Download Subject: Mathematics Lesson: Isomorphism and Theorems on

Isomorphism and Theorems on Isomorphism
Subject: Mathematics
Lesson: Isomorphism and Theorems on Isomorphism
Lesson Developer: Umesh Chand
Department / College: Assistant Professor, Department
of Mathematics, Kirorimal College
University of Delhi
Institute of Lifelong Learning, University of Delhi
pg. 1
Table of Contents:
Chapter: Isomorphism and Theorems on Isomorphism

1. Learning outcomes

2. Introduction

3. Isomorphism.
3.1 Properties of isomorphism.

4. External direct product of groups

5. Automorphism

Exercise

Summary

References / Bibliography / Further Reading.
Institute of Lifelong Learning, University of Delhi
pg. 2
1. Learning outcomes
After studying the whole content of this chapter, students will be able to understand

Isomorphism

Properties of Isomorphism.

Fundamental theorems of homomorphism.

Automorphism.
2. Introduction
Isomorphism is most important concept in Algebra. It is an extension of
homomorphism. With the help of isomorphism, we realizes that two or more groups
defined in different term are really the same or not. The term isomorphism is derived
from the Greek words isos, "same" or "equal" and morphs "form".
Institute of Lifelong Learning, University of Delhi
pg. 3
3. Isomorphism
An isomorphism from a group G to a group G is one to one mapping from G to G
that preserved the group operation.
Value Addition: Note
Two groups G and G are called isomorphic, written as G1 ~ G2. If there is an
isomorphism from Gonto G' i.e. two groups g and g are isomorphic if there exist
a mapping  : G  G such that
(i)
 is homomorphism.
(ii)
f is one-one
(iii)
f is onto.
Example 1: Let G be the group of real numbers under addition and G  be the group of
positive real numbers under multiplication. Then G and G are isomorphic under the
mapping (x) = 2x.
Solution: Since  : G  G is well defined.
To Show:  if one one.
let (x) = (y)

2 x = 2y

log22x = log22y

x=y
To Show:  is onto.
We must find for any positive real number y some real numbe x s.t.
(x) = y i.e. 2x = y

x = log2y  G
  is onto.
To show:  is homomoirphism.
For all x,y  G
(x + y) = 2x+y
Institute of Lifelong Learning, University of Delhi
pg. 4
= 2 x2 y
= (x).(y).
So  preserving the operation.
Example 2. The mapping  from R to R given by (x) = x3 is not isomorphism. Where
group R is set of real numbers under addition.
Solution: Since (x + y) = (x + y)3  x3 + y3 = (x) + (y)

 is not homomorphism.
Hence  is not isomorphism.
I. Q. 1
Theorem 1: The relation ' ~ ' (relation of Isomorphism) is an equivalence relation.
Proof: Let G* be the collection of all groups.
To show : ~ is an equivalence relation of G.
Reflexive : To show G ~ G  G  G*.
Define a mapping g : G  G as g(x) = x  x  G, this is identity mapping which is oneone onto and homomorphism.
Hence G ~ G.
Symmetric: Let G, G  G* such that G ~ G therefore  an isomorphism
 : G  G
  is one-one and onto   is invertible i.e. 1 exist and is also one-one onto  1 : G
 G is one-one and onto.
To show: 1 is homomorphism
i.e.
1 (xy)  1 (x)1 (y)
let
1 (x)  a

(a) = x and (b) = y

xy = (a)(b)= (ab)

1(xy) = ab = 1(x) 1(y)
and
1 (y)  b
Institute of Lifelong Learning, University of Delhi
(  is homomorphism)
pg. 5

1 : G  G is one-one onto homomorphism.
 It is an isomorphism i.e. G ~ G.
Transitive: Let G, G, G  G* such that G ~ G and G ~ G
To show: G ~ G
Since
G  G  
an isomorphism
G  G  
Consider composition
 : G  G
an isomorphism
 
 : G  G
: G  G defined as
( )(x)  ((x))
 
is well defined and one-one:
( )(x)  ( )(y)  ((x))  ((y))

(x) = (y)
[  is 11]

x=y
[  is 11]
 
is well defined and one-one.
onto: let g  G
Since
 : G  G
is onto
 g  G such that
(g)  g
(1)
Since g  G and  :G  G is onto

 g  G such that (g) = g
(2)
(1) and (2) 
g  (g)  ((y))  ( )(g)
so for
g  G
 g  G such that
( )(g)  g
Institute of Lifelong Learning, University of Delhi
pg. 6

 
To show:
is onto.
 
is hom.
( )(xy)  ((x(y))  ((x)(y))
( 
is homo)
 ((x))(y)
3.1. Properties of Isomorphism:
Let  : G  G be an isomorphism then.
(1)
 carries the identity of G to the identity of G'
Proof: Let e be the identity element of G and e' be the identity element of G. We show
that (e) = e.
(e)  (e.e)  (e).(e)
[ 
is homomorphism]
(e) = (e)(e)

e(e) = (e)(e)
Using left cancellation low.
(e) = e.
(iii)
G is abelian if and only if G is abelian.
Proof: Let G be abelian.
To show: G is abelian i.e. xy = yx  x, y  G.
Since x, y  G  x = (a) and y = (b) for some a, b  G
Since G is abelian

ab = ba
 a, b  G  a, b  G

(ab) = (ba)

(a)(b) = (b)(a)

xy = yx
(  is homo)
 x, y  G.
 G is abelian.
Institute of Lifelong Learning, University of Delhi
pg. 7
Conversely let G be abelian.
To show: G is abelian i.e. ab = ba  a, b  G
Since a, b  G  (a), (b)  G
 G is abelian.

(a)(b) = (b)(a)

(ab) = (ba)
(  is homomorphism)

ab = ba
(  is one-one)

G is ablien.
(iii)
Isomorphism preserves order i.e.
O((a))  O(a)
Proof; Let O(a) = n

an = e

(an) = (e)

[(a)]n = e

O((a)) | n
(  is homomorphism)
Let O((a)) = k  k / n
(1)
Since O((a)) = k

((a))k = e

(ak) = e
(  is homomorphism)

ak = e
  is 1  1

O(a) | k  n | k
(2)
From (1) and (2) we get n = k
i.e. O(a) = O((a))
(iv)
1 is an isomorphism from G to G.
Proof: Let G and G be two group s.t. G ~ G.
so  an isomorphism  : G  G
Institute of Lifelong Learning, University of Delhi
pg. 8
since  is one-one and onto   is invertible.

1 : G  G is one-one and onto.
To show : 1 is hom. i.e. 1(xy) = 1(x)1(y).
let
1(x) = a
and
1(y) = b i.e.
(a) = x
and
(b) = y
xy = (a) (b) = (ab)

(  is homomorphism)
1(xy) = ab = 1(x) (y)
 1 : G  G is one-one and onto homomorphism.
 It is an isomorphism i.e. G ~ G
(iv)
G is cyclic, then G is cyclic.
Proof: Let G =
a
be a cyclic.
let x  G
Since  is onto   t  G s.t. (t) = x.
Since t  G and G =
a
 t = an for some m  z.
x  (t)  (a m )  ((a)) m
( 
is homomorphism).
 Any element x  G is some power of (a)

G '  (a)
Hence G' is cyclic.
I.Q 2
Theorem 2. (Fundamental theorem of group homomorphism). If f : G  G be onto
homomorphism with K = ker f then
G
 G
K
Institute of Lifelong Learning, University of Delhi
pg. 9
Proof: Define a map
:
G
 G
K
s.t.
(Ka)  f (a)  a  G .
We shall show that  is an isomorphism.
To show:  is well defined.
let
Ka = Kb

ab1  K = ker f

f(ab1) = e

f(a)f(b1) = e

f(a) f(b)1 = e

f(a) = f(b)

(Ka) = (Kb).
By retracing the steps backwards, we shall prove that  is one-one.
To show :  is onto
let g  G
Since f is onto, so  g  G s.t. f(g) = g.

(Kg) = g
For g  G  Kg 

G
K
s.t. (Kg) = g.
 is onto.
Now we will prove  is homomorphism.
(KaKb)  (Kab)  f (ab)  f (a)f (b)
 (Ka).(Kb)
Hence  is isomorphism.
I.Q. 3
Institute of Lifelong Learning, University of Delhi
pg. 10
Value Addition: Note
The above theorem is also called first theorem of isomorphism. It can also be
stated as:
Let f : G  G is a homomorphism with K = ker f then
G
 f (G)
ker f
Value Addition: Note
Let f : G  G is a homomorphism f then
G
 f (G)
ker f

O(range f ) 
O(g)
O(ker f )
Example 3: Let O(G) = 20 and O(G) = 25, let f is nontrivial homomorphism from G to
G then find the order of kernel of f.
Solution: Since ker f is subgroup of G

O(ker f) /O(G)

ker f may has elements 1, 2, 4, 5, 10, 20
Case I: If O(ker f) = 1 i.e. only identity e.

f is one-one map

O(range f) = O(f(G)) =
We know

f (G)

O(G) 20

 20
ker f
1
is subgroup of G, so O(f(G)) | O(G). But 20/25.
O(ker f)  1.
Case II: If (ker f) = 2
Institute of Lifelong Learning, University of Delhi
pg. 11

O(f (G)) 
O(g)
20

 10
O(ker f ) 2
But f(G) < G i.e. f(G) subgroup of G.
10 | 25

Of(G)) | G But

Hence O(ker f)  2.
Case III: If O(ker f) = 4  O(f(G)) = 5 and Of(G) | O(G) as 5 / 25.
So O(ker f) may be 4.
Case IV: If O(kerf) = 5

O(f(G)) = 4. Hence Of(G)
So
O(ker f)  5.
|
O(G) since 4
|
25.
Case V: If O(ker f) = 10  O(f(G)) = 2
But 2 | 25 so O(ker f)  10.
Case VI : If O(ker f) = 20  All element of G maps to identify element e of G.,

f is trivial homomorphism and this trivial homomorphism is always possible.
Hence only case III is possible so kerf has only 4 elements.
Theorem 3: (Second Theorem of Isomorphism). Let H and K are two subgroup of G and
H is normal in G then
HK
K

H
HK
HK
K
f :K 
HK
H
Proof : Clearly
Define a map
f (k)  Hk
and as H  HK  G. H will be normal in HK
s.t.
k  K
To show: f is well defined.
Let
k1 = k2.
Institute of Lifelong Learning, University of Delhi
pg. 12

Hk1  Hk 2

f (k1 )  f (k 2 )
 f is well defined.
By retracing the steps backwards, we will prove that f is one-one.
Now
f (k1k 2 )  Hk1.k 2  Hk1.Hk 2  f (k1 )f (k 2 )
 f is homomorphism.
We will show that f is onto
HK
 x  HK
H
let
Hx 

x  hk

Hx  Hhk  Hk  f (k)
So for
Hx 

f is onto.
for some h  H and k  K
HK
, kK
H
such that
f (k)  Hx
By fundamental theorem of homomorphism.
HK
K

H
ker f
Now to show
let
ker f  H  K .
k  ker f
k  ker f  f (k)  H  Hk  H  k  H

kHK

ker f = H  K.
(k  K as ker f  K)
Theorem 4: (Third theorem of Isomorphism). Let H and K are two normal subgroups of
a group G and H  K then
Institute of Lifelong Learning, University of Delhi
pg. 13
G G/H
.

K K/H
Proof: K / H is a non-empty subset of G / H
let
Hk1,Hk 2  K / H
(Hk1 )(Hk 2 )1  (Hk1 )(Hk 21 )
 Hk1k 21  K / H.

K
H
is subgroup.
Now for any
Hk 
K
G
and Hg  .
H
H
(Hg)1 (Hk)(Hg)  Hg 1HkHg
= Hg1kg 

K
H
[ g  G, h  k, K is normal in G so g1kgK]
K/H is normal subgroup of G/H.
So we will do work on
Define a map
G/H
.
K/H
f : G/H  G/K
such that
f (Ha)  Ka  a  G
f is well defined as
Ha  Hb

ab1  H  K

ab1  K

Ka = Kb

f(Ha) = f(Hb)
 f is well defined.
Now retracing the steps backward, we will prove that f is one-one.
Institute of Lifelong Learning, University of Delhi
pg. 14
Clearly f is onto as for Ka  G/K, 
Ha 
G
such that f(Ha)  Ka
H
Now we well show f is homomorphism.
f (HaHb)  f (Hab)
[ H
is Normal in G]
= Kab
= KaKb
[  K is normal in G]
= f(Ha) f(Hb).

f is homomorphism.
By first theorem of Isomorphism.
G/H G
 .
ker f K
To show : ker f = K/H
Ha  ker f  f (Ha)  K

Ka  K

aK

Ha  K/ H
 ker f = K / H
Hence
G/H
G / K.
K/H
I. Q. 4
Example 4: Any infinite cyclic group is isomorphic to (Z, +).
Solution: Let
G a
be the given infinite cyclic group
Define a map f : G  Z such that
f(ai) = i ,
let
i Z
a i = aj
Institute of Lifelong Learning, University of Delhi
pg. 15

log ai = log aj

i log a = j log a

i=j

f(ai) = f(aj)
 f is well defined.
To show : f is one-one, onto, homomorphism.
One-One:
let
f (a i )  f (a j )

i=j

a i = aj
 f is one one
Onto : for each i  Z  ai  G such that
f(ai) =i
 f is onto.
Homomorphism:
f(aiaj) = f(ai+j)
=i+j
= f(ai) + f(aj)
 f is homomorphism.
Hence G ~ Z.
I. Q. 5
Example 5. Using first theorem of Isomorphism show that:
O(A n ) 
O(Sn )
.
2
Solution: Let G = {1, 1} be a group under multiplication
Define a map f : Sn  G such that
Institute of Lifelong Learning, University of Delhi
pg. 16
if  is even
if  is odd.
1,
f ()  
 –1,
We show that f is onto homomorphism.
Claim: f is homomorphism.
Let,
,   Sn .
Case I : Both  and  are even.
then f() = 1,
f() = 1
f() = 1

(product of two even permutations is even)
f ().f ()  1  f (
Case II : Both  and  are odd.
f ()  1
and
f ()  1
f ()  1
( product of two odd permutation is even)
f ()  1  (1)(1)  f ()f ()
Case III: If one is odd other is even.
let  is odd and  is even 
f ()f ()  (1)(1)  1
f ()  1 and f (n)  1 .
and f() = 1
(Product of odd and even is odd)
 f is homomorphism.
f is onto: As for 1  G,  an even permutation   Sn s.t. f() = 1 and for
1  G  an odd permutation   Sn s.t.f() = 1
 f is onto.
By first theorem of isomorphism.
Sn
G
ker f
 ker f  f ()  1
Institute of Lifelong Learning, University of Delhi
pg. 17


is even.


 An.

ker f = An

Sn
G
An

 Sn 
O
  O(G)  2
 An 

O(Sn)
(Sn)
 2  O(An) 
O(An)
2
Example 6: A finite cyclic group of order n is isomorphic to Zn.
Solution: Let
G a
be a cyclic grop of order n.

0(a) = O(G) = n.
Also
G  {e,a,a 2 ,....a n 1}
and
Zn = {0, 1, 2, ...., n  1}
Define a map f : G  Zn such that
f (a i )  i,  i  0,1,2,...
Now, we show that f is well defined, one-one, onto and homomorphism.
Well defined: let
ai  a j

log a i  log a j

ilog a  jlog a

i=j
One-one : let

i=j

a i = aj
f (a i )  f (a j )
 f is one-one.
Institute of Lifelong Learning, University of Delhi
pg. 18
Onto : f is onto as for i  Zn,  ai  G such that
f(ai) = j
 f is onto.
Homo:
f (a ia j )  f (a i j )
=i+j
= f(ai) + f(aj)
 f is homomorphism.
f is isomorphism G onto Zn.
Hence
G  Zn .
Example 7: Any finite cyclic group of order n is isomorphic to
Solution: Let
G n
Z
n
.
be a cyclic group of order n.
Define a map f : Z  G such that
f(i) = ai.
We show that f is well defined, one-one, onto and homomorphism.
We defined : let i =j

a i = ai

f(i) = f(j)
 f is well defined.
One-one : let f(i) = f(j)

a i = aj

i log a = j log a.

i = j.
 f is one-one.
Onto: For ai  G,  i  Z, such that
Institute of Lifelong Learning, University of Delhi
pg. 19
f(i) = ai
 f is well defined.
Homomorphism:
f (i  j)  a i j  a i .a j  f (i)f (j)
By fundamental theorem of homomorphism.
Z
 G.
ker f
We show
ker f  N  n
i  ker f  f (i)  e.

ai = e

O(a) / i

n/i

iN

kerf =
Hence
n
.
Z
 G.
n
I. Q. 6
Example 8: Show that Group (Q, +) cannot isomorphic to (Q*, .), where Q* = Q  {0}.
Solution: Suppose that
Q  Q*
  an isomorphism  : G  G
Since 1  Q* and  is onto.
  a  Q such that (a) = – 1.

a a
     1
2 2

a
a
   .    1
2 2
(  is homomorphism)
Institute of Lifelong Learning, University of Delhi
pg. 20
2

   a    1
  2
  

x2 = 1
where x = (a/2).
But that is contradiction as there is no rational number such that x2 =  1.
Hence
Q  Q* .
Example 9: Let G be a group and
f :G G
such that f(x) = x1 be a homomorphism.
Then G is abelian.
Solution: Let x, y  G.
and
xy  (y1x 1 ) 1
 f (y1x 1 )
 f (y1 )f (x 1 )
( f is homomorphism)
 (y1 )1 (x 1 )1
= yx.
Example 10: Let G = SL (2, R) and M be any matrix of order 2  2 with determinant 1.
Define a mapping  : G  G such that (A) = M A M1
 A  G. Show that  is
isomorphism.
Solution: Clearly MAM1  G as
| MAM1 |  | M | | A1 | | M1 |  1.1.1.  1
We show that  is well defined, one-one, onto and homomorphism.
We defined: Let A = B

MAM1 = MBM1

(A) = (B).
One-one: let (A) = (B)

MAM1 = MBM1

M1MAM1M =M1MBM1M
Institute of Lifelong Learning, University of Delhi
pg. 21

A = B.
Onto :  is onto if  B  G  A  G such that (A) = B
i.e. if MAM1 = B i.e. if A = M1BM.
  M1MBM  G such that (M1BM) = A
  is onto.
Homomorphism: (AB) = M(AB)M1
= MA (M1M)BM1
= (MAM1(MBM1)
= (A)(B).
Hence  is isomorphism.
4. External Direct Product.
Let
G1,G 2 ,...,G n
be a finite collection of groups, then external direct product, written as
G1  G 2  ...  G n
, is defined as
G  G1  G2  ...  Gn  {(g1,g2 ,...gn ) | gi  Gi }
where
(g1,g 2 ,...,g n )(g1 ,g2 ,...,gn )

is defined to be
(gg1 ,g 2g2 ,...,g n gn )
(g1 g 2 ....g n ), (gi ,g2 ,...,gn )  G .
It is understood that each product gigi is performed with the operation of Gi. G form
group w.r.t. above composition.
Value Addition: Note
(i)
The identity element of
G  G1  G 2  G3  ...  G3
is
(e1,e2 ,...,en )
where ei is the identity of Gi.
(ii)
The inverse of the element (g1, g2, ..., gn) in
given by
G  G1  G 2  ...  G n
is
(g1,g 2 ,...,g n )1  (g11,g 21,...g n1 )
where gi1 is the inverse of gi in Gi.
Institute of Lifelong Learning, University of Delhi
pg. 22
(iii)
The order of element
(g1,g 2 ,...,g n )
= l.c.m. (O(g1, O(g2), ..., O(G(h))
Example 11: U(8)  U(10).
Solution: U(8) = {1, 3, 5, 7}
U(10) = {1, 3, 7, 9}
U(8)  U(10) = {(1, 1),(1, 3), (1, 7), (1, 9), (3, 1),(3, 3), (3,7), (3, 9),
(5, 11), (5, 3), (5, 7), (5, 9), (7,1),(7,3),(7, 7), (7, 9)}
(3, 7) (7 9) = (3 8 7, 7 10 9)
= (5, 3).
Value Addition : Remember
(i)
Let G and G2 are two cyclic groups of order m and n respectively. Then G1
 G2 is cyclic iff m and n are relatively prime.
(ii)
let G1 and G2 are two abelian groups then G1  G2 is abelian.
Example 12: Z2  Z3
Solution: Z2 = {0,1}, Z3 = {0, 1, 2}
Z2  Z3  {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}
Z2 and Z3 both are abelian so
Z2  Z3
is also abelian group of order 6.
Order of an element (1, 1) = 6
Hence
1(1, 1) = (1, 1)
4(1, 1) = (0,1)
2(1, 1) = (0, 1)
5(1, 1) = (1, 2)
3(1, 1) = (1, 0)
6(1, 1) = (0, 0)
Z2  Z3
is cyclic.
I. Q. 7
Example 13: Show that  no onto homomorphism from
Solution: Suppose f :
Z8  Z2  Z4  Z4
Z8  Z2
to
Z4  Z4 .
is onto homomorphism.
Institute of Lifelong Learning, University of Delhi
pg. 23
By fundamental theorem of homomorphism.
Z8  Z2
 Z4  Z4 .
ker f

Z  Z2 
O  8
  O( Z4  Z4 )
 ker f 

O(Z8  Z2 )
 O( Z4  Z4 )
O(ker f )

O(ker f ) 


O(Z8  Z2 )
O( Z4  Z4 )
16
1
16
 Ker f = {e}.

f is one-one
 f is isomorphism i.e.
Now O(1, 0) = 8 in
Z8  Z2  Z4  Z4
Z8  Z2 , but these is no element Z4  Z4
 There does not exist onto homomorphism from
Z8  Z2  Z4  Z4 .
Example 14: Show that  no onto homomorphism from
Solution: Suppose f :
Z16  Z2  Z4  Z4
whose order is 8
Z16  Z2  Z4  Z4
is an onto homomorphism.
By fundamental theorem of homomorphism.
Z16  Z2
 Z4  Z4 .
ker f
O 

Z16  Z2 
 O( Z4  Z4 )
ker f 

O(Z16  Z2 )
 O( Z4  Z4 )
O(ker f )

O(ker f ) 
O(Z16  Z2 )
O( Z4  Z4 )
= 2.
Institute of Lifelong Learning, University of Delhi
pg. 24
ker f  {x  Z16  Z2 | f (x)  (0,0)}
 {(0,0),(8,1)}
or
or {(0, 0), (8, 0)}
{(0, 0), (0, 1)}.
Institute of Lifelong Learning, University of Delhi
pg. 25
5. Automorphism:
An isomorphism from a group G to itself is called automorphism G.
or
A mapping f : G  G such that f is 11, onto and homomorphism is called
automorphism.
Example 15: 1 The identity map from group G to G i.e. I : G  G such that I(x) = x 
x  G is an automorphism on G.
Example 16: Consider the mapping  : C  C defined by
(a + ib) = a  ib  a, b  R
check whether  is automorphism or not
Solution: let a + ib = c + id

a + ib = c + id

a = c, b = d

a  ib = c  id

(a + ib) = (c + id)
  is well-defined and obvious  is one-one.
For all a  ib  C  an element a + ib  C such that (a + ib) = a  ib
  is onto
[(a  ib)(c  id)]  [(ac  bd)  i(ad  bc)]
= (a  ib) (c  id)
= (a + ib)(c + id)
  is an automorphism.
Example 17: Let  : R2  R2 such that (a, b) = (b, a). Show  is automorphism.
Solution: let (a, b) = (c, d)
(a, b) = (c, d)

(b, a) = (a, c)
Institute of Lifelong Learning, University of Delhi
pg. 26

(a, b) = (c, d)
  is well defined,
It is obvious  is 1  1.
Onto : for all (a, b)  R2  (b, a)  R2 such that (a, b) = (b, a)

 is onto.
Homomorphism:
((a, b) (c,d)) = (ac, bd)
= (bd, ac)
= (b,a)(d,c)
= (a,b) (c, d)
  is automorphism.
Institute of Lifelong Learning, University of Delhi
pg. 27
Exercise:
1. Prove or disprove that U(20) and U(24) are isomorphic.
2. Prove that U(10)

U(12).
3. Suppose that Z10 and Z15 are both homomorphic image of G; what can we say about
| G |.
4. Find a homomorphism U(40) to U(50) with kernel {1, 9, 17, 33} and (11) = 11.
5. Suppose that  is homomorphism from S4 onto Z2. Determine ker .
6. If  is a homomorphism from Z30 onto a group of order 5, determine the kernel of .
7. Determine all homomorphism from Z4 to
Z2  Z2 .
8. Determine all homomorphism from Zn to itself.
9. Suppose that  is a homomorphism from S4 onto Z1. Determine ker . Determine all
homomorphism from S4 to Z2.
10. Suppose that there is a homomorphism from a finite group G onto Z 10. Prove that G
has normal subgroups of index 2 and 5.
11. Let C be the complex numbers and
  a b 

M  
a,
b

R


  b a 

Prove that C and M are isomorphic under addition and that C* and M*, the
nonzero elements of M, are isomorphic under multiplication.
Institute of Lifelong Learning, University of Delhi
pg. 28
Summary:
In this lesson we have covered and defined the following topics

Isomorphism.

Fundamental theorem of Homomorphism.

Isomophism theorem for group.
References:
Joseph A. Gallian, Contemporary abstract algebra (4th edition), Narosa Publishing House.
I.N.Herstein. Topics in Algebra (2nd edition), John Willey and Sons.
Institute of Lifelong Learning, University of Delhi
pg. 29