Glauberman type for a prime divisor implies not simple non-abelian

Statement
Suppose $$G$$ is a finite group and $$p$$ is a prime number dividing the order of $$G$$. Suppose further that $$G$$ is a group of Glauberman type for $$p$$. Then, $$G$$ is not a simple non-abelian group.

Proof
Given: A group $$G$$, a prime $$p$$ dividing the order of $$G$$ such that $$O_{p'}(G)N_G(Z(J(P)) = G$$.

To prove: $$G$$ is not simple non-abelian.

Proof: We consider two cases:


 * $$O_{p'}(G)$$ is nontrivial: In this case, $$O_{p'}(G)$$ is a proper nontrivial normal subgroup of $$G$$. It is proper because $$p$$ divides the order of $$G$$.
 * $$O_{p'}(G)$$ is trivial: In this case, $$N_G(Z(J(P))) = G$$. Thus, $$Z(J(P))$$ is normal in $$G$$. Since $$p$$ divides the order of $$G$$, $$P$$ is nontrivial, hence, since the ZJ-functor, by definition, sends nontrivial $$p$$-subgroups to nontrivial $$p$$-subgroups, $$Z(J(P))$$ is nontrivial. Thus, $$Z(J(P))$$ is a nontrivial normal subgroup of $$G$$. The only way it can be the whole group is if $$P$$ is abelian. Hence, we either have a proper nontrivial normal subgroup or have that $$P$$ is abelian. In either case, we are done.