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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.

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  • $\begingroup$ Hmm, so you need to show that the inner automorphism group is the only minimal normal subgroup of the automorphism group... $\endgroup$ – Nishant Aug 6 '14 at 20:24
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Here is an outline proof.

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.

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  • $\begingroup$ Sir, N centralizes Inn(G) means what? hk=kh; where h belongs to N & k belongs to Inn(G)? $\endgroup$ – user152715 Aug 6 '14 at 20:47
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Aug 6 '14 at 21:24
  • $\begingroup$ ya true. Now the question arises to show if if G is direct product of nonabelian & simple groups. Then prove that Aut(G) has only inner automorphisms. Here we can restrict any map f belongs to Aut(Aut(G)) then restrict f to Soc(Aut(G)) in Aut(G) this is possible as Soc(Aut(G)) is a characteristic subgroup of Aut(G). Now take any g in inn(Aut(G)) subgroup of Aut(Aut(G)). As AutG is isomorphic to Inn(Aut(G)) we can choose g which has the same action of f in G. Consider k=fg-1 it acts trivially on G. Then take r in (Aut(G). What will be the action of k on r?? $\endgroup$ – user152715 Aug 6 '14 at 21:45

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