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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 ?

cqfd
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jasmine
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    It won't be an automorphism: given $x$ with order $2n > 0$, then $x^n \neq 1$ but $f(x) = (x^n)^2 = x^{2n} = 1$ shows that $f$ is not injective. (And such an $x$ exists since $G$ cannot be trivial; otherwise it would have order 1.) – Hayden Jan 12 '20 at 19:44
  • thanks u@Hayden – jasmine Jan 12 '20 at 19:46

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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.

Matt Samuel
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  • For order $2$, we don't need to use Cauchy's theorem: just pair $g$ and $g^{-1}$. Then there will be an even number of elements such that $g=g^{-1}$, and $e$ is one of them. – lhf Jan 16 '20 at 17:43