If $G$ is a nonabelian finite simple group, $Aut(G)$ certainly contains a subgroup isomorphic to $G$, namely $Inn(G)$. Must this be the only subgroup of $Aut(G)$ isomorphic to $G$?

I can prove this in many cases using the smallness of $Out(G)$, but I'm wondering if this is a general fact.

  • $\begingroup$ yes it seems is a nice question! for example, we can pick $InnG\unlhd{AutG},InnG<AutG\unlhd{G}$, then $G/AutG\cong{(G/InnG)/(AutG/InnG)}\cong{C/OutG}$ this fact lead that $OutG\cong{(C/G)AutG}$, but $G$ is an nonabelian group and has at least one pair not commutative element, so it is not easy to find another isomorphic subgoup since $G$ is not isomorphic to its automorphism group! $\endgroup$ – user136991 Mar 19 '19 at 1:50
  • $\begingroup$ As far as I know, there is no known proof of this fact that does not use the classification. $\endgroup$ – verret Mar 19 '19 at 5:51

Schreier's conjecture (a theorem provable with the classification of finite simple groups) tells you that $Out(G)$ is always solvable. Therefore $Inn(G)$ is in fact the unique subgroup isomorphic to $G$ inside $Aut(G)$.

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  • 2
    $\begingroup$ Ah, because another such subgroup would have a nontrivial solvable quotient, which is impossible. Nice! $\endgroup$ – stupid_question_bot Feb 19 '19 at 23:05

Given the Schreier conjecture (which requires the Classification of Finite Simple Groups (CFSG), given our current state of knowledge), the question is answered. Here are a few remarks which can be made without the use of CFSG, but makes use of Glauberman's $Z^{\ast}$-theorem (proved using modular character theory), and makes implicit use of the Feit-Thompson odd order theorem. Glauberman proved that if $G$ is a finite non-Abelian simple group with a Sylow $2$-subgroup $S$, then $C_{{\rm Aut}_{G}(S)}$ has a normal $2$-complement and also has an Abelian Sylow $2$-subgroup. In particular,$C_{{\rm Aut}_{G}(S)}$ is solvable.

Now suppose that $G$ is a finite simple group as in the question, and $H$ is a different subgroup of ${\rm Aut}(G)$ with $H \cong G.$ Note that $G$ itself is naturally embedded as a normal subgroup of ${\rm Aut}(G)$ since ${\rm Inn}(G) \cong G/Z(G) \cong G,$ and we consider this as the "canonical" copy of $G$ in ${\rm Aut}(G).$

Now $GH$ is a subgroup of ${\rm Aut}(G)$ and since $H \cong G,$ we know that $H$ is simple. Hence either $H \cap G = 1$ or $H \cap G = H.$ Since $H \cong G$ but $H \neq G$ we must have $H \cap G = 1.$ Thus the group GH$ is a semidirect product.

Let $S$ be a Sylow $2$-subgroup of $G$. Since $G \lhd GH$ we have $GH = GN_{GH}(S)$ by e Frattini argument. Note that now $ G \cong GH/H \cong N_{GH}(S)/N_{G}(S)$ by standard isomorphism theorems. In particular, $N_{GH}(S)/N_{G}(S)$ is simple (isomorphic to $H$ and to $G$) and $N_{G}(S)$ is a maximal normal subgroup of $N_{GH}(S).$

Now $N_{G}(S)C_{GH}(S)$ is a normal subgroup of $N_{GH}(S).$ By the maximality of $N_{G}(S)$ as a normal subgroup of $N_{GH}(S),$ we either have $C_{GH}(S) \leq N_{G}(S)$ or else $N_{GH}(S) = N_{G}(S)C_{GH}(S).$

However, in the latter case, we have $G \cong N_{GH}(S)/N_{G}(S) \cong C_{GH}(S)/C_{G}(S)$ and the last group is solvable by Glauberman's Theorem (since $C_{GH}(S)$ is already solvable), contrary to the fact that $G$ is not solvable.

Hence we must have $C_{GH}(S) \leq N_{G}(S)$ and we then obtain $ G \cong [N_{GH}(S)/N_{G}(S)] \cong [N_{GH}(S)/SC_{GH}(S)]/[N_{G}(S)/SC_{G}(S)].$

Now we may note that $N_{GH}(S)/SC_{GH}(S)$ is isomorphic to a subgroup of ${\rm Out}(S).$ Hence we may conclude (without the use of CFSG) that if $G$ is a non-Abelian simple group such that ${\rm Aut}(G)$ has a subgroup isomorphic to (but different from) ${\rm Inn}(G)$,then $G$ is isomorphic to a section of ${\rm Out}(S),$ where $S$ is a Sylow $2$-subgroup of $G$.

In fact, by using a Theorem of W. Burnside, we may conclude that $G$ is isomorphic to a section of ${\rm Out}(S/\Phi(S)),$ where $\Phi(S)$ is the Frattini subgroup of $S$, so $G$ is isomorphic to a section of ${\rm GL}(d,2),$ where $d$ is the minimum number of generators of $S$. As an easy application, it is easy to conclude that a Sylow $2$-subgroup of $G$ requires at least $4$ generators when this occurs.

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