# Showing that if the center of a group is trivial, then the center of its automorphism group is trivial

I am trying to show that if for a group $G$ we have $Z(G)=1$, then $Z(\operatorname{Aut}(G))=1$.

I have tried everything I can think of, but to no avail. The first part of the problem was to show that $\operatorname{Inn}(G) \triangleleft \operatorname{Aut}(G)$ - does this have any connection? I couldn't find one.

I know that $\operatorname{Inn}(G) \cong G/Z(G)$, so in this case $\operatorname{Inn}(G) \cong G$ and $Z(\operatorname{Inn}(G))=1$. Is there any way to get from this to $Z(\operatorname{Aut}(G))=1$?

How do we go about showing this?

Thanks.

If $f$ is an automorphism of $G$ which commutes with all other automorphisms, it does so in particular for the inner automorphisms. This implies $f(g) f(x) f(g)^{-1}=g f(x) g^{-1}$ for all $x \in G$. It is not hard to deduce that $g^{-1} f(g)$ lies in the center of $G$. And then we are done.
Let $\phi$ be an automorphism that commutes with all automorphisms. Especially, for $x,y\in G$ $$yxy^{-1}=y\phi(\phi^{-1}(x))y^{-1}=\phi(y\phi^{-1}(x)y^{-1})=\phi(y)x\phi(y)^{-1},$$ i.e. conjugation by $y$ is the same as conjugation by $\phi(y)$. As $G\to\operatorname{Inn}(G)$ is injective (has $Z(G)$ as kernel), we conclude $\phi(y)=y$, i.e. $\phi=\operatorname{id}$.