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# inner automorphisms making a normal subgroup of the group of automorphisms and index of the subgroup

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2a) Given that H is isomorphic to G.
So, we follow that the cardinality of H or the order of H is equal to that of G.
We know that if O(G) = n, then there can be n ! permutations called the automorphisms
possible.
Since O (H) = O (G) = n, we follow that O (Aut (H)) = O (Aut (G)) = n!
This allows that the cardinality of both these groups of automorphisms defined over H and G
is equal.
Suppose h is an automorphism over H fixing an element
k
a
and g is an automorphism over G
fixing
k
b
, then by defining
( )
k k
a b
h g
φ
=
, we can show
φ
is an isomorphism between Aut(H) and
Aut(G).
Therefore,
( ) ( )
Aut H Aut G
2b) take the dihedral group
3
D
having 6 elements in which 3 are rotations and 3 other are flips
These can be conveniently written as permutations or automorphisms.
On the other hand, consider the bijections on 3 symbols which are 3! = 6 in number.
These also can be arranged as automorphisms.
Thus, we have two automorphic groups namely
3 3
,D S
It is easy to show that these are isomorphic. But, the respective triangular vertices and the 3
symbol sets are not isomorphic.
3a) G is a finite abelian group and
( )
2
g g
φ
=
To show
φ
is an automorphism, we use that G has no element of order 2.
That is
for any g in G. ...... (*).
Now,
(1)
( ) ( ) ( ) ( ) ( )
( )
( )
( )
( )
( )
( ) ( ) ( ) ( )
2
2 2
fg fg fg fg f g fg f gf g f fg g ff gg f g f g
φ φ φ
= = = = = = = =
This shows that
φ
is a homomorphism.
(2) Suppose
( ) ( )
f g
φ φ
=
That is,
2 2
f g=
Since
2 2
,f g
are the members of G, the only possibility by (*) is
2 2
f g f g= =
So,
φ
is a one – to – one function
(3) Suppose
g
is any member of G, then by closure law in G, we follow that
2
g
is in G such
that
( )
2
g g
φ
=
That is, for every
2
g
in G, there corresponds g in G such that
( )
2
g g
φ
=
So,
φ
is onto.
Putting (1), (2) and (3) together, it can be confirmed that
: G G
φ
is an isomorphism.
Since the domain and the co-domain are the same group, we say that
φ
is an
automorphism.
In other words,
( )
Aug G
φ
3b) we can see that the Boolean ring concept, we can interpret that
0a a
+ =
for every a
in G as
for every g in G.
In other words, it is not necessary for G to be finite abelian for every g is such that
in G.

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So, the result (3a) is not necessary for G to be finite if at all
for every g in G.
5a) we follow that H is a subgroup of K and K is a subgroup of G.
So, by Lagrange’s theorem for finite groups, we have [K: H] = number of distinct left
cosets of H in K.
K
H
=
Similarly, K is the subgroup of G.
So, the number of distinct left cosets of K in G is the index of K in G = [G:K] =
G
K
=
Now,
[ ]
:
G K G
G H
K H H
× = =
That is,
[ ] [ ] [ ]
: : :G H G K K H=
5b)
( ) { }
|
G
C a g G ag ga
= =
On the other hand,
( ) { }
|Z G z G zg gz g G= =
Now, suppose
( )
G
C a G=
That means, a commutes with every element of G.
We know that if an element commutes with every element of G, then that element must be
in the center of the group.
In the present condition, a satisfies this condition. So, a must be in Z(G).
Conversely, suppose a is in the center of the group Z(G).
Then, by definition, it follows that a commutes with every element of G.
We know that if an element commutes with every element of G, then its centralizer is
nothing but the entire G.
In other words,
( )
G
C a G=
5c) see that
( )
G
C a
is a subgroup of Z(G). And Z(G) is a normal subgroup of G.
So, by Lagrange;s theorem, it follows that the order of
( )
G
C a
divides the order of Z(G)
for every a in G.
In other words, if a is not identity element, then order of
( )
G
C a
is not 1.
So, Z(G) has a non trivial divisor.
Therefore, the order of Z(G) is not prime.
The consequene of the result 5(a), it follows that
( )
:G Z G k=
not a prime.
8) Recollect the properties of permutations.
(i) Product of two odd permutations is even
(ii) Product of two even permutations is even
(iii) Product of an even and an odd permutation is odd.
In view of this, we can write that
( ) ( ) ( )
1 2 1 2
1 1 1
φ σ σ φ σ φ σ
= − = − × =
whenever
1
σ
is odd and
2
σ
is even.
( ) ( ) ( )
1 2 1 2
1 1 1
φ σ σ φ σ φ σ
= = − × =
when both
1 2
&
σ σ
are odd
( ) ( ) ( )
1 2 1 2
1 1 1
φ σ σ φ σ φ σ
= = × =
when both are even.
Putting these cases together, we can write that
( ) ( ) ( )
1 2 1 2
φ σ σ φ σ φ σ
=
for any pair of permutations
1 2
&
σ σ
.

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2a) Given that H is isomorphic to G. So, we follow that the cardinality of H or the order of H is equal to that of G. We know that if O(G) = n, then there can be n ! permutations called the automorphisms possible. Since O (H) = O (G) = n, we follow that O (Aut (H)) = O (Aut (G)) = n! This allows that the cardinality of both these groups of automorphisms defined over H and G is equal. Suppose h is an automorphism over H fixing an element and g is an automorphism over G fixing , then by defining , we can show is an isomorphism between Aut(H) and Aut(G). Therefore, 2b) take the dihedral group having 6 elements in which 3 are rotations and 3 other are flips about the 3 sides. These can be conveniently written as permutations or automorphisms. On the other hand, consider the bijections on 3 symbols which are 3! = 6 in number. These also can be arranged as automorphisms. Thus, we have tw ...
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