I have a question about group $G$ which satisfies Inn$(G) $ char Aut$(G)$ and $Z(G)$=$\{1\}$.

Question

Let $G$ be a group which satisfies $Z(G)=\{1\}$ and ${\rm Inn(G)} \space \mathbb{char} \space {\rm Aut(G)}$; then every automorphism of $A={\rm Aut(G)}$ is an inner automorphism. ($H \space \mathbb{char} \space G$ means that $H$ is a characteristic subgroup of $G$. Note that we can assume $G \subseteq A$, since $Z(G)=\{1\}$ so $G \cong{\rm Inn}(G)$.)

I am given a hint, and it says that $C({\rm Inn}(G)) \unlhd {\rm Aut}(A)$, so one derives $C({\rm Inn}(G)) \subseteq C(A)$. I'm stuck only on this point. Why can we say that? ($C(H)$ means the centralizer of $H$ in ${\rm Aut}(A)$.)

Answer 1

Let $I = {\rm Inn}\, G$ and $C = C_{{\rm Aut}(A)}(I)$. Then $C \unlhd {\rm Aut}(A)$ and $A \unlhd {\rm Aut}(A)$. Since $A$ is by definition the group of automorphisms of $G \cong I$, no nontrivial element of $A$ can centralize $I$; i.e. $C \cap A = 1$. Hence $[C,A] \le C \cap A = 1$; i.e. $C \le C(A)$.

Answer 2

Let $\sigma\in C(Inn(G))$,

Then we have $\sigma i_g\sigma^{-1}=i_g$ where $i_g:G\to G$ by $i_g(x)=gxg^{-1}$ for $x\in G$.

We have,

$$i_g(x)=\sigma i_g \sigma^{-1}(x)$$ $$i_g(x)=\sigma(g\sigma^{-1}(x)g^{-1})$$ $$i_g(x)=\sigma(g)\sigma(\sigma^{-1}(x))\sigma(g^{-1})$$ $$i_g(x)=\sigma(g)x\sigma (g)^{-1}=i_{\sigma(g)}$$ But since $Z(G)=1$, the map $\phi:G\to Inn(G)$ by $g\to i_g$ is an bijection. Hence, $i_g=i_{\sigma(g)}\implies g=\sigma(g)$ for all $g\in G$. Hence, $\sigma=1_A$.

Answer 3

I post my solution here:(to prove $C_{AutA}(I)=1$) http://www.artofproblemsolving.com/community/c7t435f7h622064_automorphism_group_of_automorphism_group_of_simple_group_g