Isomorphism

sheisirenebkm 5,122 views 4 slides May 27, 2015
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About This Presentation

This is a handout about Homomorphism, Isomorphism, Kernel and Cayley's Theorem.
This is a group effort.

#LoveforAbstractAlgebra


Slide Content

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ISOMORPHISM

GROUP ISOMORPHISM

Let G and G’ be groups with operations ∗and ∗′. An isomorphism ?????? from a group G to a group G’ is a
one – to – one mapping (or function) from G onto G’ that preserves the group operation. That is,
?????? �∗� =?????? � ∗′??????(�) for all a, b in G.

If there is an isomorphism from G onto G’, we say that G and G’ are isomorphic and write G≅ G’.

Steps involved in proving that a group G is isomorphic to group G’:

1. “Mapping.” Define a candidate for isomorphism; that is, define a function ?????? from G to G’.
2. “1 – 1.” Prove that ?????? is one – to – one: that is, assume ?????? � =??????(�) and prove that �=�.
3. “Onto.” Prove that ?????? is onto; that is, from any element g’ in G’, find an element g in G such
that ?????? ?????? =??????′.
4. “O.P.” Prove that ?????? is operation – preserving; that is, show that ?????? �∗� =?????? � ∗′??????(�) for
all a andb in G.

Note: We can relate Steps 2 and 4 to Kernel and Homomorphism, respectively.

HOMOMORPHISM

Let G and G’ be groups with operations ∗and ∗′. A homomorphism of G into G’ is a mapping ?????? of G
intoG’ such that for every a and b in G,

?????? �∗� =?????? � ∗

?????? � .
Examples: Determine whether the given map is a homomorphism.

1. Let ??????:�

→�

under addition given by ?????? � =�
2
.
Answer: ?????? �+� = �+�
2
=�
2
+2��+�
2
≠�
2
+�
2
=?????? � +??????(�). Thus ?????? is NOT a
homomorphism.

2. Let ??????:�

→�

under multiplication given by ?????? � =�
2
.
Answer: ?????? �� = ��
2
=�
2
�
2
=�
2
�
2
=?????? � ∙??????(�). Thus ?????? is a homomorphism.

3. Let ??????:�

→�

under multiplication given by ?????? � =−�.
Answer: ?????? �� =− �� =−��≠ −� (−�)=?????? � ∙??????(�). Thus ?????? is NOT
ahomomorphism.

Types of Homomorphisms:

1. Epimorphism – surjective homomorphism
2. Monomorphism – injective homomorphism
3. Isomorphism – bijective homomorphism
4. Automorphism – isomorphism, domain and codomain are the same group
5. Endomorphism – homomorphism, domain and codomain are the same group

KERNEL

Let ??????:??????→??????′ be a homomorphism of groups. The subgroup ??????
−1
??????

= � ?????? ?????? ?????? � =??????′ is the
kernel of ??????, denoted by Ker(??????).

Examples: Find the kernel of the following homomorphism:

1. Let ??????:�

→�

under multiplication given by ?????? � =�
2
.
Answer: The identity element of the set of images is 1 under the operation of multiplication.
If �
2
=1 then �=−1 or �=1. Thus, Ker(??????)={−1,1}.

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2. Let ??????:�→� under addition given by ?????? � =5�.
Answer: The identity element of the set of images is 0 under the operation of addition. If 5x
= 0, then x = 0. Thus, Ker(??????)={0}.

COROLLARY:

A group homomorphism ??????:??????→??????′ is one – to – one map if and only if Ker(??????) = {??????}.

In view of this corollary, we can modify our steps in showing that two groups are isomorphic.

To show ??????:??????→??????′ is an isomorphism: (modified version)

1. Show ?????? is a homomorphism.
2. Show Ker(??????) = {??????}.
3. Show ?????? maps G onto G’.

Examples: (On showing isomorphism between G and G’)

A. Let us show that the binary structure <R, +> with operation the usual addition is isomorphic to
the structure <R
+, ∙> where ∙ is the usual multiplication.

Step 1. We have to somehow convert an operation of addition to multiplication. Recall from
&#3627408462;
&#3627408463;+&#3627408464;
= &#3627408462;
&#3627408463;
(&#3627408462;
&#3627408464;
) that the addition of exponents corresponds to multiplication of two quantities.
Thus we try defining ??????:&#3627408453;→&#3627408453;
+
by ?????? &#3627408485; =??????
&#3627408485;
for &#3627408485;∈&#3627408453;. Note that ??????
&#3627408485;
>0 for all &#3627408485;∈&#3627408453; so indeed
?????? &#3627408485; ∈&#3627408453;
+
.

Step 2. If ?????? &#3627408485; =??????(&#3627408486;), then ??????
&#3627408485;
=??????
&#3627408486;
. Taking the natural logarithm, we see that &#3627408485;=&#3627408486;, so ?????? is
indeed 1-1.

Step 3. If ??????∈&#3627408453;
+
, then ln⁡(??????)∈&#3627408453; and ?????? ln?????? =??????
ln ??????
=??????. Thus ?????? is onto R
+.

Step 4.For&#3627408485;,&#3627408486;∈&#3627408453;, we have?????? &#3627408485;+&#3627408486; =??????
&#3627408485;+&#3627408486;
=??????
&#3627408485;
∙??????
&#3627408486;
=?????? &#3627408485; ∙?????? &#3627408486; .

Thus, ?????? is an isomorphism.

B. Let 2&#3627408461;= 2&#3627408475; &#3627408475;∈&#3627408461; , so that 2Z is the set of all even integers, positive, negative and zero. We
claim that <Z, +> is isomorphic to <2Z, +> where + is the usual addition.

Step 1. Define ??????:&#3627408461;→2&#3627408461; by ?????? &#3627408475; =2&#3627408475; for &#3627408475;∈&#3627408461;.

Step 2. If ?????? &#3627408474; =??????(&#3627408475;), then 2&#3627408474;=2&#3627408475; so &#3627408474;=&#3627408475;. Thus ?????? is 1-1.

Step 3. If &#3627408475;∈2&#3627408461;, then n is even so &#3627408475;=2&#3627408474; for &#3627408474;=&#3627408475;/2∈&#3627408461;. Hence ?????? &#3627408474; =2(&#3627408475;/2)=&#3627408475; so ?????? is onto
2Z.

Step 4. For&#3627408474;,&#3627408475;∈&#3627408461;, we have?????? &#3627408474;+&#3627408475; =2 &#3627408474;+&#3627408475; =2&#3627408474;+2&#3627408475;=?????? &#3627408474; +??????(&#3627408475;).

Thus, ?????? is an isomorphism.

C. Determine whether the given map ?????? is an isomorphism of the first binary structure with the
second.
1. <Z, +> with <Z, +> where ?????? &#3627408475; =2&#3627408475; for &#3627408475;∈&#3627408461;.
Answer: ?????? is 1-1, but NOT onto. Thus, ?????? is NOT an isomorphism.


2. <Z, +> with <Z, +> where ?????? &#3627408475; =&#3627408475;+1 for &#3627408475;∈&#3627408461;.
Answer:?????? is 1-1, onto, but NOT operation preserving. Thus, ?????? is NOT an isomorphism.

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3. <R, ∙> with <R, ∙> where ?????? &#3627408485; =&#3627408485;
3
for &#3627408485;∈&#3627408453;.
Answer:?????? is 1-1, onto, and operation preserving. Thus, ?????? is an isomorphism.

4. <R, +> with <R
+, ∙> where ?????? ?????? =0.5
??????
for ??????∈&#3627408453;.
Answer:?????? is 1-1, onto, and operation preserving. Thus, ?????? is an isomorphism.

CAYLEY’S THEOREM

Every group is isomorphic to a group of permutations.

Example:

Let ??????={2,4,6,8}⊆Z10, where G forms a group with respect to multiplication modulo 10. Write out
the elements of a group of permutations that are isomorphic to G, and exhibit an isomorphism from
G to this group.

Solution:

Let ??????
&#3627408462;:??????→??????′ be defined by ??????
&#3627408462; &#3627408485; =&#3627408462;&#3627408485; for each &#3627408485;∈ G. Then we have the following permutations:

??????
2=



??????
2 2 =4
??????
2 4 =8
??????
2 6 =2
??????
2 8 =6

??????
4=



??????
4 2 =8
??????
4 4 =6
??????
4 6 =4
??????
4 8 =2

??????
6=



??????
6 2 =2
??????
6 4 =4
??????
6 6 =6
??????
6 8 =8

??????
8=



??????
8 2 =6
??????
8 4 =2
??????
8 6 =8
??????
8 8 =4



Thus, the set ??????

={??????
2,??????
4,??????
6,??????
8} is a group of permutations and the mapping ??????:??????→??????′ is defined
by
??????:



?????? 2 =??????
2
?????? 4 =??????
4
?????? 6 =??????
6
?????? 8 =??????
8



is an isomorphism from G to G’.

CAYLEY TABLE

A Cayley table (or operation table) of a (finite) group is a table with rows and columns labelled by
the elements of the group and the entry ??????∗ℎ in the row labelled g and column labelled h.

CAYLEY DIGRAPHS (DIRECTED GRAPHS)/CAYLEY DIAGRAMS

 For each generating set S of a finite group G, there is a directed graph representing the
group in terms of the generators of S.
 A digraph consists of a finite number of points, called vertices of the digraph, and some arcs
(each with a direction denoted by an arrowhead) joining vertices.
 In a digraph for a group G using the generator set S we have one vertex, represented by a
dot, for each element of G.
 Each generator in S is denoted by one type of arc.

Example:

At the right is a possible digraph for&#3627408461;
6 with&#3627408454;={2,3} using  for
<2> and --- for <3>.

Page 4 of 3

University of Santo Tomas
College of Education
España, Manila

MATH 115: Abstract Algebra

Performance Task 1: Oral Presentation

Written Report on
Isomorphism


Submitted by:
CAPIOSO, RIKKI JOI A.
CHUA, ERNESTO ERIC III A.
GONZALES, MA. IRENE G.
PARK, MIN YOUNG
4MM

Submitted to:
Assoc. Prof. JOEL L. ADAMOS
Course Facilitator

Date Submitted:
08 April 2015, Wednesday