Can I say this:
Consider $g h h g = 1_G$
We know that $g^2 = 1_G$ and so we get
$ g h h g = g h^2 g = g g = g^2 = 1_G $
As every group must have a unique inverse, in order for my "consider" claim to hold, we must have that $gh = hg$ and therefore the group is abelian.
Is this correct?
EDIT: What's wrong with my proof then? I've seen the $(ab)^2 = 1_G$ proof and I understand that but why is that correct and my one wrong? What am I missing out on proving?