20160323100333chapter 3 - isomorphism and homomorphism (1)

ALGEBRAIC STRUCTURESSMA 3033

SEMESTER 2 2015/2016

CHAPTER 3 : ISOMORPHISM & HOMOMORPHISM

BY:DR ROHAIDAH HJ MASRI

1SMA3033 CHAPTER 3 Sem 2 2015/2016

3.1 ISOMORPHISM

Example 1

* a b c

a c a b

b a b c

c b c a

*’ x y z

x z x y

y x y z

z y z x

Compare

S = { a, b, c } T = { x, y, z }

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Note that, this two tables are structurally alike.

S & T are isomorphic binary structure. ( S T )

3.1 ISOMORPHISM (Cont.)

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Definition 1 (Isomorphism)

Let G be a group with operation * , and let H be a group with operation #. An isomorphism of G onto H is a mapping : G H that is one to one and onto satisfies

(a * b) = (a) # (b) for all a, b in G.

Note:An isomorphism from a group G to G itself : : G Gis called an automorphism of G.

The condition (a * b) = (a) # (b)

is described as preserves the operation.

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Some steps to show that binary structures are isomorphic :

Let < S, * > & < S’, *’ > are two binary structures.

1.Define the function that gives the isomorphism of S with S’ .2.Show that is a one to one function.3.Show that is onto S.4.Show that ( a * b ) = (a) *’ (b) for all a, b in S.

Example 2

Let 2Z = { 2n | nZ } and : Z 2Z. Show that a binary structure < Z, + > is isomorphic to the structure < 2Z, + >.

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Solution:

1.To show is 1-1.

Let m, n Z.(m) = (n) 2m = 2n

m = nThen, is 1-1.

2.To show is onto.(To show n2Z mZ such that (m) = n )

Let n2Z.(m) = n 2m = n

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m = n/2Then,

is onto.

3.To show (m + n) = (m) + (n).

Let m, n Z.LHS:

(m + n) = 2(m + n)= 2m + 2n

= (m) + (n) : RHS

Therefore, is isomorphism

Z

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Example 3

Let RP denote the positive real numbers @ R+ , and define : RP R by

(x) = log10(x) for each xRP

Show that < RP, . > is isomorphic to < R, + >.

Solution :

1.To show is 1-1.(For all x, yRP, (x)=(y) then x = y )

Let x, yRP . (x)=(y) log10(x) = log10(y) x = y.Then, is 1-1.

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2. To show is onto. (yR , xRP such that (x) = y )

Let yR.(x) = ylog10(x) = y x = 10y R.

then, is onto.

3. To show (xy) = (x) + (y).Let x, y RP.

LHS : (xy) = log10(xy) = log10(x) + log10(y) = (x) + (y) : RHS

Therefore, < Rp, . > < R, + >.

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Exercise:

1.Show that the binary structure < R, + > is isomorphic to the structure < R+, . >. Mapping : R R+ is defined by (x) = ex.

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Theorem 1

If G and H are isomorphic groups and G is abelian, then H is abelian.

Proof:

Let * be the operation of Gandlet # be the operation of H.

Suppose : G H be an isomorphism.Also, let G is Abelian.If x, y H.Then, there are elements a, b G such that,

HypothesisHypothesis

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(a) = x(b) = y.

(To show H is abelian).(To show x # y = y # x ).

x # y = (a) # (b) = (a * b )

= (b * a) = (b) # (a) = y # x.

Then, H is abelian.

Since iso preserves Since G is

abelianSince iso preserves

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Theorem 2

Suppose <G, *> has identity e for *. If : G G’ is an isomorphism of <G, * > and <G’, *’>, then (e) is the identity for the binary operation *’ on G’.

Proof:

(To show (e) is the identity in <G’, *’ > )(To show (e) *’ x’ = x’ = x’ *’ (e) )

Let x’ G’ .Since is an isomorphism is 1-1, onto & preserves

operation

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Since is onto

x’ = (x) = (e * x) = (e) *’ (x) = (e) *’ x’

x’ = (x) = (x * e) = (x) *’ (e) = x’ *’ (e)

By Eq. (1) & (2); (e) *’ x’ = x’ = x’ *’ (e).

Therefore, (e) is the identity element in G’

x’ G’ x G such that (x) = x’.

By def. identity elementSince is isomorphism(1)

(2)

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Theorem 3

Let G be a group with operation * . Let G’ be a group with operation # and : G G’ is an isomorphism. Then,

( a-1) = [(a)]-1 for all aG.

Proof:

Let aG.Since (e) is identity in G’ ;Then,

(e) = ( a * a-1 ) = (a) # ( a-1)

By Theorem 2

By def. of inverse element(1) Since

isomorphism

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Also,(e) = (a-1 * a) = (a-1) (a)

From Eq. (1) & (2); (a) # ( a-1) = (a-1) (a) = (e)

Take, (a-1) (a) = (e)

[(a-1) (a)] # [(a)]-1 = (e) # [(a)]-1 (a-1) [ (a) # [(a)]-1 ] = [(a)]-1 (a-1) (e) = [(a)]-1

(a-1) = [(a)]-1

(2)

Since G’ group G’3By G’1 & G’2

By def. of G’3

Note: If : < G, . > < G’, * > iso. then ( x . y) = (x) * (y)

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Example 4

Assume that H = { u, v, w, x, y, z } is a group with respect to multiplication and that : Z6 H is an isomorphism with

(0) = u (3) = x(1) = v (4) = y(2) = w (5) = z

Replace each the following by the appropriate letter, either u, v, w, x, y or z.

(a)xw (c) y2 v -1

(b)z-1 (d) (xy) -1 z

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Solution:

(a) xw = (3)(2) = ( 3 +6 2 ) = (5) = z

(b) z-1 = [ (5) ]-1

= (5 -1 ) = (1)

By theorem 3

How to get this??

Recall: Table of integer modulo Z

+6 0 1 2 3 4 5

0

1

2

3

4

5

0 1 2 3 4 512345

2 334

44

55

55

00

00

01

11

1 22

2

334

Try (c) & (d) !

3.2 HOMOMORPHISMS

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Definition 1 (Homomorphism)

Let < G, * > and < G’, *’ > be groups and a function from G into G’. Then is called a homomorphism of G into G’ if for all a, b in G,

( a * b ) = (a) *’ (b).

Definition 2 ( Trivial Homomorphism)

Let e’ be the identity of the group G’. Define : G G’ by (a) = e’ for all a in G.Since (a*b) = e’ = e’ *’ e’ = (a) *’ (b) for all a, b in G, we find that is a homomorphism from G to G’. Here , is called a trivial homomorphism.

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3.2 HOMOMORPHISMS

Example 1

Let rZ and let r : < Z, + > < Z, + > be defined byr(n) = r n for all nZ.

Show that r is a homomorphism.

Solution:

Let m, n in Z.LHS : r (m + n)

= r (m + n)= rm + rn= r (m) + r (n) : RHS

Then,r is a homomorphism.

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3.2 HOMOMORPHISMS

Example 2

Let

be the group under matrix multiplication. Let R* be the group of all nonzero real numbers under multiplication. Define : M R* by

Determine whether is a homomorphism.

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3.2 HOMOMORPHISMS

Solution:

LHS:

RHS

Since . & + comm. on R

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3.2 HOMOMORPHISMS

Example 3

Define a function f : < Z, + > < Z, + > by f(a) = a + 1 for all a in Z. Determine whether f is a homomorphism.

Solution:

Let a, b in Z.(To show f(a + b) = f(a) + f(b) )

LHS : f(a + b) = (a + b) + 1RHS : f(a) + f(b) = (a + 1 ) + (b + 1)

= a + b + 2.See that, LHS RHS. Then f is not a homomorphism.

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3.2 HOMOMORPHISMS

Example 4

Let R* be the group of all nonzero real numbers under multiplication. Define : R* R* by (a) = | a |. Show that is a homomorphism.

Solution:

Let a, b in R*.LHS : (ab) = |ab|

= | a || b | = (a) (b) : RHS

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3.2 HOMOMORPHISMS

3.2.1 Properties of Homomorphism

Definition 3 (Inverse Image)

Let be a function of a set X into a set Y, and let A X and B Y. The image [A] of A in Y under is

{ (a) | aA }.The set [ X ] is the range of . The inverse image -1[B] of B in X is { xX | (x)B }.

XY-1[B] B

x . . (x)

The inverse image of B

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3.2 HOMOMORPHISMS

Theorem 1

Let be a homomorphism of a group G into group G’.(a)If e is the identity element in G, then (e) is the identity element e’ in G’.(b)If aG, then (a-1) = [(a)]-1.(c)If H is a subgroup of G, then [H] is a subgroup of G’.(d)If K’ is a subgroup of G’, then -1[K’] is a subgroup of G.

Proof:

(a)Let be a homomorphism of G into G’.Then,

(a) = (ae) = (a) (e)(1)

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3.2 HOMOMORPHISMS

Multiplying on the left of (1) by [(a)]-1:Then,

[(a)]-1 (a) = (ae) = [(a)]-1 (a) (e) e’ = (e)

(b)To show (a-1) = [(a)]-1.By (a),

e’ = (e) .Then,

e’ = (e) = (a a-1) = (a)(a-1).

e’ = (a)(a-1)Then by multiplying [(a)]-1 on both sides,

(a-1) = [(a)]-1

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3.2 HOMOMORPHISMS

Definition 4 ( Kernel of )

Let be a homomorphism of a group G into a group G’. The kernel of , denoted by Ker() is defined to be the set

Ker() = { aG | (a) = e’ }.

Example 5

Determine whether and are elements of ker(), where

is the homomorphism defined in example 2.

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3.2 HOMOMORPHISMS

Solution:

Note that : 1 is the identity for the group R* under multiplication.From the definition of ker(),

but 1(3) – 1(0) = 3, so is not an element of ker().

and 2(1) – 1(1) = 1, so, is an element of ker().

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3.2 HOMOMORPHISMS

Theorem 2

Let be a homomorphism of a group G into a group G’. Then, is one to one if and only if ker() = {e}.

Note:

Monomorphism - one to one homomorphism.

Epimorphism – onto homomorphism.

20160323100333chapter 3 - isomorphism and homomorphism (1)

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