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 Z2. (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 Zp.
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 Zp. Thus A is a vector space over Zp, 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 GLn(Zp).
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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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