20160323100333chapter 3 - isomorphism and homomorphism (1)
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20160323100333chapter 3 - Isomorphism and Homomorphism (1)TRANSCRIPT
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 }
2SMA3033 CHAPTER 3 Sem 2 2015/2016
Note that, this two tables are structurally alike.
S & T are isomorphic binary structure. ( S T )
3.1 ISOMORPHISM (Cont.)
3SMA3033 CHAPTER 3 Sem 2 2015/2016
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
(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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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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3.1 ISOMORPHISM (Cont.)
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
18SMA3033 CHAPTER 3 Sem 2 2015/2016
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
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().