If $σ^2$ is the identity map from $G$ to $G$, prove that $G$ is abelian.

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.

• Adding to Mikko's comment, I see problems in a sentence nearby. How does $\sigma$ being an automorphism imply that $\sigma(gh) = gh$? Commented Jan 23, 2014 at 20:09
• Also note that you haven't used the finiteness of $G$, which isn't a good sign. The result doesn't hold for infinite groups. Commented Jan 23, 2014 at 20:12
• why $\sigma(gh)=gh$? Commented Jan 23, 2014 at 20:39
• I edited it, please check if it makes more sense? Commented Jan 23, 2014 at 21:06
• Yes, you did a mistake when deriving $\sigma(gh)=gh$. How can you conclude this? Commented Jan 23, 2014 at 21:08

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 ...

• (+1) Awesome!!... you can use the "pigeon hole principle" for the surjectivity of $x\mapsto x^{-1}\sigma(x)$!! Commented Jan 23, 2014 at 22:17
• how do you got $x^{−1}σ(x)=y^{−1}σ(y)$? Commented Jan 25, 2014 at 15:16
• Well, the last line is a well-known characteristic of abelian groups, so thinking of $\;\sigma\;$ as being the well-known involution $\;x\to x^{-1}\;$ ... Commented Jan 25, 2014 at 16:01