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Let $G$ be a group such that ${\rm Aut}(G) = 1.$ I know that $G$ is abelian and that every element of $G$ satisfies the equation $x^2 = 1.$ This can be shown by considering the mapping $x \mapsto x^{-1}$ which is an automorphism.

I want to show that the order of $G$ is $1$ or $2.$ Is it sufficient to note that since $G$ is abelian and every element of $G$ satisfies $x^2 = 1$, then $G$ has non-identity automorphism if the group contains more than two elements? I was wondering if this would suffice or do we need to prove this more rigorously?

Shaun
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  • How do you know that $\mathrm{Aut}(G)$ has a non-identity element if $G$ contains more than two elements? This is the crux of the claim, so yeah, you should explain why that's true. Like, what would the non-identity automorphism be? – Mike Pierce Apr 08 '21 at 16:43
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    See: https://math.stackexchange.com/questions/395569/let-g-be-a-finite-group-with-g2-prove-that-autg-contains-at-least-tw?rq=1 – wormram Apr 08 '21 at 16:46

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