> 
> Let $G$ be a finite group and assume that $G$ admits an automorphism $σ$ such that:
> 
> 
> (a) $σ^2(g) = g$ for all $g ∈ G$;
> 
> 
> (b) $σ(g)$ is not equal to $g$ for all $g ∈ G \setminus \{1\}$.
> 
> 
> Show that $G$ is abelian.
> 
> 
> 

The previous part of the question asked me show that $σ(g)=g^{-1}$ which I already proved.

So using $σ(g)=g^{-1}$ I showed:

$$σ(g)=g^{-1}$$

$$σ(g)g=g^{-1}g$$

$$σ(g)g=e$$

$$σ^3(g)=e \quad (\text{since $σ^2(g) = g$})$$

then

$$gσ(g) = gg^{-1}$$

$$gσ(g) = e$$

$$σ^3(g) = e$$

Is this valid and sufficient to show that $G$ is Abelian?
Thank you so much.