# Understanding the necessity of a condition in a question of group theory

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