Glauberman type not implies p-constrained

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group of Glauberman type for a prime) need not satisfy the second group property (i.e., p-constrained group)
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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.

Proof

Example of the special linear group

Further information: special linear group:SL(2,5), subgroup structure of special linear group:SL(2,5)

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 $p=2$.