§ 7.1 Isomorphism Theorems; Automorphisms: solved problems

15. Prove that a finite group whose only automorphism is the identity map must have order at most two.

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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