1

1

Isomorphism Theorems

Chapter 10

2

TheoremsBecause any isomorphism θ is an 1−1 onto mapping, its always has inverse mapping θ −1 . We can see that the inversemapping θ −1 also keeps the operations. So

Theorem 1: If group G is isomorphic to group H , then group H is isomorphic to group G.

Let θ be an isomorphism from group G to group H. For any g∈G , let h=θ(g). Then order(h) is a divisor order(g).However, g=θ −1 (h), θ −1 isomorphism too. So order(g) is adivisor order(h) . Therefore order(h) = order(g).Theorem 2: Group isomorphism keeps order of elements.

2

3

Theorem 3. Any finite group G is isomorphic to a group of permutations Proof: Let x∈G we define a mapping Px from G to G

Px: G →Gsuch that Px, maps any element g∈G to xg.So Px(g) = xgWe claim that:

i) Px is 1−1. This is because if Px(g1) =Px(g2) � xg1 =xg2 � g1 =g2

ii) Px is onto. This is because for any g∈G , Px(x −1 g)=x (x −1 g)= g

So Px is an permutation among elements of GLet H be the set of all those permutations, i.e.

H={Px: x∈G }

4

( ) ( )1

1 1

1

1

1

We want to show that is a permutation group.i) Let , then both and in Then for any

Therefore is an identity permutation.

That proves and ar

)

e

.

(

i

x x

xx x

xx

x x

H

x G P P H g G

P P g P xg x xg g

P P

P P

−

− −

−

−

−

∈ ∈

= = =����

����

( ) ( ) ( )

nverse permutation of each other.

ii) Let , then for any

That proves

From above we can see that is closure under permutationmultiplication and inv

,

( ) ( )

erse

oper

y x y yx

y x yx

x y G g G

P P g P xg y xg yx g P g

P P P

H

∈ ∈= = = =

=

����

����

ation. Therefore is a subgroup

of the group consist of all permutations among elements of .

H

G

3

5

Now we make a mapping from to by ,

i) This mapping is onto because is simply a collection

of all those permutations.ii) If

So mapping is

( )

( ) = ( )

( ) ( )

x

x y

x y

G H

x G x P

H

x y P P

P e P e

xe ye

x y

θθ

θ

θ θ

θ

∀ ∈ =

� =

� =

� =� =

1 1.

iii) Formula

means this mapping keeps operation from toFrom above we have pro

(

ved that is an isomo

) ( )

rphi

( )

sm

y x yxP P P y x yx

G H

θ θ θθ

θ

−= � =� �� �� �� �

6

: Let Then is a group. Find its isomorphic permutatiuion group

{ 1, , 1

Th

, }

e

Exam

oper

. { , *}

ation table is

ple

G i i G

H

= − −

i1−i−1−1

−1i1−i−i

1−i−1ii

−i−1i11

−i−1i1×

Change the 1, i, −1 , –i to numbers 1, 2, 3 , 4 except 1st column

2143−1

3214−i

1432i

43211

4321×

4

7

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4, , ,

1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3

1 2 3 41 2 3 4

1 22 3

(

Then we make an permutation group

is a permutation group .The mapping from to is the followi

1)

( )

ng

H

H

G H

i

θ

θ

θ

� �� �� � � � � � � �� � � �� � � � � � � � �

� � � �

=

=

= 3 44 1

1 2 3 43 4 1 2

1 2 3 44 1 2 3

From above theorem

( 1)

is an isomorpuism from to

( )i

G H

θ

θ

θ

� � � �

� � � �

� � � �

− =

− =

8

{ }1 3 1 3,

2 2: Let group

Find its isomorphic permutatiuion group The operation table is

Example 2 1, .i iG

H

− + − −=

1 3 1 3,

2 2Change to number except

the 1st column.

1, 1, 2, 3 i i− + − −

5

9

Then we get table

1 2 3 1 2 3 1 2 3, , .

1 2 3 2 3 1 3 1 2

1 2 3 1 2 31 3 1 31 2 3 2 3 12

Then we make an permutation group

is a permutation group .The mapping from to is the following

(1) i

H

H

G Hθ

θ θ θ

� �� �� � � � � �� � � �� � � � � � �

� � � �� �− + − −

� �� � � �

=

= = 1 2 33 1 22

From above theorem is an isomorpuism from to

i

G Hθ

� �� �

� � � �=

10

Example 3: Verify that there is an isomorphism from the additive

Group Z6 to the multiplicative group of Z7

Proof: Z6 ={0, 1, 2, 3, 4, 5}is an additive group.

Z7 ={0, 1, 2, 3, 4, 5, 6} is also an additive group.

The multiplicative group of Z7 is G ={1, 2, 3, 4, 5, 6}

The operation table of Z6 The operation table of G

×

6

11

6

6

Because isomorphism keeps order, we look for elements that have the largest orders in both groups. In , 1 has order 6.

1 is a generator of . Therefore we need to look for element in

whose order

Z

Z G

2 3

order(1) 1

is also 6 . Now check orders. All possible orders must be divisors of 6, they are 1, 2, 3, 6.

, because that 1 is identity for the multiplication

2 =4, 2 =8(mod 7) o =1 rde

3

r(2) 3

==�

( )

2 3

this is always true, identity ma

or

pp

der(

ing

9(mod 7) =2, 3 27(mod 7) =6 Now

3) 6.

(1)=

to identity

(Both have ord

we can

er 6

make isomorphism as

(0)=1,

,Then from operation property

(

3

2

)

)= (1+1)

θ

θθ

θ θ

= � ==

= (1) (1)=3 3(mod 7) 2(3)= (1+2)= (1) (2)=3 2 6

θ θθ θ θ θ

× × =× × =

12

(4)= (1+3)= (1) (3)=3 6(mod 7) 4(5)= (1+4)= (1) (4)=3 4(mod 7) 5

However, the isomorphism is not unique, we can find that So we can make another isomorphism as order(5) 6

( this is always 0)=1,

.

t

θ θ θ θθ θ θ θ

θ

=

=

× × =× ×

( ),

Then from operation property(2)= (1+1)= (1) (1)=5 5(mod 7)

rue, identity mapping to identity

(Both has order

4(3)= (1+2)= (1) (2)=5 4(mod 7) 6(4)= (1+3)= (1) (3)=5 6(mod 7) 2

(1)

(5)= (1+4)

=5 6)

=

θ θ θ θθ θ θ θθ θ θ θθ

θ

θ

× × =× × =× × =

(1) (4)=5 2(mod 7) 3θ θ× × =

7

13

( ) ( ) ( )( ) ( ) ( )

6 3

6

3

1 2 3 1 2 3 1 2 31 2 3 2 3 1 1 3 2

1 2 3 1 2 3 1 2 33 1 2 3 2 1 2 1 3

Prove that there is no isomorohism between and

In 1 has o={0, 1 rder 6.

But in

E

, 2, 3, 4, 5}

{ , ,

xample 4:

Pro

�

� �

�,

�

order(

of

�, �, �}

:

Z

e

S

Z

S e

e

α

α= = =

= = =

=

3

6 3

) 1, order( ) order(�) 3,order(�) order(�) order(�) 2,

No element in has order 6.

Therefore there is no isomorohism between and

S

Z S

α= = == = =

14

Home works{ }

{ }

{ }

8

1 0 1 0 1 0 1 0, , ,

0 1 0 1 0 1 0 1

1. In the multiplication group is

Make its multiplication table and find its isomorphic permutatiuion group

2. Let group

gro

1, 3, 5, 7

1, ,

p

P

,

u

1

Z G

H

G i i

H� � � � � � � �− −

− −� � � � � � � �

=

= − −

=

rove that there is no isomorohism between and G H