0 Sets and Inductionnleger/3330_lecture_notes_blank.pdf · 0 Sets and Induction Sets A set is an...

0 Sets and Induction

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.

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, . . .}

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}

Defining a Set

Example: Express the following set using

set builder notation.

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

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

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.

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.

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.

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

Special Subsets

For every set S,

(i) Ø ⊆ S(ii) S ⊆ S

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.

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.

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}

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}.

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 )

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.

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.

Sets and Proofs

Example: Let S and T be sets. Prove that

S ∩ T ⊆ S ∪ T .

Sets and Proofs

Example: Let S and T be sets. Prove that

S ⊆ T if and only if S ∩ T = S.

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.

Example: Show that if n is a positive

integer, then

1 + 2 + 3 + · · ·n =n(n+ 1)

Proof: Let P (n) be the proposition

1 + 2 + 3 + · · ·n =n(n+ 1)

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

Base Step:

Example: Conjecture a formula for the

sum of the first n positive odd integers.

Then prove your conjecture using mathe-

matical induction.

1 + 3 =

1 + 3 + 5 =

1 + 3 + 5 + 7 =

1 + 3 + 5 + 7 + 9 =

Conjecture: For all positive integers n,

the following proposition P (n) holds

1 + 3 + 5 + · · ·+ 2n− 1 =

Proof:

Example: Use mathematical induction to

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

Proof: Let P (n) be the proposition

2 | (n2 + 5n).

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

Base Step:

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}.

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)}.

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.

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+.

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+.

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

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.

Binary Operations

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

real entries. Then, the operation ∗, defined

(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.

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.

• 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.

Example: Let ∗ be a binary operation on

Z defined by

a ∗ b = 2(a + b)

• is ∗ commutative?

• is ∗ associative?

Example: Multiplication of 2×2 matrices

• is associative.

)∗(e f

))∗(

)∗((

• is not commutative.

Example: Let M be the symmetric differ-

ence operator given by

A M B = (A−B) ∪ (B −A)

• is M commutative?

• is M associative?

2 Groups

Definition of a group

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

(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.

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.

Abelian Group

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

is called an abelian group.

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.

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.

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:

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:

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:

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:

Groups

Example: Consider the set Z together

with the binary operation ∗ given by

a ∗ b = 2(a + b).

Does (Z, ∗) form a group?

Groups

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

a ∗ b = 2ab.

Does (G, ∗) form a group?

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.

Lemma: Assume n ∈ Z+. Then, a = b if

and only if a− b = nk for some k ∈ Z.

Proof:

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.

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

This proves y1 = y2, and we conclude that

the inverse of an element is unique.

Inverse Element

Notation

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

element x ∈ G is denoted by x−1.

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.

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

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

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.

(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

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

Therefore, (x ∗ y)−1 = z = y−1 ∗ x−1.

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.

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.

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.

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.

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

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

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+.

• However these conditions are not suffi-

cient to make Z+ a group with respect

to ∗, since there is no left identity ele-

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.

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

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

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) =∞.

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

···

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

1 00 1

)(−1 0

0 −1

1 00 1

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).

Examples

• gcd(24,40) =

• gcd(32,100) =

• gcd(12,91) =

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.

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.

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.

Examples

• Find gcd(198, 252) using the Euclidean

algorithm

• Find gcd(414, 662) using the Euclidean

algorithm

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-

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.

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.

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.

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〉.

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

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.

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:

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).

Cyclic Groups

Theorem 4.7 If G is a cyclic group, then

G is abelian.

Proof:

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 ∗.

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

subgroup of G = GL(2,R).

a b0 d

)| a, b, d ∈ R and ad 6= 0

subgroup of G = GL(2,R).

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.

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.

Groups of Order 1

There is only one group of order 1, namely

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

• The group G has only one subgroup

H = G = {e}.

• The group G is cyclic. The element e

is a generator for G.

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

• 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.

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

• 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.

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

∗ e a b c

e e a b c

∗ e a b c

e e a b c

∗ e a b c

e e a b c

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

• 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.

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

• Klein’s 4-group has five subgroups:

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

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)

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

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.

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.

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}

Subgroups

Theorem 5.2 If G is a cyclic group, then

every subgroup of G is cyclic.

Proof:

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:

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).

Group of Unit Quaternions

Consider the general linear group of degree

two over the complex numbers

GL(2,C) ={( z1 z2

)| 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

),( 0 1

−1 0

),( 0 −1

0 −i

),( −i 0

(0 −i

−i 0

Group of Unit Quaternions

Let J =(

0 −i

), K =

−1 0

), L =

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

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.

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).

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 ).

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)}

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)}

Direct Products

Example

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

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

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).

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.

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.

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 .

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.

Examples

• f : R→ R, defined by

f(x) = x2

• f : GL(2,R) → R, defined by

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.

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}

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 ) =

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].

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.

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 .

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

Examples

• f : R→ R, defined by

f(x) = x2

• f : GL(2,R) → R, defined by

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.

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.

Example

Prove or disprove:

The function f : Z→ Z defined by

f(n) = 2n− 3

is injective.

Example

Prove or disprove:

The function f : Z→ Z defined by

f(n) = 2n− 3

is surjective.

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

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.

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

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))

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) =

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) =

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.

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.

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.

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)

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:

1 2 3 4

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

1 2 3 42 4 1 3

1 2 3 4

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

1 2 3 43 2 4 1

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 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

1 2 3 42 4 1 3

(f ◦ g)(2) =((

1 2 3 42 4 1 3

1 2 3 43 2 4 1

1 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

1 2 3 42 4 1 3

Also, we have

(f ◦ g)(3) =((

1 2 3 42 4 1 3

1 2 3 43 2 4 1

1 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

1 2 3 42 4 1 3

(f ◦ g)(4) =((

1 2 3 42 4 1 3

1 2 3 43 2 4 1

1 2 3 42 4 1 3

) ((1 2 3 43 2 4 1

1 2 3 42 4 1 3

Composition of Permutations

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

are given by

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

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

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)

Cycle Notation

Example: Consider f, g ∈ S7 given by

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

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).

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).

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

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.

Product of Cycles

Example: Consider f ∈ S9 given by

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)

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.

Order of a Permutation

Example: Consider f ∈ S9 given by

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

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)

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.

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.

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.

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.

Alternating Groups

The alternating group of degree n, de-

noted An, is defined to be the subset of Sn

consisting of all even permutations.

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.

Alternating Groups

Theorem 8.5: If n ≥ 2, then An is a

subgroup of Sn, and

|An| =n!

2=|Sn|

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)

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.

S3 as a group of symmetries

Rotations:

Reflections:

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.

The dihedral group D4

Rotations:

Reflections:

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)}

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.

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.

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.

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.

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.

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].

Equivalence Classes

Let R be the relation on S = Z defined by

aR b iff a ≡ b (mod 4)

[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.

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.

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.

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.

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)}

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.

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.

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.

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)}

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)}

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)}

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)}

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).

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|.

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|.

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

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.

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|.

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|

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|

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.

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] =∞.

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].

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.

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.

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

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.

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.

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.

ap−1 ≡ 1 (mod p)

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.

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)) =

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.

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}

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.

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)].

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]

(1,2,3)

(1,3,2)

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)}

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)}

Conjugacy Classes

Important Fact: All conjugates of a have

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

lem 5)

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)]

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.

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

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

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.

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.

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.

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)

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)}.

Normal Subgroups

Notation

We write H � G to indicate that H is a

normal subgroup of G.

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.

Normal Subgroups

Theorem 11.2 Let G be a group. Then

any subgroup of Z(G) is a normal subgroup

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).

Normal Subgroups

Theorem 11.3: Let H be a subgroup of

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

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.

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.

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.

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.

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.

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)

= 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-

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

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

forms a group with the following multipli-

cation table

∗ H(1) H(1,2)

H(1,2)

Quotient Groups

Example

Let G = Q8, and consider the normal sub-

H = Z(Q8) = {I,−I}.

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

forms a group with the following multipli-

cation table

∗ HI HJ HK HL

Quotient Groups

Example

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

subgroup

H = 〈n〉 = nZ.

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.

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.

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)

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.

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.

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.

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.

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.

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.

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)

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)=

o(ϕ(x)) =4

gcd(4,2)=

and we see that o(ϕ(x)) divides o(x).

• 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.

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+, ·).

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.

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

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

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

iG : G→ G,

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

the trivial automorphism.

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.

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.

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.

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.

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-

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

for all x, y ∈ H.

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.

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.

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}

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.

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-

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.

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.

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.

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

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

ρ(ab) = Hab

= (Ha)(Hb)

= ρ(a)ρ(b).

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 .

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

ker(ϕ) = ϕ−1({eK})

= {g ∈ G | ϕ(g) = eK}.

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.

ker(ϕ) = {m ∈ Z | ϕ(m) = 0}

= {m ∈ Z | m = 0}

Theorem 13.1 Let ϕ : G → K be a ho-

momorphism. Then, ker(ϕ) is a normal

subgroup of G .

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

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

ker(ϕ) = {f ∈ Sn | ϕ(f) = 0}

= {f ∈ Sn | f is even}

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

and by Theorem 13.1, An � Sn .

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).

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.

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.

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.

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.

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 .

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.

Finite Abelian Groups and Direct Prod-

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.

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

ϕ((a1, b1)) = ϕ((a2, b2))

⇒ a1b1 = a2b2

⇒ a−12 a1 = b2b

−11 ∈ B

⇒ a−12 a1 ∈ A ∩B

⇒ e = a−12 a1 = b2b

⇒ 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.

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} .

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 .

Corollary 14.7: Let G be an abelian group,

and consider the prime factorization of |G|given by

|G| = pr11 p

r22 · · · p

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

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

G ∼= G(p1)×G(p2)× · · · ×G(pk)

G(pi) = {x ∈ G | xprii = e}

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

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

where 0 ≤ ri < |Gi|, and

ϕ(xi) = (eG1, . . . , eGi−1

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

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

Proof: Let

H = {xr11 x

r22 · · ·x

rmm | 0 ≤ ri < |Gi|}.

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.

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) .

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

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)=

pw= pi−w.

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

o(ypt) =

gcd(pj, pt)=

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 ,

⇒ ypt

= (xc)pt

⇒ (yx−c)pt

⇒ (y1)pt

where y1 = x−cy ∈ Ay .

Therefore, Ay1 = Ay and

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

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.

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

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.

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).

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

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)}

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.

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.

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.

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)],

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.

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.

(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.

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.

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.

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.

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

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

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.

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

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-

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

commutative ring under the operations

of addition and multiplication mod n.

Subrings

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

which is a ring under the same operations

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.

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.

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

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.

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.

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.

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.

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)

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.

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.

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.

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))

A similar argument shows 0 = 0 · a for all

a ∈ R. This completes the proof.

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.

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.

Integral Domains

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

integral domain if the following conditions

1. D is a commutative ring.

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

3. D has no zero divisors.

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).

Integral Domains

Theorem 16.5: The ring (Zn,⊕,�) is an

integral domain if and only if n is prime.

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.

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.

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))

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

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

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.

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.

Fields

Theorem 16.7: Every field F is an integral

domain.

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.

Fields

Theorem 16.8: Every finite integral do-

main D is a field.

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.

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