Suppose $N\unlhd G$ with $N\cong G/N\cong S$, $S$ a nonabelian finite simple group. Assume that $\operatorname{Out}(S)$ is solvable.

Question

I am given a group $G$ and a normal subgroup $N$ such that both $N$ and $G/N$ are isomorphic to $S$, a nonabelian finite simple group. I am also told to assume that $\operatorname{Out}(S)$ is solvable. Here is what I have shown so far:

1. $C_G (N) = \{ g \in G : gn = ng \ \forall n \in N \}$ is a normal subgroup of $G$ (by showing that it is the kernel of a group homomorphism from $G$ to $\operatorname{Aut}(N)$)
2. $N \cap C_G (N) = 1$ (by showing that it is an abelian normal subgroup of $N$)
3. $\operatorname{Inn}(N) \cong S$ (by defining a homomorphism from $N$ to $\operatorname{Inn}(N)$, with trivial kernel)

Now I am asked to show that

• $C_G(N) \neq 1$, using the facts that $\operatorname{Out}(S)$ is solvable, $\operatorname{Inn}(N)$ is a normal subgroup of the image of a the homomorphism I defined in (1), and the Jordan-Holder theorem.
• $G = NC_G(N)$, I suspect using the fact that $C_G(N)$ contains a non-identity element?

Any direction would be great. If, for instance, I can show that if $C_G(N) = 1$ then $\operatorname{Out}(S) \cong G/N$, I will be done. But I don't see how to do this. The only other thing I can think of would be to show that $C_G(N)$ is composition factor of $\operatorname{Out}(S)$, which would mean it would be cyclic of prime order and therefore have at least 2 elements.