# Automorphisms of finite almost simple groups

Let $P$ be a finite nonabelian simple group. Let $G$ satisfy

$$P\leqslant G \leqslant {\rm Aut}(P),$$

where $P$ is identified with $\rm{Inn}(P)$. I am trying to see if

$${\rm Aut}(G)\cong N_{{\rm Aut}(P)}(G)$$

Here is my attempt.

Firstly, since ${\rm Z}(G)=1$, we have $G\cong{\rm Inn}(G)$ sitting inside ${\rm Aut}(G)$. Now, $P$ is characteristic in $G$, which implies the existence of a homomorphism $\varphi: {\rm Aut}(G)\to {\rm Aut}(P)$. Clearly, $\varphi$ is injective iff $G$ has no proper automorphisms that act identically on $P$. Suppose that this is the case. Then ${\rm Aut}(G)\subseteq N_{{\rm Aut}(P)}(G)$. Conversely, every nontrivial element of ${\rm Aut}(P)$ that normalizes $G$ must induce a nontrivial automorphism on it, and we have the reverse inclusion.

If the above is correct, it remains to answer

Question. Is it true that $G$ has no nontrivial automorphisms that act identically on $P$?

Any thoughts on this would be appreciated.

• BTW this isn't true if you assume only that $Z(P) = Z(G) \cap P$ and $P/Z(P) \leq G/Z(G) \leq \Aut(P)$, so the quasi-simple case is definitely weird. – Jack Schmidt Jan 25 '14 at 4:28
• I think you just need to prove $F^*(\Aut(G)) = P$, though again, this need not be true in the quasi-simple case. – Jack Schmidt Jan 25 '14 at 4:39

Indeed, your first statement is true even for the much larger class of groups for which the socle has no abelian factors. The second statement holds whenever $P \unlhd G$ is a normal subgroup such that the action of $G$ on $P$ by conjugation is faithful (which includes your case).
For suppose $\varphi$ is an automorphism of $G$ that acts trivially on $P$. For any $x \in P$, we have $x^{\varphi(g)} := \varphi(g) x \varphi(g)^{-1} = \varphi(g x g^{-1})$ since $\varphi(x)=x$ by assumption. The latter is then equal to $\varphi(x^g)$. However, since $P$ is normal in $G$, $x^g \in P$, so $\varphi(x^g) = x^g$. We've shown that for any $x \in P$ and $g \in G$, $x^{\varphi(g)} = x^g$, that is, $g \varphi(g)^{-1}$ acts trivially on $P$. But the action of $G$ on $P$ is faithful, so we must have $g \varphi(g)^{-1} = 1$.