P-solvable not implies Glauberman type for p

Statement
It is possible to have a prime number $$p$$ and a finite group $$G$$ such that $$G$$ is $$p$$-solvable (i.e., is a p-solvable group) but not a group of Glauberman type for $$p$$.

In fact, this is possible for the primes $$p = 2,3$$.

Related facts

 * strongly p-solvable implies Glauberman type for p: In particular, this shows that for $$p \ge 5$$, $$p$$-solvable implies Glauberman type for $$p$$.
 * Glauberman type not implies p-constrained
 * Glauberman type not implies p-stable

Example at the prime two
Consider the prime $$p = 2$$ and the group $$G = S_4$$, the symmetric group on the set $$\{ 1,2,3,4 \}$$. $$G$$ is a solvable group, and hence is $$2$$-solvable. However, $$G$$ is not of Glauberman type for the prime two, because if we take $$P$$ as the $$2$$-Sylow subgroup generated by $$\{ (1,2,3,4), (1,3) \}$$, then $$Z(J(P)) = \{, (1,3)(2,4)\}$$. In this case, we have:

$$O_{2'}(G)N_G(Z(J(P))) = P \ne G$$.