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The question at hand is:

Let G be a finite group and $\alpha$ an involutory automorphism of G, which doesn't fixate any element aside from the trivial one.
1) Prove that $ g \mapsto g^{-1}g^{\alpha} $ is an injection
2) Prove that $\alpha$ maps every element to his inverse
3) Prove that G is abelian

I think I've found 1), I assume $ g_1^{-1}g_1^{\alpha} = g_2^{-1}g_2^{\alpha} $ and from this I get $ (g_2g_1^{-1})^\alpha = g_2g_1^{-1} $, so $g_2 = g_1$ (is this correct?)
For 2) I'm sort of stumped though, not sure how to start proving that, so any help please?

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You should consider accepting some answers to your questions. –  tomasz Aug 6 '12 at 16:06
    
Apologies, I'm new here. I'll do it now, thanks. –  Sirzh Aug 6 '12 at 16:13

1 Answer 1

up vote 2 down vote accepted

1) is correct as you've written. 3) is an easy consqeuence of 2), so I'll leave that part to you.

For 2) we need to use the fact that $\alpha$ is involutory... Note that $\varphi:g\mapsto g^{-1}g^\alpha$ is injective, so it is bijective (because $G$ is finite). Consider an arbitrary $g$, and $h:=\varphi^{-1}(g)$. What is $\varphi^2(h)$ (in terms of $h$ first, then in terms of $g$)?

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Thanks! I got it now! –  Sirzh Aug 6 '12 at 16:10

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