Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $g$)?

share|cite|improve this answer
Thanks! I got it now! – Sirzh Aug 6 '12 at 16:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.