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This is an already answered question in this website:
Show that a finite group with certain automorphism is abelian

I have solved the question by myself and got the result but one thing I noticed is that there was not need of the given function $f$ to be onto. We only used fixed point free monomorphism , $f^2=I$ property and that $G$ is finite. Then why is it given to be automorphism? Am I getting something wrong?

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    $\begingroup$ By your assumption $f^{-1}=f$ is invertible, hence an automorphism. $\endgroup$ Apr 6, 2021 at 15:29

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It's kind of automatic: if two functions $f,g:S\rightarrow S$ (where $S$ is any set) are such that $f\circ g$ is bijective, then $f$ is surjective and $g$ is injective. Hence $f^2$ being the identity implies $f$ is also bijective.

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To prove that the map $\varphi\colon g\mapsto f(g)g^{-1}$ (hinted in the link) is surjective, you use solely the condition $f(g)=g\Longrightarrow g=e$ (which in fact ensures that $\varphi$ is injective; the rest is done by the finiteness of $G$). To prove the main result (which relies on the surjectivity of $\varphi$), you use the fact that $f$ is a morphism, while the condition $f^2=Id$, also used to prove the claim, tells you that $f$ is also bijective. That's why assuming $f$ an automorphism of order $2$ brings you in the right setting to get $G$ abelian.

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