# How to prove a group is abelian

by Johnson04
Tags: abelian, prove
 P: 18 1. The problem statement, all variables and given/known data Let $$G$$ be a finite group which possesses an automorphism $$\sigma$$ such that $$\sigma(g)=g$$ if and only if $$g=1$$. If $$\sigma^2$$ is the identity map from $$G$$ to $$G$$, prove that $$G$$ is abelian. 2. Relevant equations Show that every element of $$G$$ can be written in the form $$x^{-1}\sigma(x)$$ and apply $$\sigma$$ to such an expression. 3. The attempt at a solution My first question is how to obtain $$x=x^{-1}\sigma(x)\mbox{, }\forall x\in G$$. If $$x=x^{-1}\sigma(x)\mbox{, }\forall x\in G$$ is obtained, then by applying $$\sigma$$ to the identity, and denote $$\sigma(x)$$ by $$y$$, we get $$y=y^{-1}x$$. Then take inverse on both sides to get $$y^{-1}=x^{-1}y\Rightarrow xy^{-1}=y$$. So $$y^{-1}x=xy^{-1}$$. It seems the binary operation in the group commutes, but since $$y$$ is actually the image of $$x$$ under the automorphism $$\sigma$$. So my second question is how to reach the conclusion from this result. Thanks in advance for any help!