Group Theory Automorphism question

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

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 $g$)?