# Automorphisms of Finite Groups

Here is another problem from Lang.

Let G be a finite group. N a normal subgroup. We want to ask what structure must G have in order for all the elements of Aut(G) to send N to N. It is assumed that the order of N is relatively prime to the order of G/N. I have worked on this problem for a while and must be missing something pretty basic.

In other words, let G be finite, N a normal subgroup of G with the order of N relatively prime to the order of G/N. Prove that if g is in Aut(G)), then g(N)=N.

morphism
Homework Helper
Let's use f instead of g to avoid confusion. First, note that |f(N)|=|N| since f is a bijection. So in particular, |f(N)| and |G/N| are coprime. Next consider an arbitrary element f(n) of f(N) and set k=o(f(n)). Observe that (f(n)N)^k = (f(n)N)^|G/N| = N. So... ?

Let's use f instead of g to avoid confusion. First, note that |f(N)|=|N| since f is a bijection. So in particular, |f(N)| and |G/N| are coprime. Next consider an arbitrary element f(n) of f(N) and set k=o(f(n)). Observe that (f(n)N)^k = (f(n)N)^|G/N| = N. So... ?
I guess I would write it a little bit differently. But, yeah, your solution is right. I can't believe how a simply problem can catch me off guard like that.

Assume n is in N and n=/=e. (The case for when n=e is trivial) Let k be the order of f(n). Yes, I think i see: f(n)^k =e and so N = (f(n)N)^k. Therefore, k divides |G/N|, and we get your identity

N = (f(n)N)^k = (f(n)N)^|G/N|.

But, also we have e=f(n)^k=f(n^k). Since f is injective, n^k=e. k must be the order of n (for t<=k and n^t =e implies f(n)^t=e implies k divides t implies k=t). Therefore, k divides |N| also. Therefore, k=1 and your identity reads N=f(n)N whence f(n) is in N. As f is bijective, f(N) must equal N.

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oh, I forgot to thank you. Thanks for your help.

morphism