# Elements of $\operatorname{Aut}(\operatorname{Aut}(G))$ acting as an identity on $\operatorname{Inn}(G)$

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?

• Aug 13 '14 at 9:44

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