Is the product of two normal subgroups of a finite group equal to the group if the intersection of the normal subgroups is trivial?

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 $Out(S)$ 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 $Aut(N)$)
2. $N \cap C_G (N) = 1$ (by showing that it is an abelian normal subgroup of $N$)
3. $Inn(N) \cong S$ (by defining a homomorphism from $N$ to $Inn(N)$, with trivial kernel)

Now I am asked to show that

• $C_G(N) \neq 1$, using the facts that $Out(S)$ is solvable, $Inn(N)$ 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 $Out(S) \cong G/N$, 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 $Out(S)$, which would mean it would be cyclic of prime order and therefore have at least 2 elements.

• what does the title of the question have to do with the body? – Jorge Fernández Hidalgo Sep 26 '16 at 0:40
• Sorry - I got very off track in writing this post. It would be useful in proving that $G = N C_G (N)$, since both $N$ and $C_G(N)$ are both normal subgroups of $G$ and their intersection is trivial. – mathiest Sep 26 '16 at 0:43
• Oh I see, the statement in the title is not necessarily true though. – Jorge Fernández Hidalgo Sep 26 '16 at 0:43
• Sigh- I figured. Thanks. – mathiest Sep 26 '16 at 0:47
• In the first sentence you say $G$ is isomorphic to $S$, which implies $G$ is simple, so $N=\{1\}$ or $G$. Did you instead mean to say that $N$ is isomorphic to $S$? – stewbasic Sep 26 '16 at 0:47