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Let an element $f$ of $\operatorname{Aut}(\operatorname{Aut}(G))$ acts as an identity on $\operatorname{Inn}(G)$ then does it act as an identity on $\operatorname{Aut}(G)$?

I have taken an element of $\operatorname{Aut}(G)$ say $h$ then $h$ is equal to $g\cdot{k}$ where $g$ is in $\operatorname{Aut}(G)$ and $k$ in $\operatorname{Inn}(G)$. Next what?

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No. Take $G$ abelian of order more than 3. Then $\operatorname{Inn}(G)=1$ and $\operatorname{Aut}(\operatorname{Aut}(G))\neq 1$.

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  • $\begingroup$ Now take G non abelian & simple $\endgroup$ – Ri-Li Aug 12 '14 at 18:34
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    $\begingroup$ @user152715 You made no restrictions on the group in your question. Not even finiteness. If you meant specific kinds of groups, you should edit your question appropriately. $\endgroup$ – zibadawa timmy Aug 12 '14 at 18:40

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