 6= β ◦ α. That is, composition in M(S) is notcommutative.

The following theorem summarizes the two properties of composition in m(S).

Theorem 4.2Let S be a nonempty set.

(a) Composition in M(S) is associative.(b) ιS is the identity element in M(S).

Proof.(a) Let x ∈ S. Then for any α, β, γ ∈M(S), we have

[α ◦ (β ◦ γ)](x) = α(β ◦ γ(x))= α(β(γ(x)))= (α ◦ β)(γ(x))= [(α ◦ β) ◦ γ](x)

Thus, α ◦ (β ◦ γ) = (α ◦ β) ◦ γ.(b)If x ∈ S and α ∈M(S) then

(α ◦ ιS)(x) = α(ιS(x)) = α(x)

so that α ◦ ιS = α. Similarly, ιS ◦ α = α. Hence, ιS is the identity element ofM(S) under the operation of composition.

Example 4.2Is is true that if a nonempty set S is closed under a binary operation ∗ thenevery subset A of S is also closed under ∗?

64

Solution.If a set is closed under an operation then this does not mean that everysubset is also closed under this operation. For example, R is closed underthe operation of subtraction whereas N is not. Thus, composition is a bi-nary operation on a subset ofM(S) if and only if the subset is closed undercomposition.

Now, let’s look at the set I(S) of all invertible mappings from S to S. Clearly,I(S) is a subset of M(S).

Theorem 4.3(a) I(S) is closed under composition.(b) Composition is associative on I(S).(c) ιS is the identity element of I(S).(d) Every element of I(S) is invertible.

Proof.(a) Let α and β be two invertible mappings from S to S. Then by Theorem2.5(b), α ◦ β is also invertible. That is, α ◦ β ∈ I(S). Hence, I(S) is closedunder composition.(b) Since I(S) is closed with respect to composition and composition onM(S) is associative there then it is certainly associative when restricted onI(S).(c) Since ι−1

S = ιS then ιS ∈ I(S). Since ιS is the identity element of M(S)then ιS is the identity element of I(S).(d) Follows from the definition of I(S).

The following theorem can be used to test whether a mapping is one-to-one.

Theorem 4.4Let S be a nonempty set and α ∈M(S). Then α is one-to-one if and only ifthere exists a β ∈M(S) such that β ◦ α = ιS.

Proof.Suppose first that α is one-to-one. Pick an element x0 ∈ S and defineβ : S → S as follows

β(x) =

{y if α(y) = xx0 if α(y) 6= x,∀y ∈ S

65

We show that β ∈ M(S). Indeed, if β(x) = y and y′ ∈ S is such thatβ(x) = y′ then α(y) = α(y′ and since α is one-to-one then y = y′. Thatis, y is the unique element such that β(x) = y. If β(x) = x0 then x0

is the unique element such that β(x) = x0 since x0 is fixed. It remainsto show that β ◦ α = ιS. To see this, let x ∈ S and α(x) = y. Then(β ◦ α)(x) = β(α(x)) = β(y) = x = ιS(x). Thus, β ◦ α = ιS.Conversely, suppose that β ∈M(S) such that β ◦α = ιS. Since ιS is one-to-one then β◦α is one-to-one and therefore by Theorem 2.2(b) , α is one-to-one.

For testing onto mappings we have

Theorem 4.5Let S be a nonempty set and α ∈M(S). Then α is onto if and only if thereexists a β ∈M(S) such that α ◦ β = ιS.

Proof.Suppose first that α is onto. Then for each y ∈ S there is an x ∈ S suchthat α(x) = y. Define β : S → S by β(y) = x. Then (α ◦ β)(y) = α(β(y)) =α(x) = y so that α ◦ β = ιS.Conversely, suppose that β ∈M(S) is such that α ◦ β = ιS. Since ιS is ontothen α ◦ β is onto. By Theorem 2.1 (b), α is also onto.

Many important operations involve composition on special sets of invertiblemappings. We close this section by giving two examples. The first exampleexhibits an example of a set of mappings where composition is commutative.

Example 4.3Let P denote the Cartesian plane. Let Gp be the set of all rotations abouta fixed point p. If two rotations differ by a multiple of 360◦ then we saythat they are equal. If α and β are two elements of Gp then α ◦ β is therotation obtained by first applying β and then applying α. Thus, Gp is closedunder composition. By Theorem 4.2, composition is associative. An identityelement of Gp is the rotation of 0◦. Each rotation has an inverse: rotation ofthe same magnitude in the opposite direction. Finally, as an operation onGp, composition is commutative.

Example 4.4Let L be the set of all linear mappings αa,b : R −→ R defined by αa,b(x) =

66

ax+ b where a and b are two real numbers with a 6= 0.

(i) L is a set of invertible mappings from R into R.Indeed, if αa,b = ax+ b with a 6= 0 then α−1

a,b(x) = α 1a,− b

a(x) = 1

ax− b

a.

(ii) L is closed under compositionTo see this, pick two members L, say αa,b = ax + b and αc,d(x) = cx + d,where a 6= 0 and c 6= 0. Note that ac 6= 0. Moreover,

(αa,b ◦ αc,d)(x) = αa,b(αc,d(x))= αa,b(cx+ d) = a(cx+ d) + b= (ac)x+ ad+ b = αac,ad+b(x)

Thus, αa,b ◦ αc,d ∈ L.

(iii) Composition is associative on LSince L is a subsetM(R) and composition is associative onM(R) then com-position is also associative on L.

(iv) α1,0 is the identity element of LTo see this, let αa,b ∈ L. Then by (ii)

αa,b ◦ α1,0 = αa,b

andα1,0 ◦ αa,b = αa,b.

Remark 4.1The identity element can be found as follows: If αe,f is the identity elementof L then for any αa,b ∈ L we must have

αe,f ◦ αa,b = αa,b ◦ αe,f = αa,b.

This and (ii) imply that ea = a and eb+ f = b. Thus, e = 1 and f = 0.

Remark 4.2Note that, αa,0(x) = ax, a > 1 is magnification since it magnifies the distanceof each point from the origin by a factor of a. Also, α1,b(x) = x + b, b > 0 isa translation of x, b units to the right. Finally, note that αa,b = α1,b ◦ αa,0so that for a > 1 and b > 0, αa,b corresponds to a magnification followed bya translation.

67

Review Problems

Exercise 4.1Consider the set S = {a, b}.

(a) Find the four elements of M(S) = {π, ρ, σ, θ}.(b) Construct the Cayley table for composition.(c) What is the identity element?(d) Is ◦ commutative as an operation on M(S)?(e) Which elements of M(S) are invertible?

Exercise 4.2Consider the set Gp = {ρ1, ρ2, ρ3, ρ4}, where ρ1, ρ2, ρ3, and ρ4 denote clock-wise rotation through 0◦, 90◦, 180◦, 270◦.

(a) Construct the Cayley table for composition as an operation on Gp.(b) Is there an identity element?(c) Does each element have ab inverse?

Exercise 4.3Let B and C be subsets of L.

B = {αa,0 : a ∈ R and a 6= 0}C = {α1,b : b ∈ R}

(a) Verify that B is closed under the operation of composition. Is ◦ associa-tive? Commutative? Is there an identity element?(b) Verify that C is closed under the operation of composition. Is ◦ associa-tive? Commutative? Is there an identity element?(c) Verify that each mapping in L is the composition of a mapping in B anda mapping in C.

Exercise 4.4Let S = R− {0, 1}. Consdier the mappings from S to S defined by

α1(x) = x , α2(x) = 1x

, α3(x) = 1− x,α4(x) = 1− 1

x, α5(x) = 1

1−x , α6(x) = xx−1

68

(a) Verify that ◦ is a binary composition on {α1, · · · , α6} by constructing aCayley table.(b) Is there an identity element?(c) Show that each of the six elements has an inverse.(d) Is ◦ commutative?(e) Is ◦ associative?

Exercise 4.5Is composition a binary operation on the set of all continuous functions fromR to R?

Exercise 4.6Is composition a binary operation on the set of all differentiable functionsfrom R to R?

Exercise 4.7Prove by induction that the composition of any finite number of invertiblemappings is invertible.

Exercise 4.8Let f : S → T and g : T → U be two mappings such that g ◦ f is invertible.Show that f is one-to-one and g is onto.

Exercise 4.9Let S and T be nonempty sets. Prove the following: There exists a one-to-onemapping f : S → T if and only if there exists an onto mapping g : T → S.

Exercise 4.10Prove the following: (i) f : A→ B is one-to-one if and only if f ◦ g = f ◦ himplies g = h for all maps g, h : B → A.(ii) If A has at least two elements, then f : A → B is onto if and only ifg ◦ f = h ◦ f implies g = h for all maps g, h : B → A.

Exercise 4.11Prove that if f : R → R is either strictly increasing or strictly decreasingthen f is invertible.

69

Exercise 4.12Suppose that f and g are the functions

f = {(−3, 6), (2, 12), (4, 0), (0,−14)}g = {(0, 0), (2, 0), (−3,−3), (12, 4), (1, 2), (15, 4)}.

Find the elements of f ◦ g as points in the form (x, y).

Exercise 4.13Make up an example in which f ◦ g = g ◦ f.

Exercise 4.14Let S and T be two nonempty sets.

(a) Let α : S → T be a bijection. Show that φ : M(S) → M(T ) de-fined by φ(f) = α ◦ f ◦ α−1 is a bijection between M(S) and M(T ).(b) Show that β :M(T )→M(S) defined by β(g) = α−1 ◦g◦α is a bijection.(c) Prove that β ◦ φ = ιM(S).

70

5 Definition and Examples of Groups

In Section 3, properties of binary operations were emphasized-closure, iden-tity, inverses, associativity. Now these properties will be studied from aslightly different viewpoint by considering systems (S, ∗) that satisfy all fourof the properties. Such mathematical systems are called groups.A group may be defined as follows.

Definition 5.1A set G is a group with respect to a binary operation ∗ if the followingproperties are satisfied:

(i) (x ∗ y) ∗ z = x ∗ (y ∗ z) for all elements x, y, and z of G (the Asso-ciative Law);(ii) there exists an element e of G (the identity element of G) such thate ∗ x = x = x ∗ e, for all elements x of G;(iii) for each element x of G there exists an element x′ of G (the inverse ofx) such that x ∗ x′ = e = x′ ∗ x (where e is the identity element of G).

A group G is Abelian (or commutative) if x ∗ y = y ∗ x for all elements xand y of G.

Remark 5.1The phrase ”with respect” should be noted. For example, the set Z is agroup with respect to addition but not with respect to multiplication (it hasno inverses for elements other than ±1.)

Example 5.1We can obtain some simple examples of groups by considering appropriatesubsets of the familiar number systems.

(a) The set of even integers is an Abelian group with respect to addition.(b) The set N of positive integers is not a group with respect to additionsince it has no identity element.(c) The set Z+ = N ∪ {0} is not a group with respect to addition since noelement other than zero has an inverse.(d) The set of all nonzero rational numbers is an abelian group under multi-plication.

71

Example 5.2For any nonempty set S the collection of all invertible mappings from S toS is a group with respect to composition. This is a consequence of Theorem4.3.

The following examples give some indication of the great variety there is ingroups.

Example 5.3(a) Let p be a fixed point in the plane P and Gp denote the set of all rotationsof the plane about the point p. By Example 4.3, Gp is an Abelian group withrespect to composition.(b) The set of 2-by-2 matrices with respect to addition is an Abelian group.(Show this)

Example 5.4By Theorem 4.3, the set L of all linear mappings αa,b, with a 6= 0, from Rinto R is a group with respect to composition.

Next, we look at a group given by a Cayley table. In this case, it is easy tolocate the identity and inverses of elements.

Example 5.5Let G = {e, a, b, c} with multiplication as defined by the table below.

· e a b ce e a b ca a b c eb b c e ac c e a b

From the table, we observe that

(i) G is closed under this multiplication.(ii) e is the identity element.(iii) e−1 = e, b−1 = b, c−1 = a, and a−1 = c.(iv) the multiplication is commutative.It can be checked that the multiplication is associative. Thus, (G, .) is anabelian group.

72

Next, we record some simple consequences of the definition of a group in thefollowing theorem.

Theorem 5.1Let G be a group with respect to a binary operation ∗.

(i) The identity element is unique. That is, if e, f ∈ G are such thate ∗ a = f ∗ a = a and a ∗ e = a ∗ f = a for all a ∈ G then e = f.(ii) Every element in G has a unique inverse. That is, if a, b, c are elementsin G such that a ∗ b = a ∗ c = e and b ∗ a = c ∗ a = e, where e is the identityelement of G then b = c.

Proof.(i) Since a ∗ e = a for all a ∈ G then in particular f ∗ e = f. Similarly, sincef ∗ a = a for all a ∈ G then in particular f ∗ e = e. Thus, e = f.(ii) With a, b, and c as stated, we have

b = b ∗ e (e is the identity)= b ∗ (a ∗ c) (since a ∗ c = e)= (b ∗ a) ∗ c (∗ is associative)= e ∗ c (since b ∗ a = e)= c (e is the identity)

Remark 5.2By the theorem, it makes sense to speak of the identity element of a group,and the inverse element. It is customary to use a−1 for the inverse of anelement a.

Definition 5.2The order of a group is the number of elements in the group. It is denotedby |G|. If |G| is finite then the group is called a finite group. Otherwise,the group is called infinite group.

Review Problems

In Problems 5.1 - 5.6 decide if each of the given sets is a group with respectto the indicated operation. If it is not a group, state all of the conditions inDefinition 5.1 which fail to hold.

73

Exercise 5.1The set of all rational numbers with respect to addition.

Exercise 5.2The set {−1, 1} with respect to multiplication.

Exercise 5.3The set {−1, 0, 1} with respect to addition.

Exercise 5.4The set {10n : n ∈ Z} with respect to addition.

Exercise 5.5The set {2n : n ∈ Z} with respect to multiplication.

Exercise 5.6The set {2m3n : m,n ∈ Z} with respect to multiplication.

Exercise 5.7For f, g ∈M(R), define addition by (f + g)(x) = f(x) + g(x) for all x ∈ R.Show that M(R) is a group with respect to addition.

Exercise 5.8Let H be the set of all functions f from R to R that satisfy f(x) 6= 0 for allx ∈ R. Define, on H, the operation of multiplication by (fg)(x) = f(x)g(x)for all x ∈ R. Show that H is a group with respect to multiplication.

Exercise 5.9Let G be a set of complex numbers given by G = {1,−1, i,−i}, where i2 = −1,and consider the operation of multiplication of complex numbers in G.

(a) Construct the Cayley table of G.(b) Use (a) to show that G is an Abelian group with respect to multiplication.

Exercise 5.10Consider the Cayley table of a set G with a binary operation ∗.

74

* a b c da b c a bb c d b ac a b c dd a b d d

Is (G, ∗) a group?

Exercise 5.11For an arbitrary set A, the power set of A is the set P(A) = {X : X ⊆ A}.Let A = {a, b, c}. Show that the power set P(A) is not a group with respectto the operation of union.

Exercise 5.12Let A = {a, b, c}. Show that the power set P(A) is not a group with respectto the operation of intersection.

Exercise 5.13Let A be a nonempty set. Define addition on P(A) by

X + Y = (X ∪ Y )− (X ∩ Y ).

Prove that P(A) is a group with respect to this operation.

Exercise 5.14Let G be a nonempty set that is closed under an associative binary operation∗. Prove that G is a group with respect to ∗ is and only if the equationsa∗x = b and y ∗a = b have solutions x and y for all choices of a and b in G.

Exercise 5.15Prove that if a, x, and y are elements of a group G such that xa = ya thenx = y.

Exercise 5.16An element x in a group G is called idempotent if x2 = x. Prove that theidentity element e is the only idempotent element in a group G.

Exercise 5.17Prove that if x = x−1 in a group G then G is Abelian.

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Exercise 5.18Prove that if G is a group, a ∈ G, and a ∗ b = b for some b ∈ G, then a mustbe the identity element of G.

Exercise 5.19Prove that if |S| > 1 then M(S) is not a group with respect to composition.

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6 Permutation Groups

Let S be a nonempty set and M(S) be the collection of all mappings fromS into S. In this section, we will emphasize on the collection of all invertiblemappings from S into S. The elements of this set will be called permutationsbecause of Theorem 2.4 and the next definition.

Definition 6.1Let S be a nonempty set. A one-to-one mapping from S onto S is called apermutation.

Consider the collection of all permutations on S. Then this set is a groupwith respect to composition.

Theorem 6.1The set of all permutations of a nonempty set S is a group with respect tocomposition. This group is called the symmetric group on S and will bedenoted by Sym(S).

Proof.By Theorem 2.4, the set of all permutations on S is just the set I(S) of allinvertible mappings from S to S. According to Theorem 4.3, this set is agroup with respect to composition.

Definition 6.2A group of permutations , with composition as the operation, is called apermutation group on S.

Example 6.11. Sym(S) is a permutation group.2. The collection L of all invertible linear functions from R to R is a permu-tation group with respect to composition.(See Example 4.4.) Note that L isa proper subset of Sym(R) since we can find a function in Sym(R) whichis not in L, namely, the function f(x) = x3. This example shows that, ingeneral, a permutation group on S needs not contain all the permutationson S.

77

Example 6.2Let S = {1, 2, 3}. There are six permutations on S. We will represent thesepermutations using the two-row form as follows:

ρ1 =

(1 2 31 2 3

), ρ2 =

(1 2 32 3 1

), ρ3 =

(1 2 33 1 2

)ρ4 =

(1 2 32 1 3

), ρ5 =

(1 2 33 2 1

), ρ6 =

(1 2 31 3 2

)In composing permutations we always follow the same convention we use incomposing any other mappings: read from right to left. Thus,(

1 2 33 1 2

)◦(

1 2 33 2 1

)=

(1 2 32 1 3

)That is, 1→ 3→ 2, 2→ 2→ 1, 3→ 1→ 3.

Example 6.3Let S = {1, 2, 3}. Then Sym(S) = {ρ1, ρ2, ρ3, ρ4, ρ5, ρ6}, where the ρ′s aredefined in Example 6.2. Let’s construct the Cayley table for this group.

◦ ρ1 ρ2 ρ3 ρ4 ρ5 ρ6

ρ1 ρ1 ρ2 ρ3 ρ4 ρ5 ρ6

ρ2 ρ2 ρ3 ρ1 ρ5 ρ6 ρ4

ρ3 ρ3 ρ1 ρ2 ρ6 ρ4 ρ5

ρ4 ρ4 ρ6 ρ5 ρ1 ρ3 ρ2

ρ5 ρ5 ρ4 ρ6 ρ2 ρ1 ρ3

ρ6 ρ6 ρ5 ρ4 ρ3 ρ2 ρ1

Notice that the permutation

ρ1 =

(1 2 31 2 3

)is the identity mapping of Sym(S). Moreover,(

1 2 32 3 1

)−1

=

(1 2 33 1 2

)Thus, the inverse of an element is obtained by reading from the bottom entryto the top entry rather than from top to bottom:if 1 appears beneath 3 in ρ2

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then 3 appears beneath 1 in ρ−12 .

We will denote the above group by S3. In general, if S = {1, 2, · · · , n} thenthe symmetric group on S will be denoted by Sn.

The number of elements of Sn is found in the following theorem.

Theorem 6.2The order of Sn is n!, where 0! = 1! = 1 and n! = n(n− 1)(n− 2) · · · 2 · 1.

Proof.The proof involves the following counting principle: If a decision consists oftwo steps, if the first step can be done in r different ways and the second stepcan be done in s different ways then the decision can be made in rs differentways.The problem of computing the number of elements of Sn is the same as theproblem of computing the number of different ways the integers 1, 2, · · · , ncan be placed in the n blanks indicated(with each number used only once)(

1 2 3 · · · n· · ·

)Filling the blanks from the left, we see that the first blank can be filled withn different ways. Once this is completed, the second blank can be filled inn − 1 ways, the third in n − 2 ways and so on. Thus, by the principle ofcounting, there are n(n− 1)(n− 2) · · · 2 · 1 = n! ways of filling the blanks. Inconclusion, |Sn| = n!

Now, since S1 = {(1)} then S1 with respect to composition is commuta-tive. Similarly, since (1)(12) = (12)(1) then S2 = {(1), (12)} is also Abelian.Unfortunately, this is not true anymore for |S| > 2.

Theorem 6.3Sn is non-Abelian for n ≥ 3.

Proof.All that we need to do here is to find two permutations α and β in Sn withn ≥ 3 such that α ◦ β 6= β ◦ α. Indeed, consider the permutations

α =

(1 2 3 4 5 · · · n1 3 2 4 5 · · · n

)and β =

(1 2 3 4 5 · · · n3 2 1 4 5 · · · n

)79

Then

α◦β =

(1 2 3 4 5 · · · n2 3 1 4 5 · · · n

)and β ◦α =

(1 2 3 4 5 · · · n3 1 2 4 5 · · · n

)so that α ◦ β 6= β ◦ α

Cycle Notation for PermutationsThe cycle notation for permutations can be thought as a condensed way towrite permutations. Here is how it works.Let α ∈ Sn be the permutation

α(a1) = a2, α(a2) = a3, · · · , α(ak) = a1

and α(ai) = ai for i = k + 1, · · · , n, where a1, a2, · · · , an ∈ {1, 2, 3, · · · , n}.That is, α follows the circle pattern shown in Figure 6.1

Figure 6.1

Such a permutation is called a cycle of length k or simply a k-cycle. We willwrite

α = (a1a2a3 · · · ak) (1)

Let us elaborate a little further on the notation employed in (1). The cyclenotation is read from left to right, it says α takes a1 into a2, a2 into a3, etc.,and finally ak, the last symbol, into a1, the first symbol. Moreover, α leavesall the other elements not appearing in the representation (1) fixed.

Example 6.4The permutation

α =

(1 2 3 4 5 6 71 6 3 7 5 4 2

)80

can be represented as a 4-cycle

α = (2647).

Note that one can write the same cycle in many ways using this type ofnotation.

α = (2647)= (6472)= (4726)= (7264)

Remark 6.1A k-cycle can be written in k different ways, since

(a1a2 . . . ak) = (a2a3 . . . aka1) = · · · = (aka1 . . . ak−1).

Example 6.5It is easy to write the inverse of a cycle. Since α(ak) = ak+1 impliesα−1(ak+1) = ak, we only need to reverse the order of the cyclic pattern.For example,

(2647)−1 = (7462).

Example 6.6Multiplication of cycles is performed by applying the right permutation first.Consider the product in S5

(12)(245)(13)(125)

Reading from right to left

1 7−→ 2 7−→ 2 7−→ 4 7−→ 4

so 1 7−→ 4.Now

4 7−→ 4 7−→ 4 7−→ 5 7−→ 5

so 4 7−→ 5.Next

5 7−→ 1 7−→ 3 7−→ 3 7−→ 3

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so 5 7−→ 3.Then

3 7−→ 3 7−→ 1 7−→ 1 7−→ 2

so 3 7−→ 2.Finally

2 7−→ 5 7−→ 5 7−→ 2 7−→ 1

so 2 7−→ 1. Since all the elements of A = {1, 2, 3, 4, 5} have been accountedfor, we have

(12)(245)(13)(125) = (14532).

Remark 6.2A 1-cycle of Sn is the identity of Sn and is denoted by (1). Of course,(1) = (2) = (3) = · · · = (n).

Now not all permutations are cycles; for example, the permutation

α =

(1 2 3 4 5 6 7 82 3 1 5 6 7 4 8

)is not a cycle. However, one can check easily that

α = (123)(4567)

This suggests how we may extend the idea of cycles to cover all permutations.

Definition 6.3If α and β are two cycles, they are called disjoint if the elements movedby one are left fixed by the other, i.e., their cycle representations containdifferent elements of the set S = {1, 2, 3 · · · , n}.

Example 6.7The cycles (124) and (356) are disjoint whereas the cycles (124) and (346)are not since they habe the number 4 in common.

Theorem 6.4If α and β are disjoint cycles then αβ = βα.

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Proof.Indeed, since the cycles α and β are disjoint, each element moved by α isfixed by β and vice versa. Let α = (a1a2 · · · as) and β = (b1b2 · · · bt) where{a1, a2, · · · , as} ∩ {b1, b2, · · · , bt} = ∅.

(i) Let 1 ≤ k ≤ s. Then

(αβ)(ak) = α(β(ak)) = α(ak) = ak+1

and(βα)(ak) = β(α(ak)) = β(ak+1) = ak+1.

(ii) Let 1 ≤ k ≤ t. Then

(αβ)(bk) = α(β(bk)) = α(bk+1) = bk+1

and(βα)(bk) = β(α(bk)) = β(bk) = bk+1.

(iii) Let 1 ≤ m ≤ n and m 6∈ {a1, a2, · · · , as, b1, b2, · · · , bt}. Then

(αβ)(m) = α(β(m)) = α(m) = m

and(βα)(m) = β(α(m)) = β(m) = m.

It follows from (i), (ii), and (iii) that αβ = βα.

Theorem 6.5Every permutation of Sn is either a cycle or can be written uniquely, exceptfor order of cycles or the different ways a cycle is written, as a product ofdisjoint cycles.

Proof.The proof is by induction on n. If n = 1 then there is only one permutation,and it is the cycle (1).Assume that the result is valid for all sets with fewer than n elements. Wewill prove that the result is valid for a set with n elements.Let σ ∈ Sn. If σ = (1) then we are done. Otherwise there exists a positiveinteger m such that σm−1(1) 6= 1 and σm(1) = 1. For example, if σ =

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(145) ∈ S5 then m = 3 since σ(1) = 4, σ2(1) = σ(σ(1)) = σ(4) = 5, andσ3(1) = σ(σ2(1)) = σ(5) = 1.Let

Q = {1, σ(1), σ2(1), . . . , σm−1(1)}.

If Q = {1, 2, . . . , n} then σ is the cycle σ = (1σ(1)σ2(1) . . . σm−1(1)). IfQ 6= S then

σ = (1σ(1)σ2(1) . . . σm−1(1))τ

where τ is a permutation on the set S −Q = {t ∈ S : t 6∈ Q}. Since this sethas order smaller than n, then by the induction hypothesis τ can be writtenas a product of disjoint cycles, say, τ = τ1τ2 . . . τk. Thus,

σ = (1σ(1)σ2(1) . . . σm−1(1))τ1τ2 . . . τk

But this says that σ is expressed as a product of disjoint cycles. This com-pletes the induction step, and establishes the result for all n

Example 6.8Let S = {1, 2, ..., 8} and let

α =

(1 2 3 4 5 6 7 82 4 6 5 1 7 3 8

)Start the first cycle with 1 and continue until we get back to 1, and thenclose the first cycle. Then start the second cycle with the smallest numbernot in the first cycle, continue until we get back to that number, and thenclose the second cycle, and so on to obtain

α = (1245)(367)(8)

It is customary to omit such cycles as (8), i.e., elements left fixed by α, andwrite α simply as

α = (1245)(367)

Review Problems

Exercise 6.1

Asuume α =

(1 2 3 41 4 3 2

)and β =

(1 2 3 43 1 4 2

).

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Compute each of the following.

(a) β ◦ α (b) α ◦ β (c) α−1 (d) β−1

(e) β−1 ◦ α−1 (f) α−1 ◦ β−1 (g) (β ◦ α)−1 (h) (α ◦ β)−1

Exercise 6.2Write each of the following as a single cycle or a product of disjoint cycles.

(a)

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

)(b) (12)(13)(14)

(c) (13)−1(24)(235)−1 (d) (145)(1235)(13)

Exercise 6.3(a) Write all of the elements of S4 both in two-row form and using cyclenotation.(b) Which elements of S4 are their own inverses?

Exercise 6.4For which values of k will every k-cycle be its own inverse?

Exercise 6.5Show that every non identity element of Sn is either a 2-cycle or can bewritten as the product of 2-cycles.

Exercise 6.6Prove that if S contains at least three elements then Sym(S) is non-Abelian.

Exercise 6.7For m a positive integer, we define αm = α ◦ αm−1. If α is a cycle then αm

maps each integer in the cycle onto the integer located m places farther alongin the cycle. For instance, if α = (123456789) then α2 = (13572468). Findα3 and α4.

Exercise 6.8For σ ∈ Sn, define the order of σ to be the smallest positive integer suchthat σm = (1). Prove that if σ ∈ Sn has order m then τστ−1 has order m forall τ ∈ Sn.

Exercise 6.9

Compute the order of τ =

(1 2 3 4 5 6 7 8 9 10 117 2 11 4 6 8 9 10 1 3 5

). Let

σ = (387) ∈ S11. Find the order of στσ−1.

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Exercise 6.10Let S be a nonempty set and X a subset of S. Let G = {σ ∈ Sym(S) :σ(X) = X}. Prove that G is a permutation group.

Exercise 6.11Let S be a nonempty set and X a subset of S. Let G = {σ ∈ Sym(S) : σ(t) =t∀t ∈ X}. Prove that G is a permutation group.

Exercise 6.12Let S be a nonempty set and G ⊆ Sym(S). Let τ be a fixed element ofSym(S). Prove that

τGτ−1 = {σ ∈ Sym(S) : σ = τµτ−1 for some µ ∈ G}.

Exercise 6.13Let α be a fixed element of Sn. Show that the function φ : Sn → Sn definedby φ(τ) = ατα−1 is one-to-one and onto.

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7 Subgroups

In this section we discuss the concept of a subgroup and look at an importantsubgroup of Sn, the so-called alternating group.

7.1 Definition and Examples of Subgroups

You may have noticed that we sometimes had groups contained within largergroups. For example, the group (Z,+) is contained within the group (R,+).In this situation, we will say that (Z,+) is a subgroup of (R,+).

Definition 7.1If a subset H of a group G is closed under the binary operation of G and ifH is itself a group with respect to the induced binary operation, then H is asubgroup of G. We write H < G.

Definition 7.2A group G is considered to be a subgroup of itself. We say that G is theimproper subgroup of G. All other subgroups are proper subgroups.The subgroup {e} consisiting of the identity element of G is called the trivialsubgroup of G. All other subgroups are nontrivial.

Example 7.11. The set I of even integers is a nontrivial subgroup of Z with respect toaddition. The identity element of (I,+) is 0.2. {−1, 1} is a nontrivial subgroup of (R−{0}, ·) with identity element 1.

You might have noticed that in the above examples the identity of the sub-group always appeared to be the identity of the group. The following theoremshows that this is always the case.

Theorem 7.1Let (G, ∗) be a group with identity element e and H < G. Then e ∈ H ande is the identity element of H.

Proof.Since H is a group itself then it has an identity element which we denote by

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eH . We will show that eH = e. Since eH ∈ G then eH is invertible. That is,eH ∗ e−1

H = e−1H ∗ eH = e. Thus,

eH = eH ∗ e (e is the identity element of G)= eH ∗ (eH ∗ e−1

H ) (since eH ∗ e−1H = e)

= (eH ∗ eH) ∗ e−1H (∗ is associative)

= eH ∗ e−1H (since eH is the identity of H)

= e (since eH ∗ e−1H = e)

The following result shows that the inverse of an element in H is the same asthe inverse of that element in G.

Theorem 7.2Let (G, ∗) be a group with identity element e and H < G. If a ∈ H thena−1 ∈ G. If c is the inverse of a in H then c = a−1.

Proof.Let a ∈ H. Then a ∈ G and so a−1 ∈ G. Let c ∈ H such that c ∗ a = a ∗ c =eH .(That is, c is the inverse of a in H) We will show that c = a−1. Indeed,since c ∗ a = a ∗ c = eH = e then c is an inverse of a in G. By Theorem 3.2,c = a−1. This completes the proof of the theorem.

It is convenient to have a routine step-by-step procedure for determiningwhether a subset of a group is a subgroup of G. The following theorem givessuch a procedure.

Theorem 7.3A nonempty subset H of a group G is a subgroup of G if and only if

(i) H is closed under the binary operation of G, i.e., if a, b ∈ H then a∗b ∈ H;(ii) if a ∈ H then a−1 ∈ H.

Proof.Suppose that H is a subgroup of G. The property (i) follows directly fromthe definition of a subgroup. Condition (ii) is just Theorem 7.2.Assume now that H is a nonempty subset of G satisfying conditions (i) and(ii). By (i), ∗ is a binary operation on H. Since ∗ is associative in G then it isstill associative when restricted to H. Now, if a ∈ H, then by condition (ii),a−1 ∈ H so that by (i) a ∗ a−1 = a−1 ∗ a = e ∈ H. Hence, e is the identityof H. By condition (ii), every element of H has an inverse in H. Thus, Hsatisfies the conditions of a group and so is a subgroup of G.

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Example 7.2In M(R) under addition, consider the subset C(R) of all continuous func-tions. Clearly, C(R) is closed under addition (for the sum of continuousfunctions is continuous). For a continuous function f, the function −f (whichplays the role of its inverse) is continuous. Thus, C(R) is a subgroup ofM(R)under addition.

If the set H in Theorem 7.3 is finite then condition (ii) can be omittedaltogether.

Theorem 7.4A finite subset of a group that is closed under the group operation is asubgroup of that group.

Proof.Consider the Cayley table of H. By closure, each element in the table belongsto H. Each element of H appears exactly once in each row or column of thetable. That is, each row/column is a rearrangement of the elements of H. Toillustarte, suppose that b ∈ H and in the row of b we can find two elements xand y in H such that b∗x = a and b∗y = a, where a ∈ H. Then b∗x = b∗y.But this implies that x = e ∗ x = (b−1 ∗ b) ∗ x = b−1 ∗ (b ∗ x) = b−1 ∗ (b ∗ y) =(b−1∗b)∗y = e∗y = y. Thus, for a given element, a ∈ H, the row and columncorresponding to a must each contain a. Thus some member of H multipliesa to give the result a. The only element that could do this is the identityelement e. Thus, e ∈ H. Similarly, if e is in H then it must appear once inthe row and once in the column corresponding to a. Therefore there is someelement of H which multiplies a to give e. That can only be a−1. Thus e mustbe in H and for every a ∈ H, a−1 ∈ H. Therefore, H is a subgroup of G.

A variation of Theorem 7.3, where the conditions (i) and (ii) can be com-bined, is given by the next result.

Theorem 7.5Let H be a nonempty subset of (G, ∗). Then H is a subgroup of G if andonly if for all a, b ∈ H, we have a ∗ b−1 ∈ H.

Proof.Assume first that H is a subgroup of G. Let a, b ∈ H. Then by Theorem 7.3(ii), b−1 ∈ H. By Theorem 7.3(i), a ∗ b−1 ∈ H.

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Conversely, assume that for all a, b ∈ H we have a ∗ b−1 ∈ H. Since H isnonempty then it contains an element a. But then a ∗ a−1 = e ∈ H. Now, ifa ∈ H then e∗a−1 = a−1 ∈ H. Finally, if a, b ∈ H then a∗b = a∗(b−1)−1 ∈ H.Hence, by Theorem 14, H is a subgroup of G.

Example 7.3Let G be a nonempty group and a ∈ G. Define the set

H = {x ∈ G : ax = xa}.

Then H is a subgroup of G. To see this, note first that e ∈ H since ae = ea.Thus, H is nonempty. Now, let x, y ∈ H. Then

a(xy) = (ax)y= (xa)y= x(ay)= x(ya) = (xy)a

It follows that H is closed under multiplication. Now, if x ∈ H then

ax−1 = e(ax−1)= (x−1x)(ax−1)= x−1[(xa)x−1]= x−1[(ax)x−1]= x−1[a(xx−1)]= x−1(ae)= x−1a

Thus, x−1 ∈ H. It follows from Theorem 7.5 that H is a subgroup of G.

Next, we discuss two other types of subgroups of permutation groups. LetS 6= ∅ and G a permutation group on S. Note that G ⊆ Sym(S). Let T bea subset of S. We define

GT = {α ∈ G : α(t) = t ∀t ∈ T}

andG(T ) = {α ∈ G : α(T ) = T}.

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Example 7.4Let S = {1, 2, 3, 4}, G = S4, and T = {1, 2}. Then

GT = {(1), (34)}

andG(T ) = {(1), (12), (34), (12)(34)}

Theorem 7.6Let S,G,GT and G(T ) be defined as above. Then

(a) GT and G(T ) are subgroups of G.(b) GT is a subgroup of G(T ).

Proof.(a)Since ιS(t) = t for all t ∈ T then ιS ∈ GT so that GT is nonempty. Letα, β ∈ GT . Then for any t ∈ T we have

αβ−1(t) = α(β−1(t))= α(t) (Since β(t) = t implies β−1(t) = t)= t (Since α(t) = t)

By Theorem 7.5, GT is a subgroup of G.The proof that G(T ) is a subgroup of G is similar; simply replace t by T.

(b) We just need to show that GT ⊂ G(T ). Indeed, if α ∈ GT then α(t) = tfor all t ∈ T. But this implies that α(T ) = T. Hence α ∈ G(T ).

7.2 The Alternating Group

In this section we consider a class of subgroups known as the alternatinggroups. We start with the following definition.

Definition 7.3Any 2-cycle (ab) in Sn is called a transposition.

That is, a transposition is a permutation (a b) of a set S that interchangestwo elements a and b of S and fixes the remaining elements.

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Theorem 7.7Every element of Sn, n ≥ 2, can be expressed as a finite composition oftranspositions.

Proof.Let σ ∈ Sn with σ 6= (1). By Theorem 6.5, σ can be written as a finite productof disjoint cycles. But any k − cycle can be written as a finite product oftranspositions. That is,

(a1a2 . . . ak) = (a1ak)(a1ak−1) · · · (a1a2)

Thus, σ can be written as a finite product of transpositions. If σ = (1) thensince n ≥ 2 then (1) = (ab)(ab) where a, b ∈ {1, 2, . . . , n}.

Example 7.5One can verify easily that

(123) = (13)(12)

and(123) = (23)(12)(13)(23).

Note that the previous example shows that a permutation can be written intwo different ways as a product of transpositions. However, the number oftranspositions in both forms is even. Indeed, we have

Theorem 7.8If a permutation is expressed as a product of p transpositions and as a productof q transpositions then either p and q are both even, or p and q are bothodd.

Proof.Consider the product

P = P (x1, x2, . . . , xn) = Π1≤i<j≤n(xi − xj).

For any σ ∈ Sn define

σ(P ) = P (σ(x1), σ(x2), . . . , σ(xn)) = Π1≤i<j≤n(xσ(i) − xσ(j)).

92

Note that if γ = (kl) with k < l (if k > l then use the fact that (lk) = (kl))then

γ(P ) = −P

To see this, note that the only factors that change sign when swapping xkand xl are (xk − xl), (xk − xi), and (xi − xl), where k < i < l, and there isan odd number of them.Now assume that σ ∈ Sn can be written as a product of an even numberof transpositions σ = γ1γ2 . . . γ2q and as an odd number of transpositionsσ = δ1δ2 . . . δ2r+1. Then

σ(P ) = γ1(γ2(. . . (γ2q(P )))) = (−1)2qP = P

andσ(P ) = δ1(δ1(. . . (δ2r+1(P )))) = (−1)2r+1P = −P.

Thus, P = −P which is a contradiction.

Definition 7.4A permutation that can be written as a product of an even number of trans-positions is called an even permutation, and the one that can be written asa product of an odd number of transpositions is called an odd permutation.

We close this section we the following result.

Theorem 7.9For n ≥ 2, the set An of all even permutations in Sn is a subgroup of Sn oforder n!

2.

Proof.Since (1) = (12)(12) ∈ An then An 6= ∅. Let σ = γ1γ2 . . . γ2k and τ =δ1δ2 . . . δ2l. Then

στ−1 = γ1γ2 . . . γ2kδ−12l δ

−12l−1 . . . δ

−11 = γ1γ2 . . . γ2kδ2lδ2l−1 . . . δ1 ∈ An.

Thus, by Theorem 7.5, An is a subgroup of Sn.To prove that |An| = n!

2we just need to prove that Sn has the same number

of even permutations as odd permutations. Let On be the set of all oddpermutations of Sn. Define the mapping f : An → On by f(σ) = σ(12).

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f is one-to-oneSuppose that f(σ1) = f(σ2). Then σ1(12) = σ2(12). Thus, σ1(12)(12)−1 =σ2(12)(12)−1. That is, σ1 = σ2.

f is ontoLet σ ∈ On. Then α = σ(12) ∈ An. Moreover,

f(α) = σ(12)(12) = σ.

By Theorem 7.8, Sn = An ∪ On and An ∩ On = ∅. Thus, |An| = |On| and|Sn| = |An|+ |On| = 2|An|. By Theorem 6.2, |Sn| = n! and hence |An| = n!

2.

Definition 7.5An is called the alternating group on {1, 2, . . . , n}.

Example 7.6If S = {1, 2, 3, 4} then

A4 = {(1), (124), (142), (132), (123), (143), (134), (234), (243), (12)(34), (13)(24), (14)(23)}.

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Review Problems

Exercise 7.1Decide in each case whether the given subset is a subgroup of S5. Justify youranswers.

(a) {(1), (134), (143)} (b) {(1), (123), (234)}(c) {(1), (12)34)} (d) {(1), (1234), (1432)}

Exercise 7.2Decide in each case whether the given subset is a subgroup of {1,−1, i,−i}under multiplication. Justify your answers.

(a) {1,−1} (b) {1, i}(c) {i,−i} (d) {1,−i}

Exercise 7.3Find a subset of (Z,+) that is closed under addition but is not a subgroup of(Z,+).

Exercise 7.4Let S = {1, 2, 3} and G = S3. Write all the elements of GT and G(T ) ifT = {2, 3}.

Exercise 7.5Prove that if H and K are subgroups of a group G then H ∩ K is also asubgroup of G.

Exercise 7.6Let H = {(1), (12)} and K = {(1), (123), (132)} be subgroups of S3. Showthat H ∪K is not a subgroup of S3.

Exercise 7.7Prove that if G is a group with identity e, and x ∈ G such that x ∗ x = xthen x = e.

Exercise 7.8Assume that G is a group with operation ∗ and that a ∈ G. Let

C(a) = {x ∈ G : xa = xa}

Prove that C(a) is a subgroup of G. We call C(a) the centralizer of a in G.

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Exercise 7.9Assume that G is a group with operation ∗ and let

Z(G) = {a ∈ G : a ∗ x = x ∗ a∀x ∈ G}.

Prove that Z(G) is a subgroup of G. We call Z(G) the center of G.

Exercise 7.10Let H be a subgroup of a group G, and a a fixed element of G. Let

K = {aha−1 : for some h ∈ H}.

Prove that K is a subgroup of G.

Exercise 7.11Prove that H = {h ∈ G : h−1 = h} is a subgroup of G if G is Abelian.

Exercise 7.12Let H and K be subgroups of an Abelian group G and let

HK = {hk : h ∈ H and k ∈ K}.

Prove that HK is a subgroup of G.

Exercise 7.13Find two subgroups H and K of S3 such that HK is not a subgroup of S3.

Exercise 7.14Let H be a nonempty subset of a group G. Prove that H is a subgroup of Gif and only if a−1 ∗ b ∈ H for all a, b ∈ H.

Exercise 7.15Let n ∈ Z. Prove that the set nZ = {nx : x ∈ Z} is a subgroup of (Z,+).

Exercise 7.16Prove that 2Z ∪ 3Z is not a subgroup of (Z,+).

Exercise 7.17Prove that if H is a subgroup of K and K is a subgroup of G then H is asubgroup of G.

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Exercise 7.18Let G = R× R with binary operation (a, b) + (c, d) = (a+ c, b+ d).

(a) Show that (G,+) is a group.(b) Let A = {(a, 0) : a ∈ R} and B = {(0, b) : b ∈ R} be two subsets of G.Prove that A and B are subgroups of G.(c) Show that A ∪B is not a subgroup of G.

Exercise 7.19Solve for x in S4 : 142)2 · x = (234)−1

Exercise 7.20Prove that On is not a subgroup of Sn.

Exercise 7.21Show that a k-cycle is even if and only if k is odd.

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8 Symmetry Groups

In this section, we are interested in the symmetries of planar figures. We canidentify a symmetry as a transformation of the plane that moves the figureso that it falls back on itself. The only transformations that we’ll considerare those that preserve distance, called isometries. There are four kinds ofplanar isometries: translations, rotations, reflections, and glide reflections.In this section we will just consider rotations and reflections.

Rotations: A rotation fixes one point in the plane and turns the rest ofit some angle around that point.Translations: A translation is a mapping that sends all points the samedistance in the same direction.Reflections: A reflection fixes one line in the plane, called the axis of reflec-tion, and exchanges points on one side of the axis with points on the otherside of the axis at the same distance from the axis.Glide-reflection: Any product of a translation and a reflection.

Next, we will associate to each figure a group which characterizes the sym-metry of the figure. Let P denote the set of all points in the plane thenSym(P ) is the set of all permutations from P to P. Let M be the set of allisometries.

Theorem 8.1M is a subgroup of Sym(P ).

Proof.We will show that M satisfies the conditions of Theorem 7.5. Indeed, sincedist(ιP (p), ιP (q)) = dist(p, q) for any points p, q ∈ P then ιP is an isometryand therefore belongs to M. Thus, M 6= ∅.Next, let α, β ∈ M. We will show that α ◦ β−1 ∈ M. Since α and β arepermutations on P then α ◦ β−1 is also a permutation on P. Moreover, ifp, q ∈ P then

dist((α ◦ β−1)(p), (α ◦ β−1)(q)) = dist(α(β−1(p)), α(β−1(q)))= dist(β−1(p), β−1(q)) (since α ∈ M)= dist(β(β−1(p)), β(β−1(q))) (since β ∈ M)= dist(ιP (p), ιP (q))= dist(p, q)

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Thus, α ◦ β−1 ∈M. By Theorem 7.5, M is a subgroup of Sym(P ).

Now, let T be a subset of P. Define the set

M(T ) = {α ∈M : α(T ) = T}.

By Theorem 7.6(a), M(T ) is a subgroup of M.

Definition 8.1M(T ) is the group of all symmetries leaving T invariant. We call this groupthe symmetry group of T.

Example 8.1In this example we describe the symmetry group of a square with vertices{a, b, c, d} consisting of rotations and reflections. The eight symmetries ofthe square are

µ1 = identity permutation =

(a b c da b c d

)µ2 = Rotation clockwise 90◦ around p =

(a b c db c d a

)µ3 = Rotation clockwise 180◦ around p =

(a b c dc d a b

)µ4 = Rotation clockwise 270◦ around p =

(a b c dd a b c

)µ5 = Reflection through H =

(a b c dd c b a

)µ6 = Reflection through V =

(a b c db a d a

)µ7 = Reflection through D1 =

(a b c da d c b

)µ8 = Reflection through D2 =

(a b c dc b a d

)Thus,

M(T ) = {µ1, µ2, µ3, µ4, µ5, µ6, µ7, µ8}.

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Example 8.2Below is the Cayley table for M(T )m where T is the square in the previousexample.

◦ µ1 µ2 µ3 µ4 µ5 µ6 µ7 µ8

µ1 µ1 µ2 µ3 µ4 µ5 µ6 µ7 µ8

µ2 µ2 µ3 µ4 µ1 µ7 µ8 µ6 µ5

µ3 µ3 µ4 µ1 µ2 µ6 µ5 µ8 µ7

µ4 µ4 µ1 µ2 µ3 µ8 µ7 µ5 µ6

µ5 µ5 µ8 µ6 µ7 µ1 µ3 µ4 µ2

µ6 µ6 µ7 µ5 µ8 µ3 µ1 µ2 µ4

µ7 µ7 µ5 µ8 µ6 µ2 µ4 µ1 µ3

µ8 µ8 µ6 µ7 µ5 µ4 µ2 µ3 µ1

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Review Problems

Exercise 8.1Determine the symmetry group of an equilateral triangle.

Exercise 8.2Determine the symmetry group of an isosceles triangle.

Exercise 8.3Determine the symmetry group of a rectangle.

Exercise 8.4Determine the symmetry group of a parallelogram.

Exercise 8.5Determine the symmetry group of a regular pentagon.

Exercise 8.6Determine the symmetry group of a regular hexagon.

Exercise 8.7Construct the Cayley table for the symmetry group of a rectangle.

Exercise 8.8Construct the Cayley table for the symmetry group of an equilateral triangle.

Exercise 8.9Consider the symmetry group of the square. Let ι = µ1, α = µ2, and β = µ6

as described in Example 8.1.

(a) Compute α4 and β2.(b) Show that {µ1, · · · , µ8} = {ι, α, α2, α3, β, β ◦ α, β ◦ α2, β ◦ α3}.

Exercise 8.10Let G be the symmetry group of an equilateral triangle. Let ι be the identityelement, α be the rotation by 120◦, and β be the reflection in the vertical axisof symmetry. Show that

G = {ι, α, α2, β, β ◦ α, β ◦ α2}.

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9 Equivalence Relations

In the study of mathematics, we deal with many examples of relations be-tween elements of various sets. For example, in working with the integers, weencounter relations such as ”x is less than y”. Notice the importance of theordering of the elements of the set in this relation. That is, ”x less than y”is not the same as ”y less than x.” A relation such as the one just mentionedcan be described by the following definition

Definition 9.1A relation ∼ on a nonempty set S is a subset of the Cartesian product S×S.Thus, if (a, b) ∈∼ then we write a ∼ b.

Example 9.11. On the set Z of integers, ∼= {(x, 2x) : x ∈ Z} is a relation on Z. Notethat a ∼ b if and only if b = 2a.2. A mapping between two nonempty sets is a relation.

We now consider some properties which a given relation ∼ on a set S mayor may not have.

Definition 9.2Let ∼ be a relation on a nonempty set S. We say that ∼ is:

• reflexive if and only if for all a ∈ S, we have a ∼ a;• symmetric if and only if for a, b ∈ S if a ∼ b then b ∼ a;• transitive if and only if whenever a, b, c ∈ S such that a ∼ b and b ∼ cthen a ∼ c.

Example 9.2Let S be a nonempty set and P(S) be the collection of all subsets of S. Let∼ be the relation defined by

A ∼ B ⇐⇒ A ⊆ B, where A,B ∈ P(S).

Then∼ is reflexive and transitive but not symmetric since it is not always truethat if A ⊆ B then B ⊆ A.(For example, {2} ⊆ {1, 2} but {1, 2} 6⊆ {2}.)

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Just as there were different classes of functions (one-t-one, onto, and one-to-one correspondence), there are also special classes of relations. One ofthe most useful kind of relations (besides functions, which of course are alsorelations) are those called equivalence relations which we define next.

Definition 9.3A relation ∼ on a set S which is reflexive, symmetric, and transitive is calledan equivalence relation.

Example 9.31. The equality ”=” relation between real numbers or sets.2. The relation ”is similar to” on the set of all triangles.3. The relation ” ≥ ” between real numbers is not an equivalence relation,because although it is reflexive and transitive, it is not symmetric. e.g. 7 ≥ 5does not imply that 5 ≥ 7.

The following example is important in applications to combinatorics.

Example 9.4Let S be a nonempty set and G be a subgroup of Sym(S). Define the relation∼ on S by

a ∼ b⇐⇒ α(a) = b for some α ∈ G.

Then ∼ is an equivalence relation on S. Indeed,

∼ is reflexive: If a ∈ S then ιS(a) = a. Since G is a subgroup and ιS ∈ Gthen a ∼ a.∼ is symmetric: Let a, b ∈ S such that a ∼ b. Then there is an α ∈ Gsuch that α(a) = b. Since G is a group then α−1 ∈ G. Moreover, a = ιS(a) =(α−1 ◦ α)(a) = α−1(α(a)) = α−1(b). Thus, b ∼ a.∼ is transitive: Let a, b, c ∈ S such that a ∼ b and b ∼ c. Then there existpermutations α and β in G such that α(a) = b and β(b) = c. Since G is agroup then G is closed under composition and therefore β ◦α ∈ G. Moreover,(β ◦ α)(a) = β(α(a)) = β(b) = c. Hence, a ∼ c.

An important fact about an equivalence relation on a set A is that it inducesa partition of A into disjoint sets as indicated in the next theorem.

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Theorem 9.1Let A be a nonempty set. Let {Ai}i∈N be a partition of A. That is, {Ai}i∈Nis a family of subsets of A that satisfies the two conditions:

(i) A = ∪i∈NAi;(ii) For i 6= j, Ai ∩ Aj = ∅.

The relationx ∼ y ⇐⇒ x, y ∈ Ai, for some i

is an equivalence relation on A.

Proof.We need to show that ∼ is reflexive, symmetric and transitive.

∼ is reflexive: If a ∈ A then by (i), a ∈ Ai for some i ∈ N. From thedefinition of ∼ with b = a we have a ∼ a.∼ is symmetric: Let a, b ∈ A such that a ∼ b. Then a, b ∈ Ai for somei ∈ N. But then b, a ∈ Ai so that b ∼ a.∼ is transitive: Let a, b, c ∈ A such that a ∼ b and b ∼ c. Then a, b ∈ Aifor some i ∈ N and b, c ∈ Aj for some j ∈ N. By (ii), we must have i = j.Thus, a, c ∈ Ai for some i ∈ N. Hence, a ∼ c.

The converse of the above theorem is also true. Before proving this claim weintroduce the following concept.

Definition 9.4If ∼ is an equivalence relation on a nonempty set A and a ∼ b for somea, b ∈ A then we say that a and b are equivalent. For a fixed a ∈ A theset of all elements in S equivalent to a is called an equivalence class withrepresentative a. We will write [a]. In set-builder notation

[a] = {x ∈ A : x ∼ a}.

The subset of A containing exactly one element from each equivalent class iscalled a complete set of equivalence class representatives.

Exercise 9.1In the rectangular coordinate system we define the relation

(x1, y1) ∼ (x2, y2)⇐⇒ y1 = y2.

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(i ) Show that ∼ is an equivalence relation on the set of points in the plane.(ii) Describe the equivalence classes geometrically.(iii) Give a complete set of equivalence class representatives.

Solution.(i) For any (x, y) ∈ R2, (x, y) ∼ (x, y) so that ∼ is reflexive. Now, if (x1, y1) ∼(x2, y2) then y1 = y2. But equality in R is symmetric so that y2 = y1. Thus,(x2, y2) ∼ (x1, y1) and hence sim is symmetric. To show that ∼ is transitive,suppose that (x1, y1) ∼ (x2, y2) and (x2, y2) ∼ (x3, y3). Then y1 = y2 andy2 = y3. Since ’=’ is transitive in R then y1 = y3. Hence (x1, y1) ∼ (x3, y3).(ii) For a fixed (a, b) ∈ R2, the equivalence class of (a, b) is the horizontal linegoing through the point (a, b).(iii) The set of points on a line not parrallel to the x-axis.

Theorem 9.2If ∼ is an equivalence relation on a nonempty set A and a, b ∈ A are suchthat a ∼ b then [a] = [b].

Proof.The proof is by double inclusions. Let x ∈ [a]. Then x ∼ a. Since a ∼ b and∼ is transitive then x ∼ b which means that x ∈ [b]. Thus, [a] ⊆ [b]. Nowinterchange the letters a and b to show that [b] ⊆ [a]. Hence, [a] = [b]

Theorem 9.3Let A be a nonempty set and ∼ be an equivalence relation on A. Then theequivalence classes of A define a partition of A. That is,

(i) A = ∪a∈A[a];(ii) If [a] 6= [b] then [a] ∩ [b] = ∅.

Proof.By the definition of [a] we have that [a] ⊆ A. Hence, ∪a∈A[a] ⊆ A. We nextshow that A ⊆ ∪a∈A[a]. Indeed, let b ∈ A. Since ∼ is reflexive then b ∈ [b] andconsequently b ∈ ∪a∈A[a]. Hence, A ⊆ ∪b∈A[b]. It follows that A = ∪a∈A[a].This establishes (i).It remains to show that if [a] 6= [b] then [a]∩ [b] = ∅ for a, b ∈ A. Equivalently.we must show that if [a]∩[b] 6= ∅ then [a] = [b]. Since [a]∩[b] 6= ∅ then there isan element c ∈ [a]∩ [b]. This means that c ∈ [a] and c ∈ [b]. Hence, a ∼ c andb ∼ c. Since ∼ is symmetric and transitive then a ∼ b. Now, by Theorem 9.2,[a] = [b].

105

Review Problems

Exercise 9.2Assume that S = {w, x, y, z} and that w ∼ y and z ∼ y. Which of the fol-lowing must also be true if ∼ is to be an equivalence relation on S?

(a) y ∼ y (b) y ∼ z (c) w ∼ z (d) y ∼ x.

Exercise 9.3Let A = {1, 2, 3, 4, 5}.

(a) Show that P = {{∞,3}, {∈}, {4,5}} is a partition of A.(b) For the corresponding equivalence relation ∼, which of the following aretrue?

(i) 4 ∼ 5 (ii) 3 ∼ 3 (iii) 1 ∼ 2 (iv) 5 ∼ 1.

Exercise 9.4For points (x1, y1) and (x2, y2) in a plane with rectangular coordinate system,let (x1, y1) ∼ (x2, y2) means x1 = x2 or y1 = y2. Show that ∼ is not anequivalence relation on the set of points in the plane.

Exercise 9.5Define a relation ∼ on R by

a ∼ b iff |a| = |b|.

(a) Prove that ∼ is an equivalence relation on R.(b) Give a complete set of equivalence class representatives.

Exercise 9.6For x, y ∈ R let x ∼ y mean that xy > 0. Is ∼ reflexive? symmetric?transitive?

Exercise 9.7For sets S and T , let S ∼ T mean that there is an invertible mapping of Sonto T. Prove that ∼ is an equivalence relation.

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Exercise 9.8Let L be the set of all lines in the Cartesian plane. For l1, l2 ∈ L, definethe relation l1 ∼ l2 if and only if either the slopes of both lines are equal orundefined. Geometrically, l1 ∼ l2 if and only if l1 is parallel to l2.

(a) Show that ∼ is an equivalence relation on L.(b) For a fixed l ∈ L, what is the equivalence class of l?

Exercise 9.9On the set of all 2 × 2 matrices over R, define A ∼ B if there exists an in-vertible matrix P such that PAP−1 = B. Prove that ∼ defines an equivalencerelation.

Exercise 9.10Let Z be the set of integers and n ∈ Z. Let ∼ be the relation on Z defined bya ∼ b if a − b is a multiple of 4. We denote this relation by a ≡ b (mod 4)read ”a congruent to b modulo 4.”

(a) Show that R is an equivalence relation on Z.(b) Find the equivalence classes of ∼ .

Exercise 9.11Let G be a group and H a subgroup of G. Define on G the relation ∼ givenby a ∼ b if and only if ab−1 ∈ H. Prove that ∼ is an equivalence relation onG.

Exercise 9.12Let M = {(a, b) ∈ Z× Z : b 6= 0}. Define on M the relation (a, b) ∼ (c, d) ifand only if ad = bc. Prove that ∼ is an equivalence relation on M.

Exercise 9.13Let ∼ be an equivalence relation on a nonempty set A. Define A/ ∼= {[a] :a ∈ A}. Show that η : A→ A/sim given by η(a) = [a] is onto.

Exercise 9.14Let f : A → B be a given function. Define the relation a ∼ b if and only iff(a) = f(b). Show that ∼ is an equivalence relation.

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Exercise 9.15For the function f : R → R given by f(x) = x2, for all x ∈ R, describe theequivalence relation determined by f.

Exercise 9.16Let f : A → B be a function and ∼ be the equivalence relation defined inthe previous exercise. Prove that the mapping ρ : A/ ∼→ B defined byρ([a]) = f(a) is one-to-one.

Exercise 9.17Prove that every mapping α : A → B can be written in the form α = ρ ◦ ηwhere ρ is one-to-one and η is onto.

Exercise 9.18True or false: The number of equivalence relations on a set A is the numberof different partitions of A.

Exercise 9.19(a) Find all of the partitions of {x, y, z}.(b) How many different equivalence relations are there on a 3-element set?

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10 The Division Algorithm. Congruence Mod-

ulo n

In this section, we want to introduce an important equivalence relation onthe set of integers Z. This relation depends on the concept of divisibility ofintegers which we discuss next.

10.1 Divisibility. The Division Algorithm

In this section we study the divisibility of integers. Our main goal is toobtain the Division Algorithm. This is achieved by applying the well-orderingprinciple which we prove next.

Theorem 10.1 (The Well-Ordering Principle)If S is a nonempty subset of N then there is an m ∈ S such that m ≤ x forall x ∈ S. That is, S has a smallest element.

Proof.We will use contradiction to prove the theorem. That is, by assuming thatS has no smallest element we will prove that S = ∅.We will prove that n 6∈ S for all n ∈ N. We do this by induction on n.Since S has no smallest element then 1 6∈ S. Asuume that we have provedthat 1, 2, · · · , n 6∈ S. We will show that n + 1 6∈ S. If n + 1 ∈ S then since1, 2, 3, c . . . 6∈ S then n + 1 would be the smallest element of S and thiscontradicts the assumption that S has no smallest element. Thus, we musthave n+ 1 6∈ S. Hence, by the principle of mathematical induction, n 6∈ S forall n ∈ N. But this leads to S = ∅. This conclusion contradicts the hypothesisof the theorem where S is given to be nonempty. This establishes a proof ofthe theorem.

Remark 10.1The above theorem is false if N is replaced by Z,Q, or R.(See Exercise 10.6

Before establishing the Division Algorithm, we introduce the concept of di-visibility and derive some of its properties.

Definition 10.1An integer m is divisible by a nonzero integer n if and only if m = nq for

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some integer q. We also say that n divides m, n is a divisor of m, m is amultiple of n, or n is a factor of m. We write n|m. If n does not dividem we write n 6 | m. A positive integer n with only divisors 1 and n is calledprime.

Example 10.1Since 8 = 2 · 4 then 2| 8 and 4| 8. However, 4 6 | 6.

The following theorem discusses some of the properties of divisibility.

Theorem 10.2(a) If n|m then n|(−m).(b) If n|a and n|b then n|(a± b)(c) If n|m and m|p then n|p.(d) If n|m and m|n then either n = m or n = −m.

Proof.(a) Suppose that n|m. Then m = nq for some q ∈ Z. Thus, −m = (n)(−q)and hence n|(−m).(b) Suppose that n|a and n|b. Then a = nq and b = nq′ for some q, q′ ∈ Z.Thus, a± b = n(q ± q′). Hence, n|(a± b).(c) Suppose that n|m and m|p. Then m = nq and p = mq′ for some q, q′ ∈ Z.Thus, p = n(qq′). Since qq′ ∈ Z then n|p.(d) If n|m and m|n then m = nq and n = mq′ for some q, q′ ∈ Z. Thus,m = mqq′ or (1 − qq′)m = 0. Since m 6= 0 then qq′ = 1. This only true ifeither q = q′ = 1 or q = q′ = −1. That is, n = m or n = −m.

With the Well-Ordering Principle we can establish the following theorem.

Theorem 10.3 (Division Algorithm)If a and b are integers with b ≥ 1 then there exist unique integers q and rsuch that

a = bq + r, 0 ≤ r < b.

Proof.ExistenceConsider the sets

S = {a− bt : t ∈ Z}, S ′ = {x ∈ S : x ≥ 0}.

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The set S ′ is nonempty. To see this, if a ≥ 0 then a− 0t ∈ S and a− 0t ≥ 0.That is, a ∈ S ′. If a < 0 then since a− ba ∈ S and a− ba = a(1− b) ≥ 0 sothat a− ba ∈ S ′.Now, if 0 ∈ S ′ then a− qb = 0 for some q ∈ Z and so r = 0 and in this casethe theorem holds. So, assume that 0 6∈ S ′. By Theorem10.1, there exist asmallest element r ∈ S ′. That is,

a− qb = r, for some q ∈ Z.

Since r ∈ S ′ then r ≥ 0. It remains to show that r < b. If we assume thecontrary, i.e. r ≥ b, then

a− b(q + 1) = a− bq − b = r − b ≥ 0

and this implies that a− b(q + 1) ∈ S ′. But b > 0 so that

a− b(q + 1) = a− bq − b < a− bq = r

and this contradicts the definition of r as being the smallest element of S ′.Thus, we have

a = bq + r, 0 ≤ r < b.

UniquenessSuppose that

a = bq1 + r1, 0 ≤ r1 < b

anda = bq2 + r2, 0 ≤ r2 < b.

We must show that r1 = r2 and q1 = q2. Indeed, since bq1 + r1 = bq2 + r2

then b(q1 − q2) = r2 − r1. This says that b|(r2 − r1). But 0 ≤ r1 < b and0 ≤ r2 < b so that −b < −r1 < r2 − r1 < r2 < b. That is, −b < r2 − r1 < b.The only multiple of b strictly between −b and b is zero. Hence, r1 = r2. Butthen b(q1 − q2) = 0 and since b 6= 0 then q1 = q2.

Example 10.2If a = 11 and b = 4 then q = 2 and r = 3.

Remark 10.2The above theorem is still valid for b < 0. Thus, given two integers a and bwith b 6= 0, there exist unique integers q and r such that

a = bq + r, 0 ≤ r < |b|.

(See Exercise 10.7).

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10.2 Congruence Modulo n.

Divisibility leads to the concept of congruence.

Definition 10.2Let n be a positive integer. Integers a and b are said to be congruent mod-ulo n if a− b is divisible by n. This is denoted by writing a ≡ b(mod n). Wecall n the modulus. If a is not congruent b modulo n we write a 6≡ b (mod n).

Example 10.317 and 65 are congruent modulo 6, because 65− 17 = 48 is divisible by 6.

Theorem 10.4The following statements are all equivalent:

(i) a ≡ b(mod n)(ii) n|(a− b)(iii) a− b = nt for some t ∈ Z(iv) a = b+ nt for some t ∈ Z.

Proof.(i) =⇒ (ii): Suppose that a ≡ b(mod n). Then from Definition def32, n|(a−b).(ii) =⇒ (iii): Suppose that n|(a− b). Then by Definition 10.1, there exists at ∈ Z such that a− b = nt.(iii) =⇒ (iv): Suppose that a− b = nt for some t ∈ Z. Then by adding b toboth sides we get a = b+ nt which is the statement of (iv).(iv) =⇒ (i): Suppose that a = b + nt for some t ∈ Z. Then a − b = nt. ByDefinition 10.1, a− b is divisible by n and so a ≡ b(mod n).

Congruence modulo n is an equivalence relation on Z as shown in the nexttheorem.

Theorem 10.5For each positive integer n, congruence modulo n is an equivalence relationon Z.

Proof.We shall show that ≡ is reflexive, symmetric, and transitive.Reflexive: Since a− a = 0t for any t ∈ Z then a ≡ a(mod n).

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Symmetric: Let a, b ∈ Z be such that a ≡ b(mod n). Then a − b = nt forsome t ∈ Z. Multiplying both sides by −1 to obtain b − a = n(−t). Since(Z,+) is a group then −t ∈ Z and so b ≡ a(mod n).Transitive: Suppose that a ≡ b(mod n) and b ≡ c(mod n). Then a− b = ntand b− c = nt′ for some t, t′ ∈ Z. Adding these equations together to obtaina − c = n(t + t′). But Z is closed under addition so that t + t′ ∈ Z. Hence,a ≡ c(mod n).

Definition 10.3The equivalence classes for the equivalence relation ≡ are called congruenceclasses. They form a partition of Z. The set of all congruence classes isdenoted by Zn.

The following theorem shows that for each positive integer n, there are ncongruence classes and each integer is congruent to either 0, 1, 2, · · · , n − 1.Thus, the set {0, 1, 2, · · · , n − 1} is a complete set of representatives of therelation ≡.

Theorem 10.6Let n be a positive integer. Then each integer is congruent modulo n toprecisely one of the integers 0, 1, 2, · · · , n − 1. That is, there are n distinctcongruence classes, [0], [1], · · · , [n− 1].

Proof.Let a be any integer. Then by the Division Algorithm there exist uniqueintegers q and r such that

a = nq + r, 0 ≤ r < n.

This impplies that a− r = nq and so by Theorem 10.4, a ≡ r(mod n). Since0 ≤ r < n then a is congruent to at least one of the integers,0, 1, 2, · · · , n −1.We will show that a is congruent to exactly one of the integers listed. Tosee this, assume that a ≡ s(mod n) where 0 ≤ s < n. Then by Theorem 10.4,a = nt + s for some t ∈ Z. By uniqueness, we have r = s. This completes aproof of the theorem.

Remark 10.3It follows from the previous theorem that

Zn = {[0], [1], [2], · · · , [n− 1]}.

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Example 10.4For n = 4 the congruence classes are

[0] = {· · · ,−8,−4, 0, 4, 8, · · ·}[1] = {· · · ,−7,−3, 1, 5, 9, · · ·}[2] = {· · · ,−6,−2, 2, 6, 10, · · ·}[3] = {· · · ,−5,−1, 3, 7, 11, · · ·}

Thus, Z4 = {[0], [1], [2], [3]}

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Review Problems

Exercise 10.1List all the positive divisors of −101.

Exercise 10.2List all the prime numbers less than 100.

Exercise 10.3Find q and r such that a = bq + r.

(a) a = −7, b = 5.(b) a = 50, b = 6.(c) a = 11, b = 17.

Exercise 10.4How many positive integers divide (a) 3? (b) 9? (c) 27? (d) 3k, where k isa positive integer.

Exercise 10.5Assume that p is a prime and that k is a positive integer. How many positiveintegers divide (a) p? (b) p2? (c) p3? (d) pk?

Exercise 10.6Verify that each of the following statements is false.

(a) Every nonempty subset of Z contains a smallest element.(b) Every nonempty subset of Q contains a smallest element.

Exercise 10.7Show that the Division algorithm is valid for b < 0. Thus, given two integersa and b with b 6= 0, there exist unique integers q and r such that

a = bq + r, 0 ≤ r < |b|.

Exercise 10.8List the distinct congruence classes modulo 5, exhibiting at least three ele-ments in each class.

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Exercise 10.9Find all x such that 2x ≡ x(mod 5).

Exercise 10.10Find all the integers x such that −25 < x < 25 and x ≡ 3(mod 5).

Exercise 10.11Find all x such that 0 ≤ x < 6 and 2x ≡ 4(mod 6).

Exercise 10.12For which n is 25 ≡ 4(mod n)?

Exercise 10.13Assume that a ≡ b(mod n) and c ≡ d(mod n). Prove eac of the following:

(a) a± c ≡ b± d(mod n).(b) ac ≡ bd(mod n).(c) am ≡ bm(mod n), where m is a positive integer.

Exercise 10.14Prove that if a ≡ b(mod n) and n|a, then n|b.

Exercise 10.15Prove that if a+ x ≡ a+ y(mod n) then x ≡ y(mod n).

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11 Arithmetic Modulo n

For a positive integer n, the congruence modulo n relation induces a partitionon the set of integers by means of the elements of Zn given by

Zn = {[0], [1], · · · , [n− 1]}.

Also, recall from Theorem 9.2, that if a ≡ b(mod n) then [a] = [b]. Thus, forexample, if n = 6 then all of the following congruence classes are equal:

[3] = [9] = [−3] = {· · · ,−9,−3, 3, 9, · · ·}

A word of caution must me made regarding the notation [a]. In later sectionswe will consider mappings from Zm to Zn. So in order to distinguish betweenthe elements of these sets we will adopt the notation [a]m to be an elementof Zm and that of [a]n to be an element of Zn. In a context where only theelements of Zn are involved then we will keep the using the notation [a].

Next, we consider two operations on Zn: Addition and multiplication.

Definition 11.1For [a] ∈ Zn and [b] ∈ Zn we define addition by the rule

[a]⊕ [b] = [a+ b]

Example 11.1For n = 6, we have [−2]⊕[7] = [−2+7] = [5] and [3]⊕[9] = [3+9] = [12] = [0].

The operation of addition turns Zn into a finite Abelian group as shown next.

Theorem 11.1(a) ⊕ defines a binary operation on Zn. That is, Zn is closed under ⊕.(b) ⊕ is commutative.(c) ⊕ is associative.(d) [0] is the additive identity.(e) Each [a] ∈ Zn has an additive inverse [−a] ∈ Zn.(f) |Zn| = n.

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Proof.(a) We need to show that if ([a], [b]) ∈ Zn × Zn and ([c], [d]) ∈ Zn × Zn aresuch that ([a], [b]) = ([c], [d]) then [a]⊕ [b] = [c]⊕ [d]. That is, [a+b] = [c+d].Equivalently, according to Theorem 9.2, we need to show that a + b ≡ c +d(mod n). Since [a] = [c] then a ≡ c(mod n). Similarly, since [b] = [d] thenb ≡ d(mod n). But Theorem 10.4, a − c = nq and b − d = nq′ for someintegers q, q′. Thus, (a + b) − (c + d) = n(q + q′) and by Theorem 10.4,a+ b ≡ c+ d(mod n). Applying Theorem 9.2, we have [a+ b] = [c+ d].(b) The commutative property follows from the fact that addition in Z iscommutative

[a]⊕ [b] = [a+ b]= [b+ a]= [b]⊕ [a]

(c) The associative property follows from the fact that addition in Z is asso-ciative

([a]⊕ [b])⊕ [c] = [a+ b]⊕ [c]= [(a+ b) + c]= [a+ (b+ c)]= [a]⊕ [b+ c]= [a]⊕ ([b]⊕ [c])

(d) Since 0 is the identity of the group (Z,+) then [a] ⊕ [0] = [a + 0] = [a]and [0]⊕ [a] = [0 + a] = [a].(e) Since for each a ∈ Z we have a+ (−a) = (−a) + a = 0 then

[a]⊕ [−a] = [a+ (−a)]= [0]

and[−a]⊕ [a] = [(−a) + a]

= [0]

(f) This follows from the definition of Zn.

Remark 11.1With the above theorem, we have a tool now to construct finite abeliangroups of any order.

Example 11.2Let us construct the Cayley table for (Z4,⊕).

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⊕ [0] [1] [2] [3][0] [0] [1] [2] [3][1] [1] [2] [3] [0][2] [2] [3] [0] [1][3] [3] [0] [1] [2]

Definition 11.2The group (Zn,⊕) is called the group of integers modulo n.

Multiplication in Zn is defined as follows:

[a]� [b] = [ab]

Example 11.3For n = 6], we have [3]� [5] = [15] = [3].

We next state the basic properties for this operation.

Theorem 11.2(a) Zn is closed under �.(b) � is commutative.(c) � is associative.(d) [1] is the identity element.

Proof.The proofs of (a) - (d) are quite similar to those for the corresponding partsof Theorem th33, and are left as exercises.(See Exercise 11.10)

When we compare the properties listed in Theorems 11.1 and 11.2, we seethat the existence of multiplicative inverses is missing. So, in contrast to Znwith ⊕, Zn with � needs not be a group. The following example illustratesthis situation.

Example 11.4Writing Cayley table for Z4 we find

� [0] [1] [2] [3][0] [0] [0] [0] [0][1] [0] [1] [2] [3][2] [0] [2] [0] [2][3] [0] [3] [2] [1]

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Note that no element in Z4 satisfy the equation [x] � [0] = [1]. That is, [0]has no multiplicative inverse.

You might suspect that by removing the zero elements, the set Z∗n = {[1], [2], · · · , [n−1]} with ⊕ might be a group. Unfortunately, this is true for some values ofn but not for all n as shown in the following two examples.

Example 11.5Z∗6 is not a group with respect to � since Z∗6 is not closed under �. Indeed,[2]� [3] = [0] 6∈ Z∗6.

Example 11.6Constructing the Cayley table of Z∗5 with respect to � we find

� [1] [2] [3] [4][1] [1] [2] [3] [4][2] [2] [4] [1] [3][3] [3] [1] [4] [2][4] [4] [3] [2] [1]

Thus, [1]−1 = [1], [2]−1 = [3], [3]−1 = [2], and [4]−1 = [4].

In the next section, we will characterize those elements in Z∗n that havemultiplicative inverses in Z∗n and establish a condition on n for which Z∗n ={[1], [2], · · · , [n− 1]} is a group under the operation �.

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Review Problems

Exercise 11.1(a) Give five elements in [3] as an element of Z5.(b) Give five elements in [3] as an element of Z7.

Exercise 11.2Simplify each of the following expressions in Z5 and write the answer in termsof the elements of Z5.

(a) [2]⊕ [−7] (b) [17]⊕ [76] (c) [2]� [−7](d) [17]� [76] (e) [3]� ([2]⊕ [4]) (f) ([3]� [2])⊕ ([3]� [4])

Exercise 11.3Verify that [1]� [2]� [3]� [4] = [4] in Z4.

Exercise 11.4Find the elements [a] of Z6 for which the equation [a]�[x] = [0] has a nonzerosolution in Z6.

Exercise 11.5Whenever possible, find a solution for each of the following equations in thegiven Zn.

(a) [3]� [x] = [2] in Z6 (b)[6]� [x] = [4] in Z8

(c) [4]� [x] = [6] in Z8 (d)[8]� [x] = [6] in Z12

Exercise 11.6Contruct the Cayley table for (Z4,⊕).

Exercise 11.7Construct the Cayley table for (Z∗6,�). Show that Z∗6 is not a group under �.

Exercise 11.8Prove that � is distributive with respect to ⊕ in Zn.

Exercise 11.9Prove that (Z∗3,�) is a group.

Exercise 11.10Give a proof of Theorem 11.2.

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Exercise 11.11Solve the equation:[x]2 + [x]− [6] = [0] in Z11.

Exercise 11.12Solve the system of congruences: 2x ≡ 9(mod 15) and x ≡ 8(mod 11).

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12 Greatest Common Divisors. The Euclidean

Algorithm

As mentioned at the end of the previous section, we would like to establisha condition on n so that (Z∗n,�) is a group. This section provides the toolneeded for that.We start this section with the following definition.

Definition 12.1Let a and b be two integers, not both zero. A positive integer d is called agreatest common divisor of a and b if the following two conditions hold:

(1) d|a and d|b.(2) If c|a and c|b then c|d.We write d = (a, b) or d = gcd(a, b).

Example 12.1The greatest common divisor of -42 and 56 is 14. The greatest commondivisor is useful for writing fractions in lowest term. For example, −42

56= −3

4

where we cancelled 14 = (42, 56).

We next discuss a systematic procedure for finding the greatest commomdivisor of two integers, known as the Euclid’s Algorithm. For that purpose,we need the following result.

Theorem 12.1If a, b, q, and r are integers such that a = bq + r then (a, b) = (b, r).

Proof.Let d1 = (a, b) and d2 = (b, r). We will show that d1 = d2. Since d2|bq andd2|r then by Theorem 10.2, d2|(bq+r). That is d2|a. Thus, by Definition 12.1,d2|d1. Since d1|b then d1|bq. Since d1|a then d1|(a− bq). That is, d1|r. Hence,from the definition of d2, we have d1|d2. Since d1 and d2 are positive then byTheorem 10.2(d), we have d1 = d2.

The following theorem, establishes the existence and uniqueness of the great-est common divisor and provide an algorithm of how to find it.

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Theorem 12.2 (The Euclidean Algorithm)If a and b are two integers, not both zero, then there exists a unique positiveinteger d such that the two conditions (1) and (2) of Definition 12.1 aresatisfied.

Proof.Uniqueness: Let d1 and d2 be two positive integers that satisfy conditions(1) and (2). Then by (2) we can write d1|d2 and d2|d1. Since d1 and d2 areboth positive then Theorem 10.2 (d) implies that d1 = d2.

Existence: Without loss of generality, we may assume that b 6= 0. Notethat if a = 0 then d = |b|. Indeed, |b| |0 and |b| |b. Moreover, if c is a commondivisor of a and b then b = cq so that |b| = cq′. That is, c| |b|.So assume that a 6= 0. By the Division algorithm there exist unique integersq1 and r1 such that

a = bq1 + r1, 0 ≤ r1 < |b|.If r1 = 0 then as above, one can easily check that d = |b|. So assume thatr1 6= 0. Using the Division algorithm for a second time to find unique integersq2 and r2 such that

b = r1q2 + r2 0 ≤ r2 < r1.

We keep this process going and eventually we will find integers rn and rn+1

such thatrn−2 = rn−1qn + rn, 0 ≤ rn < rn−1

andrn−1 = rnqn+1.

That is, rn is the last nonzero remainder in the process. By Theorem 12.1,we see that rn = (a, b).

Example 12.2Performing the arithmetic for the Euclidean algorithm we have

1776 = (1)(1492) + 2841492 = (5)(284) + 72284 = (3)(72) + 6872 = (1)(68) + 468 = 4(17)

So (1776, 1492) = 4.

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An alternative proof for the existence of the greatest common divisor whichdoes not provide a systematic way for finding (a, b) is given next.

Theorem 12.3If a and b are two integers, not both zero, then there exist integers m and nsuch that

(a, b) = ma+ nb.

That is, (a, b) can be expressed as a linear combination of a and b.

Proof.Without loss of generality, we assume that b 6= 0. Let S = {xa+yb : x, y ∈ Z}and S ′ = {xa+ yb ∈ S : xa+ yb > 0}. Since b = 0a+ (1)b then b ∈ S. Thus,S 6= ∅. Also, S ′ 6= ∅. To see this, note that if b > 0 then b = 0a + (1)b ∈ S ′.If b < 0 then −b = 0a + (−1)b ∈ S ′. By Theorem 10.1, S ′ has a smallestelement d. Thus, d = ma+ nb > 0 for some integers m and n. We will shownext that d = (a, b). Applying the Division algorithm we can find integers qand r such that a = dq + r with 0 ≤ r < d. From this equation we see that

r = a− dq= a− (ma+ nb)q= (1−mq)a+ (−nq)b

If r > 0 then r ∈ S ′ and r < d. This contradicts the definition of d. Therefore,r = 0 and this gives a = dq and hence d|a. A similar argument holds for d|b.Finally, if c is an integer such that c|a and c|b then a = cq and b = cq′. Thus,

d = ma+ nb = mcq + ncq′ = c(mq + nq′)

This means that c|d. Thus, d = (a, b). This ends a proof of the Theorem.

Remark 12.1The integers m and n in Theorem 12.3 are not unique. Indeed,

(a, b) = ma+ nb= ma+ ab+ nb− ab= (m+ b)a+ (n− a)b = m′a+ n′b

Example 12.3From Example 12.2, we found that (1776, 1492) = 4. Let’s write this as a

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linear combination of 1776 and 1492. We use the equations in Example 12.2,beginning with 72 = (1)(68) + 4 and working backward one step at a time.

4 = 72− (1)(68).

Solve the equation 284 = (3)(72) + 68 for 68 and substitute in the previousequation and simplify to obtain

4 = 72− (1)(284− 3 · 72)= 72 · (4)− 284

Solve the equation 1492 = (5)(284)+72 for 72 and substitute in the previousequation and simplify to obtain

4 = (1492− 5 · 284) · 4− 284= 1492 · 4− 21 · 284

Finally, solve the equation 1776 = 1492+284 for 284 and substitute to obtain

4 = 1492 · 4− 21(1776− 1492)= −21 · 1776 + 25 · 1492

Theorem 12.4If a and b are integers, not both zero, then (a, b) = 1 if and only if ma+nb = 1for some integers m and n.

Proof.Suppose first that (a, b) = 1. By Theorem ??, there exist integers m and nsuch that ma+ nb = 1.Conversely, suppose that ma+nb = 1 for some integers m and n. If d = (a, b)then d|a and d|b so d|(ma+ nb). That is, d|1. Since d > 0 then by Theorem10.2(d) we must have d = 1.

Definition 12.2If a and b are integers such that (a, b) = 1 then we say that a and b arerelatively prime.

The next result, characterizes those elements of Zn that have multiplicativeinverses.

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Theorem 12.5[a] ∈ Zn has a multiplicative inverse if and only if (a, n) = 1.

Proof.Suppose first that [a] has a multiplicative inverse [b] in Zn. Then [a]� [b] =[ab] = [1]. This implies that ab ≡ 1(modn). Therefore, ab = nq + 1 for someinteger q. This last equality can be written in the form ba + (−q)n = 1. ByTheorem 12.4, (a, n) = 1.Conversely, suppose that (a, n) = 1. Then by Theorem 12.4, there exist in-tegers m and q such that ma + qn = 1. Thus, ma − 1 = (−q)n and hencema ≡ 1(mod n). In terms of �, we have [m] � [a] = [1]. Thus, [a] has amultiplicative inverse in Zn.

As a consequence of the above theorem we have

Theorem 12.6Every nonzero element of Zn has a multiplicative inverse if and only if n isa prime number. Thus, (Z∗n,�) is a group if and only if n is prime.

Proof.If n is prime then (a, n) = 1 for every 1 ≤ a < n. By Theorem 12.6, [a] hasa multiplicative inverse for all 1 ≤ a < n. Conversely, suppose that [a] has amultiplicative inverse for all 1 ≤ a < n. Again, by Theorem 12.6, (a, n) = 1for 1 ≤ a < n. This is true only if n is prime.

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Review Problems

Exercise 12.1In each part, find the greatest common divisor (a, b) and integers m and nsuch that (a, b) = ma+ nb.

(a) a = 65, b = 91.(b) a = 5088, b = 156.(c) a = −75, b = 105.

Exercise 12.2List all the prime numbers less than 100.

Exercise 12.3Find the smallest integer in the set

{x ∈ Z : x > 0 and x = 6m+ 15n, m, n ∈ Z}

Exercise 12.4List all of the positive integers that are less than 12 and relatively prime to12.

Exercise 12.5Prove that if c is a divisor of a and b then c is a divisor of ma + nb for allm,n ∈ Z.

Exercise 12.6Prove that if a and b are distinct prime then there exist integers m and nsuch that ma+ nb = 1.

Exercise 12.7Prove that (ab, c) = 1 if and only if (a, c) = (b, c) = 1.

Exercise 12.8Prove that if c is a positive integer then (ac, bc) = (a, b)c.

Exercise 12.9Prove that if a is an integer and p is a prime such that p 6 |a then (a, p) = 1.

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Exercise 12.10Prove that if (a, b) = 1 and c|a then (c, a) = 1.

Exercise 12.11Prove that if (a, n) = 1 then there is a solution to the congruence equationax ≡ 1(mod n).

Exercise 12.12For each positive integer n, let φ(n) denote the number of positive integersless than and rlatively prime to n. For example, φ(3) = 2 and φ(4) = 2. Wedefine φ(1) = 1. We call φ the Euler’s function.

(i) Find φ(n) for 1 ≤ n ≤ 10.(ii) Find a formula for φ(p) if p is a prime number.

Exercise 12.13For each positive integer n let

Z(n) = {[k] : 1 ≤ k < n and (k, n) = 1}.

(a) List the elements of Z(12).(b) Prove that (Z(n),�) is a group.(c) Show that the number of elements of Z(n) is φ(n).

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13 Least Common Multiple. The Fundamen-

tal Theorem of Arithmetic

In the previous section, we learned how to find the largest positive divisorof two integers that are not both zero. In this section, we want to find thesmallest common factor of two nonzero integers.

Theorem 13.1If a and b are nonzero integers then there is a unique positive integer m suchthat

(1) a|m and b|m;(2) if c is an integer such that a|c and b|c then m|c.

Proof.Let S = {x ∈ N : a|x and b|x}. Since a| |ab| and b| |ab| then |ab| ∈ S andtherefore S 6= ∅. By Theorem 10.1, S has a smallest element m. Thus, a|mand b|m. This proves (1).To prove (2), we assume that c is an integer such that a|c and b|c. By theDivision Algorithm there exist unique integers q and r such that

c = mq + r, 0 ≤ r < m.

Since a|c then c = aq1 for some q1 ∈ Z. Since a|m then m = aq2 for someq2 ∈ Z. Thus,

aq1 = aqq2 + r

This implies that a|r. A similar argument with b replacing a we find that b|r.If r > 0 then r ∈ S and this contradicts the definition of m. So we must haver = 0. Therefore, c = mq and m|c. This completes a proof of the theorem.

Definition 13.1The positive integer m is called the least common multiple of a and b andis denoted by [a, b].

Example 13.1If a = 25 and b = 33 then [25, 33] = 825. Similarly, [4,−6] = 12.

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Our next goal is to find a method for finding the least common multiple oftwo nonzero integers. The method is a result of the Fundamental Theoremof Arithmetic.For that purpose we need the following lemmas.

Lemma 13.1If a, b, and c are integers such that a|bc and (a, b) = 1 then a|c.

Proof.Since (a, b) = 1 then by Theorem 12.4, there are integers m and n such thatma + nb = 1. Multiply this equation by c to obtain mac + nbc = c. Sincea| bc then a| (mac+ nbc). That is, a|c.

Remark 13.1The condition (a, b) = 1 is critical. For example, if a = 6, b = 3, and c = 4.Then a|bc but neither a divides b or a divides c. Note that (a, b) = 3 6= 1.

Lemma 13.2If a is an integer and p is a prime number such that p 6 |a then (a, p) = 1.

Proof.Let d = (a, p). Then d|a and d|p. Since p is prime then either d = 1 or d = p.If d = p then p|a which contradicts the fact that p 6 |a. Thus, d = 1.

Lemma 13.3If a1, a2, · · · , an are integers and p is a prime such that p| a1a2 · · · an thenp| ai for some 1 ≤ i ≤ n.

Proof.The proof is by induction on n. The case n = 1 is trivial. So assume that thelemma holds for all positive integers up to and including n-1. We will showthat the result still holds for n. Since p| a1a2 · · · an then either p|a1a2 · · · an−1

or p 6 |a1a2 · · · an−1. If p| a1a2 · · · an−1 then by the induction hypothesis,p| ai for some 1 ≤ i ≤ n − 1. If p 6 |a1a2 · · · an−1 then by Lemma 13.2,(p, a1a2 · · · an−1) = 1. By Lemma13.1 (with a = a1a2 · · · an−1 and b = an) wehave p|an.

Theorem 13.2 (The Fundamental Theorem of Arithmetic)Every positive integer n > 1 is either a prime or can be written as a product ofprime integers, and this product is unique except for the order of the factors.

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Proof.The proof is by induction on n. The theorem is trivially true for n = 2 since2 is prime. Assume, then that it is true for the integers 2, 3, · · · , n − 1. Weshall prove that it is also true for n. If n is prime there is nothing to prove.Assume, then, that n is composite and that n has two factorizations, say

n = p1p2 · · · ps = q1q2 · · · qt (2)

Since p1 divides the product q1, q2, · · · , qt, then by Lemma 13.3, it must divideat least one factor. Relabel q1, q2, · · · , qt so that p1| q1. Then p1 = q1 sinceboth p1 and q1 are primes. In (2), we may cancel p1 on both sides to obtain

n

p1

= p2p3 · · · ps = q2q3 · · · qt < n.

By the induction hypothesis, the two factorizations of np1

must be identicalexcept for the order of the factors. Therefore, s = t and the factorizationsin (2) are also identical, except for the order of factors. This ends a proofof the theorem.

Remark 13.2In the factorization of an integer n > 1, a particular prime p may occur morethan once. If the distinct prime factors of n are p1 < p2 < · · · < ps and if pioccurs as a factor ki times, we can write

n = pk11 pk22 · · · pkss

We shall call this the standard form for n.

Lemma 13.4Let m and n be two integers with the following prime factorization

m = pk11 pk22 · · · pkss and n = pt11 p

t22 · · · ptss

Then m|n if and only if ki ≤ ti for 1 ≤ i ≤ s.

Proof.If p is a prime and a and b are integers such that α is the highest power ofp dividing a and β is the highest power of p dividing b then a = pαq andb = pβq′ for some q, q′ ∈ Z. Thus, ab = pα+βq′′, where q′′ ∈ Z. Hence, α+β isthe highest power of p dividing ab. So if m|n then m = nu for some integer

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u. Thus, for each 1 ≤ i ≤ s, the highest power of pi dividing n is the sum ofthe highest powers of pi dividing m and u. Thus, ki ≤ ti for 1 ≤ i ≤ s.Conversely, suppose that ki ≤ ti for all 1 ≤ i ≤ s. Let ui = ti − ki for1 ≤ i ≤ s. Let w = pu11 p

u22 · · · puss . Then n = mw and so m|n.

We have the following result which expresses [a, b] in terms of the factor-izations of a and b.

Theorem 13.3If two nonzero integers a and b have the factorizations

a = pk11 pk22 · · · pkss and b = pt11 p

t22 · · · ptss

then[a, b] = pu11 p

u22 · · · puss

where ui is the maximum of ki and ti for each 1 ≤ i ≤ s.

Proof.Let m = pu11 p

u22 · · · puss where ui is the maximum of ki and ti. We will show

that m = [a, b]. Since ki ≤ ui and ti ≤ ui for all 1 ≤ i ≤ s then byLemma 13.4, a|m and b|m. Now, if c is an integer such that a|c and b|c.Write c = pw1

1 pw22 · · · pws

s . Then by Lemma 13.4, ki ≤ wi and ti ≤ wi for all1 ≤ i ≤ s. Thus, ui ≤ wi for all 1 ≤ i ≤ s. By Lemma 4, m|c. It follows thatm = [a, b].

Finding the least common multiple Similarly, Theorem 13.2 can be usedto find the greatest common divisor of two integers.

Theorem 13.4If two nonzero integers a and b have the factorizations

a = pk11 pk22 · · · pkss and b = pt11 p

t22 · · · ptss

then(a, b) = pu11 p

u22 · · · puss

where ui is the minimum of ki and ti for each 1 ≤ i ≤ s.

Proof.Left as an exercise for the reader. See Exercise 13.3.

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Remark 13.3The above theorem gives a way to find (a, b) if we know the factorizations ofa and b. But factorizing a number is a very hard problem and so, the abovemethod is ineffective. Instead, the Euclidean Algorithm is more practical.

Example 13.2Consider the prime factorizations of the integers 31752 and 126000 :

31752 = 23 · 34 · 72 and 126000 = 24 · 32 · 53 · 7.

Then from Theorems 13.3 and 13.4 we have

[31752, 126000] = 24 · 34 · 53 · 72 = 7938000

and(31752, 126000) = 23 · 32 · 7 = 504.

134

Review Problems

Exercise 13.1Determine the standard form for each of the following integers.

(a) 684 (b) 1375

Exercise 13.2Compute the greatest common divisor and the least common multiple usingthe standard forms.

(a) 10,105 (b) -2860,-2310 (c) -39, 54.

Exercise 13.3Prove Theorem 13.4.

Exercise 13.4Prove that if a and b are positive integers then

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

Exercise 13.5Prove that if (a, b) = 1, a|m, and b|m then ab|m.

Exercise 13.6Determine φ(pk) if p is a prime number and k is a positive integer, where φis the Euler function.

Exercise 13.7Prove that if a|bc then a|(a, b)c.

Exercise 13.8Prove that [a, b, c] = [[a, b], c] = [a, [b, c]].

Exercise 13.9(a) Prove that (a, b, c) = ((a, b), c) = (a, (b, c)).(b) Prove that (a, b, c) = ma+ nb+ pc for some integers m,n, p.

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Exercise 13.10Let p > 1 be an integer with the property that if p|ab then p|a or p|b. Provethat p must be a prime number.

Exercise 13.11Let n be an integer. Prove that

√n ∈ Q if and only if n = k2 for some k ∈ Z.

Exercise 13.12Prove that 3

√2 is irrational.

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14 Elementary Properties of Groups

In this section, we prove more theorems about groups. In what follows thegroup binary operation will be referred to as multiplication and thus we willwrite ab.Several simple consequences of the definition of a group are recorded in thefollowing two theorems.

Theorem 14.1For any group G, the following properties hold:

(i) If a, b, c,∈ G and ab = ac then b = c. (left cancellation law)(ii) If a, b, c,∈ G and ba = ca then b = c. (right cancellation law)(iii) If a ∈ G then (a−1)−1 = a. The inverse of the inverse of an element isthe element itself.(iv) If a, b ∈ G then (ab)−1 = b−1a−1. That is the inverse of a product is theproduct of the inverses in reverse order.

Proof.(i) Suppose that ab = ac. Then

b = eb = (a−1a)b= a−1(ab) = a−1(ac)= (a−1a)c = ec = c

(ii) Suppose that ba = ca. Then

b = be = b(aa−1)= (ba)a−1 = (ca)a−1

= c(aa−1) = ce = c

(iii) If a ∈ G then since aa−1 = a−1a = e then a is an inverse of a−1. Sinceinverses are unique then (a−1)−1 = a.(iv) Let x be the inverse of ab. Then (ab)x = e By associativity, we havea(bx) = aa−1. By (i), we have bx = a−1. But bx = ea−1 = b(b−1a−1) so thatby applying (i) again we obtain x = b−1a−1. Therefore, (ab)−1 = b−1a−1.

Theorem 14.2If G is a group and a, b ∈ G then each of the equations ax = b and xa = bhas a unique solution. In the first, the solution is x = a−1b whereas in thesecond x = ba−1.

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Proof.Consider first the equation ax = b. We want to isolate the x on the left side.Indeed, this can be done as follows.

x = ex = (a−1a)x= a−1(ax) = a−1b

To prove uniqueness, suppose that y is another solution to the equationax = b. Then, ay = b = ax. By the left cancellation property we have x = y.Finally, the proof is similar for the equation xa = b.

Remark 14.1Suppose G is a finite group and a, b ∈ G. The product ax is in the rowlabelled by a of the Cayley table. By Theorem 14.2, the equation ax = bhas a unique solution means that b appears only once in the row of a of thetable. Thus, each element of a finite group appears exactly once in each rowof the table. Similarly, because there is a unique solution of xa = b, eachelement appears exactly once in each column of the Cayley table. It followsthat each row is a rearrangement of the elements of the group.

Next, we discuss the extension of the associative property to products withany number of factors. More specifically, we will prove the so-called gener-alized associative law which states that in a set with associative operation,a product of factors is unchanged regardless of how parentheses are insertedas long as the factors and their order of appearance in the product are un-changed.

Definition 14.1For elements a1, a2, · · · , an (n ≥ 2) of a group G define

a1a2 · · · an−1an = (a1a2 · · · an−1)an.

Theorem 14.3 (Generalized Associative Law)For any integer 1 ≤ m < n we have

(a1a2 · · · am)(am+1 · · · an) = a1a2 · · · an.

Proof.For n ≥ 2, let S(n) be the statement

(a1a2 · · · am)(am+1 · · · an) = a1a2 · · · an : where a1, a2, · · · , an ∈ G and 1 ≤ m < n.

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We prove that S(n) is true for all n ≥ 2 by induction on n. For n = 2the statement is true since (a1)(a2) = a1a2 (the only possible value for m ism = 1.) So assume that the statement is valid for 2, 3, · · · , n − 1. We willshow that S(n) is also true. Suppose that 1 ≤ m < n. Then either m = n−1or 1 ≤ m < n− 1. If m = n− 1 then

(a1a2 · · · am)(am+1 · · · an) = (a1a2 · · · an−1)an = a1a2 · · · an.

So suppose that 1 ≤ m < n− 1. Then

(a1a2 · · · am)(am+1 · · · an) = (a1a2 · · · am)[(am+1 · · · an−1)an]= [(a1a2 · · · am)(am+1 · · · an−1)]an= (a1a2 · · · an−1)an= a1a2 · · · an

Thus, S(n) is true for all n ≥ 2 and this completes a proof of the theorem.

Next, we introduce the concept of integral exponents of elements in a group.The concept plays an important role in the theory of cyclic groups.

Definition 14.2For any a ∈ G we define

a0 = ean = an−1a, for n ≥ 1a−n = (a−1)n for n ≥ 1.

Remark 14.2By Theorem 14.3, an = a · a · a · · · a where the product contains n copies ofa. Also, note that aan−1 = an−1a = an.

The familiar laws of exponents hold in a group.

Theorem 14.4Let a be an element of a group G and m and n denote integers. Then

(i) ana−n = e.(ii) aman = am+n

(iii) (am)n = amn.

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Proof.(i) The identity is trivial for n = 0. So suppose that n > 0. We use inductionon n. Since aa−1 = e then the result is true for n = 1. Suppose the identityholds for 1, 2, · · · , n− 1. We must show that it is valid for n. Indeed,

ana−n = (an−1a)[(a−1)−(n−1)a−1]= (aan−1)[(a−1)−(n−1)a−1]= [(aan−1)(a−1)−(n−1)]a−1

= [a(an−1a−(n−1))]a−1

= (ae)a−1 = aa−1 = e

Thus, the identity is true for all positive integers n.Now, suppose that n < 0 then

ana−n = (a−1)−n(a−1)−(−n) = e.

(ii) We have to show that for a fixed integer m the identity holds for allintegers n. That is, the identity holds for n = 0, n > 0, and n < 0. If m = 0then aman = a0an = ean = an = a0+n for all integers n. So, suppose firstthat m > 0.

n = 0

am+n = am+0 = am

aman = ama0 = ame = am

n > 0By induction on n > 0. The case n = 1 follows from Definition 14.2. Supposethat the identity has been established for the numbers 1, 2, · · · , n−1. We willshow that it is still true for n. Indeed,

aman = am(an−1a)= (aman−1)a= am+n−1a= am+n (by Definition 14.2)

By induction, it follows that the identity is true for all n > 0.

n < 0, n = −mSince n = −m then n + m = 0 and in this case we have am+n = a0 = e. By

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(i), we have aman = ama−m = e. It follows that aman = am+n.

n < 0, n > −mThen m = (m+ n) + (−n) where both m+ n and (−n) are positive.

aman = a(m+n)+(−n)an

= (am+na−n)an

= am+n(a−nan)= am+ne = am+n

n < 0, n < −mIn this case, −n = m+ [−(m+n)] where both m and −(n+m) are positive.

aman = am(a−1)−n

= am(a−1)m−(m+n)

= am[(a−1)m(a−1)−(m+n)]= am[a−m(a−1)−(m+n)]= (ama−m)(a−1)−(m+n)

= (a−1)−(m+n) = am+n

Similar arugument holds for a fixed m < 0 and all integers n. This completesa proof of (ii).

(iii) Similar to (ii) and is left as an exercise. See Exercise 14.21.

Remark 14.3In the case of an group G written with the binary operation +, for n ∈ Z+

and a ∈ G, one writes na instead of an, where na = a + ... + a (n times),and (−n)a = −(na) = n(−a). The laws corresponding to (ii) and (iii) ofTheorem 14.4 become

ma+ na = (m+ n)a

andn(ma) = (mn)a.

where m,n ∈ Z.

The set of all integral exponents of an element a in a group G forms asubgroup of G as shown in the next theorem.

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Theorem 14.5Let G be a group and a ∈ G. Then the set

< a >= {an : n ∈ Z}

is a subgroup of G.

Proof.The set < a > is nonempty since a0 = e ∈< a > . Now, let x, y ∈< a > .Then x = an and y = am for some integers n and m. Thus,

xy−1 = an(am)−1

= ana−m (by Theorem 14.1(iv))= an−m (by Theorem 14.4(ii))

Since n − m ∈ Z then xy−1 ∈< a > . Thus, by Theorem 7.5, < a > is asubgroup of G, as required.

Definition 14.3The subgroup < a > is called the subgroup generated by a. Any subgroupH of G that can be written as K =< a > is called a cyclic subgroup. Theelement a is called a generator. In particular, G is a cyclic group if thereis an element a ∈ G such that G =< a > .

Example 14.11. The group Z of integers under addition is a cyclic group, generated by 1(or -1). Thus, (Z,+) is a cyclic group of infinite order.2. Let n be a positive integer. The set Zn of congruence classes of integersmodulo n is a cyclic group of order n with respect to the operation of additionwith generator [1].

The following theorem shows that when powers of a are equal then the cyclicgroup < a > is of finite order.

Theorem 14.6Let G be a group and a ∈ G be such that ar = as for some integers r and swith r 6= s.

(i) There is a smallest positive integer n such that an = e.(ii) at = e if and only if n|t.(iii) The elements e, a, a2, · · · , an−1 are distinct and

< a >= {e, a, a2, · · · , an−1}.

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Proof.(i) Let S = {p ∈ Z+ : ap = e}. Assume that r > s. (If s > r then interchangethe letters r and s in the following sentences) Since ar = as then ara−s =asa−s = e. Thus, r − s ∈ S. By Theorem 10.1, there is a smallest positiveinteger n such that an = e.(ii) Suppose first that at = e for some integer t. By the Division Algorithmthere exist integers q and r such that t = nq + r where 0 ≤ r < n. Thus,e = at = anq+r = (an)qar = ear = ar. By the definition of n we must haver = 0. That is, t = nq and consequently n|t. Conversely, suppose that n|t.Then t = nq for some integer q. Thus, at = (an)q = e.(iii) First we prove that e, a, a2, · · · , an−1 are all distinct. To see this, supposethat au = av with 0 ≤ u < n and 0 ≤ v < n. Without loss of generality wemay assume that u ≥ v. Since au = av then au−v = e. By (ii), n|(u− v). But0 ≤ u− v ≤ u < n. Thus, we must have u− v = 0 or u = v.We prove < a >= {e, a, a2, · · · , an−1} by double-inclusions. It is triviallytrue that {e, a, a2, · · · , an−1} ⊆< a > . Now, let at ∈< a >. By the divisionalgorithm, there exist integers q and r such that t = nq + r with 0 ≤ r < n.Thus, at = (an)qar = ar with 0 ≤ r < n. That is, at ∈ {e, a, a2, · · · , an−1}.This completes a proof of the theorem

Definition 14.4Let a be an element of a group G. The order of a is the smallest positiveinteger n, if it exists, for which an = e. If such an integer does not exist thenwe say that a has an infinite order. We denote the order of a by o(a).

Example 14.21. In Z4, o([2]) = 2.2. In (Q∗, ·) the number 2 has infinite order since 2n 6= 1 for all positiveinteger n.

We end this section with a theorem that gives the relationship between theorder of a group and the order of an element.

Theorem 14.7If G is a group and a ∈ G then o(a) = | < a > |.Proof.If o(a) = n then by Theorem 14.6 (iii) we have < a >= {e, a, a2, · · · , an−1}.Thus, | < a > | = n = o(a). If o(a) is infinite then by Theorem 14.6(i) theintegral powers of a are all distinct. Thus, < a > is infinite and o(a) = | <a > |

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Review Problems

Exercise 14.1Solve the equation (12)x = (123) in S3.

Exercise 14.2Solve the equation x(132) = (13) in S3.

Exercise 14.3Find the order of the element (12)(34) in S4.

Exercise 14.4Consider the group Z16 under addition. List all the elements of the subgroup< [6] > .

Exercise 14.5Consider the group Z∗13 under multiplication. List all the elements of thesubgroup < [4] > .

Exercise 14.6(a) Determine the elements in the subgroup < (1234) > of S4.(b) Determine the elements in the subgroup < (12345) > of S5.(c) What is the order of the subgroup < (12 · · ·n) > of Sn?

Exercise 14.7Determine the elements of the subgroup < µ3 > of the group of symmetriesof the square.

Exercise 14.8Let α denote the clockwise rotation of the plane through 90◦ about a fixedpoint p.(α ∈ G in Example 5.3(a)) What is the order of < α >?

Exercise 14.9Prove that axb = c has a unique solution in a group.

Exercise 14.10(a) Prove that if a and b are elements in an Abelian group G, with o(a) = mand o(b) = n then (ab)mn = e.(b) Give an example in which the above statement is false if the group is notAbelian.

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Exercise 14.11Let a and b be elements of a group G. Prove that ab = ba if and only ifa−1b−1 = b−1a−1.

Exercise 14.12Assume m,n ∈ Z. Find a necessary and sufficient condition for < m >⊆<n > .

Exercise 14.13Prove that a nonidentity element of a group has order 2 if and only if it isits own inverse.

Exercise 14.14Prove that every group of even order has an element of order 2.

Exercise 14.15Prove that a group G is Abelian if anf only if (ab)−1 = a−1b−1 for all a, b ∈ G.

Exercise 14.16Prove that if a group G has no subgroup other than G and {e}, then G iscyclic.

Exercise 14.17Prove that if G is a group and a, b ∈ G then o(a−1ba) = o(b).

Exercise 14.18Prove that if a, b ∈ G then o(ab) = o(ba).

Exercise 14.19Let a be an element of a group G. Prove that o(a−1) = o(a).

Exercise 14.20Prove that a group G is Abelian if each of its nonidentity elements has order2.

Exercise 14.21Prove Theorem 14.4(iii).

Exercise 14.22In S3, list all the elements of < (123) > . Find o((123)).

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Exercise 14.23Prove that if G is a group then the mapping λ : G→ G defined by λ(x) = axis one-to-one and onto.

Exercise 14.24There is only one way to complete the following Cayley table so as to get agroup. Find it. Why is it unique?

* a b ca bbc

Exercise 14.25Let G be a cyclic group, G =< a > . Prove that G is Abelian.

Exercise 14.26Let G be a cyclic group of order n,G =< a > . Prove that as = at if and onlyif s ≡ t(mod n).

Exercise 14.27Let G =< a > be a cyclic group and H be a subgroup of G. Prove thatH =< ak > where k is the smallest positive integer such that ak ∈ H. Thus,every subgroup of a cyclic group is cyclic.

Exercise 14.28(a) Let G =< a > be a finite cyclic group of order n. Prove that < ad >=<am > where d = (m,n). Thus, the distinct subgroups of G are those subgroups< ad > where d is a divisor of n.(b) Find the distinct subgroups of a cyclic group of order 12.

Exercise 14.29(a) Let G =< a > be a cyclic group of order n. Prove that G =< am > ifand only if (m,n) = 1.(b) Suppose that G =< a > is a cyclic group of order 10. Find all possiblegenerators of G.

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15 Generated Groups. Direct Product

In this section, we discuss two procedures of building (sub)groups, namely,the (sub)groups generated by a subset of a given group and the direct productof two or more groups.

15.1 Finitely and Infinitely Generated Groups

The concept of generators can be extended from cyclic groups < a > tomore complicated situations where a subgroup is generated by more than oneelement. It is easy to see that a cyclic group with generator a is the smallestsubgroup containing the set S = {a}. Can we extend groups generated bysets with more than one element? The answer is affirmative as suggested bythe following theorem.

Theorem 15.1Let G be a group and C be the collection all subgroups of G. Then ∩H∈CH isa subgroup of G.

Proof.Let K = ∩H∈CH. Since e belongs to all the subgroups of G then e ∈ K sothat K 6= ∅. Now, let a, b ∈ K. Then a ∈ H and b ∈ H for all subgroups H ofG. But then a ∈ H and b−1 ∈ H,∀H ∈ C. Since every H in C is closed undermultiplication then ab−1 ∈ H,∀H ∈ C. That is, ab−1 ∈ K. By Theorem 7.5,K is a subgroup of G.

Theorem 15.2Let G be a group and S ⊆ G. Let C be the collection of all subgroups of Gcontaining S. Then the set < S >= ∩H∈CH satisfies the following:

(i) < S > is a subgroup of G containing S.(ii) For every H ∈ C, < S >⊆ H.Thus, < S > is the smallest subgroup of G containing S.

Proof.(i) Note that since S ⊆ G then G ∈ C. Thus, C 6= ∅. The fact that < S >is a subgroup of G follows from Theorem 15.1. Since S ⊆ H for all H ∈ Cthen S ⊆ ∩H∈CH. That is, S ⊆< S > .(ii) Let H ∈ C. If x ∈< S > then x ∈ ∩H∈CH and in particular, x ∈ H.Hence, < S >⊆ H.

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Theorem 15.3< S > is the only subgroup of G that satisfies conditions (i) and (ii) ofTheorem 15.2

Proof.Suppose that K is a subgroup of G satisfying conditions (i) and (ii) of theprevious theorem. We will show that K =< S > . Since < S > is a subgroupcontaining S then by (ii), we have < S >⊆ K. Also, since K is a subgroupof G containing S then by condition (ii) again, we have K ⊆< S > . Thus,< S >= K.

Definition 15.1If G is a group and S ⊆ G then the subgroup < S > is called the subgroupof G generated by S. If G =< S >, then we say S generates G; andthe elements in S are called generators. For S = {a1, a2, · · · , an} we write< S >=< a1, a2, · · · , an > .

Example 15.1Consider the subgroup of S4 generated by S = {(1432), (24)}. Then

< S >= {(1), (1432), (24), (14)(23), (1234), (12)(34), (13)(24), (13)}.

The following theorem characterizes the elements of < S > .

Theorem 15.4Let G be a group and S ⊆ G. Then < S > consists of finite products of ele-ments of S and inverses of elements of S.. That is, if K = {as11 a

s22 · · · a

skk , ai ∈

S, si = ±} then < S >= K.

Proof.The proof is by double inclusions. By closure, K ⊆< S > . It remainsto show that < S >⊆ K. Since each element of S is in K then S ⊆ K.We will show that K is a subgroup of G. Indeed, if x = as11 a

s22 · · · a

skk and

y = bt11 bt22 · · · btmm are two elements in K then

xy−1 = (as11 as22 · · · a

skk )(bt11 b

t22 · · · btmm )−1

= (as11 as22 · · · a

skk )(b−tmm · · · b−t22 b−t11 )

= as11 as22 · · · a

skk b−tmm · · · b−t22 b−t11

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Thus, xy−1 is a product of elements of S and/or inverses of elements ofS. Hence, xy−1 ∈ K. By Theorem 7.5, K is a subgroup of G. By Theo-rem 15.2(ii), < S >⊆ K. Hence, < S >= K.

The next theorem provides a condition under which two groups generatedby different sets are equal.

Theorem 15.5Let T1 and T2 be subsets of a group G. Then

< T1 >=< T2 > if and only if T1 ⊆< T2 > and T2 ⊆< T1 >

Proof.Suppose first that < T1 >=< T2 > . Since T1 ⊆< T1 > and < T1 >=< T2 >then T1 ⊆< T2 > . Similarly, T2 ⊆< T1 > .Conversely, suppose that T1 ⊆< T2 > and T2 ⊆< T1 > . By Theorem 15.2(ii),we have < T1 >⊆< T2 > and < T2 >⊆< T1 > . Hence, < T1 >=< T2 > .

Example 15.2rm Since {3} ⊆< 9, 12 > (3 = 12 + (−9)) and {9, 12} ⊆< 3 > then by theprevious theorem we have < 9, 12 >=< 3 > .

Example 15.3rm In S4 we have that

(124) = (123)(12)(34)(123)(234) = (132)(12)(34) = (123)−1(12)(34)(123) = (124)(234)

(12)(34) = (234)(124)−1

Thus, we conclude that {(124), (134)} ⊆< (123), (12)(34) > and {(123), (12)(34)} ⊆<(124), (134) > . Hence, by Theorem 15.5, we have < (124), (234) >=<(123), (12)(34) > .

15.2 Direct Product of Groups.

In this section we keep building examples of groups. By defining a binaryoperation on the Cartesian product of two groups we obtain the group knownas the direct product of the two groups.

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Let H and K be two arbitrary groups and let H × K be the Cartesianproduct of H and K. In set-builder notation

H ×K = {(h, k) : h ∈ H and k ∈ K}.

Equality in H × K is defined by (h, k) = (h′, k′) if and only if h = h′ andk = k′.Define multiplication on H ×K as follows:

(h1, k1)(h2, k2) = (h1h2, k1k2)

The set H × B is a group under the above multiplication as shown in thenext theorem.

Theorem 15.6Multiplication on H ×K satisfies the following properties:

(i) H ×K is closed under multiplication.(ii) Multiplication is associative.(iii) (eH , eK) is the identity element.(iv) For each (h, k) ∈ H ×K we have (h, k)−1 = (h−1, k−1).

Proof.(i) We must show that the rule on (H×K)× (H×K) −→ H×K defined by((h, k), (h′, k′)) −→ (hh′, kk′) is a mapping. Indeed, if ((h1, k1), (h′1, k

′1)) =

((h2, k2), (h′2, k′2)) then (h1, k1) = (h2, k2) and (h′1, k

′1) = (h′2, k

′2). Thus, h1 =

h2, k1 = k2, h′1 = h′2, and k′1 = k′2. Thus, h1h

′1 = h2h

′2 and k1k

′1 = k2k

′2.

Hence, (h1h′1, k1k

′1) = (h2h

′2, k2k

′2). So multiplication is a binary operation.

(ii) Multiplication on H ×K is associative since multiplication is associativeas operation on H and K.

(h1, k1)[(h2, k2)(h3, k3)] = (h1, k1)(h2h3, k2k3)= (h1(h2h3), k1(k2k3))= ((h1h2)h3, (k1k2)k3)= (h1h2, k1k2)(h3, k3)= [(h1, k1)(h2, k2)](h3, k3)

(iii) Let (h, k) ∈ H × K. Since heH = eHh = h and keK = eKk = k then(heH , keK) = (eHh, eKk) = (h, k). Hence, (h, k)(eH , eK) = (eH , eK)(h, k) =

150

(h, k). That is, (eH , eK) is the identity element of H ×K.(iv) Let (h, k) ∈ H × K. Since hh−1 = h−1h = eH and kk−1 = k−1k = eKthen (hh−1, kk−1) = (eH , eK). That is, (h, k)(h−1, k−1) = (eH , eK). Similarly,(h−1, k−1)(h, k) = (eH , eK) so that (h, k) is invertible with inverse (h−1, k−1)

Definition 15.2 By the above theorem, H×K is a group, called the directproduct of the groups H and K.

Example 15.4If Z3 = {[0], [1], [2]} and S2 = {(1), (12)} then

Z3 × S2 = {([0], (1)), ([1], (1)), ([2], (1)), ([0], (12)), ([1], (12)), ([2], (12))}

For example,

([1], (12))([2], (1)) = ([1]⊕ [2], (12)(1)) = ([0], (12)).

Some of the subgroups of H ×K can be constructed as follows.

Theorem 15.7The sets

H × {eK} = {(h, eK) : h ∈ H}and

{eH} ×K = {(eH , k)) : k ∈ K}are subgroups of H ×K.

Proof.We prove that H × {eK} is a subgroup of H × K and we leave the secondpart of the theorem as an exercise for the reader. See Exercise ??Since (eH , eK) ∈ H × {eK} then H × {eK} 6= ∅. Let (h1, eK) and (h2, eK) betwo elements of H × {eK}. Then

(h1, eK)(h2, eK)−1 = (h1, eK)(h−12 , eK) = (h1h

−12 , eK) ∈ H × {eK}

since h1h−12 ∈ H. Hence, by Theorem 7.5, H ×{eK} is a subgroup of H ×K

Remark 15.1By the Principle of Counting, |H ×K| = |H| · |K|.

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Review Problems

Exercise 15.1Show that < 24,−36, 54 > + < 6 > .

Exercise 15.2Show that < (123), (456) >=< (123)(456), (465) > in S6.

Exercise 15.3Determine the elements of < µ2, µ5 > as a subgroup of the group of symme-tries of the square (Example 8.1).

Exercise 15.4The subgroup < (1432), (24) > of S4 has order 8. List its elements.

Exercise 15.5What is < ∅ >?

Exercise 15.6Find a necessary and sufficient condition on a susbet S of a group G forS =< S > .

Exercise 15.7Prove that if a, b ∈ Z then < a, b >=< d >, where d = (a, b).

Exercise 15.8Prove that < [a] >= Zn if and only if a and n are relatively prime.

Exercise 15.9Prove that if (a, n) = d then < [a] >=< [d] > in Zn.

Exercise 15.10Prove that < [a] >=< [b] > in Zn if and only if (a, n) = (b, n).

Exercise 15.11Prove that H ×K is Abelian if and only if H and K are Abelian.

Exercise 15.12Prove that if H is a group, then the set D = {(h, h) : h ∈ H} is a sub-group of H ×H. This is called the diagonal subgroup of H ×H. What is it,geometrically, for H = R, with addition as the binary operation?

152

Exercise 15.13Prove that if A is a subgroup of H and B is a subgroup of K then A× B isa subgroup of H ×K.

Exercise 15.14Simplify the following expression in Z4 × S4.

([2], (123))−1([1], (24))([2], (123)).

Exercise 15.15 ??Let H and K be groups. Prove that {eH} ×K is a subgroup of H ×K.

Exercise 15.16What is the order of S3 × Z2.

Exercise 15.17Construct a Cayley table for Z2 × Z7. Show that the group is cyclic.

Exercise 15.18Find two cyclic groups H and K such that H ×K is not cyclic.

Exercise 15.19(a) List the elements of S3 × Z2.(b) List the elements of the cyclic subgroup < ((12), [1]) > of S3 × Z2.

Exercise 15.20Let H and K be two groups. Show that the projection maps π1 : H×K → Hdefined by π1(h, k) = h and π2 : H × K → K, defined by π2(h, k) = ksatisfy the relation πi((h1, k1)(h2, k2)) = πi(h1, k1)πi(h2, k2) where i = 1, 2and (h1, k1), (h2, k2) ∈ H ×K.

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16 Cosets

We recall from Section 11, the equivalence relation congruence modulo n onthe set of integers. So if a ≡ b(mod n) then n|(a − b) which means thata − b = nq for some q ∈ Z. Since < n >= {nt : t ∈ Z} then it follows thata− b ∈< n >, where < n > is a subgroup of Z. Thus, we have

a ≡ b(mod n)⇐⇒ a− b ∈< n > .

Now −b is the inverse of b in the additive group Z; so by replacing Z bya group G and < n > by a subgroup H of G the above relation suggestsconsidering the relation defined by

a ∼ b⇐⇒ ab−1 ∈ H

Theorem 16.1If H is a subgroup of a group G, the relation ∼ defined above is an equivalencerelation on G. The equivalence class with representative a is the set

[a] = {ha : h ∈ H}.

Proof.Given a ∈ G we have aa−1 = e ∈ H, so that a ∼ a; hence ∼ is reflexive.If a, b ∈ G with a ∼ b, then ab−1 ∈ H; but ba−1 = (ab−1)−1 ∈ H; thusb ∼ a and ∼ is symmetric. Finally, if a, b, c ∈ G with a ∼ b and b ∼ c, thenab−1, bc−1 ∈ H; thus ab−1bc−1 ∈ H, i.e., ac−1 ∈ H, and so a ∼ c; thus, ∼ istransitive. This shows that ∼ is an equivalence relation on G.Now, let a ∈ G. If b ∈ [a] then b ∼ a so that ba−1 ∈ H. If we let h = ba−1 thenb = ha and therefore b ∈ {ha : h ∈ H}. This proves that [a] ⊆ {ha : h ∈ H}.On the other hand, if b ∈ {ha : h ∈ H} then b = ha for some h ∈ H orh = ba−1 so that b ∼ a. Thus, b ∈ [a] and {ha : h ∈ H} ⊆ [a]. This ends aproof of the theorem.

Definition 16.1The set {ha : h ∈ H} is called the right coset of H to which a belongs andwill be denoted by Ha.

Remark 16.1It follows from Theorem 16.1 and Definition 16.1 that the family {Ha}a∈Gform a partition of G.

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Example 16.1Consider the setA4 of all even permutations of S4. The setH = {(1), (123), (132)}is a subgroup of A4. The right cosets of H in A4 are:

H = {(1), (123), (132)}H(12)(34) = {(12)(34), (134), (234)}H(13)(24) = {(13)(24), (243), (124)}H(14)(23) = {(14)(23), (142), (143)}

Lemma 16.1Let H be a subgroup of a group G and a, b ∈ G. Then the following statementsare all equivalent.

(i) ab−1 ∈ H.(ii) a = hb for some h ∈ H.(iii) a ∈ Hb.(iv) Ha = Hb.

Proof.(i) =⇒ (ii): If ab−1 ∈ H then h = ab−1 for some h ∈ H and therefore a = hb.(ii) =⇒ (iii): If a = hb for some h ∈ H then by the definition of Hb we havea ∈ Hb.(iii) =⇒ (iv): If a ∈ Hb then a ∼ b so that the equivalence class of a is equalto the equivalence class of b.(See Theorem 9.2). That is, Ha = Hb.(iv) =⇒ (i): If Ha = Hb then since a = ea ∈ Ha then a ∈ Hb. That is,a = hb for some h ∈ H and therefore ab−1 = h ∈ H. This completes a proofof the lemma.

Remark 16.2The above lemma provides a way in computing all the right cosets when Gis finite. To compute all the right cosets of a subgroup H in a finite groupG, first write H, and then choose any element a ∈ G such that a 6∈ H, andcompute Ha. Next, choose any element b ∈ G such that b 6∈ H ∪ Ha, andcompute Hb. Continue in this way until all the elements of G have beenexhausted.

Looking back at Example 16.1, you notice that H and all the right cosets ofH in G have the same number of elements. This is not accidental.

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Lemma 16.2If H is a finite subgroup of a group G then for any a ∈ G, |H| = |Ha|.

Proof.To show that two finite sets have the same number of elements is equivalentto establishing a bijection mapping between the two sets. So define α :H −→ Ha by α(h) = ha. This is a well-defined mapping. For if h1 = h2

then h1a = h2a or α(h1) = α(h2). To see that α is one-to-one, suppose thatα(h1) = α(h2). Then h1a = h2a. By the right cancellation law, we haveh1 = h2. Thus, α is one-to-one. Now to show that α is onto, let b ∈ Ha.Then by Lemma 16.1, Hb = Ha so that a ∈ Hb and thus a = hb for someh ∈ H. Hence, h−1 = ba−1 ∈ H and α(ba−1) = b. This ends a proof of thelemma.

Remark 16.3Left cosets result from considering the equivalence relation

a ∼ b⇐⇒ a−1b ∈ H.

A left coset has the form

aH = {ah : h ∈ H}.

One can easily proof versions of Theorem 16.1 and Lemma 16.1 for left cosets.

Example 16.2The left cosets in Example 16.1 are:

H = {(1), (123), (132)}(12)(34)H = {(12)(34), (243), (143)}(13)(24)H = {(13)(24), (142), (234)}(14)(23)H = {(14)(23), (134), (124)}

Remark 16.4Note that, other than the subgroup H, no left coset is a right coset.

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Review Problems

Exercise 16.1Determine the right cosets of < [4] > in Z8.

Exercise 16.2Determine the left cosets of < [3] > in Z12.

Exercise 16.3In the group of symmetries of the square, determine the right cosets of <µ7 > .

Exercise 16.4Determine the left cosets of < (123) > in S3.

Exercise 16.5Prove that if H is a subgroup of an Abelian group G then left cosets are equalto right cosets.

Exercise 16.6Let G = S3 and H =< (13) > .

(a) Determine the right cosets of H in G.(b) Determine the left cosets of H in G.(c) Verify that the collection of right cosets is different from the collection ofleft cosets.

Exercise 16.7Prove that each right coset of A×{e} in A×B contains precisely one elementfrom {e} ×B.

Exercise 16.8Compute the right cosets of < (12), [1] > in S3 × Z2.

Exercise 16.9Compute the left cosets of < (12), [1] > in S3 × Z2.

Exercise 16.10Let H and K be two subgroups of a group G. For a, b ∈ G define the relation

a ∼ b⇐⇒ hak = b for some h ∈ H and k ∈ K.

Prove that ∼ is an equivalence relation on G.

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Exercise 16.11Consider the equivalence relation of the previous exercise. Let a ∈ G. Provethat

[a] = HaK = {hak : h ∈ H and k ∈ K}.We call HaK a double coset of H and K in G.

Exercise 16.12Prove that if Hak and Hbk are double cosets of H and K in G then they areeither equal or disjoint.

Exercise 16.13Let H and K be subgroups of a group G and a ∈ G. Let φ : HaK → a−1HaKbe defined by φ(hak) = a−1hak. Show that φ is one-to-one and onto.

Exercise 16.14Let H be a subgroup of a group G. Prove that Hh = H = hH for all h ∈ H.

Exercise 16.15If H is a subgroup of a group G, prove that gHg−1 is a subgroup of G forany g ∈ G.

Exercise 16.16For an arbitrary subgroup H of a group G, define the mapping φ from theset of all left cosets of H in H to the set of all right cosets of H in G byφ(aH) = Ha−1. Prove that φ is one-to-one and onto.

Exercise 16.17Let H be a subgroup of a group H with the property that ghg−1 ∈ H for allg ∈ G and h ∈ H. Prove that Ha = aH for any a ∈ G.

Exercise 16.18Suppose that H is a subgroup of a group G such that gH = Hg for all g ∈ G.Prove that ghg−1 ∈ H for all g ∈ G and h ∈ H.

Exercise 16.19The center of a group G is defined by

Z(G) = {a ∈ G : ax = xa,∀x ∈ G}.

Prove that aZ(G) = Z(G)a for all a ∈ G.

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Exercise 16.20For an arbitrary subgroup H of a group G, the normalizer of H in G is theset

N (H) = {x ∈ G : xHx−1 = H}

(a) Prove that N (H) is a subgroup of G.(b) Prove that gN (H) = N (H)g for all g ∈ G.

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17 Lagrange’s Theorem

A very important corollary to the fact that the left cosets of a subgrouppartition a group is Lagrange’s Theorem. This theorem gives a relation-ship between the order of a finite group and the order of any subgroup (inparticular, if |G| <∞ and H ⊆ G is a subgroup, then |H| | |G|).

Theorem 17.1 (Lagrange’s Theorem)Let G be a finite group, and let H be a subgroup of G. Then the order of Hdivides the order of G.

Proof.By Theorem 16.1, the right cosets of H form a partition of G. Thus, eachelement of G belongs to at least one right coset of H in G, and no elementcan belong to two distinct right cosets of H in G. Therefore every element ofG belongs to exactly one right coset of H. Moreover, each right coset of Hcontains |H| elements (Lemma 16.2). Therefore, |G| = n|H|, where n is thenumber of right cosets of H in G. Hence, |H| | |G|. This ends a proof of thetheorem.

Example 17.1If |G| = 14 then the only possible orders for a subgroup are 1, 2, 7, and 14.

Definition 17.1The number of different right cosets of H in G is called the index of H inG and is denoted by [G : H].

It follows from the above definition and the proof of the above theorem that

|G| = [G : H]|H|.

Example 17.2Since |S3| = 3! = 6 and |(12)| = | < (12) > | = 2 then [S3, < (12) >] = 6

2=

3.

The rest of this section is devoted to consequences of Lagrange’s theorem;we begin with the order of an element.

Corollary 17.1If G is a finite group and a ∈ G then o(a) | |G|.

160

Proof.Since < a > is a subgroup of G then | < a > | | |G|. By Theorem 15.7,o(a) = | < a > |. Hence, o(a) | |G|.

The next two results classify all groups of order n for infinitely many valuesof n.

Corollary 17.2If |G| = p, where p is prime then the only subgroups of G are {e} and G.

Proof.Suppose the contrary, that is G has a subgroup H such that H 6= {e} andH 6= G. By Theorem 17.1, |H|||G| with 1 < |H| < p. This contradicts thefact that p is prime.

Corollary 17.3If G is a goup of prime order then it is cyclic. That is, G =< a > where ais any nonidentity element of G.

Proof.Let a ∈ G with a 6= e. Then < a > 6= {e}. By the previous corollary, G =<a > .

Example 17.3The previous corollary tells that groups of prime order are always cyclic.What about groups of prime-squared order? The group

Z2 × Z2 = {([0], [0]), ([0], [1]), ([1], [0]), ([1], [1])}

has order 4 = 22. Since each element has order 2 then by Theorem 15.7,Z2 × Z2 is not cyclic.

Example 17.4Lagrange’s Theorem greatly simplifies the problem of determining all thesubgroups of a finite group. For example, consider the group (Z6,⊕). Asidefrom {[0]} and Z6 any subgroup of Z6 must have order 2 or 3. There is onlyone subgroup of order 2, < [3] > . Also, there is only one subgroup of order3, < [2] > . A subgroup lattice shows the subgroups of Z6 and the inclusionrelation between them.

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Another important corollary to Lagrange’s Theorem is Fermat’s little theo-rem.

Theorem 17.2 (Fermat’s Little Theorem)If p is a prime number and p 6 |n then

np−1 ≡ 1(mod p).

Proof.Since p is prime then (Z∗p,�) is a group of order p − 1.(See Theorem 12.6).Hence, the order of each element of Z∗p is a divisor of p − 1. If p 6 |n then[n] ∈ Z∗p. Since o([n]) = | < [n] > | | (p − 1) then [n]p−1 = [1]. That is,[np−1] = [1]. Hence np−1 ≡ 1(mod p).

The above theorem suggests a test of primality for p. Take a number n suchthat p 6 |n and raise it to the (p − 1)st power and find its remainder whendivided by p. If the remainder is not 1 then we can conclude that p is not aprime number.

The Converse of Lagrange’s TheoremThe converse of Lagrange’s theorem is not true in general. That is, if n isa divisor of G then it does not necessarily follow that G has a subgroup oforder n.

Example 17.5The set of all even permutations

A4 = {(1), (12)(34), (13)(24), (14)(23), (123), (132), (124),(142), (134), (143), (234), (243)}

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is a subgroup of S4 (and therefore a group itself) of order 12 (See Theorem7.9). Note thatA4 has three elements of order 2 (namely, (12)(34), (13)(24), (14)(23))and 8 elements of order 3. ((123), (132), (124), (142), (134), (143), (234), (243))We will show that A4 has no subgroup of order 6.Let H be a subgroup of A4 of order 6. Then (1) ∈ H. Since A4 containsonly 3 elements of order 2 then H must contain at least one element oforder 3 of the form (abc). Then by closure, (acb) = (abc)(abc) ∈ H. If Halso contains an element of the form (ab)(cd)( or of the form (abd)) thenby closure (abc)(ab)(cd) = (ac)(bd) ∈ H and (acb)(ab)(cd) = (bcd). Thus,(bdc) = (bcd)−1 ∈ H. In either case, H has more than six elements. Thus,A4 has no subgroup of order 6.

The converse of Lagrange’s theorem is valid for cyclic groups. To prove thisresult we need the following two theorems.

Theorem 17.3Let G be a finite cyclic group of order n and generator a. That is,

G = {e, a, a2, · · · , an−1}

Every subgroup of G is cyclic. That is, a subgroup of a cyclic group is alsocyclic.

Proof.Let H be a subgroup of G. Then elements of H are of the form ak with1 ≤ k < n. Let t be the smallest positive integer such that at ∈ H. We shallprove that H =< at > . Indeed, let am ∈ H. By the Division Algorithmthere exist unique integers q and r such that m = tq + r where 0 ≤ r < t. Itfollows that am = (at)qar or ar = am(at)−q. But am ∈ H and at ∈ H then byclosure ar ∈ H. Since t is the smallest positive integer such that at ∈ H thenwe must have r = 0. Hence, am = (at)q or am ∈< at > . Clearly, < at >⊆ Hsince at ∈ H and H is a group.

Theorem 17.4Let G be as in the statement of Theorem 17.3. If 1 ≤ k < n then ak generatesa subgroup of order n

(k,n).

Proof.Let d = (k, n). By Theorem 14.6(i), | < ak > | is the smallest positive integer

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such that ak|<ak>| = e. By Theorem 14.6 (ii), n| (k| < ak > |). That is,

k| < ak > | = nq for some integer q. Hence, k|<ak>|d

= nqd

so that nd| k|<a

k>|d

.Since (n

d, kd) = 1 then by Lemma 13.1, we have n

d| | < ak > |. On the other

hand, (ak)nd = (an)

kd = e so that | < ak > | | n

d. Hence, by Theorem 10.2 (d),

| < ak > | = nd.

Theorem 17.5Let G be a cyclic group of order n and generator a. For each positive divisord of n, G has exactly one subgroup of order d.

Proof.Existence: Let d be a positive divisor of n. Then there exists a positiveinteger q < n such that n = dq. Thus, q|n and (n, q) = q. By Theorem 17.4,aq generates a subgroup of G of order n

(n,q)= n

q= d. Thus, G has at least

one subgroup of order d.

Uniqueness: Suppose that G has two subgroups of order d, say H andK. We will show that H = K. Let 1 ≤ m < n be the smallest positive inte-ger such that am ∈ H and 1 ≤ k < n be the smallest positive integer suchthat ak ∈ K. As in the proof of Theorem 17.3, we establish that H =< am >and K =< ak > . By Theorem ?? | < am > | = n

(n,m)and | < ak > | = n

(n,k).

Thus, n(n,m)

= n(n,k)

= d or (n,m) = (n, k). Now, by the Division Algorithm,

n = mq + r with 0 ≤ r < m. Since an = e ∈ H then ar = (am)−q ∈ H. Fromthe definition of m we see that r = 0. Hence, n = mq and m|n. It follows that(n,m) = m. A similar argument shows that (n, k) = k and therefore m = k.Hence, < am >=< ak >, i.e. H = K. This ends a proof of the theorem.

Remark 17.1The converse of Lagrange’s theorem holds also for finite Abelian groups. Thistopic will not be covered in this book.

Example 17.6Consider the group (Z12,⊕). Since |Z12| = 12 then the positive divisors of12 are 1, 2, 3, 4, 6, and 12. The subgroup lattice below shows the differentsubgroups of Z12 =< [1] > .

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165

Review Problems

Exercise 17.1Find [S3 :< (12) >].

Exercise 17.2Find [Z10 :< [2] >].

Exercise 17.3Let G denote the group of symmetries of the square.(Example 8.1) Find [G :<µ8 >].

Exercise 17.4A group G has subgroups of orders 4 and 10, and |G| < 50. What can youconclude about |G|?

Exercise 17.5Assume that G is a group with a subgroup H such that |H| = 6, [G : H] > 4,and |G| < 50. What are the possibilities about |G|?

Exercise 17.6Assume that G is a group with a subgroup H such that |G| < 45, |H| > 10,and [G : H] > 3. Find |G|, |H|, and [G : H].

Exercise 17.7Find all of the subgroups of Z6. Also construct the subgroup lattice.

Exercise 17.8Assume that G is a cyclic group of order n, that G =< a >, that k|n, andH =< ak > . Find [G : H].

Exercise 17.9Assume that A is a subgroup of a finite group G and B is a subgroup of afinite group H. We have seen that A × B is a subgroup of G × H. Express[G×H : A×B] in terms of [G : A] and [H : B].

Exercise 17.10Assume that G is a finite group and D denotes the diagonal subgroup ofG×G.. Find [G×G : D].

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Exercise 17.11Let G be a group of order p2. Prove that if G is not cyclic then ap = e forevery nonidentity element of G.

Exercise 17.12Prove that if H is a subgroup of a group G, a, b ∈ G and a, b 6∈ H thenab ∈ H.

Exercise 17.13Prove that if A and B are finite subgroups of a group G such that (|A|, |B|) =1 then A ∩B = {e}.

Exercise 17.14If H is a subgroup of G and [G : H] = 2 then G is Abelian.

Exercise 17.15Prove that if H is a subgroup of a finite group G then the number of rightcosets of H in G is equal to the number of left cosets of H in G.

Exercise 17.16Prove that the number of generators of a finite cyclic group of order n isφ(n), where φ is Euler’s function.

Exercise 17.17Let G be a group. Define the relation

a ∼ b⇐⇒ a = gbg−1, for some g ∈ G.

(a) Prove that ∼ is an equivalence relation.(b) Prove that [a] = {b ∈ G : a = gbg−1 for some g ∈ G}.(c) Prove that |[a]| = 1 if and only if a ∈ Z(G), where Z(G) is the center ofG, ı.e. Z(G) = {a ∈ G : ax = xa∀x ∈ G}.(d) Prove that G = Z(G) ∪ [a] where the union is taken over all a ∈ G suchthat |[a]| ≥ 2.

Exercise 17.18Let G be a group, and let a, b ∈ G be elements such that ab = ba. If the ordersof a and b are relatively prime, then o(ab) = o(a)o(b).

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Exercise 17.19Let G be a group, and let a, b ∈ G be elements such that ab = ba. Prove if(o(a), o(b)) = 1 then o(ab) is the least common multiple of o(a) and o(b).

Exercise 17.20Prove that in a finite group G or order n, we have an = e for all a ∈ G.

Exercise 17.21Show that S4 has no element of order 6.

Exercise 17.22Let G be a finite group, let H and K be subgroups of G such that K is alsoa subgroup of H. Find [G : K] in terms of [G : H] and [H : K].

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18 Group Isomorphisms

We start this section by constructing the Cayley tables of the groups (Z3,⊕)and (< (123) >, ◦).

⊕ [0] [1] [2][0] [0] [1] [2][1] [1] [2] [0][2] [2] [0] [1]

◦ (1) (123) (132)(1) (1) (123) (132)(123) (123) (132) (1)(132) (132) (1) (123)

By setting the correspondence [0] ↔ (1), [1] ↔ (123), [2] ↔ (132) we seethat the two tables differ only in the names of the symbols, and not in theirpositions. If we name the mentioned correspondence by the letter f , i.e. bywriting f([0]) = (1), f([1]) = (123) and f([2]) = (132), and then constructinga Venn diagram of this mapping we see that f is a bijection mapping withthe property f([a]⊕ [b]) = f([a])◦f([b]). This example, leads to the followingdefinition.

Definition 18.1Let (G, ∗) and (H,#) be two groups. We say that G and H are isomorphicif there is a bijective map θ : G −→ H which respects the group structure.That is to say, for every g and h in G,

θ(g ∗ h) = θ(g)#θ(h) (3)

The map θ is called an isomorphism. We write G ≈ H. If G = H then wecall θ an automorphism on G.

In words, you can first multiply in G and take the image in H, or you cantake the images in H first and multiply there, and you will get the sameanswer either way. This is referred to as ”operation preserving”.Notice that the operation on the left is occurring in G while the operationon the right is occurring in H.

Definition 18.2A function θ satisfying (3) is called a homomorphism. If G = H then wesay that θ is an endomorphism on G. A one-to-one homomorphism is calleda monomorphism. An onto homomorphism is called an epimorphism.

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Example 18.1Consider the mapping f : R+ −→ R defined by f(x) = log10 x. We knowfrom calculus that this mapping is a bijective map with the property

log10 (xy) = log10 x+ log10 y.

Thus, f is an isomorphism.

Remark 18.1The use of ∗ and # in Definition 18.1 is intended to emphasize the fact thatthe group operations may be different. Now that this point has been made,we revert to our convention of using multiplicative notation. Thus, we willwrite

θ(ab) = θ(a)θ(b).

Isomorphism is a very important idea in abstract mathematics. It allows usto establish properties of one basic structure, and then immediately deducethem for a whole range of isomorphic ’look-alikes’.

Theorem 18.1If G ≈ H and G is Abelian then H is also Abelian.

Proof.Let θ : G −→ H be an isomorphism. Let a, b ∈ H. Since θ is onto thenθ(x) = a and θ(y) = b for some unique x, y ∈ G. Thus,

ab = θ(x)θ(y) = θ(xy) = θ(yx) = θ(y)θ(x) = ba.

Hence, H is Abelian.

More properties shared by isomorphic groups will be discussed in Section19.

The next theorem list some technical facts about homomorphisms.

Theorem 18.2 (Basic properties of homomorphisms)Let θ : G −→ H be a homomorphism between two groups G and H. Then

(i) θ(eG) = eH .(ii) θ(a−1) = [θ(a)]−1.

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(iii) θ(ak) = [θ(a)]k, where k ∈ Z.(iv) The set θ(G) = {θ(a) : a ∈ G} is a subgroup of H.(v) If θ is one-to-one then G ≈ θ(G).(vi) If a ∈ G then o(θ(a)) | o(a).

Proof.(i)Since θ(eG)θ(eG) = θ(eG)eH then by the left cancellation rule we haveθ(eG) = eH .(ii) Let a ∈ G. The equation θ(a)x = eH has a unique solution x = [θ(a)]−1.But θ(a)θ(a−1) = θ(aa−1) = θ(eG) = eH . Thus, [θ(a)]−1 = θ(a−1).(iii) We consider the cases, k = 0, k > 0, and k < 0. The case k = 0 is just(i). For the case k > 0, we use induction on k ≥ 1. The result is true fork = 1. Assume it is true up to k − 1. Then

θ(ak) = θ(ak−1a) = θ(ak−1)θ(a) = (θ(a))k−1θ(a) = [θ(a)]k.

Hence, the result is true for all k > 0. If k < 0 then

θ(ak) = θ((a−1)−k) = [θ(a−1)]−k = [(θ(a))−1]−k = [θ(a)]k.

(iv) Since θ(eG) = eH then eH ∈ θ(G) so that θ(G) 6= ∅. Now, if θ(a), θ(b) ∈θ(G) then

θ(a)[θ(b)]−1 = θ(a)θ(b−1) = θ(ab−1) ∈ θ(G)

since by closure, ab−1 ∈ G. Hence, by Theorem 7.5, θ(G) is a subgroup of H.(v) By the definition of θ(G), θ : G→ θ(G) is onto. Hence, if θ is one-to-onethen θ is a bijective homomorphism, i.e. G ≈ θ(G).(vi) Since ao(a) = eG then θ(ao(a)) = θ(eG) or [θ(a)]o(a) = eH . By Theorem14.6(ii), o(θ(a)) | o(a).

Remark 18.2If you want to show that a function is not a homomorphism, do a quick check:Does it send the identity to the identity? If not, then the above theoremshows that it’s not a homomorphism. For example, if f : (Z,+) −→ (Z,+)is defined by f(x) = x+ 1 f is not an homomorphism since f(0) = 1 6= 0.

Remark 18.3The properties in the above theorem are not part of the definition of a ho-momorphism. To show that f is a homomorphism, all you need to showis that f(ab) = f(a)f(b). The properties in the theorem are automaticallytrue of any homomorphism. For example, the mapping f : (Z,+) −→ (Z,+)defined by f(x) = x2 satisfies f(0) = 0 but still f is not a homomorphism.

171

Review Problems

Exercise 18.1Let A = {2m : m ∈ Z}. Prove that Z ≈ A.

Exercise 18.2Let B = {2m3n : m,n ∈ Z}. Prove that Z× Z ≈ B.

Exercise 18.3Assume that H = {u, v, w, x, y, z} is a group with respect to multiplicationand that θ : Z6 −→ H is an isomorphism with

θ([0]) = u , θ([1]) = v , θ([2]) = wθ([3]) = x , θ([4]) = y , θ([5]) = z

Replace each of the following by the appropraite letter, either u, v, x, y, or z.

(a) xw (b) w−1 (c) v5 (d) zv−1x.

Exercise 18.4Given two groups (G, ∗) and (H,#). Prove that G×H ≈ H ×G.

Exercise 18.5Prove that (Z4,⊕) ≈ (Z∗5,�).

Exercise 18.6Let θ : Z6 → Z2 × Z3 be defined by θ([a]6) = ([a]2, [a]3).

(a) Show that θ is well-defined.(b) Prove that Z6 ≈ Z2 × Z3.

Exercise 18.7Prove that if (m,n) = 1 then Zmn ≈ Zm × Zn.

Exercise 18.8Consider G = {1, i,−1,−i} under multiplication. Prove that G ≈ Z4, with(Z4,⊕).

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Exercise 18.9Let G be a nonempty group. For each a ∈ G, define fa : G → G byfa(x) = ax.

(a) Prove that fa is well-defined.(b) Prove that fa is a permutation on G.(c) Prove that the set G′ = {fa : a ∈ G} is a group of permutations.

Exercise 18.10Let G and G′ be as in the previous exercise. Define φ : G→ G′ by φ(a) = fa.

(a) Prove that φ is well-defined.(b) Prove that φ is one-to-one and onto.(c) Prove that G ≈ G′.

Exercise 18.11Let G be an arbitrary group. Prove or disprove that the mapping φ(a) = a−1

is an isomorphism from G to G.

Exercise 18.12For each a in a goup G, define a mapping τa : G → G by τa(x) = axa−1.Prove that τa is an isomorphism from G to G.

Exercise 18.13For an arbitrary positive integer n, prove that any two cyclic groups of ordern are isomorphic.

Exercise 18.14Let G1, G2, H1, H2 be nonempty groups, and suppose that θ1 : G1 → H1 andθ2 : G2 → H2 are group isomorphisms. Define φ : G1 × G2 → H1 × H2 byφ(x1, x1) = (θ1(x1), θ2(x2)).

(a) Prove that φ is well-defined.(b) Prove that φ is an isomorphism.

Exercise 18.15Let G be a group, and let S be any set for which there exists a one-to-oneand onto function φ : G → S. Define an operation on S by setting x1 · x2 =φ(phi−1(x1)φ−1(x2)). Prove that S is a group under this operation, and thatφ is actually a group somorphism.

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Exercise 18.16Prove that Z4 is not isomorphic to Z6.

Exercise 18.17Prove that S3 is not isomorphic to Z6.

Exercise 18.18Let φ : G→ H be an isomorphism and N a subgroup of G with the propertygng−1 ∈ N for all g ∈ G and n ∈ N. Prove that hyh−1 ∈ φ(N) for y ∈ φ(N)and h ∈ H.

Exercise 18.19Suppose that G =< a > . Let H be a group and suppose that θ, φ : G → Hare group isomorphisms. Prove that if θ(a) = φ(a) then θ(x) = φ(x) for allx ∈ G.

Exercise 18.20Let H be a subgroup of a group G. For a ∈ G, define

a−1Ha = {a−1ha : h ∈ H}.

(a) Show that a−1Ha is a subgroup of G.(b) Prove that H ≈ a−1Ha.

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19 More Properties of Isomorphims

In this section we discuss more properties of group isomorphisms. We startwith the following lemma.

Lemma 19.1Let θ : G −→ H and γ : H −→ K be two group isomorphisms. Then

(a) θ−1 : H −→ G is an isomorphism.(b) γ ◦ θ : G −→ K is also an isomorphism.

Proof.(a) First we show that θ−1 is one-to-one. If θ−1(h1) = θ−1(h2) then θ(θ−1(h1)) =θ(θ−1(h2)) or ιH(h1) = ιH(h2). Hence, h1 = h2. To see that θ−1 is onto, letg ∈ G. Then θ(g) ∈ H and θ−1(θ(g)) = g. Finally, we show that θ−1 is ahomomorphism. If a, b ∈ H then a = θ(u) and b = θ(v) for some u, v ∈ G,since θ is onto. Hence, if w = θ−1(ab) then θ(w) = ab = θ(u)θ(v) = θ(uv).Thus, θ−1(ab) = uv = θ−1(a)θ−1(b). Hence, θ−1 is an isomorphism.(b) To see that γ ◦ θ is one-to-one, suppose that γ ◦ θ(g1) = γ ◦ θ(g2). Thenγ(θ(g1)) = γ(θ(g2)). Since γ is one-to-one then θ(g1) = θ(g2). Since θ is one-to-one then g1 = g2. To see that γ ◦ θ is onto, let k ∈ K. Since γ is onto thenwe can find h ∈ H such that γ(h) = k. Since θ is onto then we can find ag ∈ G such that θ(g) = h. Hence, γ(θ(g)) = γ(h) = k. To see that γ ◦ θ is ahomomorphism, let a, b ∈ G. Since both θ and γ are homomorphisms then

(γ ◦ θ)(ab) = γ(θ(ab)) = γ(θ(a)θ(b)) = γ(θ(a))γ(θ(b)) = (γ ◦ θ)(a)(γ ◦ θ)(b).

Theorem 19.1Isomorphism is an equivalence relation on the collection of all groups.

Proof.Reflexive: Let G be a group. Then the identity map ιG(a) = a for all a ∈ Gis an isomorphism. Hence, G ≈ G.Symmetric: Suppose that G ≈ H for some groups G and H. Then there isan isomorphism θ : G −→ H. By Lemma 19.1(a), θ−1 : H −→ G is also anisomorphism. Hence, H ≈ G.Transitive: Suppose that G ≈ H and H ≈ K, where G,H, and K aregroups. Then there are isomorphisms θ : G −→ H and γ : H −→ K. ByLemma 19.1(b), γ ◦ θ : G −→ K is also an isomorphism. Hence, G ≈ K.

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The following theorem lists more properties shared by isomorphic groups.Thus, the simplest way to show that two groups are not isomorphic is to finda property not shared by the two groups.

Theorem 19.2Let G and H be groups such that θ : G −→ H is an isomorphism.

(a) |G| = |H|.(b) If G =< a > then H =< θ(a) > . That is, if G is cyclic, then H is alsocyclic.(c) If K is a subgroup of G of order n then θ(K) is a subgroup of H of ordern.(d) If a ∈ G and o(a) = n then o(θ(a)) = n.

Proof.(a) There’s a one-to-one onto map between the two sets. So counting theelements of one set simultaneously counts the elements of the other.(b) Suppose that G is cyclic with generator a. We will show that H =<θ(a) > . Clearly, since θ(a) ∈ H and H is a group then < θ(a) >⊆ H.Now suppose that h ∈ H then there is an g ∈ G such that θ(g) = h(sinceθ is onto). But then g = an for some n ∈ Z. Hence, by Theorem 18.2(iii),θ(g) = θ(an) = (θ(a))n. That is, h ∈< θ(a) > and so H ⊆< θ(a) > .(c) If K is a subgroup of G then by Theorem 18.2(iv), θ(K) is a subgroupof H. The mapping θ restricted to K is a one-to-one mapping from K ontoθ(K). Thus, |K| = |θ(K)|.(d) If o(a) = n then an = eG. By Theorem 18.2 (iii), we have (θ(a))n =θ(an) = θ(eG) = eH . By Theorem 14.6(ii), o(θ(a))|n. On the other hand, since(θ(a))o(θ(a)) = eH then θ(ao(θ(a))) = eH = θ(eG). Thus, ao(θ(a)) = eG (since θ isone-to-one) and by Theorem 14.6(ii), n|o(θ(a)). Hence, by Theorem 10.2(d),o(θ(a)) = n.

Example 19.11. Z2×Z2 and Z4 are not isomorphic. Both groups have 4 elements, however,every element of Z2×Z2 has order 1 or 2, while Z4 has two elements of order4(namely, [1] and [3].)

The following theorem shows that any two finite groups of the same orderare isomorphic.

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Theorem 19.3If G is a cyclic group of order n then G ≈ Zn.

Proof.Suppose that G =< a > . Define θ : G −→ Zn by θ(ak) = [k] where 1 ≤ k <n. We show that θ is well-defined. Indeed, if ak = am then ak−m = eG so thatby Theorem 14..6 (ii), n|(k−m). Thus, a ≡ m(mod n) and by Theorem 9.2,[k] = [m]. Next, we show that θ is one-to-one. Indeed, if θ(ak) = θ(am) then[k] = [m] and this implies that n|(k−m). Thus, k−m = nq for some q ∈ Z.Therefore, ak−m = anq = (an)q = eG. Hence, ak = am. Next, we show that θis onto. Let b ∈ Z. Then by the Division Algorithm, b = nq + r, 0 ≤ r < n.Hence, ab = (an)qar = ar ∈ G and θ(ab) = θ(ar) = [r]. Since b− r = nq thenn|(b− r so that b ≡ r(mod n). By Theorem 9.2, [b] = [r]. Finally, it remainsto show that θ is a group homomorphism. Indeed,

θ(akam) = θ(ak+m) = [k +m] = [k]⊕ [m] = θ(ak)θ(am).

The following result classifies all groups of prime order.

Example 19.2If p is prime then Zp × Zp is not cyclic. For if it is, then by the previoustheorem, Zp×Zp ≈ Zp2 . But every nonidentity element of Zp×Zp has orderp whereas Zp2 has an element of order p2, namely, [p2 − 1].

Corollary 19.1If |G| = p, where p is a prime number, then G ≈ Zp.

Proof.Since G has a prime order then by Corollary 17.3, G is cyclic. By the previoustheorem, G ≈ Zp.

Theorem 19.4If G is an infinite cyclic group then G ≈ Z.

Proof.Suppose that G =< a > with o(a) =∞. Define θ : Z −→< a > by θ(n) = an.Then θ is a well-defined mapping. Indeed, if n = m then an = am. We will

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show next that θ is one-to-one. Suppose that θ(n) = θ(m). Without loss ofgenerality we can assume that n < m. Then

e = a0 = an−n = ana−n = ama−n = am−n.

This contradicts the fact that o(a) = ∞ Since m − n ∈ N. Thus, we musthave n = m. To show that θ is onto, pick an x ∈ G. Then x = an for somen ∈ Z and θ(n) = an = x. It remains to show that θ is a homomorphism:θ(n + m) = an+m = anam = θ(n)θ(m). Thus, Z ≈< a > or < a >≈ Z since≈ is symmetric. This ends a proof of the theorem.

A principal problem in finite group theory is the problem of classifying groupsof finite orders. For example, the problem of determining all isomorphismclasses is settled by the following theorem whose proof is omitted.

Theorem 19.5 (Fundamental Theorem of Finite Abelian Groups)Every finite abelian group is isomorphic to a direct product of cyclic groupsin the form

Cp1r1 × Cp2r2 × · · · × Cptrt ,

where the pi are (not necessarily distinct) primes; the product is unique upto reordering of the factors.

Example 19.3There are six isomorphism classes of Abelian groups of order 360. If G isAbelian with |G| = 360 = 23.32.5, then the possible sets of prime powers areas follows:

{23, 32, 5},{23, 3, 3, 5},{2, 22, 32, 5},{2, 22, 3, 3, 5},{2, 2, 2, 32, 5},{2, 2, 2, 3, 3, 5}.

Hence there are six mutually non-isomorphic abelian groups of order 360:

Z8 × Z9 × Z5,

Z8 × Z3 × Z3 × Z5

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Z2 × Z4 × Z9 × Z5

Z2 × Z4 × Z3 × Z3 × Z5

Z2 × Z2 × Z2 × Z2 × Z9 × Z5

Z2 × Z2 × Z2 × Z2 × Z3 × Z3 × Z5

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Review Problems

Exercise 19.1Give a reason why each of the following two groups are not isomorphic.

(a) Z5,Z6.(b) Z4 × Z2, symmetry group of the square.(c) Z,Q (with both +.(d) Z8 × Z4,Z16× Z2.

Exercise 19.2Is there a noncyclic group of order 59?Why?

Exercise 19.3Is there a noncyclic group of order 39?Why?

Exercise 19.4Give an example of a noncyclic group of order 49.

Exercise 19.5Find two groups of order 9 that are not isomorphic.

Exercise 19.6If p is a prime, then there are five isomorphism classes of Abelian group oforder p4. Give one group from each class.

Exercise 19.7An isomorphism of a group G onto itself is called an automorphism. Provethat the set of all automorphisms on G,Aut(G), is a group with respect tocomposition.

Exercise 19.8Prove that if G is Abelian then the mapping θ : G→ G given by θ(a) = a−1

is an automorphism on G.

Exercise 19.9Prove that if [a] is a generator of Zn, and θ : Zn → Zn is defined by θ([k]) =[ka], then θ ∈ Aut(Zn).

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Exercise 19.10Prove that Aut(Z) ≈ Z2.

Exercise 19.11Prove that for p prime, Aut(Zp) ≈ Zp−1.

Exercise 19.12Consider R2 under addition. Show that the mapping θ : R2 → R2 defined byθ(a, b) = (b, a) is an automorphism.

Exercise 19.13Prove that any two groups of order 3 are isomorphic.

Exercise 19.14Prove that Aut(Zn) ≈ Z∗n.

Exercise 19.15Let G be agroup and for a ∈ G define θa : G → G by θa(x) = axa−1. Provethat θa ∈ Aut(G).

Exercise 19.16Let G be a group and define Inn(G) = {thetaa : a ∈ G}, where θa is themapping defined in the previous exercise. Prove that Inn(G) is a subgroupof Aut(G).

Exercise 19.17Let θ : G → H be an isomorphism. Let a ∈ G and k be a positive integer.Show that the equation xk = a has the same number of solutions in G as doesthe equation xk = θ(a) in H.

Exercise 19.18(a) Solve the equation x4 = 1 in R∗.(b) Solve the equation x4 = 1 in C∗, where C∗ denote the group of all nonzerocomplex numbers.(c) Is R∗ isomorphic to C∗?

Exercise 19.19Let G be a finite group and θ ∈ Aut(G) such that θ2 = ιG and θ(x) 6= eG forall x 6= eG. Prove that G is Abelian.

Exercise 19.20Suppose that G and H are isomorphic groups. Prove that if every element ofG is its own inverse then every element of H is its own inverse.

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20 Cayley’s Theorem

We have already met (i.e. Section 6) the symmetric group Sym(S), the groupof all permutations on a set S. It was one of our first examples of a group. Infact it is a very important group, partly because of Cayley’s theorem whichwe discuss in this section.Cayley’s theorem represents a group as a subgroup of a permutation group(upto an isomorphism). This is often advantageous, because permutation groupsare fairly concrete objects. For example, it’s straightfroward to write pro-grams to do arithmetic in finite permutation groups.

Theorem 20.1 (Cayley’s Theorem)Any group G is isomorphic to a subgroup of Sym(G).

Proof.Given g ∈ G, we define a map λg : G → G by λg(x) = gx for all x ∈ G.This is a well-defined mapping. Indeed, if x = y then gx = gy so thatλg(x) = λg(y). Next, we show that λg is one-to-one. To see this, supposethat λg(x) = λg(y). Then gx = gy and by the left-cancellation propertyx = y. To see that λg is onto, let y ∈ G. Then g−1y ∈ G and λg(g

−1y) = y.Hence, λg ∈ Sym(G).Next, We define Λ : G → Sym(G) by Λ(g) = λg. This is a well-definedmapping. For if g1 = g2 then g1x = g2x for all x ∈ G, that is, λg1(x) = λg2(x)for all x ∈ G and hence λg1 = λg2 , i.e. Λ(g1) = Λ(g2).Now, given g1, g2 ∈ G we have

λg1g2(x) = (g1g2)x = g1(g2x) = λg1(g2x) = λg1λg2(x) for all x ∈ G.

Thus, Λ(g1g2) = λg1g2 = λg1λg2 = Λ(g1)Λ(g2), and so Λ is a homomorphism.Finally, we show that Λ is one-to-one. Indeed, if Λ(g1) = Λ(g2) then λg1(x) =λg2(x) for all x ∈ G. In particular, λg1(e) = λg2(e). That is, g1e = g2e org1 = g2. By Theorem 18.2(v), G ≈ Λ(G).

Corollary 20.1Every finite group G of order n is isomorphic to a subgroup of Sn.

Proof.Write G in set-builder notation as G = {a1, a2, · · · , an}. The proof of The-

orem 20.1 assigns to the element ai the map λai =

(a1 · · · anaia1 · · · aian

).

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Now, aia1, · · · , aian is just the ith row of the Cayley table for G, and thusis simply a rearrangement of a1, · · · , an, say aθi(1), · · · , aθi(n) where θi ∈ Sn;

so λai =

(a1 · · · anaθi(1) · · · aθi(n)

). If we now replace ai by i, we map λai to(

1 · · · nθi(1) · · · θi(n)

)= θi. By Theorem 20.1, {θ1, θ2, · · · , θn} ≈ Λ(G) ≈ G.

That is, G is isomorphic to a subgroup of Sn.

Example 20.1Consider the following Cayley table of a group G = {e, a, b, c}.

V e a b ce e a b ca a e c bb b c e ac c b a e

We then have

λe =

(e a b ce a b c

), θ1 =

(1 2 3 41 2 3 4

)= (1)

λa =

(e a b ca e c b

), θ2 =

(1 2 3 42 1 4 3

)= (12)(34)

λb =

(e a b cb c e a

), θ3 =

(1 2 3 43 4 1 2

)= (13)(24)

λc =

(e a b cc b a e

), θ4 =

(1 2 3 44 3 2 1

)= (14)(23)

Hence G is isomorphic to the subgroup

{(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} = 〈(1 2)(3 4), (1 3)(2 4)〉

of S4.

183

Review Problems

Exercise 20.1Write the permutation associated with each element of Z5 by the isomorphismΛ in the proof of Cayley’s Theorem.

Exercise 20.2Write the permutation associated with each element of the symmetry groupof the square by the isomorphism Λ in the proof of Cayley’s Theorem.

Exercise 20.3Write the permutation associated with each element of group G = {1, i,−1,−i}(under multiplication) of the by the isomorphism Λ in the proof of Cayley’sTheorem.

Exercise 20.4Write the permutation associated with each element of a cyclic group G =<a > of order 4 by the isomorphism Λ in the proof of Cayley’s Theorem.

Exercise 20.5For each a in a group g, define δa : G→ G by δa(x) = xa.

(a) Prove that δa is a permutation on G.(b) Prove that H = {δa : a ∈ G} is a group with respect to composition.(c) Define φ : G → H by φ(a) = δa. Determine whether or not φ is alwaysan ismorphism.

Exercise 20.6For each a in a group g, define ρa : G→ G by ρa(x) = xa−1.

(a) Prove that ρa is a permutation on G.(b) Prove that H = {ρa : a ∈ G} is a group with respect to composition.(c) Define φ : G → H by φ(a) = ρa. Determine whether or not φ is alwaysan ismorphism.

Exercise 20.7Assume that G is a group, and that λa : G→ G and ρa : G→ G are definedby λa(x) = ax and ρa(x) = xa−1 for each x ∈ G. For each a ∈ G, defineγa : G→ G by γa = ρa ◦ λa. Prove that γa ∈ Aut(G).

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Exercise 20.8Let γa be defined as in Exercise 20.7, and define φ : G→ Sym(G) by φ(a) =γa for each a ∈ G. Prove that φ(a) = ιG if and only if ax = xa for all x ∈ G.

Exercise 20.9Let θ : G→ G is a one-to-one and onto mapping such that λgθ = θλg for allg ∈ G. Prove that there is an a ∈ G such that θ = ρa.

Exercise 20.10Prove that for each positive integer n, there are finitely many isomorphismclasses of groups of order n.

185

21 Homomorphisms and Normal Subgroups

Recall that an isomorphism is a function θ : G −→ H such that θ is one-to-one, onto and such that θ(ab) = θ(a)θ(b) for all a, b ∈ G. We shall see thatan isomorphism is simply a special type of function called a group homomor-phism. We will also see a relationship between group homomorphisms andnormal subgroups.

Definition 21.1A function θ from a group G to a group H is said to be a homomorphismprovided that for all a, b ∈ G we have that

θ(ab) = θ(a)θ(b).

If θ : G −→ H is a one-to-one homomorphism, we call θ a monomor-phism and if θ : G −→ H is an onto homomorphism, then we call θ anepimorphism. Of course, a bijective homomorphism is an isomorphism.

Example 21.1Define θ : Z −→ Zn by θ(a) = [a]. Then

θ(a+ b) = [a+ b] = [a]⊕ [b] = θ(a)⊕ θ(b),

so that θ is a homomorphism. Note that θ(n) = θ(2n) with n 6= 2n so thatθ is not one-to-one. However, θ is onto.

Example 21.2Define θ : Z −→ Z by θ(a) = 2a. Then

θ(a+ b) = 2(a+ b) = 2a+ 2b = θ(a) + θ(b),

so that θ is a homomorphism. Note that θ is not onto since there is no integern that satisfies θ(n) = 3. However, θ is one-to-one since θ(n) = θ(m) implies2n = 2m and this in turn implies that n = m.

We have seen that the range of a homomorphism is a subgroup of thecodomain. (Theorem 18.2(iv)). The following subset determines a subgroupof the domain of a homomorphism.

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Definition 21.2Let θ : G −→ H be a group homomorphism. Then the kernel of θ is the set

Ker θ = {g ∈ G : θ(g) = eH}.

Example 21.3In Example 21.1, a ∈ Ker θ iff [a] = θ(a) = [0], i.e iff a = nq for someq ∈ Z > Thus,Ker θ = {nq : q ∈ Z}. In Example 21.2, a ∈ Ker θ iff2a = θ(a) = 0, i.e. iff a = 0. Hence, Ker θ = {0}.

As point it out earlier the kernel is a subgroup of the domain.

Theorem 21.1Let θ : G −→ H be a homomorphism. Then

(i) Ker θ is a subgroup of G.(ii) For any x ∈ Ker θ and g ∈ G we have gxg−1 ∈ Ker θ.

Proof.(i) By Theorem 18.2(i), θ(eG) = eH so that eG ∈ Ker θ. Hence, Ker θ 6= ∅.Now, let x, y ∈ Ker θ. Then

θ(xy−1) = θ(x)θ(y−1) = θ(x)(θ(y))−1 = eHe−1H = eH .

Thus, xy−1 ∈ Ker θ. By Theorem 7.5, Ker θ is a subgroup of G.

(ii) Let x ∈ Ker θ and g ∈ G. Then

θ(gxg−1) = θ(g)θ(x)θ(g−1) = θ(g)eH(θ(g))−1 = θ(g)(θ(g))−1 = eH .

Thus, gxg−1 ∈ Ker θ.

Theorem 21.1(ii) is one of the common properties that kernels share: Theyare all normals in the sense of the following definition.

Definition 21.3Let H be a subgroup of a group G. Then H is normal iff ghg−1 ∈ H for allg ∈ G and h ∈ H. We write H / G.

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Example 21.4Let H be any subgroup of an Abelian group G. Since hg = gh for all g ∈ Gand all h ∈ H then ghg−1 = h ∈ H for all g ∈ G and h ∈ H. That is, H /G.

Example 21.5Let G = S3 and H =< (12) >= {(1), (12)}. Since (123)(12)(123)−1 = (23) 6∈H then H is not a normal subgroup of G.

Lemma 21.1The following statements are equivalent:

(i) gng−1 ∈ N for all n ∈ N and g ∈ G;(ii) g−1ng ∈ N for all n ∈ N and g ∈ G;

Proof.(i) → (ii): Suppose that gng−1 ∈ N for all n ∈ N and g ∈ G. In particular,g−1n(g−1)−1 ∈ N since g−1 ∈ G. But (g−1)−1 = g so that g−1ng ∈ N.(ii) → (i): Suppose that g−1ng ∈ N for all n ∈ N and g ∈ G. Since(g−1)−1 = g then gng−1 = (g−1)−1ng−1 ∈ N.

The following lemma shows that the homomorphic image of a normal sub-group is normal for onto maps.

Lemma 21.2Let θ : G→ H be an epimorphism and N / G. Then θ(N) / H.

Proof.From Theorem 18.2 (iv), we know that θ(N) is a subgroup of H. Let y ∈ θ(N)and h ∈ H. Then y = θ(x) ∈ θ(N) for some x ∈ N and h = θ(g) for someg ∈ G (since θ is onto). But N/G so that gxg−1 ∈ N. Thus, θ(gxg−1) ∈ θ(N).But θ(gxg−1) = θ(g)θ(x)θ(g−1) = hyh−1 ∈ θ(N). Hence, θ(N) / H.

The following theorem describes a commonly used way for testing whether ahomomorphism is one-to-one or not.

Theorem 21.2Let θ : G −→ H be a homomorphism. Then θ is one-to-one if and only ifKer θ = {eG}.

188

Proof.Suppose first that θ is one-to-one. Let x ∈ Ker θ. Then θ(x) = eH = θ(eG).Hence, x = eG. Thus, Ker θ ⊆ {eG}. Since θ(eG) = eH then {eG} ⊆ Ker θ. Itfollows that Ker θ = {eG}. Conversely, suppose that Ker θ = {eG}. Supposethat θ(x) = θ(y). Then eH = θ(x)(θ(y))−1 = θ(x)θ(y−1) = θ(xy−1). Thus,xy−1 ∈ Ker θ. But then xy−1 = eG so that x = y.

189

Review Problems

Exercise 21.1Define θ : Z6 → Z3 by θ([a]6) = [a]3.

(a) Prove that θ is well-defined.(b) Prove that θ is a homomorphism.(c) Find Ker θ.

Exercise 21.2(a) Prove that θ : Z3 → Z3 defined by θ([a]3) = [a]6 is not well-defined.(b) For which pairs m,n is β : Zn → Zm, given by β([a]n) = [a]m, well-defined?

Exercise 21.3(a) Prove that every homomorphic image of an Abelian group is Abelian.(b) Prove that every homomorphic image of a cyclic group is cyclic.

Exercise 21.4Let G denote the subgroup {1,−1, i,−i} of complex numbers (operation mul-tiplication). Define θ : Z→ G by θ(n) = in. Show that θ is a homomorphismand determine Ker θ.

Exercise 21.5There is a unique homomorphism θ : Z6 → S3 such that θ([1]) = (123).Determine θ([k]) for each [k] ∈ Z6. Which elements are in Ker θ?

Exercise 21.6Prove that N / G if and only if gN = Ng for all g ∈ G.

Exercise 21.7Prove that if N is a subgroup of G such that [G : N ] = 2 then N / G.

Exercise 21.8Prove that An / Sn for all n ≥ 2.

Exercise 21.9Consider the subgroup H = A3 = {(1), (123), (132)} of S3. Let x = (12) andh = (123). Show that xh 6= hx and xH = Hx. This shows that the equalityxH = Hx does not mean that xh = hx for all x ∈ G and h ∈ H.

190

Exercise 21.10Prove that if C denote the collection of all normal subgroups of a group G.prove that N = ∩H∈CH is also a normal subgroup of G.

Exercise 21.11Prove that if N / G then for any subgroup H of G, we have H ∩N / H.

Exercise 21.12Find all normal subgroups of S3.

Exercise 21.13Let H be a subgroup of G and K / G.

(a) Prove that HK is a subgroup of G, where HK = {hk : h ∈ H and k ∈K}.(b) Prove that HK = KH.(c) Prove that K / HK.

Exercise 21.14Prove that if H and K are normal subgroups of G then HK is a normalsubgroup of G.

Exercise 21.15Prove that if H and K are normal subgroups of G such that H ∩K = {eG}then hk = kh for all h ∈ H and k ∈ K.

Exercise 21.16The center of the group G is defined by

Z(G) = {g ∈ G : xg = gx∀x ∈ G}.

Prove that Z(G) / G.

Exercise 21.17Let G and H be groups. Prove that G×{eH} is a normal subgroup of G×H.

Exercise 21.18Let N be a normal subgroup of G, and let a, b, c, d ∈ G. prove that if aN = cNand bN = dN then abN = cdN.

191

Exercise 21.19Let G be a non-abelian group of order 8. Prove that G has at least oneelement of order 4. Hence prove that G has a normal cyclic subgroup oforder 4.

Exercise 21.20Suppose that θ : G → H is a homomorphism. Let K = Ker θ and a ∈ G.Prove that aK = {x ∈ G : θ(x) = θ(a)}.

Exercise 21.21Let S be any set, and let B be any proper subset of S. Let H = {θ ∈ Sym(S) :θ(B) = B}. Prove that H is a subgroup of Sym(S) that is not normal.

Exercise 21.22A group G is called simple if {eG} and G are the only normal subgroups ofG. Prove that a cyclic group of prime order is simple.

Exercise 21.23Let U and V be nonabelian simple groups. Show that G = U×V has preciselyfour different normal subgroups.

192

22 Quotient Groups

Let’s look closely on the construction of the group (Zn,⊕). We know thatthe elements in Zn are equivalence classes of the equivalence relation definedon Z by

a ∼ b if and only if n|(a− b).

Also, an element of Zn is a right coset. Indeed, if 0 ≤ k < n then

< n > +k = [k] = {nq + k : q ∈ Z}.

The operation ⊕ is defined by

[a]⊕ [b] = [a+ b]

or using right cosets

(< n > +a)⊕ (< n > +b) =< n > +(a+ b).

Note that the cyclic group < n > is a normal subgroup of Z since Z is Abelian(See Example 21.4). We are going to use the above ideas to construct newgroups where G is a group replacing Z and N is a normal subgroup of Gplaying the role of < n > . More precisely, we have the following theorem.

Theorem 22.1Let N be a normal subgroup of a group G and let G/N be the set of all rightcosets of N in G. Define the operation on G/N ×G/N by

(Na)(Nb) = N(ab).

Then (G/N, ·) is a group, called the quotient group of G by N.

Proof.· is a well-defined operationWe must show that if (Na1, Nb1) = (Na2, Nb2) then N(a1b1) = N(a2b2). Tosee this, since (Na1, Nb1) = (Na2, Nb2) then Na1 = Na2 and Nb1 = Nb2.Since a1 = ea1 ∈ Na1 then a1 ∈ Na2 so that a1 = na2 for some n ∈ N.Similarly, b1 = n′b2 for some n′ ∈ N. Therefore, a1b1 = na2n

′b2. Since N / Gthen a2n

′a−12 ∈ N, say a2n

′a−12 = n′′ ∈ N. Hence, a2n

′ = n′′a2 so thata1b1 = nn′′a2b2. But nn′′ ∈ N so that a1b1 ∈ N(a2b2). Since N(a2b2) is anequivalence class and a1b1 ∈ N(a2b2) then N(a1b1) = N(a2b2) (Theorem 9.2).

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· is associativeLet a, b, c ∈ G. Then

Na(NbNc) = Na(Nbc) = N(a(bc)) = N((ab)c) = N(ab)Nc = (NaNb)(Nc)

where we used the fact that multiplication in G is associative.

Ne = N is the identity elementIf a ∈ G then (Na)(Ne) = N(ae) = Na and (Ne)(Na) = N(ea) = Na.

Every element of G/N is invertibleIf a ∈ G then (Na)(Na−1) = N(aa−1) = Ne and (Na−1)(Na) = N(a−1a) =Ne so that (Na)−1 = Na−1.

Remark 22.1Note that for a finite group G, the number of elements of G/N is just theindex of N in G, i.e. [G : N ]. That is, |G/N | = [G : N ]. By Lagrange’s

theorem, |G| = [G : N ]|N | so that [G : N ] = |G||N | . Hence, |G/N | = |G|

|N | .

Remark 22.2If N is not a normal subgroup of G, then the operation defined in the previoustheorem will not be well-defined. To see this, consider the subgroup N =<(12) >= {(1), (12)} of S3. Since (123)(12)(123)−1 = (23) 6∈ N then N isnot a normal subgroup of S3. However, N(123) = N(23) = {(123), (23)}and N(132) = N(13) = {(13), (132)}. But N(123)(132) 6= N(23)(13) sinceN(123)(132) = N and N(23)(13) = N(123).

Example 22.1Let G = S3 and N =< (123) >= {(1), (123), (132)}. One can verify easilythat N / G, G/N = {N,N(12)}.

Example 22.2Quotient groups can be used to show that A4 has no subgroup of order 6 andthus showing that the converse of Lagrange’s Theorem is false in general. Tosee this, assume that N is a subgroup of A4 of order 6. Then [G : N ] = 2 andthereforeN/A4 (See Exercise 21.7). Hence, A4/N makes sense. Moreover, foreach a ∈ N, (Na)−1 = Na so that Na2 = N and hence a2inN. One can showthat (123)2 = (132), (132)2 = (123), (124)2 = (142), (142)2 = (124), (134)2 =

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(143), (143)2 = (134), (234)2 = (243), and (243)2 = (234). This yields morethan six different elements of A4 in N. That is, |N | > 6, a contradiction.Thus, A4 has no subgroup of order 6.

By Theorem 21.1(ii), the kernel of a homomorphism is a normal subgroup.Conversely, quotient groups enable us to show that every normal subgroupis the kernel of some homomorphism.

Theorem 22.2Let G be a group and let N / G. The mapping η : G −→ G/N defined byη(a) = Na for all a ∈ G is a homomorphism such that Ker η = N. We callη the natural homomorphism from G onto G/N.

Proof.First we show that η is well-defined. Indeed, if a = b then Na = Nb, i.e.η(a) = η(b). Since η(ab) = N(ab) = NaNb = η(a)η(b) then η is a homo-morphism. η is onto since any member of G/N is of the form Na for somea ∈ G. That is, η(a) = Na. It remains to show that Ker η = N. The proofis by double inclusions. If a ∈ Ker η then η(a) = Ne. Thus, Na = Ne sothat a = ne = n for some n ∈ N. Hence, a ∈ N and Ker η ⊆ N. Conversely,for all n ∈ N we have η(n) = Nn = N = Ne so that N ⊆ Ker η.

The operation of multiplication on the quotient group was defined in termsof right cosets. The following theorem shows that when working with cosetsof a normal subgroup N , it is immaterial whether we use Na or aN.

Example 22.3If G = Z and N =< n > then G/N = Zn and η : Z −→ Zn is given byη(a) =< n > +a = [a] and Ker η =< n > .

Theorem 22.3Let N be a normal subgroup of a group G. Then aN = Na for all a ∈ G.

Proof.Let x ∈ aN. Then x = an for some n ∈ N. SinceN is normal then ana−1 ∈ N.Thus, ana−1 = n′ ∈ N and therefore an = n′a. That is, x ∈ Na. Hence,aN ⊆ Na. A similar argument shows that Na ⊆ aN.

We close this section with some properties of quotient groups. More proper-ties of quotient groups are discussed in the exercises.

195

Lemma 22.1If G is Abelian and N / G then G/N is also Abelian.

Proof.We have to show that for anyNa,Nb ∈ G/N we have (Na)(Nb) = (Nb)(Na).Now, since G is Abelian then ab = ba. Thus,

(Na)(Nb) = Nab = Nba = (Nb)(Na).

That is, G/N is Abelian.

Lemma 22.2If G is a group such that G/Z(G) is cyclic then G is Abelian, where

Z(G) = {g ∈ G : xg = gx ∀x ∈ G}.

Proof.First we show that Z(G) / G. Indeed, if g ∈ G and x ∈ Z(G) then gxg−1 =gg−1x = x ∈ Z(G). Since G/Z(G) is a cyclic group then G/Z(G) =<Z(G)g > for some g ∈ G. If a, b ∈ G then Z(G)a and Z(G)b belong toG/Z(G). Thus, Z(G)a = Z(G)gn and Z(G)b = Z(G)gm for some integersn and m. Hence, a = xgn and b = ygm for some x, y ∈ Z(G). This impliesthat ab = (xgn)(ygm) = xygn+m = yxgmgn = ygmxgn = ba. Hence, G isAbelian.

Lemma 22.3Let G be a group of order p2 where p is a prime number. Assuming that thecenter of G is nontrivial, then G is Abelian.

Proof.Let G be a group such that |G| = p2. Since e ∈ C(G) then C(G) 6= ∅. ByLagrange’s theorem either |C(G)| = p2 or |C(G)| = p. If |C(G)| = p2 thenC(G) = G and so G is Abelian. If |C(G)| = p then |G/C(G)| = p and soG/C(G) is cyclic (See Corollary 3). By the previous lemma, G is Abelian.

196

Review Problems

Exercise 22.1Find the order of each of the following quotient groups.

(i) Z8/ < [4] > .(ii) < 2 > / < 8 > .

Exercise 22.2Let G = A4 and H = {(1), (12)(34), (13)(24), (14)(23)} / G. Write out thedistinct elements of G/H and make a Cayley table for G/H.

Exercise 22.3Construct the Cayley table for Z12/ < [4] > .

Exercise 22.4Let G be a cyclic group. Prove that for every subgroup H of G, G/H iscyclic.

Exercise 22.5Assume N / G.

(a) Prove that if [G : N ] is prime, then G/N is cyclic.(b) Prove or disprove the converse of the statement in part (a).

Exercise 22.6Determine the order of Z12 × Z4/ < [3], [2] > .

Exercise 22.7Prove that if N / G and a ∈ G then o(Na)|o(a).

Exercise 22.8Prove that G/N is Abelian if and only if aba−1b−1 ∈ N for all a, b ∈ G.

Exercise 22.9Let N be a normal subgroup of G and a ∈ G. Prove that the order of Na inG/N is the smallest positive integer n such that an ∈ N.

197

Exercise 22.10Let θ : G→ H be a homomorphism and K = Ker θ. Show that φ : G/K →θ(G) defined by φ(Ka) = θ(a) is an isomorphism. Also, show that φ ◦ η = θ,where η is the canonical homomorphism.

Exercise 22.11Let H and K be subgroups of a group G such that K/G. Prove that H/H∩Kand HK/K make sense, where HK = {hk : h ∈ H and k ∈ K}.

Exercise 22.12Let H and K be as in the previous exercise. Show that φ : H → HK/Kdefined by φ(h) = Kh an epimorphism with Ker φ = H ∩K.

Exercise 22.13Assume that H,K4G and K4H. Prove that H/K / G/K.

Exercise 22.14Let H,K, and G as in the previous exercise. Show that φ : G/K → G/H,defined by φ(Kg) = Hg is an epimorphism with Ker φ = H/K.

Exercise 22.15If N is a subgroup of G such that the product of two right cosets of N in Gis again a right coset of N in G, prove that N / G.

198

23 Isomorphism Theorems

Theorem 22.2 shows that each quotient group of a group G is the homomor-phic image of G. The theorem below shows that the converse is also true.That is, each homomorphic image is isomorphic to a quotient group.

Theorem 23.1 (The Fundamental Homomorphism Theorem)Let θ : G −→ H be a homomorphism. Then

G/K ≈ θ(G)

where K = Ker θ. Moreover, φ◦η = θ where η is the natural homomorphismintroduced in Theorem 22.2

Proof.By Theorem 21.1, K = Ker θ /G so that by Theorem 22.1, G/K is a group.Define φ : G/K −→ θ(G) by φ(Ka) = θ(a). This is a well-defined mapping.To see this, suppose that Ka = Kb. Since a = ea ∈ Ka then a ∈ Kb so thata = kb for some k ∈ K. Hence, θ(a) = θ(kb) = θ(k)θ(b) = eHθ(b) = θ(b).Thus, φ(Ka) = φ(Kb).Next, we show that φ is a homomorphism. Let Ka,Kb ∈ G/K. Then

φ(KaKb) = φ(Kab) = θ(ab) = θ(a)θ(b) = φ(Ka)φ(Kb).

To show that φ is one-to-one we use Theorem 21.2. That is we show thatKer φ = {K}. If Ka ∈ Ker φ then φ(Ka) = eH . Thus, θ(a) = eH so thata ∈ K. This implies that Ka = K.Now, if y ∈ θ(G) then y = θ(a) = φ(Ka) for some a ∈ G. This shows that φis onto. Hence, φ is an ismomorphism and thus G/K ≈ θ(G).Finally, note that φ ◦ η and θ have the same domain, codomain and for allg ∈ G, (φ◦ η)(g) = φ(η(g)) = φ(Ng) = θ(g). Thus, φ◦ η = θ. This completesa proof of the theorem.

Example 23.1Looking at Example 22.3, we see that the mapping θ : Z −→ Zn defined byθ(a) = [a] satisfies the assumptions of Theorem 23.1. Hence, Z/ < n >≈ Zn

As a consequence of Theorem 23.1, we have the following two theorems.

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Theorem 23.2 (First Isomorphism Theorem)Let H and K be subgroups of a group G such that K / G. Then the set

HK = {hk : h ∈ H and k ∈ K}

is a subgroup of G such that K / HK and

H/H ∩K ≈ HK/K.

Proof.First we show that HK is a subgroup of G. Since eG = eGeG and eG ∈ H∩Kthen eG ∈ HK so that HK 6= ∅. Now, if a, b ∈ HK then a = h1k1 andb = h2k2. Thus, ab−1 = h1k1k

−12 h−1

2 = h1kh−12 , where k = k1k

−12 ∈ K.

Since K / G then h2kh−12 ∈ K so that h2kh

−12 = k′ for some k′ ∈ K. Thus,

kh−12 = h−1

2 k′. This shows that ab−1 = h1h−12 k′ ∈ HK. Hence, by Theorem

7.5, HK is a subgroup of G.Next, we show that K /HK. Since k = eHk ∈ HK then K ⊆ HK. Considerthe product (hk)k′(hk)−1 = hkk′k−1h−1 = hk′′h−1 ∈ K since K / G. Hence,K / HK.Now, consider the mapping θ : H −→ HK/K defined by θ(h) = Kh. This isa well-defined mapping: if h1 = h2 then Kh1 = Kh2. This mapping is ontofrom its definition. To see that θ is a homomorphism, we take h1, h2 ∈ Hand find

θ(h1h2) = Kh1h2 = (Kh1)(Kh2) = θ(h1)θ(h2).

Finally, we show that Ker θ = H ∩K. Indeed, if x ∈ H ∩K then x ∈ K andx ∈ H. Thus, Kx = K. That is, θ(x) = K and so x ∈ Ker θ. Conversely, ifx ∈ Ker θ then Kx = K and this implies that x ∈ K. Since the kernel is asubset of H then x ∈ H. Thus, x ∈ H ∩K and so Ker θ = H ∩K. ApplyingTheorem 23.1, we obtain H/(H ∩K) ≈ HK/K.

Theorem 23.3 (Second Isomorphism Theorem)Assume that H,K /G with K /H. Then H/K /G/K and (G/K)/(H/K) ≈G/H.

Proof.First we prove that H/K is a subgroup of G/K. Since K ∈ H/K thenH/K 6= ∅. Since (Kh1)(Kh2)−1 = (Kh1)(Kh−1

2 ) = Kh1h−12 ∈ H/K then by

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Theorem 7.5, H/K is a subgroup of G/K. To see that H/K / G/K we pickelements Kh ∈ H/K and Kg ∈ G/K and find that

(Kg)(Kh)(Kg)−1 = KghKg−1 = Kghg−1 ∈ H/K

since ghg−1 ∈ H (H / G.)Next, we define θ : G/K −→ G/H by θ(Kg) = Hg. This is well-defined mapfor if Kg = Kg′ then g = kg′ for some k ∈ K ⊆ H. Hence, g ∈ Hg′ andHg = Hg′. From the definition of θ, we have that θ is onto.θ is a homomorphism: If Kg1, Kg2 ∈ G/K then θ(Kg1Kg2) = θ(Kg1g2) =Hg1g2 = (Hg1)(Hg2) = θ(Kg1)θ(Kg2).Ker θ = H/K : If Kg ∈ Ker θ then θ(Kg) = H, i.e. Hg = H so that g ∈ H.Thus, Kg ∈ H/K and this shows that Ker θ ⊆ H/K. Now, if Kh ∈ H/Kthen θ(Kh) = Hh = H so that Kh ∈ Ker θ. Hence, H/K ⊆ Ker θ and thisshows that Ker θ = H/K. By Theorem 23.1, we have

(G/K)/(H/K) ≈ G/H.

201

Review Problems

Exercise 23.1Find all the homomorphic images of S3.

Exercise 23.2Prove that if G is any group with identity e then G/{e} ≈ G.

Exercise 23.3For any groups A and B, prove the following.

(a) A ≈ A× {e} / A×B.(b) A×B

A×{e} ≈ B.

Exercise 23.4Prove that if G is a simple Abelian group then G ≈ Zp for some prime numberp.

Exercise 23.5Prove that Z18

<[3]>≈ Z3.

Exercise 23.6Prove that if θ is a homomorphism of G onto H, B / H, and A = {g ∈ G :θ(g) ∈ B}, then A / G.

Exercise 23.7Give an example to show that if A and B are subgroups of a group G thenAB need not be a subgroup of G.

Exercise 23.8Suppose that N / G. Let C be the collection of all subgroups of G containingH. Prove that the map φ defined by φ(S) = S/H, where S ∈ C is one-to-oneand onto.

Exercise 23.9Let G be a group with normal subgroups H and K such that G = HK andH ∩K = {e}. Prove that G ≈ H ×K.

202

Exercise 23.10Let G be a group.

(a) Prove that fa : G→ G, given by fa(x) = axa−1 is an isomorphism.(b) Prove that the set Inn(G) = {fa : a ∈ G} is a group under composition.(c) Prove that Inn(G) ≈ GZ(G), where Z(G) is the center of G.

Exercise 23.11Prove that 6Z

12Z ≈2Z4Z . Hint:Use First Isomorphism Theorem.

Exercise 23.12Prove that if H is a normal subgroup of G of prime index p then for allsubgroups K of G either (i) K is a subgroup of H or (ii) G = HK and[K : K ∩H] = p.

Exercise 23.13Let G be a finite group, H and N are subgroups of G such that N /G. Provethat if |H| and [G : N ] are relatively prime then H is a subgroup of N.

Exercise 23.14Let G be a finite group and let H and K be subgroups of G with K /G. Provethat

|HK| = |H| |K||H ∩K

.

Exercise 23.15Let G be a group and let H and K be normal subgroups of G with K ⊆ H.Suppose that G/K is cyclic. Prove that G/H and H/K are cyclic.

Exercise 23.16Suppose that K / G and H is a subgroup of G such that H ∩ K = {eG}.Prove that H is isomorphic to a subgroup of G/K. Prove that if G = HKthen G/K ≈ H.

Exercise 23.17Suppose that φ : G → H is an epimorphism and N / G. Prove that thereexists a homomorphism from G/N onto H/φ(N).

203

Exercise 23.18Let G and H be groups and let K / G and K ′ / H.

(a) Prove that K ×K ′ / G×H.(b) Prove that G×H

K×K′ ≈ G/K ×H/K ′.

Exercise 23.19Let M and n be relatively prime. Show that φ : Z × Zm × Zn, given byφ(a) = ([a]m, [a]n) is a homomorphism, onto, and Ker φ =< mn > .

Exercise 23.20Prove that if m and n are relatively prime then Zmn ≈ Zm × Zn.

204

24 Rings: Definition and Basic Results

In this section, we introduce another type of algebraic structure, called ring.A group is an algebraic structure that requires one binary operation. A ringis an algebraic structure that requires two binary operations that satisfy someconditions listed in the following definition.

Definition 24.1A ring is a nonempty set R with two binary operations (usually written asaddition and multiplication) such that for all a, b, c ∈ R,

(1) R is closed under addition: a + b ∈ R. (2) Addition is associative:(a+ b) + c = a+ (b+ c).(3) Addition is commutative: a+ b = b+ a.(4) R contains an additive identity element, called zero and usually denotedby 0 or 0R: a+ 0 = 0 + a = a.(5) Every element of R has an additive inverse: a+ (−a) = (−a) + a = 0.(6) R is closed under multiplication: ab ∈ R.(7) Multiplication is associative: (ab)c = a(bc).(8) Multiplication distributes over addition: a(b+ c) = ab+ac and (a+ b)c =ac+ bc.If ab = ba for all a, b ∈ R then we call R a commutative ring.

In other words, a ring is a commutative group with the operation + and anadditional operation, multiplication, which is associative and is distributivewith respect to +.

Remark 24.1Note that we don’t require a ring to be commutative with respect to multipli-cation, or to have multiplicative identity, or to have multiplicative inverses.A ring may have these properties, but is not required to. These additionalproperties will be discussed in the next section.

Example 24.1The sets Z,Q,R, with the usual operations of multiplication and additionform commutative rings.

Example 24.2The set Zn with the operations of multiplication and addition modulo n forms

205

a commutative ring for any n ∈ N. The only properties left to establish arethe distributive laws. We will show that [a]� ([b]⊕ [c]) = [a]� [b]⊕ [a]� [c].The proof of ([a]⊕ [b])� [c] = [a]� [c]⊕ [b]� [c] is similar.

[a]� ([b]⊕ [c]) = [a]� [b+ c]= [a(b+ c)]= [ab+ ac]= [ab]⊕ [ac]= [a]� [b]⊕ [a]� [c]

Example 24.3The set of even integers together with the usual addition and multiplicationin Z is a commutative ring.

We next turn to discussing some basic properties of rings.

Theorem 24.1The following hold in any ring R.

(i) a+ b = a+ c implies b = c for all a, b, c ∈ R.(ii) a0 = 0a = 0 for all a ∈ R;(iii) a(−b) = (−a)a = −(ab) for all a, b ∈ R;(iv) −(−a) = a.(v) −(a+ b) = (−a) + (−b).(vi) (−a)(−b) = ab for all a, b ∈ R;(vii) a(b − c) = ab − ac and (a − b)c = ac − bc for all a, b, c ∈ R, where wea− b stands for a+ (−b).(viii) (−1)a = −a if R has a multiplicative identity,i.e. 1Ra = a1R = a forall a ∈ R.

Proof.(i) b = 0+b = ((−a)+a)+b = (−a)+(a+b) = (−a)+(a+c) = ((−a)+a)+c =0 + c = c.(ii) a0 = a(0 + 0) = a0 + a0. Thus, 0 + a0 = a0 = a0 + a0 so that by theright cancellation property of the additive group we have a0 = 0. A similarargument holds for 0a = 0.(iii) Since ab+(−a)b = (a+(−a))b = 0b = 0 then (−a)b is the additive inverseof ab. That is, −(xab) = (−a)b. Similarly, since ab + a(−b) = a(b + (−b)) =a0 = 0 then −(ab) = a(−b).

206

(iv) Follows from the definition of additive inverse.(v) Since (a+b)+((−a)+(−b)) = [(a+b)+(−a)]+(−b) = [(−a)+(a+b)]+(−b) = [((−a)+a)+b]+(−b) = (0+b)+(−b) = 0+(b+(−b)) = 0+0 = 0 then(−a) + (−b) is the additive inverse of a+ b. That is, −(a+ b) = (−a) + (−b).(vi) Using (iii), we have (−a)(−b) = −[a(−b)] = −[−(ab)] = ab.(vii) We prove that a(b− c) = ab− ac. The prove that (a− b)c = ac− bc issimilar.

a(b− c) = a(b+ (−c))= ab+ a(−c) by Definition 24.1(8)= ab− ac by (ii)

(viii) Follows from (iii).

As we pointed out earlier in the section, the multiplicative operation in aring does not necessarily have to satisfy either the commutative law or havean identity element. The following definition introduces the terminology usedwhen multiplication is either commutative or has an identity element.

Definition 24.2Let R be a ring such that ab = ba for all a, b ∈ R. Then we call R a com-mutative ring. If e ∈ R is such that ae = ea = a for all a ∈ R then e iscalled a unity for the ring and the ring is called unitary ring.

Example 24.41. Z,Q, and R are commutative rings with unity 1.2. The ring of even integers is a commutative ring with no unity.3. The ring of integers modulo n is a commutative ring with unity [1].4. The ring M of 2× 2 matrices is noncommutative with unity the matrix(

1 00 1

).

When a unity in a ring exists then it must be unique.

Theorem 24.2If R is a ring and e1 and e2 are unity elements then we must have e1 = e2.

Proof.Since e1 is a unity then e1a = a for all a ∈ R. In particular, letting a = e2

to obtain e1e2 = e2. Similarly, since e2 is a unity then ae2 = a for all a ∈ R.

207

Letting a = e1 to obtain e1e2 = e1. Thus, e1 = e2.

As in the case of a group, the existence of the unity element leads to thediscussion of multiplicative inverses.

Definition 24.3Let R be a ring with unity e. We say that x is a multiplicative inverse ofan element a ∈ R if ax = xa = e.

Multiplicative inverses are unique according to the following theorem.

Theorem 24.3Let R be a ring with unity e. Let a ∈ R and suppose that x and y aremultiplicative inverses of a. Then x = y.

Proof.Since x and y are multiplicative inverses of a then we have ax = xa = e anday = ya = e. Thus, x = ex = (ya)x = y(ax) = ye = y.

Notation Let R be a ring with unity. If a ∈ R has a multiplicative in-verse then we will denote the inverse by a−1.

Example 24.5In a ring R, it is possible that some of the elements have multiplicativeinverses whereas others don’t. For example, in the ring Z10, [1] and [9] aretheir own multiplicative inverses, [3] and [7] are inverses of each other. Theremaining elements of Z10 have no multiplicative inverses.

208

Review Problems

Exercise 24.1Compute [3]� ([4]⊕ [5]) in Z6.

Exercise 24.2Let M be the collection of all 2 by 2 matrices with entry in R. Show that(M,+, ·) is a non commutative ring, where addition and multiplication asdefined in Exercises 3.15 and 3.16.

Exercise 24.3Let Z[

√2] = {a + b

√2 : a, b ∈ Z}. Show that Z[

√2] with the usual addition

and multiplication in Z is a commutative ring.

Exercise 24.4Let M(R) be the collection of all mappings from R to R. Define addition by(f + g)(x) = f(x) + g(x) and multiplication by (fg)(x) = f(x)g(x). Showthat (M(R),+, ·) is a commutative ring.

Exercise 24.5Let E be the set of even integers. Prove that with the usual addition, andwith the multiplication mn = 1

2mn, E is a ring. Is there a unity?

Exercise 24.6Find the elements of Z8 that have multiplicative inverses.

Exercise 24.7Let R and S be two rings. Show that the Cartesian product R × S togetherwith the operations

(a, b) + (c, d) = (a+ c, b+ d)(a, b)(c, d) = (ac, bd)

is a ring.

Exercise 24.8The addition table and part of the multiplication table for the ring R ={a, b, c} are given below. Use the distributive laws to complete the multi-plication table.

209

+ a b ca a b cb b c ac c a b

· a b ca a a ab a cc a

Exercise 24.9Let (R,+) be an Abelian group. Define multiplication by ab = 0 for alla, b ∈ R. Show that (R,+, ·) is a commutative ring.

Exercise 24.10In the ring of integers, if ab = ac, with a 6= 0, then b = c. Is this true for allrings?

Exercise 24.11Prove that in a ring R we have

(x+ y)(z + t) = xz + xt+ yz + yt

for all x, y, z, t,∈ R.

Exercise 24.12Let R be a ring in which a2 = a for all a ∈ R. Prove that R is commutativeand that a+ a = 0 for all a ∈ R. (Hint: Consider (a+ a)2 and (a+ b)2.)

Exercise 24.13Prove that (a+ b)2 = a2 + 2ab+ b2 for all a, b in a ring R if and only if R iscommutative.

Exercise 24.14Prove that if R is a commutative ring, a, b ∈ R, and n ∈ N then

(a+ b)n = an +

(n1

)an−1b+

(n2

)an−2b2 + · · ·+

(n

n− 1

)abn−1 + bn

where (np

)=

n!

p!(n− p)!

210

Exercise 24.15An element a of a ring R is said to be nilpotent if an = 0 for some positiveinteger n. Prove that if R is commutative and if a and b are nilpotent, thenso is a + b.

Exercise 24.16Show that the set

R =

{(0 a0 b

): a, b ∈ R

}with the usual matrix addition and multiplication is a ring.

Exercise 24.17A (real) polynomial is a formal expression of the form

p(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a)0,

where a0, · · · , an ∈ R and x is a variable. Polynomials can be added andmultiplied as usual. With these operations, show that the set R[x] of allpolynomials is a ring.

Exercise 24.18Suppose that R is a ring in which all elements a satisfy a2 = a.(Such a ringis called a Boolean ring.)

(i) Prove that a = −a for all a ∈ R.(ii) Prove that R is commutative.

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25 Integral Domains. Subrings

In Section 24 we defined the terms unitary rings and commutative rings.These terms together with the concept of zero divisors discussed below areused to define a special type of ring known as an integral domain.Let R be a ring. Then, by Theorem 24.1(ii), we have a0 = 0a = 0 for alla ∈ R. This shows that if a product is zero then one of the factors is 0.The converse is not always true. For example, in Z10, [2] and [5] are nonzeroelements with [2]� [5] = [0]. The following definition singles out those ringswhere a product of two (additive) nonidentity elements is the zero element.

Definition 25.1Let R be a commutative ring. An element a ∈ R, a 6= 0 is called a zerodivisor in R if there exists an element b ∈ R, b 6= 0 such that ab = 0.

Example 25.11. The ring of integers Z has no zero divisors.2. The elements [2] and [5] are zero divisors in Z10.

Remark 25.1Definition 25.1 is restricted to elements in a commutative ring. It is possibleto have noncommutative rings where ab = 0 but ba 6= 0. Indeed, the ring ofall 2× 2 matrices is noncommutative. Moreover, we have(

0 10 0

)(1 00 0

)=

(0 00 0

)and (

1 00 0

)(0 10 0

)=

(0 10 0

)The following definition shows that zero divisors can not exist in an integraldomain.

Definition 25.2A commutative ring with unity e 6= 0 and no zero divisors is called an inte-gral domain.

Remark 25.2The requirement e 6= 0 means that the ring has at least two elements, thezero element and the unity element.

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Example 25.21. Z,Q, and R are all integral domains.2. The set E of evens integers is not an integral domain since it has no unityelement.3. Z10 is not an integral domain since [2] and [5] are zero divisors.4. The ring M of all 2 × 2 matrices is not an integral domain for tworeasons: first, the ring is noncommutative, and second, it has zero divisors.(See Remark 25.1

Example 25.2(4), shows that Z10 is a commutative ring with unity but is notan integral domain since [2] and [5] are zero divisors. Note that 10 = 2× 5.We can generalize this fact to any composite number n. So if n = rs wherer, s > 1, then [r] � [s] = [rs] = [n] = [0] so that [r] and [s] are zero divisorsof Zn. That is, Zn is not an integral domain.The next result provides a condition on n so that Zn is an integral domain.

Theorem 25.1For n > 1, Zn has no zero divisors if and only if n is prime.

Proof.Suppose first that n is prime. Let [a]� [b] = [0] in Zn with [a] 6= [0]. We willshow that [b] = [0] in Zn. Since [a]� [b] = [0] then [ab] = [0] and this impliesthat n|ab. Since [a] 6= [0] then n 6 |a. Since n is prime then by Lemma 13.3,we must have n|b. That is, [b] = [0]. Therefore, Zn has no zero divisors, andis an integral domain.Conversely, suppose that Zn is an integral domain. Assume that n is notprime. As pointed out in the discussion preceding the theorem, Zn is not anintegral domain, a contradiction. Hence, n must be prime.

An important consequence of the absence of zeros in an integral domainis that the cancellation law for multiplication must hold.

Theorem 25.2If a, b, and c are elements in integral domain D such that a 6= 0 and ab = ac,then b = c.

Proof.Since ab = ac then a(b−c) = 0 with a 6= 0 in D. But D is an integral domainso we must have b− c = 0 or b = c.

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The converse of the previous theorem is also true.

Theorem 25.3If D is a commutative ring with unity e 6= 0 such that the cancellation prop-erty holds then D is an integral domain.

Proof.Suppose that for all a, b, c ∈ D, ab = ac and a 6= 0 implies b = c. We willshow that D has no zero divisors. Let a, b ∈ D be such that ab = 0 witha 6= 0. Since a0 = 0 then ab = a0. By the cancellation law, b = 0. This showsthat D has no zero divisors, so D is an integral domain.

The notion of subring is the abvious analogue of the notion of subgroup.

Definition 25.3A subring of a ring R is any subset S ⊆ R which forms a ring with respectto the operation of R.

Example 25.3The set E of even integers is a subring of all integers. The set of integersis a subring of the ring of rational numbers. The set of rational numbers isa subring of the ring of all real numbers. The set of all real numbers is asubring of the ring of complex numbers.

As in groups, we can reduce the number of axioms one has to check whenproving that something is a subring.

Theorem 25.4Let R be a ring and S a subset of R. Then S is a subring of R if and only if

(i) S 6= ∅;(ii) For all a, b ∈ S we have a− b ∈ S and ab ∈ S.

Proof.Suppose first that S is a subring of R. Then S being a ring itself, it mustcontain the zero element of R. Thus, S 6= ∅. Now, let a, b ∈ S. Since S is aring then (S,+) is a group so that a− b ∈ S. Also, S is closed with respectto multiplication so that ab ∈ S.

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Conversely, suppose that S is a subset of R satisfying conditions (i) and(ii). Since S is nonempty and a − b ∈ S for all a, b ∈ S then by Theorem7.5, (S,+) is a group. By (ii), S is closed with respect to multiplication.Since multiplication is associative in R and S is closed then multiplicationis associative when restricted to S. Thus, S is a ring and hence a subring ofR.

Example 25.4Consdier the subset of the ring M of all 2× 2 matrices:

S =

{(a b0 c

): a, b, c ∈ R

}We will show that S is a subring of M. Since the zero matrix is in S thenS 6= ∅. Since (

a b0 c

)−(d e0 f

)=

(a− d b− e

0 c− f

)∈ S

and (a b0 c

)(d e0 f

)=

(ad ae+ bf0 cf

)∈ S

then by Theorem 25.4, S is a subring of M.

215

Review Problems

Exercise 25.1Find the zero divisors of Z6.

Exercise 25.2Verify that ([2], [0]) is a zero divisor in Z3 × Z3.

Exercise 25.3Which elements of Z× Z are zero divisors?

Exercise 25.4Prove that if an element a in a ring R has a multiplicative inverse in R, thena is not a zero divisor in R.

Exercise 25.5Show that a zero divisor can not have a multiplicative inverse.

Exercise 25.6Let R be a commutative ring with unity e. Let a ∈ R be such that an = 0 forsome n ∈ N. Prove that a is either 0 or a zero divisor.

Exercise 25.7Let D be an integral domain. Show that if a ∈ D such that a2 = a and a 6= ethen a is a zero divisor.

Exercise 25.8Let R be a commutative ring. For each a ∈ R let Ha = {x ∈ R : ax = 0}.Show that for all x, y ∈ Ha, we have xy ∈ Ha.

Exercise 25.9Show that Z[

√2] (Exercise 24.3) is an integral domain.

Exercise 25.10Show that the ring M(R) of all mappings from R to R is not an integraldomain.(See Exercise 24.4.)

Exercise 25.11State and prove a theorem giving a necessary and sufficient condition for asubset of an integral domain to be an integral domain.

216

Exercise 25.12Prove that if D is an integral domain and a2 = e then a = ±e.

Exercise 25.13Let R and S be integral domains. Prove that R×S is also an integral domain.

Exercise 25.14Let R be a ring with unity e. Let S be the collection of all elements in R withmultiplicative inverse. Prove that (S, ·) is a group.

Exercise 25.15Let R be a commutative ring with unity e such that every nonzero element ofR has a multiplicative inverse. Show that R is an integral domain.

Exercise 25.16Let C(R) denote the collection of all continuous functions from R to R. Showthat C(R) is a subring of M(R).

Exercise 25.17Prove that {(a, a) : a ∈ R} is a subring of R×R.

Exercise 25.18Let R be a ring with identity e and S a subring of R such that e ∈ S. Provethat if u is a unit in S then u is a unit in R. Show by an example that theconverse is false.

Exercise 25.19The center of a ring R is defined to be {c ∈ R : cr = rc ∀r ∈ R}. Prove thatthe center of a ring is a subring. What is the center ofa commutative ring?

Exercise 25.20Let C be the collection of all subrings of a ring R. Prove that ∩H∈CH is asubring of R.

Exercise 25.21Show that the set

S =

{(a bc 0

)}is not a subring of the ring M of all 2× 2 matrices.

217

26 Ideals and Quotient Rings

In this section we develop some theory of rings that parallels the theory ofgroups discussed in ealrier sections of the book. We shall see that the conceptof an ideal in a ring is analogous to that of a normal subgroup in a group.Using ideals we will construct quotient rings.

Definition 26.1A subring I of a ring R is an ideal if whenever r ∈ R and a ∈ I, then ra ∈ Iand ar ∈ I.

Example 26.11. The subrings I = {0} and I = R are always ideals of a ring R.2. The set E of even integers is an ideal of Z.3. The set I = {[0], [2], [4]} is an ideal of Z6.

Remark 26.1Note that if R is a ring with unity element e and I is an ideal of R then fromthe above definition, for any r ∈ R we have re = er = r ∈ I. That is, R = I.

Example 26.1(1) can be generalized to the set of all multiples of any fixedinteger n as shown in the next lemma.

Lemma 26.1Let R be a commutative ring with unity element e. The set

(a) = {ar : r ∈ R}

is an ideal of R.

Proof.First, we will show that (a) is a subring of R. Since a = ae then a ∈ (a) andso (a) 6= ∅. Let ax ∈ (a) and ay ∈ (a). Then ax− ay = a(x− y) ∈ (a) sincex− y ∈ R. Thus, ax− ay ∈ (a). Also, (ax)(ay) = a(xy) ∈ (a) since xy ∈ R.Thus, by Theorem 25.4, (a) is a subring of R. Finally, for any ar ∈ (a) and allt ∈ R we have t(ar) = t(ra) = (tr)a ∈ (a) and (ar)t = a(rt) = (rt)a ∈ (a).Thus, (a) is an ideal of R.

Definition 26.2Let R be a commutative ring with unity and a ∈ R. Then the ideal (a) iscalled the principal ideal generated by a in R.

218

Theorem 26.1Every ideal in the ring Z is a principal ideal.

Proof.Let I be an ideal in R. If I = {0} then there is nothing to prove since{−} = (0). So assume that I 6= {0}. Since I 6= {0} then there exists anm ∈ I such that m 6= 0. Since I is an ideal and −1 ∈ Z then −m ∈ I.Thus, both m and −m belong to I so that I contains positive integers. LetM = {m ∈ I : m > 0}. By Theorem 10.1, M contains a smallest element,call it n.Now, for any k ∈ I, by the Division algorithm there exist integers q and rsuch that

k = nq + r 0 ≤ r < n.

Thus, r = k − nq. Since I is a subring of R then by closure r ∈ I. By thedefinition of n we must have r = 0. This implies that k = nq and so I ⊆ (n).Since (n) ⊆ I then we conclcude that I = (n). This establsihes a proof ofthe theorem.

In analogy to congruence in Z and Zn = Z/ < n > we now will build aring R/I for any ideal I in any ring R. For a, b ∈ R, we say a is congruent tob modulo I [and write a ≡ b(mod I)] if and only if a− b ∈ I. Note that whenI = (n) ⊆ Z is the principal ideal generated by n, then a−b ∈ I ⇐⇒ n|(a−b),so this is our old notion of congruence.

Theorem 26.2Let I be an ideal of a ring R. Then congruence modulo I is an equivalencerelation on R. For any a ∈ R we have

[a] = {r ∈ R : a ≡ r(mod I)} = {a+ i : i ∈ I}.

Proof.Reflexive: a− a = 0 ∈ I since I is a subring.Symmetric: Assume that a ≡ b(mod I). Then a− b ∈ I. Since I is a subring,its additive inverse, b− a is also in I, so b ≡ a(mod I).Transitive: Assume a ≡ b(mod I) and b ≡ c(mod I). Then a − b ∈ I andb−c ∈ I. Hence, by closure, a−c = (a−b)+(b−c) ∈ I. That is, a ≡ c(mod I).Now, for any a ∈ R, the equivalence class of a is the set

[a] = {r ∈ R : a ≡ r(mod I)}.

219

But a ≡ r(mod I) if and only if a − r ∈ I and this is equivalent to sayingthat a− r = i ∈ I. That is,

[a] = {a+ i : i ∈ I}.

We denote the equivalence class of a ∈ R by a+ I.

Remark 26.2Note that by Theorem 9.2, a ≡ b(mod I)⇐⇒ a− b ∈ I ⇐⇒ a+ I = b+ I.

Next, consider the set R/I of all cosets a + I where a ∈ R. On this set, wedefine addition as follows:

(a+ I) + (b+ I) = (a+ b) + I.

Theorem 26.3The set R/I is an Abelian group with respect to addition.

Proof.First, we show that addition is well-defined. Suppose that (a + I, b + I) =(c+I, d+I). Then a+I = c+I and b+I = d+I. Thus, a−c ∈ I and b−d ∈ I.By the closure of addition in I we have (a−c)+(b−d) = (a+b)−(c+d) ∈ I.Hence, (a+ b) + I = (c+ d) + I.Addition is commutative since (a+ I) + (b+ I) = (a+ b) + I = (b+ a) + I =(b+ I) + (a+ I) since addition is R is commutative. Similarly, since additionin R is associative then addition is R/I is associative. The coset I is the zeroelement and the additive inverse of a+ I is (−a) + I. Hence, (R/I,+) is anAbelian group.

In order to turn R/I into a ring we need to introduce a multiplication onR/I. For a+ I ∈ R/I and b+ I ∈ R/I we define

(a+ I)(b+ I) = ab+ I.

Theorem 26.4(a) R/I is closed with respect to multiplication.(b) Multiplication is associative.(c) Multiplication is distributive with respect to addition.

220

Proof.(a) Suppose that (a + I, b + I) = (c + I, d + I). Then a + I = c + I andb + I = d + I. Thus, a − c ∈ I and b − d ∈ I. It follows that a = c + x andb = d + y for some x, y ∈ I. Thus, ab = (c + x)(d + y) = cd + cy + dx + xy.Since I is an ideal then cy, dx, xy ∈ I. By the closure of addition in I wehave w = cy + dx + xy ∈ I. Hence, ab = cd + w with w ∈ I and this meansthat ab− cd ∈ I so that ab ≡ cd(mod I) and consequently ab+ I = cd+ I.(b)Multiplication is associative.

(a+ I)[(b+ I)(c+ I)] = (a+ I)(bc+ I)= a(bc) + I= (ab)c+ I (since multiplication is associative in R)= (ab+ I)(c+ I)= [(a+ I)(b+ I)](c+ I)

(c) We will verify the left distributive law. The proof of the right distributivelaw is similar.

(a+ I)[(b+ I) + (c+ I)] = (a+ I)(b+ c+ I)= a(b+ c) + I= ab+ ac+ I (by the distributive law in R)= (ab+ I)(ac+ I)= (a+ I)(b+ I) + (a+ I)(c+ I)

It follows from Theorem 26.4 and Theorem 26.3 that R/I is a ring.

Definition 26.3If I is an ideal of a ring R then with the operations of addition and mul-tiplication defined above, R/I is a ring called the quotient ring of R byI.

221

Review Problems

Exercise 26.1Consider the ring

R =

{(a b0 c

): a, b, c ∈ Z

}.

Show that the set

I =

{(0 b0 0

): b ∈ Z

}.

is an ideal of R.

Exercise 26.2Prove that xR[x] is an ideal of R[x].

Exercise 26.3Show that the set Ir = {f ∈ M(R) : f(r) = 0}, where r ∈ R is fixed, is anideal of M(R). Can we replace the 0 by any number and still get an ideal?

Exercise 26.4Show that the set

I =

{(a b0 0

): b ∈ R

}.

is not an ideal of the ring of all 2× 2 matrices.

Exercise 26.5Let R be a commutative ring with unity element e. Let c1, c2, · · · , cn ∈ R.Show that the set I = {r1c1 + r2c2 + · · ·+ rncn : r1, r2, · · · , rn ∈ R} is an idealof R.

Exercise 26.6Let I be an ideal of a ring R. Prove that if a ≡ b(mod I) and c ≡ d(mod I)then a+ c ≡ b+ d(mod I) and ac ≡ bd(mod I).

Exercise 26.7Prove that every subring of Z is an ideal of Z.

Exercise 26.8Prove that if R is a commutative ring with unity, and a ∈ R, then (a) is thesmallest ideal of R containing a.

222

Exercise 26.9Let m and n be nonzero integers. Prove that (m) ⊆ (n) if and only if n|m.

Exercise 26.10An element a in a commutative ring is said to be nilpotent if an = 0 forsome positive integer n. Prove that the set of all nilpotent elements in acommutative ring R is an ideal of R.

Exercise 26.11Prove that a nonempty susbet I ofa ring R is an ideal of R if and only if Isatisfies:

(i) if a, b ∈ I then a− b ∈ I;(ii) if r ∈ R and a ∈ I then ar ∈ I and ra ∈ I.

Exercise 26.12Let R be a commutatvie ring with unity and I an ideal of R. Prove that theset of all a ∈ I such that an ∈ I for some n ∈ N is an ideal of R.

Exercise 26.13Prove that an arbitrary intersection of ideals of a ring R is an ideal of R.

Exercise 26.14Find two ideals I1 and I2 of the ring Z such that I1 ∪ I2 is not an ideal.

Exercise 26.15An ideal P of a commutative ring R is called prime ideal if P 6= R and ifa, b ∈ R then either a ∈ P or b ∈ P. Prove that if n is a positive integer then(n) is a prime ideal of Z if and only if n is prime.

Exercise 26.16Prove that if I is an ideal of a ring R then R/I is commutative if and onlyif ab− ba ∈ I for all a, b ∈ R.

Exercise 26.17Prove that if R is commutative and I is an ideal of R then R/I is commuta-tive.

Exercise 26.18Prove that if R has unity e, then I + e is a unity for R/I.

223

Exercise 26.19Assume that R is a commutative ring with unity and P 6= R is an ideal ofR. Prove that R is a prime ideal if and only if R/I is an integral domain.

Exercise 26.20Prove that if I is a subring of a ring R, and the operations of addition andmultiplication on the collection of all right cosets R/I are well-defined thenI is an ideal of R.

224