Generalized Glauberman-Thompson normal p-complement theorem

This article gives the statement, and possibly proof, of a normal p-complement theorem: necessary and/or sufficient conditions for the existence of a Normal p-complement (?). In other words, it gives necessary and/or sufficient conditions for a given finite group to be a P-nilpotent group (?) for some prime number $p$.
View other normal p-complement theorems

Statement

Suppose $p$ is an odd prime number and $W$ is a characteristic p-functor. Suppose the following is true: in any strongly p-solvable group, $W$ is a characteristic p-functor whose normalizer generates whole group with p'-core. In other words, if $G$ is strongly p-solvable, then we have the following where $P$ is a $p$-Sylow subgroup of $G$:

$G=O_{p'}(G)N_{G}(W(P))$

Then, $W$ is a characteristic p-functor that controls normal p-complements for all finite groups. In other words, for any finite group $G$, it is true that if $P$ is a $p$-Sylow subgroup and $N_{G}(W(P))$ possesses a normal p-complement (i.e., it is a p-nilpotent group), then $G$ also possesses a normal p-complement (i.e., $G$ is also a p-nilpotent group).

Facts used

1. Characterization of minimal counterexamples to a characteristic p-functor controlling normal p-complements: If $G$ is a counterexample of minimum order, then $O_{p'}(G)$ is trivial, $O_{p}(G)$ is nontrivial but proper in $P$, and $G=O_{p,p',p}(G)$. In other words, $G/O_{p}(G)$ has a normal $p$-complement.

Proof

Proof idea

The goal is to prove that no minimal counterexample exists, so we begin with the characterization of minimal counterexamples given in fact (1) and obtain a contradiction by showing that this is not a counterexample.

Note that Step (2) is needed only for the case $p=3$. For other primes, it is unnecessary.

Note that Fact (1), which we use to get up to Step (3), does not use anything about the ZJ-functor. The reasoning from beyond Step (4) also uses nothing about the ZJ-functor. The crucial step that uses the ZJ-functor is going from Step (3) to Step (4), which requires Fact (2). The proof of that relies on the behavior of the ZJ-functor.

Proof details

Given: An odd prime $p$ (fixed throughout the proof). A characteristic $p$-functor $W$ (fixed throughout the proof) such that for any strongly p-solvable group, $W$ is a characteristic p-functor whose normalizer generates whole group with p'-core. A minimal counterexample $G$ for $p$, i.e., a finite group $G$ of minimum possible order such that $G$ does not have a normal $p$-complement but $N_{G}(W((P))$ does for some Sylow subgroup $P$ of $G$.

To prove: A contradiction. We'll show that $G$ does have a normal p-complement.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Suppose $G$ is a minimal counterexample. Then, $O_{p'}(G)=1$ and $G=O_{p,p',p}(G)$, so in particular, $G$ is $p$-solvable.	Fact (1)
2	The $2$-Sylow subgroups of $G$ are abelian				PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
3	$G$ is strongly $p$-solvable			Steps (1), (2)	[SHOW MORE] If $p\neq 3$, then $p$-sovable is the same as strongly $p$-solvable. If $p=3$, then, since the $2$-Sylow subgroups of $G$ are abelian, no subquotient of $G$ can have a quaternion Sylow subgroup. In particular, no subquotient of $G$ can be isomorphic to $SL(2,3)$. Thus, $G$ is strongly $p$-solvable.
4	$O_{p'}(G)N_{G}(W(P))=G$ for any $p$-Sylow subgroup $P$ of $G$. Another way of putting this is that $W$ is a characteristic p-functor whose normalizer generates whole group with p'-core in $G$.		The given condition on $W$	Step (3)	Step-given combination direct
5	$W(P)$ is normal in $G$, or equivalently, $N_{G}(W(P)))=G$.			Steps (1), (4)	This follows from the definition of being of Glauberman type, and the fact that $O_{p'}(G)=1$ by assumption (Step (1))
6	$G$ has a normal $p$-complement, and we have the desired contradiction.			Step (5)	By assumption, $N_{G}(W(P)))$ has a normal $p$-complement. Thus, by Step (5), so does $G$.