# $\operatorname{Soc}(\operatorname{Aut}( G))$ is isomorphic to $G$, for $G$ a nonabelian, simple group.

Prove that $\operatorname{Soc}(\operatorname{Aut}(G))$ is isomorphic to $G$, for $G$ a nonabelian, simple group.

Here, $\operatorname{Soc}(G)$ is the subgroup generated by all the minimal normal subgroup of $G$. Now, as $G$ is simple we have $\operatorname{Inn}(G)$ is isomorphic to $G$. Then $\operatorname{Inn}(G)$ is isomorphic to $\operatorname{Soc}(\operatorname{Aut}(G))$, but what should be the map? Because the $\operatorname{Aut}(G)$ is not clearly visible.

• Hmm, so you need to show that the inner automorphism group is the only minimal normal subgroup of the automorphism group... – Nishant Aug 6 '14 at 20:24

Since ${\rm Inn}(G) \cong G$ is simple, ${\rm Inn}(G)$ is clearly a minimal normal subgroup of ${\rm Aut}(G)$. So we have to prove that there are no other minimal normal subgroups of ${\rm Aut}(G)$.
If $N$ were another one, then $N \cap {\rm Inn}(G) = 1$, so $N$ centralizes ${\rm Inn(G)}$. But that implies that $N$ acts trivially on $G$ (exercise), so $N=1$, contradiction.
• Yes that's right! And $k$ is conjugation by some element $g \in G$, so for all $x \in g$, $hk(x) = kh(x)$, i.e. $h(gxg^{-1}) = gh(x)g^{-1}$. So $g^{-1}h(g)$ centralizes $h(x)$ for all $x \in G$, and hence $g^{-1}h(g) = 1$, so $h(g)=g$ and $h$ is the identity map. – Derek Holt Aug 6 '14 at 21:24