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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 $N$.

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  • $\begingroup$ I edited you LaTeX: I took it out of the title (LaTeX in the title means that the front page can load slowly for some people, and isn't really necessary here, unlike if you were trying to write an integral or something), and you should use \operatorname as opposed to \text for operators. $\endgroup$ – user1729 Aug 17 '13 at 20:22
  • $\begingroup$ @anon Yes, that was a typo. $\endgroup$ – PVAL-inactive Aug 17 '13 at 22:14
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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 $G=H\times N$.

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