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This question already has an answer here:

Let $G$ be a finitely generated group. Show that if Aut($G$) is the trivial group, then so is $G$.

I know that if Aut($G$) is the trivial group then $G$ must be abelian but I'm not sure how to use that fact here. What's a good way to show that $G$ is the trivial group, given that Aut($G$) is also trivial?

Note: This question is different from |G|>2 implies G has non trivial automorphism because there is no proof provided in that question that the opposite is also true, i.e. that $|G| \leq 2 \Rightarrow$ Aut($G$) is trivial.

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marked as duplicate by Dietrich Burde, Daniel Fischer Apr 28 '16 at 19:23

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    $\begingroup$ That's not quite true: $G=\mathbb{Z}/2\mathbb{Z}$ has a trivial automorphism group. This is the only other case though, at least for finite groups. $\endgroup$ – carmichael561 Apr 28 '16 at 18:48
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    $\begingroup$ Once you know that $G$ is abelian, you can appeal to the structure theorem for f.g. abelian groups. $\endgroup$ – Qiaochu Yuan Apr 28 '16 at 18:50
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    $\begingroup$ Hint: for abelian groups, the map $g \mapsto g^{-1}$ is an automorphism. $\endgroup$ – Henry Swanson Apr 28 '16 at 18:52
  • $\begingroup$ @carmichael561 are you sure that's the only exception? The other answers seem to suggest otherwise. $\endgroup$ – user3749214 Apr 28 '16 at 18:56
  • $\begingroup$ @DietrichBurde I edited my question to explain why I believe this is not a duplicate. I may be wrong, however. Can you please explain why you think otherwise? Thanks! $\endgroup$ – user3749214 Apr 28 '16 at 19:13
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$Aut(S_2)$ is trivial but $S_2$ is non-trivial.

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