Showing a group $G$ with a certain automorphism $\sigma$ is Abelian

Question

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.

Answer 1

Since your argument only ever considers one element $g\in G$, I don't see how it can show that $G$ is abelian.

Instead, use the fact that $\sigma$ is a homomorphism: $$ (gh)^{-1}=\sigma(gh)=\sigma(g)\sigma(h)=g^{-1}h^{-1} $$ Since $(gh)^{-1}=h^{-1}g^{-1}$, this shows that $$ h^{-1}g^{-1}=g^{-1}h^{-1}$$ for all $g,h\in G$, which implies that $G$ is abelian.

Answer 2

You seem to be confused about what $\sigma^3(g)$ means. It means $\sigma(\sigma(\sigma(g)))$, not $\sigma(g)\sigma(g)\sigma(g)$. So you can't say that $\sigma(g)g=\sigma^3(g)$ since $g=\sigma^2(g)$.

Also, it's unclear how you are claiming that what you have written would say that $G$ is abelian. To show that you need to show that $gh=hg$ for any $g,h\in G$. (What you have written would, if correct, imply that $G$ is trivial and in particular abelian, but I'm not sure this is what you had in mind.)

Instead, try just using the fact that $\sigma$ is a homomorphism. That is, you know $\sigma(g)=g^{-1}$ for any $g$, and also that $\sigma(gh)=\sigma(g)\sigma(h)$ for any $g$ and $h$. See what you can learn when you combine these two facts.