*Solution:*
Assume that the only automorphism of G is the identity mapping.
Then all inner automorphisms of G are trivial, so G must be abelian.
Next, the function µ : G: - > G
defined by µ(x) = x^{ -1}, for all x in G,
is an automorphism, so by hypothesis it must be trivial.
This forces x = x^{ -1} for all x in G.
If G is written additively,
then G has a vector space structure over the field **Z**_{2}.
(Since every element of G has order 2,
it works to define 0 **·** x = 0
and 1 **·** x = x,
for all x in G.)
With this vector space structure,
any group homomorphism is a linear transformation
(and vice versa),
so the automorphism group of G is a group of invertible matrices.
Therefore Aut (G) is nontrivial,
unless G is zero or one-dimensional.

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**16.**
Let p be a prime number,
and let A be a finite abelian group in which every element has order p.
Show that Aut(A) is isomorphic to
a group of matrices over **Z**_{p}.

*Solution:*
The first step in seeing this is to recognize
that if every element of A has order p,
then the usual multiplication n **·** a,
defined for integers n in **Z**, and elements a in A,
actually defines a scalar multiplication on A,
for scalars in **Z**_{p}.
Thus A is a vector space over **Z**_{p},
and as such it has some dimension, say n.
Every automorphism of A is a linear transformation,
so Aut(A) is isomorphic to the general linear group
GL_{n}(**Z**_{p}).

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**17.**
Let G be a group and let N be a normal subgroup of G of finite index.
Suppose that H is a finite subgroup of G
and that the order of H is relatively prime to the index of N in G.
Prove that H is contained in N.

*Solution:*
Let
: G -> G/N be the natural projection. Then
(H) is a subgroup of G/N, so its order must be a divisor of |G/N|.
On the other hand, |
(H) | must be a divisor of |H|.
Since gcd (|H|,[G:N]) = 1, we must have |
(H) | = 1, which implies that H
ker ( ) = N.

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**18.**
Let G be a finite group and let K be a normal subgroup of G
such that gcd (|K|,[G:K]) = 1.
Prove that K is a characteristic subgroup of G.

*Note:*
Recall the definition given in Exercise 7.6.6 of the text.
The subgroup K is a
**characteristic subgroup** of G if
µ(K) K
for all µ in Aut (G).

*Solution:*
Let µ be any automorphism of G.
Then µ(K) is a subgroup of G, with |K| elements.
Since gcd (|K|,[G:K]) = 1,
we can apply the result in the previous problem,
which implies that µ(K)
K.

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**19.**
A subgroup H of a finite group G is called a
**Hall subgroup** of G if its index in G
is relatively prime to its order.
That is, if gcd (|H|,[G:H]) = 1.
Prove that if H is a Hall subgroup of G
and N is any normal subgroup of G,
then (a) H
N is a Hall subgroup of N
and (b) HN / N is a Hall subgroup of G / N.

*Solution:*
Assume that |H| is relatively prime to |G:H|.
We will use the fact that any divisor of |H|
must be relatively prime to any divisor of |G:H|.
It follows from Theorem
7.1.2 that

|HN| / |N| = |H| / |H N| and |HN| / |H| = |N| / |H N| .

To prove part (a) we have the following computations.

|H N| · |H:H N| = |H|

|N:HN| · |G:HN|

= ( |N| / |HN| ) · ( |G| / |HN| )

= ( |HN| ) / ( |H| ) · ( |G| ) / ( |HN| )

= ( |G| ) / ( |H| )

= |G:H|

To prove part (b) we have the following computations.

|HN/N| · |HN|

= ( |HN| ) / ( |N| ) · |HN|

= ( |H| ) / ( |HN| ) · |HN|

= |H|

|G/N : HN/N| · |HN:H|

= ( |G| ) / ( |N| ) · ( |N| ) / ( |HN| ) · ( |HN| ) / ( |H| )

= ( |G| ) / ( |H| )

= |G:H|

This completes the proof.

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