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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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user3749214
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    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. – carmichael561 Apr 28 '16 at 18:48
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    Once you know that $G$ is abelian, you can appeal to the structure theorem for f.g. abelian groups. – Qiaochu Yuan Apr 28 '16 at 18:50
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    Hint: for abelian groups, the map $g \mapsto g^{-1}$ is an automorphism. – Henry Swanson Apr 28 '16 at 18:52
  • @carmichael561 are you sure that's the only exception? The other answers seem to suggest otherwise. – user3749214 Apr 28 '16 at 18:56
  • @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! – user3749214 Apr 28 '16 at 19:13
  • @user3749214 The other answers all suggest using a group of order $2$, and up to isomorphism, there is just $\Bbb Z / 2 \Bbb Z \cong S_2 \cong {e, a} \cong {\rm many\ other\ names}$. – pjs36 Apr 28 '16 at 19:16
  • When @carmichael561 said "only" he meant up to isomorphism, because obviously an isomorphic copy of a counterexample is also a counterexample. The groups in the answers are isomorphic to $\mathbb Z/2\mathbb Z$. – Andreas Blass Apr 28 '16 at 19:17
  • @DietrichBurde then is my theorem as stated simply not true? Or are groups isomorphic to $C_2$ not finitely generated, meaning the group $G$ in my question must be trivial? – user3749214 Apr 28 '16 at 19:25
  • Indeed, your title claim is not true. Counterexample: $G=C_2$ has $Aut(G)=id$ trivial, but $C_2$ is not trivial. And that a group of at most $2$ elements has trivial automorphism group is obvious (and has been proved at MSE, too). – Dietrich Burde Apr 28 '16 at 19:26

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$Aut(S_2)$ is trivial but $S_2$ is non-trivial.

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