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## Homework Statement

Let [tex]G[/tex] be a finite group which possesses an automorphism [tex]\sigma[/tex] such that [tex]\sigma(g)=g[/tex] if and only if [tex]g=1[/tex]. If [tex]\sigma^2[/tex] is the identity map from [tex]G[/tex] to [tex]G[/tex], prove that [tex]G[/tex] is abelian.

## Homework Equations

Show that every element of [tex]G[/tex] can be written in the form [tex]x^{-1}\sigma(x)[/tex] and apply [tex]\sigma[/tex] to such an expression.

## The Attempt at a Solution

My first question is how to obtain [tex]x=x^{-1}\sigma(x)\mbox{, }\forall x\in G[/tex].

If [tex]x=x^{-1}\sigma(x)\mbox{, }\forall x\in G[/tex] is obtained, then by applying [tex]\sigma[/tex] to the identity, and denote [tex]\sigma(x)[/tex] by [tex]y[/tex], we get [tex]y=y^{-1}x[/tex]. Then take inverse on both sides to get [tex]y^{-1}=x^{-1}y\Rightarrow xy^{-1}=y[/tex]. So [tex]y^{-1}x=xy^{-1}[/tex]. It seems the binary operation in the group commutes, but since [tex]y[/tex] is actually the image of [tex]x[/tex] under the automorphism [tex]\sigma[/tex]. So my second question is how to reach the conclusion from this result.

Thanks in advance for any help!

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