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 1ir}
 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 = x1
identity
= x  (x + x )
complement
= (x  x) + (x  x )
distributivity
= (x  x) + 0
complement
= xx
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 )
= aa
= 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,
abR
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 22 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 x1.
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.
ab = ac 
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 a1  F.
a1  (a  b) = a1  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
 a1 = 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 a1  R (a  R), then
f(a1) = [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
an = (a1)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, a1  H.
Proof. (if) closure : from (a)
associativity : from G
identity : a  a1 = 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
 a1 = 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(a1) = [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) x1 = [f(a)] 1 = f(a1)  f(S) ( a1  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
(an = (a1)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) x1 = am  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, …,
am1};
(b) as = at iff s  t (mod m).
51
Proof. (1) a0, a1, a2, …, am1 are all distinct.
If ai = aj for some i, j, 0  i  j  m  1,
then aji = aj  ai = 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, …, am1} and < a >
has order m.
as = at  ast = 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
aji = a j ai = 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  aqt = as  (at)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 = {00, 01, 02} = {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 = { r20, 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 = (bh2h11) h3 = b(h2h11h3)  bH,
 aH  bH.
Similarly, aH  bH.
So, aH = bH.
(d)
Analogous to (c).
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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).
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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.
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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.
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