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0 Sets and Induction
Sets
A set is an unordered collection of objects,
called elements or members of the set. A
set is said to contain its elements. We write
a ∈ A to denote that a is an element of the
set A. The notation a /∈ A denotes that a
is not an element of the set A.
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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, . . .}
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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}
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Defining a Set
Example: Express the following set using
set builder notation.
S = {. . . ,−12,−8,−4,0,4,8,12, . . .}
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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
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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.
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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.
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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.
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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
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Special Subsets
For every set S,
(i) Ø ⊆ S(ii) S ⊆ S
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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.
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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.
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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}
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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}.
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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 )
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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.
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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.
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Sets and Proofs
Example: Let S and T be sets. Prove that
S ∩ T ⊆ S ∪ T .
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Sets and Proofs
Example: Let S and T be sets. Prove that
S ⊆ T if and only if S ∩ T = S.
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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.
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Mathematical Induction
Example: Show that if n is a positive
integer, then
1 + 2 + 3 + · · ·n =n(n+ 1)
2
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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:
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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 =
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Mathematical Induction
Conjecture: For all positive integers n,
the following proposition P (n) holds
1 + 3 + 5 + · · ·+ 2n− 1 =
Proof:
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Mathematical Induction
Example: Use mathematical induction to
prove that 2 divides n2 + 5n for all n ≥ 1.
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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:
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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}.
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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)}.
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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.
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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+.
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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+.
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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.
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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.
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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.
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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.
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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.
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Commutativity and Associativity
Example: Let ∗ be a binary operation on
Z defined by
a ∗ b = 2(a + b)
• is ∗ commutative?
• is ∗ associative?
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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
)
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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?
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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.
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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.
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Abelian Group
A group (G, ∗) for which ∗ is commutative
is called an abelian group.
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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.
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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.
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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:
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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:
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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:
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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:
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Groups
Example: Consider the set Z together
with the binary operation ∗ given by
a ∗ b = 2(a + b).
Does (Z, ∗) form a group?
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Groups
Example: Consider the set G = R − {0}together with the binary operation ∗ given
by
a ∗ b = 2ab.
Does (G, ∗) form a group?
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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.
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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:
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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
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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.
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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.
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Inverse Element
Notation
If (G, ∗) is a group, then the inverse of an
element x ∈ G is denoted by x−1.
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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.
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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.
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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
).
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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.
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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
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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.
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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.
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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.
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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.
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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.
Page 78
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.
Page 79
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+.
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• 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 · · ·... ... ... ... ... ... . . .
Page 81
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.
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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
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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
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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) =∞.
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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
···
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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
)
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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
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Greatest Common Divisor
Examples
• gcd(24,40) =
• gcd(32,100) =
• gcd(12,91) =
8
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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
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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
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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
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The Euclidean Algorithm
Examples
• Find gcd(198, 252) using the Euclidean
algorithm
• Find gcd(414, 662) using the Euclidean
algorithm
12
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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
Page 94
Greatest Common Divisor
Example: Express gcd(198,252) = 18 as
a linear combination of 252 and 198.
14
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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
Page 96
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
Page 100
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
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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
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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
Page 103
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
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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
Page 106
Cyclic Groups
Theorem 4.7 If G is a cyclic group, then
G is abelian.
Proof:
22
Page 107
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
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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
Page 109
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
Page 110
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
Page 111
Trivial Subgroups
For any group G, the subgroups H = {e}and H = G are called trivial subgroups.
4
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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
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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
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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
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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
Page 116
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
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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
Page 118
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
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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
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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
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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
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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
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Subgroups
Theorem 5.2 If G is a cyclic group, then
every subgroup of G is cyclic.
Proof:
16
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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
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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
Page 130
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
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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
Page 132
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
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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
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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
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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
Page 136
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
Page 137
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
Page 138
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
Page 142
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
Page 143
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
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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
Page 145
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
Page 146
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
Page 147
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
Page 148
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
Page 149
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
Page 150
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
Page 151
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
Page 152
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
Page 153
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
Page 154
Example
Prove or disprove:
The function f : Z→ Z defined by
f(n) = 2n− 3
is injective.
12
Page 155
Example
Prove or disprove:
The function f : Z→ Z defined by
f(n) = 2n− 3
is surjective.
13
Page 156
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
Page 157
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
Page 158
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
Page 159
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
Page 160
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
Page 161
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
Page 162
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
Page 163
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
Page 164
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
Page 165
(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)
Page 166
for all x ∈ X. This proves that every ele-
ment of SX has an inverse.
Thus, (SX , ◦) is a group.
Page 167
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
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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)
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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
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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
).
Page 171
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
Page 172
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
Page 173
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
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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
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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
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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
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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
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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.
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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)
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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.
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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
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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
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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
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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
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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
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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
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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
Page 188
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
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Alternating Groups
Theorem 8.5: If n ≥ 2, then An is a
subgroup of Sn, and
|An| =n!
2=|Sn|
2
18
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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
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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
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S3 as a group of symmetries
Rotations:
Reflections:
23
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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
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The dihedral group D4
Rotations:
Reflections:
25
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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
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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
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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
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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
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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
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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
Page 201
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.
Page 202
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
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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
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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
Page 205
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.
Page 206
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
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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.
Page 208
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
Page 209
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
Page 210
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
Page 211
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.
Page 212
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
Page 213
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.
Page 214
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
Page 215
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
Page 216
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
Page 217
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
Page 218
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
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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
Page 220
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
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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
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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
Page 223
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
Page 224
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|.
Page 225
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
Page 226
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
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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
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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
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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
Page 230
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.
Page 231
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
Page 232
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
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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
Page 234
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
Page 235
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
Page 236
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
Page 237
(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.
Page 238
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
Page 239
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.
Page 240
That is,
rp−1 ≡ 1 (mod p)
This completes the proof.
Page 241
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
Page 242
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
Page 243
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
Page 244
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
Page 245
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
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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
Page 247
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)].
Page 248
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
Page 249
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
Page 250
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
Page 251
Conjugacy Classes
Important Fact: All conjugates of a have
the same order as a. (Homework 4, Prob-
lem 5)
25
Page 252
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
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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
Page 254
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.
Page 255
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
Page 256
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
Page 257
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.
Page 258
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
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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
Page 260
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).
Page 261
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.
Page 262
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
Page 263
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
Page 264
Normal Subgroups
Notation
We write H � G to indicate that H is a
normal subgroup of G.
6
Page 265
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
Page 266
Normal Subgroups
Theorem 11.2 Let G be a group. Then
any subgroup of Z(G) is a normal subgroup
of G.
8
Page 267
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.
Page 268
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
Page 269
Normal Subgroups
Theorem 11.3: Let H be a subgroup of
G such that [G : H] = 2. Then H is normal
in G.
10
Page 270
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.
Page 271
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
Page 272
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
Page 273
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
Page 274
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.
Page 275
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
Page 276
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
Page 277
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.
Page 278
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.
Page 279
(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.
Page 280
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
Page 281
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
Page 282
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
Page 283
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
Page 284
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|.
Page 285
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.
Page 286
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
Page 287
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
Page 288
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
Page 289
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
Page 290
f g f ◦ g ϕ(f) ϕ(g) ϕ(f ◦ g)
even even
even odd
odd even
odd odd
Page 291
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
Page 292
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.
Page 293
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
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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).
Page 295
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.
Page 296
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
Page 297
Therefore,
ϕ(xn) = [ϕ(xm)]−1
= [[ϕ(x)]m]−1
= [ϕ(x)]−m
= [ϕ(x)]n.
This completes the proof of (ii).
Page 298
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.
Page 299
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
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• 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
Page 301
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
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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
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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
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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
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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
Page 306
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.
Page 307
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
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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
Page 309
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
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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)).
Page 311
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.
Page 312
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
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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.
Page 314
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.
Page 315
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
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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
Page 317
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
Page 318
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,⊕).
Page 319
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
Page 320
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
Page 321
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
Page 322
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.
Page 323
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
Page 324
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.
Page 325
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
Page 326
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.
Page 327
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.
Page 328
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
Page 329
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
Page 330
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
Page 332
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
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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
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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
Page 335
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
Page 336
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
Page 337
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.
Page 338
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
Page 339
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
Page 340
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.
Page 341
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.
Page 342
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
Page 343
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
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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
Page 345
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
Page 346
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
Page 347
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
Page 348
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
Page 349
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:
Page 350
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 .
Page 351
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.
Page 352
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
Page 353
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 .
Page 354
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
Page 355
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
Page 356
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.
Page 357
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
Page 358
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
Page 359
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.
Page 360
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
Page 361
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
Page 362
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
Page 363
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.
Page 364
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
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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
Page 366
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
Page 367
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
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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
Page 369
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
Page 370
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
Page 371
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
Page 372
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.
Page 373
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
Page 374
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
Page 375
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
Page 376
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
Page 377
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
Page 378
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
Page 379
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
Page 380
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
Page 381
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
Page 382
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
Page 383
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
Page 384
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
Page 385
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
Page 386
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.
Page 387
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
Page 388
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
Page 389
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
Page 390
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
Page 391
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
Page 392
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
Page 393
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
Page 394
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
Page 395
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
Page 396
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
Page 397
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
Page 398
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
Page 399
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
Page 400
Integral Domains
Theorem 16.5: The ring (Zn,⊕,�) is an
integral domain if and only if n is prime.
20
Page 401
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
Page 402
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.
Page 403
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
Page 404
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
Page 405
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
Page 406
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
Page 407
Fields
Theorem 16.7: Every field F is an integral
domain.
26
Page 408
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
Page 409
Fields
Theorem 16.8: Every finite integral do-
main D is a field.
28
Page 410
Fields
Proof: Let D = {d1, d2, . . . , dn} be a finite
integral domain. To show that D is a field
we must show that every nonzero element
of D has a multiplicative inverse.
Assume a ∈ D and a 6= 0. By Theorem
16.6, the elements ad1, ad2, . . . , adn are all
distinct. Since D has exactly n elements,
it follows that
D = {ad1, ad2, . . . , adn}.
In particular the unity e ∈ D, can be ex-
pressed in the form
e = adj,
for some dj ∈ D where dj 6= 0. This shows
that a has a multiplicative inverse. Hence
D is a field.
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