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NOTE: I am aware of the link at For a Simple Group G, Z(Aut(G)) Is Trivial if and only if G is Non-Abelian. I am looking for proof verification of $G$ non-abelian $\implies Z(\mathrm{Aut}(G)) = \{1\}$.

Suppose $G$ is not abelian. Then as $Z(G) \trianglelefteq G$, we get $Z(G)$ must be $\{1\}$. Now, note that $G/Z(G) \cong \mathrm{Inn}(G)$ but since $Z(G) = \{1\}$, we get $G \cong{\rm Inn}(G)$.

I will claim without proof that $Z(\mathrm{Aut}(G)) \trianglelefteq \mathrm{Inn}(G)$. Since $G$ is simple and $G \cong \mathrm{Inn}(G)$, then $\mathrm{Inn}(G)$ is simple.

Thus, $Z(\mathrm{Aut}(G)) = \{1\}$ or $\mathrm{Inn}(G)$. If $Z(\mathrm{Aut}(G)) = \mathrm{Inn}(G)$, then this will mean $G$ is abelian, a contradiction. Thus, $Z(\mathrm{Aut}(G)) = \{1\}$.

Shaun
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Grigor Hakobyan
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    How can we verify your proof when you claim something without proof? Also it's not clear. I am referring to the claim that the center of the automorphism group only consists of inner automorphisms. – Martin Brandenburg Oct 17 '23 at 15:49
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    Let $\tau \in Z(Aut(G)), \sigma \in Aut(G)$. Then $\tau \sigma = \sigma \tau \implies \tau = \sigma \tau \sigma^{-1} \implies \tau(x) = \sigma(x) x \sigma(x)^{-1} \implies \tau \in Inn(G)$. Oh wait a minute. I see the issue: $\sigma(x)$ DEPENDS on $x$. Maybe that's my error – Grigor Hakobyan Oct 17 '23 at 15:55
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    You need to prove that if an automorphism centralizes ${\rm Inn}(G)$ then it acts trivially on $G$. – Derek Holt Oct 17 '23 at 16:00

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