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 $$P/\Phi(P)$$.

Facts used

 * 1) uses::Cayley's theorem
 * 2) uses::Bryant-Kovacs theorem
 * 3) uses::Burnside's theorem on coprime automorphisms and Frattini subgroup
 * 4) uses::Normal Hall implies permutably complemented

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 $$G$$.