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 its 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?
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 $g$)?