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0 Sets and Induction

Sets

A set is an unordered collection of objects,

called elements or members of the set. A

set is said to contain its elements. We write

a ∈ A to denote that a is an element of the

set A. The notation a /∈ A denotes that a

is not an element of the set A.

1

Defining a Set

The roster method is a way of defining a

set by listing all of its members.

Examples

• S = {a, b, c}

• S = {1,2,3,4}

• S = {1,4,7,10,13,16, . . .}

2

Defining a Set

Another way to describe a set is using set

builder notation.

Examples

• S = {1,2,3,4}, or

S = {x | x is a positive integer ≤ 4}

• S = {1,4,7,10,13,16, . . .}, or

S = {x | x = 1 + 3k for some nonnegative integer k}

3

Defining a Set

Example: Express the following set using

set builder notation.

S = {. . . ,−12,−8,−4,0,4,8,12, . . .}

4

Important Sets

Notation

Z = {. . . ,−2,−1,0,1,2, . . .}, the set of integers

Z+ = {1,2,3, . . .}, the set of positive integers

Q = {pq| p ∈ Z, q ∈ Z, and q 6= 0}, the set of ratio-

nal numbers

R, the set of real numbers

R+, the set of positive real numbers

C, the set of complex numbers

5

Special Sets

A set with no elements is called the empty

set, or null set, and is denoted by Ø. The

empty set can also be denoted by { }.

A set with one element is called a singleton

set. For example, S = {1} is a singleton

set containing only the number 1.

6

Subsets

The set A is a subset of B if and only if

every element of A is also an element of B.

That is, A is a subset of B iff

∀x (x ∈ A → x ∈ B).

We use the notation A ⊆ B to indicate that

A is a subset of B.

7

Proper Subsets

We say that A is a proper subset of B if

and only if A ⊆ B and A 6= B. That is, A

is a proper subset of B iff

∀x (x ∈ A → x ∈ B) ∧ ∃x (x ∈ B ∧ x /∈ A).

We use the notation A ( B to indicate that

A is a proper subset of B.

8

Subsets

Examples

• If S = {1,2,3} and T = {1,2,3,4,5},then S ⊆ T .

• If S = {π,√

2} and T = {5,√

2}, then

S * T .

• Z+ ⊆ Z

9

Special Subsets

For every set S,

(i) Ø ⊆ S(ii) S ⊆ S

10

Equality of Sets

Two sets are equal if and only if they have

the same elements. Therefore, if A and B

are sets, then A = B if and only if

∀x (x ∈ A ↔ x ∈ B).

That is, A = B iff A ⊆ B and B ⊆ A.

11

Equality of Sets

Example

The sets {1,3,5} and {3,5,1} are equal,

because they have the same elements. Note

that the order in which the elements of a

set are listed does not matter.

12

Intersection and Union

Let S and T be sets.

• The intersection of S and T , denoted

S ∩ T , is the set of elements which be-

long to both S and T . That is,

S ∩ T = {x | x ∈ S and x ∈ T}

• The union of S and T , denoted S ∪ T ,

is the set of elements which belong to

S or T . That is,

S ∪ T = {x | x ∈ S or x ∈ T}

13

Intersection and Union

Example

Let S = {1,2,3,4,5} and T = {2,4,6}.Then,

S ∩ T = {2,4}.S ∪ T = {1,2,3,4,5,6}.

14

Sets and Logic

Statements involving sets and their logical

equivalents.

x /∈ S ¬(x ∈ S)

x ∈ S ∪ T (x ∈ S) ∨ (x ∈ T )

x ∈ S ∩ T (x ∈ S) ∧ (x ∈ T )

S ⊆ T ∀x (x ∈ S → x ∈ T )

S = T ∀x (x ∈ S ↔ x ∈ T )

15

Sets and Proofs

To show A ⊆ B

Assume x ∈ A, then show x ∈ B.

To show A * B

Show there exists an x ∈ A such that x 6= B.

16

Sets and Proofs

To show A = B

Step 1. Assume x ∈ A, then show x ∈ B.

Step 2. Assume x ∈ B, then show x ∈ A.

17

Sets and Proofs

Example: Let S and T be sets. Prove that

S ∩ T ⊆ S ∪ T .

18

Sets and Proofs

Example: Let S and T be sets. Prove that

S ⊆ T if and only if S ∩ T = S.

19

Mathematical Induction

Let P (n) is a propositional function with

domain Z+. To prove that P (n) is true

for all positive integers n, we complete two

steps:

1. (Base Case) Verify that P (1) is true.

2. (Inductive Step) Prove the conditional

statement P (k) → P (k + 1) is true for

all positive integers k.

To complete the inductive step, we assume

P (k) is true (this assumption is called the

induction hypothesis), then show P (k+1)

must also be true.

20


Example: Show that if n is a positive

integer, then

1 + 2 + 3 + · · ·n =n(n+ 1)

2

21


Proof: Let P (n) be the proposition

1 + 2 + 3 + · · ·n =n(n+ 1)

2.

We want to show P (n) is true for all n ≥ 1.

Base Step:

22


Example: Conjecture a formula for the

sum of the first n positive odd integers.

Then prove your conjecture using mathe-

matical induction.

1 =

1 + 3 =

1 + 3 + 5 =

1 + 3 + 5 + 7 =

1 + 3 + 5 + 7 + 9 =

23


Conjecture: For all positive integers n,

the following proposition P (n) holds

1 + 3 + 5 + · · ·+ 2n− 1 =

Proof:

24


Example: Use mathematical induction to

prove that 2 divides n2 + 5n for all n ≥ 1.

27


Proof: Let P (n) be the proposition

2 | (n2 + 5n).

We want to show P (n) is true for all n ≥ 1.

Base Step:

28

1 Binary Operations

Cartesian Product

Let A and B be sets. The Cartesian prod-

uct of A and B, denoted by A × B, is the

set of all ordered pairs (a, b), where a ∈ A

and b ∈ B. Hence,

A×B = {(a, b) | a ∈ A and b ∈ B}.

1

Cartesian Product

Example:

If A = {1,2} and B = {a, b, c}, then

A×B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}.

2

Binary Operations

If S is a set, then a binary operation ∗on S is a function that associates to each

ordered pair (s1, s2) ∈ S × S an element of

S which we denote by s1 ∗ s2.

That is, ∗ is a function from S × S to S.

3

Binary Operations

Examples

• Addition defines a binary operation on

Z+ since n + m ∈ Z+ for all (n,m) ∈Z+ × Z+.

• Multiplication defines a binary opera-

tion on Z+ since n · m ∈ Z+ for all

(n,m) ∈ Z+ × Z+.

• Subtraction does not define a binary

operation on Z+ since there exists (n,m) ∈Z+ × Z+ such that n−m /∈ Z+.

• Division does not define a binary oper-

ation on Z+ since there exists (n,m) ∈Z+ × Z+ such that n/m /∈ Z+.

4

Binary Operations

Examples

• Subtraction is a binary operation on each

of the sets Z, R, and Q.

• Division is a binary operation on each

of the sets Q+ and R+.

5

Binary Operations

Example

Let X be a set, and let P(X) be the set

of all subsets of X. Then the operations

of union and intersection are binary opera-

tions on P(X).

For example, if X = {1,2}, then

P(X) = {Ø, {1}, {2}, {1,2}}

and the operations of union and intersec-

tions are binary operations on S.

The set P(X) is called the power set of

X.

6

Binary Operations

Example

Let X be a set, and let P(X) be the set

of all subsets of X. Then the operation M

defined by

A M B = (A−B) ∪ (B −A)

is a binary operation on P(X).

The set A M B is called the symmetric

difference of A and B.

7

Binary Operations

Let S be the set of all 2× 2 matrices with

real entries. Then, the operation ∗, defined

by

(a11 a12

a21 a22

)∗(

b11 b12

b21 b22

)=

a11b11 + a12b21 a11b12 + a12b22

a21b11 + a22b21 a21b12 + a22b22

is a binary operation on S. The corre-

sponds to multiplication of matrices.

8

Commutativity and Associativity

• A binary operation ∗ on S is called

commutative iff a ∗ b = b ∗ a for all

a, b ∈ S.

• A binary operation ∗ on S is called

associative iff (a ∗ b) ∗ c = a ∗ (b ∗ c) for

all a, b, c ∈ S.

9


• Subtraction on Z is neither commuta-

tive nor associative. For example,

1− 2 6= 2− 1

(1− 2)− 3 6= 1− (2− 3)

• Division on R+ is neither commutative

nor associative. For example,

1/2 6= 2/1

(1/2)/3 6= 1/(2/3)

• Addition and multiplication are both as-

sociative and commutative on the sets

Z, Q, R.

10


Example: Let ∗ be a binary operation on

Z defined by

a ∗ b = 2(a + b)

• is ∗ commutative?

• is ∗ associative?

11


Example: Multiplication of 2×2 matrices

• is associative.

((a b

c d

)∗(e f

g h

))∗(

i j

k l

)=

(a b

c d

)∗((

e f

g h

)∗(

i j

k l

))

• is not commutative.

(0 1

1 0

)∗(

1 2

3 4

)=

(3 4

1 2

)(

1 2

3 4

)∗(

0 1

1 0

)=

(2 1

4 3

)

12


Example: Let M be the symmetric differ-

ence operator given by

A M B = (A−B) ∪ (B −A)

• is M commutative?

• is M associative?

13

2 Groups

Definition of a group

A set G together with a binary operation ∗is called a group if the following conditions

hold

(i) (closure) x ∗ y ∈ G for all x, y ∈ G.

(ii) (associativity) (x ∗ y) ∗ z = x ∗ (y ∗ z)

for all x, y, z ∈ G.

(iii) (identity element) There is an ele-

ment e ∈ G such that x ∗ e = e ∗ x = x for

all x ∈ G.

(iv) (inverse elements) For each element

x ∈ G there is an element y ∈ G such that

x ∗ y = y ∗ x = e.

14

Definition of a group

Remarks

• We use the notation (G, ∗) to represent

the group with elements in G under the

operation ∗.

• Condition (i) states that ∗ is a binary

operation on G. We say G is closed

with respect to ∗.

• The element e in condition (iii) is called

an identity element of G. In particular,

this means that all groups are nonempty.

• Condition (iv) states that every element

x in G has an inverse element y.

15

Abelian Group

A group (G, ∗) for which ∗ is commutative

is called an abelian group.

16

Groups

Example: (Z,+)

The set Z under addition is a group.

(i) x + y ∈ Z for all x, y ∈ Z.

(ii) (x+y)+z = x+(y+z) for all x, y, z ∈ Z.

(iii) x + 0 = 0 + x = x for all x ∈ Z.

(iv) for each x ∈ Z there is an element

−x ∈ Z such that x+(−x) = (−x)+x = 0.

17

Groups

Example: (Q+, ·)

The set Q+ under multiplication is a group.

(i) x · y ∈ Q+ for all x, y ∈ Q+.

(ii) (x · y) · z = x · (y · z) for all x, y, z ∈ Q+.

(iii) x · 1 = 1 · x = x for all x ∈ Q+.

(iv) for each x ∈ Q+ there is an element1x ∈ Q+ such that x · 1

x = 1x · x = 1.

18

Groups

Example: (Rn,+)

The set of ordered n-tuples (a1, a2, . . . , an)

of real numbers forms a group under the

operation ∗ given by

(a1, a2, . . . , an)∗(b1, b2, . . . , bn) = (a1+b1, a2+b2, . . . , an+bn)

identity element:

inverse element:

19

Groups

Example: (GL(2,R), ·)

The set of invertible 2×2 matrices with real

entries forms a group under matrix multi-

plication.

This group is called the general linear group

of degree 2 over R, and is denoted by

(GL(2,R).

identity element:

inverse element:

20

Groups

Example: ({f | f : R→ R},+)

The set of real-valued functions with do-

main R forms a group under the operation

of pointwise addition of functions.

(f + g)(x) = f(x) + g(x)

identity element:

inverse element:

21

Groups

Example: (P(X),M)

The set of subsets of a set X forms a group

under the operation of symmetric differ-

ence of sets:

A M B = (A−B) ∪ (B −A)

identity element:

inverse element:

22

Groups

Example: Consider the set Z together

with the binary operation ∗ given by

a ∗ b = 2(a + b).

Does (Z, ∗) form a group?

23

Groups

Example: Consider the set G = R − {0}together with the binary operation ∗ given

by

a ∗ b = 2ab.

Does (G, ∗) form a group?

24

Additive group of integers modulo n

Theorem (The Division Algorithm)

Let a be an integer and n a positive inte-

ger. Then there are unique integers q and

r, with 0 ≤ r < n, such that a = qn + r.

Notation:

The integer r in the division algorithm is

called the remainder of a mod n, and

will be denoted by a.

25


Lemma: Assume n ∈ Z+. Then, a = b if

and only if a− b = nk for some k ∈ Z.

Proof:

26


Example: (Zn,⊕)

Let n be a positive integer, and consider

the set Zn = {0,1,2, . . . , n− 1}.

Zn forms a group under the binary opera-

tion ⊕ defined by

x⊕ y = x + y

where x + y is the unique number in Zn

such that x + y ≡ x + y (mod n)

This group is called the additive group of

integers modulo n.

identity element:

inverse element:27

3 Fundamental Theorems About

Groups

Uniqueness of the Identity Element

Theorem 3.1 If (G, ∗) is a group, then

there is only one identity element in G.

Proof: Assume e1 and e2 are identity ele-

ments of G. Since e1 is an identity element,

we know

e1 ∗ e2 = e2.

Since e2 is an identity element, we know

e1 ∗ e2 = e1.

This proves e1 = e2.

1

Uniqueness of Inverses

Theorem 3.2 If (G, ∗) is a group and x

is any element of G, then x has only one

inverse in G.

Proof: Assume x ∈ G and assume y1 and

y2 are inverses of x. Then,

y1 = y1 ∗ e

= y1 ∗ (x ∗ y2)

= (y1 ∗ x) ∗ y2

= e ∗ y2

= y2

This proves y1 = y2, and we conclude that

the inverse of an element is unique.

2

Inverse Element

Notation

If (G, ∗) is a group, then the inverse of an

element x ∈ G is denoted by x−1.

3

Inverse Element

Theorem 3.3 If (G, ∗) is a group and

x ∈ G , then (x−1)−1 = x.

Proof: Let y = (x−1)−1. Then, y is the

unique element of G such that

x−1 ∗ y = y ∗ x−1 = e.

Note that the previous equation is satisfied

when y = x, since x−1 is the inverse of x.

Therefore, (x−1)−1 = y = x.

4

Alternate Proof: Assume x ∈ G. We have,

(x−1)−1 = (x−1)−1 ∗ e

= (x−1)−1 ∗ (x−1 ∗ x)

= ((x−1)−1 ∗ x−1) ∗ x

= e ∗ x

= x.

Inverse Element

Examples

• Consider the group (Z,+). Then, for

all n ∈ Z, we have −(−n) = n.

• Consider the group GL(2,R). Then, for

all(a bc d

)∈ GL(2,R), we have

((a bc d

)−1)−1

=(a bc d

).

5

Inverse of a Product

Theorem 3.4 If (G, ∗) is a group and

x, y ∈ G, then

(x ∗ y)−1 = y−1 ∗ x−1.

Proof: Assume x, y ∈ G. Let z = y−1∗x−1.

Then,

(x ∗ y) ∗ z = (x ∗ y) ∗ (y−1 ∗ x−1)

= x ∗ (y ∗ (y−1 ∗ x−1))

= x ∗ ((y ∗ y−1) ∗ x−1)

= x ∗ (e ∗ x−1)

= x ∗ x−1

= e.

A similar argument shows z ∗ (x ∗ y) = e.

Therefore, (x ∗ y)−1 = z = y−1 ∗ x−1.

6

Alternate proof:

(x ∗ y)−1 = (x ∗ y)−1 ∗ e

= (x ∗ y)−1 ∗ (x ∗ x−1)

= (x ∗ y)−1 ∗ ((x ∗ e) ∗ x−1)

= (x ∗ y)−1 ∗ ((x ∗ (y ∗ y−1)) ∗ x−1)

= (x ∗ y)−1 ∗ (((x ∗ y) ∗ y−1) ∗ x−1)

= (x ∗ y)−1 ∗ ((x ∗ y) ∗ (y−1 ∗ x−1))

= ((x ∗ y)−1 ∗ (x ∗ y)) ∗ (y−1 ∗ x−1)

= e ∗ (y−1 ∗ x−1)

= y−1 ∗ x−1

Inverse Element

Theorem 3.5 Assume (G, ∗) is a group

and let x, y ∈ G. If either x ∗ y = e or

y ∗ x = e, then y = x−1.

Proof: First assume x ∗ y = e. We will

solve for y by multiplying both sides of the

equation on the left by x−1. We have,

x−1 ∗ (x ∗ y) = x−1 ∗ e

(x−1 ∗ x) ∗ y = x−1

e ∗ y = x−1

y = x−1

Similarly, if y ∗ x = e, then right multipli-

cation by x−1 shows y = x−1.

7

Cancellation Laws

Theorem 3.6 Assume (G, ∗) is a group

and let x, y ∈ G. Then:

(i) if x ∗ y = x ∗ z, then y = z; and

(ii) if y ∗ x = z ∗ x, then y = z.

Proof: See homework.

8

Identities and Inverses

Notation

Let G be a set, and assume ∗ is an asso-

ciative binary operation on G. Then:

• if e ∈ G and x ∗ e = x for all x ∈ G,

then we say e is a right identity with

respect to ∗.

• if x ∈ G and e ∗ x = x for all x ∈ G,

then we say e is a left identity with

respect to ∗.

• if x, y ∈ G and x ∗ y = x, then we say

y is a right inverse of x.

• if x, y ∈ G and y ∗ x = x, then we say

y is a left inverse of x.

9

Sufficient Conditions for a Group

Theorem 3.8 Let G be a set, and assume

∗ is an associative binary operation on G.

If there exists a right identity e ∈ G with

respect to ∗, and if every element x ∈ G

has a right inverse, then (G, ∗) is a group.

10

Proof: First we will show that right identity

e ∈ G is also a left identity. Given x ∈ G,

let y denote the right inverse of x, and let

z denote the right inverse of y. Then,

e ∗ x = (e ∗ x) ∗ e

= (e ∗ x) ∗ (y ∗ z)

= e ∗ (x ∗ (y ∗ z))

= e ∗ ((x ∗ y) ∗ z)

= e ∗ (e ∗ z)

= (e ∗ e) ∗ z

= e ∗ z

= (x ∗ y) ∗ z

= x ∗ (y ∗ z)

= x ∗ e

= x

This shows that e ∗ x = x for any x ∈ G.

Hence, e is an (two-sided) identity element.

Finally we will show that for any x ∈ G, if y

is the right inverse of x, then y is also a left

inverse of x. If z denotes the right inverse

of y, we have

y ∗ x = (y ∗ x) ∗ e

= (y ∗ x) ∗ (y ∗ z)

= y ∗ (x ∗ (y ∗ z))

= y ∗ ((x ∗ y) ∗ z)

= y ∗ (e ∗ z)

= (y ∗ e) ∗ z

= y ∗ z

= e.

This shows that for all x ∈ G there exists a

y ∈ G such that x∗y = y∗x = e. Therefore,

(G, ∗) is a group.

Insufficient Conditions for a Group

Example: Consider the set Z+ with the

binary operation given by

x ∗ y = x.

• ∗ is associative since

(x ∗ y) ∗ z = x ∗ y = x = x ∗ (y ∗ z)

• the element 1 ∈ Z+ is a right identity

element since x ∗ 1 = x for all x ∈ Z+.

(In fact, any element y ∈ Z+ is an right

identity element.)

• every element y ∈ Z+ has the element

1 ∈ Z+ as a left inverse element since

1 ∗ y = 1 for all y ∈ Z+.

11

• However these conditions are not suffi-

cient to make Z+ a group with respect

to ∗, since there is no left identity ele-

ment:

y ∗ x 6= x unless y = x

• Note that multiplication table for Z+

with respect to ∗ has repeated ele-

ments in its rows (See Homework 3,

Problem 3):

∗ 1 2 3 4 5 · · ·

1 1 1 1 1 1 · · ·

2 2 2 2 2 2 · · ·

3 3 3 3 3 3 · · ·

4 4 4 4 4 4 · · ·

5 5 5 5 5 5 · · ·... ... ... ... ... ... . . .

4 Powers of an Element; Cyclic

Groups

Notation

When considering an abstract group (G, ∗),

we will often simplify notation as follows

• x ∗ y will be expressed as xy

• (x ∗ y) ∗ z will be expressed as xyz

• x ∗ (y ∗ z) will be expressed as xyz

In other words, we will omit the symbol ∗,and use product notation, unless a more

appropriate notation is called for (e.g., the

symbol + will be used in the case of the

group (Z,+)). Also, when associativity al-

lows, we can omit parentheses.

1

Powers of an Element

Notation

Let G be a group, and let x be an element

of G. We define powers of x as follows:

x0 = e

xn = xxx · · ·x︸︷︷︸n factors

x−n = x−1x−1x−1 · · ·x−1︸︷︷︸n factors

2


Theorem 4.1 Let G be a group and let

x ∈ G. Let m,n be integers. Then:

(i) xmxn = xm+n

(ii) (xn)−1 = x−n

(iii) (xm)n = xnm = (xn)m

3

Order of an Element

Definitions

If G is a group and x ∈ G, then x is said to

be of finite order if there exists a positive

integer n such that xn = e.

If such an integer exists, then the smallest

positive n such that xn = e is called the

order of x and denoted by o(x).

If x is not of finite order, then we say that

x is of infinite order and write o(x) =∞.

4

Order of an Element

Examples

• Let G = (Z3,⊕). Then o(1) = 3, since

1 6= 0

1⊕ 1 6= 0

1⊕ 1⊕ 1 = 0

• Let G = (Z,+). Then o(1) =∞, since

1 6= 0

1 + 1 6= 0

1 + 1 + 1 6= 0

1 + 1 + 1 + 1 6= 0

···

5

Order of an Element

Examples

• Let G = (Q+, ·). Then o(2) =∞, since

21 6= 1

22 6= 1

23 6= 1

24 6= 1

···

• Let G = GL(2,R), then o (( −1 00 −1 )) = 2,

since (−1 0

0 −1

)16=(

1 00 1

)(−1 0

0 −1

)2=(

1 00 1

)

6

Greatest Common Divisor

Let a and b be integers, not both zero.

The largest integer d such that d | a and

d | b is called the greatest common divisor

of a and b. The greatest common divisor

of a and b is denoted by gcd(a, b).

7


Examples

• gcd(24,40) =

• gcd(32,100) =

• gcd(12,91) =

8

Relatively Prime

The integers a and b are relatively prime if

their greatest common divisor is 1.

Examples

• The integers 12 and 25 are relatively

prime, since gcd(12,25) = 1

• The integers 27 and 75 are not rela-

tively prime, since gcd(27,75) = 3 6= 1.

9

The Euclidean Algorithm

Suppose that a and b are positive integers

with a ≥ b. Successively applying the divi-

sion algorithm, we obtain

a = b q0 + r1, 0 ≤ r1 < b,

b = r1 q1 + r2, 0 ≤ r2 < r1,

r1 = r2 q2 + r3, 0 ≤ r3 < r2,

r2 = r3 q3 + r4, 0 ≤ r4 < r3,

···

rn−2 = rn−1 qn−1 + rn, 0 ≤ rn < rn−1,

rn−1 = rn qn.

Theorem: gcd(a, b) = rn, where rn is the

last nonzero remainder in the sequence above.

10


Lemma Let a = bq + r, where a, b, q, and

r are integers. Then gcd(a, b) = gcd(b, r).

Proof: Assume d | a and d | b. Therefore

d | r where r = a − bq. This shows that all

common divisors of a and b are common

divisors of b and r.

Next, assume d | b and d | r. Therefore d | awhere a = bq + r. This shows that all com-

mon divisors of b and r are common divisors

of a and b.

Therefore the common divisors of a and b

are the same as the common divisors of b

and r. Therefore, their greatest common

divisors are the same.

11


Examples

• Find gcd(198, 252) using the Euclidean

algorithm

• Find gcd(414, 662) using the Euclidean

algorithm

12


Theorem 4.2 If a and b are integers, not

both zero, then there exist integers x and

y such that gcd(a, b) = ax + by.

Proof: Solving for the last nonzero remain-

der in the Euclidean Algorithm, we have

rn = rn−2 − rn−1 qn−1

Using the preceding step of the Euclidean

Algorithm, we may express the term rn−1

in terms of rn−2 and rn−3, thereby obtain-

ing

rn = rn−2 − (rn−3 − rn−2 qn−2) qn−1

After a sequence of n − 1 substitutions,

we obtain an expression for gcd(a, b) = rn

in terms of a and b.13


Example: Express gcd(198,252) = 18 as

a linear combination of 252 and 198.

14


Theorem 4.3 If a | bc and gcd(a, b) = 1

and, then a | c.

Proof: Since gcd(a, b) = 1, there exist

integers x and y such that

ax + by = 1.

Multiplying on both sides by c yields

axc + byc = c

a(xc) + bc(y) = c

Since a divides each term on the left-hand

side, a also divides the sum of the two

terms. Therefore, a divides c.

15


Theorem 4.4 Let G be a group and let

x ∈ G. Let m,n, d be integers. Then:

(i) o(x) = o(x−1)

(ii) If o(x) = n and xm = e, then n |m

(iii) If o(x) = n and gcd(m,n) = d, then

o(xm) = n/d.

16

Cyclic Groups

Definition

A group G is called cyclic if there is an

element x ∈ G such that

G = 〈x〉 = {xn | n ∈ Z}.

The element x is called a generator for

G, and the cyclic group generated by x is

denoted by 〈x〉.

17

Cyclic Groups

Example

• The group G = (Z1,⊕) is the trivial

group {0} consisting of one (identity)

element. It is the cyclic group gener-

ated by x = 0:

(Z1,⊕) = 〈0〉

• For all n ≥ 2, the group G = (Zn,⊕) is a

finite cyclic group generated by x = 1:

(Zn,⊕) = 〈1〉

= {0,1, 1⊕ 1, . . . , 1⊕ 1⊕ 1⊕ · · · ⊕ 1︸︷︷︸n− 1 terms

}

18

Cyclic Groups

Example

• The group G = (Z,+) is an infinite

cyclic group generated by x = 1:

(Z,+) = 〈1〉

= {. . . , (−1) + (−1), (−1), 0, 1, 1 + 1, 1 + 1 + 1, . . .}

• The group G = (Q,+) is not cyclic,

since there does not exist q ∈ Q such

that ever rational number r ∈ Q has the

form r = nq for some n ∈ Z.

19

Cyclic Groups

Theorem 4.5 Let G = 〈x〉. If o(x) =∞,

then xj 6= xk for j 6= k, and consequently

G is infinite. If o(x) = n, then xj = xk iff

j ≡ k (mod n), and consequently the dis-

tinct elements of G are e, x, x2, . . . , xn−1.

Proof:

20

Cyclic Groups

Definition The order of a group G, denoted

by |G|, is the number of elements in G.

Corollary 4.6 If G = 〈x〉, then |G| = o(x).

21

Cyclic Groups

Theorem 4.7 If G is a cyclic group, then

G is abelian.

Proof:

22

5 Subgroups

Definition of a Subgroup

A subset H of a group (G, ∗) is called a

subgroup of G if the elements of H form

a group under ∗.

1

Subgroups

Examples

• (2Z,+) is a subgroup of (Z,+)

• (Z,+) is a subgroup of (Q,+)

• (Q,+) is a subgroup of (R,+)

• (Q+, ·) is a subgroup of (R+, ·)

•{(

a b−b a

)| a, b ∈ R and a2 + b2 6= 0

}is a

subgroup of G = GL(2,R).

•{(

a b0 d

)| a, b, d ∈ R and ad 6= 0

}is a

subgroup of G = GL(2,R).

2

Subgroups

Theorem If H is a subgroup of G and e

is the identity element for G, then e ∈ H

and e is the identity element for H.

3

Proof: Assume H is a subgroup of G.

Then H is group and must have an identity

element eH. We claim eH = e where e is

the identity element for G. Since eH is an

identity element for H, we know,

eH ∗ eH = eH .

Since H ⊆ G, we know eH ∈ G. There-

fore, eH has an inverse element e−1H ∈ G.

Multiplying on both sides of the previous

equation by e−1H yields:

e−1H ∗ (eH ∗ eH) = e−1

H ∗ eH(e−1

H ∗ eH) ∗ eH = e

e ∗ eH = e

eH = e

Trivial Subgroups

For any group G, the subgroups H = {e}and H = G are called trivial subgroups.

4

Groups of Order 1

There is only one group of order 1, namely

G = {e}. The group table for G is given by

∗ e

e e

• The group G has only one subgroup

H = G = {e}.

• The group G is cyclic. The element e

is a generator for G.

5

Groups of Order 2

There is only one group of order 2. If we

write G = {e, a}, then the group table for

G is given by

∗ e a

e e a

a a

• The group G has two subgroups, namely

H = {e} and H = G = {e, a}.

• The group G is cyclic. The element a

is a generator for G.

6

Groups of Order 3

There is only one group of order 3. If we

write G = {e, a, b}, then the group table

for G is given by

∗ e a b

e e a b

a a

b b

• The group G has two subgroups, namely

H = {e} and H = G = {e, a, b}.

• The group G is cyclic. Both elements

a and b are generators for G.

7

Groups of Order 4

There are two groups of order 4. If we

write G = {e, a, b, c}, then the possible group

tables for G are

∗ e a b c

e e a b c

a a

b b

c c

∗ e a b c

e e a b c

a a

b b

c c

∗ e a b c

e e a b c

a a

b b

c c

∗ e a b c

e e a b c

a a

b b

c c

8

Cyclic Group of Order 4

If we write G = {e, a, b, c}, then the follow-

ing group table for G represents a cyclic

group of order 4.

∗ e a b c

e e a b c

a a

b b

c c

• The group G has three subgroups, namely

H = {e}, H = {e, b}, and H = {e, a, b, c}.

• The group G is cyclic. Both elements

a and c are generators for G.

9

Klein’s 4-group

If we write G = {e, a, b, c}, then the fol-

lowing group table for G represents a non-

cyclic group of order 4 called Klein’s 4-

group.

∗ e a b c

e e a b c

a a

b b

c c

• Klein’s 4-group has five subgroups:

H = {e}, H = {e, a}, H = {e, b},H = {e, c}, and H = {e, a, b, c}.

10

Subgroup Lattice

A subgroup lattice is a diagram which de-

picts the subgroups of a group G in a way

that indicates all subset relations among

subgroups.

Example (Klein’s 4-group)

11

Groups of Order n

The previous examples beg the question:

How many groups of order n exist for a

given integer n?

The answer for n ≤ 99 is given below.

+ 0 1 2 3 4 5 6 7 8 9

0 0 1 1 1 2 1 2 1 5 2

10 2 1 5 1 2 1 14 1 5 1

20 5 2 2 1 15 2 2 5 4 1

30 4 1 51 1 2 1 14 1 2 2

40 14 1 6 1 4 2 2 1 52 2

50 5 1 5 1 15 2 13 2 2 1

60 13 1 2 4 267 1 4 1 5 1

70 4 1 50 1 2 3 4 1 6 1

80 52 15 2 1 15 1 2 1 12 1

90 10 1 4 2 2 1 231 1 5 2

12

Subgroups

Theorem 5.1 Let H be a nonempty sub-

set of a group G. Then H is a subgroup

of G if and only if the following two con-

ditions are satisfied:

(i) for all a, b ∈ H, ab ∈ H, and

(ii) for all a ∈ H, a−1 ∈ H.

Remark: Condition (i) is expressed by say-

ing that H is closed under the binary oper-

ation on G, and condition (ii) is expressed

by saying that H is closed under inverses.

13

Proof:

Subgroups

Theorem Let G be a group, and let a ∈ G.

Then 〈a〉, the set of elements generated by

integer powers of a, is a subgroup of G.

Proof: Let H = 〈a〉. Clearly H is nonempty

since a ∈ H. Also for all aj, ak ∈ H, we have

ajak = aj+k ∈ H. Also, for any aj ∈ H,

we have (aj)−1 = a−j ∈ H. Therefore, by

Theorem 5.1, H = 〈a〉 is a subgroup of G.

14

Subgroups

Examples

• Let G = (Z,+) and let n ∈ Z. Then,

H = 〈n〉 = nZ is a subgroup of G.

• Let G = (Z6,⊕). Then the following

are subgroups of G:

〈0〉 = {0}

〈1〉 = {0,1,2,3,4,5}

〈2〉 = {0,2,4}

〈3〉 = {0,3}

〈4〉 = {0,2,4}

〈5〉 = {0,1,2,3,4,5}

15

Subgroups

Theorem 5.2 If G is a cyclic group, then

every subgroup of G is cyclic.

Proof:

16

Subgroups

Theorem 5.4 Let H and K be subgroups

of a group G. Then:

(i) H ∩K is a subgroup of G; and

(ii) H ∪K is a subgroup of G if and only

if H ∪K = H or H ∪K = K.

Proof:

17

Subgroups

Theorem 5.5 Let G = 〈x〉 be a cyclic

group of order n. Then:

(i) For any positive integer m, G has a

subgroup of order m if and only if m

divides n.

(ii) If m divides n, then G has a unique

subgroup of order m.

(iii) The elements xr and xs generate the

same subgroup of G if and only if

(r, n) = (s, n).

18

Group of Unit Quaternions

Consider the general linear group of degree

two over the complex numbers

GL(2,C) ={( z1 z2

z3 z4

)| zk ∈ C and z1z4 − z2z3 6= 0

}

The group of unit quaternions, denoted

Q8, is the subgroup of GL(2,C) consisting

of the following 8 elements:

{(1 0

0 1

),

(−1 0

0 −1

),( 0 1

−1 0

),( 0 −1

1 0

)(

i 0

0 −i

),( −i 0

0 i

),(

0 i

i 0

),

(0 −i

−i 0

)}

19

Group of Unit Quaternions

Let J =(

i 0

0 −i

), K =

( 0 1

−1 0

), L =

(0 i

i 0

).

The group of unit quaternions has the form

Q8 = {I, −I, J, −J, K, −K, L, −L},

and the following relations hold

J2 = K2 = L2 = −I

JK = L

KL = J

LJ = K

KJ = −L

LK = −J

JL = −K

20

6 Direct Products

Direct Product of Groups

Let G and H be groups. Then the set of

ordered pairs

G×H = {(g, h) | g ∈ G, h ∈ H}

forms a group under componentwise mul-

tiplication:

(g1, h1) ∗ (g2, h2) = (g1 g2, h1h2)

The group G × H is called the direct

product of G and H.

1

Direct Product of Groups

Remarks

• The identity element of G×H is (eG, eH)

where eG and eH are the respective

identity elements of G and H.

• The inverse of (g, h) ∈ G × H is the

ordered pair (g−1, h−1).

2

Generalized Direct Product

Let G1, G2, G3, . . . , Gn be groups. Then

the set of ordered n-tuples

G1 × · · · ×Gn = {(g1, g2, . . . , gn) | gi ∈ Gi}

forms a group under componentwise mul-

tiplication.

• The identity element of G1 × · · · × Gn

is (eG1, eG2

, . . . , eGn) where eGiis the

identity element for Gi.

• The inverse of (g1, g2, . . . , gn) is the

element (g−11 , g−1

2 , . . . , g−1n ).

3

Direct Products

Example

Let G = H = (Z2,⊕). Then,

G×H = Z2 × Z2

is the group of ordered pairs

{(0,0), (0,1), (1,0), (1,1)}

under componentwise addition mod 2.

• Z2 × Z2 is a finite, non-cyclic group of

order 4 with the following non-trivial

subgroups

〈(0,1)〉 = {(0,0), (0,1)}

〈(1,0)〉 = {(0,0), (1,0)}

〈(1,1)〉 = {(0,0), (1,1)}

4

Direct Products

Example

Let G = (Z2,⊕) and H = (Z3,⊕). Then,

G×H = Z2 × Z3

is the group of ordered pairs

{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)}

under componentwise addition.

• Z2×Z3 is a finite, cyclic group of order

6 with generator (1,1), and with the

non-trivial subgroups

〈(0,1)〉 = {(0,0), (0,1), (0,2)}

〈(1,0)〉 = {(0,0), (1,0)}

5

Direct Products

Example

Let G1 = G2 = · · · = Gn = (R,+). Then,

G1 × · · · ×Gn = R× · · · × R︸︷︷︸n-factors

= Rn

is the group of ordered n-tuples of real

numbers under componentwise addition.

• The identity element of Rn is (0,0, . . . ,0).

• The inverse of (x1, x2, . . . , xn) is the

element (−x1,−x2, . . . ,−xn).

6

Direct Products

Theorem 6.1 Let G = G1 × · · · ×Gn.

(i) If gi ∈ Gi for 1 ≤ i ≤ n, and each gi

has finite order, then o((g1, . . . , gn))

is the least common multiple of the

numbers o(g1), o(g2), . . . , o(gn).

(ii) If each Gi is a cyclic group of finite

order, then G is cyclic iff |Gi| and

|Gj| are relatively prime for i 6= j.

7

Proof:

Direct Products

Examples

• Consider the element (1,3) ∈ Z6 × Z8.

Since o(1) = 6 in the group Z6, and

o(3) = 8 in the group Z8, we have

o((1,3)) = lcm(6,8) = 24

• The groups Z14×Z15 and Z8×Z9×Z5

are cyclic.

• The groups Z14×Z16 and Z8×Z9×Z6

are not cyclic.

8

7 Functions

Function, Domain, and Codomain

Let A and B be nonempty sets. A function

f from A to B, denoted

f : A→ B,

is a subset of the Cartesian product A×B

such that for each a ∈ A, there is a unique

element b ∈ B such that (a, b) belongs to f .

We say that A is the domain of f and B is

the codomain of f .

1

Range of a Function

If f : A → B and the ordered pair (a, b)

belongs to f , then we write

f(a) = b

and we say b the image of a under f .

The range of f , denoted f(A), is the set

of all images of elements of A.

2

Examples

• f : R→ R, defined by

f(x) = x2

• f : GL(2,R) → R, defined by

f((

a bc d

))= ad− bc

• f : G → G, defined by

f(x) = a ∗ x

where G is a group and a ∈ G.

• F : C(R,R) → R, defined by

F (g) =∫ 1

0g(x) dx

where C(R,R) is the set of continuous

functions from R to R.

3

Image and Preimage

If f : A→ B and S ⊆ A, then the image of

S under f is the set

f(S) = {b ∈ B | b = f(s) for some s ∈ S}

If T ⊂ B, then the preimage of T is the set

f−1(T ) = {a ∈ A | f(a) ∈ T}

4

Example

Let f : R→ R be defined by f(x) = x2.

• domain of f :

• codomain of f :

• range of f :

• If S = [−2,3), then f(S) =

• If T = (−1,4], then f−1(T ) =

5

Example

Let f : R→ R be defined by f(x) = x2.

• The domain of f is R.

• The codomain of f is R.

• The range of f is R+ ∪ {0}.

• If S = [−2,3), then f(S) = [0,9).

• If T = (−1,4], then f−1(T ) = [−2,2].

6

One-To-One Function

A function f : A→ B is said to be one-to-

one, or injective, if and only if the follow-

ing condition holds:

∀x∀y (x 6= y → f(x) 6= f(y))

or equivalently

∀x∀y (f(x) = f(y) → x = y)

where the domain for x and y is the set A.

If f is one-to-one, we say f is an injection.

7

Onto Function

A function f : A → B is said to be onto,

or surjective, if and only if the following

condition holds:

∀b ∃a (f(a) = b)

where the domain for b is the set B, and

the domain for a is the set A.

If f is onto, we say that f is a surjection.

Note that f is onto if and only if f(A) = B.

That is, f is onto if and only if the range

of f is equal to the codomain of f .

8

Examples

Let A = {1,2,3,4} and B = {a, b, c}.

Determine if the given functions are one-

to-one, onto, both or neither.

1. f : A→ B defined by

f(1) = a

f(2) = b

f(3) = c

f(4) = b

2. f : B → A defined by

f(a) = 3

f(b) = 1

f(c) = 4

9

Examples

• f : R→ R, defined by

f(x) = x2

• f : GL(2,R) → R, defined by

f((

a bc d

))= ad− bc

• f : G → G, defined by

f(x) = a ∗ x

where G is a group and a ∈ G.

• F : C(R,R) → R, defined by

F (g) =∫ 1

0g(x) dx

where C(R,R) is the set of continuous

functions from R to R.

10

Proof Strategy

To show f is injective

Show that if f(x) = f(y), then x = y.

To show f is not injective

Show that there exist x and y such that

f(x) = f(y) and x 6= y.

To show f is surjective

Show that for each element y in the codomain

there exists an element x in the domain

such that f(x) = y.

To show f is not surjective

Show there exists an element y in the codomain

such that y 6= f(x) for any x in the domain.

11

Example

Prove or disprove:

The function f : Z→ Z defined by

f(n) = 2n− 3

is injective.

12

Example

Prove or disprove:

The function f : Z→ Z defined by

f(n) = 2n− 3

is surjective.

13

Bijection

A function f : A→ B is said to be a bijec-

tion, or one-to-one correspondence if f

is both one-to-one and onto.

Examples

• f : Z→ Z defined by f(n) = n + 1

• f : R→ R defined by f(x) = 2x

• f : Z+ → Z+ defined by f(n) ={n + 1 if n is odd

n− 1 if n is even

14

Inverse Function

Let f : A→ B be a bijection. The inverse

function of f , denoted by f−1, is the func-

tion that assigns to an element b ∈ B the

unique element a ∈ A such that f(a) = b.

That is,

f−1(b) = a if and only if f(a) = b

If f has an inverse function, we say that f

is invertible.

Note that we use the notation f−1(T ) to

denote the preimage of a set T ⊆ B, even

when f is not invertible.

15

Example

Determine if each function invertible. If so,

find f−1.

1. f : Z→ Z defined by f(n) = n + 1

2. f : R→ R defined by f(x) = 2x + 3

3. f : R→ R defined by f(x) = x2

16

Composition of Functions

Let g : A→ B and f : B → C be functions.

The composition of the functions f and

g, denoted by f ◦ g, is a function from A to

C, defined by

(f ◦ g) (a) = f(g(a))

17

Examples

Let A = {1,2,3,4}, B = {a, b, c}, C = {w, x, y, z},and let f and g be the functions below

g : A→ B defined by

g(1) = a

g(2) = b

g(3) = c

g(4) = b

f : B → C defined by

f(a) = x

f(b) = w

f(c) = z

Then, f ◦ g : A→ C is given by

(f ◦ g)(1) =

(f ◦ g)(2) =

(f ◦ g)(3) =

(f ◦ g)(4) =

18

Composition of Functions

Example

Let g : R → R and f : R → R be defined by

g(x) = 12(x− 3) and f(x) = 2x + 3. Find

(a) (f ◦ g)(x) =

(b) (g ◦ f)(x) =

19

Inverse Functions and Composition

Let f : A → B and g : B → A. Then, g is

the inverse function for f if and only if the

following two conditions hold.

(i) (f ◦ g)(x) = x for all x ∈ B.

(ii) (g ◦ f)(x) = x for all x ∈ A.

In particular, if f is invertible, then

(i) f(f−1(x)) = x for all x ∈ B.

(ii) f−1(f(x)) = x for all x ∈ A.

20

Identity Function

Let A be a set. The identity function on

A is the function iA : A→ A defined by

iA(x) = x

That is, iA is the function that assigns each

element of A to itself.

The identity function iA is one-to-one and

onto, so it is a bijection.

21

Set of Bijections from X to X

Let X be a nonempty set, and define

SX = {f : X → X | f is a bijection}

Theorem 7.1: (SX , ◦) is a group.

Proof:

(i) (Closure) See Homework 5, Problem 7

(ii) (Associativity) For all f, g, h ∈ SX, and

for all x ∈ X,

((f ◦ g) ◦ h)(x) = (f ◦ g)(h(x))

= f(g(h(x)))

= f((g ◦ h)(x))

= (f ◦ (g ◦ h))(x)

That is, (f ◦g)◦h = f ◦(g ◦h), which proves

◦ is associative.22

(iii) (identity element) The identity func-

tion iX : X → X is a bijection, hence be-

longs to SX, and has the property

(f ◦ iX)(x) = f(iX(x))

= f(x)

= iX(f(x))

= (iX ◦ f)(x)

for all x ∈ X. This proves iX is an identity

element for SX.

(iv) (inverse elements) For all f ∈ SX,

there is an element g ∈ SX, namely g =

f−1, such that

(f ◦ f−1)(x) = f(f−1(x))

= iX(x)

= f−1(f(x))

= (f−1 ◦ f)(x)

for all x ∈ X. This proves that every ele-

ment of SX has an inverse.

Thus, (SX , ◦) is a group.

8 Symmetric Groups

Permutations

If X is a nonempty set, then a permutation

of X is a bijection f : X → X.

Recall that the set SX of all permutations

of X forms a group under composition of

functions. This group, denoted (SX , ◦), is

called the symmetric group on X.

1

Symmetric Group of Degree n

Let X = {1,2,3, . . . , n}.

The symmetric group on X is called the

symmetric group of degree n, and is de-

noted by Sn.

Given a permutation f ∈ Sn, we represent

f in array form as follows:

1 2 3 · · · n

f(1) f(2) f(3) · · · f(n)

2

Composition of Permutations

Example

Consider f, g ∈ S4 defined by

f(1) = 2 g(1) = 3

f(2) = 4 g(2) = 2

f(3) = 1 g(3) = 4

f(4) = 3 g(4) = 1

In array form we have:

f =(

1 2 3 4

f(1) f(2) f(3) f(4)

)=(

1 2 3 42 4 1 3

)

g =(

1 2 3 4

g(1) g(2) g(3) g(4)

)=(

1 2 3 43 2 4 1

)

3

To compute the composition f ◦g, the fol-

lowing diagram is helpful.

1 2 3 4

3 2 4 1

1 4 3 2

Therefore,

f ◦ g =(

1 2 3 41 4 2 3

).

Observe that

(f ◦ g)(1) =((

1 2 3 42 4 1 3

)◦(

1 2 3 43 2 4 1

))(1)

=(

1 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

)(1)

)

=(

1 2 3 42 4 1 3

)(3)

= 1

(f ◦ g)(2) =((

1 2 3 42 4 1 3

)◦(

1 2 3 43 2 4 1

))(2)

=(

1 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

)(2)

)

=(

1 2 3 42 4 1 3

)(2)

= 4

Also, we have

(f ◦ g)(3) =((

1 2 3 42 4 1 3

)◦(

1 2 3 43 2 4 1

))(3)

=(

1 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

)(3)

)

=(

1 2 3 42 4 1 3

)(4)

= 3

(f ◦ g)(4) =((

1 2 3 42 4 1 3

)◦(

1 2 3 43 2 4 1

))(4)

=(

1 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

)(4)

)

=(

1 2 3 42 4 1 3

)(1)

= 2

Composition of Permutations

Example: Determine f ◦g where f, g ∈ S7

are given by

f =(

1 2 3 4 5 6 73 4 2 1 5 6 7

)

g =(

1 2 3 4 5 6 77 1 3 4 5 2 6

)

Solution:

1 2 3 4 5 6 7

7 1 3 4 5 2 6

4

Cycles

An element f ∈ Sn is called a cycle if there

exist distinct integers

i1, i2, . . . , ir ∈ {1,2, . . . , n}

such that

f(i1) = i2

f(i2) = i3

···

f(ir−1) = f(ir)

f(ir) = i1

and f(j) = j otherwise. In this case, we

express f in cycle notation as follows:

f = (i1, i2, i3, . . . , ir)

5

Cycle Notation

Example: Consider f, g ∈ S7 given by

f =(

1 2 3 4 5 6 73 4 2 1 5 6 7

)

g =(

1 2 3 4 5 6 77 1 3 4 5 2 6

)

Then, f and g are both cycles of length 4.

In cycle notation, we have

f = (1, 3, 2, 4),

g = (1, 7, 6, 2).

6

Cycle Notation

• A cycle of length r is called an r-cycle.

• A 2-cycle is called a transposition.

• The identity permutation is usually de-

noted by the 1-cycle (1).

7

Disjoint Cycles

Two cycles (x1, x2, . . . , xr) and (y1, y2, . . . , ys)

are said to be disjoint if r ≥ 2, s ≥ 2, and

{x1, x2, . . . , xr} ∩ {y1, y2, . . . , ys} = Ø

Examples

• (1,3), (2,4) ∈ S4

• (1,5,2), (3,7,8,4) ∈ S8

• (2,6,9,4), (5,7) ∈ S9

8

Product of Cycles

While not all permutations f ∈ Sn are cy-

cles, all permutations can be expressed as

a product (composition) of disjoint cycles.

Theorem 8.1: Let f ∈ Sn. Then, there

exist disjoint cycles f1, f2, . . . , fm ∈ Sn

such that

f = f1 ◦ f2 ◦ · · · ◦ fm.

9

Product of Cycles

Example: Consider f ∈ S9 given by

f =(

1 2 3 4 5 6 7 8 94 1 7 9 6 5 8 3 2

).

Note that f is composed of three disjoint

cycles as indicated below

f(1) = 4 f(3) = 7 f(5) = 6

f(4) = 9 f(7) = 8 f(6) = 5

f(9) = 2 f(8) = 3

f(2) = 1

Therefore, we write

f = (1,4,9,2)(3,7,8)(5,6)

10

Order of a Permutation

Theorem: Assume f = f1 ◦ f2 ◦ · · · ◦ fmwhere f1, f2, . . . , fm are disjoint cycles.

Then o(f) is the least common multiple

of o(f1), o(f2), . . . , o(fm). That is,

o(f) = lcm (o(f1), o(f2), . . . , o(fm)) .

Note: If f is an r-cycle, then o(f) = r.

11

Order of a Permutation

Example: Consider f ∈ S9 given by

f =(

1 2 3 4 5 6 7 8 94 1 7 9 6 5 8 3 2

).

As shown in the previous example, f can

be expressed as a composition of disjoint

cycles as follows:

f = (1,4,9,2)(3,7,8)(5,6).

Therefore,

o(f) = lcm (4, 3, 2) = 12

12

Cycles and Transpositions

Every cycle f = (i1, i2, . . . , ir) can be ex-

pressed as a product of transpositions as

follows:

f = (i1, ir) ◦ (i1, ir−1) ◦ · · · ◦ (i1, i3) ◦ (i1, i2)

Examples

• (1,3,2,4) = (1,4) (1,2) (1,3)

• (3,7,9,5,8) = (3,8) (3,5) (3,9) (3,7)

11

Permutations and Transpositions

Theorem 8.3: Every permutation f ∈ Sn,

where n ≥ 2, can be expressed as a product

of transpositions.

Proof: By Theorem 8.1,

f = f1 ◦ f2 ◦ · · · ◦ fm

where f1, f2, . . . , fm ∈ Sn are disjoint cy-

cles. Since each cycle fi can be expressed

as a product of transpositions (as described

above), this proves f is a product of trans-

positions.

12

Even/Odd Permutations

We say that a permutation f ∈ Sn is even

if it can be written as the product of an

even number of transpositions, and we say

f is odd if it can be written as the product

of an odd number of transpositions.

Theorem 8.4: No permutation is both

even and odd.

13

Product of Cycles

Example: Consider the permutation

f = (1,3,2,4)(1,7,6,2) ∈ S7

Express f as a product of disjoint cycles.

14

Even/Odd Permutations

Example: Consider the permutation

f = (1,3,2,4)(1,7,6,2) ∈ S7

Express f as a product of transpositions,

then determine if f is even or odd.

15

Alternating Groups

The alternating group of degree n, de-

noted An, is defined to be the subset of Sn

consisting of all even permutations.

16

Alternating Groups

Example: List all elements in S3 using cy-

cle notation, then express each element as

a product of transpositions. Finally, deter-

mine the elements of An.

17

Alternating Groups

Theorem 8.5: If n ≥ 2, then An is a

subgroup of Sn, and

|An| =n!

2=|Sn|

2

18

The symmetric group S3

◦ (1) (1,2) (1,3) (2,3) (1,2,3) (1,3,2)

(1) (1) (1,2) (1,3) (2,3) (1,2,3) (1,3,2)

(1,2) (1,2) (1) (1,3,2) (1,2,3) (2,3) (1,3)

(1,3) (1,3) (1,2,3) (1) (1,3,2) (1,2) (2,3)

(2,3) (2,3) (1,3,2) (1,2,3) (1) (1,3) (1,2)

(1,2,3) (1,2,3) (1,3) (2,3) (1,2) (1,3,2) (1)

(1,3,2) (1,3,2) (2,3) (1,2) (1,3) (1) (1,2,3)

19

S3 as a group of symmetries

If we associate the elements of the set

X = {1,2,3} with the vertices of an equi-

lateral triangle, then each element of the

permutation group S3 can be associated

with a symmetry of the triangle as follows.

22

S3 as a group of symmetries

Rotations:

Reflections:

23

Dihedral Groups

In general, the symmetries a regular n-gon

can be associated with a subgroup of Sn

called the dihedral group of order 2n. This

group of symmetries consists of n rotations

and n reflections, and is denoted by Dn.

If n = 3, the dihedral group D3 is identical

to S3. If n ≥ 4, the dihedral group Dn is a

nontrivial subgroup of Sn. We consider the

case n = 4 below.

24

The dihedral group D4

Rotations:

Reflections:

25

The dihedral group D4

Therefore, the elements of D4 are

{(1), (1,2,3,4), (1,3)(2,4), (1,4,3,2),

(1,2)(3,4), (2,4), (1,4)(2,3), (1,3)}

26

9 Equivalence Relations; Cosets

Relations

A relation on a nonempty set S is a nonempty

subset R of S × S.

Notation

If (x, y) ∈ R, we write xRy.

1

Relations

Example

Let S be a nonempty set, then any function

f ⊆ S × S is a relation. For example, if

S = R, then

f = {(x, x2) | x ∈ R} ⊆ R× R

is the relation corresponding to the func-

tion f(x) = x2.

2

Relations

Example

Let S = {1,2,3}. Then,

R = {(1,2), (1,3), (2,3)} ⊆ S × S

is a relation on S corresponding to the

“less than” relation. That is, for x, y ∈ S

xRy iff x < y.

3

Equivalence Relation

A relation R on S is called an equiva-

lence relation if the following three prop-

erties hold:

1. (Reflexivity)

xRx, for all x ∈ S.

2. (Symmetry)

if xRy, then y Rx.

3. (Transitivity)

if xRy and y R z, then xR z.

4


Example

Let R be the relation on S = Z defined by

aR b iff a ≡ b (mod n)

where n is a fixed positive integer. Prove

that R is an equivalence relation.

5

Proof: Recall that a ≡ b (mod n) if and

only if b− a = nk for some integer k.

(Reflexivity)

(i) xRx, since x− x = 0 = n · 0.

(Symmetry)

(ii) Assume xRy. Then, x − y = nk for

some integer k. Therefore, y − x = n(−k),

where −k is an integer. Therefore, y Rx.

(Transitivity)

(iii) Assume xRy and y R z. Then, x−y =

nk for some integer k, and y − z = nj for

some integer j. Then,

x− z = (x− y) + (y − z) = nk + nj

That is, x− z = n(k + j) where k + j is an

integer. Therefore, xR z.

Equivalence Classes

Let R be an equivalence relation on a nonempty

set S. Given s ∈ S, we define

[s] = {x ∈ S | xR s}.

That is, [s] is the subset of elements in S

which are related to s.

The set [s] is called the equivalence class

of s under R. An element x ∈ S is called

a representative of [s] if x ∈ [s].

6

Equivalence Classes

Let R be the relation on S = Z defined by

aR b iff a ≡ b (mod 4)

Then,

[0] = {. . . ,−8,−4, 0, 4, 8, . . .}

[1] = {. . . ,−7,−3, 1, 5, 9, . . .}

[2] = {. . . ,−6,−2, 2, 6, 10, . . .}

[3] = {. . . ,−5,−1, 3, 7, 11, . . .}

where 0, 1, 2, and 3 are representatives of

their respective equivalence classes.

7

Equivalence Classes

Theorem 9.1 Let R be an equivalence

relation on S. Then, every element of S

is in exactly one equivalence class under R.

That is, the equivalence classes partition S

into a family of mutually disjoint nonempty

subsets.

Conversely, given any partition of S into

mutually disjoint nonempty subsets, there

is an equivalence relation on S whose equiv-

alence classes are precisely the subsets in

the given partition of S.

8

Proof: For every s ∈ S, we have s ∈ [s].

This shows every element of S is in at least

one equivalence class.

Next, suppose s ∈ [s1] and s ∈ [s2]. We

want to show [s1] = [s2]. Since s1 Rs (by

symmetry) and sR s2, it follows by transi-

tivity that s1 Rs2. Therefore, by a home-

work exercise, [s1] = [s2]. This shows that

each element in S is contained in exactly

one equivalence class.


Theorem 9.2: Assume G is a group, and

let H be a subgroup G. Then the relation

≡H defined by

x ≡H y iff xy−1 ∈ H

is an equivalence relation.

9

Proof:

For any x ∈ G, we have xx−1 = e ∈ H.

Therefore, x ≡H x, which means ≡H is

reflexive.

Next, assume x ≡H y. Then, xy−1 ∈ H,

and since H is closed under inverses, we

also have

yx−1 = (y−1)−1x−1 = (xy−1)−1 ∈ H

That is, yx−1 ∈ H, which means y ≡H x.

This proves ≡H is symmetric.

Finally, assume x ≡H y and y ≡H z. This

means, xy−1 ∈ H and yz−1 ∈ H. Since H is

closed under multiplication, we also have

xz−1 = (xy−1)(yz−1) ∈ H

Therefore, x ≡H z, which proves ≡H is

transitive.

Cosets

Assume G is a group, and let H be a sub-

group of G. Given any fixed element a ∈ G,

the set

aH = {x ∈ G | x = ah for some h ∈ H}

is called a left coset of H in G.

Similarly, the set

Ha = {x ∈ G | x = ha for some h ∈ H}

is called a right coset of H in G.

10

Cosets

Example: Let G = S3, and consider the

subgroup H = {(1), (1,2)}. Then, for the

fixed element a = (1,2,3) ∈ G, we have

the left coset

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

and the right coset

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

11

Cosets


let H be a subgroup of G. If a ∈ G and [a]

denotes the equivalence class of a under

the equivalence relation ≡H. Then,

[a] = Ha

That is, the equivalence classes of ≡H are

precisely the right cosets of H.

12

Proof: Assume G is a group, and let H be

a subgroup of G. If a ∈ G, we have

x ∈ [a] iff x ≡H a

iff xa−1 ∈ H

iff xa−1 = h for some h ∈ H

iff x = ha for some h ∈ H

iff x ∈ Ha.

Therefore, [a] = Ha.

Cosets

Corollary 9.4: Assume G is a group, and

let H be a subgroup of G. Then for all

a, b ∈ G,

Ha = Hb if and only if ab−1 ∈ H.

13

Proof: Assume G is a group, and let H be

a subgroup of G. For all a, b ∈ G, we have

Ha = Hb iff [a] = [b]

iff a ≡H b

iff ab−1 ∈ H

That is, Ha = Hb if and only if ab−1 ∈ H.

Cosets


let H be a subgroup of G. If a ∈ G and [a]

denotes the equivalence class of a under

the equivalence relation H ≡, defined by

x H ≡ y iff x−1y ∈ H,

then [a] = aH. That is, the equivalence

classes of H ≡ are precisely the left cosets

of H. Furthermore, we have

[a] = [b] iff aH = bH iff a−1b ∈ H.

14

Cosets


subgroup H = {(1), (1,2)}. Then, the left

cosets of H are

(1)H = {(1)(1), (1)(1,2)}

(1,2)H = {(1,2)(1), (1,2)(1,2)}

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

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

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

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

15

Cosets


subgroup H = {(1), (1,2)}. Then, the left

cosets of H are

(1)H = {(1)(1), (1)(1,2)} = {(1), (1,2)}

(1,2)H = {(1,2)(1), (1,2)(1,2)} = {(1), (1,2)}

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

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

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

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

16

Cosets


subgroup H = {(1), (1,2)}. Then, the right

cosets of H are

H(1) = {(1)(1), (1,2)(1)}

H(1,2) = {(1)(1,2), (1,2)(1,2)}

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

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

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

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

17

Cosets


subgroup H = {(1), (1,2)}. Then, the right

cosets of H are

H(1) = {(1)(1), (1,2)(1)} = {(1), (1,2)}

H(1,2) = {(1)(1,2), (1,2)(1,2)} = {(1), (1,2)}

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

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

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

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

18

Cosets

Remark: If G is an abelian group, and H

is a subgroup of G. Then the left cosets

and right cosets of H are identical. That

is aH = Ha for all a ∈ G.

Example: Let G = (Z,+), and consider

the subgroup H = 4Z. The left and right

cosets of H are given by

0 + 4Z = {. . . ,−8,−4, 0, 4, 8, . . .} = 4Z + 0

1 + 4Z = {. . . ,−7,−3, 1, 5, 9, . . .} = 4Z + 1

2 + 4Z = {. . . ,−6,−2, 2, 6, 10, . . .} = 4Z + 2

3 + 4Z = {. . . ,−5,−1, 3, 7, 11, . . .} = 4Z + 3

Also note that

a+4Z = b+4Z iff a−b ∈ 4Z iff a ≡ b (mod 4).

19

10 Counting the Elements of a

Finite Group

Lagrange’s Theorem

Theorem 10.1 (Lagrange’s Theorem)

Let G be a finite group, and assume H is a

subgroup of G. Then, |H| divides |G|.

1


Lemma 10.2: Let G be a group, and as-

sume H is a subgroup of G. Then, for all

a, b ∈ G, there is a one-to-one correspon-

dence between the elements of the right

coset Ha and the right coset Hb.

In particular, if G is a finite group then we

write |Ha| = |Hb|.

2


Proof: Let H be a subgroup of G, and

assume a, b ∈ G. We consider the function

f : Ha→ Hb defined by

f(ha) = hb

We want to show f is one-to-one and onto.

First, observe that

f(h1a) = f(h2a)

iff h1b = h2b

iff (h1b)b−1 = (h2b)b

−1

iff h1 = h2

iff h1a = h2a

This proves f is one-to-one. Next given

any element hb ∈ Hb, we have f(x) = hb

when x = ha. Hence, f is onto.

3


Proof of Lagrange’s Theorem: Assume H

is a subgroup of a finite group G. Let ≡H

denote the equivalence relation given by

a ≡H b iff ab−1 ∈ H.

By Theorem 9.1, the equivalence classes

under ≡H partition G into mutually dis-

joint nonempty subsets, and by Theorem

9.3, these equivalence classes are just the

right cosets of H. Since G is finite, there

are finitely many distinct right cosets which

we denote by Ha1, Ha2, . . . , Hak. Thus,

|G| = |Ha1 ∪ Ha2 ∪ · · · ∪ Hak|

= |Ha1|+ |Ha2|+ · · ·+ |Hak|

= |H|+ |H|+ · · ·+ |H|︸︷︷︸k terms

= k|H|.

4

Note that the third equality uses the fact

(Lemma 10.2) that all right cosets have the

same size; that is,

|Hai| = |He| = |H|

for all i = 1,2, . . . , k.

Since we’ve shown |G| = k|H|, we conclude

that |H| divides |G|.

Index of a Subgroup

If G is any group (not necessarily finite)

and H is any subgroup, then the number of

distinct right cosets of H in G is called the

index of H in G. We denote this number

by [G : H].

In particular, if G is a finite group, the proof

of Lagrange’s Theorem shows that

|G| = [G : H] · |H|

5

Index of a Subgroup

Example

Let G = S3 and consider H = {(1), (1,2)}.Then, H has 3 distinct right cosets, namely,

H(1) = {(1), (1,2)}

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

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

Therefore, [G : H] = 3, that is, the index

of H in G is 3. Also, note that |H| divides

|G|. Specifically, we have

|G| = 6 = 3 · 2 = [G : H] · |H|

6

Index of a Subgroup

Example

Let G = (Z,+), and consider the subgroup

H = 4Z. Then H has 4 distinct right

cosets, namely,

4Z + 0 = {. . . ,−8,−4, 0, 4, 8, . . .}

4Z + 1 = {. . . ,−7,−3, 1, 5, 9, . . .}

4Z + 2 = {. . . ,−6,−2, 2, 6, 10, . . .}

4Z + 3 = {. . . ,−5,−1, 3, 7, 11, . . .}

Therefore, [G : H] = 4, that is, the index

of H in G is 4.

In this case, Lagrange’s theorem does not

apply since G has infinite order.

7

Index of a Subgroup

Example

Let G = (R,+), and consider the subgroup

H = Z. Then H has infinitely many dis-

tinct right cosets since H + x 6= H + y for

all x, y ∈ [0,1) such that x 6= y.

In this case, we write [G : H] =∞.

8

Index of a Subgroup

Theorem 10.3 Let H be a subgroup of

G. Then the number of distinct left cosets

of H in G is [G : H].

9

Proof: Assume S is the set of all right

cosets of H and assume T is the set of all

left cosets of H. We will show that the

function f : T → S defined by

f(aH) = Ha−1

is a bijection. First, by Corollary 9.4 and

Theorem 9.5, we have

f(aH) = f(bH)

iff Ha−1 = Hb−1

iff a−1b ∈ H

iff aH = bH

This proves f is one-to-one. Next given

any element Ha ∈ S, we have f(x) = Ha

when x = a−1H ∈ T . Hence, f is onto.

This proves f is a bijection.

Applications of Lagrange’s Theorem

Theorem 10.4 Let G be a finite group

and let x ∈ G. Then o(x) divides |G|.Consequently, x|G| = e for every x ∈ G.

10


Proof: Assume G is a finite group and let

x ∈ G. Consider the subgroup H = 〈x〉.By Lagrange’s Theorem, |H| divides |G|.Since |H| = |〈x〉| = o(x), this proves o(x)

divides |G|.

Finally, if we write |G| = o(x) · k for some

integer k ∈ Z+, we have

x|G| = xo(x)·k = (xo(x))k = ek = e.

This completes the proof.

11


Theorem 10.5 Let G be a finite group

and assume that |G| is prime. Then G is

cyclic, and any element of G other than

the identity element e is a generator for

G.

12


Proof: Assume |G| = p, where p > 1 is

a prime number. Then given any x ∈ G,

we have |〈x〉| divides p. If x 6= e, we know

{e, x} ⊆ 〈x〉. Hence |〈x〉| ≥ 2. Since the

only divisors of p are 1 and p, we conclude

that |〈x〉| = p. This proves that G is cyclic

and that 〈x〉 = G for all x 6= e.

13


Example: Consider the set

G = Zp − {0} = {1,2,3, . . . , p− 1},

where p is a prime number. Then, G

forms a group under multiplication mod-

ulo p. That is, G is a group with respect

to the binary operation � defined by

a� b = ab.

where ab is the remainder of ab mod p.

14

Proof that (Zp − {0},�) is a group.

(i) (Closure) Given a, b ∈ Zp−{0}, we want

to show a� b ∈ Zp − {0}. Clearly, we have

a� b = ab ∈ Zp

It suffices to show ab 6= 0. If not, then

ab = np for some integer p. That is, p | ab.Since p is prime, this means p | a or p | b.Therefore, a = 0 or b = 0, which contra-

dicts the assumption a, b ∈ Zp − {0}. This

proves closure.

(ii) (Associativity) See Homework

(iii) (Identity element) There exists an ele-

ment e ∈ Zp−{0}, namely e = 1, such that

e� a = 1� a = a = a

anda� e = a� 1 = a = a

Therefore, Zp−{0} has an identity element.

(iv) (Inverse elements) We want to show

that for any a ∈ Zp − {0}, there exists

b ∈ Zp − {0} such that a � b = 1. Since

gcd(a, p) = 1, it follows by Theorem 4.2,

that there exist x, y ∈ Z such that

ax + py = 1

Therefore,

a� x = ax = 1− py = 1.

Thus, b = x is the inverse element of a.


Theorem 10.6 (Fermat’s Theorem)

Let p be a prime number and suppose a is

an integer such that p does not divide a.

Then,

ap−1 ≡ 1 (mod p)

15

Proof: Let p be a prime number and sup-

pose a is an integer such that p does not di-

vide a. By the division algorithm, we have

a = qp + r where q ∈ Z and

r ∈ {1,2,3, . . . , p− 1}

Therefore,

ap−1 ≡ (qp + r)p−1 (mod p)

≡ rp−1 (mod p)

where the last congruence follows from the

binomial theorem.

Finally, since G = Zp − {0} forms a group

under multiplication modulo p, it follows

from the previous example and Theorem

10.5 that

rp−1 = r � · · · � r︸︷︷︸(p− 1)-times

= r|G| = e.

That is,

rp−1 ≡ 1 (mod p)


Centralizer of an Element

Assume G is a group. Given a fixed ele-

ment g ∈ G, we define the centralizer of

g to be the set

Z(g) = {x ∈ G | xg = gx}.

That is, Z(g) is the set of elements in G

that commute with g.

Note that all integer powers of g are con-

tained in Z(g). In particular this includes

g−1, g0 = e, and g1 = g.

Previously we proved that Z(g) is a sub-

group of G.

16

Centralizer of an Element

Example

Consider the group G = S3. Then,

Z((1)) =

Z((1,2)) =

Z((1,3)) =

Z((2,3)) =

Z((1,2,3)) =

Z((1,3,2)) =

17

Center of a Group

Assume G is a group. The center of G,

denoted by Z(G), is the set of elements in

G that commute with all elements of G.

Explicitly, we have

Z(G) = {x ∈ G | xg = gx for all g ∈ G}.

That is, x ∈ Z(G) if and only if x ∈ Z(g)

for all g ∈ G.

Note that a group G is abelian if and only

if Z(G) = G.

18

Center of a Group

Examples

• If G is abelian, then Z(G) = G.

• If G = S3, then Z(G) = {(1)}

• If G = Q8, then Z(G) = {I,−I}

19

Conjugacy Classes

Assume G is a group, and consider the

relation R on G defined by

aR b iff a = xbx−1 for some x ∈ G.

If aR b, we say that a is conjugate to b.

Lemma 10.7: The above relation R is

an equivalence relation on G.

Proof: Homework 7, Problem 4.

20

Conjugacy Classes

The equivalence class [a] of an element

a ∈ G under R is called the conjugacy

class of a and consists of all conjugates

of a. That is,

[a] = {xax−1 | x ∈ G}

Lemma 10.7: Let G be a finite group

and let a ∈ G. Then the number of distinct

conjugates of a in G is [G : Z(a)].

21

Proof: For all x, y ∈ G, we have

xax−1 = yay−1 iff ax−1y = x−1ya

iff x−1y ∈ Z(a)

iff x Z(a)≡ y

iff xZ(a) = yZ(a).

Therefore, the number of distinct conju-

gates of a corresponds to the number of

distinct left cosets of Z(a) which, by The-

orem 10.3, is [G : Z(a)].

Conjugacy Classes

Example Consider the group G = S3.

Complete the table below. Then write G as

the disjoint union of its conjugacy classes.

a Z(a) [G : Z(a)] [a]

(1)

(1,2)

(1,3)

(2,3)

(1,2,3)

(1,3,2)

22

Conjugacy Classes

Let’s compute the conjugates of a = (1,2) :

(1)(1,2)(1)−1 = (1,2)

(1,2)(1,2)(1,2)−1 = (1,2)

(1,3)(1,2)(1,3)−1 = (2,3)

(2,3)(1,2)(2,3)−1 = (1,3)

(1,2,3)(1,2)(1,2,3)−1 = (2,3)

(1,3,2)(1,2)(1,3,2)−1 = (1,3)

Therefore, the conjugacy class of a (the set

of distinct conjugates of a) is:

[a] = {(1,2), (1,3), (2,3)}

23

Conjugacy Classes

Next compute the conjugates of a = (1,2,3) :

(1)(1,2,3)(1)−1 = (1,2,3)

(1,2)(1,2,3)(1,2)−1 = (1,3,2)

(1,3)(1,2,3)(1,3)−1 = (1,3,2)

(2,3)(1,2,3)(2,3)−1 = (1,3,2)

(1,2,3)(1,2,3)(1,2,3)−1 = (1,2,3)

(1,3,2)(1,2,3)(1,3,2)−1 = (1,2,3)

Therefore, the conjugacy class of a (the set

of distinct conjugates of a) is:

[a] = {(1,2,3), (1,3,2)}

24

Conjugacy Classes

Important Fact: All conjugates of a have

the same order as a. (Homework 4, Prob-

lem 5)

25

Conjugacy Classes

Example Consider the group G = S3.

Complete the table below. Then write G as

the disjoint union of its conjugacy classes.

a Z(a) [G : Z(a)] [a]

(1) S3 1 {(1)}

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

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

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

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

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

G = {(1)} ∪ {(1,2), (1,3), (2,3)} ∪ {(1,2,3), (1,3,2)}

= [(1)] ∪ [(1,2)] ∪ [(1,2,3)]

26

Class Equation

Theorem 10.9 (Class equation): Let G

be a finite group, and let {a1, a2, . . . , ak}consist of one element from each conju-

gacy class containing at least two elements.

Then,

|G| = |Z(G)|+[G : Z(a1)]+· · ·+[G : Z(ak)].

27

Proof: Since G is the disjoint union of its

conjugacy classes, we have

G = [a1] ∪ · · · ∪ [ak] ∪ {ak+1} · · · ∪ {ak+s}

where {ak+1}, . . . , {ak+s} are the conjugacy

classes with one element. Therefore, by

Lemma 10.8, we have:

|G| = [G : Z(a1)] + · · ·+ [G : Z(ak)] + s.

It suffice to show that s = |Z(G)|. By

Lemma 10.8,

[a] = {a} iff [G : Z(a)] = 1

iff Z(a) = G

iff a ∈ Z(G)

Therefore, the conjagacy classes with a sin-

gle element correspond precisely to the ele-

ments of Z(G). This completes the proof.

Class Equation

Example Consider the group G = S3. In

the example above, we showed that

G = {(1)} ∪ {(1,2), (1,3), (2,3)} ∪ {(1,2,3), (1,3,2)}

= [(1)] ∪ [(1,2)] ∪ [(1,2,3)]

Let a1 = (1,2) and a2 = (1,2,3) represent

the conjugacy classes of size greater than

1, and note that Z(G) = {(1)}. Then,

|G| = |Z(G)|+ [G : Z(a1)] + [G : Z(a2)]

= |{(1)}|+ [G : Z((1,2))] + [G : Z((1,2,3))]

= 1 + 3 + 2

= 6

28

11 Normal Subgroups

Conjugate Subgroups

Theorem 11.4: Let G be a group, and

assume H is a subgroup of G. Then, for

each fixed element g ∈ G, the set

gHg−1 = {ghg−1 | h ∈ H}

is a subgroup of G with the same number

of elements as H.

We say that gHg−1 is a conjugate sub-

group of H.

1

Proof: The proof that gHg−1 is a subgroup

is Exercise 9, Homework 4. The proof that

gHg−1 has the same number of elements

as H is also left as an exercise.

Normal Subgroups

Assume H is a subgroup of G. We say that

H is a normal subgroup if ghg−1 ∈ H for

all h ∈ H and for all g ∈ G.

2

Normal Subgroups

Theorem 11.1: Let H be a subgroup of

G. Then the following are equivalent:

(i) H is a normal subgroup;

(ii) gHg−1 = H for all g ∈ G;

(iii) gH = Hg for all g ∈ G, i.e., the left

cosets and right cosets of H are iden-

tical.

3

Proof:

First, we will prove statements (i) and (ii)

are equivalent.

Assume (ii) holds. Then for all g ∈ G, we

have gHg−1 = H. In particular, gHg−1 ⊆H, which gives (i).

Conversely, assume (i) holds. Then for all

g ∈ G, we have gHg−1 ⊆ H, and we have

g−1H(g−1)−1 ⊆ H

iff H(g−1)−1 ⊆ gH

iff H ⊆ gHg−1

Therefore, gHg−1 = H for all g ∈ G, which

proves (ii).

Next, we will prove statements (ii) and (iii)

are equivalent. For all g ∈ G, we have

gHg−1 = H iff gH = Hg

where the second equality is obtained from

the first after multiplication by g on the

right.


Normal Subgroups

Example

Let G = S3. Then

H = 〈(1,2,3)〉 = {(1), (1,2,3), (1,3,2)}

is a normal subgroup since the left cosets

and right cosets of H are identical.

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

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

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

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

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

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

4

Normal Subgroups

Example

Let G = S3. Then

H = 〈(1,2)〉 = {(1), (1,2)}

is not a normal subgroup since the left

cosets and right cosets of H are not iden-

tical. For example, the left coset

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

does not match the right coset

H(1,3) = {(1,3), (1,3,2)}.

5

Normal Subgroups

Notation

We write H � G to indicate that H is a

normal subgroup of G.

6

Normal Subgroups

Example

Let G and H be groups, then

{eG} ×H � G×H, and

G× {eH} � G×H.

Observe that for all (eG, x) ∈ {eG} ×H and

for all (g, h) ∈ G×H,

(g, h)(eG, x)(g, h)−1 = (geGg−1, hxh−1)

= (eG, hxh−1)

∈ {eG} ×H.

This proves {eG}×H is a normal subgroup.

A similar calculation shows that G × {eH}is normal.

7

Normal Subgroups

Theorem 11.2 Let G be a group. Then

any subgroup of Z(G) is a normal subgroup

of G.

8

Proof: Let H be a subgroup of Z(G).

Then each element of H commutes with

all elements of G. Therefore, for all g ∈ G,

and for all h ∈ H, we have

ghg−1 = hgg−1 = h ∈ H.

This proves H �G.

Normal Subgroups

Corollary: If G is abelian, then every sub-

group of G is normal.

Proof: If G is abelian, then Z(G) = G, so

every subgroup of G is a subgroup of Z(G).

9

Normal Subgroups

Theorem 11.3: Let H be a subgroup of

G such that [G : H] = 2. Then H is normal

in G.

10

Proof: We want to show that for all g ∈ G,

gH = Hg. Since [G : H] = 2, there are

exactly two left cosets and two right cosets

of H. These cosets must be H and G−H.

If g ∈ H, then

gH = H = Hg.

If g /∈ H, then

gH = G−H = Hg.

Therefore, for all g ∈ G, gH = Hg.

Normal Subgroups

Example

Let G = Q8. Then the subgroups

〈J〉 = {I,−I, J,−J},

〈K〉 = {I,−I,K,−K},

〈L〉 = {I,−I, L,−L},

are all normal in G, since

[Q8 : 〈J〉] = [Q8 : 〈K〉] = [Q8 : 〈L〉] = 2.

11

Normal Subgroups

Example

Let G = Sn. Then the alternating group

An is normal in G, since it follows from

Theorem 8.5 that [Sn : An] = 2.

12

Normal Subgroups

Corollary 11.5: Let G be a group, and

assume H is a subgroup of G. If G has

only one subgroup of size |H|, then H is

normal in G.

13

Proof: By Theorem 11.4, gHg−1 is a sub-

group of G of size |H|. If G has only one

subgroup of size |H|, then gHg−1 = H.

Therefore, by Theorem 11.1, H is normal.

Normal Subgroups

Example

Let G = A4, and consider the subgroup

H = {(1), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}.

We have |H| = 4 and the group G has no

other subgroups of order 4. This follows

from the fact that every element of G not

contained in H is a 3-cycle and these el-

ements cannot be contained in a subgroup

of order 4. Therefore, by Corollary 11.5,

H is normal in G.

14

Quotient Groups

Theorem 11.6: If H � G, then the set

of right cosets of H in G, denoted

G/H = {Ha | a ∈ G},

is group with respect to the operation ∗,defined by

Ha ∗Hb = H(ab).

Furthermore, |G/H| = [G : H].

Notation

We say that G/H is the quotient group

of G by H.

15

Proof: First we must show that ∗ is a

well-defined binary operation when H is a

normal subgroup. That is, we must show

Ha1 ∗Hb1 = Ha2 ∗Hb2

when Ha1 = Ha2 and Hb1 = Hb2. We have

Ha1 ∗Hb1 = Ha2 ∗Hb2

iff H(a1b1) = H(a2b2)

iff (a1b1)(a2b2)−1 ∈ H

iff a1b1b−12 a−1

2 ∈ H

iff a1b1b−12 a−1

1 a1a−12 ∈ H

iff [a1(b1b−12 )a−1

1 ](a1a−12 ) ∈ H.

Since the last statement is satisfied when

Ha1 = Ha2 and Hb1 = Hb2 (that is, when

a1a−12 ∈ H and b1b

−12 ∈ H), it follows that

∗ is well-defined.

Next we must verify the four conditions of

a group.

(i) (Closure) Given Ha,Hb ∈ G/H, we

have Ha ∗ Hb = H(ab) ∈ G/H. This

proves closure.

(ii) (Associativity) Given Ha,Hb,Hc ∈ G/H,

we have

(Ha ∗Hb) ∗Hc = H(ab) ∗Hc

= H[(ab)c]

= H[a(bc)]

= Ha ∗H(bc)

= Ha ∗ (Hb ∗Hc)

Therefore, ∗ is associative.

(iii) (identity element) There exists an el-

ement He ∈ G/H such that

He ∗Ha = H(ea) = Ha; and

Ha ∗He = H(ae) = Ha,

for all Ha ∈ G/H.

(iv) (inverse elements) The inverse of the

element Ha ∈ G/H is the element

Ha−1 since

Ha ∗Ha−1 = H(aa−1)

= He

= H(a−1a)

= Ha−1 ∗Ha.

This proves that G/H is a group. The

fact that |G/H| = [G : H] follows immedi-

ately from the definition of G/H.

Quotient Groups

Example

Let G = S3, and consider the normal sub-

group

H = 〈(1,2,3)〉 = {(1), (1,2,3), (1,3,2)}.

Then,

G/H = {H(1), H(1,2)}

forms a group with the following multipli-

cation table

∗ H(1) H(1,2)

H(1)

H(1,2)

16

Quotient Groups

Example

Let G = Q8, and consider the normal sub-

group

H = Z(Q8) = {I,−I}.

Then,

G/H = {HI, HJ, HK, HL}

forms a group with the following multipli-

cation table

∗ HI HJ HK HL

HI

HJ

HK

HL

17

Quotient Groups

Example

Let G = (Z,+), and consider the normal

subgroup

H = 〈n〉 = nZ.

Then,

G/H = {nZ+ 0, nZ+ 1, . . . , nZ+ (n−1)}

forms a group of size [Z : nZ] = n.

We will prove later that Z / nZ is equivalent

to the group Zn.

18

Quotient Groups

Theorem 11.7 Let G be a finite abelian

group and suppose p is a prime number

that divides |G|. Then G has a subgroup

of order p.

19

Proof:

We will proof the result using (strong) in-

duction on the size of the group G.

Base Case: If |G| = 1, then the result is

true since no prime p divides |G|.

Inductive Step: Assume the theorem is true

when G is abelian and |G| < m. We want to

show that this implies the theorem is true

when G is abelian and |G| = m.

Let G be an abelian group of order m, and

let x 6= e be an element of G. Since G is

abelian, H = 〈x〉 is a normal subgroup of

G, and by Lagrange’s Theorem

|G| = |G/H| · |H|.

Now, assume p is prime and p divides |G|.Then, p divides |H|, or p divides |G/H|. In

the former case, if we write |H| = kp, then

〈xk〉 is a subgroup of G of order p, since

o(xk) =o(x)

gcd(o(x), k)=

|H|gcd(|H|, k)

=|H|k

= p.

On the other hand, if p divides |G/H|, then

by the inductive hypothesis, the abelian group

G/H has a subgroup of order p. Since p is

prime this subgroup is cyclic and can be ex-

pressed in the form 〈Hg〉, where o(Hg) = p.

It follows from a homework exercise that p

divides o(g); that is o(g) = kp for some in-

teger k. Therefore, 〈gk〉 is a subgroup of G

of order p, and the proof is complete.

12 Homomorphisms

Definition

Let G and H be groups. A function

ϕ : G→ H

is called a homomorphism if

ϕ(ab) = ϕ(a)ϕ(b)

for all a, b ∈ G.

A one-to-one homomorphism is called a

monomorphism. An onto homomorphism

is called an epimorphism.

1

Homomorphism

Example

Let G = (Z,+) and let H = (2Z,+). Then

the function ϕ : G→ H defined by

ϕ(n) = 2n

is a homomorphism since

ϕ(n+m) = 2(n+m)

= 2n+ 2m

= ϕ(n) + ϕ(m)

for all n,m ∈ Z.

2

Homomorphism

Example

Let G = (Z,+) and let H = (Zn,⊕). Then


ϕ(m) = m,

where m is the remainder of m modulo n,

is a homomorphism since

ϕ(m1 +m2) = m1 +m2

= m1 ⊕m2

= ϕ(m1)⊕ ϕ(m2)

for all m1,m2 ∈ Z.

3

Homomorphism

Example

Let G = (Sn, ◦) and let H = (Z2,⊕), and

consider the function ϕ : G→ H defined by

ϕ(f) =

0 if f is even

1 if f is odd

To determine if ϕ is a homomorphism, we

consider the following chart.

4

f g f ◦ g ϕ(f) ϕ(g) ϕ(f ◦ g)

even even

even odd

odd even

odd odd

Homomorphism

Example

Let G = (Sn, ◦) and let H = (Z2,⊕), and

consider the function ϕ : G→ H defined by

ϕ(f) =

0 if f is even

1 if f is odd

To determine if ϕ is a homomorphism, we

consider the following chart.

5

f g f ◦ g ϕ(f) ϕ(g) ϕ(f ◦ g)

even even even 0 0 0

even odd odd 0 1 1

odd even odd 1 0 1

odd odd even 1 1 0

In all cases,

ϕ(f ◦ g) = ϕ(f)⊕ ϕ(g).

Therefore, ϕ is homomorphism.

Homomorphism

Theorem 12.4: Let ϕ : G → H be a

homomorphism. Then

(i) ϕ(eG) = eH

(ii) ϕ(xn) = [ϕ(x)]n for all x ∈ G and for

all n ∈ Z.

(iii) if o(x) = n, then o(ϕ(x)) divides n.

6

Proof of (i):

Since ϕ is a homomorphism, we have

ϕ(eG) = ϕ(eG ∗ eG) = ϕ(eG) ∗ ϕ(eG)

That is,

ϕ(eG) = ϕ(eG) ∗ ϕ(eG).

Multiplying both sides of this equation on

the left by ϕ(eG)−1, we obtain

eH = ϕ(eG).

Proof of (ii):

Case 1: If n = 0, then by part (i)

ϕ(x0) = ϕ(eG) = eH = (ϕ(x))0,

for all x ∈ G.

Case 2: We will use induction to prove the

statement holds for all n > 0.

Base Case: When n = 1, we have

ϕ(x1) = ϕ(x) = ϕ(x)1.

Inductive Step: Assume

ϕ(xk) = ϕ(x)k.

for some integer k. We want to show the

statement also holds for n = k + 1.

We have,

ϕ(xk+1) = ϕ(x · xk)

= ϕ(x)ϕ(xk)

= ϕ(x)ϕ(x)k

= ϕ(x)k+1

This proves, ϕ(xk+1) = ϕ(x)k+1 and it fol-

lows by induction that

ϕ(xn) = ϕ(x)n

for all x ∈ G, and for all n ∈ Z+.

Case 3: If n < 0, then write n = −mwhere m > 0, and observe that

ϕ(xn)ϕ(xm) = ϕ(xn · xm)

= ϕ(x−m · xm)

= ϕ(eG)

= eH

Therefore,

ϕ(xn) = [ϕ(xm)]−1

= [[ϕ(x)]m]−1

= [ϕ(x)]−m

= [ϕ(x)]n.

This completes the proof of (ii).

Proof of (iii):

If x ∈ G and o(x) = n, then by parts (i)

and (ii) we have

[ϕ(x)]n = ϕ(xn) = ϕ(eG) = eH .

That is, [ϕ(x)]n = eH, and it follows by

Theorem 4.4(ii) that o(ϕ(x)) divides n.

Homomorphism

Example

Let G = (Z12,⊕) and let H = (Z4,⊕).

Consider the homomorphism ϕ : G → H

defined by

ϕ(x) = x,

where x is the remainder of x mod 4.

• To illustrate Theorem 12.4(iii), let x = 10.

Then ϕ(x) = 2,

o(x) =12

gcd(12,10)=

12

2= 6,

o(ϕ(x)) =4

gcd(4,2)=

4

2= 2,

and we see that o(ϕ(x)) divides o(x).

7

• To illustrate Theorem 12.4(ii), let x = 10,

and let n = 3. Then

x3 = 10⊕ 10⊕ 10 = 6 ∈ Z12

Therefore,

ϕ(x3) = ϕ(6) = 2 ∈ Z4

On the other hand,

ϕ(x)3 = ϕ(x)⊕ϕ(x)⊕ϕ(x) = 2⊕2⊕2 = 2 ∈ Z4.

This verifies that

ϕ(x3) = 2 = ϕ(x)3

Isomorphism

A homomorphism ϕ : G → H is called an

isomorphism if ϕ is one-to-one and onto.

In this case we say G is isomorphic to H

and write G ∼= H.

8

Isomorphism

Example

Let G = (R,+) and let H = (R+, ·). Then


ϕ(x) = ex

is an isomorphism since ϕ is one-to-one

and onto, and

ϕ(x+ y) = ex+y

= exey

= ϕ(x)ϕ(y)

for all x, y ∈ R.

Therefore, we write (R,+) ∼= (R+, ·).

9

Isomorphism

If G and H are finite groups, and

ϕ : G→ H

is an isomorphism, then the condition

ϕ(ab) = ϕ(a)ϕ(b)

means that the multiplication table for H

is obtained from the multiplication table for

G by replacing each entry in the latter table

with its image under ϕ.

10

Isomorphism

Example

Consider the Klein 4-group V = {e, a, b, c}.The multiplication table for V is given by

∗ e a b c

e e a b c

a a e c b

b b c e a

c c b a e

11

Consider the function ϕ : V → Z2 × Z2,

defined by

ϕ(e) = (0,0) ϕ(b) = (0,1)

ϕ(a) = (1,0) ϕ(c) = (1,1)

Then the image under ϕ of the entries in

the multiplication table for V produces the

multiplication table for Z2 × Z2 as shown

below

ϕ(e) = (0,0) ϕ(b) = (0,1)

ϕ(a) = (1,0) ϕ(c) = (1,1)

∗ ϕ(e) ϕ(a) ϕ(b) ϕ(c)

ϕ(e) ϕ(e) ϕ(a) ϕ(b) ϕ(c)

ϕ(a) ϕ(a) ϕ(e) ϕ(c) ϕ(b)

ϕ(b) ϕ(b) ϕ(c) ϕ(e) ϕ(a)

ϕ(c) ϕ(c) ϕ(b) ϕ(a) ϕ(e)

⊕ (0,0) (1,0) (0,1) (1,1)

(0,0) (0,0) (1,0) (0,1) (1,1)

(1,0) (1,0) (0,0) (1,1) (0,1)

(0,1) (0,1) (1,1) (0,0) (1,0)

(1,1) (1,1) (0,1) (1,0) (0,0)

Since the group tables are equivalent via

the map ϕ, and since ϕ is one-to-one and

onto, it follows that ϕ is an isomorphism.

Automorphism

Let G be a group. An isomorphism

ϕ : G→ G

is called an automorphism. The identity

map

iG : G→ G,

defined by iG(g) = g for all g ∈ G, is called

the trivial automorphism.

12

Automorphism

Example

Let G = (Z,+). The function ϕ : G → G

defined by

ϕ(n) = −n

is an automorphism since ϕ is one-to-one

and onto, and

ϕ(n+m) = −(n+m)

= (−n) + (−m)

= ϕ(n) + ϕ(m)

for all n,m ∈ Z.

13

Isomorphism

Theorem 12.5: Let ϕ : G → H be an

isomorphism. Then

(i) o(x) = o(ϕ(x)), for all x ∈ G;

(ii) G and H have the same cardinality;

(iii) G is abelian if and only if H is abelian.

14

Proof of (i): Assume ϕ : G → H is an

isomorphism. Then for all x ∈ G and for

all n ∈ Z, we have

xn = eG iff ϕ(xn) = ϕ(eG)

(since ϕ is one-to-one)

iff ϕ(x)n = eH .

(by Theorem 12.4)

Therefore, x has finite order if and only if

o(ϕ(x)) has finite order, and the smallest

positive integer n for which xn = eG is the

same as the smallest positive integer n for

which ϕ(x)n = eH.

Therefore, o(x) = o(ϕ(x)).

Proof of (ii): Assume ϕ : G → H is an

isomorphism. Then, there is a one-to-one

correspondence (bijection) between the el-

ements of G and H via the map ϕ. This

means G and H have the same cardinality.

Proof of (iii): Homework.

Homomorphisms

Theorem 12.1:

(i) Let ϕ : G → H and ψ : H → K be

homomorphisms. Then ψ ◦ϕ : G→ K

is a homomorphism.

(ii) If ϕ and ψ are isomorphisms, then

ψ ◦ ϕ is an isomorphism.

(iii) If ϕ : G→ H is an isomorphism, then

ϕ−1 : H → G is an isomorphism.

15

Proof of (i):

For all a, b ∈ G, we have

(ψ ◦ ϕ)(ab) = ψ[ϕ(ab)]

= ψ[ϕ(a)ϕ(b)]

= ψ[ϕ(a)]ψ[ϕ(b)]

= [(ψ ◦ ϕ)(a)][(ψ ◦ ϕ)(b)].

Hence, ψ ◦ ϕ is a homomorphism.

Proof of (ii):

If ϕ and ψ are isomorphisms, then by part

(i), ψ ◦ ϕ is a homomorphism. Also, ψ ◦ ϕis bijective, since the composition of bijec-

tive functions is bijective. Therefore, ψ ◦ϕis an isomorphism.

Proof of (iii):

Let ϕ : G → H is an isomorphism, and let

ϕ−1 : H → G be its inverse function. First

note that ϕ−1 is a bijection. Also, for all

x, y ∈ H

ϕ−1(xy) = ϕ−1(x)ϕ−1(y)

iff ϕ[ϕ−1(xy)] = ϕ[ϕ−1(x)ϕ−1(y)]

iff xy = ϕ[ϕ−1(x)]ϕ[ϕ−1(y)]

iff xy = xy

Since the last equation is true, the first

equation is also true. That is

ϕ−1(xy) = ϕ−1(x)ϕ−1(y)

for all x, y ∈ H, which proves that ϕ−1 is

an isomorphism.

Equivalence of Groups

Theorem 12.1 shows that ∼= defines an

equivalence relation on the set of all groups.

Indeed,

(Reflexivity) Given any group G, the iden-

tity map iG : G → G is an isomorphism.

This shows G ∼= G. Hence ∼= is reflexive.

(Symmetry) If G ∼= H, then there exists

an isomorphism ϕ : G → H. By Theorem

12.1 (iii), ϕ−1 : H → G is an isomorphism.

Therefore H ∼= G . This proves ∼= is sym-

metric.

(Transitive) If G ∼= H and H ∼= K, then

by Theorem 12.1 (ii), G ∼= K. Therefore,∼= is transitive.

16


Example

Let G = (R,+) and H = (R+, ·), and con-

sider the isomorphism ϕ : G→ H given by

ϕ(x) = ex.

By Theorem 12.1 (iii), the function

ϕ−1(x) = lnx

is an isomorphism from H to G. Indeed,

the condition that ϕ−1 : H → G is a homo-

morphism corresponds to the familiar prop-

erty

ln(xy) = ln(x) + ln(y)

for all x, y ∈ H.

17


Theorem 12.2: Let G be a cyclic group

of order n. Then, G ∼= (Zn,⊕). Consequently,

any two cyclic groups of order n are iso-

morphic to each other.

18

Proof: Let G = 〈g〉 = {e, g1, g2, . . . , gn−1},where o(g) = n. We define ϕ : Zn → G by

ϕ(j) = gj

for all 0 ≤ j ≤ n− 1. We want to show ϕ

is an isomorphism. Clearly ϕ is one-to-one

and onto. Also, for all j, k ∈ Zn we have

ϕ(j ⊕ k) = gj⊕k

= gj+k

= gjgk

= ϕ(j)ϕ(k).

This proves ϕ is an isomorphism. There-

fore, G ∼= (Zn,⊕).


Theorem 12.3: Let G be an infinite

cyclic group. Then, G ∼= (Z,+). Consequently,

any two infinite cyclic groups are isomor-

phic to each other.

19

Image and Inverse Image

Notation

Let ϕ : G→ K be a function.

• If H ⊆ G, then the image of H under

ϕ is the set

ϕ(H) = {k ∈ K | k = ϕ(h) for someh ∈ H}

• If J ⊆ K, then the inverse image (or

pre-image) of J under ϕ is the set

ϕ−1(J) = {h ∈ H | ϕ(h) ∈ J}

20

Homomorphisms

Theorem 12.6: Let ϕ : G → K be a

homomorphism. Then:

(i) If H is a subgroup of G. then ϕ(H)

is a subgroup of K.

(ii) If J is a subgroup of K, then ϕ−1(J)

is a subgroup of G.

(iii) If J � K, then ϕ−1(J) � G.

(iv) If ϕ is onto and H �G, then ϕ(H)� K.

21

Proof of (i):

Assume H is a subgroup of G. We want

to show ϕ(H) is a subgroup of K. By

Theorem 5.1, it suffices to show that ϕ(H)

is closed under the group operation and

closed under inverses.

First, let x, y ∈ ϕ(H). Then x = ϕ(h1) and

y = ϕ(h2) where h1, h2 ∈ H. Therefore,

xy = ϕ(h1)ϕ(h2)

= ϕ(h1h2),

where h1h2 ∈ H since H is a subgroup of

G. This proves xy ∈ ϕ(H); that is, ϕ(H)

is closed under the group operation.

Next, let let x ∈ ϕ(H). Then x = ϕ(h) for

some h ∈ H. Therefore,

x−1 = (ϕ(h))−1

= ϕ(h−1),

where h−1 ∈ H since H is a subgroup of

G. This proves x−1 ∈ ϕ(H); that is, ϕ(H)

is closed under inverses.

This proves ϕ(H) is a subgroup of K.

Proofs of (ii) and (iii): Homework

Proof of (iv): Assume ϕ is onto and as-

sume H �G. We want to show ϕ(H) � K.

Assume x ∈ ϕ(H) and assume k ∈ K.

Then x = ϕ(h) for some h ∈ H, and since

ϕ is onto k = ϕ(g) for some g ∈ G. There-

fore,

kxk−1 = ϕ(g)ϕ(h)ϕ(g)−1

= ϕ(g)ϕ(h)ϕ(g−1)

= ϕ(gh)ϕ(g−1)

= ϕ(ghg−1),

where ghg−1 ∈ H since H is normal in

G. This proves kxk−1 ∈ ϕ(H). Therefore

ϕ(H) is a normal subgroup of K.


Cayley’s Theorem

Theorem 12.7 (Cayley’s Theorem)

If G is a group, then G is isomorphic to a

subgroup of SG, the symmetric group on

the set G.

22

Proof: Recall that for any fixed g ∈ G,

the function fg : G→ G defined by

fg(x) = gx

is a bijection (Homework 5, Problem 6).

This means fg ∈ SG, the symmetric group

on the set G.

We will show that the function ϕ : G→ SG,

defined by

ϕ(g) = fg

is a one-to-one homomorphism.

ϕ is one-to-one: We have

ϕ(g1) = ϕ(g2) ⇒ fg1 = fg2

⇒ fg1(e) = fg2(e)

⇒ g1e = g2e

⇒ g1 = g2

This proves ϕ is one-to-one.

ϕ is a homomorphism: For all g1, g2 ∈ G

ϕ(g1g2) = fg1g2

= fg1 ◦ fg2

= ϕ(g1) ◦ ϕ(g2)

where the second equality follows from the

fact that for all x ∈ G

fg1g2(x) = g1g2x = fg1(g2x) = fg1(fg2(x)).

This proves that ϕ is a homomorphism, and

since ϕ is one-to-one, it follows that G

is isomorphic to its image ϕ(G) which, by

Theorem 12.6, is a subgroup of SG.


Cayley’s Theorem

Remark (Finite Case)

If G is a finite group, then each row in the

group table for G corresponds to left mul-

tiplication by some element g ∈ G. There-

fore, the rows in the group table for G

correspond exactly to the permutations fg

in the proof of Caley’s Theorem. This idea

is illustrated in the next example.

23

Cayley’s Theorem

Consider the group of quaternions

Q8 = {I,−I, J,−J, K,−K, L,−L}

If we relabel the elements of Q8 so that

I 7→ 1 K 7→ 5

−I 7→ 2 −K 7→ 6

J 7→ 3 L 7→ 7

−J 7→ 4 −L 7→ 8

then Q′8 = {1,2,3,4,5,6,7,8} represents

an isomorphic copy of Q8. We will use the

proof of Caley’s Theorem to find a sub-

group H of S8 such that Q′8∼= H.

24

The group table for Q8 is given below.

∗ I −I J −J K −K L −L

I I −I J −J K −K L −L

−I −I I −J J −K K −L L

J J −J −I I L −L −K K

−J −J J I −I −L L K −K

K K −K −L L −I I J −J

−K −K K L −L I −I −J J

L L −L K −K −J J −I I

−L −L L −K K J −J I −I

Relabelling the rows in the table determines

the permutations in the isomorphic sub-

group H.

∗ 1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

13 Homomorphisms and Normal

Subgroups

Definition

Let G be a group and let H be a normal

subgroup of G. The function ρ : G→ G/H

defined by

ρ(a) = Ha

is called the canonical homomorphism from

G onto G/H.

To see that ρ is a homomorphism observe

that

ρ(ab) = Hab

= (Ha)(Hb)

= ρ(a)ρ(b).

1

Canonical Homomorphism

Example

Let G = Q8 and let H = Z(Q8) = {I,−I}.Recall that

G/H = {HI, HJ, HK, HL}

is isomorphic to the Klein 4-group V . The

canonical homomorphism ρ : G → G/H is

given by

ρ(I) = ρ(−I) = HI

ρ(J) = ρ(−J) = HJ

ρ(K) = ρ(−K) = HK

ρ(L) = ρ(−L) = HL

We say that V is a homomorphic image

of Q8 .

2

Kernel of a Homomorphism

Let ϕ : G → K be homomorphism. The

kernel of ϕ is the set of all elements in G

that map to the identity element eK. That

is,

ker(ϕ) = ϕ−1({eK})

= {g ∈ G | ϕ(g) = eK}.

3


Example

Let G = (Z,+) and let K = (Zn,⊕), and

consider the homomorphism ϕ : G → K

defined by

ϕ(m) = m,

where m is the remainder of m modulo n.

Then,

ker(ϕ) = {m ∈ Z | ϕ(m) = 0}

= {m ∈ Z | m = 0}

= nZ.

4


Theorem 13.1 Let ϕ : G → K be a ho-

momorphism. Then, ker(ϕ) is a normal

subgroup of G .

5

Proof: Since {eK} � K , it follows by The-

orem 12.6(iii) that

ker(ϕ) = ϕ−1({eK}) � G.

Or, by a direct argument, observe that for

any g ∈ ker(ϕ) and for any x ∈ G, we have

ϕ(xgx−1) = ϕ(x)ϕ(g)ϕ(x−1)

= ϕ(x) eK ϕ(x)−1

= ϕ(x)ϕ(x)−1

= eK,

which shows that xgx−1 ∈ ker(ϕ) . There-

fore, ker(ϕ) � G.


Example

Let G = (Sn, ◦) and let K = (Z2,⊕), and


given by

ϕ(f) =

0 if f is even

1 if f is odd

Then,

ker(ϕ) = {f ∈ Sn | ϕ(f) = 0}

= {f ∈ Sn | f is even}

= An.

That is, ker(ϕ) is the alternating group An,

and by Theorem 13.1, An � Sn .

6


Theorem 13.2 (Fundamental Theorem

on Group Homomorphisms)

Let ϕ : G → K be a homomorphism from

G onto K. Then

K ∼= G/ ker(ϕ).

More generally, if ϕ is not onto, then

ϕ(G) ∼= G/ ker(ϕ).

7

Proof: Let N = ker(ϕ). We will show

that the function ϕ : G/ ker(ϕ) → K de-

fined by

ϕ(Na) = ϕ(a)

is an isomorphism. First we must check

that ϕ is well-defined.

ϕ is well-defined: For all a, b ∈ G we have

Na = Nb iff ab−1 ∈ N = ker(ϕ)

iff ϕ(ab−1) = eK

iff ϕ(a)ϕ(b−1) = eK

iff ϕ(a)ϕ(b)−1 = eK

iff ϕ(a) = ϕ(b)

iff ϕ(Na) = ϕ(Nb)

Therefore Na = Nb implies ϕ(Na) = ϕ(Nb) ,

which shows that ϕ is well-defined.

ϕ is one-to-one and onto: The argument

above also shows that if ϕ(Na) = ϕ(Nb) ,

then Na = Nb , which proves that ϕ is

one-to-one.

Since ϕ is onto, given any k ∈ K , there

exists a ∈ G such that

ϕ(Na) = ϕ(a) = k.

Therefore, ϕ is onto.

ϕ is an homomorphism: Finally, to show

that ϕ is a homomorphism note that

ϕ(NaNb) = ϕ(Nab)

= ϕ(ab)

= ϕ(a)ϕ(b)

= ϕ(a)ϕ(b).



Example

Let G = (Z,+) and let K = (Zn,⊕), and


defined by

ϕ(m) = m,

where m is the remainder of m modulo n.

Since ϕ is onto, and ker(ϕ) = nZ we have,

by Theorem 13.2

Zn ∼= Z / nZ.

8


Example

Let G = (Sn, ◦) and let K = (Z2,⊕), and


given by

ϕ(f) =

0 if f is even

1 if f is odd

Since ϕ is onto, and ker(ϕ) = An we have,

by Theorem 13.2

Z2∼= Sn /An.

9

14 Direct Products and Finite

Abelian Groups

The goal of this section is to prove the

following theorem on the structure of finite

abelian groups.

Theorem 14.2: Every finite abelian group

G is isomorphic to a direct product of cyclic

groups whose orders are prime powers.

1

Direct Products and Finite Abelian Groups

Example Assume G is an abelian group

such that |G| = 180 = 22 · 32 · 5. It follows

from Theorem 14.2 that G is isomorphic

to one of the following groups

G ∼= Z4 × Z9 × Z5

G ∼= Z2 × Z2 × Z9 × Z5

G ∼= Z4 × Z3 × Z3 × Z5

G ∼= Z2 × Z2 × Z3 × Z3 × Z5

For example, if G is the abelian group

Z180, then G ∼= Z4 × Z9 × Z5.

2

Groups of Prime Power Order

If p is a prime number and G is a group,

then G has p-power order if |G| = pk for

some positive integer k .

A group G is called a p-group if for every

x ∈ G , o(x) is a power of p .

3

Groups of Prime Power Order

Theorem: A finite abelian group has p-

power order if and only if it is a p-group.

Proof:

First, if G has p-power order, then by The-

orem 10.4, o(x) divides |G| for all x ∈ G.

Therefore, G is a p-group.

Conversely, assume G is a p-group, and

assume for the sake of contradiction that G

does not have p-power order. Then there

exists a prime q 6= p such that q | |G| . By

Theorem 11.7, G has a subgroup H = 〈x〉such that o(x) = q , which contradicts the

fact that G is a p-group.


4

Finite Abelian Groups and Direct Prod-

ucts

Theorem 14.1: If A and B are normal

subgroups of G such that

(i) A ∩B = {e}, and

(ii) G = AB = {ab | a ∈ A and b ∈ B},

then G ∼= A×B.

5

Proof: We will show that ϕ : A × B → G

defined by

ϕ((a, b)) = ab

is an isomorphism. First, it follows by as-

sumption (ii) that ϕ is onto. Next, observe

that

ϕ((a1, b1)) = ϕ((a2, b2))

⇒ a1b1 = a2b2

⇒ a−12 a1 = b2b

−11 ∈ B

⇒ a−12 a1 ∈ A ∩B

⇒ e = a−12 a1 = b2b

−11

⇒ a1 = a2 and b1 = b2

⇒ (a1, b1) = (a2, b2)

Therefore, ϕ is one-to-one. To show that

ϕ is a homomorphism, we need the follow-

ing lemma:

Lemma: Under the assumptions above, if

a ∈ A and b ∈ B , then ab = ba .

To prove the lemma, observe that ab = ba

if and only if bab−1a−1 = e . Since A and

B are normal subgroups, we have bab−1 ∈A and ab−1a−1 ∈ B . Therefore,

bab−1a−1 = (bab−1)a−1 ∈ A, and

bab−1a−1 = b(ab−1a−1) ∈ B.

This proves bab−1a−1 ∈ A∩B and it follows

that bab−1a−1 = e .

Finally, we have

ϕ((a1, b1) ∗ (a2, b2)) = ϕ((a1a2, b1b2))

= (a1a2)(b1b2)

= a1(a2b1)b2

= a1(b1a2)b2

= (a1b1)(a2b2)

= ϕ((a1, b1)) ∗ ϕ((a2, b2))

Therefore ϕ is a homomorphism, and the

proof is complete.


ucts

Lemma 14.7: If G is an abelian group

and |G| = mn where gcd(m,n) = 1 , then

G ∼= A × B where A = {x ∈ G | xm = e}and B = {x ∈ G | xn = e} .

6

Proof: Since gcd(m,n) = 1, there exist

integers r and s such that rn + sm = 1 .

Therefore, for any x ∈ G, we have

x = x1 = xrn+sm = xrnxsm

where xrn ∈ A and xrn ∈ B since

(xrn)m = (xrm)n = (xr)|G| = e.

It follows that

G = AB = {ab | a ∈ A and b ∈ B}.

It is easy to check that A and B are sub-

groups of G , and they are both normal

since G is abelian. Also, A ∩ B = {e} ,

since x ∈ A ∩B implies

x = xrnxsm = (xn)r(xm)s = ee = e.

Therefore, by Theorem 14.1, G ∼= A×B .


ucts

Corollary 14.7: Let G be an abelian group,

and consider the prime factorization of |G|given by

|G| = pr11 p

r22 · · · p

rkk

where p1, p2, . . . , pk are distinct primes and

r1, r2, . . . , rk are positive integers. Then,

G ∼= G(p1)×G(p2)× · · · ×G(pk)

where

G(pi) = {x ∈ G | xprii = e}

is a pi-group for all i = 1,2, . . . , k.

7


ucts

Lemma A: Let G1 = 〈g1〉, G2 = 〈g2〉, . . . ,

Gm = 〈gm〉 be finite cyclic groups. If G is

a finite abelian group and

ϕ : G → G1 ×G2 × · · · ×Gm

is an isomorphism, then every element g ∈G has a unique representation of the form

g = xr11 x

r22 · · ·x

rmm

where 0 ≤ ri < |Gi|, and

ϕ(xi) = (eG1, . . . , eGi−1

, gi, eGi+1, . . . , eGn)

for each i = 1,2, . . .m.

8

Proof: Let

H = {xr11 x

r22 · · ·x

rmm | 0 ≤ ri < |Gi|}.

Then,

ϕ(H) = {ϕ(xr11 x

r22 · · ·x

rmm ) | 0 ≤ ri < |Gi|}

= {ϕ(x1)r1ϕ(x2)r2 · · ·ϕ(xm)rm | 0 ≤ ri < |Gi|}

= {(gr11 , g

r22 , · · · , grmm ) | 0 ≤ ri < |Gi|}

= G1 ×G2 × · · · ×Gm.

Since ϕ is a bijection, there is a one-to-one

correspondence between the elements of H

and G1 ×G2 × · · · ×Gm , and since

G = ϕ−1(G1 ×G2 × · · · ×Gm) = H

it follows that G = H. This completes the

proof.


ucts

Lemma B: Let G be a finite abelian p-

group, and let A = 〈x〉 be a cyclic sub-

group of maximal order (i.e. o(x) ≥ o(g)

for all g ∈ G). Then for every element Ay

in G/A , there exists y1 ∈ G such that

Ay = Ay1 and o(Ay) = o(Ay1) = o(y1) .

9

Proof: Let |G| = pa and o(x) = pi where

1 ≤ i ≤ a . If Ay ∈ G/A , then o(Ay) = pt

where t ≤ a − i . In particular, we have

ypt ∈ A , which means

ypt

= xn

for some integer n < pi.

Let pw be the highest power of p that

divides n . Then, we have

o(xn) =pi

gcd(pi, n)=

pi

pw= pi−w.

On the other hand, if o(y) = j , then

o(ypt) =

pj

gcd(pj, pt)=

pj

pt= pj−t.

Therefore, i − w = j − t , which means,

w = t + i − j . By assumption, i = o(x) ≥o(y) = j . Therefore, w ≥ t and it follows

that pt divides n . If we write n = cpt ,

then

ypt

= xn

⇒ ypt

= (xc)pt

⇒ (yx−c)pt

= e

⇒ (y1)pt

= e

where y1 = x−cy ∈ Ay .

Therefore, Ay1 = Ay and

o(y1) = pt = o(Ay) = o(Ay1).



ucts

Lemma C: If G is a finite abelian p-group,

then G is isomorphic to a direct product

of cyclic groups of p-power order.

10

Proof: We will prove the statement by

(strong) induction on n where |G| = pn.

Base Case: If n = 1 , then |G| = p . There-

fore, G is cyclic and the result holds.

Inductive Step: Let |G| = pn and assume

the statement is true for all groups of order

pk where k < n .

Let x be an element in G of maximal or-

der (i.e. o(x) ≥ o(g) for all g ∈ G), and

consider the subgroup A = 〈x〉 . The quo-

tient group G/A is a p-group for which

the inductive hypothesis applies, therefore

G/A ∼= G1 ×G2 × · · · ×Gm

where each Gi is cyclic. By Lemmas A and

B, there exist y1, y2, . . . , ym ∈ G such that

o(yi) = o(Ayi) = |Gi| for all i = 1,2, . . . ,m

and each Ag ∈ G/A has the unique rep-

resentation

Ag = (Ay1)r1(Ay2)r2 · · · (Aym)rm

where 0 ≤ ri < |Gi| .

We claim that G = AB where

B = {yr11 y

r22 · · · y

rmm | 0 ≤ ri < |Gi|}.

Indeed, for any g ∈ G we have

Ag = (Ay1)r1(Ay2)r2 · · · (Aym)rm

= A(yr11 y

r22 · · · y

rmm )

= Ab

where b ∈ B. Therefore gb−1 ∈ A , which

implies g = ab for some a ∈ A. Therefore

G = AB , as claimed.

To show that A ∩ B = {e} , observe that

for b = yr11 y

r22 · · · y

rmm ∈ B , we have

b ∈ A iff Ab = Ae

iff A(yr11 y

r22 · · · y

rmm ) = Ae

iff (Ay1)r1(Ay2)r2 · · · (Aym)rm = Ae

iff ri = 0 for all i = 1,2, . . . ,m

iff b = e.

Therefore, by Theorem 14.1, G ∼= A × B .

Since B ∼= G/A , the proof is complete.

Fundamental Theorem on Finite Abelian

Groups

Theorem 14.2: Every finite abelian group

G is isomorphic to a direct product of cyclic

groups whose orders are prime powers.

Proof: The result follows directly from

Corollary 14.7 and Lemma C above.

11

15 Sylow Theorems

Normalizer

Let G be a group, and let H be a subgroup

of G. The normalizer of H in G is the

subset

N(H) = {g ∈ G | gHg−1 = H}.

It is easy to show that N(H) is a sub-

group of G, and H is a normal subgroup

of N(H).

1

Normalizer

Example: Let G = S3, and let H = 〈(1,2)〉.We have N(H) = {(1), (1,2)} = H, since

(1)H (1)−1 = {(1), (1,2)} = H

(1,2)H (1,2)−1 = {(1), (1,2)} = H

(1,3)H (1,3)−1 = {(1), (2,3)} 6= H

(2,3)H (2,3)−1 = {(1), (1,3)} 6= H

(1,2,3)H (1,2,3)−1 = {(1), (2,3)} 6= H

(1,3,2)H (1,3,2)−1 = {(1), (1,3)} 6= H

2

Normalizer

Example: Let G = D4, and let H = 〈(1,3)〉.Then,

N(H) = {(1), (1,3), (2,4), (1,3)(2,4)}

as the following table illustrates

g gHg−1 gHg−1 = H ?

(1) {(1), (1,3)} X

(1,3) {(1), (1,3)} X

(2,4) {(1), (1,3)} X

(1,3)(2,4) {(1), (1,3)} X

(1,2)(3,4) {(1), (2,4)}

(1,4)(2,3) {(1), (2,4)}

(1,2,3,4) {(1), (2,4)}

(1,4,3,2) {(1), (2,4)}

3

p-Sylow subgroup

Let G be a group. If p is a prime number

such that pn divides |G| but pn+1 does

not divide |G| , then any subgroup of order

pn in G is called a p-Sylow subgroup.

In other words, if |H| = pn where pn is the

largest power of p that divides |G|, then H

is called a p-Sylow subgroup of G.

4

Converse of Lagrange’s Theorem

For finite abelian groups, the following con-

verse to Lagrange’s Theorem holds.

Theorem: Let G be a finite abelian group.

If m divides |G|, then there exists a sub-

group H of G such that |H| = m.

For nonabelian groups, this is false. For

example, the group A4 has order 12 and

has no subgroups of order m = 6.

The First Sylow Theorem (below) provides

a partial converse to the Lagrange’s The-

orem for a general group. That is the the-

orem above holds for any group if m = pk

where p is a prime number.

5

First Sylow Theorem

Theorem 15.1: Let G be a finite group,

p be a prime number, and k be a positive

integer.

(i) If pk divides |G|, then G has a sub-

group of order pk. In particular, G

has a p-Sylow subgroup.

(ii) Let H be any p-Sylow subgroup of

G. If K is any of order pk in G, then

for some g ∈ G we have K ⊆ gHg−1.

In particular, K is contained in some

p-Sylow subgroup of G.

6

First Sylow Theorem

Proof of (i): Let G be a finite group.

Base Case: If |G| = 2 , the result is trivial.

Inductive Step: Assume the theorem holds

for all groups of order less than |G|. Fur-

ther, assume pk divides |G|.

If G has a proper subgroup H such that pk

divides |H|, then by the inductive hypoth-

esis H has a subgroup of order pk and the

result holds.

If not, then p divides [G : H] for every

proper subgroup H. It follows that p di-

vides Z(G) since

|G| = |Z(G)|+[G : Z(a1)]+· · ·+[G : Z(ak)],

7

where all terms other than Z(G) are di-

visible by p. Therefore, by Theorem 11.7,

Z(G) has a subgroup A of order p.

Since A ⊆ Z(G) , it follows by Theorem

11.2, that A � G . By the inductive hy-

pothesis, G/A has a subgroup J of order

pk−1 , and

J ∼= ρ−1(J) /A

where ρ : G → G/A is the canonical ho-

momorphism.

Therefore ρ−1(J) has order pk, and the

proof is complete.

Second Sylow Theorem

Theorem 15.2: All p-Sylow subgroups

of G are conjugate to each other. Con-

sequently, a p-Sylow subgroup is normal if

and only if it is the only p-Sylow subgroup.

8

Third Sylow Theorem

Theorem 15.3: Let np denote the num-

ber of p-Sylow subgroups in G , and let

H be any one of these p-Sylow subgroups.

Then:

(i) np = [G : N(H)].

(ii) np ≡ 1 (mod p)

Note: If we write G = pnm where p does

not divide m, then |H| = pn and

m = [G : H]

= [G : N(H)] · [N(H) : H]

= np · [N(H) : H]

That is, (i) implies np divides m.

9

Implications of Sylow Theorems

Theorem 15.4: Let G be a finite group,

and let p be a prime number. Then G is

a p-group if and only if |G| = pk for some

for some integer k > 0.

10

Implications of Sylow Theorems

Theorem 15.5: Let G be a finite group

of order pq, where p and q are primes and

p < q. If p does not divide q − 1, then G

is cyclic.

11

Simple Group

We say that a group G is simple, if G

contains no non-trivial normal subgroups.

Theorem 15.3(ii) can be used to prove groups

of a certain order are not simple. Indeed,

if np = 1 then there is only one p-Sylow

subgroup, and by Corollary 11.5 it must be

normal.

For example, if |G| = 15 = 3 · 5 , then

n5 ≡ 1 (mod 5)

⇒ n5 = 1,6,11,16, . . .

Since distinct 5-Sylow subgroups have only

the identity element in common, if n5 ≥ 6 ,

then G contains at least (5 − 1) ∗ 6 = 24

elements, which is impossible. Therefore,

G has a normal subgroup of order 5.

12

Number of Groups of Order n

+ 0 1 2 3 4 5 6 7 8 9

0 0 1 1 1 2 1 2 1 5 2

10 2 1 5 1 2 1 14 1 5 1

20 5 2 2 1 15 2 2 5 4 1

30 4 1 51 1 2 1 14 1 2 2

40 14 1 6 1 4 2 2 1 52 2

50 5 1 5 1 15 2 13 2 2 1

60 13 1 2 4 267 1 4 1 5 1

70 4 1 50 1 2 3 4 1 6 1

80 52 15 2 1 15 1 2 1 12 1

90 10 1 4 2 2 1 231 1 5 2

13

Cyclic groups of order pq ( q 6≡ 1 (mod p))

+ 0 1 2 3 4 5 6 7 8 9

0 0 1 1 1 2 1 2 1 5 2

10 2 1 5 1 2 1 14 1 5 1

20 5 2 2 1 15 2 2 5 4 1

30 4 1 51 1 2 1 14 1 2 2

40 14 1 6 1 4 2 2 1 52 2

50 5 1 5 1 15 2 13 2 2 1

60 13 1 2 4 267 1 4 1 5 1

70 4 1 50 1 2 3 4 1 6 1

80 52 15 2 1 15 1 2 1 12 1

90 10 1 4 2 2 1 231 1 5 2

3× 5 5× 7 7× 11

3× 11 5× 13 7× 13

3× 17 5× 17

3× 23 5× 19

3× 29

14

16 Rings and Fields

A ring is a set R equipped with two binary opera-

tions + and · satisfying the following axioms

(i) (closure of R with respect to +)

x + y ∈ R for all x, y ∈ R.

(ii) (associativity of +) (x+ y) + z = x+ (y + z)

for all x, y, z ∈ R.

(iii) (additive identity) There is an element 0 ∈ R

such that x + 0 = 0 + x = x for all x ∈ R.

(iv) (additive inverses) For each element x ∈ R

there is an element (−x) ∈ R such that x +

(−x) = (−x) + x = 0.

(v) (commutativity of +) x + y = y + x for all

x, y ∈ R.

(vi) (closure of R with respect to ·)x · y ∈ R for all x, y ∈ R.

(vii) (associativity of ·) (x · y) · z = x · (y · z) for all

x, y, z ∈ R.

(viii) (distributive laws) x · (y + z) = x · y +x · z and

(x + y) · z = x · z + y · z for all x, y, z ∈ R.

1

Definition of a Ring

Remarks

• Conditions (i)-(v) imply that R is an

abelian group under the operation +.

• If the “multiplication” operation · in

R is commutative, then we say that R

is a commutative ring. Otherwise, we

say that R is a noncommutative ring.

• We use the notation (R,+, ·) to repre-

sent the ring with elements in R under

the operations of + and ·.

• Note that + and · are intended to

represent generalized binary operations,

and will not always correspond to the

usual operations of addition and multi-

plication.2

Rings

Examples

• The sets Z, Q, R, and C under the usual

operations of addition and multiplica-

tion are all commutative rings.

• The set 2Z of even integers under the


plication is a commutative ring.

• The set of n×n matrices with entries in

Z, denoted Mn(Z), is a non-commutative

ring with respect to the operations of

matrix addition and matrix multiplica-

tion.

• The set Zn, where n ∈ Z+, is a (finite)

commutative ring under the operations

of addition and multiplication mod n.

3

Subrings

A subring S of a ring R is a subset of R

which is a ring under the same operations

as R.

Examples

• 2Z is a subring of Z under the usual op-

erations of addition and multiplication.

• Mn(Z) is a subring of Mn(Q) under the

operations of matrix addition and ma-

trix multiplication.

4

Subrings

Theorem 16.1: Assume (R,+, ·) is a ring.

A nonempty subset S of R is a subring of R

if and only if the following conditions hold:

(i) For all x, y ∈ S, x+ y ∈ S and x · y ∈ S.

(ii) For all x ∈ S, (−x) ∈ S.

5

Subrings

Proof: First, assume S is a subring of R.

Clearly property (i) holds. (This follows

from ring axioms (i) and (vi).) For each

x ∈ S, let y ∈ S denote the additive inverse

of x in the ring S. Then,

y = y + 0

= y + (x + (−x))

= (y + x) + (−x)

= 0 + (−x)

= (−x)

where (−x) denotes the additive inverse of

x in the ring R. This proves (−x) = y ∈ S

which establishes property (ii).

Conversely suppose S is a nonempty sub-

set of R for which (i) and (ii) hold. It

6

follows by Theorem 5.1, that ring axioms

(i) through (iv) hold for the set S. Also, by

property (ii), ring axiom (vi) is satisfied for

the set S. Finally, since R is a ring, each of

the ring axioms (v), (vii), and (viii) hold

for all elements in R, and therefore must

hold on the set S as well.

This proves that S satisfies all ring axioms

(i) through (viii). Hence, S is a ring.

Rings with Unity

Let R be a ring. If there exists an element

e ∈ R such that x ·e = e ·x = x for all x ∈ R,

then e is called a unity (or multiplicative

identity element), and we say that R is a

ring with unity.

7

Rings with Unity

Examples

• The element e = 1 ∈ Z is a unity in the

ring (Z,+, ·) since x · 1 = 1 · x = x for

all x ∈ Z. Therefore, (Z,+, ·) is a com-

mutative ring with unity.

• (2Z,+, ·) is a commutative ring without

unity.

• The identity matrix In ∈ Mn(Z) is a

unity in the ring Mn(Z) since A · In =

In ·A = A for all A ∈Mn(Z). Therefore,

Mn(Z) is a noncommutative ring with

unity.

8

Rings with Unity

Theorem 16.2: If (R,+, ·) is a ring with

unity, then the unity element in R is unique.

Proof: Assume e1 and e2 are both unity

elements in R. Then, e1 · e2 = e2 since e1

is a unity element. On the other hand,

e1 · e2 = e1 since e2 is a unity element.

Therefore, e1 = e1 · e2 = e2 which proves

the unity element is unique.

9

Rings with Unity

Assume R be a ring with unity e, and let

a ∈ R. If there is an element x ∈ R such

that a ·x = x ·a = e, then we say that x is a

multiplicative inverse of a, and a is called

a unit (or an invertible element) in R.

10

Rings with Unity

Theorem 16.3: Assume (R,+, ·) is a ring

with unity. If an element a ∈ R has a mul-

tiplicative inverse, then the multiplicative

inverse is unique (and is denoted a−1).

Proof: (See Theorem 3.2)

11

Rings with Unity

Example

Consider the ring (Z10,⊕,�). Since 1 ∈ Z10

is a unity in the ring, we have the following:

• The element 1 ∈ Z10 is a unit since

1� 1 = 1 · 1 = 1 = 1.


3� 7 = 3 · 7 = 21 = 1.


7� 3 = 7 · 3 = 21 = 1.


9� 9 = 9 · 9 = 81 = 1.

12

Multiplication by Zero

Theorem 16.4: Assume (R,+, ·) is a ring

with additive identity 0. Then, for all a ∈ R

0 · a = a · 0 = 0.

13


Proof: Assume a ∈ R. Then,

a · 0 = a · (0 + 0)

= a · 0 + a · 0.

Therefore,

a · 0 + (−(a · 0)) = a · 0 + a · 0 + (−(a · 0)).

It follows that

0 = a · 0.

A similar argument shows 0 = 0 · a for all

a ∈ R. This completes the proof.

14


Alternate Proof: Assume a ∈ R. Then,

a · 0 = a · 0 + 0

= a · 0 + (a · 0 + (−(a · 0)))

= (a · 0 + a · 0) + (−(a · 0))

= (a · (0 + 0)) + (−(a · 0))

= (a · 0) + (−(a · 0))

= 0.

A similar argument shows 0 = 0 · a for all

a ∈ R. This completes the proof.

15

Zero Divisors

Let R be a ring with additive identity 0,

and let a ∈ R. If a 6= 0, and if there exists

an element b 6= 0 in R such that a · b = 0 or

b · a = 0, then we say a is a zero divisor.

16

Zero Divisors

Example

Consider the ring (Z10,⊕,�). Then:

• The element 2 ∈ Z10 is a zero divisor

since 2� 5 = 2 · 5 = 10 = 0.


since 4� 5 = 4 · 5 = 20 = 0.


since 5� 2 = 5 · 2 = 10 = 0.


since 6� 5 = 6 · 5 = 30 = 0.


since 8� 5 = 8 · 5 = 40 = 0.

17

Integral Domains

Let D be a ring. We say that D is an

integral domain if the following conditions

hold:

1. D is a commutative ring.

2. D has a unity e and e 6= 0.

3. D has no zero divisors.

18

Integral Domains

Examples

• The rings Z, Q, R, and C under the


plication are all integral domains.

• The ring (2Z,+, ·) is not an integral do-

main since 2Z does not contain a unity.

• The ring (Z10,⊕,�) is not an integral

domain since it has zero divisors.

• The ring M2(Z) is not an integral do-

main since it is not commutative. Note

also that M2(Z) contains zero divisors

(Homework 10 #5).

19

Integral Domains

Theorem 16.5: The ring (Zn,⊕,�) is an

integral domain if and only if n is prime.

20

Integral Domains

Proof: First note that (Zn,⊕,�) is a com-

mutative ring with unity e = 1. Therefore

(Zn,⊕,�) is an integral domain if and only

if it has no zero divisors.

First, assume n is prime, and assume for

the sake of contradiction that a ∈ Zn is a

zero divisor. Then, a 6= 0 and there exists

b 6= 0 in Zn such that

0 = a� b = ab

Therefore, n divides ab, and since n is prime,

it follows that n divides a, or n divides

b. However this is a contradiction since

a, b ∈ {1,2, . . . , n − 1}. Therefore, if n is

prime, then (Zn,⊕,�) is an integral domain.

21

Next we will show that if (Zn,⊕,�) is an in-

tegral domain, then n is prime. We will use

a proof by contraposition. Assume n is not

prime. Then there exist a, b ∈ {2,3, . . . , n−1} such that n = ab. Therefore, a � b = 0

where a, b ∈ Zn and a and b are nonzero.

Therefore, (Zn,⊕,�) has zero divisors and

is not an integral domain.


Integral Domains

Theorem 16.6: Assume D is an integral

domain. If a, b, and c are elements of D

such that a 6= 0 and ab = ac, then b = c.

22

Integral Domains

Proof: Assume D is an integral domain,

and assume a, b, and c are elements of D

such that a 6= 0 and ab = ac. Then,

a(b + (−c)) = ab + a(−c)

= ab + a((−e)c)

= ab + (−e)ac

= ab + (−(ac))

= 0

Since a 6= 0 and D has no zero divisors, it

follows that b + (−c) = 0. Therefore b = c.

23

Fields

Let F be a ring. We say that F is a field

if the following conditions hold:

1. F is a commutative ring.

2. F has a unity e and e 6= 0.

3. Every nonzero element of F has a mul-

tiplicative inverse.

24

Fields

Examples

• The rings Q, R, and C under the usual

operations of addition and multiplica-

tion are all fields.

• The ring (Z,+, ·) is not a field since not

every element of Z has a multiplicative

inverse in Z.

• The ring (Zp,⊕,�) is a field if and only

if p is prime.

25

Fields

Theorem 16.7: Every field F is an integral

domain.

26

Fields

Proof: Assume F is a field. Therefore,

F is a commutative ring with unity 6= 0.

We want to show F has no zero divisors.

Assume for the sake of contradiction that

there exist nonzero elements a, b ∈ F such

that ab = 0. Since F is a field, the element

a 6= 0 has a multiplicative inverse a−1 ∈ F .

Therefore,

b = eb = (a−1a)b = a−1(ab) = a−1(0) = 0.

This is a contradiction since we assumed

b 6= 0. Therefore F has no zero divisors

and it follows that F is an integral domain.

27

Fields

Theorem 16.8: Every finite integral do-

main D is a field.

28

Fields

Proof: Let D = {d1, d2, . . . , dn} be a finite

integral domain. To show that D is a field

we must show that every nonzero element

of D has a multiplicative inverse.

Assume a ∈ D and a 6= 0. By Theorem

16.6, the elements ad1, ad2, . . . , adn are all

distinct. Since D has exactly n elements,

it follows that

D = {ad1, ad2, . . . , adn}.

In particular the unity e ∈ D, can be ex-

pressed in the form

e = adj,

for some dj ∈ D where dj 6= 0. This shows

that a has a multiplicative inverse. Hence

D is a field.

29

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