For nonabelian finite simple $G$, does $Aut(G)$ have a unique subgroup isomorphic to $G$?

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.

• 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! – user136991 Mar 19 at 1:50
• As far as I know, there is no known proof of this fact that does not use the classification. – verret Mar 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 $$Aut(G)$$.

• Ah, because another such subgroup would have a nontrivial solvable quotient, which is impossible. Nice! – stupid_question_bot Feb 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.