A set is any well-defined collection of objects, each of which is called
a member or an element of the set. The notation x ∈ A means that the
object x is a member of the set A. The notation x /∈ A means that x is not
a member of A.
Relations
Definition
r-tuple :
(a1, a2, …, ar)
ai : i-th coordinate (component)
ordered sequence
any two coordinates are not
necessarily distinct
Cartesian Product :
A1 A2 … Ar = {(a1, a2, …, ar) | ai Ai for 1ir}
Ai : set
1
A1 A2 … Ar = Ar, if A1 =A2 = … =Ar =A
Ex. A = {0, 1}, B = {a, b}.
A B = {(0, a), (0, b), (1, a), (1, b)}.
A2 = {(0, 0), (0, 1), (1, 0), (1, 1)}.
Relation : A subset of A1 A2 … Ar is called an
r-ary relation on A1, A2, …, Ar.
| A1|| A2 |... | Ar |
there are 2
relations on
A1, A2, …, Ar
Ex. A = {2, 3, 4} and B = {2, 3, 4, 5, 6}. The relation R
is defined as follows : a R b if a divides b. Hence,
R = {(2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4)}.
Binary Relations
A binary relation from set A to set B is any subset R of A × B.
An element a ∈ A is related to b ∈ B in the relation R if (a, b) ∈ R,
often written aRb. If (a, b) /∈ R, write aR/ b.
2
Representation of a binary relation :
R = {(2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4)}.
1. Relation matrix
A = {2, 3, 4}
2 3 4 5 6
2 1
3 0
4 0
0
1
0
1
0
0
1
0
0
1
1
0
B = {2, 3, 4, 5, 6}
2. Graphical representation
Let R be a binary relation on A.
R is reflexive if x A (x R x).
3
Ex. “=” and “” are reflexive.
R is irreflexive if x A (x
x).
Ex. “” and “<” are irreflexive.
R is symmetric if x, y A (x R y y R x).
Ex. “=” is symmetric.
R is asymmetric if x, y A (x R y y x).
Ex. “<” is asymmetric.
R is antisymmetric if x, y A (x R y and y R x
x = y).
Ex. “” and “” are antisymmetric.
R is transitive if x, y, z A (x R y and y R z
x R z).
Ex. “=” is transitive.
Let R1 be a relation from A1 to A2 and R2 be a relation
from A2 to A3. The composition of R1 and R2, denoted
4
by R1。R2, is a relation from A1 to A3.
R1。R2 = {(x, y)| x R1 z and z R2 y for some z A2}.
Ex. R1 = {(1, 2), (3, 4), (2, 4), (4, 2)}.
R2 = {(2, 4), (2, 3), (4, 1)}.
R1。R2 = {(1, 4), (1, 3), (3, 1), (2, 1), (4, 4), (4, 3)}.
k
Generally, R
R
R is written as Rk.
...
R0 : the identity relation, i.e., {(x, x)| x A}.
+
R =
i 1
R i is called the transitive closure of R.
R* = R0 R+ is called the reflexive transitive closure
of R.
R+ = R。R* = R*。R.
R = R+ if R is transitive.
R = R* if R is both reflexive and transitive.
A
If R is a binary relation on A, then R+ =
5
i 1
Ri .
Equivalence Relations
A binary relation R on A is an equivalence relation if
it is reflexive, symmetric and transitive.
Ex. “=” is an equivalence relation.
Ex. The relation R defined below is an equivalence
relation.
1
2
3
4
1
1
0
4 0
1
1
0
0
0
1
0
1
0
0
1
1
1
2
3
{1, 2} and {3, 4} are called
equivalence classes.
Let R be an equivalence relation on T. A subset E of T
is an equivalence class with respect to R and T if
y E (x R y);
1.
x,
2.
x E, y T
– E (x y).
6
A method to construct equivalence classes :
Ex. R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (3, 4), (3, 6),
(4, 3), (4, 4), (4, 6), (5, 5), (6, 3), (6, 4), (6, 6)}.
Initially : {1}, {2}, {3}, {4}, {5}, {6}
Scan R :
(1, 2) R
{1, 2}, {3}, {4}, {5}, {6}
(3, 4) R
{1, 2}, {3, 4}, {5}, {6}
(3, 6) R
{1, 2}, {3, 4, 6}, {5}
The set of equivalence classes with respect to R and T
is a partition of T. ({S1, S2, …, Sk} is a partition of T if
k
i 1
Si = T and Si Sj = for all i j.)
Partial Ordering, Total Ordering
Partial ordering : a relation on A is called a partial
ordering if it is reflexive, antisymmetric and transitive, where
7
A is called a partially ordered set
(poset for short).
A partial ordering is commonly denoted by
.
Ex. “”, ””, “” and “” are partial orderings.
When A is finite, a partial ordering on A can be
conveniently depicted by means of an ordering
diagram (or a Hasse diagram).
Each element is a vertex.
A vertex ai appears below another vertex aj (ai aj)
iff ai
aj.
An edge connects ai with aj iff ai
no ak such that ai
ak
aj and there is
aj .
Ex. A = {2, 3, 4, 6, 8, 12, 36, 60}. A partial ordering
defined on A is : i | j iff i is a divisor of j.
8
minimal elements : 2, 3
maximal elements : 8, 36, 60
upper (lower) bound of 4, 6 : 12, 36, 60 (2)
least (greatest) upper (lower) bound of 4, 6 : 12 (2)
If we “stretch” the ordering diagram in such a way
that all vertices are aligned in a single column, with
all descending paths preserved, we get a topological
order of the elements of A. (The topological order for
a poset is not unique)
9
The elements a1, a2, …, an of a poset A are in a
topological order iff there exists no i and j,
1 j < i n, for which aj
ai.
Total ordering : a partial ordering
on A is called a
total ordering if for all ai, aj A,
either ai
aj or aj
Ex. “” and “” are total ordering.
10
ai.
The ordering diagram for a total ordering is a chain.
11
Boolean Algebra
Definition
K : a set of distinct elements
+, : two binary operators
(K, , +) is a Boolean algebra iff the following holds:
1. Closure under and +
For all a, b K, a b K and a + b K.
2. Commutativity of and +
For all a, b K, a b = b a and a + b = b + a.
3. Distributivity of and +
For all a, b, c K, a (b + c) = a b + a c and
a + (b c) = (a + b) (a + c).
12
4. Identity and zero elements
K contains two elements 1 (identity) and 0 (zero) :
a 1 = a and a + 0 = a for all a K.
5. Complement
For every a K, there exists a ( a) such that
a a = 0 and a + a = 1.
a is the complement of a.
6. There are at least two distinct elements a and b
(a b) in K.
Ex. Let K = {1, 2, 3, 5, 6, 10, 15, 30} be the set of all
positive integer divisors of 30. For any a, b K,
define a + b (a b) to be the l.c.m. (g.c.d.) of a, b,
and a = 30/a. Then, with 1 as the zero and 30 as
the identity, (K, , +) is a Boolean algebra.
13
Proof of a + (b c) = (a + b) (a + c) :
Let a = 2 k 3k 5k , b = 2m 3m 5m , c = 2 n 3n 5n .
1
3
2
1
2
3
1
2
3
Then b c = 2 s 3s 5s , where si = min{mi, ni}. So,
1
3
2
a + (b c) = 2t 3t 5t , where ti = max{ki, min{mi, ni}}.
1
2
3
Also, (a + b) (a + c) = 2u 3u 5u , where ui = min{max{
1
2
3
ki, mi}, max{ki, ni}}. Since ki, mi and ni are all either
0 or 1, ti = ui.
Ex. ({true, false}, , ) is a Boolean algebra.
1. Distributivity may be verified by the truth table
method.
2. The identity and zero are true and false,
respectively.
Let α and β be two Boolean expressions. α and β are
said to be duals of each other, if one can be derived
from the other by using the following substitution.
1. Replace all occurrences of by + and + by .
2. Replace all occurrences of 0 by 1 and 1 by 0.
14
Ex. (a + b) and a b are duals of each other.
(a b c) + (c d) + (a f) and (a + b + c) (c + d) (a + f)
are duals of each other.
Note that dual Boolean expressions appear in the
definitions of closure, commutativity, distributivity,
identity, zero and complement.
Theorem. (Principle of Duality) If S is a theorem
about a Boolean algebra, and S can be proved with
closure, commutativity, distributivity, identity, zero,
complement and some properties derived from them,
then it’s dual is likewise a theorem.
15
Ex. Proof of x + x = x, where (K, , +) is a Boolean
algebra and x K.
x = x+0
zero
= x + (x x )
complement
= (x + x) (x + x )
distributivity
= (x + x) 1
complement
= x+x
identity
Proof of x x = x
x = x1
identity
= x (x + x )
complement
= (x x) + (x x )
distributivity
= (x x) + 0
complement
= xx
zero
16
Theorem. Let (K, , +) be a Boolean algebra.
(1) The identity and zero are unique.
(2) a a = a and a + a = a for every a K.
(3) a 0 = 0 and a + 1 = 1 for every a K.
(4) a is unique for every a K.
(5) (a ) = a for every a K.
(6) The identity and zero are distinct. Also, 1 = 0
and 0 = 1.
(7) a (a + b) = a and a + (a b) = a for every a, b
K.
(8) a b = a c and a b = a c
a + b = a + c and a + b = a + c
b = c.
b = c.
(9) a (b c) = (a b) c and a + (b + c) = (a + b) + c
for every a, b, c K.
(10) (DeMorgan’s law)
a b = a
for every a, b K.
17
+ b and
a b = a b
Proof. (1) Suppose 1 and 1’ are two identities.
1 = 1’ 1 = 1’.
(3) a 0 = (a 0) + 0
= (a 0) + (a a )
= a (0 + a )
= aa
= 0.
(4) Suppose a and a ' are complements of a.
a a ' = a a ' + 0 = ( a a ' ) + ( a a)
= a ( a ' + a) = a 1 = a .
Similarly, a ' a = a ' .
Thus, a = a ' .
(5) a = a 1 = a ( a + (a ) ) = a a + a (a )
= 0 + a (a ) = a (a ) .
Similarly, (a ) = (a ) a.
Thus, a = (a ) .
18
(6) Let a K and a 1.
If 1 = 0, then a + 1 = a + 0, which implies
1 = a, a contradiction.
1 = 1 1 = 0.
(7) a (a + b) = (a a) + (a b) = a + a b
= a 1 + a b = a (1 + b) = a 1 = a.
(8) b = 1 b = (a + a ) b = a b + a b
= a c + a c = (a + a ) c = 1 c = c.
(9) a + (a (b c)) = (a + a) (a + (b c))
= a (a + (b c)) = a.
a + ((a b) c) = (a + (a b)) (a + c)
= a (a + c) = a.
Similarly, a + (a (b c)) = a + ((a b) c) =
a + (b c).
Thus, from (8), a (b c) = (a b) c.
19
(10) (a b) + ( a + b ) = ((a b) + a ) + b
= ((a + a ) (b + a )) + b = (1 (b + a )) + b
= (b + a ) + b = ( a + b) + b = a + (b + b ) = a + 1
= 1.
(a b) ( a + b ) = ((a b) a ) + ((a b) b )
= 0 + 0 = 0.
Thus, a + b is the complement of a b.
20
Rings
Definition
R : a set of distinct elements
+, : two binary operators
(R, +, ) is a ring if for all a, b, c R, the following
are satisfied :
1. Closure under + and
a + b R,
abR
2. Associativity of +
a + (b + c) = (a + b) + c
3. Commutativity of +
a+b = b+a
21
4. Identity for +
There exists z R such that a + z = z + a = a
for every a R.
5. Inverse under +
For each a R, there exists b R with
a + b = b + a = z.
6. Associativity of
A (b c) = (a b) c
7. Distributivity of over +
a (b + c) = (a b) + (a c)
(b + c) a = (b a) + (c a)
Ex. Under ordinary addition and multiplication, Z,
Q, R, C are rings. Their additive identity is 0,
and the additive inverse of x is – x.
The identity z for + is often referred to as the zero of
the ring.
22
Let (R, +, ) be a ring.
1. If a b = b a for all a, b R, then R is called a
commutative ring.
2. R is said to have no proper divisor of zero if for
any a, b R, a b = z a = z or b = z.
3. If there exists u R such that a u = u a = a for all
a R, we call u the unity, or multiplicative identity,
of R. R is then called a ring with unity.
Ex. Let M2(z) denote the set of all 22 matrices with
integer components. We define
a b e
c d g
f a e b f
;
h c g d h
a b e
c d g
f ae bg af bh
.
h ce dg cf dh
(M2(z), +, ) is a ring.
23
0 0
(a) additive identity z =
.
0
0
a b
a b
(b) additive inverse of
is c d .
c
d
(c) (M2(z), +, ) is not commutative.
1 2 3 7 5 7 10 13 3 7 1 2
1 1 1 0 4 7 1 1 1 0 1 1 .
(d) (M2(z), +, ) has proper divisors of zero.
1 1 2 1 0 0
1 1 2 1 0 0 .
Let R be a ring with unity u. If a, b R and a b =
b a = u, then b (a) is called a multiplicative inverse
of a (b).
24
Integral Domain
Let R be a ring. Then, R is called an integral domain
if the following hold.
1. R is commutative.
2. R has a unity u (u ≠ z). (u ≠ z means that an
integral
domain has at least two elements.)
3. R has no zero divisor.
Field
Let R be a ring. Then R is called a field if the following
hold.
1. R is commutative.
2. R has a unity u (u ≠ z).
3. There is a multiplicative inverse of a (≠ z) for every
a R.
25
Properties of Rings
Theorem. For any ring (R, +, ),
(a) the zero (additive identity) z is unique;
(b) the additive inverse of each a R is unique.
Proof.
(a) Let z1 and z2 be two zeros. Then,
z1 = z1 + z2 = z2.
(b) Let b and c be two additive inverse of a.
a+b = b+a = z
and
a + c = c + a = z.
Then, b = b + z = b + (a + c) = (b + a) + c
= z + c = c.
As a result of the uniqueness of the additive inverse,
we denote the additive inverse of a by – a.
26
Theorem. (Cancellation Laws of Addition)
For a, b, c R,
(a) a + b = a + c
⇒
(b) b + a = c + a ⇒
b=c;
b = c.
A general ring does not satisfy the cancellation laws of
multiplication.
Theorem. a z = z a = z for any a R.
Proof. z + a z = a z = a (z + z) = a z + a z
z = a z.
Theorem. Suppose that (R, +, ) is a ring.
For any a, b R,
(a) – (– a) = a ;
(b) a (– b) = (– a) b = – (a b) ;
(c) (– a) (– b) = a b.
27
Proof.
(a) a + (– a) = z. So, a is the additive inverse of – a.
(b) a b + a (– b) = a (b + (– b)) = a z = z.
So, a (– b) is the additive inverse of a b.
(c) From (b), (– a) (– b) = – (a (– b)) = – (– (a b)).
From (a), – (– (a b)) = a b.
Theorem. Given a ring (R, +, ),
(a) if R has a unity, it is unique;
(b) if R has a unity and x R, the multiplicative
inverse of x is unique.
Proof. Left as an exercise.
As a result of the theorem, we denote the
multiplicative inverse (if it exists ) of x by x1.
28
Theorem. Let (R, +, ) be a commutative ring with unity.
Then, R is an integral domain if and only if for a, b, c R,
a z, a b = a c b = c.
(Hence, a commutative ring with unity that satisfies the
cancellation law of multiplication is an integral domain.)
Proof. (if) Let a, b R with a, b = z.
If a z, then b = z since a b = z = a z.
So, R has no proper divisor of zero.
(only if ) Let a, b c R, a z, and a b = a c.
ab = ac
a b + (– (a c)) = z
a (b + (– c)) = z
b + (– c) = z
b = – (– c) = c.
The cancellation law of multiplication does not imply
the existence of multiplicative inverse. For example, the
integral domain (R, +, ) satisfies the cancellation law of
multiplication, but contains only two elements, 1 and
– 1, which have multiplicative inverses.
29
Theorem. If (F, +, ) is a field, then it is an integral domain.
Proof. Let a, b F with a b = z.
If a z; then a1 F.
a1 (a b) = a1 z.
u b = z.
b = z.
An integral domain is not necessarily a field.
Theorem. A finite integral domain (D, +, ) is a field.
Proof. D is finite D = {d1, d2, …, dn}, where di’s
are distinct.
Let a D and a z.
D is an integral domain a d1, a d2, …, a dn
are all distinct.
Hence, {d1, d2, …, dn} = {a d1, a d2, …, a dn}.
u D u = a dk = dk a for some k
a1 = dk D.
30
Subring
For a ring (R, +, ), a nonempty subset S of R is said to
be a subring of R, if (S, +, ) is a ring.
Ex. The set of all even integers is a subring of (z, +, ).
In fact, for any n z+, nz = {nx | x z} is a subring
of (z, +, ).
Ex. (z, +, ) is a subring of (Q, +, ).
Theorem. Given a ring (R, +, ), a nonempty subset S
of R is a subring of R iff
1. for all a, b S, a + b S and a b S;
2. for all a S, a S.
Proof. See the textbook.
31
Theorem. For any ring (R, +, ), if S R and S ,
then
1. (S, +, ) is a subring of R iff for a, b S,
a + ( b) S and a b S;
2. if S is finite, then (S, +, ) is a subring of R iff
for a, b S, a + b S and a b S.
Ideal
A subset I of a ring R is an ideal of R if the following
hold:
1. I is a subring of R;
2. x I and r R imply x r I and r x I.
The Integer Modulo n
Let n z+ and n 1. For a, b z, we say that a is
congruent to b modulo n, denoted by a b (mod n), if
a = b + kn, for some k z.
32
Ex. 17 2 (mod 5); 7 49 (mod 6).
Define a R b iff a b (mod n).
Theorem. The relation R (congruence modulo n) is an
equivalence relation on z.
Proof. Left as an exercise.
Congruence modulo n partitions z into n equivalence
classes.
[0] = {0+nx | x z} = {…, 2n, n, 0, n, 2n, …}.
[1] = {1+nx | x z} = {…, 2n+1, n+1, 1, n+1, 2n+1, …}.
[2] = {2+nx | x z} = {…, 2n+2, n+2, 2, n+2, 2n+2, …}.
.
.
.
.
.
.
[n 1] = {(n 1)+ nx | x z} = {…, n 1, 1, n 1, 2n 1,
3n 1, …}.
33
Let zn = {[0], [1], [2], …, [n 1]}.
For [a], [b] zn, define + and‧as follows:
[a] + [b] = [a + b] and [a] [b] = [ab].
Ex. For n = 7, [2] + [6] = [8] = [1] and
[2] [6] = [12] = [5].
Theorem. For n z+ and n 2, (zn, +, ) is a
commutative ring with unity [1].
Proof. Left as an exercise.
Ex. z5 and z6.
[i] is denoted by i
+
0
1
2
3
4
.
0
1
2
3
4
0
1
2
3
0
1
2
3
1
2
3
4
2
3
4
0
3
4
0
1
4
0
1
2
0
1
2
3
0
0
0
0
0
1
2
3
0
2
4
1
0
3
1
4
0
4
3
2
4
4
0
1
2
3
4
0
4
3
2
1
z5
z5 is a field since every nonzero element has a
multiplicative inverse.
34
+
0
1
2
3
4
5
.
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
1
2
3
4
5
0
2
3
4
5
0
1
3
4
5
0
1
2
4
5
0
1
2
3
5
0
1
2
3
4
0
1
2
3
4
5
0
0
0
0
0
0
0
1
2
3
4
5
0
2
4
0
2
4
0
3
0
3
0
3
0
4
2
0
4
2
0
5
4
3
2
1
z6
z6 is not a field.
Theorem. zn is a field iff n is a prime.
Proof. (if) Suppose 0 < a < n. gcd(a, n) = 1.
there exist integers s, t with as + nt = 1
as 1 (mod n)
[a] [s] = [as] = [1]
[s] is the multiplicative inverse of [a].
So, zn is a field.
(only if ) Assume n = n1n2 is not a prime.
[n1] [0] and [n2] [0].
But, [n1] [n2] = [n1n2] = [n] = [0].
So, zn is not an integral domain
zn is not a field.
35
Theorem. In zn, [a] has a multiplicative inverse iff
gcd(a, n) = 1.
Proof. Similar to the proof above.
Ring Homomorphism and Isomorphism
Let (R, +, ) and (S, , ) be rings. A function
f : R S is called a ring homomorphism if for all
a, b R,
(a) f(a + b) = f(a) f(b);
(b) f(a b) = f(a) f(b).
Ex. Consider (z, +, ) and (z6, +, ).
Define f : z z6 by f(x) = [x].
For any x, y z,
f(x + y) = [x + y] = [x] + [y] = f(x) + f(y);
f(x y) = [x y] = [x] [y] = f(x) f(y).
So, f is a ring homomorphism.
36
Let f : (R, +, ) ( S, , ) be a ring homomorphism.
If f is one-to-one and onto, then f is called a ring
isomorphism and we say that R and S are isomorphic
rings.
For (R, +, ) and a R, we define
1. 0a = z, 1a = a, (n + 1)a = na + a, and ( n)a = n( a),
where n 1;
2. a0 = u, a1 = a, and an+1 = an a.
Theorem. If f : (R, +, ) (S, , ) is a ring
homomorphism, then
(a) f(zR) = zs , where zR and zs are the zeros of R and
S;
(b) f( a) = f(a) for any a R;
(c) f(na) = nf(a) for any a R and n Z;
(d) f(an) = [f(a)]n for any a R and n Z+;
(e) if A is a subring of R, f(A) is a subring of S.
37
Proof. (a) zs f(zR) = f(zR) = f(zR + zR) = f(zR) f(zR)
zs = f(zR).
(b) zs = f(zR) = f(a + ( a)) = f(a) f( a)
f( a) is the additive inverse of f(a)
f( a) = f(a).
(c) By induction on n ( 0),
n = 0,
f(0a) = f(zR) = zs = of(a);
n = k,
f(ka) = kf(a);
n = k + 1,
f((k + 1)a) = f(ka + a) = f(ka) f(a)
= kf(a) f(a) = (k + 1)f(a).
When n 0, f(( n)a) = f(n( a)) = nf( a)
= n( f(a)) = ( n)f(a).
(d) Left as an exercise.
(e) For any x = f(a) f(A), y = f(b) f(A) (a, b A)
x y = f(a) f(b) = f(a + b) f(A) (since a + b A)
x y = f(a) f(a) = f(a b) f(A) (since a b A)
f( a) = f(a) = x f(A) (since a A)
f(a) is a subring of S.
38
Theorem. If f : (R, +, ) (S, , ) is a ring
homomorphism from R onto S, then
(a) if R has unity uR, f(uR) is the unity of S;
(b) if R has unity uR and a1 R (a R), then
f(a1) = [f(a)] 1 S;
(c) if R is commutative, then S is commutative;
(d) if I is an ideal of R, then f(I) is an ideal of S.
Proof. (a), (b) ,(c) left as an exercise.
(d) I is a subring of R f(I) is a subring of S.
Let x f(I) and y S.
x = f(a) for a I and y = f(b) for b R
x y = f(a) f(b)
= f(a b) f(I) (since a b I).
Similarly, y x f(I).
f(I) is an ideal of S.
39
Ex. Let C be the set of complex numbers and S be
the set of real matrices of the form
a b
b a .
(C, +, ) is a field and (S, +, ) is a ring.
a b
Define f : C S by f(a + bi) =
.
b
a
1. f((a + bi) + (x + yi)) = f((a + x) + (b + y)i)
b y
a x
a b
x
=
=
+
y
b a
( b y ) a x
y
x
= f(a + bi) + f(x + yi).
2. f((a + bi) (x + yi)) = f((ax by) + (bx + ay)i)
bx ay
ax by
a b x
=
=
b a y
( bx ay ) ax by
= f(a + bi) f(x + yi).
40
y
x
3. f is one-to-one and onto.
f is a ring isomorphism.
We can compute (4 + 5i)(2 3i) through matrix
operations as follows.
(4 + 5i)(2 3i) =
f-1f((4 + 5i)(2 3i))
=
f-1(f(4 + 5i)f(2 3i))
=
4 5 2 3
f-1 5 4 3 2
=
23 2
f-1 2 23
=
23 2i.
41
Groups
Definition
G : a nonempty set
:
a binary operation
(G, ) is called a group if the following hold.
1. Closure
For a, b G, a b G.
2. Associativity
For a, b, c G, a (b c) = (a b) c.
3. Identity
There exists e G with a e = e a = a for all a G.
4. Inverses
For each a G, there exists b G with a b =
b a = e.
42
Let (G, ) be a group. If a b = b a for all a, b G,
then G is called a commutative, or abelian, group.
Ex. Under ordinary addition, each of Z, Q, R, C is an
abelian group. None of these are groups under
multiplication since 0 has no multiplicative inverse.
Ex. If (R, +, ) is a ring, then (R, +) is an abelian
group.
Theorem. For any group G,
(a) the identity of G is unique;
(b) the inverse of each element of G is unique;
(c) if a, b, c G and a b = a c, then b = c;
(d) if a, b, c G and b a = c a, then b = c;
(e) G is abelian iff (ab)2 = a2 b2 for all a, b G.
Proof. Left as an exercise.
43
a-1 : inverse of a.
Define a0 = e, a1 = a, an+1 = an a for n 1, and
an = (a1)n.
Ex. G = (z6, +) is an abelian group.
Let H = {[0], [2], [4]}. H is a subset of G.
+
[0]
[2]
[4]
[0]
[0]
[2]
[4]
[2]
[2]
[4]
[0]
[4]
[4]
[0]
[2]
(H, +) is a group.
Let G be a group and H G. If H is a group under
the binary operation of G, we call H a subgroup of G.
{e} is said to be the trivial subgroup of G.
44
Theorem. If H is a nonempty subset of a group G, then
H is a subgroup of G iff
(a)
for all a, b H, a b = H;
(b)
for all a H, a1 H.
Proof. (if) closure : from (a)
associativity : from G
identity : a a1 = e H
inverse : from (b)
(only if) trivial
Theorem. If G is a group and H G with H finite,
then H is a subgroup of G iff H is closed under the
binary operation of G.
45
Proof. (if) Let a H = {h1, h2, …, hn}, where n = |H|
is finite.
a H = {a h1, a h2, ..., a hn} = H
a hi = a for some i
hi = e
a hj = e for some j.
(hj a)2 = (hj (a hj)) a = (hj e) a = hj a
hj a = e = a hj
a1 = hj H
H is a subgroup of G.
(only if) trivial
Theorem. Let (G, ) and (H, ) be groups. Define the
binary operation on G H by (g1, h1) (g2, h2) = (g1
g2, h1 h2). Then, (G × H, ) is a group, called the
direct product of G and H.
Proof. Left as an exercise.
46
Ex. Consider the groups (z2, +) and (z3, +). Define
on z2 × z3 by (a1, b1) (a2, b2) = (a1 + a2, b1 + b2). Then
(z2 × z3, ) is a group with identity ([0], [0]). The inverse,
for example, of ([1], [2]) is ([1], [1]).
Group Homomorphism
If (G, ) and (H, ) are groups and f : G H, f is
called a group homomorphism if for all a, b G,
f(a b) = f(a) f(b).
Ex. Let G = (z, +) and H = (z4, +). Define f : G H
by f(x) = [x] = {x + 4k | k z}.
For any x, y G,
f(x + y) = [x + y] = [x] + [y] = f(x) + f(y)
f is a group homomorphism.
47
Theorem. Let (G, ), (H, ) be groups with respective
identities eG, eH. If f : G H is a homomorphism, then
(a) f(eG) = eH;
(b) f(a1) = [f(a)]1 for any a G;
(c) f(S) is a subgroup of H for any subgroup S of G.
Proof.
(a) eH f(eG) = f(eG) = f(eG eG) = f(eG) f(eG)
eH = f(eG).
(b) Left as an exercise.
(c) Let a, b S. Then, x = f(a) f(S) and
y = f(b) f(s).
(i)
x y = f(a) f(b) = f(a b) f(S)
( a b S)
(ii) x1 = [f(a)] 1 = f(a1) f(S) ( a1 S)
f(S) is a subgroup of H.
48
If f : (G, ) (H, ) is a homomorphism, we call f an
isomorphism if it is one-to-one and onto. In this case,
G and H are said to be isomorphic groups.
Ex. Define f : (R+, ) (R, +) by f(x) = log10(x).
f is one-to-one and onto.
For a, b R+, f(a b) = log10(a b) = log10 a +
log10 b = f(a) + f(b)
f is an isomorphism.
Ex.
G = ({1, 1, i, i}, ) is a group.
H = (z4, +) is a group.
Define f : G H by f(1) = [0], f( 1) = [2],
f(i) = [1] and f( i) = [3]. f is an isomorphism.
For example, f(i ( i)) = f(1) = [0] = [1] + [3] =
f(i) + f( i).
Further, ({1, 1}, ) is a subgroup of G, and
(f({1, 1}), ) = ({[0], [2]}, +) is a subgroup of H.
49
i1 = i, i2 = 1, i3 = i
and i4 = 1
every element of G is a power of i.
It is said that i generates G and denoted by G = < i >.
Cyclic Groups
A group G is cyclic if there is a G such that for all x
G, x = an for some n Z. In this case, G is denoted by
G = < a >, and a is said to be a generator of G.
(Note that G = {an | n Z}.)
Ex.
Consider the group (Z, ).
Z = 1 and Z = 1 .
For example, 3 = (1)3 = 1 1 1.
3 = (1)3 = ( 1)3
(an = (a1)n)
= ( 1) ( 1) ( 1).
3 = ( 1)3 = ( ( 1))3 = (1)3
= 1 1 1.
3 = ( 1)3= ( 1) ( 1) ( 1).
50
Theorem. Let G be a group, a G, and S = {ak | k Z}.
Then, S is a subgroup of G. This subgroup is called
the subgroup generated by a and denoted by a .
Proof. Let x = am S and y = an S.
(1) x y = am an = am+n S.
(2) x1 = am S.
S is a subgroup.
If G is a group and a G, the order of a, denoted by
o(a), is | a |. If | a | is infinite, we say that a has
infinite order.
Theorem. Let a be an element in a group G, and
suppose an = e for some positive integer n. If m is the
least positive integer such that am = e, then
(a) a has order m and a = {a0 = e = am, a1, a2, …,
am1};
(b) as = at iff s t (mod m).
51
Proof. (1) a0, a1, a2, …, am1 are all distinct.
If ai = aj for some i, j, 0 i j m 1,
then aji = aj ai = e. A contradiction!
(2) For any k, ak = ar for some 0 r m 1.
k = mq + r, 0 r m 1.
ak = a mq+r = amqar = ar.
(1), (2) < a > = {e, a, a2, …, am1} and < a >
has order m.
as = at ast = e
s t 0 (mod m)
s t (mod m).
Theorem. Let G be a cyclic group.
(a) If G is infinite, then G is isomorphic to (z, +).
(b) If |G| = n, then G is isomorphic to (zn, +).
52
Proof. (a) Let G = <a> = { ak | k Z}.
(1) ai aj for all i, j.
if ai = aj for i j, then
aji = a j ai = e. (assume j > i)
G is finite, a contradiction !
(2) Define f : G Z by f(ak) = k.
f(am an) = f(am+n) = m + n = f(am) + f(an).
(3) f is one-to-one and onto
f is an isomorphism
(b) Left as an exercise.
Theorem. Any subgroup of a cyclic group is cyclic.
53
Proof. Let G = < a > be a cyclic group and H be a
subgroup of G.
Assume H {e}. Let t be the smallest positive integer
such that at H.
(1) < at > H
by the closure property
(2) < at > H
Assume as H, where s = qt + r, q, r Z, and
0 r t.
ar = as aqt = as (at)q H
a contradiction !
(1), (2) H = < at >.
54
Cosets and Lagrange’s Theorem
Suppose that H is a subgroup of G. For any a G, the
set a H = {a h | h H} (H a = {h a | h H}) is a left
coset (right coset) of H in G.
Ex. Suppose G = (z12, ) and H = {[0], [4], [8]}.
[0] H = {[0], [4], [8]} = H.
[4] H = {[0], [4], [8]} = H.
[8] H = {[0], [4], [8]} = H.
[1] H = [5] H = [9] H = {[1], [5], [9]}.
[2] H = [6] H = [10] H = {[2], [6], [10]}.
[3] H = [7] H = [11] H = {[3], [7], [11]}.
H ([1] H) ([2] H) ([3] H) is a partition of G.
55
Ex. G = {0, 1, 2, r1, r2, r3}, where
1 2 3
0 =
1
2
3
1 2 3
1 =
3
1
2
1 2 3
2 =
2
3
1
1 2 3
r1 =
2
1
3
1 2 3
r2 =
1
3
2
1 2 3
r3 =
,
3
2
1
is a group. H = {0, 1, 2,} is a subgroup of G.
0H = {00, 01, 02} = {0, 1, 2} = H.
1H = 2H = H.
r1H = r2H = r3H ={r1, r2, r3}.
H r1H is a partition of G.
K = {0, r1} is a subgroup of G.
Kr2 = {0r2, r1r2} = {r2, 1}.
r2K = { r20, r2r1} = { r2, 2}.
Kr2 r2K.
56
Theorem. If H is a subgroup of a finite group G, then
for any a, b G,
(a) |aH| = |H|;
(b) |Ha| = |H|;
(c) aH = bH or aH bH = ;
(d) Ha = Hb or Ha Hb = .
Proof. (a) Let hi, hj H.
hi hj ahi ahj.
(b)
Analogous to (a).
(c)
Assume aH bH .
|aH| = |H|.
Let c = ah1 = bh2, where h1, h2 H.
If x = ah3 aH, where h3 H,
then x = (bh2h11) h3 = b(h2h11h3) bH,
aH bH.
Similarly, aH bH.
So, aH = bH.
(d)
Analogous to (c).
57
Theorem. Let H be a subgroup of a finite group G.
(a) The distinct left cosets of H in G form a
partition of G.
(b) The distinct right cosets of H in G form a
partition of G.
Proof. (a) (i) e H, where e is the identity of G.
(ii) aH = bH or aH bH , where a, b G.
(iii) For each g G,
g gH.
(i), (ii), (iii) distinct left cosets of H in G
form a partition of G.
(b) Analogous to (a).
58
Theorem. (Lagrange’s Theorem) Let H be a subgroup
of a finite group G. Then, |H| divides |G|.
Proof. (i) |aH| = |H| for all a G.
(ii) Distinct left cosets of H in G form a
partition of G.
(i), (ii) |H| divides |G|.
Corollary. If G is finite and a G, then o(a) divides |G|.
Corollary. Any group of prime order is cyclic.
Lagrange’s theorem is useful to finding all the
subgroups of a finite group.
59
Ex. G = {0, 1, 2, r1, r2, r3}, where
1 2 3
0 =
1
2
3
1 2 3
1 =
3
1
2
1 2 3
2 =
2
3
1
1 2 3
r1 =
2
1
3
1 2 3
r2 =
1
3
2
1 2 3
r3 =
,
3
2
1
is a group. Find all of the subgroups of G. *****
|G| = 6
Any subgroup of G has 1, 2, 3 or 6
elements.
2, 3 are prime The subgroups of G having 2 or 3
elements are cyclic.
# of elements = 1 :
{0}.
# of elements = 6 :
G.
# of elements = 2 or 3 :
< 1 > = < 2 > = {0, 1, 2}.
< r1 > = {0, r1}.
< r2 > = {0, r2}.
< r3 > = {0, r3}.
Subgroups of sizes 2 and 3 are cyclic.
60
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