P-constrained not implies p-solvable

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., p-constrained group) need not satisfy the second group property (i.e., p-solvable group)
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Definition

It is possible to have a finite group $G$ and a prime number $p$ such that $G$ is a p-constrained group but not a p-solvable group.

Related facts

Converse

p-solvable implies p-constrained

Opposite facts

p-constrained implies not simple non-abelian

Proof

Let $G$ be the wreath product of Z2 and A5 defined as the wreath product with base group cyclic group:Z2 and acting group alternating group:A5, where we use the natural permutation action of the acting group on a set of five elements. More explicitly, $G$ is the external semidirect product of elementary abelian group:E32 and alternating group:A5 where the latter acts on the former by coordinate permutations induced by the permutations on a set of five elements.

The group $G$ has order $2^{5} \cdot 60 = 1920 = 2^{7} \cdot 3 \cdot 5$

Let $p = 2$ .

We note that:

$G$ is $p$ -constrained: Indeed, $O_{p^{'}} (G)$ is trivial, and $O_{p^{'}, p} (G)$ is the base of the semidirect product, i.e., a normal subgroup isomorphic to elementary abelian group:E32. In particular, this is contained in any $p$ -Sylow subgroup $P$ , so $P \cap O_{p^{'}, p} (G)$ is also the normal subgroup that forms the base of the semidirect product. The subgroup is a self-centralizing normal subgroup, because it is an abelian normal subgroup and the induced action by the quotient is faithful. Thus, we get the condition $C_{G} (P \cap O_{p^{'}, p} (G)) \leq O_{p, p} (G)$ .
$G$ is not $p$ -solvable: For this, note that alternating group:A5, the simple non-abelian composition factor of $G$ , is neither a $p$ -group nor a $p^{'}$ -group.