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Let $G$ be a group of order $\ge3$

If $Z(\operatorname{Aut}(G))$ is trivial then so is $Z(G)$?

By assumption, we can choose any $g$ and $g'$ in $G$ s.t. $g\neq g'$ and $g,g'\neq e$

Since any inner automorphisms are not in $Z(\operatorname{Aut}(G))$, there exists $x$ s.t. $c_g c_g'(x)\neq c_g' c_g(x)$ i.e. $gg'xg'^{-1}g^{-1}\neq g'gxg^{-1}g'^{-1}$

So, $gg'\neq g'g$ for any $g,g'$, hence $g\notin Z(G)$.

It is my solution. Is it right?

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  • $\begingroup$ Your proof is not correct. You pick $g \in G$ distinct from the identity, and you want to show that $g \notin Z(G)$. In order to show this, you pick $g'$ with the property that $c_g$ and $c_{g'}$ do not commute: but there is no justification on existence of $g'$. $\endgroup$ – Crostul Apr 2 '16 at 10:21
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    $\begingroup$ $Q_8$ is a counterexample. ${\rm Aut}(G) \cong S_4$. $\endgroup$ – Derek Holt Apr 2 '16 at 11:46
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Not completely. We have the homomorphism $G\to\operatorname{Aut}(G)$, $g\mapsto c_g$ and $Z(G)$ is its kernel. So in order to use that $c_g\notin Z(\operatorname{Aut}(G))$ you seem to assume already that $g\notin Z(G)$.

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