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

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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. Cayley's theorem
  2. Bryant-Kovacs theorem
  3. Burnside's theorem on coprime automorphisms and Frattini subgroup
  4. Normal Hall implies permutably complemented


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.