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Assume that $G$ is a finite group, $f:G \to G$ is an isomorphism, and $f\circ f={id}_G$.
For each $g \in G-\{e_g\}$, We have $f(g) \neq g$, Prove that $G$ is an abelian group.

The question also provides a hint :

First, Prove that $G=\{g^{-1}f(g):g\in G\}$.

My problem solving the question:

I don't know how to prove the thing that the hint wants! Otherwise i'm done. I know how to show that $G$ is abelian if $G=\{g^{-1}f(g):g\in G\}$ is proved.

1 Answers 1

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Hint: Prove that $\phi: g \mapsto g^{-1}f(g)$ is injective. Since $G$ is finite, $\phi$ is surjective.

Note that $\phi$ is not necessarily a homomorphism, but that is not needed.

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    How do we see that $\phi$ is injective? I don't see anything obvious here. – 2017-02-15
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    @Omnomnomnom Then you forgot to use the assumption that $f$ has no no non-trivial fixed points. – 2017-02-15
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    @MooS Ah! That will do it. Need to read more closely next time – 2017-02-15