Glauberman type is not quotient-closed

Statement
It is possible to have a prime number $$p$$, a finite group $$G$$, and a normal subgroup $$N$$ of $$G$$, such that $$G$$ is a group of Glauberman type for $$p$$, but $$G/N$$ is not.

Facts used

 * 1) uses::Glauberman type for a prime divisor implies not simple non-abelian

Example of $$SL(2,5)$$ and $$PSL(2,5)$$
We take $$p = 2$$, $$G = SL(2,5)$$, and $$N = Z(G)$$.

If $$P$$ is a $$2$$-Sylow subgroup of $$G$$, then $$P$$ is isomorphic to a quaternion group with $$Z(J(P)) = Z(G)$$. In particular, $$O_{2'}(G)N_G(Z(J(P))) = N_G(Z(G)) = G$$. Thus, $$G$$ has Glauberman type for the prime $$p = 2$$.

On the other hand, the quotient $$G/N$$ is a simple non-abelian group isomorphic to alternating group:A5, so by fact (1), it is not of Glauberman type.