Strongly p-solvable implies normalizer of D*-subgroup generates whole group with p'-core for odd p

General statement
Suppose $$G$$ is a finite group and $$p$$ is an odd prime number. If $$G$$ is a strongly p-solvable group, then the D*-subgroup functor in $$G$$ is a characteristic p-functor whose normalizer generates whole group with p'-core.

Statement for p'-core-free finite groups
Suppose $$G$$ is a finite group and $$p$$ is an odd prime number. Suppose $$O_{p'}(G)$$ is trivial, i.e., $$G$$ has no nontrivial normal $$p'$$-subgroup. If $$G$$ is a strongly p-solvable group, then the following equivalent conditions are satisfied:


 * 1) For one (and hence every) $$p$$-Sylow subgroup $$P$$ of $$G$$, $$D^*(P)$$ is a normal subgroup of $$G$$.
 * 2) For one (and hence every) $$p$$-Sylow subgroup $$P$$ of $$G$$, $$D^*(P)$$ is a characteristic subgroup of $$G$$.

Facts used

 * 1) uses::Strongly p-solvable implies p-stable
 * 2) uses::p-solvable implies p-constrained
 * 3) uses::p-constrained and p-stable implies normalizer of D*-subgroup generates whole group with p'-core for odd p

Proof
The proof follows by combining facts (1)-(3) (also note that strongly p-solvable implies p-solvable by definition).