TITLE: If $σ^2$ is the identity map from $G$ to $G$, prove that $G$ is abelian.
QUESTION [6 upvotes]: Let $G$ be a finite group which possesses an automorphism $σ$ such that $σ(g)=g$ if and only if $g=1$. If $σ^2$ is the identity map from $G$ to $G$, prove that $G$ is abelian.
This is what I got 
Let $G$ be a finite group which possesses an automorphism $σ$ such that $σ(g)=g$ if and only if $g=1$. 
Assume that $σ^2$ is the identity map from $G$ to $G$, we will show that $G $is abelian.
Let $g,h∈G$, since $σ^2$ is the identity map from $G$ to $G$
$σ^2 (g)=g$
$σ^2 (h)=h$
And
$σ^2 (gh)=gh=σ^2 (g) σ^2 (h)$
Since $σ∈Aut(G)$, $σ$ is isomorphism (homophism and bijective) , so  $σ(gh)=σ(g)σ(h)=gh$. Thus, $gh=1=hg$. Hence, $G$ is abelian.
Did I do it correctly, I keep feeling I missed something.

REPLY [6 votes]: Hints:
$$\forall\,x,y,\in G\;,\;\;x^{-1}\sigma(x)=y^{-1}\sigma(y)\implies yx^{-1}=\sigma(y)\sigma(x)^{-1}=\sigma(yx^{-1})\stackrel{\text{given!}}\iff x=y$$
Thus, the map $\;x\mapsto x^{-1}\sigma(x)\;$ is bijective (why?) and thus
$$\forall a\in G\;\exists\,x_a\in G\;\;s.t.\;\; a=x_a^{-1}\sigma(x_a)$$
Try to take it from here...and remember: if $\;a\mapsto a^{-1}\;$ is a homomorphism, then $\;G\;$ is abelian ...