# 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.