Download Relation: A subset of A1 ´ A2 ´ … ´ Ar is called an

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
• 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)}.
1
Relation : A subset of A1 × A2 × … × Ar is called an
r-ary relation on A1, A2, …, Ar.
| A |× | A 2 |× ... × | A r |
• there are 2 1
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 iff a divides b. Hence,
R = {(2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4)}.
• Binary Relations
Representation of a binary relation :
R = {(2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4)}.
1. Relation matrix
2
3
4
5
6
A = {2, 3, 4}
2 ⎡1
3 ⎢⎢ 0
4 ⎢⎣ 0
0
1
1
0
0
0
B = {2, 3, 4, 5, 6}
0
1
0
1⎤
1 ⎥⎥
0 ⎥⎦
2
2. Graphical representation
Let R be a binary relation on A (i.e., from A to A).
• R is reflexive iff ∀ x ∈ A (x R x).
Ex. “=” and “⊇” are reflexive.
• R is irreflexive iff ∀ x ∈ A (x
x).
Ex. “⊂” and “<” are irreflexive.
• R is symmetric iff ∀ x, y ∈ A (x R y ⇒ y R x).
Ex. “=” is symmetric.
• R is asymmetric iff ∀ x, y ∈ A (x R y ⇒ y x).
Ex. “<” is asymmetric.
3
• R is antisymmetric iff ∀ x, y ∈ A (x R y and y R x
⇒ x = y).
Ex. “≤” and “⊆” are antisymmetric.
• R is transitive iff ∀ 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
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)}.
6 4 7k 4 8
Generally, R o R o . . . o R is written as Rk.
4
R0 : the identity relation, i.e., {(x, x)| x ∈ A}.
+
R =
∞
R i is called the transitive closure of R.
U
i =1
*
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 −1
i
• If R is a binary relation on A, then R = U R .
+
i =1
• Equivalence Relations
A binary relation R on A is an equivalence relation iff
it is reflexive, symmetric and transitive.
Ex. “=” is an equivalence relation.
Ex. The relation R defined below is an equivalence
relation.
5
1
2
3
4
1
0
1
0
0
1
0
1
0⎤
0 ⎥⎥
1⎥
⎥
1⎦
1 ⎡1
2 ⎢⎢ 1
3 ⎢0
⎢
4 ⎣0
{1, 2} and {3, 4} are called
equivalence classes.
Let R be an equivalence relation on A. A subset E of A
is an equivalence class with respect to R and A iff
y ∈ E (x R y);
1.
∀ x,
2.
∀ x ∈ E, ∀ y ∈ A
– E (x y).
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}
6
The set of equivalence classes with respect to R and A
is a partition of A. ({S1, S2, …, Sk} is a partition of A iff
k
Si = A and Si ∩ Sj = φ for all i ≠ j.)
U
i =1
• Partial Ordering, Total Ordering
Partial ordering : a relation on A is called a partial
ordering if it is reflexive, antisymmetric and transitive, where
A is called a partially ordered set
(poset for short).
A partial ordering is commonly denoted by p .
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).
7
• Each element is a vertex.
• A vertex ai appears below another vertex aj (ai ≠ aj)
iff ai p aj.
• An edge connects ai with aj iff ai p aj and there is
no ak such that ai p ak p 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.
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)
8
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)
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 p ai.
9
Total ordering : a partial ordering p on A is called a
total ordering if for all ai, aj ∈ A,
either ai p aj or aj p ai.
Ex. “≤” and “≥” are total ordering.
The ordering diagram for a total ordering is a chain.
(do Exercise # 9)
10
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).
11
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.
12
Proof of a + (b ⋅ c) = (a + b) ⋅ (a + c) :
Let a = 2k 3k 5k , b = 2m 3m 5m , c = 2n 3n 5n .
1
3
2
1
2
3
1
2
3
Then b ⋅ c = 2s 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.
13
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.
14
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
15
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 ⇒
b = c.
a + b = a + c and a + b = a + 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 + b and a
for every a, b ∈ K.
16
+ 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) An immediate consequence of the definition
of complement (refer to page 12).
17
(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.
18
(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.
19
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
20
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.
21
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.
22
⎡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), and a, b are two units of R.
23
• 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.
24
• 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 inverses 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.
25
Theorem. (Cancellation Laws of Addition)
For a, b, c ∈ R,
(a) a + b = a + c ⇒
b=c;
(b) b + a = c + a ⇒ 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.
26
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.
27
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 (Z, +, ⋅) satisfies the cancellation law of
multiplication, but contains only two elements, 1 and
– 1, which have multiplicative inverses.
28
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.
(do Exercise # 10)
29
• 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. S is a ring iff z ∈ S.
z = a + (− a) ∈ S.
30
Theorem. For any ring (R, +, ⋅), if S ⊆ R and S ≠ φ,
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.
31
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, …}.
32
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
0
1
2
3
4
0
0
0
0
0
0
1
2
3
4
1
2
3
4
2
3
4
0
3
4
0
1
4
0
1
2
0
1
2
3
1
2
3
4
0
0
0
0
1
2
3
4
2
4
1
3
3
1
4
2
4
3
2
1
Z5
Z5 is a field since every nonzero element has a
multiplicative inverse.
33
+
0
1
2
3
4
5
.
0
1
2
3
4
5
0
0
1
2
3
4
5
0
0
0
0
0
0
0
1
2
3
4
5
1
2
3
4
5
2
3
4
5
0
3
4
5
0
1
4
5
0
1
2
5
0
1
2
3
0
1
2
3
4
1
2
3
4
5
0
0
0
0
0
1
2
3
4
5
2
4
0
2
4
3
0
3
0
3
4
2
0
4
2
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.
34
Theorem. In Zn, [a] has a multiplicative inverse
(equivalently, [a] is a unit) 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.
35
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.
36
Proof. (a) zS ⊕ f(zR) = f(zR) = f(zR + zR) = f(zR) ⊕ f(zR)
⇒ zS = f(zR).
(b) f(a) ⊕ f(− a) = f(a + (− a)) = f(zR) = zS
⇒ 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 = 0f(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 (also by induction on n).
(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(b) = f(a ⋅ b) ∈ f(A) (since a ⋅ b ∈ A)
− x = − f(a) = f(− a) ∈ f(A) (since − a ∈ A)
⇒ f(A) is a subring of S.
37
Theorem. If f : (R, +, ⋅) → (S, ⊕, ~) is a ring
homomorphism and onto, where |S| > 1, 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 f(S) is commutative;
(d) if I is an ideal of R, then f(I) is an ideal of S.
Proof. (a), (b) and (c) are 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
(since f is onto)
⇒ 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.
38
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
= ⎢
⎥ = ⎢− b a⎥ + ⎢− y
−
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).
39
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 ⎜ ⎢ − 5 4⎥ ⎢ 3 2 ⎥ ⎟⎟
⎦ ⎣
⎦⎠
⎝⎣
=
⎛ ⎡ 23 − 2⎤ ⎞
f ⎜ ⎢ 2 23 ⎥ ⎟⎟
⎦⎠
⎝⎣
=
23 − 2i.
-1 ⎜
-1 ⎜
40
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. Inverse
For each a ∈ G, there exists b ∈ G with a ⋅ b =
b ⋅ a = e.
41
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.
42
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 be a nonempty subset of G.
If H is a group under the binary operation of G, then
H is called a subgroup of G.
{e} is said to be the trivial subgroup of G.
43
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. Suppose that G is a group and H is a nonempty
subset of G. If H is finite, then H is a subgroup of G iff
H is closed under the binary operation of G.
44
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 = a ⋅ e 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.
45
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]).
(do Exercise # 11)
• 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].
For any x, y ∈ G,
f(x + y) = [x + y] = [x] + [y] = f(x) + f(y)
⇒ f is a group homomorphism.
46
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(an) = [f(a)]n for any a ∈ G and n ∈ Z;
(d) 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), (c) Left as an exercise.
(d) 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)
(Q a ° b ∈ S)
(ii) x−1 = [f(a)] −1 = f(a−1) ∈ f(S) (Q a−1 ∈ S)
⇒ f(S) is a subgroup of H.
47
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.
48
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).
49
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). (⇒ m | n)
50
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 (from (1))
⇔ s − t ≡ 0 (mod m) (from (2))
⇔ 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, +).
51
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.
52
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 >.
53
• 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.
54
Ex. G = {π0, π1, π2, r1, r2, r3}, where
⎛1 2 3⎞
π0 = ⎜
⎟
1
2
3
⎠
⎝
⎛ 1 2 3⎞
π1 = ⎜
⎟
3
1
2
⎠
⎝
⎛ 1 2 3⎞
π2 = ⎜
⎟
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 = {π0, π1, π2} is a subgroup of G.
π0H = {π0π0, π0π1, π0π2} = {π0, π1, π2} = H.
π1H = π2H = H.
r1H = r2H = r3H ={r1, r2, r3}.
H ∪ r1H is a partition of G.
K = {π0, r1} is a subgroup of G.
Kr2 = {π0r2, r1r2} = {r2, π1}.
r2K = { r2π0, r2r1} = { r2, π2}.
⇒ Kr2 ≠ r2K.
55
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 = φ;
(d) Ha = Hb or Ha ∩ Hb = φ.
Proof. (a) Let hi, hj ∈ H.
hi ≠ hj ⇒ ahi ≠ ahj.
(b)
Analogous to (a).
(c)
Assume aH ∩ bH ≠ φ.
∴ |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).
56
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 = φ, 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).
57
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.
58
Ex. G = {π0, π1, π2, r1, r2, r3}, where
⎛1 2 3⎞
⎜
⎟
1
2
3
⎠
⎝
π1 =
⎛ 1 2 3⎞
⎜
⎟
3
1
2
⎝
⎠
π2 =
r1 = ⎛⎜ 1 2 3 ⎞⎟
r2 =
⎛1 2 3⎞
⎟
⎜
1
3
2
⎠
⎝
r3 =
π0 =
⎝ 2 1 3⎠
⎛ 1 2 3⎞
⎜
⎟
2
3
1
⎠
⎝
⎛ 1 2 3⎞ ,
⎜
⎟
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 :
{π0}.
# of elements = 6 :
G.
# of elements = 2 or 3 :
< π1 > = < π2 > = {π0, π1, π2}.
< r1 > = {π0, r1}. < r2 > = {π0, r2}.
< r3 > = {π0, r3}.
Subgroups of sizes 2 and 3 are cyclic.
(do Exercise # 12)
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