# Every p'-group is the p'-part of the automorphism group of a p-group

## Statement

Suppose $p$ is a prime number and $G$ is a finite group whose order is no a multiple of $p$. Then, there exists a $p$-group $P$ such that $\operatorname{Aut}(P)$ has a normal $p$-Sylow subgroup, with $G$ as a complement to the normal Sylow subgroup.

Further, the normal $p$-Sylow subgroup is the kernel of the action of $\operatorname{Aut}(P)$ on $P/\Phi(P)$.

## Proof

Given: A finite group $G$, a prime $p$ not dividing the order of $G$.

To prove: There exists a $p$-group $P$ such that $\operatorname{Aut}(P)$ has a normal $p$-Sylow subgroup with complement $G$.

Proof: There exists $n > 1$ for which $G$ can be embedded as a subgroup of $GL(n,p)$ (this follows from fact (1), and the fact that the symmetric group on $n$ letters embeds inside $GL(n,p)$). By fact (2), we can find a $p$-group $P$ such that the image of $\operatorname{Aut}(P)$ in $\operatorname{Aut}(P/\Phi(P))$ is $G$.

By fact (3), the kernel of the map from $\operatorname{Aut}(P)$ to $\operatorname{Aut}(P/\Phi(P))$ is a $p$-group. Since the kernel is a $p$-group and the quotient is a $p'$-group, the kernel is a normal $p$-Sylow subgroup. By fact (4), it has a permutable complement in $\operatorname{Aut}(P)$, and this complement is isomorphic to $G$.