Glauberman type not implies p-constrained

Statement
It is possible to have a prime number $$p$$ and a finite group $$G$$ such that $$G$$ is a group of Glauberman type for $$p$$ but is not a p-constrained group.

Example of the special linear group
Let $$G = SL(2,5)$$ and $$p = 2$$. Then, $$O_{p',p}(G) = Z(G)$$ is cyclic of order two, and $$G/Z(G)$$ is a simple group isomorphic to alternating group:A5. If $$P$$ is a $$2$$-Sylow subgroup of $$G$$, then $$P$$ is isomorphic to a quaternion group and $$P \cap O_{p',p}(G) = Z(G)$$, so $$C_G(P \cap O_{p',p}(G)) = G \not \subseteq O_{p',p}(G)$$. Thus, $$G$$ is not $$p$$-constrained.

On the other hand, $$Z(J(P)) = Z(G)$$, so $$O_{p'}(G)N_G(Z(J(P))) = N_G(Z(G)) = G$$, so $$G$$ is of Glauberman type with respect to the prime $$p = 2$$.