LECTURE NOTES ON ALGEBRAIC STRUCTURES

RENE E. LEONIDA

Department of Mathematics
College of Natural Sciences and Mathematics
Mindanao State University
General Santos City

June 2017
 TABLE OF CONTENTS

TITLE PAGE 1

1 INTRODUCTION 1
1.1 Equivalence Relation . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 The Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 INTRODUCTION TO GROUPS 12
2.1 Definition of a Group . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 Some Examples of Groups . . . . . . . . . . . . . . . . . . . . 13
2.3 Elementary Properties of Groups . . . . . . . . . . . . . . . . 18
2.4 The Order of an Element . . . . . . . . . . . . . . . . . . . . . 21
2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 PERMUTATION GROUPS 25
3.1 Permutation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.2 Permutation Groups . . . . . . . . . . . . . . . . . . . . . . . 25
3.3 Cycles and Tranpositions . . . . . . . . . . . . . . . . . . . . . 27

4 SUBGROUPS AND NORMAL SUBGROUPS 32

4.1 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.2 Testing Sets for Subgroup . . . . . . . . . . . . . . . . . . . . 33
4.3 Subgroup Generated by a Subset . . . . . . . . . . . . . . . . 38
4.4 Product of Subgroups . . . . . . . . . . . . . . . . . . . . . . . 41
4.5 Cylic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.6 Cosets and Lagrange’s Theorem . . . . . . . . . . . . . . . . . 47
4.7 Normal Subgroups . . . . . . . . . . . . . . . . . . . . . . . . 53
4.8 Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . 55

5 HOMOMORPHISMS AND ISOMORPHISMS OF GROUPS 58

5.1 Homomorphisms of Groups . . . . . . . . . . . . . . . . . . . . 58
5.2 Isomorphism and Correspondence Theorems . . . . . . . . . . 68

BIBLIOGRAPHY 73
 CHAPTER 1

INTRODUCTION

1.1 Equivalence Relation

Definition 1.1.1 Let A and B be nonempty sets. Then R is called a binary

relation from a A into B if R is a subset of A × B.

Example 1.1.2 Consider the set of integers Z. Let R = {(m, n) ∈ Z × Z :

m < n}. Then R is a binary relation on Z.

Example 1.1.3 Let R = {(x, y) : x, y ∈ R, x2 + y 2 = 1}. Then R is a binary

relation on R. R is the set of all points in the Euclidean plane consisting the
circle with center (0, 0) and radius 1.

Let R be a relation from a set A into a set B. If (x, y) ∈ R, then

sometimes we say that x is related to y with respect to R or simply x is
related to y. If A = B, then we speak of a binary relation on A.

Definition 1.1.4 The binary relation ∼ on A is said to be an equivalence

relation on A if
(1) for all a ∈ A, a ∼ a.
(2) for all a, b ∈ A, a ∼ b implies b ∼ a.
(3) for all a, b, c ∈ A, a ∼ b and b ∼ c implies a ∼ c.

The first of these properties is called reflexivity, the second, symmetry,

and the third, transitivity.

Example 1.1.5 Let A be any set. For all a, b ∈ S, define a ∼ b if and only if
a = b. Then ∼ defines an equivalence relation on A.

Proof : (1) Let a ∈ A. Then a = a. Thus, a ∼ a.

(2) Let a, b ∈ A such that a ∼ b. Then a = b. Thus, b = a. Hence,
b ∼ a.
(3) Let a, b, c ∈ A such that a ∼ b and b ∼ c. Then a = b and b = c.
Thus, a = c. Hence, a ∼ c.
Therefore, ∼ defines an equivalence relation on A. 
 2

Example 1.1.6 Let Z be the set of all integers. Given a, b ∈ Z, define a ∼ b

if and only if a − b is an even integer. Then ∼ is an equivalence relation on Z.

Proof : (1) Let a ∈ Z. Since a − a = 0 and 0 is an even integer, it follows that

a − a is an even integer. Thus, a ∼ a.
(2) Let a, b ∈ Z such that a ∼ b. Then a − b is an even integer. Thus,
−(a − b) is an even integer. This implies that b − a is an even integer. Hence,
b ∼ a.
(3) Let a, b, c ∈ Z such that a ∼ b and b ∼ c. Then a − b and b − c
are even integers. Thus, (a − b) + (b − c) is an even integer. This implies that
a − c is an even integer. Hence, a ∼ c.
Therefore, ∼ is an equivalence relation on Z. 

Example 1.1.7 Let n > 1 be a fixed integer. For all a, b ∈ Z, define a ∼ b if

and only if a − b is a multiple of n. Then ∼ is an equivalence relation on Z.

Proof : Note that a − b is a multiple of n if and only if there exists k ∈ Z such

that a − b = nk.
(1) Let a ∈ Z. Since a − a = 0, it follows that there exists 0 ∈ Z such
that a − a = n(0). Thus, a ∼ a.
(2) Let a, b ∈ Z such that a ∼ b. Then there exists k ∈ Z such that
a − b = nk. Thus, b − a = −nk = n(−k). Let l = −k ∈ Z. This implies that
there exist l ∈ Z such that b − a = nl. Hence, b ∼ a.
(3) Let a, b, c ∈ Z such that a ∼ b and b ∼ c. Then there exist k, l ∈ Z
sucht that a−b = nk and b−c = nl. Thus, (a−b)+(b−c) = nk+nl = n(k+l).
Let p = k + l ∈ Z. This implies that there exists p ∈ Z such that a − c = np.
Hence, a ∼ c.
Therefore, ∼ is an equivalence relation on Z. 

Example 1.1.8 For all a, b ∈ R, define a ∼ b if and only if a ≥ b. Then ∼ is

not an equivalence relation on R.

Proof : (1) Consider 2, 1 ∈ R. Then 2 ∼ 1 since 2 ≥ 1. But 1  2 since 1  2.

Thus, there exists 2, 1 ∈ R such that 2 ∼ 1 but 1  2. Therefore, ∼ is not an
equivalence relation on R. 

Example 1.1.9 For all a, b ∈ R, define a ∼ b if and only if a = ±b. Is ∼ an

equivalence relation on R?

Definition 1.1.10 If A is a set and if ∼ is an equivalence relation on A, then

the equivalence class of a ∈ A, denoted by [a], is the set
 3

[a] = {x ∈ A : x ∼ a}.

Example 1.1.11 Let A be any set. For all a, b ∈ A, define a ∼ b if and only
if a = b. Then ∼ defines an equivalence relation on A. Find the equivalence
classes.

Solution: The equivalence class of a ∈ A is

[a] = {x ∈ A : x ∼ a}
= {x ∈ A : x = a}
= {a}.

Hence, there is only one equivalence class, namely [a] = {a}. 

Example 1.1.12 For all a, b ∈ Z, define a ∼ b if and only if a − b is an even

integer. Then ∼ is an equivalence relation on Z. Find the equivalence classes.

Solution: The equivalence class of a ∈ Z is

[a] = {x ∈ Z : x ∼ a}
= {x ∈ Z : x − a is an even integer}
= {x ∈ Z : x − a = 2k, where k ∈ Z}
= {x ∈ Z : x = a + 2k, where k ∈ Z}.

If a = 0, then [0] = {x ∈ Z : x = 2k, where k ∈ Z}.

If a = 1, then [1] = {x ∈ Z : x = 1 + 2k, where k ∈ Z}.
If a = 2, then [2] = {x ∈ Z : x = 2 + 2k, where k ∈ Z} = [0].
If a = 3, then [3] = {x ∈ Z : x = 3 + 2k, where k ∈ Z} = [1].
If a = 4, then [4] = {x ∈ Z : x = 4 + 2k, where k ∈ Z} = [0].

Thus, if a is even, we have [0] = {x ∈ Z : x = 2n, n ∈ Z} and if a is odd, we

have [1] = {x ∈ Z : x = 1 + 2n, n ∈ Z}.

Hence, there are two distinct equivalence classes, namely [0] and [1]. 

Example 1.1.13 Let S = Z and let n > 1 be a fixed integer. For all a, b ∈ S,
define a ∼ b if and only if a − b is a multiple of n. Then ∼ is an equivalence
relation on S = Z. Find the equivalence classes.
 4

Solution: The equivalence class of a ∈ S is

[a] = {x ∈ Z : x ∼ a}
= {x ∈ Z : a − x is a multiple of n}
= {x ∈ Z : x = a + nk, where k ∈ Z}
If a = 0, then [0] = {x ∈ Z : x = nk, where k ∈ Z}.
If a = 1, then [1] = {x ∈ Z : x = 1 + nk, where k ∈ Z}.
If a = 2, then [2] = {x ∈ Z : x = 2 + nk, where k ∈ Z}.
If a = 3, then [3] = {x ∈ Z : x = 3 + nk, where k ∈ Z}.

If a = n − 1, then [n − 1] = {x ∈ Z : x = (n − 1) + nk, where k ∈ Z}.

If a = n, then [n] = {x ∈ Z : x = n) + nk, where k ∈ Z} = [0].

Hence, there are n distinct equivalence classes, namely [0], [1], [2],...,[n − 1]. 
Theorem 1.1.14 Let ∼ be an equivalence relation on the set A. Then
(i) for all x ∈ A, [x] 6= ∅.
(ii) if y ∈ [x], then [x] = [y], where x, y ∈ A.
(iii) for all[x, y ∈ A, either [x] = [y] or [x] ∩ [y] = ∅.
(iv) A = [x], that is, A is the union of all equivalence classes with
x∈A
respect to ∼.
Proof : (i) Let x ∈ A. By Definition 1.1.4(1), x ∼ x. Thus, by Definition
1.1.10, x ∈ [x]. Hence, [x] 6= ∅.

(ii) Let x, y ∈ A such that y ∈ [x]. Then y ∼ x and by symmetry, x ∼ y.

We will show that [x] ⊆ [y] and [y] ⊆ [x]. Let u ∈ [y]. Then u ∼ y. Since
y ∼ x, by transitivity, u ∼ x. Hence, u ∈ [x], which implies that [y] ⊆ [x].
Similarly, [x] ⊆ [y]. Therefore, [x] = [y].

(iii) Let x, y ∈ A. If [x] ∩ [y] = ∅, then we are done. Assume that

[x] ∩ [y] 6= ∅. Then there exists u ∈ [x] ∩ [y]. Thus, u ∈ [x] and u ∈ [y].
This implies that u ∼ x and u ∼ y and by symmetry applied to u ∼ x, we
have x ∼ u and u ∼ y. By transitivity, x ∼ y. Hence, x ∈ [y]. By (ii), [x] = [y].

(iv) Let x ∈ A. By Definition [ 1.1.4(1), x ∼ x. Thus, by Definition

S
1.1.10, x ∈ [x]. This implies that x ∈ [x]. Hence, A ⊆ x∈A [x]. But, for
[ x∈A [
all x ∈ A, [x] ⊆ A. Thus, [x] ⊆ A. Therefore, A = [x]. 
x∈A x∈A
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1.2 Mappings

Definition 1.2.1 Let S and T be sets. A mapping from S to T is a subset

σ of S × T such that for every s ∈ S there exists a unique t ∈ T such that
(a, b) ∈ σ.

The notation σ : S → T indicates that σ is a mapping from S to

T . If (s, t) ∈ σ, it is customary to write t = σ(s). We often refer to t as the
image of s under σ.

Definition 1.2.2 A mapping σ : S → T is said to be injective or one-one if

for all s1 , s2 ∈ S sucht that s1 6= s2 , then σ(s1 ) 6= σ(s2 ).

Equivalently, a mapping σ is injective if and only if for all s1 , s2 ∈ S

such that σ(s1 ) = σ(s2 ), then s1 = s2 .

Definition 1.2.3 A mapping σ : S → T is said to be surjective or onto T , if

for every t ∈ T there exists s ∈ S such that σ(s) = t.

If we call the subset σ(S) = {y ∈ T : y = σ(s) for some s ∈ S} the

image of S under σ. then σ is onto if σ(S) = T .

Definition 1.2.4 The two mappings σ and τ of S into T are said to be equal
if σ(s) = τ (s) for every s ∈ S.

Definition 1.2.5 If σ : S → T and τ : T → U , then the composition of σ

and τ is the mapping τ ◦ σ : S → U defined by (τ ◦ σ)(s) = τ (σ(s)) for every
s ∈ S.

Example 1.2.6 Let S = {a, b, c}, let σ : S → S be defined by

σ(a) = b, σ(b) = c, σ(c) = a,

and let τ : S → S be defined by

τ (a) = a, τ (b) = c, τ (c) = b.

Then
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(τ ◦ σ)(a) = τ (σ(a)) = τ (b) = c,

(τ ◦ σ)(b) = τ (σ(b)) = τ (c) = b,
(τ ◦ σ)(c) = τ (σ(c)) = τ (a) = a.

and

(σ ◦ τ )(a) = σ(τ (a)) = σ(a) = b,

(σ ◦ τ )(b) = σ(τ (b)) = σ(c) = a,
(σ ◦ τ )(c) = σ(τ (c)) = σ(b) = c.

Note that τ ◦ σ 6= σ ◦ τ .

Lemma 1.2.7 If σ : S → T and τ : T → U , and µ : U → V , then (µ◦τ )◦σ =

µ ◦ (τ ◦ σ).

Proof : Let s ∈ S. Then

((µ ◦ τ ) ◦ σ)(s) = (µ ◦ τ )(σ(s)) = µ(τ (σ(s)))

and

(µ ◦ (τ ◦ σ))(s) = µ((τ ◦ σ)(s)) = µ(τ (σ(s))).

Therefore, (µ ◦ τ ) ◦ σ = µ ◦ (τ ◦ σ). 

Definition 1.2.8 If S is a nonempty set, then A(S) is the set of all one-one
mappings of S onto itself.

Theorem 1.2.9 Let A(S) be the set of all one-one mappings of S onto itself.
Then
1. For all σ, τ ∈ A(S), τ ◦ σ ∈ A(S).
2. For all σ, τ, µ ∈ A(S), (µ ◦ τ ) ◦ σ = µ ◦ (τ ◦ σ).
3. There exists ι ∈ A(S) such that σ ◦ ι = ι ◦ σ = σ, for all σ ∈ A(S).
4. For all σ ∈ A(S), there exists σ −1 ∈ A(S) such that σ ◦ σ −1 =
σ −1 ◦ σ = ι.

Example 1.2.10 If S = {a, b, c}, find A(S). Construct a multiplication table.

 7

1.3 The Integers

In this section, we make the assumption that all symbols, written as

lowercase Latin letters will be integers.

Theorem 1.3.1 (Division Algorithm) Let a, b ∈ Z with b 6= 0. Then there

exists unique integers q and r such that a = qb + r, where 0 ≤ r ≤ |b|.

Proof : See [6], page 10.

Corollary 1.3.2 For any two integers a and b with b > 0, there exist unique
integers q and r such that a = qb + r, where 0 ≤ r < b.

Proof : Let a, b ∈ Z with b > 0. By Theorem 1.3.1, there exists unique integers
q and r such that a = qb + r, where 0 ≤ r ≤ |b|. Since b > 0, it follows that
|b| = b. Hence, a = qb + r, where 0 ≤ r < b. 

Definition 1.3.3 Let a, b ∈ Z with b 6= 0. Then b is said to divide a or b is a

divisor of a, written b|a, if there exists q ∈ Z such that a = bq. When b does
not divide a, we sometimes write b - a.

Example 1.3.4 1. 2|10 since there exists 5 ∈ Z such that 10 = 5 · 2.

2. 3|(−6) since there exists −2 ∈ Z such that −6 = −2 · 3.

3. 4 - 9 since for all q ∈ Z, 9 6= q · 4.

4. 8 - 4 since for all q ∈ Z, 4 6= q · 8.

Lemma 1.3.5 Let a, b, c ∈ Z.

1. If a|1, then a = ±1.
2. If a|b and b|a, then a = ±b.
3. If a|b and a|c, then a|(bm + cn) for all m, n ∈ Z.

Proof : 1. Let a|1. Then there exists q ∈ Z such that aq = 1. Thus, aq = (1)(1)
or aq = (−1)(−1). Hence, a = 1 or a = −1.

2. Let a|b and b|a. Then there exist p, q ∈ Z such that b = ap and
a = bq. Thus, a · 1 = a = bq = (ap)q = a(pq). This implies that pq = 1.
Hence, q|1. By 1, q = ±1. Therefore, a = ±b.
 8

3. Let a|b and a|c. Then there exist p, q ∈ Z such that b = ap and
c = aq. Thus, bm = apm and cn = aqn, for all m, n ∈ Z. Hence, bm + cn =
apm + aqn = a(pm + qn), for all m, n ∈ Z. Therefore, a|(bm + cn) for all
m, n ∈ Z. 
Definition 1.3.6 The positive integer c is said to be a greatest common
divisor of a and b if
1. c|a and c|b.
2. for all c0 ∈ Z, c0 |a and c0 |b, then c0 |c.
We shall use the notation (a, b) for the greatest common divisor of a and
b. Since the greatest common divisor is positive, we have (a, b) = (a, −b) =
(−a, b) = (−a, −b).

Example 1.3.7 (2, 8) = 2, (9, 12) = 3, (−4, 18) = 2, (−12, −28) = 4.

Lemma 1.3.8 If a greatest common divisor of a and b exists, then it must be
unique.
Proof : Suppose c1 and c2 are greatest common divisors a and b. Then c1 |c2
and c2 |c1 . By Lemma 1.3.5(2), c1 = ±c2 . But a greatest common divisor must
be positive. Hence, c1 = c2 . Therefore, the greatest common divisor of a and
b is unique. 
Lemma 1.3.9 If a and b are integers, not both zero, then (a, b) exists; moreover,
we can find integers m0 and n0 such that (a, b) = m0 a + n0 b.
Proof : Let M = {ma + nb ∈ Z : m, n ∈ Z}. Since one of a or b is not 0, M
contains nonzero integers. Since x = ma+nb ∈ M , −x = (−m)a+(−n)b ∈ M .
So, M contains positive integers. Moreover, M contains a smallest positive
integer, say c and it has the form c = m0 a + n0 b. We claim that c = (a, b).

1. If x = ma + nb ∈ M , then by the Euclidean algorithm, x = tc + r,

where 0 ≤ r < c. Thus,
r = x − tc = ma + nb − t(m0 a + n0 b) = (m − tm0 )a + (n − tn0 )b,
which means that r ∈ M . Since 0 ≤ r and r < c, by the choice of c, we
must have r = 0. Hence, x = tc, which shows that c|x for any x ∈ M . But
a = 1a + 0b ∈ M and b = 0a + 1b ∈ M . Therefore, c|a and c|b.
2. If c0 is an integer such that c0 |a and c0 |b, then c0 |(m0 a + n0 b). Hence,
c0 |c.
Therefore, c = (a, b). 
 9

Definition 1.3.10 The integers a and b are relatively prime if (a, b) = 1.

Corollary 1.3.11 If a and b are relatively prime, we can find integers m and
n such that ma + nb = 1.

Definition 1.3.12 The integer p > 1 is a prime number if its only divisors
are ±1, ±p.

Another way of putting this is to say that an integer p > 1 is a prime

number if and only if given any integer n then either (p, n) = 1 or p|n.

Lemma 1.3.13 If a is relatively prime to b but a|bc, then a|c.

Proof : Suppose that a is relatively prime to b. By Corollary 1.3.11, we can

find integers m and n such that ma + nb = 1. Thus, mac + nbc = c. Note that
a|mac and since a|bc, a|nbc. Hence, a|(mac + nbc). Therefore, a|c. 

Corollary 1.3.14 If a prime number divides the product of certain integers,

then it must divide at least one of these integers.

Theorem 1.3.15 Any positive integer a > 1 can be factored in a unique way
as a = pα1 1 pα2 2 · · · pαt t , where p1 > p2 > · · · > pt are prime numbers and where
each αi > 0.

Definition 1.3.16 Let n > 0 be a fixed integer. We define a ≡ b mod n if

n|(a − b).

The relation is referred to as congruence modulo n, n is called the

modulus of the relation, and we read a ≡ b mod n as ”a is congruent to b
modulo n.”

For example, 9 ≡ 5 mod 2, 21 ≡ −9 mod 10, −17 ≡ 8 mod 5.

Lemma 1.3.17 .
1. The relation congruence modulo n defines an equivalence relation on
the set of integers.
2. This equivalence relation has n distinct equivalence classes.
3. If a ≡ b mod n and c ≡ d mod n, then (a + c) ≡ (b + d) mod n and
ac ≡ bd mod n.
4. If ab ≡ ac mod n and (a, n) = 1, then b ≡ c mod n.
 10

Proof : 1. (i) Let a be an integer. Since a − a = 0 and n|0, it follows that

n|(a − a). Thus, a ≡ a mod n.
(ii) Let a, b ∈ Z such that a ≡ b mod n. Then n|(a − b). Thus,
n|[−(a − b)]. This implies that n|(b − a). Hence, b ≡ a mod n.
(iii) Let a, b, c ∈ Z such that a ≡ b mod n and b ≡ c mod n. Then
n|(a − b) and n|(b − c). Thus, n|[(a − b) + (b − c)]. This implies that n|(a − c).
Hence, a ≡ c mod n.
Therefore, the relation congruence modulo n defines an equivalence
relation on the set of integers.

2. Let the equivalence class, under this relation, of a be denoted by

[a]; we call it the the congruence class (mod n) of a. Given any integer a,
by the Euclidean algorithm, a = kn + r, where 0 ≤ r < n. This implies that
a ∈ [r] and by Theorem 1.1.14(ii), [a] = [r]. Thus, there are at most n distinct
congruence classes; namely, [0], [1], [2],...,[n − 1]. These congruence classes are
distinct, for if [i] = [j] with, say 0 ≤ i < j < n, then n|(j − i), where j − i is a
positive integer less than n which is impossible. Therefore, there are exactly
n distinct congruence classes: [0], [1], [2],...,[n − 1].

3. Suppose that a ≡ b mod n and c ≡ d mod n. Then n|(a − b)

and n|(c − d). Thus, n|[(a − b) + (c − d)] and n|[(a − b)c + (c − d)b], that is,
n|[(a + c) − (b + d)] and n|(ac − bd). Hence, (a + c) ≡ (b + d) mod n and
ac ≡ bd mod n.
4. Suppose that ab ≡ ac mod n. Then n|(ab − ac), that is, n|a(b − c).
Since (a, n) = 1, by Lemma 1.3.13, n|(b − c). Hence, b ≡ c mod n. 

Let Zn be the set of the congruence classes mod n; that is,

Zn = {[0], [1], [2], ..., [n − 1]}.

Given two elements, [a] and [b] in Zn , let us define

[a] +n [b] = [a + b] ( or [a] + [b] mod n = [a + b]);

[a] ·n [b] = [ab] (or [a] · [b] mod n = [ab].

These operations are called ”addition modulo n” and ”multiplication modulo

n”.

Lemma 1.3.18 The operations +n and ·n used in Zn are well-defined.

 11

Proof : 1. Let [a], [b] ∈ Zn . Then [a] +n [b] = [a + b] ∈ Zn . Next, let

[a], [b], [c], [d] ∈ Zn such that [a] = [c] and [b] = [d]. Then a ≡ c mod n
and b ≡ d mod n. Thus, there exists s, t ∈ Z such that a − c = ns and
b − d = nt, which means that

(a + b) − (c + d) = (a − c) + (b − d) = ns + nt = n(s + t).

This implies that (a + b) ≡ (c + d) mod n, that is, [a + b] = [c + d]. Hence,

[a] +n [b] = [c] +n [d]. This shows that +n is well-defined.

2. Let [a], [b] ∈ Zn . Then [a] ·n [b] = [ab] ∈ Zn . Next, let [a], [b], [c], [d] ∈
Zn such that [a] = [c] and [b] = [d]. Then a ≡ c mod n and b ≡ d mod n.
Thus, there exists s, t ∈ Z such that a − c = ns and b − d = nt, which means
that

ab − cd = ab − bc + bc − cd = b(a − c) + c(b − d) = bns + cnt = n(bs + ct).

This implies that (ab) ≡ (cd) mod n, that is, [ab] = [cd]. Hence, [a] ·n [b] =
[c] ·n [d]. This shows that ·n is well-defined. 

These operations in Zn have the following properties:

Lemma 1.3.19 Let [a], [b], [c] in Zn . Then

(1) [a] +n [b] = [b] +n [a]
(2) [a] ·n [b] = [b] ·n [a]
(3) ([a] +n [b]) +n [c] = [a] +n ([b] +n [c])
(4) ([a] ·n [b]) ·n [c] = [a] ·n ([b] ·n [c])
(5) [a] ·n ([b] +n [c]) = [a] ·n [b] +n [a] ·n [c]
(6) [0] +n [a] = [a]
(7) [1] ·n [a] = [a].

The set Zn plays an important role in algebra and number theory. It

is called the set of integers mod n.
 CHAPTER 2

INTRODUCTION TO GROUPS

2.1 Definition of a Group

Definition 2.1.1 A group is an ordered pair (G, ∗), where G is a nonempty

set and ∗ is a binary operation on G such that the following properties hold:
(G1) For all a, b, c ∈ G, a ∗ (b ∗ c) = (a ∗ b) ∗ c.
(G2) There exists e ∈ G such that for all a ∈ G, a ∗ e = a and e ∗ a = a.
(G3) For all a ∈ G, there exists a−1 ∈ G such that a ∗ a−1 = e and
−1
a ∗ a = e.

Theorem 2.1.2 Let (G, ∗) be a group.

(i) There exists a unique element e ∈ G such that for all a ∈ G, e∗a = a
and a ∗ e = a.
(ii) For all a ∈ G, there exists a unique a−1 ∈ G such that a−1 ∗ a = e
and a ∗ a−1 = e.

Proof : (i) Let all a ∈ G. By (G2), there exists e ∈ G such that e ∗ a = a and
a ∗ e = a. Suppose that there exists e0 ∈ G such that e0 ∗ a = a and a ∗ e0 = a.
Then, in particular, e ∗ e0 = e0 and e = e ∗ e0 . Thus, e = e ∗ e0 = e0 . This shows
that the identity element e is unique.
(ii) Let a ∈ G. By (G3), there exists a−1 ∈ G such that a−1 ∗ a = e
and a ∗ a−1 = e. Suppose that there exists a−1 −1
1 ∈ G such that a1 ∗ a = e and
a ∗ a−1
1 . Then, a
−1
= a−1 ∗ e = a−1 ∗ (a ∗ a−1
1 ) = (a
−1
∗ a) ∗ a−1 −1
1 = e ∗ a1 = a1 .
−1

Hence, a−1 is unique. 

The unique element e ∈ G that satisfies (G2) is called the identity

element of the group (G, ∗). Let a ∈ G. Then the unique element a−1 ∈ G
that satisfy (G3) is called the inverse of a.

Definition 2.1.3 Let (G, ∗) be a group. If for every a, b ∈ G, a ∗ b = b ∗ a,

then (G, ∗) is called an abelian or commutative group. A group (G, ∗) is called
nonabelian if it is not abelian.
 13

2.2 Some Examples of Groups

An ordered pair (G, ∗) is a group if and only if the following properties hold:
(i) G is a nonempty set.
(ii) For all a, b ∈ G, a ∗ b ∈ G.
(G1) For all a, b, c ∈ G, a ∗ (b ∗ c) = (a ∗ b) ∗ c.
(G2) There exists e ∈ G such that for all a ∈ G, a ∗ e = a and e ∗ a = a.
(G3) For all a ∈ G, there exists a−1 ∈ G such that a ∗ a−1 = e and
−1
a ∗ a = e.

Example 2.2.1 Consider Z, the set of integers, together with the operation
+, the usual addition. Then (Z, +) is an abelian group.

Proof : (i) We know that 0 ∈ Z. Thus, Z is nonempty.

(ii) Let a, b ∈ Z. Then a + b ∈ Z.

(G1) Let a, b, c ∈ Z. Then (a + b) + c = a + (b + c).

(G2) There exists 0 ∈ Z such that for all a ∈ Z, a+0 = a and 0+a = a.

(G3) Let a ∈ Z. Then there exists −a ∈ Z such that a + (−a) = 0 and

−a + a = 0.

Let a, b ∈ Z. Then a + b = b + a.

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

Example 2.2.2 As in Example 2.2.1, (Q, +) and (R, +) are also abelian
groups, where + is the usual addition.

Example 2.2.3 Consider Q\{0}, the set of nonzero rational numbers, together
with the binary operation ·, the usual multiplication. Then (Q\{0}, ·) is an
abelian group.

Proof : (i) We know that 1 ∈ Q\{0}. Thus, Q\{0} is nonempty.

(ii) Let a, b ∈ Q\{0}. Then a · b ∈ Q\{0}.

 14

(G1) Let a, b, c ∈ Q\{0}. Then (a · b) · c = a · (b · c).

(G2) There exists 1 ∈ Q\{0} such that for all a ∈ Q\{0}, a · 1 = a and
1 · a = a.

1
(G3) Let a ∈ Q\{0}. Then a 6= 0. Thus, a
6= 0. Hence, there exists
1
a
∈ Q\{0} such that a · a1 = 1 and a1 · a = 1.

Let a, b ∈ Q\{0}. Then a · b = b · a.

Therefore, (Q\{0}, ·) is an abelian group. 

Example 2.2.4 As in Example 2.2.3, (R\{0}, ·) is an abelian group, where ·

is the usual multiplication.

Example 2.2.5 Let n be a fixed integer. Consider Zn , the set of integers

modulo n, together with the binary operation +n . Then (Zn , +n ) is an abelian
group.

Proof : (i) By Lemma 1.3.17, [0] ∈ Zn . Thus, Zn is nonempty.

(ii) By Lemma 1.3.18, +n in Zn is well-defined. Thus, for all [a], [b] ∈

Zn , [a] +n [b] ∈ Zn .

(G1) Let [a], [b], [c] ∈ Zn . Then a, b, c ∈ Z. Thus (a + b) + c = a + (b + c)

since (Z, +) is a group. Hence,

([a] +n [b]) +n [c] = [a + b] +n [c]

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

(G2) There exists [0] ∈ Zn such that for all [a] ∈ Zn ,

[a] +n [0] = [a] and [0] +n [a] = [a].

(G3) Let [a] ∈ Zn . Then a ∈ Z. Thus, −a ∈ Z such that a + (−a) = 0

and −a + a = 0. Since −[a] = [−a], it follows that −[a] ∈ Zn . Hence, for all
[a] ∈ Zn , there exists −[a] = [−a] ∈ Zn such that
 15

[a] +n (−[a]) = [a] +n [−a] = [a + (−a)] = [0]

and

−[a] +n [a] = [−a] +n [a] = [−a + a] = [0].

Finally, let [a], [b] ∈ Zn . Then a, b ∈ Z. Thus, a + b = b + a since (Z, +)

is an abelian group. Hence,

[a] +n [b] = [a + b] = [b + a] = [b] +n [a].

Therefore, (Zn , +n ) is an abelian group. 

Example 2.2.6 Let E = {2n : n ∈ Z}. Then (E, +), where + is the usual
addition, is an abelian group.

Proof : (i) Since 0 ∈ Z, it follows that 2(0) ∈ E. Thus, E is nonempty.

(ii) Let 2n1 , 2n2 ∈ E. Then n1 , n2 ∈ Z. Thus, n1 + n2 ∈ Z. Hence,

2(n1 + n2 ) ∈ E. But, 2n1 + 2n2 = 2(n1 + n2 ). Therefore, 2n1 + 2n2 ∈ E.

(G1) Let 2n1 , 2n2 , 2n3 ∈ E. Then n1 , n2 , n3 ∈ Z. Thus, (n1 + n2 ) + n3 =

n1 + (n2 + n3 ). Hence,

(2n1 + 2n2 ) + 2n3 = 2(n1 + n2 ) + 2n3

= 2[(n1 + n2 ) + n3 ]
= 2[n1 + (n2 + n3 )]
= 2n1 + 2(n2 + n3 )
= 2n1 + (2n2 + 2n3 ).

(G2) There exists 2(0) ∈ E such that for all 2n ∈ E,

2n + 2(0) = 2(n + 0) = 2n and 2(0) + 2n(0 + n) = 2n.

(G3) Let 2n ∈ E. Then n ∈ Z. Thus, −n ∈ Z. Hence, 2(−n) ∈ E.

Therefore, there exists −(2n) = 2(−n) ∈ E such that

2n + (−(2n)) = 2n + 2(−n) = 2(n + (−n)) = 2(0)

and

−(2n) + 2n = 2(−n) + 2n = 2(−n + n) = 2(0).

 16

Finally, let 2n1 , 2n2 ∈ E. Then n1 , n2 ∈ Z. Thus, n1 + n2 = n2 + n1 .

Hence,
2n1 + 2n2 = 2(n1 + n2 ) = 2(n2 + n1 ) = 2n2 + 2n1 .
Therefore, (E, +) is an abelian group. 

Example 2.2.7 Let n > 1 be an integer. Define Un = {a ∈ Z : 0 <

a < n, (a, n) = 1}. Show that (Un , ·n ) is an abelian group. Note that
U7 = {1, 2, 3, 4, 5, 6} and U8 = {1, 3, 5, 7}.
√ √ √ √
Example
√ 2.2.8 Let Q[ 2] = {a + b 2|a, b ∈ Q}. For all a + b 2, c + d 2∈
Q[ 2], define
√ √ √
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2.
√
Then (Q[ 2], +) is an abelian group.
√ √ √
Proof : (i) Since 0 ∈ Q, it follows that 0 + 0 2 ∈ Q[ 2]. Thus, Q[ 2] is
nonempty.
√ √ √
(ii) Let a + b 2, c + d 2 ∈ Q[ 2]. √Then a,√ b, c, d ∈ Q. Thus,
a + c,√b + d ∈ Q. √ Hence, (a + c) +√(b + d) 2 ∈ Q[ 2]. √ By definition,
√
(a +
√ b 2) + (c + d 2) = (a + c) + (b + d) 2. Therefore, (a + b 2) + (c + d 2) ∈
Q[ 2].
√ √ √ √
(G1) Let a + b 2, c + d 2, e + f 2 ∈ Q[ 2]. Then a, b, c, d, e, f ∈ Q.
Thus, (a + c) + e = a + (c + e) and (b + d) + f = b + (d + f ). Hence,
√ √ √ √ √
((a + b 2) + (c + d 2)) + (e + f 2) = ((a + c) + (b + d) 2) + (e + f 2)
√
= ((a + c) + e) + ((b + d) + f ) 2
√
= (a + (c + e)) + (b + (d + f )) 2
√ √
= (a + b 2) + ((c + e) + (d + f ) 2)
√ √ √
= (a + b 2) + ((c + e 2) + (d + f 2)).
√ √ √ √
(G2) There exists 0 + 0 2 ∈ Q[ 2] such that for all a + b 2 ∈ Q[ 2],
√ √ √ √
(a + b 2) + (0 + 0 2) = (a + 0) + (b + 0) 2 = a + b 2
and
√ √ √ √
(0 + 0 2) + (a + b 2) = (0 + a) + (0 + b) 2 = a + b 2.
 17

√ √
(G3) Let a √+ b 2 ∈√ Q[ 2]. Then a, b ∈ Q. Thus, √ −a, −b ∈ Q.
Hence,
√ (−a) +
√ (−b) 2 ∈ Q[ 2]. Therefore, there exists −(a + b 2) = (−a) +
(−b) 2 ∈ Q[ 2] such that
√ √ √ √ √
(a + b 2) + (−(a + b 2)) = (a + b 2) + ((−a) + (−b) 2) = (0 + 0 2).
and
√ √ √ √ √
−(a + b 2) + (a + b 2) = ((−a) + (−b) 2) + (a + b 2) = (0 + 0 2).
√ √ √
Let a + b 2, c + d 2 ∈ Q[ 2]. Then a, b, c, d ∈ Q. Thus, a + c = c + a
and b + d = d + b. Hence
√ √ √ √
(a + b√ 2) + (c + √
d 2) = (a + c) + (b + d) 2 = (c + a) + (d + b) 2 =
(c + d 2) + (a + b 2).
√
Therefore, (Q[ 2], +) is an abelian group. 

Example 2.2.9 Consider Sn , the set of all one-one mappings of S onto S,

together with the binary operation ◦. Then Sn , ◦) is a nonabelian group.

If the set S contains n elements, then the group Sn has n! elements. This
highly important example will be called the symmetriuc group of degree n.

Example 2.2.10 Let S = {a, b, c} and let S3 = {ι, σ, τ, µ, δ, }, where

ι(a) = a, ι(b) = b, ι(c) = c; σ(a) = a, σ(b) = c, σ(c) = b;

τ (a) = c, τ (b) = b, τ (c) = a; µ(a) = b, µ(b) = a, µ(c) = c;

δ(a) = b, δ(b) = c, δ(c) = a; (a) = c, (b) = a, (c) = b.

Then (S3 , ◦) is a nonabelian group with 6 elements. Construct a multiplication

table for (S3 , ◦).

Example 2.2.11 Let

  
a b
GL(2, R) = : a, b, c, d ∈ R, ad − bc 6= 0 .
c d
Define a binary operation ∗ on GL(2, R) by
     
a b u v au + bw av + bs
∗ =
c d w s cu + dw cv + ds
 18

for all
   
a b u v
, ∈ GL(2, R).
c d w s
This binary operation is the usual matrix multiplication. Then (GL(2, R), ∗)
is a nonabelian group. This group is known as the general linear group of
degree 2.

Example 2.2.12 Let (Z, ∗), where a ∗ b = a − b for all a, b ∈ Z. Then (Z, ∗)
is not a group.

Proof : Consider 0, 1, 2 ∈ Z. Then (0 ∗ 1) ∗ 2 = (0 − 1) − 2 = −3 and 0 ∗ (1 ∗ 2) =

0 − (1 − 2) = 1. Thus, there exist 0, 1, 2 ∈ Z such that (0 ∗ 1) ∗ 2 6= 0 ∗ (1 ∗ 2).
Therefore, (Z, ∗) is not a group. 

Example 2.2.13 Consider the ordered pair (Z+ , ∗), where a ∗ b = ab for all
a, b ∈ Z+ . Then (Z+ , ∗) is not a group.

Proof : There exists 1 ∈ Z+ such that a ∗ 1 = a1 = a and 1 ∗ a = 1a = a for

all a ∈ Z+ . Now, consider 2 ∈ Z+ . Suppose there exists 2−1 ∈ Z+ such that
2−1 ∗ 2 = 1 and 2 ∗ 2−1 = 1. Then 2−1 · 2 = 1 and 2 · 2−1 = 1. Thus, 2−1 = 21 .
This is a contradiction since 12 6= Z+ . Hence, 2−1 does not exist. Therefore,
(Z+ , ∗) is not a group. 

2.3 Elementary Properties of Groups

Theorem 2.3.1 Let (G, ∗) be a group.

(i) (a−1 )−1 = a for all a ∈ G.
(ii) (a ∗ b)−1 = b−1 ∗ a−1 for all a, b ∈ G.
(iii) (Cancellation Law) For all a, b, c ∈ G, if either a ∗ c = b ∗ c or
c ∗ a = c ∗ b, then a = b.
(iv) For all a, b ∈ G, the equations a ∗ x = b and y ∗ a = b have unique
solutions in G for x and y.

Proof : (i) let a ∈ G. By (G2), there exists a−1 ∈ G such that a−1 ∗ a = e.
Since a−1 ∈ G, by (G2), there exists (a−1 )−1 ∈ G such that a−1 ∗ (a−1 )−1 = e.
Hence,
(a−1 )−1 = (a−1 )−1 ∗ e = (a−1 )−1 ∗ (a−1 ∗ a) = ((a−1 )−1 ∗ a−1 ) ∗ a = e ∗ a = a.
 19

(ii) Let a, b ∈ G. Then a ∗ b ∈ G. By (G2), there exist (a ∗ b)−1 ∈ G

such that (a ∗ b)−1 ∗ (a ∗ b) = e. Now,
(a ∗ b) ∗ (b−1 ∗ a−1 ) = ((a ∗ b) ∗ b−1 ) ∗ a−1
= (a ∗ (b ∗ b−1 )) ∗ a−1
= (a ∗ e) ∗ a−1
= a ∗ a−1
= e.
Hence,
(a ∗ b)−1 = (a ∗ b)−1 ∗ e
= (a ∗ b)−1 ∗ [(a ∗ b) ∗ (b−1 ∗ a−1 )]
= [(a ∗ b)−1 ∗ (a ∗ b)] ∗ (b−1 ∗ a−1 )
= e ∗ (b−1 ∗ a−1 )
= b−1 ∗ a−1 .
(iii) Let a, b, c ∈ G. Suppose that a ∗ c = b ∗ c. Then
a = a ∗ e = a ∗ (c ∗ c−1 )
= (a ∗ c) ∗ c−1
= (b ∗ c) ∗ c−1
= b ∗ (c ∗ c−1 )
=b∗e
= b.
Similarly, if c ∗ a = c ∗ b, we can show that a = b.

(iv) Let a, b ∈ G. Suppose that a ∗ x = b. Note that a−1 ∗ b ∈ G.

Substituting a−1 ∗ b for x in the equation a ∗ x = b, we get
a ∗ (a−1 ∗ b) = (a ∗ a−1 ) ∗ b = e ∗ b = b.
This implies that a−1 ∗ b is a solution of the equation a ∗ x = b.
Now, suppose that c is any solution of a ∗ x = b. Then a ∗ c = b. Hence,
c = e ∗ c = (a−1 ∗ a) ∗ c
= a−1 ∗ (a ∗ c)
= a−1 ∗ b.
This shows that the solution is unique. Similar arguments holds for the
equation y ∗ a = b. 
 20

Corollary 2.3.2 Let (G, ∗) be a group and a ∈ G. If a ∗ a = a, then a = e.

Proof : Let a ∈ G. Suppose that a ∗ a = a. By (G2), there exists e ∈ G such
that a ∗ e = a. Thus, a ∗ a = a ∗ e. By the cancellation law, we have a = e. 
Example 2.3.3 Show that is every element of a group (G, ∗) is its own inverse,
then (G, ∗) is abelian.
Proof : Let a, b ∈ G. Then a ∗ b ∈ G. Thus, a−1 = a, b−1 = b, and (a ∗ b)−1 =
a ∗ b. Hence,
a ∗ b = (a ∗ b)−1 = b−1 ∗ a−1 = b ∗ a.
Therefore, (G, ∗) is abelian. 

Let (G, ∗) be a group, and a, b, c ∈ G. By the associative law, a∗(b∗c) =

(a ∗ b) ∗ c. Hence, we can define
a ∗ b ∗ c = a ∗ (b ∗ c) = (a ∗ b) ∗ c.
Let a, b, c ∈ G. Then
a ∗ b ∗ c ∗ d = (a ∗ b ∗ c) ∗ d
= (a ∗ (b ∗ c)) ∗ d
= a ∗ ((b ∗ c) ∗ d)
= a ∗ (b ∗ (c ∗ d))
= (a ∗ b) ∗ (c ∗ d)
= ((a ∗ b) ∗ c) ∗ d.

Let (G, ∗) be a group, and n ∈ Z. we define the integral power an as

follows:

a0 = e
an = a ∗ an−1 if n > 0
an = (a−1 )−n if n < 0.

In additive notatiion, we would have: Let (G, ∗) be a group, and n ∈ Z.

we define na as follows:

0a = e
na = a + (n − 1)a if n > 0
na = −n(−a) if n < 0.
 21

Example 2.3.4 If (G, ∗) is a group such that (a ∗ b)2 = a2 ∗ b2 for all a, b ∈ G,

then (G, ∗) is abelian.

Proof : Let a, b ∈ G such that (a∗b)2 = a2 ∗b2 . Then (a∗b)∗(a∗b) = (a∗a)∗(b∗b).

By (G1), we have a∗(b∗a)∗b = a∗(a∗b)∗b. By the cancellation law, b∗a = a∗b.
Hence, (G, ∗) is abelian. 

Definition 2.3.5 A group (G, ∗) is called a finite group if G has only a finite
number of elements. The order of a group (G, ∗), written |G|, is the number
of elements of G.

Example 2.3.6 The groups (Zn , +n ) and (S3 , ◦) are finite groups.

Definition 2.3.7 A group with an infinite number of elements is called an

infinite group.

Example 2.3.8 The groups (Q, +),(R, +), (Q\{0}, ·), and (R\{0}, ·) are infinte
groups.

2.4 The Order of an Element

Let G be a finite group and a ∈ G. Then a2 = a ∗ a and by induction,

we can show that am ∈ G for all m ≥ 1. Thus, {a, a2 , ..., am , ...} ⊆ G. Since
G is finite, all elements of the set {a, a2 , ..., am , ...} cannot be distinct. Hence,
there exists positive integer p, q with p > q such that ap = aq . This implies
that ap−q = e, where p − q > 0. Write n = p − q > 0. Therefore, an = e for
some positive integer n.
Also, if G is an infinite group and a ∈ G, then it may still possible that
n
a = e for some positive integer n.

Definition 2.4.1 Let (G, ∗) be a group and a ∈ G. If there exists a positive

integer n such that an = e, then the smallest positive integer m such that
am = e is called the order of a. If no such positive integer n exists, then we
say that a is of infinite order.

Notation We denote the order of an element a of a group (G, ∗) by ◦(a).

 22

Remark 2.4.2 If (G, ∗) is a group, then ◦(e) = 1.

Example 2.4.3 Consider the group (S3 , ◦). Then ◦(σ) = ◦(τ ) = ◦(µ) = 2
and ◦(δ) = ◦() = 3.

Proof :
σ 1 = σ, σ 2 = σ ◦ σ = ι, σ 3 = σ ◦ σ 2 = σ ◦ ι = σ, σ 4 = σ ◦ σ 3 = σ ◦ σ = ι;

τ 1 = τ , τ 2 = τ ◦ τ = ι, τ 3 = τ ◦ τ 2 = τ ◦ ι = τ , τ 4 = τ ◦ τ 3 = τ ◦ τ = ι;

µ1 = µ, µ2 = µ ◦ µ = ι, µ3 = µ ◦ µ2 = µ ◦ ι = µ, µ4 = µ ◦ µ3 = µ ◦ µ = ι;

δ 2 = , δ 3 = δ ◦ δ 2 = δ ◦  = ι, δ 4 = δ ◦ δ 3 = δ, δ 5 = δ ◦ δ 4 = , δ 6 = δ ◦ δ 5 = ι;

and
2 = δ, 3 =  ◦ 2 =  ◦ δ = ι, 4 =  ◦ 3 = , 5 =  ◦ 4 = δ, 6 =  ◦ 5 = ι;
Hence, ◦(σ) = ◦(τ ) = ◦(µ) = 2 and ◦(δ) = ◦() = 3. 

Example 2.4.4 Consider the group (Z6 , +6 ). Let [2], [3], [5] ∈ Z6 . Then
◦([2]) = 3, ◦([3]) = 2, and ◦([5]) = 6.

Proof : We have
1[2] = [2], 2[2] = [4], 3[2] = [0], 4[2] = [2], 5[2] = [4], 6[2] = [0].
Hence, ◦([2]) = 3.

1[3] = [3], 2[3] = [0], 3[3] = [3], 4[3] = [0], 5[3] = [3], 6[3] = [0].

Hence, ◦([3]) = 2.

1[5] = [5], 2[5] = [4], 3[5] = [3], 4[5] = [2], 5[5] = [1], 6[5] = [0].

Hence, ◦([5]) = 5. 
 
0 −1
Example 2.4.5 Consider the group (GL(2, R), ∗) and the elements
  −1 0
1 1
and . Find the order of these elements.
0 1
 23

Solution: We have
 2      
0 −1 0 −1 0 −1 1 0
= ∗ = .
−1 0 −1 0 −1 0 0 1
 
0 −1
Thus, ◦ = 2.
−1 0
Next,
 2      
1 1 1 1 1 1 1 2
= ∗ =
0 1 0 1 0 1 0 1
and
 3    2      
1 1 1 1 1 1 1 1 1 2 1 3
= ∗ = ∗ = .
0 1 0 1 0 1 0 1 0 1 0 1
By induction, we can show that
 n  
1 1 1 n
= .
0 1 0 1
 
0 −1
Hence, the order of is infinite. 
−1 0

Example 2.4.6 Consider the group (R\{0}, ·). We have (−1)2 = 1, which
implies that ◦(−1) = 2. Let x ∈ R\{0} such that x 6= 1 and x 6= −1. Then
for all positive integers n, xn 6= 1, which implies that ◦(x) is infinite. Hence,
(R\{0}, ·) has elements of finite order and has elements of infinite order.

Example 2.4.7 Consider the group (R, +). If x ∈ R such that x 6= 0, then
for all positive integers n, nx 6= 0, which implies that ◦(x) is infinite. Hence,
all nonidentity elements of the group (R, +) is of infinite order.

Theorem 2.4.8 Let (G, ∗) be a group and a ∈ G such that ◦(a) = n. If

am = e for some integer m, then n|m.

Proof : Let a ∈ G such that ◦(a) = n. Then an = e. Suppose am = e for some

integer m. By the division algorithm, there exists integers q and r such that
m = nq + r, where 0 ≤ r < n. Thus, r = m − nq and

ar = am−nq = am ∗ (an )−q = e ∗ e−q = e.

 24

If 0 < r < n such that ar = e, then this contradicts the minimality of n.

Hence, r = 0, which implies that m = nq. Therefore, n|m. 

Definition 2.4.9 A group (G, ∗) is called a torsion group if every element

of G is of finite order. If every nonidentity element of G is of infinite order,
then G is called a torsion-free group.

Example 2.4.10 The groups (Z6 , +6 ), (U7 , ·7 ), and (S3 , ◦) are torsion groups.

Example 2.4.11 The groups (Z, +), (Q, +), (R, +), and (R+ , ·) are torsion-free
groups.

Example 2.4.12 The groups (Q\{0}, ·) and (R\{0}, ·) are neither a torsion
nor a torsion-free groups.

Example 2.4.13 Consider the group of symmetries of the square: Sym =

{r360 , r90 , r180 , r270 , h, v, d1 , d2 }. Find the order of each element.

2.5 Exercises

1. Determine if the following ordered pairs G form a group.

(a) (Z, ∗), where a ∗ b = ab for all a, b ∈ Z.
(b) (Z+ ∪ {0}, ∗), where a ∗ b = a + b for all a, b ∈ Z+ ∪ {0}.
(c) (Q\{−1}, ∗), where a ∗ b = a + b + ab for all a, b ∈ Q\{−1}.
(d) (Q, ∗), where a ∗ b = ab
2
for all a, b ∈ Q.
(e) (R, ∗), where a ∗ b = |a||b| for all a, b ∈ R.
2. If G is an abelian group, prove that (a ∗ b) = an ∗ bn for all a, b ∈ G and for
all positive integers n.

3. In S3 , show that there are four elements x satisfying x2 = e and three

elements y satisfying y 3 = e.
 CHAPTER 3

PERMUTATION GROUPS

3.1 Permutation

Definition 3.1.1 Let X be a nonempty set. A permutation π of X is a

one-one function from X onto X.
Example 3.1.2 Let X be a nonempty set. Define π : X → X by π(x) = x for
all x ∈ X. Then π is a one-one function of X onto X. Thus, π is a permutation
of X. Note that π is called the identity permutation and is, usually, denoted
by ιX or e.
Example 3.1.3 Let X = {a, b, c}. Define α : X → X such that α(a) = b,
α(b) = a, and α(c) = c. Then α is a one-one function of X onto X. Thus, α
is a permutation of X.
Example 3.1.4 Consider R, the set of real numbners. Define α : R → R by
α(x) = 3x + 5 for all x ∈ R. Then α is a permutation of R.
Proof : Let x1 , x2 ∈ R such that α(x1 ) = α(x2 ). Then 3x1 + 5 = 3x2 + 5. Thus,
3x1 = 3x2 . This implies that x1 = x2 . Hence, α is injective.

Let y ∈ R. Then y−53

∈ R. Take x = y−5
3
. Thus, there exists x ∈ R
such that α(x) = 3x + 5 = 3 y−5


3
+ 5 = y. Hence, α is surjective.

Therefore, α is a permutation of R. 

3.2 Permutation Groups

Definition 3.2.1 A group (G, ∗) is called a permutation group on a nonempty

set X if the elements of G are permutations of X and the operations ∗ is the
composition of two functions.
Example 3.2.2 Let X = {1, 2}. Define α : X → X by α(1) = 1, α(2) = 2.
Then α is a permutation of X. Next, define β : X → X by β(1) = 2. Then
β is a one-one function of X onto X. Thus, β is a permutation of X. Let
SX = {α, β}. Then α ◦ α = α, α ◦ β = β, β ◦ α = β, and β ◦ β = α. Hence,
(SX , ◦) is a group. Note that α is the identity and β −1 = β.
 26

Our study of permutation groups will focus on permutations groups on

finite sets, that is, X is a finite set.
Let us fix some notations which will be useful when working with
permutations.
Let In = {1, 2, ..., n}, n ≥ 1. Let π be a permutation groups. Then

π = {(1, π(1)), (2, π(2)), ..., (n, π(n))}.

It is sometimes convenient to describe a permutation by means of the following

notational device:
 
1 2 3 ··· n
π=
π(1) π(2) π(3) · · · π(n)
This notation is due to Cauchy and is called the two-row notation.

Example 3.2.3 Let n = 4 and π be the permutation on I4 defined by π(1) =

2, π(2) = 4, π(3) = 3, and π(4) = 1. Then using the two-row notation we can
write
 
1 2 3 4
π= .
2 4 3 1

Example 3.2.4 Let n = 7 and π and σ be two permutation on I7 defined by

 
1 2 3 4 5 6 7
π= .
1 3 4 6 7 2 5
and
 
1 2 3 4 5 6 7
σ= .
2 5 3 1 7 6 4
Then computing for π ◦ σ, we get
 
1 2 3 4 5 6 7
π= .
3 7 4 1 5 2 6

Let Sn denote the set of all permutations on In , n ≥ 1.

Example 3.2.5 Consider S3 = {ι, σ, τ, µ, δ, }, the group of all permutations

on I3 = {1, 2, 3}. Then using the two-row notation, we have
 27
     
1 2 3 1 2 3 1 2 3
ι= , σ= , τ= ,
1 2 3 1 3 2 3 2 1
     
1 2 3 1 2 3 1 2 3
µ , δ= , =
2 1 3 2 3 1 3 1 2

Theorem 3.2.6 (i) If n ≥ 3, then (Sn , ◦) is noncommutative.

(ii) |Sn | = n!.

Definition 3.2.7 The group (Sn , ◦) is called the symmetric group on In .

3.3 Cycles and Tranpositions

Definition 3.3.1 Let π be an element of Sn . Then π is called a k-cycle,

written (i1 i2 · · · ik ) if
 
i1 i2 · · · ik−1 ik
π= ,
i2 i3 · · · ik i1

that is, π(ij ) = ij+1 , j = 1, 2, ..., j − 1, π(ik ) = i1 , and π(a) = a for ant other
element of In .

Note that if π = (i1 i2 · · · ik ), then

π = (i1 i2 · · · ik ) = (i2 i3 · · · ik i1 ) = · · · = (ij ij+1 · · · ik i1 · · · ij−1 )

 
1 2 3 4 5 6 7 8
Example 3.3.2 Let π = ∈ S8 . Write π as a
2 5 1 8 3 7 6 4
product of cycles.
 
1 2 3 4 5 6 7 8
Solution: π = = (1 2 5 3) ◦ (4 8) ◦ (6 7). 
2 5 1 8 3 7 6 4
A k-cycle is called a transposition when k = 2.
 28

Example 3.3.3 The group S3 can be written in three ways:

S3 = {ι, σ, τ, µ, δ, }
           
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
= , , , , ,
1 2 3 1 3 2 3 2 1 2 1 3 2 3 1 3 1 2
= {e, (23), (13), (12), (123), (132)}.

Definition 3.3.4 Let π1 , π2 , ..., πk ∈ Sn . Then π1 π2 · · · πk are called disjoint

if for all i, 1 ≤ i ≤ k and for all a ∈ In , πi (a) 6= a implies πj (a) = a for all
j 6= i, 1 ≤ j ≤ k.

In other words, π1 , π2 , ..., πk ∈ Sn are disjoint if for all 1 ≤ j ≤ k and

for all a ∈ In , if πi moves a, then all other permutations πj must fix a, that is,
πj (a) = a for all j 6= i, 1 ≤ j ≤ k.

Theorem 3.3.5 Let π, λ ∈ Sn such that π and λ are disjoint. Then π ◦ λ =

λ ◦ π.

Proof : Let π and λ be disjoint permutations on In . Let a ∈ S such that

π(a) 6= a. Then λ(a) = a. Let π(a) = b. Then
(π ◦ λ)(a) = π(λ(a)) = π(a) = b.
Also,
(λ ◦ π)(a) = λ(π(a)) = λ(b).
If π(b) = b, then π(b) = b = π(a), so a = b. Thus, π(a) = b = a, a
contradiction. Hence, π(b) 6= b, so λ(b) = a. This implies that
(λ ◦ π)(a) = λ(π(a)) = λ(b) = b.
Therefore, (π ◦ λ)(a) = (λ ◦ π)(a).
Suppose that π(a) = a. If λ(a) = a, then (π ◦ λ)(a) = a = (λ ◦ π)(a).
Suppose that λ(a) 6= a. By similar argument as before, (π ◦ λ)(a) = (λ ◦ π)(a).
Thus, π ◦ λ = λ ◦ π. 
 
1 2 3 4 5 6 7 8
Consider π = ∈ Sn . Then π can be written
2 5 1 8 3 7 6 4
as a product of disjoint cycles. That is, π = (1 2 5 3) ◦ (4 8) ◦ (6 7).
 29

Theorem 3.3.6 Any nonidentity permutation π of Sn (n ≥ 2) can be uniquely

expressed (up to the order of the factors) as a product of disjoint cycles, where
each cycle is of length at least two.

Corollary 3.3.7 Let n ≥ 2. Any permutationπ of Sn can be expressed as a

product of transposition.

Proof : This fact is immediate from the following equations:

e = (1) = (1 2) ◦ (1 2)
and for k ≥ 2
(i1 i2 · · · ik ) = (i1 ik ) ◦ (i1 ik−1 ) ◦ · · · ◦ (i1 i2 ),
where {i1 , i2 , ..., ik } ⊆ In . 

Remark 3.3.8 Let n ≥ 2 and σ ∈ Sn be a cycle. Show that σ is a k-cycle if

and only if ◦(σ) = k.

Remark 3.3.9 Let σ ∈ Sn and σ = σ1 ◦ σ2 ◦ · · · ◦ σk be a product of disjoint

cycles. Suppose that ◦(σi ) = ni , i = 1, 2, ..., n. Then ◦(σ) = lcm(n1 , n2 , ..., nk ).

Let π ∈ Sn . Since Sn is a finite group, it follows that ◦(π) is finite. In

order to find the order of π, we need to compute π, π 2 , π 3 , ..., until we find the
first positive integer k such that π k = e. Finding such a positive integer could
be a tedious task. However, we can effectively make use of the decomposition
of π as a product of disjoint cycles, compute the order of each cycle, which
is the length of the cycle (Remarm 3.1.16) and from the order of the cycles
deduce the order of π (see Remark 3.1.17).

Let π ∈ Sn , n ≥ 2. In order to express π as a product of disjoint cycles,

first consider 1, π(1), π 2 (1), π 3 (1), ... and find the smallest positive integer r
such that π r (1) = 1. Let
σ1 = (1 π(1) π 2 (1) · · · π r−1 (1)).
Then σ1 is a cycle of length r. Let i be the first element of In not appearing
in σ1 . Consider i, π(i), π 2 (i), π 3 (i), ... and find the smallest positive integer s
such that π s−1 (i) = i. Let
 30

σ2 = (i π(i) π 2 (i) · · · π s−1 (i)).

Then σ2 is a cycle of length s. It can be shown that σ1 and σ2 are disjoint

cycles. If

{1, π(1), π 2 (1), ..., π r−1 (1)} ∪ {i, π(i), π 2 (i), ..., π s−1 (i)} =
6 In ,

then consider the first element of In not appearing in

{1, π(1), π 2 (1), ..., π r−1 (1)} ∪ {i, π(i), π 2 (i), ..., π s−1 (i)} =
6 In ,

and continue the process to construct the cycle σ3 . Since In is finite, the above
process must stop with some cycle σm . Then π = σ1 ◦ σ2 ◦ · · · ◦ σm .

Example 3.3.10 Consider the permutation

 
1 2 3 4 5 6 7
π=
6 3 5 2 4 7 1

on I7 . Then π(1) = 6, π 2 (1) = π(6) = 7, π 3 = π(7) = 1. Thus, σ1 = (1 6 7) is

a cycle of length 2. Next, 2 is the first element of I7 not appearing in (1 6 7).
Now, π(2) = 3, π 2 (2) = π(3) = 5, π 3 (2) = π(5) = 4, π 4 (2) = π(4) = 2. Hence,
σ2 = (2 3 5 4) is a cylce of length 4. Clearly, σ1 and σ2 are disjoint. Therefore,
π = σ1 ◦ σ2 = (1 6 7)(2 3 5 4). Therefore, ◦(π) = lcm(3,4)=12.

While writing permutation as a product of disjoint cycles, it is customary

not to write cycles of length one in the product. Thus, if some element of In
does not appear in any of the cycles, then it is assumed to be fixed. For
example, if π = (1 2 5) ◦ (4 6) ∈ S7 , then since 3 and 7 neither appear in π,
they are fixed, that is, π(3) = 3 and π(7) = 7.

Given a permutation π ∈ Sn , n ≥ 2, we can write π as product of

transpositions. However, the representation of π as a product of transposition
need not be unique. For example, (1 2 3) = ((1 3) ◦ (1 2) = (2 1) ◦ (2 3) =
(3 2) ◦ (3 1). However, the number of transpositions in any representation of
a permutation is either even or odd, but not both.

Theorem 3.3.11 Let n ≥ 2 and π ∈ Sn . Suppose that

π = σ1 ◦ σ2 ◦ · · · ◦ σr = τ1 ◦ τ2 ◦ · · · ◦ τs ,
 31

where σi , τj ∈ Sn are transpositions, i = 1, 2, ..., r and j = 1, 2, ..., s. Then both

r and s are either even or odd.

Definition 3.3.12 Let π ∈ Sn . If π is a product of an even number of

transpositions, then π is called an even permutation; otherwise π is called
an odd permutation.

Corollary 3.3.13 Let π ∈ Sn be a k-cycle. Then π is an even permutation if

and only if k is odd.
Proof : Let π = (1 2 · · · k). Then π = (1 k) ◦ (1 k − 1) ◦ · · · ◦ (1 2), that is, π
is a product of k − 1 transpositions. If π is an even transposition, then k − 1
is even. Thus, k is odd.
Conversely, suppose that k is odd. Then k − 1 is even. Thus, π is an
even permutation. 
Let An denote the subset of Sn consisting of all even permutations,
n ≥ 2.

Theorem 3.3.14 For n ≥ 2, the pair (An , ◦) is a group, called the alternating
group on In .
Proof : Since e = (1 2)(1 2), e ∈ An . Thus, An 6= ∅.
Let π1 , π2 ∈ An . By Theorem 3.1.19, π1 π2 is even, that is, π1 π2 ∈ An .
Let π ∈ An . Then there exists π −1 such that ππ −1 = e. Since π and e
are even, π −1 is even. Hence, π −1 ∈ An .
Therefore, (An , ◦) is a group. 
 
1 2 3 4 5 6 7 8
Example 3.3.15 Express the permutation σ = on
2 3 8 5 6 4 7 1
I8 as a product of disjoint cycles and then as a product of transpositions. Is
σ an even permutation?

Example 3.3.16 Write all elements of S4 . Determine A4 .

Example 3.3.17 Prove that every transposition is its own inverse.

Example 3.3.18 Find the order of (1 2 3 4) ◦ (5 6 7) in S7 .

 CHAPTER 4

SUBGROUPS AND NORMAL SUBGROUPS

4.1 Subgroups

Definition 4.1.1 Let (G, ∗) be group and H a nonempty subset of G. Then

(H, ∗) is called a subgroup of (G, ∗) if (H, ∗) is a group.

Theorem 4.1.2 Let (G, ∗) be group and (H, ∗) a subgroup of (G, ∗).
(i) The identity elements of (H, ∗) and (G, ∗) are the same.
(ii) If h ∈ H, then the inverse of h in H and the inverse of h in G is
the same.

Proof : (i) Let eH denote the identity of H and e denote the identity of G.
Then

eH ∗ eH = eH ∗ e.

By the cancellation law, eH = e. Hence, the identity elements of G and H are

the same.
(ii) Let h ∈ H. Let h0 denote the inverse of h in H and h−1 denote the
inverse of h in G. Then

h ∗ h0 = e = h0 ∗ h and h ∗ h−1 = e = h−1 ∗ h.

Thus,

h0 − h0 ∗ e = h0 ∗ (h ∗ h−1 ) = (h0 ∗ h) ∗ h−1 = e ∗ h−1 = h−1 .

Therefore, the inverse of h in H and the inverse of h in G is the same. 

Remark 4.1.3 If (G, ∗) is a group, then ({e}, ∗) and (G, ∗) are subgroups of
(G, ∗). These subgroups are called trivial.

Example 4.1.4 Consider the following list of groups.

(i) ({0}, +), (Z, +), (Q, +) ,(R, +).
(ii) ({1}, ·), (Q\{0}, ·) ,(R\{0}, ·).
Each group is the subgroup of the group listed to its right.
 33

Notation We would like to change our notation slightly. We shall generally

use the notation G instead of (G, ∗) for a group and we write ab for a ∗ b. We
shall refer to ab as the product of a and b. This notation is usually called
mutiplicative notation.

Theorem 4.1.5 (Subgroup Criterion) Let G be a group and H a nonempty

subset of G. Then H is a subgroup of G if and only if for all a, b ∈ H,
ab−1 ∈ H.

Proof : Suppose H is a subgroup of G. Then H is a group. Let a, b ∈ H. Thus,

b ∈ H implies that there exists b−1 ∈ H. Hence, ab−1 ∈ H since H is closed
under the binary operation.
Conversely, suppose that for all a, b ∈ H, ab−1 ∈ H. Since H is
nonempty, there exists a ∈ H. Thus, a, a ∈ H implies that e = aa−1 ∈ H.
This means that H contains the identity. Next, let b ∈ H. Then e, b ∈ H
implies that b−1 = eb−1 ∈ H. Hence, every element in H has an inverse in H.
Finally, Let a, b ∈ H. Then a, b−1 ∈ H. This implies that ab = a(b−1 )−1 ∈ H,
that is, H is closed under the binary operation. Therefore, H is a group.
Consequently, H is a subgroup of G. 

Corollary 4.1.6 Let G be a group and H be a finite nonempty subset of G.

Then H is a subgroup of G if and only if for all a, b ∈ H, ab ∈ H.

Proof : If H is a subgroup of G, then for all a, b ∈ H, ab ∈ H.

Conversely, suppose that for all a, b ∈ H, ab ∈ H. Let h ∈ H. Then
h, h , ..., hn , ... ∈ H. Thus, {h, h2 , ..., hn , ...} ⊆ H. Since H is finite, the
2

elements of {h, h2 , ..., hn , ...} cannot be distinct. This implies that there exists
integers r and s such that 0 ≤ r < s and hr = hs . Hence, s − r ≥ 1 and
e = hs−r = hhs−r−1 ∈ H, which means that h−1 ∈ H. Now, let a, b ∈ H.
Then a, b−1 ∈ H. By hypothesis, ab−1 ∈ H. Therefore, by Theorem 3.1.5, H
is a subgroup of G. 

4.2 Testing Sets for Subgroup

Subgroup Criterion Let G be a group. Then H is a subgroup of G if and

only if the following properties hold:
(i) H is nonempty.
(ii) H ⊆ G.
(iii) For all a, b ∈ H, ab−1 ∈ H.
 34

Subgroup Criterion Let G be a group. Then H is a subgroup of G if and

only if the following properties hold:
(i) H is nonempty.
(ii) H ⊆ G.
(iii) For all a, b ∈ H, ab ∈ H.
(iv) For all a ∈ H, a−1 ∈ H.

Theorem 4.1.6 Let G be a group and H a finite set. Then H is a subgroup

of G if and only if the following properties hold:
(i) H is nonempty.
(ii) H ⊆ G.
(iii) For all a, b ∈ H, ab ∈ H.

Example 4.2.1 Consider the group (Z, +) and the subset 5Z = {5n : n ∈ Z}
of Z. Then 5Z is a subgroup of (Z, +).

Proof : (i) Since 0 ∈ Z, it follows that 5(0) ∈ 5Z. Thus 5Z is nonempty.

(ii) Let 5n ∈ 5Z. Then n ∈ Z. Thus, 5n ∈ Z. Hence, 5Z ⊆ Z.

(iii) Let 5n1 , 5n2 ∈ 5Z. Then n1 , n2 ∈ Z. Thus, n1 + n2 ∈ Z. Hence,

5(n1 + n2 ) ∈ 5Z. Therefore, 5n1 + 5n2 = 5(n1 + n2 ) ∈ 5Z.

By the Subgroup Criterion, 5Z is a subgroup of (Z, +). 

√ √ √
Example 4.2.2 Let Z[ 2] = {a + b 2 : a, b ∈ Z}. Then Z[ 2] is a subgroup
of (R, +).
√ √ √
Proof : (i) Since 0 ∈ Z, it follows that 0 + 0 2 ∈ Z[ 2]. Thus Z[ 2] is
nonempty.
√ √
√ (ii) Let a + b 2 ∈ √ 2]. Then a, b ∈ Z. Thus, a, b ∈ R. Hence,
Z[
a + b 2 ∈ R. Therefore, Z[ 2] ⊆ R.
√ √ √
(iii) Let a + b 2, c + d 2 ∈ Z[ 2]. Then
√ a, b, c, d ∈√Z. Thus, a + c, b +
d ∈ √ Z. Hence,
√ (a + c) + (b +√d) 2 √∈ Z[ 2]. Therefore,
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 ∈ Z[ 2].
√
By the Subgroup Criterion, Z[ 2] is a subgroup of (R, +). 
 35
  
a b
Example 4.2.3 Let H = : a, b, d ∈ R, ad 6= 0 . Then H is a subgroup
0 d
of GL(2, R).
 
1 0
Proof : (i) Since 1, 0 ∈ R and 1 · 1 6= 0, it follows that ∈ H. Thus H is
0 1
nonempty.
 
a b
(ii) Let ∈ H. Then a, b, d ∈ R and ad 6= 0. Thus, a, b, 0, d ∈ R
0 d  
a b
and ad − b0 6= 0. Hence, ∈ GL(2, R). Therefore, H ⊆ GL(2, R).
0 d
   
a b e f
(iii) Let , ∈ H. Then a, b, d, e, f, h ∈ R and, ad 6= 0 and
0 d 0 h
eh 6= 0. Thus, 1e 6= 0 and h1 6= 0. Hence, ae , afeh +be d
, ∈ R and a · hd 6= 0. This
 a af +be     h −1  ae af +be

a b e f
implies that e eh
d ∈ H. Therefore, = e eh
d ∈ H.
0 h
0 d 0 h 0 h

By the Subgroup Criterion, H is a subgroup of GL(2, R). 

√ √ √
Example 4.2.4 Let Z[ 2] = {a + b 2 : a, b ∈ Z}. Then Z[ 2] is not a
subgroup of (Z, +).
√ √ √ √
Proof√: Consider
√ 0 + 2 2 = 2 √2 ∈ Z[ 2]. Then 0, 2 ∈ Z. But 2∈/ Z. √
Thus,
0+2 2 = 2 2 ∈ / Z. Hence, Z[ 2] * Z. By the Subgroup Criterion, Z[ 2] is
not a subgroup of (Z, +). 
Example 4.2.5 Let H = {[0], [3], [6], [8], [9]}. Then H is not a subgroup of
(Z12 , +12 ).
Proof : Consider [3], [8] ∈ H. Then [3] +12 [8] = [11]. But [11] ∈ / H. Thus,
there exist [3], [8] ∈ H such that [3] +12 [8] ∈
/ H. By Corollary 4.1.6, H is not
a subgroup of (Z12 , +12 ). 
Lemma 4.2.6 Let G be a group and Z(G) = {x ∈ G|ax = xa for all a ∈ G}.
Then Z(G) is an abelian subgroup of G. Z(G) is called the center of G.
Proof : (i) Since ae = a = ea for all a ∈ G, it follows that e ∈ Z(G). Thus,
Z(G) is nonempty.

(ii) Let x ∈ Z(G). By the definition of Z(G), x ∈ G. Thus, Z(G) ⊆ G.

(iii) Let x, y ∈ Z(G). Then ax = xa and ay = ya for all a ∈ G. Thus,

 36

ay −1 = eay −1 = y −1 yay −1 = y −1 ayy −1 = y −1 ae = y −1 a

for all a ∈ G. Hence,

a(xy −1 ) = (ax)y −1 = (xa)y −1 = x(ay −1 ) = x(y −1 a) = (xy −1 )a

for all a ∈ G. Therefore, xy −1 ∈ Z(G).

By the Subgroup Criterion, Z(G) is a subgroup of G.

Let x, y ∈ Z(G). Then x, y ∈ G. Thus, yx = xy. Therefore, Z(G) is

an abelian subgroup of G. 

Lemma 4.2.7 Let G \ be a group and {Hi |i ∈ I} be any nonempty family of

subgroups of G. Then Hi is a subgroup of G.
i∈I

Proof : (i) Since Hi is

\a subgroup of G for all i ∈ I, it follows that e ∈ Hi for
T
all i ∈ I. Thus, e ∈ Hi . Hence, i∈I Hi is nonempty.
i∈I
\
(ii) Let x ∈ Hi . Then x ∈ Hi for all i ∈ I. Since Hi is a subgroup
i∈I \
of G for all i ∈ I, it follows that x ∈ G. Hence, Hi ⊆ G.
i∈I
\
(iii) Let a, b ∈ Hi . Then a, b ∈ Hi for all i ∈ I. Since Hi is a
i∈I
subgroup\of G for all i ∈ I, it follows that ab−1 ∈ Hi for all i ∈ I. Hence,
a, b−1 ∈ Hi .
i∈I
\
By the Subgroup Criterion, Hi is a subgroup of G. 
i∈I

Example 4.2.8 If a ∈ G, define N (a) = {x ∈ G|xa = ax}. Show that N (a)

is a subgroup of G. N (a) is usually called the normalizer of a in G.

Proof : (i) Since ea = a = ae, it follows that e ∈ N (a). Thus, N (a) is

nonempty.

(ii) Let x ∈ N (a). By definition of N (a), x ∈ G. Thus, N (a) ⊆ G.

(iii) Let x, y ∈ N (a). Then xa = ax and ya = ay. Thus,

 37

y −1 a = y −1 ae = y −1 ayy −1 = y −1 yay −1 = eay −1 = ay −1 .

Hence,

(xy −1 )a = x(y −1 a) = x(ay −1 ) = (xa)y −1 = (ax)y −1 = a(xy −1 ).

Therefore, xy −1 ∈ N (a).

By the Subgroup Criterion, N (a) is a subgroup of G. 

Example 4.2.9 Let H be a subgroup of a group G and let g ∈ G. Prove that

gHg −1 = {ghg −1 |h ∈ H} is a subgroup of G and |gHg −1 | = |H|.

Proof : (i) Since e ∈ H, it follows that geg −1 ∈ gHg −1 . Hence, gHg −1 is

nonempty.

(ii) Let ghg −1 ∈ gHg −1 . Then h ∈ H. Thus, h ∈ G. Hence,

−1
ghg ∈ G. Therefore, gHg −1 ⊆ G.

(iii) Let ghg −1 , gkg −1 ∈ gHg −1 . Then h, k ∈ H. Thus, k −1 ∈ H

and hk −1 ∈ H. Hence, g(hk −1 )g −1 ∈ gHg −1 . Therefore, (ghg −1 )(gkg −1 )−1 =
ghg −1 (g −1 )−1 k −1 g −1 = ghg −1 gk −1 g −1 = ghek −1 g −1 = g(hk −1 )g −1 ∈ gHg −1 .

By the Subgroup Criterion, gHg −1 is a subgroup of G.

Let f : H → gHg −1 defined by f (h) = ghg −1 .

(i) Let h, k ∈ H such that h = k. Then ghg −1 = gkg −1 . Thus,

f (h) = f (k). Hence, f is well-defined.

(ii) Let h, k ∈ H such that f (h) = f (k). Then ghg −1 = gkg −1 . By the
cancellation law, h = k. Thus, f is injective.

(iii) Let y ∈ gHg −1 . Then y = ghg −1 for some h ∈ H. Thus, there

exists h ∈ H such that f (h) = ghg −1 = y. Hence, f is surjective.

Therefore, H is isomorphic to gHg −1 . Consequently, |H| = |gHg −1 |.



Remark 4.2.10 Let H and K be subgroups of group G. If K ⊆ H, then K

is a subgroup of H.
 38

Example 4.2.11 Find all the subgroups of the group (Z, +).

Example 4.2.12 Find all the subgroups of the group (Sym, ◦).

4.3 Subgroup Generated by a Subset

Definition 4.3.1 Let G be a group\and S ⊆ G. Let S = {H | H is a subgroup

of G and S ⊆ H}. Define hSi = H, that is, hSi is the intersection of all
H∈S
subgroups H of G such that S ⊆ H.
(i) The subgroup hSi of G is called the subgroup generated by S.
(ii) If G = hSi, then S is called a set of generators for G.

Remark 4.3.2 Let G be a group and S ⊆ G. If either S = ∅ or S = {e},

then hSi = {e}. Moreover, hGi = G.

Let S = {H | H is a subgroup of G and S ⊆ H}, where S 6= ∅. Then

(S, ≤) is a partially ordered set, where ≤ denotes the set inclusion relation.
Hence, hSi is the smallest subgroup of G which contains S.
The elements of S are the generators of the subgroup hSi. If S =
{a1 , a2 , ..., an }, we write ha1 , a2 , ..., an i instead of hSi. If G = ha1 , a2 , ..., an i,
(ai ∈ G), then G is said to be finitely generated. If a ∈ G, the subgroup
hai is called the cylic group or cyclic subgroup generated by a.

Theorem 4.3.3 Let G be a group and S a nonempty subset of G. Then

hSi = {an1 1 an2 2 · · · ant t |ai ∈ S, ni ∈ Z}.

Proof : Let

A = {an1 1 an2 2 · · · ant t |ai ∈ S, ni ∈ Z}.

Then A ⊆ hSi. We show that A is a subgroup of G containning S.

Since S 6= ∅, there exists a ∈ S. Thus, a = a1 ∈ A, which means that
S ⊆ A. Now, let

a = an1 1 an2 2 · · · ant t , b = bn1 1 an2 2 · · · ans s ∈ A.

 39

Then
ab−1 = (an1 1 an2 2 · · · ant t )(bn1 1 bn2 2 · · · bns s )−1
= a1n1 an2 2 · · · ant t b−n
s
s
· · · b2−n2 b1−n1 ∈ A.
Hence, A is a subgroup of G. By definition of hSi, it is the smallest subgroup
of G containing S. Therefore, hSi ⊆ A. Accordingly,
hSi = {an1 1 an2 2 · · · ant t |ai ∈ S, ni ∈ Z}. 

Corollary 4.3.4 If G is a group and a ∈ G.Then hai = {an |n ∈ Z}.

Proof : By Theorem 4.3.3,

hai = {an1 an2 · · · ant |ni ∈ Z}
= {an1 +n2 ···+nt |ni ∈ Z}
= {an |n ∈ Z}.
In additive notation, we would have hai = {an | n ∈ Z}.

Example 4.3.5 Let G = (S3 , ◦) and S = {δ}. Find hSi.

Solution:
hSi = hδi
= {δ n | n ∈ Z}
= {δ 0 , δ 1 , δ 2 , δ 3 , δ 4 , δ 5 , ...}
= {ι, δ, , ι, δ, , ...}
= {ι, δ, }.
Therefore, hSi = hδi = {ι, δ, }. 

Example 4.3.6 Let G = (S3 , ◦) and S = {δ, }. Find hSi.

Solution:
hSi = hδ, i
= {δ n1 n2 | n1 , n2 ∈ Z}
= {δ 0 0 , δ 1 0 , δ 0 1 , δ 1 1 , δ 0 2 , δ 2 0 , ...}
= {ι, δ, , ι, δ, , ...}
= {ι, δ, }.
Therefore, hSi = hδ, i = {ι, δ, } = hδi. 
 40

Example 4.3.7 Let G = (S3 , ◦). Find hσ, τ i.

Solution:
hσ, τ i = {τ n1 σ n2 τ n3 | n1 , n2 , n3 ∈ Z}
= {τ 0 σ 0 τ 0 , τ 0 σ 1 τ 0 , τ 1 σ 0 τ 0 , τ 1 σ 1 τ 1 , τ 1 σ 1 τ 0 , τ 0 σ 1 τ 1 , ...}
= {ι, σ, τ, µ, δ, }.

Therefore, hσ, τ i = S3 . 

Example 4.3.8 Let G = (S3 , ◦). Find hµ, δi.

Example 4.3.9 Let G = (Z, +) and S = {2}. Find hSi.

Solution:
hSi = h2i
= {2n | n ∈ Z}
= {..., −4, −2, 0, 2, 4, 6, ...Z}
= E.

Therefore, hSi = h2i = {2n | n ∈ Z} = E. 

Example 4.3.10 Let G = (Z, +) {4, 6} ⊆ G. Find h4, 6i.

Solution:
h4, 6i = {4n1 + 6n2 | n1 , n2 ∈ Z}
= {2(2n1 + 3n2 ) | n1 , n2 ∈ Z}
= {2n | n ∈ Z}
= E.

Therefore, h4, 6i = h2i = E. 

Example 4.3.11 Let G = (Z, +). Find h8, 12i.

Example 4.3.12 Let G = (Z, +). Find h4, 7i.

Example 4.3.13 Let G = (Sym, ◦). Find hr90 i, hr90 , r270 i, hhi, and hh, vi.
 41

4.4 Product of Subgroups

Definition 4.4.1 Let H and K be nonempty subsets of a group G. The

product of H and K is defined to be the set

HK = {hk|h ∈ H, k ∈ K}.

Let H1 , H2 ,...,Hn be nonempty subsets of a group G. The product of

H1 , H2 ,...,Hn is defined to be the set

H1 H2 · · · Hn = {h1 h2 · · · hn |hi ∈ Hi , i = 1, 2, ..., n}.

Example 4.4.2 Consider G = (Sym, ◦). Let H = {r360 , d1 } and K =

{r360 , h}. Then H and K are subgroups of G. We have

HK = {r360 r360 , r360 h, d1 r360 , d1 h} = {r360 , h, d1 , r90 },

and

KH = {r360 r360 , r360 d1 , hr360 , hd1 } = {r360 , d1 , h, r270 },

Note that hd1 = r270 ∈

/ HK, which means that HK is not a subgroup of Sym.
Also, d1 h = r90 ∈
/ KH, which means that KH is not a subgroup of Sym.

The last example shows that in general the product of subgroups need
not be a subgroup. In the following theorem, we give necessary and sufficient
condition for the product to be a subgroup.

Theorem 4.4.3 Let H and K be subgroups of a group G. Then HK is a

subgroup of G if and only if HK = KH.

Proof : Suppose that HK is a subgroup of G. Let kh ∈ KH. Then k ∈ K

and h ∈ H. Now, h = he ∈ HK and k = ek ∈ HK. This implies that
kh ∈ HK. Thus, KH ⊆ HK. On the other hand, let hk ∈ HK. Then
(hk)−1 ∈ HK, which implies that (hk)−1 = h1 k1 for some h1 ∈ H and k1 ∈ K.
Thus, h−1 −1
1 ∈ H and k1 ∈ K. Hence,

hk = (h1 k1 )−1 = k1−1 h−1

1 ∈ KH.
 42

This implies that HK ⊆ KH. Therefore, HK = KH.

Conversely, suppose that HK = KH. Clearly, HK is nonempty and

HK ⊆ G. Let h1 k1 , h2 k2 ∈ HK. Then h1 , h2 ∈ H and k1 , k2 ∈ K. Thus,

(h1 k1 )(h2 k2 )−1 = h1 k1 k2−1 h−1

2
−1
= h1 k3 h2 , k1 k2−1 = k3
= h1 h4 k4 , k3 h−1
2 = h4 k4 since HK = KH
= h5 k4 , h1 h4 = h5 .

Hence, (h1 k1 )(h2 k2 )−1 ∈ HK. By the Subgroup Criterion, HK is a subgroup

of G. 

Corollary 4.4.4 If H and K are subgroups of an abelian group G, then HK

is a subgroup of G.

Proof : If G is abelian, then HK = KH. By Theorem 4.4.3, HK is a subgroup

of G. 

Example 4.4.5 Let G = (Sym, ◦) and consider the subgroups H = {r360 , h}

and K = {r360 , v}. Then HK is a subgroup of G = (Sym, ◦).

Proof : We have
HK = {r360 r360 , r360 v, hr360 , hv} = {r360 , v, h, r180 },
and
KH = {r360 r360 , r360 h, vr360 , vh} = {r360 , h, v, r180 }.
Thus, HK = KH. By Theorem 4.4.3, HK is a subgroup of G = (Sym, ◦).

Example 4.4.6 Let G = (Sym, ◦) and consider the subgroups H = {r360 , r180 }
and K = {r360 , v}. Is HK a subgroup of G = (Sym, ◦)?

Example 4.4.7 Let G = (Sym, ◦) and consider the subgroups H = {r360 , r180 }
and K = {r360 , h}. Is HK a subgroup of G = (Sym, ◦)?

Example 4.4.8 Let G = (Sym, ◦) and consider the subgroups H = {r360 , r180 }
and K = {r360 , d1 }. Is HK a subgroup of G = (Sym, ◦)?

Example 4.4.9 Let G = (Sym, ◦) and consider the subgroups H = {r360 , d1 }

and K = {r360 , d2 }. Is HK a subgroup of G = (Sym, ◦)?
 43

Example 4.4.10 Let G = (S3 , ◦) and consider the subgroups H = {ι, σ} and
K = {ι, τ }. Is HK a subgroup of G = (S3 , ◦)?
Example 4.4.11 Let G = (S3 , ◦) and consider the subgroups H = {ι, σ} and
K = {ι, δ, σ}. Is HK a subgroup of G = (S3 , ◦)?
Theorem 4.4.12 Let H and K be subgroups of a group G. Then HK is a
subgroup of G if and only if HK = hH ∪ Ki.
Proof : Suppose that HK is a subgroup of G. Let h ∈ H and k ∈ K. Then
h = he ∈ HK and k = ek ∈ HK. Thus, H ∪ K ⊆ HK. But hH ∪ Ki is the
smallest subgroup of G containing H ∪ K. Hence, hH ∪ Ki ⊆ HK.
Let hk ∈ HK. Then h ∈ H and k ∈ K. Since H ⊆ hH ∪ Ki
and K ⊆ hH ∪ Ki, h, k ∈ hH ∪ Ki. But hH ∪ Ki is a subgroup, hence,
hk ∈ hH ∪ Ki. Thus, HK ⊆ hH ∪ Ki. Therefore, HK = hH ∪ Ki.
Conversely, suppose that HK = hH ∪ Ki. Since hH ∪ Ki is a subgroup
of G, it follows that HK is a subgroup of G.
Example 4.4.13 Let G = (Sym, ◦) and consider the subgroups H = {r360 , d1 }
and K = {r360 , d2 }. Is HK a subgroup of G = (Sym, ◦)?
Proof : We have
HK = {r360 r360 , r360 d2 , d1 r360 , d1 d2 } = {r360 , d2 , d1 , r180 },
H ∪ K = {r360 , d1 , d2 },
and
KH = {r360 r360 , r360 h, vr360 , vh} = {r360 , h, v, r180 }.
Thus, HK = KH. By Theorem 4.4.3, HK is a subgroup of G = (Sym, ◦).

  
a 0
Example 4.4.14 Let G = GL(2, R) and let H = ∈ G|a 6= 0 . Then
0 a
H is a subgroup of G.

Example 4.4.15 Find all the subgroups of the group Z12 .

Example 4.4.16 Find all the subgroups of the group S3 .

Example 4.4.17 If H is a subgroup of G, then the centralizer of H is the set

C(H) = {x ∈ G|xh = hx for all h ∈ H}. Prove that C(H) is a subgroup of
G.
 44

4.5 Cylic Groups

Definition 4.5.1 A group G is called a cyclic group if there exists a ∈ G such

that G = hai.

We recall that hai = {an |n ∈ Z}.

Theorem 4.5.2 Every cyclic group is abelian.

Proof : Let G = hai and let x, y ∈ G. Then x = am and y = an for some

m, n ∈ Z. Thus,
xy = am an = am+n = an+m = an am = yx.
Hence, G is abelian. 

Example 4.5.3 (Z, +) is a cyclic group.

Proof : Consider 1 ∈ Z. Then h1i = {1(n) : n ∈ Z} = {n : n ∈ Z} = Z. Hence,

Z = h1i. 

Example 4.5.4 Let n ∈ Z. Then nZ is a cyclic group.

Proof : Consider n ∈ nZ. Then hni = {nk : k ∈ Z} = nZ. Hence, nZ = hni.



Example 4.5.5 (Zn , +n ) is a cyclic group.

Proof : Consider [1] ∈ Zn . Then h[1]i = {[1]n : n ∈ Z} = {[n] : n ∈ Z} = Zn .

Hence, Zn = h[1]i. 

Example 4.5.6 Let a be a symbol and n a positive integer. Define ∗ by

means of the following operation table.
∗ a0 a1 a2 · · · an−2 an−1
a0 a0 a1 a2 · · · an−2 an−1
a1 a1 a2 a3 · · · an−1 a0
a2 a2 a3 a4 ··· a0 a1
.. .. .. .. .. .. ..
. . . . . . .
an−2 an−2 an−1 a0 ··· a n−4
an−3
an−1 an−1 a0 a1 · · · an−3 an−2
 45

Then ({a0 , a1 , ..., an−2 , an−1 }, ∗) is a cyclic group generated by a1 .

Example 4.5.7 Consider the set G = {e, a, b, c}. Define ∗ on G by means of

the following operation table.
∗ e a b c
e e a b c
a a e c b
b b c e a
c c b a e

From the multiplication table, it follows that (G, ∗) is an abelian group. But
G is not a cyclic group since

hei = {e}, hai = {e, a}, hbi = {e, b}, hci = {e, c}

and each of these subgroups is not equal to G. This group is known as the
Klein 4-group.

Theorem 4.5.8 Let hai be a cyclic group of order n. Then hai = {e, a, a2 , ..., an−1 }.

Proof : Note that hai = {ai |i ∈ Z}. Since hai is finite, there exists i, j ∈ Z
(j > i) such that ai = aj . Thus, aj−i = e and j − i > 0. Let m be the
smallest positive integer such that am = e. We claim that the elements of the
set S = {e, a, a2 , ..., am−1 } are distinct.
Suppose that that there exists integers s, t with 0 ≤ s < t < m such
that a 6= at . Then at−s = e with 0 < t − s < m. This contradicts the
s

minimality of m. Thus, the elements of the set S = {e, a, a2 , ..., am−1 } are
distinct and S ⊆ hai.
Now, let ak ∈ hai. By the division algorithm, there exists integers q, r
such that k = qm + r, where 0 ≤ r < m. Hence, ak = (aq )m ar = ear =
ar ∈ S. Thus, hai ⊆ S. This means that S = hai. Therefore, m = n and
hai = {e, a, a2 , ..., an−1 }. 

The following corollaries immediately follows from the proof of Theorem

4.2.8.

Corollary 4.5.9 Let hai be a finite cyclic group. Then ◦(a) = |hai|.

Corollary 4.5.10 A finite group G is a cyclic group if and only if there exists
an element a ∈ G such that ◦(a) = |G|.
 46

Theorem 4.5.11 Every subgroup of a cyclic group is cyclic.

Proof : Let H be a subgroup of a cyclic group G = hai. If H = {e}, then H =

hei. Suppose that H 6= {e}. Then there exists b ∈ H such that b 6= e. Since
b ∈ G, b = am for some integer m 6= 0. Since H is a group, a−m = b−1 ∈ H.
This implies that either m is −m is positive. Thus, H contains at least one
element which is a positive power of a. Let n be the smallest positive integer
such that an ∈ H. We claim that H = han i.
Since an ∈ H, han i ⊆ H. Let h ∈ H. Then h = ak for aome integer k.
By the division algorithm, there exist integers q, r such that k = nq + r, where
0 ≤ r < n. Thus, ar = ak−nq = ak (an )−q . Since ak , an ∈ H, we have ar ∈ H.
If r > 0, then it contradicts the minimality of n. Hence, r = 0 and k = nq.
This implies that h = ak = (an )q ∈ han i, that is H ⊆ han i. Hence, H = han i.
Therefore, H is cyclic. 

Example 4.5.12 (Q, +) is not cyclic.

Proof : Suppose Q is cyclic. Then Q = h pq i, where p and q are relatively prime.

This means that every element of Q is generated by pq . Consider 2q p
∈ Q. Then
p p 1
there exists integer n such that n q = 2q . This implies that n = 2 . This is a
contradiction. Therefore, Q is not cyclic. 

Example 4.5.13 Let G be an infinite cyclic group generated by a. Show that

(a) ar = at if and only if r = t, where r, t are integers.
(b) G has exactly two generators.

Proof : (a) Let ar = at , where r, t ∈ Z. Suppose that r 6= t. Without loss of

generality, we may assume that r > t. Then r − t > 0 and ar−t = e. Thus,
◦(a) = r − t, that is, the order of G is finite. This is a contradiction since the
order of G is infinite.
Conversely, if r = t, then ar = at .

(b) Suppose that G = hbi for some b ∈ G. Since a ∈ G = hbi and

b ∈ G = hai, we have a = br and b = at for some integers r and t. Thus,

a = br = (at )r = art .

By (a), rt = 1. This implies that either r = t = 1 or r = t = −1. Hence,

either b = a or b = a−1 . Since 1 6= −1, by (a), a 6= a−1 . Therefore, G has
exactly two generators. 
 47

Example 4.5.14 Let G = hai be a cyclic group of order 30. Find all cyclic
subgroups of G.
Example 4.5.15 . Find all cyclic subgroups of S3 . Is S3 a cyclic group?
Example 4.5.16 Show that U8 is not a cyclic group. Show that U9 is a cyclic
group.

4.6 Cosets and Lagrange’s Theorem

Definition 4.6.1 Let H be a subgroup of a group G and a ∈ G. The sets

aH = {ah|h ∈ H} and Ha = {ha|h ∈ H} are called the left and right cosets
of H in G, respectively. The element a is called a representative of aH and
Ha.

If G is abelian, then of course aH = Ha. Observe that eH = H = He

and that a = ae ∈ H and a = ea ∈ H.

Example 4.6.2 Consider the group S3 = {ι, σ, τ, µ, δ, }. Then H = {ι, σ} is

a subgroup of S3 . Find all the left and right cosets of H in S3 .
Solution: We find the left cosets of H in G.
ιH = σH = H
τH = {τ ◦ ι, τ ◦ σ} = {τ, δ}
µH = {µ ◦ ι, µ ◦ σ} = {µ, }
δH = {δ ◦ ι, δ ◦ σ} = {δ, τ }
H = { ◦ ι,  ◦ σ} = {, µ}.
Hence, the left cosets of H in G are H, τ H, and µH.

We find the right cosets of H in G.

Hι = Hσ = H
Hτ = {ι ◦ τ, σ ◦ τ } = {τ, }
Hµ = {ι ◦ µ, σ ◦ µ} = {µ, δ}
Hδ = {ι ◦ δ, σ ◦ δ} = {δ, µ}
H = {ι ◦ , σ ◦ } = {, τ }.
Therefore, the right cosets of H in G are H, Hτ , and Hµ. 
 48

Example 4.6.3 Consider the group G = (Z12 , +12 ). Then H = {[0], [4], [8]}
is a subgroup of S3 . Find all the left and right cosets of H in G.

Solution: We find the left cosets of H in G.

[0] +12 H = {[0], [4], [8]} = H

[1] +12 H = {[1] +12 [0], [1] +12 [4], [1] +12 [8]} = {[1], [5], [9]}
[2] +12 H = {[2] +12 [0], [2] +12 [4], [2] +12 [8]} = {[2], [6], [10]}
[3] +12 H = {[3] +12 [0], [3] +12 [4], [3] +12 [8]} = {[3], [7], [11]}
[4] +12 H = {[4] +12 [0], [4] +12 [4], [4] +12 [8]} = {[4], [8], [0]} = H
[5] +12 H = {[5] +12 [0], [5] +12 [4], [5] +12 [8]} = {[5], [9], [1]} = [1] +12 H
[6] +12 H = {[6] +12 [0], [6] +12 [4], [6] +12 [8]} = {[6], [10], [2]} = [2] +12 H
[7] +12 H = {[7] +12 [0], [7] +12 [4], [7] +12 [8]} = {[7], [11], [3]} = [3] +12 H
[8] +12 H = {[8] +12 [0], [8] +12 [4], [8] +12 [8]} = {[8], [0], [4]} = H
[9] +12 H = {[9] +12 [0], [9] +12 [4], [9] +12 [8]} = {[9], [1], [5]} = [1] +12 H
[10] +12 H = {[10] +12 [0], [10] +12 [4], [10] +12 [8]} = {[10], [6], [2]} = [2] +12 H
[11] +12 H = {[11] +12 [0], [11] +12 [4], [11] +12 [8]} = {[11], [7], [3]} = [3] +12 H

Hence, the left cosets of H in G are

H = {[0], [4], [8]},

[1] +12 H = {[1], [5], [9]},
[2] +12 H = {[2], [6], [10]},
[3] +12 H = {[3], [7], [11]}.

Since G = (Z12 , +12 ) is abelian, it follows that the left cosets and the
right cosets of H in G are equal. Therefore, the right cosets of H in G are

H = {[0], [4], [8]},

H +12 [1] = {[1], [5], [9]},
H +12 [2] = {[2], [6], [10]},
H +12 [3] = {[3], [7], [11]}.

Example 4.6.4 Consider the group (Z, +) and the subgroup H = h5i =
{5n|n ∈ Z}. Find all the left and right cosets of H in (Z, +).
 49

Solution: We find the left cosets of H in (Z, +).

0+H = 0 + {5n|n ∈ Z} = {5n|n ∈ Z} = H

1+H = 1 + {5n|n ∈ Z} = {5n + 1|n ∈ Z}
2+H = 2 + {5n|n ∈ Z} = {5n + 2|n ∈ Z}
3+H = 3 + {5n|n ∈ Z} = {5n + 3|n ∈ Z}
4+H = 4 + {5n|n ∈ Z} = {5n + 4|n ∈ Z}
5+H = 5 + {5n|n ∈ Z} = {5n + 5|n ∈ Z} = {5n|n ∈ Z} = H
6+H = 6 + {5n|n ∈ Z} = {5n + 6|n ∈ Z} = {5n + 1|n ∈ Z} = 1 + H.

Hence, the left cosets of H in (Z, +) are

H = {5n|n ∈ Z},
1+H = {5n + 1|n ∈ Z}},
2+H = {5n + 2|n ∈ Z},
3+H = {5n + 3|n ∈ Z},
4+H = {5n + 4|n ∈ Z}.

Since (Z, +) is abelian, it follows that the left cosets and the right cosets
of H in (Z, +) are equal. Therefore, the right cosets of H in (Z, +) are

H = {5n|n ∈ Z},
H + 1 = {5n + 1|n ∈ Z},
H + 2 = {5n + 2|n ∈ Z},
H + 3 = {5n + 3|n ∈ Z},
H + 4 = {5n + 4|n ∈ Z}.

Theorem 4.6.5 Let H be a subgroup of a group G and a, b ∈ G. Then

(i) aH = bH if and only if b−1 a ∈ H.
(ii) Ha = Hb if and only if ab−1 ∈ H.

Proof : (i) Suppose that aH = bH. Since a = ae ∈ aH, there exists h0 ∈ H

such that a = bh0 . Thus, b−1 a = h0 ∈ H.
Conversely, suppose that b−1 a ∈ H. Then h0 = b−1 a for some h0 ∈ H.
Thus, a = bh0 . Now, let ah ∈ aH. Then ah = bh0 h ∈ bH, which implies that
aH ⊆ bH. Next, let bh ∈ bH. Since a = bh0 , it follows that b = ah0−1 . Hence,
bh = ah0−1 h ∈ aH, which implies that bH ⊆ aH. Therefore, aH = bH.
(ii) The proof is similar to (i). 
 50

Theorem 4.6.6 Let H be a subgroup of a group G. Then for all a, b ∈ G,

either aH = bH or aH ∩ bH = ∅.

Proof : Let a, b ∈ G. If aH ∩ bH = ∅, then we are done. Suppose that

aH ∩ bH 6= ∅. Then there exists c ∈ aH ∩ bH, that is, c ∈ aH and c ∈ bH.
Thus, c = ah1 and c = ah2 for some h1 , h2 ∈ H. Hence, ah1 = ah2 , which
implies that b−1 a = h2 h−1 −1
1 . Therefore, b a ∈ H. By Theorem 4.6.5(i),
aH = bH.

Theorem 4.6.7 Let H be a subgroup of a group G. Then the elements of H

are in one-one correspondence with the elements of any left (right) coset of H
in G.

Proof : Let a ∈ G and aH be a left coset of H in G. Define f : H → aH by

f (h) = ah for all h ∈ H.
Let h1 , h2 ∈ H such that h1 = h2 . Then ah1 = ah2 , that is, f (h1 ) =
f (h2 ). Thus, f is well-defined.
Let Let h1 , h2 ∈ H such that f (h1 ) = f (h2 ). Then ah1 = ah2 , that is,
h1 = h2 . Hence, f is one-one.
Let y ∈ aH. Then y = ah for some h ∈ H. Thus, there exists h ∈ H
such that f (h) = ah. Hence, f is onto.
Therefore, f maps H onto aH. Similarly, we can show that the elements
of H are in one-one correspondence with the elements of Ha. 

The next result is a consequence of Theorem 4.6.7.

Corollary 4.6.8 Let H be a subgroup of a group G. Then for all a ∈ G,

|H| = |aH| = |Ha|.

Theorem 4.6.9 Let H be a subgroup of a group G. Then there is a one-one

correspondence of the set of all left cosets of H in G onto the set of all right
cosets of H in G.

Proof : Let L = {aH|a ∈ G} be the set of all left cosets of H in G and

R = {Ha|a ∈ G} be the set of all right cosets of H in G. Define f : L → R
by f (aH) = Ha−1 for all aH ∈ L. Then for all aH ∈ L, f (aH) = Ha−1 ∈ R.
Let aH, bH ∈ L such that aH = bH. By Theorem 4.6.5(i), b−1 a ∈ H,
which implies that b−1 (a−1 )−1 ∈ H. By Theorem 4.6.5(ii), Hb−1 = Ha−1 .
Thus, f (aH) = f (bH). Hence, f is well-defined.
Let aH, bH ∈ L such that f (aH) = f (bH). Then Ha−1 = Hb−1 .
 51

By Theorem 4.6.5(ii), a−1 (b−1 )−1 ∈ H, that is, a−1 b ∈ H. Thus, b−1 a =
(a−1 b)−1 ∈ H. By Theorem 4.6.5(i), aH = bH. Hence, f is one-one.
Let Ha ∈ R. Then a ∈ G and a−1 ∈ G. Thus, there exists a−1 H ∈ L
such that f (a−1 H) = H(a−1 )−1 = Ha. Hence, f is onto.
Therefore, there is a one-one correspondence of the set of all left cosets
of H in G onto the set of all right cosets of H in G. 

Definition 4.6.10 Let H be a subgroup of a group G. Then the number of

distinct left (right) cosets, written [G : H], of H in G is called the index of H
in G.

By Theorem 4.6.8, the number of left cosets and the number of right
cosets of a subgroup H of a group G are the same. Thus, [G : H] is well-defined.

If G is finite, then [G : H] is finite. The following example is one, where

G is infinite and [G : H] is finite.

Example 4.6.11 Let n be a fixed positive integer. Consider the cyclic group
(hni, +) of (Z, +). Let k + hni be a left coset of hni in Z. By the division
algorithm, there exist integers q and r such that k = qn + r, where 0 ≤ r < n.
Then k − r = qn ∈ hni. By Theorem 4.6.5, k + hni = r + hni. Now, suppose
that i + hni = j + hni, where 0 ≤ i, j < n. Then i − j ∈ hni by Theorem 4.6.5.
Thus, n|(i − j). Since n > (i − j), we have either i − j = 0 or i = j. Hence,
the distinct left cosets of hni in Z are 0 + hni, 1 + hni,...,n − 1 + hni.

Theorem 4.6.12 (Lagrange) Let H be a subgroup of a finite group G. Then

the order of H divides the order of G. In particular, |G| = [G : H]|H|.

Proof : Since G is a finite group, the number of left cosets of H in G is finite.

Let {a1 H, a2 H, ..., ar H} be the set of all distinct left cosets of H in G. Then
[r
G= H and ai H ∩ aj H = ∅ for all i 6= j, 1 ≤ i, j ≤ r. Hence, [G : H] = r
i=1
and

|G| = |a1 H| + |a2 H| + · · · + |ar H|.

 52

By Corollary 4.6.8, |H| = |ai H| for all i, 1 ≤ i ≤ r. Thus,

|G| = |H| + |H| + · · · + |H| (r terms)

= r|H|
= [G : H]|H|.

Therefore, the order of H divides the order of G. 

Corollary 4.6.13 Let G be a group of finite order n. Then the order of any
element a of G divides n and an = e.

Proof : Let a ∈ G and ◦(a) = k. Let H = hai. By Corollary 4.5.9, |H| =

|hai| = ◦(a) = k. By Theorem 4.6.12, k divides n. Thus, there exists integer
q such that n = kq. Hence,

an = akq = (ak )q = eq = e.

Let G be a finite group of order n and a ∈ G. Then ◦(a) divides n by

Corollary 4.6.13. Thus, to find the find ◦(a), we only need to check ak , where
k is a positive divisor of n.

Example 4.6.14 Consider the group Z20 and [6] ∈ Z20 . Find ◦([6]).

Solution: We have |Z20 | = 20 and the positive divisors of 20 are: 1, 2, 4, 5, 10,

and 20. Thus,

1[6] = [6] 6= [0], 2[6] = [12] 6= [0], 4[6] = [4] 6= [0], 5[6] = [10] 6= [0], 10[6] = [0].

Hence, ◦([6]) = 10. 

Corollary 4.6.15 Let G be a group of prime order. Then G is cyclic.

Proof : Suppose that |G| is prime. Then |G| ≥ 2. Thus, there exists a ∈ G
such that a 6= e. Let H = hai. Then {e} ⊆ H and by Lagrange’s Theorem,
|H| divides |G|. But |G| is prime and |H| =
6 1. Thus, |H| = |G|. Since H ⊆ G,
it follows that H = G. Therefore, G is cyclic. 

Example 4.6.16 Let H be a subgroup of a group G. Show that for all a ∈ G,

aH = H if and only if a ∈ H.
 53

Proof : Suppose that aH = H. Then a = ae ∈ aH = H.

Conversely, suppose that a ∈ H. Let ah ∈ aH, where h ∈ H. Thus,
ah ∈ H, that is, aH ⊆ H. Next, let h ∈ H. Then a−1 h ∈ H. Hence,
h = a(a−1 h) ∈ aH, that is, H ⊆ aH. Therefore, aH = H. 

Example 4.6.17 In S3 , for each subgroup, find the left and right cosets.

Example 4.6.18 Let H = {r360 , h} be a subgroup of the group of symmetries

of the square. List all the left and right cosets of H.

4.7 Normal Subgroups

Definition 4.7.1 Let G be a group. A subgroup H of G is said to be a normal

subgroup of G if aH = Ha for all a ∈ G.

From the definition of a normal subgroup, it follows that for any group
G, G and {e} are normal subgroups of G.

If H is a normal subgroup of G, this does not always mean that ah = ha

for all h ∈ H and for all a ∈ G as shown by the following example.

Example 4.7.2 Consider the subgroup H5 = {ι, δ, } of S3 and let δ ∈ H5 .

Then

σ◦δ =τ and δ ◦ σ = µ.

Hence,

σ ◦ δ 6= δ ◦ σ

even though

σH = Hσ.

Theorem 4.7.3 Let H be a subgroup of a group G. Then H is a normal

subgroup of G if and only if for all a ∈ G, aHa−1 ⊆ H.
 54

Proof : Suppose that H is a normal subgroup of G. Let a ∈ G and let aha−1 ∈

aHa−1 , where h ∈ H. Since H is a normal subgroup of G, aH = Ha. Also,
since ah ∈ aH, we have ah ∈ Ha. Thus, ah = h0 a for some h0 ∈ H. Hence,
aha−1 = h0 ∈ H, which implies that aHa−1 ⊆ H.
Conversely, suppose that aHa−1 ⊆ H for all a ∈ G. Let ah ∈ aH,
where h ∈ H. Then aha−1 ∈ aHa−1 ⊆ H, that is, aha−1 ∈ H. Thus,
aha−1 = h0 for some h0 ∈ H. This implies that ah = h0 a ∈ Ha. Hence,
aH ⊆ Ha. Similarly, we can show that Ha ⊆ aH. Therefore, aH = Ha.
Consequently, H is a normal subgroup of G.
The next result is an immediate consequence of Theorem 4.7.3.

Corollary 4.7.4 Let H be a subgroup of a group G. Then H is a normal

subgroup of G if and only if for all a ∈ G and for all h ∈ H, aha−1 ∈ H.
Theorem 4.7.5 Let H and K be normal subgroups of a group G. Then
(i) H ∩ K is a normal subgroup of G.
(ii) HK = KH is a normal subgroup of G.

Proof : (i) Since the intersection of subgroups is a subgroup, H ∩ K is a

subgroup of G. Let g ∈ G and let gag −1 ∈ g(H ∩ K)g −1 , where a ∈ H ∩ K.
Then a ∈ H and a ∈ K. Since H and K are normal subgroups, gag −1 ∈ H and
gag −1 ∈ K. Thus, gag −1 ∈ H ∩ K, which means that g(H ∩ K)g −1 ⊆ H ∩ K.
By Theorem 4.7.3, H ∩ K is a normal subgroup of G.
(ii) First we show that HK = KH. Let hk ∈ HK, where h ∈ H
and k ∈ K. Since K is a normal subgroup and h ∈ G, we have hK = Kh.
Thus, hk ∈ hK. Since Kh ⊆ KH, we have hk ∈ HK. Hence, HK ⊆ KH.
Similarly, KH ⊆ HK and so HK = KH. By Theorem 4.4.3, HK is a
subgroup of G. Let g ∈ G. Since H and K are normal subgroups, gHg −1 ⊆ H
and gKg −1 ⊆ K. Now,
g(HK)g −1 = g(HeK)g −1
= g(Hg −1 gK)g −1
= (gHg −1 )(gKg −1 )
⊆ HK.
Therefore, by Theorem 4.7.3, HK is a normal subgroup of G. 
Remark 4.7.6 If H and K are subgroups of G such that H is normal in G,
then H is normal subgroup of K.
 55

4.8 Quotient Groups

Theorem 4.8.1 Let H be a normal subgroup of a group G. Denote the set of

all left cosets {aH|a ∈ G} by G/H and define ∗ on G/H by for all aH, bH ∈
G/H,
(aH) ∗ (bH) = abH.
Then (G/H, ∗) is a group.

Proof : Let aH, bH, a0 H, b0 H ∈ G/H such that aH = a0 H and bH = b0 H. Then

a = a0 h1 and b = b0 h2 for some h1 , h2 ∈ H. Thus,
(a0 b0 )−1 (ab) = b0−1 a0−1 ab
= b0−1 a0−1 a0 h1 b0 h2
= b0−1 h1 b0 h2
= (b0−1 h1 b0 )h2 .

Since H is a normal subgroup, (b0−1 h1 b0 )h2 ∈ H, that is, (a0 b0 )−1 (ab) ∈ H. By
Theorem 4.3.3, abH = a0 b0 H. Hence, ∗ is well-defined.
(G1) Let aH, bH, cH ∈ G/H. Then

(aH ∗ bH) ∗ cH = abH ∗ cH

= (ab)cH
= a(bc)H
= aH ∗ bcH
= aH ∗ (bH ∗ cH).
Hence, ∗ is associative.
(G2) Since e ∈ G, there exists eH ∈ G/H such that
aH ∗ eH = aeH = aH = eaH = eH ∗ aH.
for all aH ∈ G/H.
(G3) Let aH ∈ G/H. Then a ∈ G and a−1 ∈ G. Thus, there exists
−1
a H ∈ G/H such that
aH ∗ a−1 H = aa−1 H = eH = a−1 aH = a−1 H ∗ aH.
Therefore, (G/H, ∗) is a group. 

Definition 4.8.2 Let G be a group and H be a normal subgroup of G. The

group G/H is called the quotient group of G by H.
 56

Corollary 4.8.3 If G is a finite group and H be a normal subgroup of G,

|G|
then |G/H| = .
|H|

Proof : Since G/H has as its elements the left cosets of H in G, and since there
|G| |G|
are [G : H] = such cosets, we have |G/H| = . 
|H| |H|

Example 4.8.4 Consider the subgroup (hni, +) of the group (Z, +), where n
is a fixed positive integer. Since Z is commutative, hni is a normal subgroup
of Z. Hence, Z/hni is a group, where

(a + hni) + (b + hni) = (a + b) + hni.

for all a + hni, b + hni ∈ Z/hni. The distinct left coset of hni in Z are

0 + hni, 1 + hni, 2 + hni, ..., n − 1 + hni.

Therefore,

Z/hni = {0 + hni, 1 + hni, 2 + hni, ..., n − 1 + hni}.

Example 4.8.5 Consider the subgroup H = {[0], [4]} of the group Z8 . Then
H is a normal subgroup of Z8 . Find Z8 /H.

Solution: Since |H| = 2 and |Z8 | = 8, we have |Z8 /H| = [Z8 : H] = 4. Now,

[0] +8 H = [4] +8 H = H

[1] +8 H = {[1] + [0], [1] + [4]} = {[1], [5]} = [5] +8 H

[2] +8 H = {[2] + [0], [2] + [4]} = {[2], [6]} = [6] +8 H

[3] +8 H = {[3] + [0], [3] + [4]} = {[3], [7]} = [7] +8 H

Therefore, Z8 /H = {[0] +8 H, [1] +8 H, [2] +8 H, [3] +8 H}. 

Example 4.8.6 Consider the subgroup H5 = {ι, δ, } of the group S3 . Then

H5 is a normal subgroup of S3 . Find S3 /H5 .

Example 4.8.7 Consider Z4 × Z6 , the direct product of Z4 and Z6 . Let

H = h([0], [1])i = {([0], [0]), ([0], [1]), ([0], [2]), ([0], [3]), ([0], [4]), ([0], [5])}.
 57

Since Z4 × Z6 is commutative, H is a normal subgroup of Z4 × Z6 . Find

(Z4 × Z6 )/H.

Example 4.8.8 Show that every subgroup of an abelian group is normal.

\
Example 4.8.9 . Let H be a subgroup of a group G. Then W = gHg −1
g∈G
is a normal subgroup of G.

Example 4.8.10 Let H be a subgroup of a group G. Suppose that the

product of two left cosets of H in G is again a left coset of H in G. Prove that
H is a normal subgroup of G.

Example 4.8.11 Prove that Z(G) is a normal subgroup of a group G.

Example 4.8.12 Let G be a group, K a subgroup of G, and H a normal

subgroup of G. Prove that H ∩ K is a normal subgroup of K.

Proof : Let khk −1 ∈ kH ∩ Kk −1 . Then k ∈ K and h ∈ H ∩ K. Thus,

khk −1 ∈ H since H is normal in G, and khk −1 ∈ K since K is a subgroup of
G. Hence, khk −1 ∈ H ∩ K. This implies that kH ∩ Kk −1 ⊆ H ∩ K. Therefore,
H ∩ K is a normal subgroup of K. 
 CHAPTER 5

HOMOMORPHISMS AND ISOMORPHISMS OF GROUPS

5.1 Homomorphisms of Groups

Definition 5.1.1 A mapping f from a group G into a group G0 is said to be

a homomorphism if for all a, b ∈ G, f (ab) = f (a)f (b).

Notice that on the left side of the relation, namely, in the term f (ab),
the product ab is computed in G using the product of elements of G, whereas
on the right side of the relation, namely, f (a)f (b), the product is that of
elements of G0 .

Example 5.1.2 Let G and G0 be groups. Define f : G → G0 by f (a) = e0 for

all a ∈ G, where e0 is the identity element in G0 . Then f is a homomorphism
of G into G0 . This is called the trivial homomorphism.
Proof : Let a, b ∈ G. Then ab ∈ G. Thus, f (a) = e0 , f (b) = e0 , and f (ab) = e0 .
Hence, f (ab) = e0 = e0 e0 = f (a)f (b). Therefore, f is a homomorphism. 

Example 5.1.3 Let G be a group. Define f : G → G by f (a) = a for all

a ∈ G. Then f is a homomorphism of G into G.
Let a, b ∈ G. Then ab ∈ G. Thus, f (a) = a, f (b) = b, and f (ab) = ab. Hence,
f (ab) = ab = f (a)f (b). Therefore, f is a homomorphism. 

Theorem 5.1.4 Let f be a homomorphism of a group G into a group G0 .

Then
(i) f (e) = e0 .
(ii) f (a−1 ) = f (a)−1 for all a ∈ G.
(iii) If H is a subgroup of G, then f (H) = {f (h)|h ∈ H} is a subgroup
0
G.
(iv) If H 0 is a subgroup of G0 , then f −1 (H 0 ) = {g ∈ G|f (g) ∈ H 0 } is a
subgroup of G, and if H 0 is a normal subgroup of G0 , then f −1 (H 0 ) is a normal
subgroup of G.
(v) If G is abelian, then f (G) is abelian.
(vi) If a ∈ G is such that ◦(a) = n, then ◦(f (a)) divides n.
 59

Proof : (i) We have

f (e)f (e) = f (ee), since f is a homomorphism

= f (e)
= e0 f (e).

By the cancellation law, f (e) = e0 .

(ii) Let a ∈ G. Then f (a) ∈ G0 . Thus, there exists a−1 ∈ G such that
aa−1 = e and there exists f (a)−1 ∈ G0 such that f (a)f (a)−1 = e0 . Hence,

f (a−1 ) = f (a−1 )e0

= f (a−1 )[f (a)f (a)−1 ]
= [f (a−1 )f (a)]f (a)−1
= f (aa−1 )f (a)−1 , since f is a homomorphism
= f (e)f (a)−1
= e0 f (a)−1 , by (i)
= f (a)−1 .

(iii) Let H be a subgroup of G. Then e ∈ H and by (i), e0 = f (e).

Thus, e0 = f (e) ∈ f (H), which implies that f (H) 6= ∅.
Let x, y ∈ f (H). Then there exists a, b ∈ H such that x = f (a) and
y = f (b). Thus ab−1 ∈ H and f (ab−1 ) ∈ f (H). Hence,

xy −1 = f (a)f (b)−1
= f (a)f (b−1 ), by (ii)
= f (ab−1 ), since f is a homorphism.

This implies that xy −1 ∈ f (H). Therefore, f (H) is a subgroup of G0 .

(iv) Suppose that H 0 is a subgroup of G0 . Then e0 ∈ H 0 . By (i),

f (e) = e0 . Thus, e ∈ f −1 (H 0 ). Hence, f −1 (H 0 ) is nonempty.
Let a, b ∈ f −1 (H 0 ). Then f (a), f (b) ∈ H 0 , and f (a)f (b)−1 ∈ H 0 since
H is a subgroup of G0 . But
0

f (ab−1 ) = f (a)f (b−1 ), since F is a homomorphism

= f (a)f (b)−1 , by (ii).

Hence, f (ab−1 ) ∈ H 0 . This implies that ab−1 ∈ f −1 (H 0 ).

Therefore, f −1 (H 0 ) is a subgroup of G.
 60

Suppose H 0 is a normal subgroup of G0 . Let g ∈ G and a ∈ f −1 (H 0 ).

Then f (g) ∈ G0 and f (a) ∈ H 0 . Thus, f (g)f (a)f (g)−1 ∈ H 0 since H 0 is a
normal subgroup of G0 . But

f (gag −1 ) = f (g)f (a)f (g −1 ), since f is a homomorphism

= f (g)f (a)f (b)−1 , by (ii).

Hence, f (gag −1 ) ∈ H 0 . This implies that gag −1 ∈ f −1 (H 0 ). Therefore, f −1 (H 0 )

is a normal subgroup of G.

(v) Suppose G is abelian. Let x, y ∈ f (G). Then there exists a, b ∈ G

such that x = f (a) and y = f (b). Since G is abelian, it follows that ab = ba.
and f (a), f (b) ∈ f (G). Since f is a homomorphism, we have

xy = f (a)f (b)
= f (ab), since f is a homomorphism
= f (ba)
= f (b)f (a), since f is a homomorphism
= yx.

Therefore, f (G) is abelian.

(vi) Let a ∈ G such that ◦(a) = n. Then an = e. Thus,

(f (a))n = f (an ), since f is a homomorphism

= f (e)
= e0 , by (i).

By Theorem 2.2.13, ◦(f (a)) divides n. 

Definition 5.1.5 Let f be a homomorphism of a group G into a group G0 .

The kernel of f , written Kerf , is defined to be the set

Kerf = {a ∈ G|f (a) = e0 }.

Example 5.1.6 Define the function f from (Z, +) into (Zn , +n ) by f (a) = [a]
for all a ∈ Z. It follows from the definition that f maps Z onto Zn . Let a, b ∈ Z.
Then
 61

f (a + b) = [a + b] = [a] +n [b] = f (a) +n f (b).

Thus, f is a homomorphism of Z into Zn . We have

kerf = {a ∈ Z|f (a) = [0]}

= {a ∈ Z|[a] = [0]}
= {a ∈ Z|n|a}
= {a ∈ Z|a = qn, q ∈ Z}
= {a = qn|q ∈ Z}.

The above example shows that a nontrivial finite group may be an

image of an infinite group under a homomorphism. By Theorem 5.1.4(v), a
nonabelian group cannot be an image under a homomorphism of an abelian
group. The next example show that two finite groups G and G0 having the
same number of elements need not have a homomorphism from G onto G0 .

Example 5.1.7 The groups Z4 × Z4 and Z8 × Z2 are abelian and each of

order 16. Suppose there exists a homomorphism f of Z4 × Z4 onto Z8 × Z2 .
Now, a = ([7].[0]) ∈ Z8 × Z2 and ◦(a) = 8. Since f is onto, there exists
b ∈ Z4 × Z4 such that f (b) = a. By Theorem 4.1.4(vi), ◦(f (b)) divides ◦(b).
Since ◦(f (b)) = 8 and Z4 × Z4 has elements of order 1,2, and 4 only, ◦(f (b))
cannot divide ◦(b). This is a contradiction. Hence, there does not exist any
homomorphism from Z4 × Z4 onto Z8 × Z2 .

Definition 5.1.8 Let G and G0 be groups. A homomorphism f : G → G0 is

called an epimorphism if f is onto G0 and f is called a monomorphism if
f is one-one. If there is an epimorphism f from G onto G0 , then G0 is called
a homomorphic image of G.

The homomorpgism in Example 5.1.7 is an epimorphism, but not a

monomorphism.

Example 5.1.9 Let R∗ be a group of all nonzero real numbers under multiplication.
Define f : R∗ → R∗ by f (a) = |a|. Then f is neither an epimorphism nor a
monomorphism.
 62

Proof : Let a, b ∈ R∗ . Then f (ab) = |ab| = |a||b| = f (a)f (b). Thus, f is a

homomorphism.
Consider 1, −1 ∈ R∗ . Then f (1) = |1| = 1 and f (−1) = | − 1| = 1.
Hence, there exist 1, −1 ∈ R∗ with 1 6= −1 but f (1) = f (−1). Hence, f is not
injective.
Consider −1 ∈ R∗ . Suppose f is surjective. Then there exists x ∈ R∗
such that f (x) = −1. Thus, |x| = −1. This is a contradiction since a positive
real number is not equal to a negative real number. Hence, f is not surjective.
Therefore, f is neither an epimorphism nor a monomorphism. 

Theorem 5.1.10 Let f be a homomorphism of a group G into a group G0 .

Then f is one-one if and only if Kerf = {e}.

Proof : Suppose f is one-one. Let a ∈ Kerf . Then f (a) = e0 . By Theorem

5.1.4(i), f (e) = e0 . Thus, f (a) = f (e). Since f is one-one, a = e. hence,
Kerf = {e}.
Conversely, suppose that Kerf = {e}. Let a, b ∈ G such that f (a) =
f (b). Then f (a)f (b)−1 = e0 . Since f is a homomorphism,

f (ab−1 ) = f (a)f (b−1 ) = f (a)f (b)−1 = e0 .

This implies that ab−1 ∈ Kerf = {e}. Hence, ab−1 = e, that is, a = b.
Therefore, f is one-one. 

Theorem 5.1.11 Let f be a homomorphism of a group G into a group G0 .

Then Kerf is a normal subgroup of G.

Proof : By Theorem 5.1.4(i), f (e) = e0 . Thus, e ∈ Kerf , which implies that

Kerf 6= ∅. Let a, b ∈ Kerf . Then f (a) = e0 and f (b) = e0 . Since f is a
homomorphism, we have

f (ab−1 ) = f (a)f (b−1 ) = f (a)f (b)−1 = e0 e0−1 = e0 .

Hence, ab−1 Kerf . This means that Kerf is a subgroup of G. Now, let g ∈ G
and k ∈ Kerf . Then f (k) = e0 . Since f is a homomorphism,

f (gkg −1 ) = f (g)f (k)f (g −1 ) = f (g)e0 f (g)−1 = f (g)f (g)−1 = e0 .

Therefore, gkg −1 ∈ Kerf . By Corollary 4.4.4, Kerf is a normal subgroup of

G. 
 63
  
a b
Example 5.1.12 Let GL(2, R) = |a, b, c, d ∈ R, ad − bc 6= 0 be a
c d
nonabelian group and let R∗ be the group of all nonzero
 real numbners under

a b
multiplication. Define f : GL(2, R) → R∗ by f = ad − bc for all
  c d
a b
∈ GL(2, R). Show that f is an epimorphism but not a monomorphism.
c d
Find Kerf .

The previous example shows that there may exist a homomorphism of

a noncommutative group onto a commutative group.

Theorem 5.1.13 Let H be a normal subgroup of G. Define a function π from

G onto the quotient group G/H by π(a) = aH for all a ∈ G. Then π is an
epimorphism of G onto G/H and Kerπ = H.

Proof : From the definition of π, it follows that π is a function from G onto

G/H. Let a, b ∈ G. Then π(a) = aH and π(b) = bH. Thus, π(ab) = (ab)H =
(aH)(bH) = π(a)π(b). Hence, π is a homomorphism. Now,
Kerπ = {a ∈ G|π(a) = H} = {a ∈ G|aH = H} = {a ∈ G|a ∈ H}.
Therefore, Kerπ = H. 

The map π : G → G/H, where H is a normal subgroup of a group G,

is called the canonical epimorphism.

Definition 5.1.14 A homomorphism f of a group G into a group G0 is called

an isomorphism of G onto G0 if f is one-one and onto G0 . In this case, we
write G ≡ G0 and say that G and G0 are isomorphic. An isomorphism of a
group G onto G is called an automorphism.

Theorem 5.1.15 Let f be an isomorphism of a group G onto a group G0 .

Then
(i) f −1 : G0 → G is an isomorphism.
(ii) G is abelian if and only if G0 is abelian.
(iii) For all a ∈ G, ◦(a) = ◦(f (a)).
(iv) G is a torsion group if and only if G0 is a torsion group.
(v) G is cyclic if and only if G0 is cyclic.
 64

Proof : (i) Since f is one-one and onto G0 , it follows that f −1 is one-one and onto
G. Let x, y ∈ G0 . Since f is onto G0 , there exist a, b ∈ G such that f (a) = x and
f (b) = y. Thus, a = f −1 (x), b = f −1 (y), and xy = f (a)f (b) = f (ab). Hence,
f −1 (xy) = ab = f −1 (x)f −1 (y), which implies that f −1 is a homomorphism.
Therefore, f −1 is an isomorphism.

(ii) Suppose that G is abelian. Let x, y ∈ G0 . Since f is onto G0 , there

exist a, b ∈ G such that f (a) = x and f (b) = y. Thus,

xy = f (a)f (b) = f (ab) = f (ba) = f (b)f (a) = yx.

Hence, G0 is abelian.
Conversely, suppose that G0 is abelian. Let a, b ∈ G. Then

f (ab) = f (a)f (b) = f (b)f (a) = f (ba).

Since f is one-one, ab = ba. Hence, G is abelian.

(iii) Let a ∈ G. By induction, it follows that for all positive integers n,

f (a ) = (f (a))n . Since f is one-one, an = e if and only if (f (a))n = f (an ) =
n

f (e) = e0 . Thus, a is of finite order if and only if f (a) is of finite order. Suppose
that ◦(a) = m and ◦(f (a)) = n. Since am = e, (f (a))m = e0 . By Theorem
2.2.11, n|m. Also, (f (a))n = e0 implies that an = e. Hence, m|n. Since m and
n are both positive integers, m = n.

(iv) This follows from (iii).

(v) Suppose G is cyclic. Then G = hai for some a ∈ G. Since f (a) ∈ G0 ,

hf (a)i ⊆ G0 . Let b ∈ G0 . Since f is onto G0 , there exists c ∈ G such that
f (c) = b. Now, c = an for some integer n. Thus,

b = f (c) = f (an ) = (f (a))n ∈ hf (a)i.

Hence, G0 = hf (a)i, which means that G0 is cyclic. The converse follows since
f −1 is an isomorphism. 

Example 5.1.16 (R, +) is isomorphic to (R+ , ·).

Proof : Define f : R → R+ by f (a) = ea for all a ∈ R. Let a, b ∈ R such that

a = b. Then ea = eb , that is, f (a) = f (b). Thus, f is well-defined.
 65

Let a, b ∈ R such that f (a) = f (b). Then ea = eb , that is, a = b.

Hence, f is one-one.

Let y ∈ R+ . Then ln y ∈ R. Choose a = ln y. Thus, a ∈ R and

f (a) = ea = eln y = y. Hence, f is surjective.

Therefore, f is an isomorphism. Accordingly, (R, +) is isomorphic to

+
(R , ·). 

Example 5.1.17 (Z, +) is not isomorphic to (Q, +).

Proof : (Z, +) is cyclic since Z = h1i. By Example 3.2.13, (Q, +) is not cyclic.
By Theorem 5.1.15(v), (Z, +) is not isomorphic to (Q, +). 

Example 5.1.18 (Q, +) is not isomorphic to (Q∗ , ·).

Proof : Every nonidentity element of (Q, +) is of infinite order. There exists a

nonidentity element −1 ∈ Q∗ which is of finite order, that is, ◦(−1) = 2 since
(−1)2 = 1. By Theorem 5.1.15(iii), (Q, +) is not isomorphic to (Q∗ , ·). 

Example 5.1.19 Let n be a positive integer. Define f : Zn → Z/hni by

f ([a]) = a + hni for all [a] ∈ Zn . Then f is an isomorphism of Zn onto Z/hni.

Proof : Let [a], [b] ∈ Zn such that [a] = [b]. Then n|(a − b), which implies that
a − b = nq for some integer q. Thus, a − b ∈ hni, that is, a + hni = b + hni.
Hence, f ([a]) = f ([b]). Therefore, f is well-defined.

Let [a], [b] ∈ Zn such that f ([a]) = f ([b]). Then a + hni = b + hni,
which implies that a − b ∈ hni. Thus, a − b = nq for some integer q, that is,
n|(a − b). Hence, [a] = [b]. Therefore, f is injective.

Let a + hni ∈ Z/hni. Then there exists [a] ∈ Zn such that f ([a]) =
a + hni. Thus, f is surjective.

Let [a], [b] ∈ Zn . Then f ([a]) = a + hni and f ([b]) = b + hni. Thus,

f ([a] +n [b]) = [a + b] = (a + b) + hni = (a + hni) + (b + hni) = f ([a]) + f ([b]).

 66

Hence, f is a homomorphism.

Therefore, f is an isomorphism of Zn onto Z/hni. 

Example 5.1.20 Consider the sets G = {e, a, b, c} and G0 = {1, −1, i, −i}.
Define ∗ and · on G and G0 , respectively, by means of the following tables.
∗ e a b c
e e a b c
a a e c b
b b c e a
c c b a e

· 1 -1 i −i
1 1 -1 i −i
-1 -1 1 −i i
i i −i 1 -1
−i −i i -1 1
Then G and G0 are groups. Note that since aa = e, bb = e, cc = e, no element
of G has order 4. Thus, G is not cyclic. G0 = hii, that is, G0 is a cyclic group
generated by i. Therefore, G and G0 are not isomorphic.

Theorem 5.1.21 Every finite cyclic group of order n is isomorphic to (Zn , +n )

and every infinite cyclic group is isomorphic to (Z, +).

Proof : Let (hai, ∗) be a cyclic group of order n. Let G = hai. Define the
function f : G → Zn by f (a1 ) = [i] for all ai ∈ G. Then ai = aj if and
only if aj−i = e if and only if n|(j − i) if and only if [i] = [j] if and only if
f (ai ) = f (aj ). Thus, f is well-defined and one-one. Now,
f (ai aj ) = f (ai+j = [i + j] = [i] +n [j] = f (ai ) +n f (aj ).
Hence, f is a homomorphism. Since f is one-one and G and Zn are finte with
the same number of elements, f is onto Zn . Therefore, G ∼
= Zn .

Let G = hai be an infinite cyclic group. Define the function f : G → Z

by f (a ) = i for all i ∈ Z. Then ai = aj if and only if aj−i = e if and only
i

if j − i = 0 (since a is of infinte order) if and only if i = j if and only if

f (ai ) = f (aj ). Thus, f is injective. It is clear from the definition of f that f
is surjective. Now,
 67

f (ai aj ) = f (ai+j = i + j = f (ai ) + f (aj ).

Hence, f is a homomorphism. Therefore, G ∼
= Z. 

Corollary 5.1.22 Any two cyclic groups of the same order are isomorphic.

From the above corollary, it follows that there is only one (up to
isomorphism) cyclic group having a prescribed order.

Theorem 5.1.23 There are only two groups of order 4 (up to isomorphism),
a cyclic group of order 4 and K4 (Klein 4-group).

Proof : Let G be a group of order 4 which is not cyclic. (Example 5.1.20 shows
that such a group exists.) Then no element of G can have order 4, for if a ∈ G
has order 4, then e, a, a2 , a3 would be distinct elements of G and thus G would
be cyclic. This is contrary to the assumption that G is not cyclic.
Let G = {e, a, b, c}. Since the order of every element of G divides the
order of G, a, b, and c have order 2. If ab = a, then b = e, a contradiction.
Thus, ab 6= a. Similarly, ab 6= b. Suppose ab = e. Then aab = ae. Since
a2 = e, b = a, a contradiction. Hence, ab = c. Similarly, ba = c. This means
that ab = ba. By similar arguments, we have ac = b = ca and bc = a = cb.
Therefore, G is a commutative group and its operation table is given
by the table in Example 5.1.20. Consequently, there is essentially one group
of order 4 which is not cyclic. This is the Klein 4-group.
By Corollary 5.1.22, all cyclic groups of the same orders are isomorphic.
Therefore, we have exactly two nonisomorphic groups of order 4, namely, the
Klein 4-group and the cyclic group of order 4. 

Theorem 5.1.24 There are only two groups of order 6 (up to isomorphism).

Example 5.1.25 determine whether the indicated function f is a homomorphism

from the first group into the second group. If f is a homomorphism, determine
its kernel.

(a) (R+ , ·), (R+ , ·); f (a) = a2 for all a ∈ R+ .

(b) (R, +), (R+ , ·); f (a) = 2a for all a ∈ R.
(c) (R\{0}, ·), (R+ , ·); f (a) = |a| for all a ∈ R\{0}.
(d) (Z, +), (Z, +); f (a) = a + 1 for all a ∈ Z.
 68

Example 5.1.26 Let f : G → G0 be an epimorphism of groups. If H is a

normal subgroup of G, then show that f (H) is a normal subgroup of G0 .

Example 5.1.27 Let f : G → H be a homomorphism of groups. Prove that

f −1 : H → G is an isomorphism.

Example 5.1.28 Let G, H, and K be groups. Prove that

(a) G × H ∼= H × G.
(b) If G = H and H ∼
∼ = K, then G ∼
= K.

Example 5.1.29 Show that (Q, +) is not isomorphic to (R, +).

5.2 Isomorphism and Correspondence Theorems

Theorem 5.2.1 Let f be a homomorphism of a group G onto a group G0 ,

H a normal subgroup of G such that H ⊆ Kerf , and g be the canonical
homomorphism of G into G/H. Then there exists a unique homomorphism h
of G/H onto G0 such that f = h ◦ g. Furthermore, h is one-one if and only if
H = Kerf .

Proof : Define h : G/H → G0 by h(aH) = f (a) for all aH ∈ G/H.

Let aH, bH ∈ G/H such that aH = bH. Then b−1 a ∈ H ⊆ Kerf ,

which implies that f (b−1 a) = e0 . Thus, f (b−1 )f (a) = e0 , that is, f (a) = f (b).
Hence, f is well-defined.

Let a ∈ G. Then (h◦g)(a) = h(g(a)) = h(aH) = f (a). Thus, h◦g = f .

Let a0 ∈ G0 . Since f is onto G0 , there exists a ∈ G such that f (a) = a0 .

Hence, there exists aH ∈ G/H such that h(aH) = f (a) = a0 . Thus, h is onto
G0 .

Let aH, bH ∈ G/H. Then h(aH) = f (a) and f (bH)) = f (b). Thus,

h((aH)(bH)) = h((ab)H) = f (ab) = f (a)f (b) = h(aH)h(bH).

 69

Hence, h is a homomorphism.

To prove that h is unique, suppose that f = h0 ◦ g for some h0 from

G/H onto G0 . Then

h(aH) = f (a) = (h0 ◦ g)(a) = h0 (g(a)) = h0 (aH)

for all aH ∈ G/H. Thus, h = h0 . Hence, h is unique.

Therefore, h is a homomorphism of G/H onto G0 such that f = h ◦ g.

Suppose that h is one-one. Let a ∈ Kerf . Then f (a) = e0 and so

h(aH) = f (a) = e0 . Since h(eH) = e0 , we have h(aH) = h(eH). Since h
ius one-one, aH = eH, that is, a ∈ H. Hence, Kerf ⊆ H. By hypothesis,
H ⊆ Kerf . Therefore, H = Kerf .
Conversely, suppose that H = Kerf . Let aH, bH ∈ G/H such that
f (aH) = f (bH). Then f (a) = f (b), that is, f (b−1 a) = e0 . Thus, b−1 a ∈
Kerf = H. This implies that aH = bH. Hence, h is one-one. 

From the above theorem, it follows that if H = Kerf , the h is an

isomorphism and hence G/Kerf is isomorphic to G0 , that is, every homomorphism
of a group G onto a group G0 induces an isomorphism of G/Kerf onto G0 .
This result plays a fundamental role in group theory. It is known as the
fundamental theorem of homomorphisms of groups. This result is also called
the first isomorphism theorem for groups.

Theorem 5.2.2 (First Isomorphism Theorem) Let f be a homomorphism

of a group G into a group G0 . Then f induces an isomorphism

G/Kerf ∼
= f (G).

Proof : Let K = Kerf . Define h : G/K → f (G) by h(aK) = f (a) for all
aK ∈ G/K.

Let aK, bK ∈ G/K such that aK = bK. Then b−1 a ∈ K = Kerf ,

which implies that f (b−1 a) = e0 . Since f is a homomorphism, f (b)−1 f (a) = e0 ,
that is, f (a) = f (b). Hence, h(aK) = h(bK). This shows that h is well-defined.

Let aK, bK ∈ G/K. Then h(aK) = f (a) and h(bK) = f (b). Thus,
h((aK)(bK)) = h((ab)K) = f (ab) = f (a)f (b) = h(aK)h(bK). Hence, h is a
 70

homomorphism.

Let aK, bK ∈ G/K such that h(aK) = h(bK). Then f (a) = f (b).
Since f is a homomorphism, f (b)−1 f (a) = e0 , that is,f (b−1 a) = e0 . Thus,
b−1 a ∈ K = Kerf . This implies that , aK = bK. Hence, h is one-one.

Let y ∈ f (G). Then y = f (b) for some b ∈ G. Hence, there exists

bK ∈ G/K such that h(bK) = f (b) = y. Thus, h is onto f (G).

Therefore, G/Kerf ∼
= f (G). 

Recall that a group G0 is called a homomorphic image of a group G

if there exists a homomorphism of G onto G0 .

Example 5.2.3 The group S3 has (up to isomorphism) only three homormorphic
images. This follows from the fact that S3 has only three normal subgroups,
namely, S3 , {ι}, H5 = {ι, δ, }. Thus, the homomorphic images are S3 , Z1 ,
and Z2 since S3 ∼= S3 /{ι}, Z1 ∼
= S3 /S3 , and Z2 ∼
= S3 /H5 .

Theorem 5.2.4 Let G0 be a homomorphic image of a group G.

(i) If G is cyclic, then G0 is cyclic.
(ii) If G is abelian, then G0 is abelian.
(iii) If G contains an element of order n and |G| is finite, then G0
contains an element of order n.

Proof : Exercise!

Theorem 5.2.5 (Second Isomorphism Theorem) Let H and K be subgroups

of a group G with K normal in G. Then
H/(H ∩ K) ∼
= HK/K.

Proof : Define f : H → HK/K by f (h) = hK for all h ∈ H. Then for all

h ∈ H, h ∈ HK Thus, f (h) = hK ∈ HK/K.

Let h1 , h2 ∈ H such that h1 = h2 . Then h−1

2 h1 = e, which implies
that h−1
2 h1 ∈ K. Thus, h1 K = h2 K, that is, f (h1 ) = f (h2 ). Hence, f is
well-defined.
 71

Let (hk)K ∈ HK/K. Then h ∈ H and k ∈ K. Note that kK = K.

Thus, there exists h ∈ H such that f (h) = hK = h(kK) = (hk)K. Hence, f
is onto HK/K. This implies that f (H) = HK/K.

Let h1 , h2 ∈ H. Then f (h1 ) = h1 K and f (h2 ) = h2 K. Thus, f (h1 h2 ) =

h1 h2 K = (h1 K)(h2 K = f (h1 )f (h2 ). Hence, f is a homomorphism.

We have
Kerf = {h ∈ H|f (h) = K}, K is the identity in HK/K
= {h ∈ H|hK = K}
= {h ∈ H|h ∈ K}
= H ∩ K.
By The first isomorphism theorem, H/(H ∩ K) ∼
= HK/K. 

Theorem 5.2.6 (Third Isomorphism Theorem) Let H and K be normal

subgroups of a group G such that K is a subgroup of H. Then H/K is a
normal subgroup of G/K and
(G/K)/(H/K) ∼
= G/H.
Proof : K = eK ∈ H/K. Thus, H/K 6= ∅.
Let h1 K, h2 K ∈ H/K. Then h1 , h2 ∈ H. Since H is a subgroup of G,
h1 , h−1
2 ∈ H. Hence, (h1 K)(h2 K)
−1
= (h1 K)(h−1 −1
2 K) = (h1 h2 )K ∈ H/K.
Let hK ∈ H/K and gK ∈ G/K. Then h ∈ H and g ∈ G. Since H is
normal in G, ghg −1 ∈ H. Hence,
(gK)(hK)(gK)−1 = (gK)(hK)(g −1 K) = (ghg −1 )K ∈ H/K.
Therefore, H/K is a normal subgroup of G/K.

Define f : G/K → G/H by f (gK) = gH for all gK ∈ G/K.

Let g1 K, g2 K ∈ G/K such that g1 K = g2 K. Then g2−1 g1 ∈ K. Since K
is a subgroup of H, g2−1 g1 ∈ H. Thus, g1 H = g2 H. Hence, f (g1 K) = f (g2 K),
which shows that f is well-defined.
We have
Kerf = {gK ∈ G/K|f (gK) = H}, H is the indentity in G/H
= {gK ∈ G/K|gH = H}
= {gK ∈ G/K|g ∈ H}
= H/K.
 72

We have
f (G/K) = {f (gK)|gK ∈ G/K}
= {gH|g ∈ G}
= G/H.

Let g1 K, g2 K. Then f (g1 K) = g1 H and f (g2 K) = g2 H. Thus,

f ((g1 K)(g2 K)) = f ((g1 g2 )K) = (g1 g2 )H = (g1 H)(g2 )H) = f (g1 K)f (g2 )K).
Hence, f is a homomorphism.

By the first isomorphism theorem,

(G/K)/(H/K) ∼
= G/H.

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[1] Beachy, John A. and William D. Blair, Abstract Algebra: A Study Guide
for Beginners, Northern Illinois Univ. ,2013.

[2] Herstein, I. N., Abstract Algebra, 3rd edition, Prentice-Hall, New Jersey,
1996.

[3] Hungerford, Thomas W., Algebra, Springer-Verlag, New York, 1974.

[4] Judson, Thomas W., Abstract Algebra: Theory and applications, S.F.
Austin State Univ., 2012.

[5] Lalonde, Scott, Notes on Abstract Algebra, Dartmouth Collge, 2013.

[6] Leonida, Rene E., Lecture Notes on Logic and Set Theory, Mindano State
Univ., General Santos City, 2016.

[7] Leonida, Rene E., Lecture Notes on Abstract Algebra for Teachers,
Mindano State Univ., General Santos City, 2017.

[8] Malik, D.S., Mordeson,John N., and Sen, M.K., Introduction to Abstract
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[9] Sethuraman, B.B., A Gentle Introduction to Abstract Algebra, California

State Univ., 2012.