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 $G = S L (2, 5)$ 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 $C_{G} (P \cap O_{p^{'}, p} (G)) = G \subseteq̸ O_{p^{'}, p} (G)$ . 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 $p = 2$ .