0 Sets and Inductionnleger/3330_lecture_notes_blank.pdf · 0 Sets and Induction Sets A set is an...
Transcript of 0 Sets and Inductionnleger/3330_lecture_notes_blank.pdf · 0 Sets and Induction Sets A set is an...
0 Sets and Induction
Sets
A set is an unordered collection of objects,
called elements or members of the set. A
set is said to contain its elements. We write
a ∈ A to denote that a is an element of the
set A. The notation a /∈ A denotes that a
is not an element of the set A.
1
Defining a Set
The roster method is a way of defining a
set by listing all of its members.
Examples
• S = {a, b, c}
• S = {1,2,3,4}
• S = {1,4,7,10,13,16, . . .}
2
Defining a Set
Another way to describe a set is using set
builder notation.
Examples
• S = {1,2,3,4}, or
S = {x | x is a positive integer ≤ 4}
• S = {1,4,7,10,13,16, . . .}, or
S = {x | x = 1 + 3k for some nonnegative integer k}
3
Defining a Set
Example: Express the following set using
set builder notation.
S = {. . . ,−12,−8,−4,0,4,8,12, . . .}
4
Important Sets
Notation
Z = {. . . ,−2,−1,0,1,2, . . .}, the set of integers
Z+ = {1,2,3, . . .}, the set of positive integers
Q = {pq| p ∈ Z, q ∈ Z, and q 6= 0}, the set of ratio-
nal numbers
R, the set of real numbers
R+, the set of positive real numbers
C, the set of complex numbers
5
Special Sets
A set with no elements is called the empty
set, or null set, and is denoted by Ø. The
empty set can also be denoted by { }.
A set with one element is called a singleton
set. For example, S = {1} is a singleton
set containing only the number 1.
6
Subsets
The set A is a subset of B if and only if
every element of A is also an element of B.
That is, A is a subset of B iff
∀x (x ∈ A → x ∈ B).
We use the notation A ⊆ B to indicate that
A is a subset of B.
7
Proper Subsets
We say that A is a proper subset of B if
and only if A ⊆ B and A 6= B. That is, A
is a proper subset of B iff
∀x (x ∈ A → x ∈ B) ∧ ∃x (x ∈ B ∧ x /∈ A).
We use the notation A ( B to indicate that
A is a proper subset of B.
8
Subsets
Examples
• If S = {1,2,3} and T = {1,2,3,4,5},then S ⊆ T .
• If S = {π,√
2} and T = {5,√
2}, then
S * T .
• Z+ ⊆ Z
9
Special Subsets
For every set S,
(i) Ø ⊆ S(ii) S ⊆ S
10
Equality of Sets
Two sets are equal if and only if they have
the same elements. Therefore, if A and B
are sets, then A = B if and only if
∀x (x ∈ A ↔ x ∈ B).
That is, A = B iff A ⊆ B and B ⊆ A.
11
Equality of Sets
Example
The sets {1,3,5} and {3,5,1} are equal,
because they have the same elements. Note
that the order in which the elements of a
set are listed does not matter.
12
Intersection and Union
Let S and T be sets.
• The intersection of S and T , denoted
S ∩ T , is the set of elements which be-
long to both S and T . That is,
S ∩ T = {x | x ∈ S and x ∈ T}
• The union of S and T , denoted S ∪ T ,
is the set of elements which belong to
S or T . That is,
S ∪ T = {x | x ∈ S or x ∈ T}
13
Intersection and Union
Example
Let S = {1,2,3,4,5} and T = {2,4,6}.Then,
S ∩ T = {2,4}.S ∪ T = {1,2,3,4,5,6}.
14
Sets and Logic
Statements involving sets and their logical
equivalents.
x /∈ S ¬(x ∈ S)
x ∈ S ∪ T (x ∈ S) ∨ (x ∈ T )
x ∈ S ∩ T (x ∈ S) ∧ (x ∈ T )
S ⊆ T ∀x (x ∈ S → x ∈ T )
S = T ∀x (x ∈ S ↔ x ∈ T )
15
Sets and Proofs
To show A ⊆ B
Assume x ∈ A, then show x ∈ B.
To show A * B
Show there exists an x ∈ A such that x 6= B.
16
Sets and Proofs
To show A = B
Step 1. Assume x ∈ A, then show x ∈ B.
Step 2. Assume x ∈ B, then show x ∈ A.
17
Sets and Proofs
Example: Let S and T be sets. Prove that
S ∩ T ⊆ S ∪ T .
18
Sets and Proofs
Example: Let S and T be sets. Prove that
S ⊆ T if and only if S ∩ T = S.
19
Mathematical Induction
Let P (n) is a propositional function with
domain Z+. To prove that P (n) is true
for all positive integers n, we complete two
steps:
1. (Base Case) Verify that P (1) is true.
2. (Inductive Step) Prove the conditional
statement P (k) → P (k + 1) is true for
all positive integers k.
To complete the inductive step, we assume
P (k) is true (this assumption is called the
induction hypothesis), then show P (k+1)
must also be true.
20
Mathematical Induction
Example: Show that if n is a positive
integer, then
1 + 2 + 3 + · · ·n =n(n+ 1)
2
21
Mathematical Induction
Proof: Let P (n) be the proposition
1 + 2 + 3 + · · ·n =n(n+ 1)
2.
We want to show P (n) is true for all n ≥ 1.
Base Step:
22
Mathematical Induction
Example: Conjecture a formula for the
sum of the first n positive odd integers.
Then prove your conjecture using mathe-
matical induction.
1 =
1 + 3 =
1 + 3 + 5 =
1 + 3 + 5 + 7 =
1 + 3 + 5 + 7 + 9 =
23
Mathematical Induction
Conjecture: For all positive integers n,
the following proposition P (n) holds
1 + 3 + 5 + · · ·+ 2n− 1 =
Proof:
24
Mathematical Induction
Example: Use mathematical induction to
prove that 2 divides n2 + 5n for all n ≥ 1.
27
Mathematical Induction
Proof: Let P (n) be the proposition
2 | (n2 + 5n).
We want to show P (n) is true for all n ≥ 1.
Base Step:
28
1 Binary Operations
Cartesian Product
Let A and B be sets. The Cartesian prod-
uct of A and B, denoted by A × B, is the
set of all ordered pairs (a, b), where a ∈ A
and b ∈ B. Hence,
A×B = {(a, b) | a ∈ A and b ∈ B}.
1
Cartesian Product
Example:
If A = {1,2} and B = {a, b, c}, then
A×B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}.
2
Binary Operations
If S is a set, then a binary operation ∗on S is a function that associates to each
ordered pair (s1, s2) ∈ S × S an element of
S which we denote by s1 ∗ s2.
That is, ∗ is a function from S × S to S.
3
Binary Operations
Examples
• Addition defines a binary operation on
Z+ since n + m ∈ Z+ for all (n,m) ∈Z+ × Z+.
• Multiplication defines a binary opera-
tion on Z+ since n · m ∈ Z+ for all
(n,m) ∈ Z+ × Z+.
• Subtraction does not define a binary
operation on Z+ since there exists (n,m) ∈Z+ × Z+ such that n−m /∈ Z+.
• Division does not define a binary oper-
ation on Z+ since there exists (n,m) ∈Z+ × Z+ such that n/m /∈ Z+.
4
Binary Operations
Examples
• Subtraction is a binary operation on each
of the sets Z, R, and Q.
• Division is a binary operation on each
of the sets Q+ and R+.
5
Binary Operations
Example
Let X be a set, and let P(X) be the set
of all subsets of X. Then the operations
of union and intersection are binary opera-
tions on P(X).
For example, if X = {1,2}, then
P(X) = {Ø, {1}, {2}, {1,2}}
and the operations of union and intersec-
tions are binary operations on S.
The set P(X) is called the power set of
X.
6
Binary Operations
Example
Let X be a set, and let P(X) be the set
of all subsets of X. Then the operation M
defined by
A M B = (A−B) ∪ (B −A)
is a binary operation on P(X).
The set A M B is called the symmetric
difference of A and B.
7
Binary Operations
Let S be the set of all 2× 2 matrices with
real entries. Then, the operation ∗, defined
by
(a11 a12
a21 a22
)∗(
b11 b12
b21 b22
)=
a11b11 + a12b21 a11b12 + a12b22
a21b11 + a22b21 a21b12 + a22b22
is a binary operation on S. The corre-
sponds to multiplication of matrices.
8
Commutativity and Associativity
• A binary operation ∗ on S is called
commutative iff a ∗ b = b ∗ a for all
a, b ∈ S.
• A binary operation ∗ on S is called
associative iff (a ∗ b) ∗ c = a ∗ (b ∗ c) for
all a, b, c ∈ S.
9
Commutativity and Associativity
• Subtraction on Z is neither commuta-
tive nor associative. For example,
1− 2 6= 2− 1
(1− 2)− 3 6= 1− (2− 3)
• Division on R+ is neither commutative
nor associative. For example,
1/2 6= 2/1
(1/2)/3 6= 1/(2/3)
• Addition and multiplication are both as-
sociative and commutative on the sets
Z, Q, R.
10
Commutativity and Associativity
Example: Let ∗ be a binary operation on
Z defined by
a ∗ b = 2(a + b)
• is ∗ commutative?
• is ∗ associative?
11
Commutativity and Associativity
Example: Multiplication of 2×2 matrices
• is associative.
((a b
c d
)∗(e f
g h
))∗(
i j
k l
)=
(a b
c d
)∗((
e f
g h
)∗(
i j
k l
))
• is not commutative.
(0 1
1 0
)∗(
1 2
3 4
)=
(3 4
1 2
)(
1 2
3 4
)∗(
0 1
1 0
)=
(2 1
4 3
)
12
Commutativity and Associativity
Example: Let M be the symmetric differ-
ence operator given by
A M B = (A−B) ∪ (B −A)
• is M commutative?
• is M associative?
13
2 Groups
Definition of a group
A set G together with a binary operation ∗is called a group if the following conditions
hold
(i) (closure) x ∗ y ∈ G for all x, y ∈ G.
(ii) (associativity) (x ∗ y) ∗ z = x ∗ (y ∗ z)
for all x, y, z ∈ G.
(iii) (identity element) There is an ele-
ment e ∈ G such that x ∗ e = e ∗ x = x for
all x ∈ G.
(iv) (inverse elements) For each element
x ∈ G there is an element y ∈ G such that
x ∗ y = y ∗ x = e.
14
Definition of a group
Remarks
• We use the notation (G, ∗) to represent
the group with elements in G under the
operation ∗.
• Condition (i) states that ∗ is a binary
operation on G. We say G is closed
with respect to ∗.
• The element e in condition (iii) is called
an identity element of G. In particular,
this means that all groups are nonempty.
• Condition (iv) states that every element
x in G has an inverse element y.
15
Abelian Group
A group (G, ∗) for which ∗ is commutative
is called an abelian group.
16
Groups
Example: (Z,+)
The set Z under addition is a group.
(i) x + y ∈ Z for all x, y ∈ Z.
(ii) (x+y)+z = x+(y+z) for all x, y, z ∈ Z.
(iii) x + 0 = 0 + x = x for all x ∈ Z.
(iv) for each x ∈ Z there is an element
−x ∈ Z such that x+(−x) = (−x)+x = 0.
17
Groups
Example: (Q+, ·)
The set Q+ under multiplication is a group.
(i) x · y ∈ Q+ for all x, y ∈ Q+.
(ii) (x · y) · z = x · (y · z) for all x, y, z ∈ Q+.
(iii) x · 1 = 1 · x = x for all x ∈ Q+.
(iv) for each x ∈ Q+ there is an element1x ∈ Q+ such that x · 1
x = 1x · x = 1.
18
Groups
Example: (Rn,+)
The set of ordered n-tuples (a1, a2, . . . , an)
of real numbers forms a group under the
operation ∗ given by
(a1, a2, . . . , an)∗(b1, b2, . . . , bn) = (a1+b1, a2+b2, . . . , an+bn)
identity element:
inverse element:
19
Groups
Example: (GL(2,R), ·)
The set of invertible 2×2 matrices with real
entries forms a group under matrix multi-
plication.
This group is called the general linear group
of degree 2 over R, and is denoted by
(GL(2,R).
identity element:
inverse element:
20
Groups
Example: ({f | f : R→ R},+)
The set of real-valued functions with do-
main R forms a group under the operation
of pointwise addition of functions.
(f + g)(x) = f(x) + g(x)
identity element:
inverse element:
21
Groups
Example: (P(X),M)
The set of subsets of a set X forms a group
under the operation of symmetric differ-
ence of sets:
A M B = (A−B) ∪ (B −A)
identity element:
inverse element:
22
Groups
Example: Consider the set Z together
with the binary operation ∗ given by
a ∗ b = 2(a + b).
Does (Z, ∗) form a group?
23
Groups
Example: Consider the set G = R − {0}together with the binary operation ∗ given
by
a ∗ b = 2ab.
Does (G, ∗) form a group?
24
Additive group of integers modulo n
Theorem (The Division Algorithm)
Let a be an integer and n a positive inte-
ger. Then there are unique integers q and
r, with 0 ≤ r < n, such that a = qn + r.
Notation:
The integer r in the division algorithm is
called the remainder of a mod n, and
will be denoted by a.
25
Additive group of integers modulo n
Lemma: Assume n ∈ Z+. Then, a = b if
and only if a− b = nk for some k ∈ Z.
Proof:
26
Additive group of integers modulo n
Example: (Zn,⊕)
Let n be a positive integer, and consider
the set Zn = {0,1,2, . . . , n− 1}.
Zn forms a group under the binary opera-
tion ⊕ defined by
x⊕ y = x + y
where x + y is the unique number in Zn
such that x + y ≡ x + y (mod n)
This group is called the additive group of
integers modulo n.
identity element:
inverse element:27
3 Fundamental Theorems About
Groups
Uniqueness of the Identity Element
Theorem 3.1 If (G, ∗) is a group, then
there is only one identity element in G.
Proof: Assume e1 and e2 are identity ele-
ments of G. Since e1 is an identity element,
we know
e1 ∗ e2 = e2.
Since e2 is an identity element, we know
e1 ∗ e2 = e1.
This proves e1 = e2.
1
Uniqueness of Inverses
Theorem 3.2 If (G, ∗) is a group and x
is any element of G, then x has only one
inverse in G.
Proof: Assume x ∈ G and assume y1 and
y2 are inverses of x. Then,
y1 = y1 ∗ e
= y1 ∗ (x ∗ y2)
= (y1 ∗ x) ∗ y2
= e ∗ y2
= y2
This proves y1 = y2, and we conclude that
the inverse of an element is unique.
2
Inverse Element
Notation
If (G, ∗) is a group, then the inverse of an
element x ∈ G is denoted by x−1.
3
Inverse Element
Theorem 3.3 If (G, ∗) is a group and
x ∈ G , then (x−1)−1 = x.
Proof: Let y = (x−1)−1. Then, y is the
unique element of G such that
x−1 ∗ y = y ∗ x−1 = e.
Note that the previous equation is satisfied
when y = x, since x−1 is the inverse of x.
Therefore, (x−1)−1 = y = x.
4
Alternate Proof: Assume x ∈ G. We have,
(x−1)−1 = (x−1)−1 ∗ e
= (x−1)−1 ∗ (x−1 ∗ x)
= ((x−1)−1 ∗ x−1) ∗ x
= e ∗ x
= x.
Inverse Element
Examples
• Consider the group (Z,+). Then, for
all n ∈ Z, we have −(−n) = n.
• Consider the group GL(2,R). Then, for
all(a bc d
)∈ GL(2,R), we have
((a bc d
)−1)−1
=(a bc d
).
5
Inverse of a Product
Theorem 3.4 If (G, ∗) is a group and
x, y ∈ G, then
(x ∗ y)−1 = y−1 ∗ x−1.
Proof: Assume x, y ∈ G. Let z = y−1∗x−1.
Then,
(x ∗ y) ∗ z = (x ∗ y) ∗ (y−1 ∗ x−1)
= x ∗ (y ∗ (y−1 ∗ x−1))
= x ∗ ((y ∗ y−1) ∗ x−1)
= x ∗ (e ∗ x−1)
= x ∗ x−1
= e.
A similar argument shows z ∗ (x ∗ y) = e.
Therefore, (x ∗ y)−1 = z = y−1 ∗ x−1.
6
Alternate proof:
(x ∗ y)−1 = (x ∗ y)−1 ∗ e
= (x ∗ y)−1 ∗ (x ∗ x−1)
= (x ∗ y)−1 ∗ ((x ∗ e) ∗ x−1)
= (x ∗ y)−1 ∗ ((x ∗ (y ∗ y−1)) ∗ x−1)
= (x ∗ y)−1 ∗ (((x ∗ y) ∗ y−1) ∗ x−1)
= (x ∗ y)−1 ∗ ((x ∗ y) ∗ (y−1 ∗ x−1))
= ((x ∗ y)−1 ∗ (x ∗ y)) ∗ (y−1 ∗ x−1)
= e ∗ (y−1 ∗ x−1)
= y−1 ∗ x−1
Inverse Element
Theorem 3.5 Assume (G, ∗) is a group
and let x, y ∈ G. If either x ∗ y = e or
y ∗ x = e, then y = x−1.
Proof: First assume x ∗ y = e. We will
solve for y by multiplying both sides of the
equation on the left by x−1. We have,
x−1 ∗ (x ∗ y) = x−1 ∗ e
(x−1 ∗ x) ∗ y = x−1
e ∗ y = x−1
y = x−1
Similarly, if y ∗ x = e, then right multipli-
cation by x−1 shows y = x−1.
7
Cancellation Laws
Theorem 3.6 Assume (G, ∗) is a group
and let x, y ∈ G. Then:
(i) if x ∗ y = x ∗ z, then y = z; and
(ii) if y ∗ x = z ∗ x, then y = z.
Proof: See homework.
8
Identities and Inverses
Notation
Let G be a set, and assume ∗ is an asso-
ciative binary operation on G. Then:
• if e ∈ G and x ∗ e = x for all x ∈ G,
then we say e is a right identity with
respect to ∗.
• if x ∈ G and e ∗ x = x for all x ∈ G,
then we say e is a left identity with
respect to ∗.
• if x, y ∈ G and x ∗ y = x, then we say
y is a right inverse of x.
• if x, y ∈ G and y ∗ x = x, then we say
y is a left inverse of x.
9
Sufficient Conditions for a Group
Theorem 3.8 Let G be a set, and assume
∗ is an associative binary operation on G.
If there exists a right identity e ∈ G with
respect to ∗, and if every element x ∈ G
has a right inverse, then (G, ∗) is a group.
10
Proof: First we will show that right identity
e ∈ G is also a left identity. Given x ∈ G,
let y denote the right inverse of x, and let
z denote the right inverse of y. Then,
e ∗ x = (e ∗ x) ∗ e
= (e ∗ x) ∗ (y ∗ z)
= e ∗ (x ∗ (y ∗ z))
= e ∗ ((x ∗ y) ∗ z)
= e ∗ (e ∗ z)
= (e ∗ e) ∗ z
= e ∗ z
= (x ∗ y) ∗ z
= x ∗ (y ∗ z)
= x ∗ e
= x
This shows that e ∗ x = x for any x ∈ G.
Hence, e is an (two-sided) identity element.
Finally we will show that for any x ∈ G, if y
is the right inverse of x, then y is also a left
inverse of x. If z denotes the right inverse
of y, we have
y ∗ x = (y ∗ x) ∗ e
= (y ∗ x) ∗ (y ∗ z)
= y ∗ (x ∗ (y ∗ z))
= y ∗ ((x ∗ y) ∗ z)
= y ∗ (e ∗ z)
= (y ∗ e) ∗ z
= y ∗ z
= e.
This shows that for all x ∈ G there exists a
y ∈ G such that x∗y = y∗x = e. Therefore,
(G, ∗) is a group.
Insufficient Conditions for a Group
Example: Consider the set Z+ with the
binary operation given by
x ∗ y = x.
• ∗ is associative since
(x ∗ y) ∗ z = x ∗ y = x = x ∗ (y ∗ z)
• the element 1 ∈ Z+ is a right identity
element since x ∗ 1 = x for all x ∈ Z+.
(In fact, any element y ∈ Z+ is an right
identity element.)
• every element y ∈ Z+ has the element
1 ∈ Z+ as a left inverse element since
1 ∗ y = 1 for all y ∈ Z+.
11
• However these conditions are not suffi-
cient to make Z+ a group with respect
to ∗, since there is no left identity ele-
ment:
y ∗ x 6= x unless y = x
• Note that multiplication table for Z+
with respect to ∗ has repeated ele-
ments in its rows (See Homework 3,
Problem 3):
∗ 1 2 3 4 5 · · ·
1 1 1 1 1 1 · · ·
2 2 2 2 2 2 · · ·
3 3 3 3 3 3 · · ·
4 4 4 4 4 4 · · ·
5 5 5 5 5 5 · · ·... ... ... ... ... ... . . .
4 Powers of an Element; Cyclic
Groups
Notation
When considering an abstract group (G, ∗),
we will often simplify notation as follows
• x ∗ y will be expressed as xy
• (x ∗ y) ∗ z will be expressed as xyz
• x ∗ (y ∗ z) will be expressed as xyz
In other words, we will omit the symbol ∗,and use product notation, unless a more
appropriate notation is called for (e.g., the
symbol + will be used in the case of the
group (Z,+)). Also, when associativity al-
lows, we can omit parentheses.
1
Powers of an Element
Notation
Let G be a group, and let x be an element
of G. We define powers of x as follows:
x0 = e
xn = xxx · · ·x︸ ︷︷ ︸n factors
x−n = x−1x−1x−1 · · ·x−1︸ ︷︷ ︸n factors
2
Powers of an Element
Theorem 4.1 Let G be a group and let
x ∈ G. Let m,n be integers. Then:
(i) xmxn = xm+n
(ii) (xn)−1 = x−n
(iii) (xm)n = xnm = (xn)m
3
Order of an Element
Definitions
If G is a group and x ∈ G, then x is said to
be of finite order if there exists a positive
integer n such that xn = e.
If such an integer exists, then the smallest
positive n such that xn = e is called the
order of x and denoted by o(x).
If x is not of finite order, then we say that
x is of infinite order and write o(x) =∞.
4
Order of an Element
Examples
• Let G = (Z3,⊕). Then o(1) = 3, since
1 6= 0
1⊕ 1 6= 0
1⊕ 1⊕ 1 = 0
• Let G = (Z,+). Then o(1) =∞, since
1 6= 0
1 + 1 6= 0
1 + 1 + 1 6= 0
1 + 1 + 1 + 1 6= 0
···
5
Order of an Element
Examples
• Let G = (Q+, ·). Then o(2) =∞, since
21 6= 1
22 6= 1
23 6= 1
24 6= 1
···
• Let G = GL(2,R), then o (( −1 00 −1 )) = 2,
since (−1 0
0 −1
)16=(
1 00 1
)(−1 0
0 −1
)2=(
1 00 1
)
6
Greatest Common Divisor
Let a and b be integers, not both zero.
The largest integer d such that d | a and
d | b is called the greatest common divisor
of a and b. The greatest common divisor
of a and b is denoted by gcd(a, b).
7
Greatest Common Divisor
Examples
• gcd(24,40) =
• gcd(32,100) =
• gcd(12,91) =
8
Relatively Prime
The integers a and b are relatively prime if
their greatest common divisor is 1.
Examples
• The integers 12 and 25 are relatively
prime, since gcd(12,25) = 1
• The integers 27 and 75 are not rela-
tively prime, since gcd(27,75) = 3 6= 1.
9
The Euclidean Algorithm
Suppose that a and b are positive integers
with a ≥ b. Successively applying the divi-
sion algorithm, we obtain
a = b q0 + r1, 0 ≤ r1 < b,
b = r1 q1 + r2, 0 ≤ r2 < r1,
r1 = r2 q2 + r3, 0 ≤ r3 < r2,
r2 = r3 q3 + r4, 0 ≤ r4 < r3,
···
rn−2 = rn−1 qn−1 + rn, 0 ≤ rn < rn−1,
rn−1 = rn qn.
Theorem: gcd(a, b) = rn, where rn is the
last nonzero remainder in the sequence above.
10
The Euclidean Algorithm
Lemma Let a = bq + r, where a, b, q, and
r are integers. Then gcd(a, b) = gcd(b, r).
Proof: Assume d | a and d | b. Therefore
d | r where r = a − bq. This shows that all
common divisors of a and b are common
divisors of b and r.
Next, assume d | b and d | r. Therefore d | awhere a = bq + r. This shows that all com-
mon divisors of b and r are common divisors
of a and b.
Therefore the common divisors of a and b
are the same as the common divisors of b
and r. Therefore, their greatest common
divisors are the same.
11
The Euclidean Algorithm
Examples
• Find gcd(198, 252) using the Euclidean
algorithm
• Find gcd(414, 662) using the Euclidean
algorithm
12
Greatest Common Divisor
Theorem 4.2 If a and b are integers, not
both zero, then there exist integers x and
y such that gcd(a, b) = ax + by.
Proof: Solving for the last nonzero remain-
der in the Euclidean Algorithm, we have
rn = rn−2 − rn−1 qn−1
Using the preceding step of the Euclidean
Algorithm, we may express the term rn−1
in terms of rn−2 and rn−3, thereby obtain-
ing
rn = rn−2 − (rn−3 − rn−2 qn−2) qn−1
After a sequence of n − 1 substitutions,
we obtain an expression for gcd(a, b) = rn
in terms of a and b.13
Greatest Common Divisor
Example: Express gcd(198,252) = 18 as
a linear combination of 252 and 198.
14
Greatest Common Divisor
Theorem 4.3 If a | bc and gcd(a, b) = 1
and, then a | c.
Proof: Since gcd(a, b) = 1, there exist
integers x and y such that
ax + by = 1.
Multiplying on both sides by c yields
axc + byc = c
a(xc) + bc(y) = c
Since a divides each term on the left-hand
side, a also divides the sum of the two
terms. Therefore, a divides c.
15
Powers of an Element
Theorem 4.4 Let G be a group and let
x ∈ G. Let m,n, d be integers. Then:
(i) o(x) = o(x−1)
(ii) If o(x) = n and xm = e, then n |m
(iii) If o(x) = n and gcd(m,n) = d, then
o(xm) = n/d.
16
Cyclic Groups
Definition
A group G is called cyclic if there is an
element x ∈ G such that
G = 〈x〉 = {xn | n ∈ Z}.
The element x is called a generator for
G, and the cyclic group generated by x is
denoted by 〈x〉.
17
Cyclic Groups
Example
• The group G = (Z1,⊕) is the trivial
group {0} consisting of one (identity)
element. It is the cyclic group gener-
ated by x = 0:
(Z1,⊕) = 〈0〉
• For all n ≥ 2, the group G = (Zn,⊕) is a
finite cyclic group generated by x = 1:
(Zn,⊕) = 〈1〉
= {0,1, 1⊕ 1, . . . , 1⊕ 1⊕ 1⊕ · · · ⊕ 1︸ ︷︷ ︸n− 1 terms
}
18
Cyclic Groups
Example
• The group G = (Z,+) is an infinite
cyclic group generated by x = 1:
(Z,+) = 〈1〉
= {. . . , (−1) + (−1), (−1), 0, 1, 1 + 1, 1 + 1 + 1, . . .}
• The group G = (Q,+) is not cyclic,
since there does not exist q ∈ Q such
that ever rational number r ∈ Q has the
form r = nq for some n ∈ Z.
19
Cyclic Groups
Theorem 4.5 Let G = 〈x〉. If o(x) =∞,
then xj 6= xk for j 6= k, and consequently
G is infinite. If o(x) = n, then xj = xk iff
j ≡ k (mod n), and consequently the dis-
tinct elements of G are e, x, x2, . . . , xn−1.
Proof:
20
Cyclic Groups
Definition The order of a group G, denoted
by |G|, is the number of elements in G.
Corollary 4.6 If G = 〈x〉, then |G| = o(x).
21
Cyclic Groups
Theorem 4.7 If G is a cyclic group, then
G is abelian.
Proof:
22
5 Subgroups
Definition of a Subgroup
A subset H of a group (G, ∗) is called a
subgroup of G if the elements of H form
a group under ∗.
1
Subgroups
Examples
• (2Z,+) is a subgroup of (Z,+)
• (Z,+) is a subgroup of (Q,+)
• (Q,+) is a subgroup of (R,+)
• (Q+, ·) is a subgroup of (R+, ·)
•{(
a b−b a
)| a, b ∈ R and a2 + b2 6= 0
}is a
subgroup of G = GL(2,R).
•{(
a b0 d
)| a, b, d ∈ R and ad 6= 0
}is a
subgroup of G = GL(2,R).
2
Subgroups
Theorem If H is a subgroup of G and e
is the identity element for G, then e ∈ H
and e is the identity element for H.
3
Proof: Assume H is a subgroup of G.
Then H is group and must have an identity
element eH. We claim eH = e where e is
the identity element for G. Since eH is an
identity element for H, we know,
eH ∗ eH = eH .
Since H ⊆ G, we know eH ∈ G. There-
fore, eH has an inverse element e−1H ∈ G.
Multiplying on both sides of the previous
equation by e−1H yields:
e−1H ∗ (eH ∗ eH) = e−1
H ∗ eH(e−1
H ∗ eH) ∗ eH = e
e ∗ eH = e
eH = e
Trivial Subgroups
For any group G, the subgroups H = {e}and H = G are called trivial subgroups.
4
Groups of Order 1
There is only one group of order 1, namely
G = {e}. The group table for G is given by
∗ e
e e
• The group G has only one subgroup
H = G = {e}.
• The group G is cyclic. The element e
is a generator for G.
5
Groups of Order 2
There is only one group of order 2. If we
write G = {e, a}, then the group table for
G is given by
∗ e a
e e a
a a
• The group G has two subgroups, namely
H = {e} and H = G = {e, a}.
• The group G is cyclic. The element a
is a generator for G.
6
Groups of Order 3
There is only one group of order 3. If we
write G = {e, a, b}, then the group table
for G is given by
∗ e a b
e e a b
a a
b b
• The group G has two subgroups, namely
H = {e} and H = G = {e, a, b}.
• The group G is cyclic. Both elements
a and b are generators for G.
7
Groups of Order 4
There are two groups of order 4. If we
write G = {e, a, b, c}, then the possible group
tables for G are
∗ e a b c
e e a b c
a a
b b
c c
∗ e a b c
e e a b c
a a
b b
c c
∗ e a b c
e e a b c
a a
b b
c c
∗ e a b c
e e a b c
a a
b b
c c
8
Cyclic Group of Order 4
If we write G = {e, a, b, c}, then the follow-
ing group table for G represents a cyclic
group of order 4.
∗ e a b c
e e a b c
a a
b b
c c
• The group G has three subgroups, namely
H = {e}, H = {e, b}, and H = {e, a, b, c}.
• The group G is cyclic. Both elements
a and c are generators for G.
9
Klein’s 4-group
If we write G = {e, a, b, c}, then the fol-
lowing group table for G represents a non-
cyclic group of order 4 called Klein’s 4-
group.
∗ e a b c
e e a b c
a a
b b
c c
• Klein’s 4-group has five subgroups:
H = {e}, H = {e, a}, H = {e, b},H = {e, c}, and H = {e, a, b, c}.
10
Subgroup Lattice
A subgroup lattice is a diagram which de-
picts the subgroups of a group G in a way
that indicates all subset relations among
subgroups.
Example (Klein’s 4-group)
11
Groups of Order n
The previous examples beg the question:
How many groups of order n exist for a
given integer n?
The answer for n ≤ 99 is given below.
+ 0 1 2 3 4 5 6 7 8 9
0 0 1 1 1 2 1 2 1 5 2
10 2 1 5 1 2 1 14 1 5 1
20 5 2 2 1 15 2 2 5 4 1
30 4 1 51 1 2 1 14 1 2 2
40 14 1 6 1 4 2 2 1 52 2
50 5 1 5 1 15 2 13 2 2 1
60 13 1 2 4 267 1 4 1 5 1
70 4 1 50 1 2 3 4 1 6 1
80 52 15 2 1 15 1 2 1 12 1
90 10 1 4 2 2 1 231 1 5 2
12
Subgroups
Theorem 5.1 Let H be a nonempty sub-
set of a group G. Then H is a subgroup
of G if and only if the following two con-
ditions are satisfied:
(i) for all a, b ∈ H, ab ∈ H, and
(ii) for all a ∈ H, a−1 ∈ H.
Remark: Condition (i) is expressed by say-
ing that H is closed under the binary oper-
ation on G, and condition (ii) is expressed
by saying that H is closed under inverses.
13
Proof:
Subgroups
Theorem Let G be a group, and let a ∈ G.
Then 〈a〉, the set of elements generated by
integer powers of a, is a subgroup of G.
Proof: Let H = 〈a〉. Clearly H is nonempty
since a ∈ H. Also for all aj, ak ∈ H, we have
ajak = aj+k ∈ H. Also, for any aj ∈ H,
we have (aj)−1 = a−j ∈ H. Therefore, by
Theorem 5.1, H = 〈a〉 is a subgroup of G.
14
Subgroups
Examples
• Let G = (Z,+) and let n ∈ Z. Then,
H = 〈n〉 = nZ is a subgroup of G.
• Let G = (Z6,⊕). Then the following
are subgroups of G:
〈0〉 = {0}
〈1〉 = {0,1,2,3,4,5}
〈2〉 = {0,2,4}
〈3〉 = {0,3}
〈4〉 = {0,2,4}
〈5〉 = {0,1,2,3,4,5}
15
Subgroups
Theorem 5.2 If G is a cyclic group, then
every subgroup of G is cyclic.
Proof:
16
Subgroups
Theorem 5.4 Let H and K be subgroups
of a group G. Then:
(i) H ∩K is a subgroup of G; and
(ii) H ∪K is a subgroup of G if and only
if H ∪K = H or H ∪K = K.
Proof:
17
Subgroups
Theorem 5.5 Let G = 〈x〉 be a cyclic
group of order n. Then:
(i) For any positive integer m, G has a
subgroup of order m if and only if m
divides n.
(ii) If m divides n, then G has a unique
subgroup of order m.
(iii) The elements xr and xs generate the
same subgroup of G if and only if
(r, n) = (s, n).
18
Group of Unit Quaternions
Consider the general linear group of degree
two over the complex numbers
GL(2,C) ={( z1 z2
z3 z4
)| zk ∈ C and z1z4 − z2z3 6= 0
}
The group of unit quaternions, denoted
Q8, is the subgroup of GL(2,C) consisting
of the following 8 elements:
{(1 0
0 1
),
(−1 0
0 −1
),( 0 1
−1 0
),( 0 −1
1 0
)(
i 0
0 −i
),( −i 0
0 i
),(
0 i
i 0
),
(0 −i
−i 0
)}
19
Group of Unit Quaternions
Let J =(
i 0
0 −i
), K =
( 0 1
−1 0
), L =
(0 i
i 0
).
The group of unit quaternions has the form
Q8 = {I, −I, J, −J, K, −K, L, −L},
and the following relations hold
J2 = K2 = L2 = −I
JK = L
KL = J
LJ = K
KJ = −L
LK = −J
JL = −K
20
6 Direct Products
Direct Product of Groups
Let G and H be groups. Then the set of
ordered pairs
G×H = {(g, h) | g ∈ G, h ∈ H}
forms a group under componentwise mul-
tiplication:
(g1, h1) ∗ (g2, h2) = (g1 g2, h1h2)
The group G × H is called the direct
product of G and H.
1
Direct Product of Groups
Remarks
• The identity element of G×H is (eG, eH)
where eG and eH are the respective
identity elements of G and H.
• The inverse of (g, h) ∈ G × H is the
ordered pair (g−1, h−1).
2
Generalized Direct Product
Let G1, G2, G3, . . . , Gn be groups. Then
the set of ordered n-tuples
G1 × · · · ×Gn = {(g1, g2, . . . , gn) | gi ∈ Gi}
forms a group under componentwise mul-
tiplication.
• The identity element of G1 × · · · × Gn
is (eG1, eG2
, . . . , eGn) where eGiis the
identity element for Gi.
• The inverse of (g1, g2, . . . , gn) is the
element (g−11 , g−1
2 , . . . , g−1n ).
3
Direct Products
Example
Let G = H = (Z2,⊕). Then,
G×H = Z2 × Z2
is the group of ordered pairs
{(0,0), (0,1), (1,0), (1,1)}
under componentwise addition mod 2.
• Z2 × Z2 is a finite, non-cyclic group of
order 4 with the following non-trivial
subgroups
〈(0,1)〉 = {(0,0), (0,1)}
〈(1,0)〉 = {(0,0), (1,0)}
〈(1,1)〉 = {(0,0), (1,1)}
4
Direct Products
Example
Let G = (Z2,⊕) and H = (Z3,⊕). Then,
G×H = Z2 × Z3
is the group of ordered pairs
{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)}
under componentwise addition.
• Z2×Z3 is a finite, cyclic group of order
6 with generator (1,1), and with the
non-trivial subgroups
〈(0,1)〉 = {(0,0), (0,1), (0,2)}
〈(1,0)〉 = {(0,0), (1,0)}
5
Direct Products
Example
Let G1 = G2 = · · · = Gn = (R,+). Then,
G1 × · · · ×Gn = R× · · · × R︸ ︷︷ ︸n-factors
= Rn
is the group of ordered n-tuples of real
numbers under componentwise addition.
• The identity element of Rn is (0,0, . . . ,0).
• The inverse of (x1, x2, . . . , xn) is the
element (−x1,−x2, . . . ,−xn).
6
Direct Products
Theorem 6.1 Let G = G1 × · · · ×Gn.
(i) If gi ∈ Gi for 1 ≤ i ≤ n, and each gi
has finite order, then o((g1, . . . , gn))
is the least common multiple of the
numbers o(g1), o(g2), . . . , o(gn).
(ii) If each Gi is a cyclic group of finite
order, then G is cyclic iff |Gi| and
|Gj| are relatively prime for i 6= j.
7
Proof:
Direct Products
Examples
• Consider the element (1,3) ∈ Z6 × Z8.
Since o(1) = 6 in the group Z6, and
o(3) = 8 in the group Z8, we have
o((1,3)) = lcm(6,8) = 24
• The groups Z14×Z15 and Z8×Z9×Z5
are cyclic.
• The groups Z14×Z16 and Z8×Z9×Z6
are not cyclic.
8
7 Functions
Function, Domain, and Codomain
Let A and B be nonempty sets. A function
f from A to B, denoted
f : A→ B,
is a subset of the Cartesian product A×B
such that for each a ∈ A, there is a unique
element b ∈ B such that (a, b) belongs to f .
We say that A is the domain of f and B is
the codomain of f .
1
Range of a Function
If f : A → B and the ordered pair (a, b)
belongs to f , then we write
f(a) = b
and we say b the image of a under f .
The range of f , denoted f(A), is the set
of all images of elements of A.
2
Examples
• f : R→ R, defined by
f(x) = x2
• f : GL(2,R) → R, defined by
f((
a bc d
))= ad− bc
• f : G → G, defined by
f(x) = a ∗ x
where G is a group and a ∈ G.
• F : C(R,R) → R, defined by
F (g) =∫ 1
0g(x) dx
where C(R,R) is the set of continuous
functions from R to R.
3
Image and Preimage
If f : A→ B and S ⊆ A, then the image of
S under f is the set
f(S) = {b ∈ B | b = f(s) for some s ∈ S}
If T ⊂ B, then the preimage of T is the set
f−1(T ) = {a ∈ A | f(a) ∈ T}
4
Example
Let f : R→ R be defined by f(x) = x2.
• domain of f :
• codomain of f :
• range of f :
• If S = [−2,3), then f(S) =
• If T = (−1,4], then f−1(T ) =
5
Example
Let f : R→ R be defined by f(x) = x2.
• The domain of f is R.
• The codomain of f is R.
• The range of f is R+ ∪ {0}.
• If S = [−2,3), then f(S) = [0,9).
• If T = (−1,4], then f−1(T ) = [−2,2].
6
One-To-One Function
A function f : A→ B is said to be one-to-
one, or injective, if and only if the follow-
ing condition holds:
∀x∀y (x 6= y → f(x) 6= f(y))
or equivalently
∀x∀y (f(x) = f(y) → x = y)
where the domain for x and y is the set A.
If f is one-to-one, we say f is an injection.
7
Onto Function
A function f : A → B is said to be onto,
or surjective, if and only if the following
condition holds:
∀b ∃a (f(a) = b)
where the domain for b is the set B, and
the domain for a is the set A.
If f is onto, we say that f is a surjection.
Note that f is onto if and only if f(A) = B.
That is, f is onto if and only if the range
of f is equal to the codomain of f .
8
Examples
Let A = {1,2,3,4} and B = {a, b, c}.
Determine if the given functions are one-
to-one, onto, both or neither.
1. f : A→ B defined by
f(1) = a
f(2) = b
f(3) = c
f(4) = b
2. f : B → A defined by
f(a) = 3
f(b) = 1
f(c) = 4
9
Examples
• f : R→ R, defined by
f(x) = x2
• f : GL(2,R) → R, defined by
f((
a bc d
))= ad− bc
• f : G → G, defined by
f(x) = a ∗ x
where G is a group and a ∈ G.
• F : C(R,R) → R, defined by
F (g) =∫ 1
0g(x) dx
where C(R,R) is the set of continuous
functions from R to R.
10
Proof Strategy
To show f is injective
Show that if f(x) = f(y), then x = y.
To show f is not injective
Show that there exist x and y such that
f(x) = f(y) and x 6= y.
To show f is surjective
Show that for each element y in the codomain
there exists an element x in the domain
such that f(x) = y.
To show f is not surjective
Show there exists an element y in the codomain
such that y 6= f(x) for any x in the domain.
11
Example
Prove or disprove:
The function f : Z→ Z defined by
f(n) = 2n− 3
is injective.
12
Example
Prove or disprove:
The function f : Z→ Z defined by
f(n) = 2n− 3
is surjective.
13
Bijection
A function f : A→ B is said to be a bijec-
tion, or one-to-one correspondence if f
is both one-to-one and onto.
Examples
• f : Z→ Z defined by f(n) = n + 1
• f : R→ R defined by f(x) = 2x
• f : Z+ → Z+ defined by f(n) ={n + 1 if n is odd
n− 1 if n is even
14
Inverse Function
Let f : A→ B be a bijection. The inverse
function of f , denoted by f−1, is the func-
tion that assigns to an element b ∈ B the
unique element a ∈ A such that f(a) = b.
That is,
f−1(b) = a if and only if f(a) = b
If f has an inverse function, we say that f
is invertible.
Note that we use the notation f−1(T ) to
denote the preimage of a set T ⊆ B, even
when f is not invertible.
15
Example
Determine if each function invertible. If so,
find f−1.
1. f : Z→ Z defined by f(n) = n + 1
2. f : R→ R defined by f(x) = 2x + 3
3. f : R→ R defined by f(x) = x2
16
Composition of Functions
Let g : A→ B and f : B → C be functions.
The composition of the functions f and
g, denoted by f ◦ g, is a function from A to
C, defined by
(f ◦ g) (a) = f(g(a))
17
Examples
Let A = {1,2,3,4}, B = {a, b, c}, C = {w, x, y, z},and let f and g be the functions below
g : A→ B defined by
g(1) = a
g(2) = b
g(3) = c
g(4) = b
f : B → C defined by
f(a) = x
f(b) = w
f(c) = z
Then, f ◦ g : A→ C is given by
(f ◦ g)(1) =
(f ◦ g)(2) =
(f ◦ g)(3) =
(f ◦ g)(4) =
18
Composition of Functions
Example
Let g : R → R and f : R → R be defined by
g(x) = 12(x− 3) and f(x) = 2x + 3. Find
(a) (f ◦ g)(x) =
(b) (g ◦ f)(x) =
19
Inverse Functions and Composition
Let f : A → B and g : B → A. Then, g is
the inverse function for f if and only if the
following two conditions hold.
(i) (f ◦ g)(x) = x for all x ∈ B.
(ii) (g ◦ f)(x) = x for all x ∈ A.
In particular, if f is invertible, then
(i) f(f−1(x)) = x for all x ∈ B.
(ii) f−1(f(x)) = x for all x ∈ A.
20
Identity Function
Let A be a set. The identity function on
A is the function iA : A→ A defined by
iA(x) = x
That is, iA is the function that assigns each
element of A to itself.
The identity function iA is one-to-one and
onto, so it is a bijection.
21
Set of Bijections from X to X
Let X be a nonempty set, and define
SX = {f : X → X | f is a bijection}
Theorem 7.1: (SX , ◦) is a group.
Proof:
(i) (Closure) See Homework 5, Problem 7
(ii) (Associativity) For all f, g, h ∈ SX, and
for all x ∈ X,
((f ◦ g) ◦ h)(x) = (f ◦ g)(h(x))
= f(g(h(x)))
= f((g ◦ h)(x))
= (f ◦ (g ◦ h))(x)
That is, (f ◦g)◦h = f ◦(g ◦h), which proves
◦ is associative.22
(iii) (identity element) The identity func-
tion iX : X → X is a bijection, hence be-
longs to SX, and has the property
(f ◦ iX)(x) = f(iX(x))
= f(x)
= iX(f(x))
= (iX ◦ f)(x)
for all x ∈ X. This proves iX is an identity
element for SX.
(iv) (inverse elements) For all f ∈ SX,
there is an element g ∈ SX, namely g =
f−1, such that
(f ◦ f−1)(x) = f(f−1(x))
= iX(x)
= f−1(f(x))
= (f−1 ◦ f)(x)
for all x ∈ X. This proves that every ele-
ment of SX has an inverse.
Thus, (SX , ◦) is a group.
8 Symmetric Groups
Permutations
If X is a nonempty set, then a permutation
of X is a bijection f : X → X.
Recall that the set SX of all permutations
of X forms a group under composition of
functions. This group, denoted (SX , ◦), is
called the symmetric group on X.
1
Symmetric Group of Degree n
Let X = {1,2,3, . . . , n}.
The symmetric group on X is called the
symmetric group of degree n, and is de-
noted by Sn.
Given a permutation f ∈ Sn, we represent
f in array form as follows:
1 2 3 · · · n
f(1) f(2) f(3) · · · f(n)
2
Composition of Permutations
Example
Consider f, g ∈ S4 defined by
f(1) = 2 g(1) = 3
f(2) = 4 g(2) = 2
f(3) = 1 g(3) = 4
f(4) = 3 g(4) = 1
In array form we have:
f =(
1 2 3 4
f(1) f(2) f(3) f(4)
)=(
1 2 3 42 4 1 3
)
g =(
1 2 3 4
g(1) g(2) g(3) g(4)
)=(
1 2 3 43 2 4 1
)
3
To compute the composition f ◦g, the fol-
lowing diagram is helpful.
1 2 3 4
3 2 4 1
1 4 3 2
Therefore,
f ◦ g =(
1 2 3 41 4 2 3
).
Observe that
(f ◦ g)(1) =((
1 2 3 42 4 1 3
)◦(
1 2 3 43 2 4 1
))(1)
=(
1 2 3 42 4 1 3
) ((1 2 3 43 2 4 1
)(1)
)
=(
1 2 3 42 4 1 3
)(3)
= 1
(f ◦ g)(2) =((
1 2 3 42 4 1 3
)◦(
1 2 3 43 2 4 1
))(2)
=(
1 2 3 42 4 1 3
) ((1 2 3 43 2 4 1
)(2)
)
=(
1 2 3 42 4 1 3
)(2)
= 4
Also, we have
(f ◦ g)(3) =((
1 2 3 42 4 1 3
)◦(
1 2 3 43 2 4 1
))(3)
=(
1 2 3 42 4 1 3
) ((1 2 3 43 2 4 1
)(3)
)
=(
1 2 3 42 4 1 3
)(4)
= 3
(f ◦ g)(4) =((
1 2 3 42 4 1 3
)◦(
1 2 3 43 2 4 1
))(4)
=(
1 2 3 42 4 1 3
) ((1 2 3 43 2 4 1
)(4)
)
=(
1 2 3 42 4 1 3
)(1)
= 2
Composition of Permutations
Example: Determine f ◦g where f, g ∈ S7
are given by
f =(
1 2 3 4 5 6 73 4 2 1 5 6 7
)
g =(
1 2 3 4 5 6 77 1 3 4 5 2 6
)
Solution:
1 2 3 4 5 6 7
7 1 3 4 5 2 6
4
Cycles
An element f ∈ Sn is called a cycle if there
exist distinct integers
i1, i2, . . . , ir ∈ {1,2, . . . , n}
such that
f(i1) = i2
f(i2) = i3
···
f(ir−1) = f(ir)
f(ir) = i1
and f(j) = j otherwise. In this case, we
express f in cycle notation as follows:
f = (i1, i2, i3, . . . , ir)
5
Cycle Notation
Example: Consider f, g ∈ S7 given by
f =(
1 2 3 4 5 6 73 4 2 1 5 6 7
)
g =(
1 2 3 4 5 6 77 1 3 4 5 2 6
)
Then, f and g are both cycles of length 4.
In cycle notation, we have
f = (1, 3, 2, 4),
g = (1, 7, 6, 2).
6
Cycle Notation
• A cycle of length r is called an r-cycle.
• A 2-cycle is called a transposition.
• The identity permutation is usually de-
noted by the 1-cycle (1).
7
Disjoint Cycles
Two cycles (x1, x2, . . . , xr) and (y1, y2, . . . , ys)
are said to be disjoint if r ≥ 2, s ≥ 2, and
{x1, x2, . . . , xr} ∩ {y1, y2, . . . , ys} = Ø
Examples
• (1,3), (2,4) ∈ S4
• (1,5,2), (3,7,8,4) ∈ S8
• (2,6,9,4), (5,7) ∈ S9
8
Product of Cycles
While not all permutations f ∈ Sn are cy-
cles, all permutations can be expressed as
a product (composition) of disjoint cycles.
Theorem 8.1: Let f ∈ Sn. Then, there
exist disjoint cycles f1, f2, . . . , fm ∈ Sn
such that
f = f1 ◦ f2 ◦ · · · ◦ fm.
9
Product of Cycles
Example: Consider f ∈ S9 given by
f =(
1 2 3 4 5 6 7 8 94 1 7 9 6 5 8 3 2
).
Note that f is composed of three disjoint
cycles as indicated below
f(1) = 4 f(3) = 7 f(5) = 6
f(4) = 9 f(7) = 8 f(6) = 5
f(9) = 2 f(8) = 3
f(2) = 1
Therefore, we write
f = (1,4,9,2)(3,7,8)(5,6)
10
Order of a Permutation
Theorem: Assume f = f1 ◦ f2 ◦ · · · ◦ fmwhere f1, f2, . . . , fm are disjoint cycles.
Then o(f) is the least common multiple
of o(f1), o(f2), . . . , o(fm). That is,
o(f) = lcm (o(f1), o(f2), . . . , o(fm)) .
Note: If f is an r-cycle, then o(f) = r.
11
Order of a Permutation
Example: Consider f ∈ S9 given by
f =(
1 2 3 4 5 6 7 8 94 1 7 9 6 5 8 3 2
).
As shown in the previous example, f can
be expressed as a composition of disjoint
cycles as follows:
f = (1,4,9,2)(3,7,8)(5,6).
Therefore,
o(f) = lcm (4, 3, 2) = 12
12
Cycles and Transpositions
Every cycle f = (i1, i2, . . . , ir) can be ex-
pressed as a product of transpositions as
follows:
f = (i1, ir) ◦ (i1, ir−1) ◦ · · · ◦ (i1, i3) ◦ (i1, i2)
Examples
• (1,3,2,4) = (1,4) (1,2) (1,3)
• (3,7,9,5,8) = (3,8) (3,5) (3,9) (3,7)
11
Permutations and Transpositions
Theorem 8.3: Every permutation f ∈ Sn,
where n ≥ 2, can be expressed as a product
of transpositions.
Proof: By Theorem 8.1,
f = f1 ◦ f2 ◦ · · · ◦ fm
where f1, f2, . . . , fm ∈ Sn are disjoint cy-
cles. Since each cycle fi can be expressed
as a product of transpositions (as described
above), this proves f is a product of trans-
positions.
12
Even/Odd Permutations
We say that a permutation f ∈ Sn is even
if it can be written as the product of an
even number of transpositions, and we say
f is odd if it can be written as the product
of an odd number of transpositions.
Theorem 8.4: No permutation is both
even and odd.
13
Product of Cycles
Example: Consider the permutation
f = (1,3,2,4)(1,7,6,2) ∈ S7
Express f as a product of disjoint cycles.
14
Even/Odd Permutations
Example: Consider the permutation
f = (1,3,2,4)(1,7,6,2) ∈ S7
Express f as a product of transpositions,
then determine if f is even or odd.
15
Alternating Groups
The alternating group of degree n, de-
noted An, is defined to be the subset of Sn
consisting of all even permutations.
16
Alternating Groups
Example: List all elements in S3 using cy-
cle notation, then express each element as
a product of transpositions. Finally, deter-
mine the elements of An.
17
Alternating Groups
Theorem 8.5: If n ≥ 2, then An is a
subgroup of Sn, and
|An| =n!
2=|Sn|
2
18
The symmetric group S3
◦ (1) (1,2) (1,3) (2,3) (1,2,3) (1,3,2)
(1) (1) (1,2) (1,3) (2,3) (1,2,3) (1,3,2)
(1,2) (1,2) (1) (1,3,2) (1,2,3) (2,3) (1,3)
(1,3) (1,3) (1,2,3) (1) (1,3,2) (1,2) (2,3)
(2,3) (2,3) (1,3,2) (1,2,3) (1) (1,3) (1,2)
(1,2,3) (1,2,3) (1,3) (2,3) (1,2) (1,3,2) (1)
(1,3,2) (1,3,2) (2,3) (1,2) (1,3) (1) (1,2,3)
19
S3 as a group of symmetries
If we associate the elements of the set
X = {1,2,3} with the vertices of an equi-
lateral triangle, then each element of the
permutation group S3 can be associated
with a symmetry of the triangle as follows.
22
S3 as a group of symmetries
Rotations:
Reflections:
23
Dihedral Groups
In general, the symmetries a regular n-gon
can be associated with a subgroup of Sn
called the dihedral group of order 2n. This
group of symmetries consists of n rotations
and n reflections, and is denoted by Dn.
If n = 3, the dihedral group D3 is identical
to S3. If n ≥ 4, the dihedral group Dn is a
nontrivial subgroup of Sn. We consider the
case n = 4 below.
24
The dihedral group D4
Rotations:
Reflections:
25
The dihedral group D4
Therefore, the elements of D4 are
{(1), (1,2,3,4), (1,3)(2,4), (1,4,3,2),
(1,2)(3,4), (2,4), (1,4)(2,3), (1,3)}
26
9 Equivalence Relations; Cosets
Relations
A relation on a nonempty set S is a nonempty
subset R of S × S.
Notation
If (x, y) ∈ R, we write xRy.
1
Relations
Example
Let S be a nonempty set, then any function
f ⊆ S × S is a relation. For example, if
S = R, then
f = {(x, x2) | x ∈ R} ⊆ R× R
is the relation corresponding to the func-
tion f(x) = x2.
2
Relations
Example
Let S = {1,2,3}. Then,
R = {(1,2), (1,3), (2,3)} ⊆ S × S
is a relation on S corresponding to the
“less than” relation. That is, for x, y ∈ S
xRy iff x < y.
3
Equivalence Relation
A relation R on S is called an equiva-
lence relation if the following three prop-
erties hold:
1. (Reflexivity)
xRx, for all x ∈ S.
2. (Symmetry)
if xRy, then y Rx.
3. (Transitivity)
if xRy and y R z, then xR z.
4
Equivalence Relation
Example
Let R be the relation on S = Z defined by
aR b iff a ≡ b (mod n)
where n is a fixed positive integer. Prove
that R is an equivalence relation.
5
Proof: Recall that a ≡ b (mod n) if and
only if b− a = nk for some integer k.
(Reflexivity)
(i) xRx, since x− x = 0 = n · 0.
(Symmetry)
(ii) Assume xRy. Then, x − y = nk for
some integer k. Therefore, y − x = n(−k),
where −k is an integer. Therefore, y Rx.
(Transitivity)
(iii) Assume xRy and y R z. Then, x−y =
nk for some integer k, and y − z = nj for
some integer j. Then,
x− z = (x− y) + (y − z) = nk + nj
That is, x− z = n(k + j) where k + j is an
integer. Therefore, xR z.
Equivalence Classes
Let R be an equivalence relation on a nonempty
set S. Given s ∈ S, we define
[s] = {x ∈ S | xR s}.
That is, [s] is the subset of elements in S
which are related to s.
The set [s] is called the equivalence class
of s under R. An element x ∈ S is called
a representative of [s] if x ∈ [s].
6
Equivalence Classes
Let R be the relation on S = Z defined by
aR b iff a ≡ b (mod 4)
Then,
[0] = {. . . ,−8,−4, 0, 4, 8, . . .}
[1] = {. . . ,−7,−3, 1, 5, 9, . . .}
[2] = {. . . ,−6,−2, 2, 6, 10, . . .}
[3] = {. . . ,−5,−1, 3, 7, 11, . . .}
where 0, 1, 2, and 3 are representatives of
their respective equivalence classes.
7
Equivalence Classes
Theorem 9.1 Let R be an equivalence
relation on S. Then, every element of S
is in exactly one equivalence class under R.
That is, the equivalence classes partition S
into a family of mutually disjoint nonempty
subsets.
Conversely, given any partition of S into
mutually disjoint nonempty subsets, there
is an equivalence relation on S whose equiv-
alence classes are precisely the subsets in
the given partition of S.
8
Proof: For every s ∈ S, we have s ∈ [s].
This shows every element of S is in at least
one equivalence class.
Next, suppose s ∈ [s1] and s ∈ [s2]. We
want to show [s1] = [s2]. Since s1 Rs (by
symmetry) and sR s2, it follows by transi-
tivity that s1 Rs2. Therefore, by a home-
work exercise, [s1] = [s2]. This shows that
each element in S is contained in exactly
one equivalence class.
Equivalence Relation
Theorem 9.2: Assume G is a group, and
let H be a subgroup G. Then the relation
≡H defined by
x ≡H y iff xy−1 ∈ H
is an equivalence relation.
9
Proof:
For any x ∈ G, we have xx−1 = e ∈ H.
Therefore, x ≡H x, which means ≡H is
reflexive.
Next, assume x ≡H y. Then, xy−1 ∈ H,
and since H is closed under inverses, we
also have
yx−1 = (y−1)−1x−1 = (xy−1)−1 ∈ H
That is, yx−1 ∈ H, which means y ≡H x.
This proves ≡H is symmetric.
Finally, assume x ≡H y and y ≡H z. This
means, xy−1 ∈ H and yz−1 ∈ H. Since H is
closed under multiplication, we also have
xz−1 = (xy−1)(yz−1) ∈ H
Therefore, x ≡H z, which proves ≡H is
transitive.
Cosets
Assume G is a group, and let H be a sub-
group of G. Given any fixed element a ∈ G,
the set
aH = {x ∈ G | x = ah for some h ∈ H}
is called a left coset of H in G.
Similarly, the set
Ha = {x ∈ G | x = ha for some h ∈ H}
is called a right coset of H in G.
10
Cosets
Example: Let G = S3, and consider the
subgroup H = {(1), (1,2)}. Then, for the
fixed element a = (1,2,3) ∈ G, we have
the left coset
aH = {(1,2,3)(1), (1,2,3)(1,2)}
and the right coset
Ha = {(1)(1,2,3), (1,2)(1,2,3)}
11
Cosets
Theorem 9.3: Assume G is a group, and
let H be a subgroup of G. If a ∈ G and [a]
denotes the equivalence class of a under
the equivalence relation ≡H. Then,
[a] = Ha
That is, the equivalence classes of ≡H are
precisely the right cosets of H.
12
Proof: Assume G is a group, and let H be
a subgroup of G. If a ∈ G, we have
x ∈ [a] iff x ≡H a
iff xa−1 ∈ H
iff xa−1 = h for some h ∈ H
iff x = ha for some h ∈ H
iff x ∈ Ha.
Therefore, [a] = Ha.
Cosets
Corollary 9.4: Assume G is a group, and
let H be a subgroup of G. Then for all
a, b ∈ G,
Ha = Hb if and only if ab−1 ∈ H.
13
Proof: Assume G is a group, and let H be
a subgroup of G. For all a, b ∈ G, we have
Ha = Hb iff [a] = [b]
iff a ≡H b
iff ab−1 ∈ H
That is, Ha = Hb if and only if ab−1 ∈ H.
Cosets
Theorem 9.5: Assume G is a group, and
let H be a subgroup of G. If a ∈ G and [a]
denotes the equivalence class of a under
the equivalence relation H ≡, defined by
x H ≡ y iff x−1y ∈ H,
then [a] = aH. That is, the equivalence
classes of H ≡ are precisely the left cosets
of H. Furthermore, we have
[a] = [b] iff aH = bH iff a−1b ∈ H.
14
Cosets
Example: Let G = S3, and consider the
subgroup H = {(1), (1,2)}. Then, the left
cosets of H are
(1)H = {(1)(1), (1)(1,2)}
(1,2)H = {(1,2)(1), (1,2)(1,2)}
(1,3)H = {(1,3)(1), (1,3)(1,2)}
(2,3)H = {(2,3)(1), (2,3)(1,2)}
(1,2,3)H = {(1,2,3)(1), (1,2,3)(1,2)}
(1,3,2)H = {(1,3,2)(1), (1,3,2)(1,2)}
15
Cosets
Example: Let G = S3, and consider the
subgroup H = {(1), (1,2)}. Then, the left
cosets of H are
(1)H = {(1)(1), (1)(1,2)} = {(1), (1,2)}
(1,2)H = {(1,2)(1), (1,2)(1,2)} = {(1), (1,2)}
(1,3)H = {(1,3)(1), (1,3)(1,2)} = {(1,3), (1,2,3)}
(2,3)H = {(2,3)(1), (2,3)(1,2)} = {(2,3), (1,3,2)}
(1,2,3)H = {(1,2,3)(1), (1,2,3)(1,2)} = {(1,3), (1,2,3)}
(1,3,2)H = {(1,3,2)(1), (1,3,2)(1,2)} = {(2,3), (1,3,2)}
16
Cosets
Example: Let G = S3, and consider the
subgroup H = {(1), (1,2)}. Then, the right
cosets of H are
H(1) = {(1)(1), (1,2)(1)}
H(1,2) = {(1)(1,2), (1,2)(1,2)}
H(1,3) = {(1)(1,3), (1,2)(1,3)}
H(2,3) = {(1)(2,3), (1,2)(2,3)}
H(1,2,3) = {(1)(1,2,3), (1,2)(1,2,3)}
H(1,3,2) = {(1)(1,3,2), (1,2)(1,3,2)}
17
Cosets
Example: Let G = S3, and consider the
subgroup H = {(1), (1,2)}. Then, the right
cosets of H are
H(1) = {(1)(1), (1,2)(1)} = {(1), (1,2)}
H(1,2) = {(1)(1,2), (1,2)(1,2)} = {(1), (1,2)}
H(1,3) = {(1)(1,3), (1,2)(1,3)} = {(1,3), (1,3,2)}
H(2,3) = {(1)(2,3), (1,2)(2,3)} = {(2,3), (1,2,3)}
H(1,2,3) = {(1)(1,2,3), (1,2)(1,2,3)} = {(2,3), (1,2,3)}
H(1,3,2) = {(1)(1,3,2), (1,2)(1,3,2)} = {(1,3), (1,3,2)}
18
Cosets
Remark: If G is an abelian group, and H
is a subgroup of G. Then the left cosets
and right cosets of H are identical. That
is aH = Ha for all a ∈ G.
Example: Let G = (Z,+), and consider
the subgroup H = 4Z. The left and right
cosets of H are given by
0 + 4Z = {. . . ,−8,−4, 0, 4, 8, . . .} = 4Z + 0
1 + 4Z = {. . . ,−7,−3, 1, 5, 9, . . .} = 4Z + 1
2 + 4Z = {. . . ,−6,−2, 2, 6, 10, . . .} = 4Z + 2
3 + 4Z = {. . . ,−5,−1, 3, 7, 11, . . .} = 4Z + 3
Also note that
a+4Z = b+4Z iff a−b ∈ 4Z iff a ≡ b (mod 4).
19
10 Counting the Elements of a
Finite Group
Lagrange’s Theorem
Theorem 10.1 (Lagrange’s Theorem)
Let G be a finite group, and assume H is a
subgroup of G. Then, |H| divides |G|.
1
Lagrange’s Theorem
Lemma 10.2: Let G be a group, and as-
sume H is a subgroup of G. Then, for all
a, b ∈ G, there is a one-to-one correspon-
dence between the elements of the right
coset Ha and the right coset Hb.
In particular, if G is a finite group then we
write |Ha| = |Hb|.
2
Lagrange’s Theorem
Proof: Let H be a subgroup of G, and
assume a, b ∈ G. We consider the function
f : Ha→ Hb defined by
f(ha) = hb
We want to show f is one-to-one and onto.
First, observe that
f(h1a) = f(h2a)
iff h1b = h2b
iff (h1b)b−1 = (h2b)b
−1
iff h1 = h2
iff h1a = h2a
This proves f is one-to-one. Next given
any element hb ∈ Hb, we have f(x) = hb
when x = ha. Hence, f is onto.
3
Lagrange’s Theorem
Proof of Lagrange’s Theorem: Assume H
is a subgroup of a finite group G. Let ≡H
denote the equivalence relation given by
a ≡H b iff ab−1 ∈ H.
By Theorem 9.1, the equivalence classes
under ≡H partition G into mutually dis-
joint nonempty subsets, and by Theorem
9.3, these equivalence classes are just the
right cosets of H. Since G is finite, there
are finitely many distinct right cosets which
we denote by Ha1, Ha2, . . . , Hak. Thus,
|G| = |Ha1 ∪ Ha2 ∪ · · · ∪ Hak|
= |Ha1|+ |Ha2|+ · · ·+ |Hak|
= |H|+ |H|+ · · ·+ |H|︸ ︷︷ ︸k terms
= k|H|.
4
Note that the third equality uses the fact
(Lemma 10.2) that all right cosets have the
same size; that is,
|Hai| = |He| = |H|
for all i = 1,2, . . . , k.
Since we’ve shown |G| = k|H|, we conclude
that |H| divides |G|.
Index of a Subgroup
If G is any group (not necessarily finite)
and H is any subgroup, then the number of
distinct right cosets of H in G is called the
index of H in G. We denote this number
by [G : H].
In particular, if G is a finite group, the proof
of Lagrange’s Theorem shows that
|G| = [G : H] · |H|
5
Index of a Subgroup
Example
Let G = S3 and consider H = {(1), (1,2)}.Then, H has 3 distinct right cosets, namely,
H(1) = {(1), (1,2)}
H(1,2,3) = {(2,3), (1,2,3)}
H(1,3,2) = {(1,3), (1,3,2)}
Therefore, [G : H] = 3, that is, the index
of H in G is 3. Also, note that |H| divides
|G|. Specifically, we have
|G| = 6 = 3 · 2 = [G : H] · |H|
6
Index of a Subgroup
Example
Let G = (Z,+), and consider the subgroup
H = 4Z. Then H has 4 distinct right
cosets, namely,
4Z + 0 = {. . . ,−8,−4, 0, 4, 8, . . .}
4Z + 1 = {. . . ,−7,−3, 1, 5, 9, . . .}
4Z + 2 = {. . . ,−6,−2, 2, 6, 10, . . .}
4Z + 3 = {. . . ,−5,−1, 3, 7, 11, . . .}
Therefore, [G : H] = 4, that is, the index
of H in G is 4.
In this case, Lagrange’s theorem does not
apply since G has infinite order.
7
Index of a Subgroup
Example
Let G = (R,+), and consider the subgroup
H = Z. Then H has infinitely many dis-
tinct right cosets since H + x 6= H + y for
all x, y ∈ [0,1) such that x 6= y.
In this case, we write [G : H] =∞.
8
Index of a Subgroup
Theorem 10.3 Let H be a subgroup of
G. Then the number of distinct left cosets
of H in G is [G : H].
9
Proof: Assume S is the set of all right
cosets of H and assume T is the set of all
left cosets of H. We will show that the
function f : T → S defined by
f(aH) = Ha−1
is a bijection. First, by Corollary 9.4 and
Theorem 9.5, we have
f(aH) = f(bH)
iff Ha−1 = Hb−1
iff a−1b ∈ H
iff aH = bH
This proves f is one-to-one. Next given
any element Ha ∈ S, we have f(x) = Ha
when x = a−1H ∈ T . Hence, f is onto.
This proves f is a bijection.
Applications of Lagrange’s Theorem
Theorem 10.4 Let G be a finite group
and let x ∈ G. Then o(x) divides |G|.Consequently, x|G| = e for every x ∈ G.
10
Applications of Lagrange’s Theorem
Proof: Assume G is a finite group and let
x ∈ G. Consider the subgroup H = 〈x〉.By Lagrange’s Theorem, |H| divides |G|.Since |H| = |〈x〉| = o(x), this proves o(x)
divides |G|.
Finally, if we write |G| = o(x) · k for some
integer k ∈ Z+, we have
x|G| = xo(x)·k = (xo(x))k = ek = e.
This completes the proof.
11
Applications of Lagrange’s Theorem
Theorem 10.5 Let G be a finite group
and assume that |G| is prime. Then G is
cyclic, and any element of G other than
the identity element e is a generator for
G.
12
Applications of Lagrange’s Theorem
Proof: Assume |G| = p, where p > 1 is
a prime number. Then given any x ∈ G,
we have |〈x〉| divides p. If x 6= e, we know
{e, x} ⊆ 〈x〉. Hence |〈x〉| ≥ 2. Since the
only divisors of p are 1 and p, we conclude
that |〈x〉| = p. This proves that G is cyclic
and that 〈x〉 = G for all x 6= e.
13
Applications of Lagrange’s Theorem
Example: Consider the set
G = Zp − {0} = {1,2,3, . . . , p− 1},
where p is a prime number. Then, G
forms a group under multiplication mod-
ulo p. That is, G is a group with respect
to the binary operation � defined by
a� b = ab.
where ab is the remainder of ab mod p.
14
Proof that (Zp − {0},�) is a group.
(i) (Closure) Given a, b ∈ Zp−{0}, we want
to show a� b ∈ Zp − {0}. Clearly, we have
a� b = ab ∈ Zp
It suffices to show ab 6= 0. If not, then
ab = np for some integer p. That is, p | ab.Since p is prime, this means p | a or p | b.Therefore, a = 0 or b = 0, which contra-
dicts the assumption a, b ∈ Zp − {0}. This
proves closure.
(ii) (Associativity) See Homework
(iii) (Identity element) There exists an ele-
ment e ∈ Zp−{0}, namely e = 1, such that
e� a = 1� a = a = a
anda� e = a� 1 = a = a
Therefore, Zp−{0} has an identity element.
(iv) (Inverse elements) We want to show
that for any a ∈ Zp − {0}, there exists
b ∈ Zp − {0} such that a � b = 1. Since
gcd(a, p) = 1, it follows by Theorem 4.2,
that there exist x, y ∈ Z such that
ax + py = 1
Therefore,
a� x = ax = 1− py = 1.
Thus, b = x is the inverse element of a.
Applications of Lagrange’s Theorem
Theorem 10.6 (Fermat’s Theorem)
Let p be a prime number and suppose a is
an integer such that p does not divide a.
Then,
ap−1 ≡ 1 (mod p)
15
Proof: Let p be a prime number and sup-
pose a is an integer such that p does not di-
vide a. By the division algorithm, we have
a = qp + r where q ∈ Z and
r ∈ {1,2,3, . . . , p− 1}
Therefore,
ap−1 ≡ (qp + r)p−1 (mod p)
≡ rp−1 (mod p)
where the last congruence follows from the
binomial theorem.
Finally, since G = Zp − {0} forms a group
under multiplication modulo p, it follows
from the previous example and Theorem
10.5 that
rp−1 = r � · · · � r︸ ︷︷ ︸(p− 1)-times
= r|G| = e.
That is,
rp−1 ≡ 1 (mod p)
This completes the proof.
Centralizer of an Element
Assume G is a group. Given a fixed ele-
ment g ∈ G, we define the centralizer of
g to be the set
Z(g) = {x ∈ G | xg = gx}.
That is, Z(g) is the set of elements in G
that commute with g.
Note that all integer powers of g are con-
tained in Z(g). In particular this includes
g−1, g0 = e, and g1 = g.
Previously we proved that Z(g) is a sub-
group of G.
16
Centralizer of an Element
Example
Consider the group G = S3. Then,
Z((1)) =
Z((1,2)) =
Z((1,3)) =
Z((2,3)) =
Z((1,2,3)) =
Z((1,3,2)) =
17
Center of a Group
Assume G is a group. The center of G,
denoted by Z(G), is the set of elements in
G that commute with all elements of G.
Explicitly, we have
Z(G) = {x ∈ G | xg = gx for all g ∈ G}.
That is, x ∈ Z(G) if and only if x ∈ Z(g)
for all g ∈ G.
Note that a group G is abelian if and only
if Z(G) = G.
18
Center of a Group
Examples
• If G is abelian, then Z(G) = G.
• If G = S3, then Z(G) = {(1)}
• If G = Q8, then Z(G) = {I,−I}
19
Conjugacy Classes
Assume G is a group, and consider the
relation R on G defined by
aR b iff a = xbx−1 for some x ∈ G.
If aR b, we say that a is conjugate to b.
Lemma 10.7: The above relation R is
an equivalence relation on G.
Proof: Homework 7, Problem 4.
20
Conjugacy Classes
The equivalence class [a] of an element
a ∈ G under R is called the conjugacy
class of a and consists of all conjugates
of a. That is,
[a] = {xax−1 | x ∈ G}
Lemma 10.7: Let G be a finite group
and let a ∈ G. Then the number of distinct
conjugates of a in G is [G : Z(a)].
21
Proof: For all x, y ∈ G, we have
xax−1 = yay−1 iff ax−1y = x−1ya
iff x−1y ∈ Z(a)
iff x Z(a)≡ y
iff xZ(a) = yZ(a).
Therefore, the number of distinct conju-
gates of a corresponds to the number of
distinct left cosets of Z(a) which, by The-
orem 10.3, is [G : Z(a)].
Conjugacy Classes
Example Consider the group G = S3.
Complete the table below. Then write G as
the disjoint union of its conjugacy classes.
a Z(a) [G : Z(a)] [a]
(1)
(1,2)
(1,3)
(2,3)
(1,2,3)
(1,3,2)
22
Conjugacy Classes
Let’s compute the conjugates of a = (1,2) :
(1)(1,2)(1)−1 = (1,2)
(1,2)(1,2)(1,2)−1 = (1,2)
(1,3)(1,2)(1,3)−1 = (2,3)
(2,3)(1,2)(2,3)−1 = (1,3)
(1,2,3)(1,2)(1,2,3)−1 = (2,3)
(1,3,2)(1,2)(1,3,2)−1 = (1,3)
Therefore, the conjugacy class of a (the set
of distinct conjugates of a) is:
[a] = {(1,2), (1,3), (2,3)}
23
Conjugacy Classes
Next compute the conjugates of a = (1,2,3) :
(1)(1,2,3)(1)−1 = (1,2,3)
(1,2)(1,2,3)(1,2)−1 = (1,3,2)
(1,3)(1,2,3)(1,3)−1 = (1,3,2)
(2,3)(1,2,3)(2,3)−1 = (1,3,2)
(1,2,3)(1,2,3)(1,2,3)−1 = (1,2,3)
(1,3,2)(1,2,3)(1,3,2)−1 = (1,2,3)
Therefore, the conjugacy class of a (the set
of distinct conjugates of a) is:
[a] = {(1,2,3), (1,3,2)}
24
Conjugacy Classes
Important Fact: All conjugates of a have
the same order as a. (Homework 4, Prob-
lem 5)
25
Conjugacy Classes
Example Consider the group G = S3.
Complete the table below. Then write G as
the disjoint union of its conjugacy classes.
a Z(a) [G : Z(a)] [a]
(1) S3 1 {(1)}
(1,2) {(1), (1,2)} 3 {(1,2), (1,3), (2,3)}
(1,3) {(1), (1,3)} 3 {(1,2), (1,3), (2,3)}
(2,3) {(1), (2,3)} 3 {(1,2), (1,3), (2,3)}
(1,2,3) {(1), (1,2,3), (1,3,2)} 2 {(1,2,3), (1,3,2)}
(1,3,2) {(1), (1,2,3), (1,3,2)} 2 {(1,2,3), (1,3,2)}
G = {(1)} ∪ {(1,2), (1,3), (2,3)} ∪ {(1,2,3), (1,3,2)}
= [(1)] ∪ [(1,2)] ∪ [(1,2,3)]
26
Class Equation
Theorem 10.9 (Class equation): Let G
be a finite group, and let {a1, a2, . . . , ak}consist of one element from each conju-
gacy class containing at least two elements.
Then,
|G| = |Z(G)|+[G : Z(a1)]+· · ·+[G : Z(ak)].
27
Proof: Since G is the disjoint union of its
conjugacy classes, we have
G = [a1] ∪ · · · ∪ [ak] ∪ {ak+1} · · · ∪ {ak+s}
where {ak+1}, . . . , {ak+s} are the conjugacy
classes with one element. Therefore, by
Lemma 10.8, we have:
|G| = [G : Z(a1)] + · · ·+ [G : Z(ak)] + s.
It suffice to show that s = |Z(G)|. By
Lemma 10.8,
[a] = {a} iff [G : Z(a)] = 1
iff Z(a) = G
iff a ∈ Z(G)
Therefore, the conjagacy classes with a sin-
gle element correspond precisely to the ele-
ments of Z(G). This completes the proof.
Class Equation
Example Consider the group G = S3. In
the example above, we showed that
G = {(1)} ∪ {(1,2), (1,3), (2,3)} ∪ {(1,2,3), (1,3,2)}
= [(1)] ∪ [(1,2)] ∪ [(1,2,3)]
Let a1 = (1,2) and a2 = (1,2,3) represent
the conjugacy classes of size greater than
1, and note that Z(G) = {(1)}. Then,
|G| = |Z(G)|+ [G : Z(a1)] + [G : Z(a2)]
= |{(1)}|+ [G : Z((1,2))] + [G : Z((1,2,3))]
= 1 + 3 + 2
= 6
28
11 Normal Subgroups
Conjugate Subgroups
Theorem 11.4: Let G be a group, and
assume H is a subgroup of G. Then, for
each fixed element g ∈ G, the set
gHg−1 = {ghg−1 | h ∈ H}
is a subgroup of G with the same number
of elements as H.
We say that gHg−1 is a conjugate sub-
group of H.
1
Proof: The proof that gHg−1 is a subgroup
is Exercise 9, Homework 4. The proof that
gHg−1 has the same number of elements
as H is also left as an exercise.
Normal Subgroups
Assume H is a subgroup of G. We say that
H is a normal subgroup if ghg−1 ∈ H for
all h ∈ H and for all g ∈ G.
2
Normal Subgroups
Theorem 11.1: Let H be a subgroup of
G. Then the following are equivalent:
(i) H is a normal subgroup;
(ii) gHg−1 = H for all g ∈ G;
(iii) gH = Hg for all g ∈ G, i.e., the left
cosets and right cosets of H are iden-
tical.
3
Proof:
First, we will prove statements (i) and (ii)
are equivalent.
Assume (ii) holds. Then for all g ∈ G, we
have gHg−1 = H. In particular, gHg−1 ⊆H, which gives (i).
Conversely, assume (i) holds. Then for all
g ∈ G, we have gHg−1 ⊆ H, and we have
g−1H(g−1)−1 ⊆ H
iff H(g−1)−1 ⊆ gH
iff H ⊆ gHg−1
Therefore, gHg−1 = H for all g ∈ G, which
proves (ii).
Next, we will prove statements (ii) and (iii)
are equivalent. For all g ∈ G, we have
gHg−1 = H iff gH = Hg
where the second equality is obtained from
the first after multiplication by g on the
right.
This completes the proof.
Normal Subgroups
Example
Let G = S3. Then
H = 〈(1,2,3)〉 = {(1), (1,2,3), (1,3,2)}
is a normal subgroup since the left cosets
and right cosets of H are identical.
(1)H = {(1), (1,2,3), (1,3,2)} = H(1)
(1,2)H = {(1,2), (1,3), (2,3)} = H(1,2)
(1,3)H = {(1,2), (1,3), (2,3)} = H(1,3)
(2,3)H = {(1,2), (1,3), (2,3)} = H(2,3)
(1,2,3)H = {(1), (1,2,3), (1,3,2)} = H(1,2,3)
(1,3,2)H = {(1), (1,2,3), (1,3,2)} = H(1,3,2)
4
Normal Subgroups
Example
Let G = S3. Then
H = 〈(1,2)〉 = {(1), (1,2)}
is not a normal subgroup since the left
cosets and right cosets of H are not iden-
tical. For example, the left coset
(1,3)H = {(1,3), (1,2,3)}
does not match the right coset
H(1,3) = {(1,3), (1,3,2)}.
5
Normal Subgroups
Notation
We write H � G to indicate that H is a
normal subgroup of G.
6
Normal Subgroups
Example
Let G and H be groups, then
{eG} ×H � G×H, and
G× {eH} � G×H.
Observe that for all (eG, x) ∈ {eG} ×H and
for all (g, h) ∈ G×H,
(g, h)(eG, x)(g, h)−1 = (geGg−1, hxh−1)
= (eG, hxh−1)
∈ {eG} ×H.
This proves {eG}×H is a normal subgroup.
A similar calculation shows that G × {eH}is normal.
7
Normal Subgroups
Theorem 11.2 Let G be a group. Then
any subgroup of Z(G) is a normal subgroup
of G.
8
Proof: Let H be a subgroup of Z(G).
Then each element of H commutes with
all elements of G. Therefore, for all g ∈ G,
and for all h ∈ H, we have
ghg−1 = hgg−1 = h ∈ H.
This proves H �G.
Normal Subgroups
Corollary: If G is abelian, then every sub-
group of G is normal.
Proof: If G is abelian, then Z(G) = G, so
every subgroup of G is a subgroup of Z(G).
9
Normal Subgroups
Theorem 11.3: Let H be a subgroup of
G such that [G : H] = 2. Then H is normal
in G.
10
Proof: We want to show that for all g ∈ G,
gH = Hg. Since [G : H] = 2, there are
exactly two left cosets and two right cosets
of H. These cosets must be H and G−H.
If g ∈ H, then
gH = H = Hg.
If g /∈ H, then
gH = G−H = Hg.
Therefore, for all g ∈ G, gH = Hg.
Normal Subgroups
Example
Let G = Q8. Then the subgroups
〈J〉 = {I,−I, J,−J},
〈K〉 = {I,−I,K,−K},
〈L〉 = {I,−I, L,−L},
are all normal in G, since
[Q8 : 〈J〉] = [Q8 : 〈K〉] = [Q8 : 〈L〉] = 2.
11
Normal Subgroups
Example
Let G = Sn. Then the alternating group
An is normal in G, since it follows from
Theorem 8.5 that [Sn : An] = 2.
12
Normal Subgroups
Corollary 11.5: Let G be a group, and
assume H is a subgroup of G. If G has
only one subgroup of size |H|, then H is
normal in G.
13
Proof: By Theorem 11.4, gHg−1 is a sub-
group of G of size |H|. If G has only one
subgroup of size |H|, then gHg−1 = H.
Therefore, by Theorem 11.1, H is normal.
Normal Subgroups
Example
Let G = A4, and consider the subgroup
H = {(1), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}.
We have |H| = 4 and the group G has no
other subgroups of order 4. This follows
from the fact that every element of G not
contained in H is a 3-cycle and these el-
ements cannot be contained in a subgroup
of order 4. Therefore, by Corollary 11.5,
H is normal in G.
14
Quotient Groups
Theorem 11.6: If H � G, then the set
of right cosets of H in G, denoted
G/H = {Ha | a ∈ G},
is group with respect to the operation ∗,defined by
Ha ∗Hb = H(ab).
Furthermore, |G/H| = [G : H].
Notation
We say that G/H is the quotient group
of G by H.
15
Proof: First we must show that ∗ is a
well-defined binary operation when H is a
normal subgroup. That is, we must show
Ha1 ∗Hb1 = Ha2 ∗Hb2
when Ha1 = Ha2 and Hb1 = Hb2. We have
Ha1 ∗Hb1 = Ha2 ∗Hb2
iff H(a1b1) = H(a2b2)
iff (a1b1)(a2b2)−1 ∈ H
iff a1b1b−12 a−1
2 ∈ H
iff a1b1b−12 a−1
1 a1a−12 ∈ H
iff [a1(b1b−12 )a−1
1 ](a1a−12 ) ∈ H.
Since the last statement is satisfied when
Ha1 = Ha2 and Hb1 = Hb2 (that is, when
a1a−12 ∈ H and b1b
−12 ∈ H), it follows that
∗ is well-defined.
Next we must verify the four conditions of
a group.
(i) (Closure) Given Ha,Hb ∈ G/H, we
have Ha ∗ Hb = H(ab) ∈ G/H. This
proves closure.
(ii) (Associativity) Given Ha,Hb,Hc ∈ G/H,
we have
(Ha ∗Hb) ∗Hc = H(ab) ∗Hc
= H[(ab)c]
= H[a(bc)]
= Ha ∗H(bc)
= Ha ∗ (Hb ∗Hc)
Therefore, ∗ is associative.
(iii) (identity element) There exists an el-
ement He ∈ G/H such that
He ∗Ha = H(ea) = Ha; and
Ha ∗He = H(ae) = Ha,
for all Ha ∈ G/H.
(iv) (inverse elements) The inverse of the
element Ha ∈ G/H is the element
Ha−1 since
Ha ∗Ha−1 = H(aa−1)
= He
= H(a−1a)
= Ha−1 ∗Ha.
This proves that G/H is a group. The
fact that |G/H| = [G : H] follows immedi-
ately from the definition of G/H.
Quotient Groups
Example
Let G = S3, and consider the normal sub-
group
H = 〈(1,2,3)〉 = {(1), (1,2,3), (1,3,2)}.
Then,
G/H = {H(1), H(1,2)}
forms a group with the following multipli-
cation table
∗ H(1) H(1,2)
H(1)
H(1,2)
16
Quotient Groups
Example
Let G = Q8, and consider the normal sub-
group
H = Z(Q8) = {I,−I}.
Then,
G/H = {HI, HJ, HK, HL}
forms a group with the following multipli-
cation table
∗ HI HJ HK HL
HI
HJ
HK
HL
17
Quotient Groups
Example
Let G = (Z,+), and consider the normal
subgroup
H = 〈n〉 = nZ.
Then,
G/H = {nZ+ 0, nZ+ 1, . . . , nZ+ (n−1)}
forms a group of size [Z : nZ] = n.
We will prove later that Z / nZ is equivalent
to the group Zn.
18
Quotient Groups
Theorem 11.7 Let G be a finite abelian
group and suppose p is a prime number
that divides |G|. Then G has a subgroup
of order p.
19
Proof:
We will proof the result using (strong) in-
duction on the size of the group G.
Base Case: If |G| = 1, then the result is
true since no prime p divides |G|.
Inductive Step: Assume the theorem is true
when G is abelian and |G| < m. We want to
show that this implies the theorem is true
when G is abelian and |G| = m.
Let G be an abelian group of order m, and
let x 6= e be an element of G. Since G is
abelian, H = 〈x〉 is a normal subgroup of
G, and by Lagrange’s Theorem
|G| = |G/H| · |H|.
Now, assume p is prime and p divides |G|.Then, p divides |H|, or p divides |G/H|. In
the former case, if we write |H| = kp, then
〈xk〉 is a subgroup of G of order p, since
o(xk) =o(x)
gcd(o(x), k)=
|H|gcd(|H|, k)
=|H|k
= p.
On the other hand, if p divides |G/H|, then
by the inductive hypothesis, the abelian group
G/H has a subgroup of order p. Since p is
prime this subgroup is cyclic and can be ex-
pressed in the form 〈Hg〉, where o(Hg) = p.
It follows from a homework exercise that p
divides o(g); that is o(g) = kp for some in-
teger k. Therefore, 〈gk〉 is a subgroup of G
of order p, and the proof is complete.
12 Homomorphisms
Definition
Let G and H be groups. A function
ϕ : G→ H
is called a homomorphism if
ϕ(ab) = ϕ(a)ϕ(b)
for all a, b ∈ G.
A one-to-one homomorphism is called a
monomorphism. An onto homomorphism
is called an epimorphism.
1
Homomorphism
Example
Let G = (Z,+) and let H = (2Z,+). Then
the function ϕ : G→ H defined by
ϕ(n) = 2n
is a homomorphism since
ϕ(n+m) = 2(n+m)
= 2n+ 2m
= ϕ(n) + ϕ(m)
for all n,m ∈ Z.
2
Homomorphism
Example
Let G = (Z,+) and let H = (Zn,⊕). Then
the function ϕ : G→ H defined by
ϕ(m) = m,
where m is the remainder of m modulo n,
is a homomorphism since
ϕ(m1 +m2) = m1 +m2
= m1 ⊕m2
= ϕ(m1)⊕ ϕ(m2)
for all m1,m2 ∈ Z.
3
Homomorphism
Example
Let G = (Sn, ◦) and let H = (Z2,⊕), and
consider the function ϕ : G→ H defined by
ϕ(f) =
0 if f is even
1 if f is odd
To determine if ϕ is a homomorphism, we
consider the following chart.
4
f g f ◦ g ϕ(f) ϕ(g) ϕ(f ◦ g)
even even
even odd
odd even
odd odd
Homomorphism
Example
Let G = (Sn, ◦) and let H = (Z2,⊕), and
consider the function ϕ : G→ H defined by
ϕ(f) =
0 if f is even
1 if f is odd
To determine if ϕ is a homomorphism, we
consider the following chart.
5
f g f ◦ g ϕ(f) ϕ(g) ϕ(f ◦ g)
even even even 0 0 0
even odd odd 0 1 1
odd even odd 1 0 1
odd odd even 1 1 0
In all cases,
ϕ(f ◦ g) = ϕ(f)⊕ ϕ(g).
Therefore, ϕ is homomorphism.
Homomorphism
Theorem 12.4: Let ϕ : G → H be a
homomorphism. Then
(i) ϕ(eG) = eH
(ii) ϕ(xn) = [ϕ(x)]n for all x ∈ G and for
all n ∈ Z.
(iii) if o(x) = n, then o(ϕ(x)) divides n.
6
Proof of (i):
Since ϕ is a homomorphism, we have
ϕ(eG) = ϕ(eG ∗ eG) = ϕ(eG) ∗ ϕ(eG)
That is,
ϕ(eG) = ϕ(eG) ∗ ϕ(eG).
Multiplying both sides of this equation on
the left by ϕ(eG)−1, we obtain
eH = ϕ(eG).
Proof of (ii):
Case 1: If n = 0, then by part (i)
ϕ(x0) = ϕ(eG) = eH = (ϕ(x))0,
for all x ∈ G.
Case 2: We will use induction to prove the
statement holds for all n > 0.
Base Case: When n = 1, we have
ϕ(x1) = ϕ(x) = ϕ(x)1.
Inductive Step: Assume
ϕ(xk) = ϕ(x)k.
for some integer k. We want to show the
statement also holds for n = k + 1.
We have,
ϕ(xk+1) = ϕ(x · xk)
= ϕ(x)ϕ(xk)
= ϕ(x)ϕ(x)k
= ϕ(x)k+1
This proves, ϕ(xk+1) = ϕ(x)k+1 and it fol-
lows by induction that
ϕ(xn) = ϕ(x)n
for all x ∈ G, and for all n ∈ Z+.
Case 3: If n < 0, then write n = −mwhere m > 0, and observe that
ϕ(xn)ϕ(xm) = ϕ(xn · xm)
= ϕ(x−m · xm)
= ϕ(eG)
= eH
Therefore,
ϕ(xn) = [ϕ(xm)]−1
= [[ϕ(x)]m]−1
= [ϕ(x)]−m
= [ϕ(x)]n.
This completes the proof of (ii).
Proof of (iii):
If x ∈ G and o(x) = n, then by parts (i)
and (ii) we have
[ϕ(x)]n = ϕ(xn) = ϕ(eG) = eH .
That is, [ϕ(x)]n = eH, and it follows by
Theorem 4.4(ii) that o(ϕ(x)) divides n.
Homomorphism
Example
Let G = (Z12,⊕) and let H = (Z4,⊕).
Consider the homomorphism ϕ : G → H
defined by
ϕ(x) = x,
where x is the remainder of x mod 4.
• To illustrate Theorem 12.4(iii), let x = 10.
Then ϕ(x) = 2,
o(x) =12
gcd(12,10)=
12
2= 6,
o(ϕ(x)) =4
gcd(4,2)=
4
2= 2,
and we see that o(ϕ(x)) divides o(x).
7
• To illustrate Theorem 12.4(ii), let x = 10,
and let n = 3. Then
x3 = 10⊕ 10⊕ 10 = 6 ∈ Z12
Therefore,
ϕ(x3) = ϕ(6) = 2 ∈ Z4
On the other hand,
ϕ(x)3 = ϕ(x)⊕ϕ(x)⊕ϕ(x) = 2⊕2⊕2 = 2 ∈ Z4.
This verifies that
ϕ(x3) = 2 = ϕ(x)3
Isomorphism
A homomorphism ϕ : G → H is called an
isomorphism if ϕ is one-to-one and onto.
In this case we say G is isomorphic to H
and write G ∼= H.
8
Isomorphism
Example
Let G = (R,+) and let H = (R+, ·). Then
the function ϕ : G→ H defined by
ϕ(x) = ex
is an isomorphism since ϕ is one-to-one
and onto, and
ϕ(x+ y) = ex+y
= exey
= ϕ(x)ϕ(y)
for all x, y ∈ R.
Therefore, we write (R,+) ∼= (R+, ·).
9
Isomorphism
If G and H are finite groups, and
ϕ : G→ H
is an isomorphism, then the condition
ϕ(ab) = ϕ(a)ϕ(b)
means that the multiplication table for H
is obtained from the multiplication table for
G by replacing each entry in the latter table
with its image under ϕ.
10
Isomorphism
Example
Consider the Klein 4-group V = {e, a, b, c}.The multiplication table for V is given by
∗ e a b c
e e a b c
a a e c b
b b c e a
c c b a e
11
Consider the function ϕ : V → Z2 × Z2,
defined by
ϕ(e) = (0,0) ϕ(b) = (0,1)
ϕ(a) = (1,0) ϕ(c) = (1,1)
Then the image under ϕ of the entries in
the multiplication table for V produces the
multiplication table for Z2 × Z2 as shown
below
ϕ(e) = (0,0) ϕ(b) = (0,1)
ϕ(a) = (1,0) ϕ(c) = (1,1)
∗ ϕ(e) ϕ(a) ϕ(b) ϕ(c)
ϕ(e) ϕ(e) ϕ(a) ϕ(b) ϕ(c)
ϕ(a) ϕ(a) ϕ(e) ϕ(c) ϕ(b)
ϕ(b) ϕ(b) ϕ(c) ϕ(e) ϕ(a)
ϕ(c) ϕ(c) ϕ(b) ϕ(a) ϕ(e)
⊕ (0,0) (1,0) (0,1) (1,1)
(0,0) (0,0) (1,0) (0,1) (1,1)
(1,0) (1,0) (0,0) (1,1) (0,1)
(0,1) (0,1) (1,1) (0,0) (1,0)
(1,1) (1,1) (0,1) (1,0) (0,0)
Since the group tables are equivalent via
the map ϕ, and since ϕ is one-to-one and
onto, it follows that ϕ is an isomorphism.
Automorphism
Let G be a group. An isomorphism
ϕ : G→ G
is called an automorphism. The identity
map
iG : G→ G,
defined by iG(g) = g for all g ∈ G, is called
the trivial automorphism.
12
Automorphism
Example
Let G = (Z,+). The function ϕ : G → G
defined by
ϕ(n) = −n
is an automorphism since ϕ is one-to-one
and onto, and
ϕ(n+m) = −(n+m)
= (−n) + (−m)
= ϕ(n) + ϕ(m)
for all n,m ∈ Z.
13
Isomorphism
Theorem 12.5: Let ϕ : G → H be an
isomorphism. Then
(i) o(x) = o(ϕ(x)), for all x ∈ G;
(ii) G and H have the same cardinality;
(iii) G is abelian if and only if H is abelian.
14
Proof of (i): Assume ϕ : G → H is an
isomorphism. Then for all x ∈ G and for
all n ∈ Z, we have
xn = eG iff ϕ(xn) = ϕ(eG)
(since ϕ is one-to-one)
iff ϕ(x)n = eH .
(by Theorem 12.4)
Therefore, x has finite order if and only if
o(ϕ(x)) has finite order, and the smallest
positive integer n for which xn = eG is the
same as the smallest positive integer n for
which ϕ(x)n = eH.
Therefore, o(x) = o(ϕ(x)).
Proof of (ii): Assume ϕ : G → H is an
isomorphism. Then, there is a one-to-one
correspondence (bijection) between the el-
ements of G and H via the map ϕ. This
means G and H have the same cardinality.
Proof of (iii): Homework.
Homomorphisms
Theorem 12.1:
(i) Let ϕ : G → H and ψ : H → K be
homomorphisms. Then ψ ◦ϕ : G→ K
is a homomorphism.
(ii) If ϕ and ψ are isomorphisms, then
ψ ◦ ϕ is an isomorphism.
(iii) If ϕ : G→ H is an isomorphism, then
ϕ−1 : H → G is an isomorphism.
15
Proof of (i):
For all a, b ∈ G, we have
(ψ ◦ ϕ)(ab) = ψ[ϕ(ab)]
= ψ[ϕ(a)ϕ(b)]
= ψ[ϕ(a)]ψ[ϕ(b)]
= [(ψ ◦ ϕ)(a)][(ψ ◦ ϕ)(b)].
Hence, ψ ◦ ϕ is a homomorphism.
Proof of (ii):
If ϕ and ψ are isomorphisms, then by part
(i), ψ ◦ ϕ is a homomorphism. Also, ψ ◦ ϕis bijective, since the composition of bijec-
tive functions is bijective. Therefore, ψ ◦ϕis an isomorphism.
Proof of (iii):
Let ϕ : G → H is an isomorphism, and let
ϕ−1 : H → G be its inverse function. First
note that ϕ−1 is a bijection. Also, for all
x, y ∈ H
ϕ−1(xy) = ϕ−1(x)ϕ−1(y)
iff ϕ[ϕ−1(xy)] = ϕ[ϕ−1(x)ϕ−1(y)]
iff xy = ϕ[ϕ−1(x)]ϕ[ϕ−1(y)]
iff xy = xy
Since the last equation is true, the first
equation is also true. That is
ϕ−1(xy) = ϕ−1(x)ϕ−1(y)
for all x, y ∈ H, which proves that ϕ−1 is
an isomorphism.
Equivalence of Groups
Theorem 12.1 shows that ∼= defines an
equivalence relation on the set of all groups.
Indeed,
(Reflexivity) Given any group G, the iden-
tity map iG : G → G is an isomorphism.
This shows G ∼= G. Hence ∼= is reflexive.
(Symmetry) If G ∼= H, then there exists
an isomorphism ϕ : G → H. By Theorem
12.1 (iii), ϕ−1 : H → G is an isomorphism.
Therefore H ∼= G . This proves ∼= is sym-
metric.
(Transitive) If G ∼= H and H ∼= K, then
by Theorem 12.1 (ii), G ∼= K. Therefore,∼= is transitive.
16
Equivalence of Groups
Example
Let G = (R,+) and H = (R+, ·), and con-
sider the isomorphism ϕ : G→ H given by
ϕ(x) = ex.
By Theorem 12.1 (iii), the function
ϕ−1(x) = lnx
is an isomorphism from H to G. Indeed,
the condition that ϕ−1 : H → G is a homo-
morphism corresponds to the familiar prop-
erty
ln(xy) = ln(x) + ln(y)
for all x, y ∈ H.
17
Equivalence of Groups
Theorem 12.2: Let G be a cyclic group
of order n. Then, G ∼= (Zn,⊕). Consequently,
any two cyclic groups of order n are iso-
morphic to each other.
18
Proof: Let G = 〈g〉 = {e, g1, g2, . . . , gn−1},where o(g) = n. We define ϕ : Zn → G by
ϕ(j) = gj
for all 0 ≤ j ≤ n− 1. We want to show ϕ
is an isomorphism. Clearly ϕ is one-to-one
and onto. Also, for all j, k ∈ Zn we have
ϕ(j ⊕ k) = gj⊕k
= gj+k
= gjgk
= ϕ(j)ϕ(k).
This proves ϕ is an isomorphism. There-
fore, G ∼= (Zn,⊕).
Equivalence of Groups
Theorem 12.3: Let G be an infinite
cyclic group. Then, G ∼= (Z,+). Consequently,
any two infinite cyclic groups are isomor-
phic to each other.
19
Image and Inverse Image
Notation
Let ϕ : G→ K be a function.
• If H ⊆ G, then the image of H under
ϕ is the set
ϕ(H) = {k ∈ K | k = ϕ(h) for someh ∈ H}
• If J ⊆ K, then the inverse image (or
pre-image) of J under ϕ is the set
ϕ−1(J) = {h ∈ H | ϕ(h) ∈ J}
20
Homomorphisms
Theorem 12.6: Let ϕ : G → K be a
homomorphism. Then:
(i) If H is a subgroup of G. then ϕ(H)
is a subgroup of K.
(ii) If J is a subgroup of K, then ϕ−1(J)
is a subgroup of G.
(iii) If J � K, then ϕ−1(J) � G.
(iv) If ϕ is onto and H �G, then ϕ(H)� K.
21
Proof of (i):
Assume H is a subgroup of G. We want
to show ϕ(H) is a subgroup of K. By
Theorem 5.1, it suffices to show that ϕ(H)
is closed under the group operation and
closed under inverses.
First, let x, y ∈ ϕ(H). Then x = ϕ(h1) and
y = ϕ(h2) where h1, h2 ∈ H. Therefore,
xy = ϕ(h1)ϕ(h2)
= ϕ(h1h2),
where h1h2 ∈ H since H is a subgroup of
G. This proves xy ∈ ϕ(H); that is, ϕ(H)
is closed under the group operation.
Next, let let x ∈ ϕ(H). Then x = ϕ(h) for
some h ∈ H. Therefore,
x−1 = (ϕ(h))−1
= ϕ(h−1),
where h−1 ∈ H since H is a subgroup of
G. This proves x−1 ∈ ϕ(H); that is, ϕ(H)
is closed under inverses.
This proves ϕ(H) is a subgroup of K.
Proofs of (ii) and (iii): Homework
Proof of (iv): Assume ϕ is onto and as-
sume H �G. We want to show ϕ(H) � K.
Assume x ∈ ϕ(H) and assume k ∈ K.
Then x = ϕ(h) for some h ∈ H, and since
ϕ is onto k = ϕ(g) for some g ∈ G. There-
fore,
kxk−1 = ϕ(g)ϕ(h)ϕ(g)−1
= ϕ(g)ϕ(h)ϕ(g−1)
= ϕ(gh)ϕ(g−1)
= ϕ(ghg−1),
where ghg−1 ∈ H since H is normal in
G. This proves kxk−1 ∈ ϕ(H). Therefore
ϕ(H) is a normal subgroup of K.
This completes the proof.
Cayley’s Theorem
Theorem 12.7 (Cayley’s Theorem)
If G is a group, then G is isomorphic to a
subgroup of SG, the symmetric group on
the set G.
22
Proof: Recall that for any fixed g ∈ G,
the function fg : G→ G defined by
fg(x) = gx
is a bijection (Homework 5, Problem 6).
This means fg ∈ SG, the symmetric group
on the set G.
We will show that the function ϕ : G→ SG,
defined by
ϕ(g) = fg
is a one-to-one homomorphism.
ϕ is one-to-one: We have
ϕ(g1) = ϕ(g2) ⇒ fg1 = fg2
⇒ fg1(e) = fg2(e)
⇒ g1e = g2e
⇒ g1 = g2
This proves ϕ is one-to-one.
ϕ is a homomorphism: For all g1, g2 ∈ G
ϕ(g1g2) = fg1g2
= fg1 ◦ fg2
= ϕ(g1) ◦ ϕ(g2)
where the second equality follows from the
fact that for all x ∈ G
fg1g2(x) = g1g2x = fg1(g2x) = fg1(fg2(x)).
This proves that ϕ is a homomorphism, and
since ϕ is one-to-one, it follows that G
is isomorphic to its image ϕ(G) which, by
Theorem 12.6, is a subgroup of SG.
This completes the proof.
Cayley’s Theorem
Remark (Finite Case)
If G is a finite group, then each row in the
group table for G corresponds to left mul-
tiplication by some element g ∈ G. There-
fore, the rows in the group table for G
correspond exactly to the permutations fg
in the proof of Caley’s Theorem. This idea
is illustrated in the next example.
23
Cayley’s Theorem
Consider the group of quaternions
Q8 = {I,−I, J,−J, K,−K, L,−L}
If we relabel the elements of Q8 so that
I 7→ 1 K 7→ 5
−I 7→ 2 −K 7→ 6
J 7→ 3 L 7→ 7
−J 7→ 4 −L 7→ 8
then Q′8 = {1,2,3,4,5,6,7,8} represents
an isomorphic copy of Q8. We will use the
proof of Caley’s Theorem to find a sub-
group H of S8 such that Q′8∼= H.
24
The group table for Q8 is given below.
∗ I −I J −J K −K L −L
I I −I J −J K −K L −L
−I −I I −J J −K K −L L
J J −J −I I L −L −K K
−J −J J I −I −L L K −K
K K −K −L L −I I J −J
−K −K K L −L I −I −J J
L L −L K −K −J J −I I
−L −L L −K K J −J I −I
Relabelling the rows in the table determines
the permutations in the isomorphic sub-
group H.
∗ 1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
13 Homomorphisms and Normal
Subgroups
Definition
Let G be a group and let H be a normal
subgroup of G. The function ρ : G→ G/H
defined by
ρ(a) = Ha
is called the canonical homomorphism from
G onto G/H.
To see that ρ is a homomorphism observe
that
ρ(ab) = Hab
= (Ha)(Hb)
= ρ(a)ρ(b).
1
Canonical Homomorphism
Example
Let G = Q8 and let H = Z(Q8) = {I,−I}.Recall that
G/H = {HI, HJ, HK, HL}
is isomorphic to the Klein 4-group V . The
canonical homomorphism ρ : G → G/H is
given by
ρ(I) = ρ(−I) = HI
ρ(J) = ρ(−J) = HJ
ρ(K) = ρ(−K) = HK
ρ(L) = ρ(−L) = HL
We say that V is a homomorphic image
of Q8 .
2
Kernel of a Homomorphism
Let ϕ : G → K be homomorphism. The
kernel of ϕ is the set of all elements in G
that map to the identity element eK. That
is,
ker(ϕ) = ϕ−1({eK})
= {g ∈ G | ϕ(g) = eK}.
3
Kernel of a Homomorphism
Example
Let G = (Z,+) and let K = (Zn,⊕), and
consider the homomorphism ϕ : G → K
defined by
ϕ(m) = m,
where m is the remainder of m modulo n.
Then,
ker(ϕ) = {m ∈ Z | ϕ(m) = 0}
= {m ∈ Z | m = 0}
= nZ.
4
Kernel of a Homomorphism
Theorem 13.1 Let ϕ : G → K be a ho-
momorphism. Then, ker(ϕ) is a normal
subgroup of G .
5
Proof: Since {eK} � K , it follows by The-
orem 12.6(iii) that
ker(ϕ) = ϕ−1({eK}) � G.
Or, by a direct argument, observe that for
any g ∈ ker(ϕ) and for any x ∈ G, we have
ϕ(xgx−1) = ϕ(x)ϕ(g)ϕ(x−1)
= ϕ(x) eK ϕ(x)−1
= ϕ(x)ϕ(x)−1
= eK,
which shows that xgx−1 ∈ ker(ϕ) . There-
fore, ker(ϕ) � G.
Kernel of a Homomorphism
Example
Let G = (Sn, ◦) and let K = (Z2,⊕), and
consider the homomorphism ϕ : G → K
given by
ϕ(f) =
0 if f is even
1 if f is odd
Then,
ker(ϕ) = {f ∈ Sn | ϕ(f) = 0}
= {f ∈ Sn | f is even}
= An.
That is, ker(ϕ) is the alternating group An,
and by Theorem 13.1, An � Sn .
6
Kernel of a Homomorphism
Theorem 13.2 (Fundamental Theorem
on Group Homomorphisms)
Let ϕ : G → K be a homomorphism from
G onto K. Then
K ∼= G/ ker(ϕ).
More generally, if ϕ is not onto, then
ϕ(G) ∼= G/ ker(ϕ).
7
Proof: Let N = ker(ϕ). We will show
that the function ϕ : G/ ker(ϕ) → K de-
fined by
ϕ(Na) = ϕ(a)
is an isomorphism. First we must check
that ϕ is well-defined.
ϕ is well-defined: For all a, b ∈ G we have
Na = Nb iff ab−1 ∈ N = ker(ϕ)
iff ϕ(ab−1) = eK
iff ϕ(a)ϕ(b−1) = eK
iff ϕ(a)ϕ(b)−1 = eK
iff ϕ(a) = ϕ(b)
iff ϕ(Na) = ϕ(Nb)
Therefore Na = Nb implies ϕ(Na) = ϕ(Nb) ,
which shows that ϕ is well-defined.
ϕ is one-to-one and onto: The argument
above also shows that if ϕ(Na) = ϕ(Nb) ,
then Na = Nb , which proves that ϕ is
one-to-one.
Since ϕ is onto, given any k ∈ K , there
exists a ∈ G such that
ϕ(Na) = ϕ(a) = k.
Therefore, ϕ is onto.
ϕ is an homomorphism: Finally, to show
that ϕ is a homomorphism note that
ϕ(NaNb) = ϕ(Nab)
= ϕ(ab)
= ϕ(a)ϕ(b)
= ϕ(a)ϕ(b).
This completes the proof.
Kernel of a Homomorphism
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.
Since ϕ is onto, and ker(ϕ) = nZ we have,
by Theorem 13.2
Zn ∼= Z / nZ.
8
Kernel of a Homomorphism
Example
Let G = (Sn, ◦) and let K = (Z2,⊕), and
consider the homomorphism ϕ : G → K
given by
ϕ(f) =
0 if f is even
1 if f is odd
Since ϕ is onto, and ker(ϕ) = An we have,
by Theorem 13.2
Z2∼= Sn /An.
9
14 Direct Products and Finite
Abelian Groups
The goal of this section is to prove the
following theorem on the structure of finite
abelian groups.
Theorem 14.2: Every finite abelian group
G is isomorphic to a direct product of cyclic
groups whose orders are prime powers.
1
Direct Products and Finite Abelian Groups
Example Assume G is an abelian group
such that |G| = 180 = 22 · 32 · 5. It follows
from Theorem 14.2 that G is isomorphic
to one of the following groups
G ∼= Z4 × Z9 × Z5
G ∼= Z2 × Z2 × Z9 × Z5
G ∼= Z4 × Z3 × Z3 × Z5
G ∼= Z2 × Z2 × Z3 × Z3 × Z5
For example, if G is the abelian group
Z180, then G ∼= Z4 × Z9 × Z5.
2
Groups of Prime Power Order
If p is a prime number and G is a group,
then G has p-power order if |G| = pk for
some positive integer k .
A group G is called a p-group if for every
x ∈ G , o(x) is a power of p .
3
Groups of Prime Power Order
Theorem: A finite abelian group has p-
power order if and only if it is a p-group.
Proof:
First, if G has p-power order, then by The-
orem 10.4, o(x) divides |G| for all x ∈ G.
Therefore, G is a p-group.
Conversely, assume G is a p-group, and
assume for the sake of contradiction that G
does not have p-power order. Then there
exists a prime q 6= p such that q | |G| . By
Theorem 11.7, G has a subgroup H = 〈x〉such that o(x) = q , which contradicts the
fact that G is a p-group.
This completes the proof.
4
Finite Abelian Groups and Direct Prod-
ucts
Theorem 14.1: If A and B are normal
subgroups of G such that
(i) A ∩B = {e}, and
(ii) G = AB = {ab | a ∈ A and b ∈ B},
then G ∼= A×B.
5
Proof: We will show that ϕ : A × B → G
defined by
ϕ((a, b)) = ab
is an isomorphism. First, it follows by as-
sumption (ii) that ϕ is onto. Next, observe
that
ϕ((a1, b1)) = ϕ((a2, b2))
⇒ a1b1 = a2b2
⇒ a−12 a1 = b2b
−11 ∈ B
⇒ a−12 a1 ∈ A ∩B
⇒ e = a−12 a1 = b2b
−11
⇒ a1 = a2 and b1 = b2
⇒ (a1, b1) = (a2, b2)
Therefore, ϕ is one-to-one. To show that
ϕ is a homomorphism, we need the follow-
ing lemma:
Lemma: Under the assumptions above, if
a ∈ A and b ∈ B , then ab = ba .
To prove the lemma, observe that ab = ba
if and only if bab−1a−1 = e . Since A and
B are normal subgroups, we have bab−1 ∈A and ab−1a−1 ∈ B . Therefore,
bab−1a−1 = (bab−1)a−1 ∈ A, and
bab−1a−1 = b(ab−1a−1) ∈ B.
This proves bab−1a−1 ∈ A∩B and it follows
that bab−1a−1 = e .
Finally, we have
ϕ((a1, b1) ∗ (a2, b2)) = ϕ((a1a2, b1b2))
= (a1a2)(b1b2)
= a1(a2b1)b2
= a1(b1a2)b2
= (a1b1)(a2b2)
= ϕ((a1, b1)) ∗ ϕ((a2, b2))
Therefore ϕ is a homomorphism, and the
proof is complete.
Finite Abelian Groups and Direct Prod-
ucts
Lemma 14.7: If G is an abelian group
and |G| = mn where gcd(m,n) = 1 , then
G ∼= A × B where A = {x ∈ G | xm = e}and B = {x ∈ G | xn = e} .
6
Proof: Since gcd(m,n) = 1, there exist
integers r and s such that rn + sm = 1 .
Therefore, for any x ∈ G, we have
x = x1 = xrn+sm = xrnxsm
where xrn ∈ A and xrn ∈ B since
(xrn)m = (xrm)n = (xr)|G| = e.
It follows that
G = AB = {ab | a ∈ A and b ∈ B}.
It is easy to check that A and B are sub-
groups of G , and they are both normal
since G is abelian. Also, A ∩ B = {e} ,
since x ∈ A ∩B implies
x = xrnxsm = (xn)r(xm)s = ee = e.
Therefore, by Theorem 14.1, G ∼= A×B .
Finite Abelian Groups and Direct Prod-
ucts
Corollary 14.7: Let G be an abelian group,
and consider the prime factorization of |G|given by
|G| = pr11 p
r22 · · · p
rkk
where p1, p2, . . . , pk are distinct primes and
r1, r2, . . . , rk are positive integers. Then,
G ∼= G(p1)×G(p2)× · · · ×G(pk)
where
G(pi) = {x ∈ G | xprii = e}
is a pi-group for all i = 1,2, . . . , k.
7
Finite Abelian Groups and Direct Prod-
ucts
Lemma A: Let G1 = 〈g1〉, G2 = 〈g2〉, . . . ,
Gm = 〈gm〉 be finite cyclic groups. If G is
a finite abelian group and
ϕ : G → G1 ×G2 × · · · ×Gm
is an isomorphism, then every element g ∈G has a unique representation of the form
g = xr11 x
r22 · · ·x
rmm
where 0 ≤ ri < |Gi|, and
ϕ(xi) = (eG1, . . . , eGi−1
, gi, eGi+1, . . . , eGn)
for each i = 1,2, . . .m.
8
Proof: Let
H = {xr11 x
r22 · · ·x
rmm | 0 ≤ ri < |Gi|}.
Then,
ϕ(H) = {ϕ(xr11 x
r22 · · ·x
rmm ) | 0 ≤ ri < |Gi|}
= {ϕ(x1)r1ϕ(x2)r2 · · ·ϕ(xm)rm | 0 ≤ ri < |Gi|}
= {(gr11 , g
r22 , · · · , grmm ) | 0 ≤ ri < |Gi|}
= G1 ×G2 × · · · ×Gm.
Since ϕ is a bijection, there is a one-to-one
correspondence between the elements of H
and G1 ×G2 × · · · ×Gm , and since
G = ϕ−1(G1 ×G2 × · · · ×Gm) = H
it follows that G = H. This completes the
proof.
Finite Abelian Groups and Direct Prod-
ucts
Lemma B: Let G be a finite abelian p-
group, and let A = 〈x〉 be a cyclic sub-
group of maximal order (i.e. o(x) ≥ o(g)
for all g ∈ G). Then for every element Ay
in G/A , there exists y1 ∈ G such that
Ay = Ay1 and o(Ay) = o(Ay1) = o(y1) .
9
Proof: Let |G| = pa and o(x) = pi where
1 ≤ i ≤ a . If Ay ∈ G/A , then o(Ay) = pt
where t ≤ a − i . In particular, we have
ypt ∈ A , which means
ypt
= xn
for some integer n < pi.
Let pw be the highest power of p that
divides n . Then, we have
o(xn) =pi
gcd(pi, n)=
pi
pw= pi−w.
On the other hand, if o(y) = j , then
o(ypt) =
pj
gcd(pj, pt)=
pj
pt= pj−t.
Therefore, i − w = j − t , which means,
w = t + i − j . By assumption, i = o(x) ≥o(y) = j . Therefore, w ≥ t and it follows
that pt divides n . If we write n = cpt ,
then
ypt
= xn
⇒ ypt
= (xc)pt
⇒ (yx−c)pt
= e
⇒ (y1)pt
= e
where y1 = x−cy ∈ Ay .
Therefore, Ay1 = Ay and
o(y1) = pt = o(Ay) = o(Ay1).
This completes the proof.
Finite Abelian Groups and Direct Prod-
ucts
Lemma C: If G is a finite abelian p-group,
then G is isomorphic to a direct product
of cyclic groups of p-power order.
10
Proof: We will prove the statement by
(strong) induction on n where |G| = pn.
Base Case: If n = 1 , then |G| = p . There-
fore, G is cyclic and the result holds.
Inductive Step: Let |G| = pn and assume
the statement is true for all groups of order
pk where k < n .
Let x be an element in G of maximal or-
der (i.e. o(x) ≥ o(g) for all g ∈ G), and
consider the subgroup A = 〈x〉 . The quo-
tient group G/A is a p-group for which
the inductive hypothesis applies, therefore
G/A ∼= G1 ×G2 × · · · ×Gm
where each Gi is cyclic. By Lemmas A and
B, there exist y1, y2, . . . , ym ∈ G such that
o(yi) = o(Ayi) = |Gi| for all i = 1,2, . . . ,m
and each Ag ∈ G/A has the unique rep-
resentation
Ag = (Ay1)r1(Ay2)r2 · · · (Aym)rm
where 0 ≤ ri < |Gi| .
We claim that G = AB where
B = {yr11 y
r22 · · · y
rmm | 0 ≤ ri < |Gi|}.
Indeed, for any g ∈ G we have
Ag = (Ay1)r1(Ay2)r2 · · · (Aym)rm
= A(yr11 y
r22 · · · y
rmm )
= Ab
where b ∈ B. Therefore gb−1 ∈ A , which
implies g = ab for some a ∈ A. Therefore
G = AB , as claimed.
To show that A ∩ B = {e} , observe that
for b = yr11 y
r22 · · · y
rmm ∈ B , we have
b ∈ A iff Ab = Ae
iff A(yr11 y
r22 · · · y
rmm ) = Ae
iff (Ay1)r1(Ay2)r2 · · · (Aym)rm = Ae
iff ri = 0 for all i = 1,2, . . . ,m
iff b = e.
Therefore, by Theorem 14.1, G ∼= A × B .
Since B ∼= G/A , the proof is complete.
Fundamental Theorem on Finite Abelian
Groups
Theorem 14.2: Every finite abelian group
G is isomorphic to a direct product of cyclic
groups whose orders are prime powers.
Proof: The result follows directly from
Corollary 14.7 and Lemma C above.
11
15 Sylow Theorems
Normalizer
Let G be a group, and let H be a subgroup
of G. The normalizer of H in G is the
subset
N(H) = {g ∈ G | gHg−1 = H}.
It is easy to show that N(H) is a sub-
group of G, and H is a normal subgroup
of N(H).
1
Normalizer
Example: Let G = S3, and let H = 〈(1,2)〉.We have N(H) = {(1), (1,2)} = H, since
(1)H (1)−1 = {(1), (1,2)} = H
(1,2)H (1,2)−1 = {(1), (1,2)} = H
(1,3)H (1,3)−1 = {(1), (2,3)} 6= H
(2,3)H (2,3)−1 = {(1), (1,3)} 6= H
(1,2,3)H (1,2,3)−1 = {(1), (2,3)} 6= H
(1,3,2)H (1,3,2)−1 = {(1), (1,3)} 6= H
2
Normalizer
Example: Let G = D4, and let H = 〈(1,3)〉.Then,
N(H) = {(1), (1,3), (2,4), (1,3)(2,4)}
as the following table illustrates
g gHg−1 gHg−1 = H ?
(1) {(1), (1,3)} X
(1,3) {(1), (1,3)} X
(2,4) {(1), (1,3)} X
(1,3)(2,4) {(1), (1,3)} X
(1,2)(3,4) {(1), (2,4)}
(1,4)(2,3) {(1), (2,4)}
(1,2,3,4) {(1), (2,4)}
(1,4,3,2) {(1), (2,4)}
3
p-Sylow subgroup
Let G be a group. If p is a prime number
such that pn divides |G| but pn+1 does
not divide |G| , then any subgroup of order
pn in G is called a p-Sylow subgroup.
In other words, if |H| = pn where pn is the
largest power of p that divides |G|, then H
is called a p-Sylow subgroup of G.
4
Converse of Lagrange’s Theorem
For finite abelian groups, the following con-
verse to Lagrange’s Theorem holds.
Theorem: Let G be a finite abelian group.
If m divides |G|, then there exists a sub-
group H of G such that |H| = m.
For nonabelian groups, this is false. For
example, the group A4 has order 12 and
has no subgroups of order m = 6.
The First Sylow Theorem (below) provides
a partial converse to the Lagrange’s The-
orem for a general group. That is the the-
orem above holds for any group if m = pk
where p is a prime number.
5
First Sylow Theorem
Theorem 15.1: Let G be a finite group,
p be a prime number, and k be a positive
integer.
(i) If pk divides |G|, then G has a sub-
group of order pk. In particular, G
has a p-Sylow subgroup.
(ii) Let H be any p-Sylow subgroup of
G. If K is any of order pk in G, then
for some g ∈ G we have K ⊆ gHg−1.
In particular, K is contained in some
p-Sylow subgroup of G.
6
First Sylow Theorem
Proof of (i): Let G be a finite group.
Base Case: If |G| = 2 , the result is trivial.
Inductive Step: Assume the theorem holds
for all groups of order less than |G|. Fur-
ther, assume pk divides |G|.
If G has a proper subgroup H such that pk
divides |H|, then by the inductive hypoth-
esis H has a subgroup of order pk and the
result holds.
If not, then p divides [G : H] for every
proper subgroup H. It follows that p di-
vides Z(G) since
|G| = |Z(G)|+[G : Z(a1)]+· · ·+[G : Z(ak)],
7
where all terms other than Z(G) are di-
visible by p. Therefore, by Theorem 11.7,
Z(G) has a subgroup A of order p.
Since A ⊆ Z(G) , it follows by Theorem
11.2, that A � G . By the inductive hy-
pothesis, G/A has a subgroup J of order
pk−1 , and
J ∼= ρ−1(J) /A
where ρ : G → G/A is the canonical ho-
momorphism.
Therefore ρ−1(J) has order pk, and the
proof is complete.
Second Sylow Theorem
Theorem 15.2: All p-Sylow subgroups
of G are conjugate to each other. Con-
sequently, a p-Sylow subgroup is normal if
and only if it is the only p-Sylow subgroup.
8
Third Sylow Theorem
Theorem 15.3: Let np denote the num-
ber of p-Sylow subgroups in G , and let
H be any one of these p-Sylow subgroups.
Then:
(i) np = [G : N(H)].
(ii) np ≡ 1 (mod p)
Note: If we write G = pnm where p does
not divide m, then |H| = pn and
m = [G : H]
= [G : N(H)] · [N(H) : H]
= np · [N(H) : H]
That is, (i) implies np divides m.
9
Implications of Sylow Theorems
Theorem 15.4: Let G be a finite group,
and let p be a prime number. Then G is
a p-group if and only if |G| = pk for some
for some integer k > 0.
10
Implications of Sylow Theorems
Theorem 15.5: Let G be a finite group
of order pq, where p and q are primes and
p < q. If p does not divide q − 1, then G
is cyclic.
11
Simple Group
We say that a group G is simple, if G
contains no non-trivial normal subgroups.
Theorem 15.3(ii) can be used to prove groups
of a certain order are not simple. Indeed,
if np = 1 then there is only one p-Sylow
subgroup, and by Corollary 11.5 it must be
normal.
For example, if |G| = 15 = 3 · 5 , then
n5 ≡ 1 (mod 5)
⇒ n5 = 1,6,11,16, . . .
Since distinct 5-Sylow subgroups have only
the identity element in common, if n5 ≥ 6 ,
then G contains at least (5 − 1) ∗ 6 = 24
elements, which is impossible. Therefore,
G has a normal subgroup of order 5.
12
Number of Groups of Order n
+ 0 1 2 3 4 5 6 7 8 9
0 0 1 1 1 2 1 2 1 5 2
10 2 1 5 1 2 1 14 1 5 1
20 5 2 2 1 15 2 2 5 4 1
30 4 1 51 1 2 1 14 1 2 2
40 14 1 6 1 4 2 2 1 52 2
50 5 1 5 1 15 2 13 2 2 1
60 13 1 2 4 267 1 4 1 5 1
70 4 1 50 1 2 3 4 1 6 1
80 52 15 2 1 15 1 2 1 12 1
90 10 1 4 2 2 1 231 1 5 2
13
Cyclic groups of order pq ( q 6≡ 1 (mod p))
+ 0 1 2 3 4 5 6 7 8 9
0 0 1 1 1 2 1 2 1 5 2
10 2 1 5 1 2 1 14 1 5 1
20 5 2 2 1 15 2 2 5 4 1
30 4 1 51 1 2 1 14 1 2 2
40 14 1 6 1 4 2 2 1 52 2
50 5 1 5 1 15 2 13 2 2 1
60 13 1 2 4 267 1 4 1 5 1
70 4 1 50 1 2 3 4 1 6 1
80 52 15 2 1 15 1 2 1 12 1
90 10 1 4 2 2 1 231 1 5 2
3× 5 5× 7 7× 11
3× 11 5× 13 7× 13
3× 17 5× 17
3× 23 5× 19
3× 29
14
16 Rings and Fields
A ring is a set R equipped with two binary opera-
tions + and · satisfying the following axioms
(i) (closure of R with respect to +)
x + y ∈ R for all x, y ∈ R.
(ii) (associativity of +) (x+ y) + z = x+ (y + z)
for all x, y, z ∈ R.
(iii) (additive identity) There is an element 0 ∈ R
such that x + 0 = 0 + x = x for all x ∈ R.
(iv) (additive inverses) For each element x ∈ R
there is an element (−x) ∈ R such that x +
(−x) = (−x) + x = 0.
(v) (commutativity of +) x + y = y + x for all
x, y ∈ R.
(vi) (closure of R with respect to ·)x · y ∈ R for all x, y ∈ R.
(vii) (associativity of ·) (x · y) · z = x · (y · z) for all
x, y, z ∈ R.
(viii) (distributive laws) x · (y + z) = x · y +x · z and
(x + y) · z = x · z + y · z for all x, y, z ∈ R.
1
Definition of a Ring
Remarks
• Conditions (i)-(v) imply that R is an
abelian group under the operation +.
• If the “multiplication” operation · in
R is commutative, then we say that R
is a commutative ring. Otherwise, we
say that R is a noncommutative ring.
• We use the notation (R,+, ·) to repre-
sent the ring with elements in R under
the operations of + and ·.
• Note that + and · are intended to
represent generalized binary operations,
and will not always correspond to the
usual operations of addition and multi-
plication.2
Rings
Examples
• The sets Z, Q, R, and C under the usual
operations of addition and multiplica-
tion are all commutative rings.
• The set 2Z of even integers under the
usual operations of addition and multi-
plication is a commutative ring.
• The set of n×n matrices with entries in
Z, denoted Mn(Z), is a non-commutative
ring with respect to the operations of
matrix addition and matrix multiplica-
tion.
• The set Zn, where n ∈ Z+, is a (finite)
commutative ring under the operations
of addition and multiplication mod n.
3
Subrings
A subring S of a ring R is a subset of R
which is a ring under the same operations
as R.
Examples
• 2Z is a subring of Z under the usual op-
erations of addition and multiplication.
• Mn(Z) is a subring of Mn(Q) under the
operations of matrix addition and ma-
trix multiplication.
4
Subrings
Theorem 16.1: Assume (R,+, ·) is a ring.
A nonempty subset S of R is a subring of R
if and only if the following conditions hold:
(i) For all x, y ∈ S, x+ y ∈ S and x · y ∈ S.
(ii) For all x ∈ S, (−x) ∈ S.
5
Subrings
Proof: First, assume S is a subring of R.
Clearly property (i) holds. (This follows
from ring axioms (i) and (vi).) For each
x ∈ S, let y ∈ S denote the additive inverse
of x in the ring S. Then,
y = y + 0
= y + (x + (−x))
= (y + x) + (−x)
= 0 + (−x)
= (−x)
where (−x) denotes the additive inverse of
x in the ring R. This proves (−x) = y ∈ S
which establishes property (ii).
Conversely suppose S is a nonempty sub-
set of R for which (i) and (ii) hold. It
6
follows by Theorem 5.1, that ring axioms
(i) through (iv) hold for the set S. Also, by
property (ii), ring axiom (vi) is satisfied for
the set S. Finally, since R is a ring, each of
the ring axioms (v), (vii), and (viii) hold
for all elements in R, and therefore must
hold on the set S as well.
This proves that S satisfies all ring axioms
(i) through (viii). Hence, S is a ring.
Rings with Unity
Let R be a ring. If there exists an element
e ∈ R such that x ·e = e ·x = x for all x ∈ R,
then e is called a unity (or multiplicative
identity element), and we say that R is a
ring with unity.
7
Rings with Unity
Examples
• The element e = 1 ∈ Z is a unity in the
ring (Z,+, ·) since x · 1 = 1 · x = x for
all x ∈ Z. Therefore, (Z,+, ·) is a com-
mutative ring with unity.
• (2Z,+, ·) is a commutative ring without
unity.
• The identity matrix In ∈ Mn(Z) is a
unity in the ring Mn(Z) since A · In =
In ·A = A for all A ∈Mn(Z). Therefore,
Mn(Z) is a noncommutative ring with
unity.
8
Rings with Unity
Theorem 16.2: If (R,+, ·) is a ring with
unity, then the unity element in R is unique.
Proof: Assume e1 and e2 are both unity
elements in R. Then, e1 · e2 = e2 since e1
is a unity element. On the other hand,
e1 · e2 = e1 since e2 is a unity element.
Therefore, e1 = e1 · e2 = e2 which proves
the unity element is unique.
9
Rings with Unity
Assume R be a ring with unity e, and let
a ∈ R. If there is an element x ∈ R such
that a ·x = x ·a = e, then we say that x is a
multiplicative inverse of a, and a is called
a unit (or an invertible element) in R.
10
Rings with Unity
Theorem 16.3: Assume (R,+, ·) is a ring
with unity. If an element a ∈ R has a mul-
tiplicative inverse, then the multiplicative
inverse is unique (and is denoted a−1).
Proof: (See Theorem 3.2)
11
Rings with Unity
Example
Consider the ring (Z10,⊕,�). Since 1 ∈ Z10
is a unity in the ring, we have the following:
• The element 1 ∈ Z10 is a unit since
1� 1 = 1 · 1 = 1 = 1.
• The element 3 ∈ Z10 is a unit since
3� 7 = 3 · 7 = 21 = 1.
• The element 7 ∈ Z10 is a unit since
7� 3 = 7 · 3 = 21 = 1.
• The element 9 ∈ Z10 is a unit since
9� 9 = 9 · 9 = 81 = 1.
12
Multiplication by Zero
Theorem 16.4: Assume (R,+, ·) is a ring
with additive identity 0. Then, for all a ∈ R
0 · a = a · 0 = 0.
13
Multiplication by Zero
Proof: Assume a ∈ R. Then,
a · 0 = a · (0 + 0)
= a · 0 + a · 0.
Therefore,
a · 0 + (−(a · 0)) = a · 0 + a · 0 + (−(a · 0)).
It follows that
0 = a · 0.
A similar argument shows 0 = 0 · a for all
a ∈ R. This completes the proof.
14
Multiplication by Zero
Alternate Proof: Assume a ∈ R. Then,
a · 0 = a · 0 + 0
= a · 0 + (a · 0 + (−(a · 0)))
= (a · 0 + a · 0) + (−(a · 0))
= (a · (0 + 0)) + (−(a · 0))
= (a · 0) + (−(a · 0))
= 0.
A similar argument shows 0 = 0 · a for all
a ∈ R. This completes the proof.
15
Zero Divisors
Let R be a ring with additive identity 0,
and let a ∈ R. If a 6= 0, and if there exists
an element b 6= 0 in R such that a · b = 0 or
b · a = 0, then we say a is a zero divisor.
16
Zero Divisors
Example
Consider the ring (Z10,⊕,�). Then:
• The element 2 ∈ Z10 is a zero divisor
since 2� 5 = 2 · 5 = 10 = 0.
• The element 4 ∈ Z10 is a zero divisor
since 4� 5 = 4 · 5 = 20 = 0.
• The element 5 ∈ Z10 is a zero divisor
since 5� 2 = 5 · 2 = 10 = 0.
• The element 6 ∈ Z10 is a zero divisor
since 6� 5 = 6 · 5 = 30 = 0.
• The element 8 ∈ Z10 is a zero divisor
since 8� 5 = 8 · 5 = 40 = 0.
17
Integral Domains
Let D be a ring. We say that D is an
integral domain if the following conditions
hold:
1. D is a commutative ring.
2. D has a unity e and e 6= 0.
3. D has no zero divisors.
18
Integral Domains
Examples
• The rings Z, Q, R, and C under the
usual operations of addition and multi-
plication are all integral domains.
• The ring (2Z,+, ·) is not an integral do-
main since 2Z does not contain a unity.
• The ring (Z10,⊕,�) is not an integral
domain since it has zero divisors.
• The ring M2(Z) is not an integral do-
main since it is not commutative. Note
also that M2(Z) contains zero divisors
(Homework 10 #5).
19
Integral Domains
Theorem 16.5: The ring (Zn,⊕,�) is an
integral domain if and only if n is prime.
20
Integral Domains
Proof: First note that (Zn,⊕,�) is a com-
mutative ring with unity e = 1. Therefore
(Zn,⊕,�) is an integral domain if and only
if it has no zero divisors.
First, assume n is prime, and assume for
the sake of contradiction that a ∈ Zn is a
zero divisor. Then, a 6= 0 and there exists
b 6= 0 in Zn such that
0 = a� b = ab
Therefore, n divides ab, and since n is prime,
it follows that n divides a, or n divides
b. However this is a contradiction since
a, b ∈ {1,2, . . . , n − 1}. Therefore, if n is
prime, then (Zn,⊕,�) is an integral domain.
21
Next we will show that if (Zn,⊕,�) is an in-
tegral domain, then n is prime. We will use
a proof by contraposition. Assume n is not
prime. Then there exist a, b ∈ {2,3, . . . , n−1} such that n = ab. Therefore, a � b = 0
where a, b ∈ Zn and a and b are nonzero.
Therefore, (Zn,⊕,�) has zero divisors and
is not an integral domain.
This completes the proof.
Integral Domains
Theorem 16.6: Assume D is an integral
domain. If a, b, and c are elements of D
such that a 6= 0 and ab = ac, then b = c.
22
Integral Domains
Proof: Assume D is an integral domain,
and assume a, b, and c are elements of D
such that a 6= 0 and ab = ac. Then,
a(b + (−c)) = ab + a(−c)
= ab + a((−e)c)
= ab + (−e)ac
= ab + (−(ac))
= 0
Since a 6= 0 and D has no zero divisors, it
follows that b + (−c) = 0. Therefore b = c.
23
Fields
Let F be a ring. We say that F is a field
if the following conditions hold:
1. F is a commutative ring.
2. F has a unity e and e 6= 0.
3. Every nonzero element of F has a mul-
tiplicative inverse.
24
Fields
Examples
• The rings Q, R, and C under the usual
operations of addition and multiplica-
tion are all fields.
• The ring (Z,+, ·) is not a field since not
every element of Z has a multiplicative
inverse in Z.
• The ring (Zp,⊕,�) is a field if and only
if p is prime.
25
Fields
Theorem 16.7: Every field F is an integral
domain.
26
Fields
Proof: Assume F is a field. Therefore,
F is a commutative ring with unity 6= 0.
We want to show F has no zero divisors.
Assume for the sake of contradiction that
there exist nonzero elements a, b ∈ F such
that ab = 0. Since F is a field, the element
a 6= 0 has a multiplicative inverse a−1 ∈ F .
Therefore,
b = eb = (a−1a)b = a−1(ab) = a−1(0) = 0.
This is a contradiction since we assumed
b 6= 0. Therefore F has no zero divisors
and it follows that F is an integral domain.
27
Fields
Theorem 16.8: Every finite integral do-
main D is a field.
28
Fields
Proof: Let D = {d1, d2, . . . , dn} be a finite
integral domain. To show that D is a field
we must show that every nonzero element
of D has a multiplicative inverse.
Assume a ∈ D and a 6= 0. By Theorem
16.6, the elements ad1, ad2, . . . , adn are all
distinct. Since D has exactly n elements,
it follows that
D = {ad1, ad2, . . . , adn}.
In particular the unity e ∈ D, can be ex-
pressed in the form
e = adj,
for some dj ∈ D where dj 6= 0. This shows
that a has a multiplicative inverse. Hence
D is a field.
29