I have found this question Let $G$ be an Abelian group with odd order. Show that $\varphi : G \to G$ such that $\varphi(x)=x^2$ is an automorphism.
My question is that what would happen if $G$ has even order? Is $f(x) =x^2$ an automorphism ?
Is $f(x) =x^2$ an automorphism
group-isomorphismgroup-theory
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Best Answer
By Cauchy's theorem, such a $G$ has an element $x$ of order $2$, so that $x^2=1$. The endomorphism $\varphi$ thus has nontrivial kernel, hence cannot be an automorphism.