Generalized Glauberman-Thompson normal p-complement theorem

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) uses::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 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: