Subgroup complement for normal subgroup of $G$ with trivial center and $\mathrm{Aut}(N)=\mathrm{Inn}(N)$

Question

Let $G$ be a finite group and $N \subseteq G$ be a normal subgroup. If $Z(N)$ is trivial and $\operatorname{Aut} N=\operatorname{Inn} N$, then show $N$ has a complement $H$ and $H$ is normal in $G$

I can see that it is required that $H= \{ x\in G \mid xy=yx\text{ for all } y\in N\}$, and if this is in fact a subgroup complement for $N$ then it is easy to check it is normal, but I am having difficulty seeing how to use the fact that there are no outer automorphisms of $N$ and that $Z(N)$ is trivial to actually show such a group is a complement for $N$.

Answer 1

There is a map $G\to{\rm Aut}(N)\to{\rm Inn}(N)\to N/Z(N)\to N$. (All but the first map is actually an isomorphism.) It first sends $g$ to the conjugation-by-$g$ automorphism. The kernel is $H=C_G(N)$.

This map restricts to the identity on $N$, so in particular it is surjective. Thus $|H|=|G/N|$ and

$$|HN|=\frac{|H|\cdot|N|}{|H\cap N|}=\cdots$$

Can you finish? (What is $H\cap N$?)

More generally, without using finiteness of $G$, one may argue that $H\to G\to N$ is a split exact sequence. Observe $g\mapsto n$ implies that $gn^{-1}$ is in the kernel, $H$, therefore $G=HN$. Moreover since $[H,N]=1$ we may conclude $G=H\times N$.