P-constraint is not subgroup-closed

This article gives the statement, and possibly proof, of a group property (i.e., p-constrained group) not satisfying a group metaproperty (i.e., subgroup-closed group property).
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Statement

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

Related facts

• p-constraint is not quotient-closed
• p-solvable implies p-constrained
• p-constrained not implies p-solvable
• Constrained for a prime divisor implies not simple non-abelian

Facts used

1. Constrained for a prime divisor 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:

1. $G$ is $p$-constrained: Indeed, $O_{p^{\prime }}(G)$ is trivial, and $O_{p^{\prime },p}(G)$ 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 $P\cap O_{p^{\prime },p}(G)$ 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 $C_G(P\cap O_{p^{\prime },p}(G))\le O_{p^{\prime },p}(G)$.
2. $G$ has a subgroup isomorphic to alternating group:A5, namely the non-normal part of its realization as a semidirect product. This subgroup is a simple non-abelian group with $p$ dividing its order (i.e., 2 divides 60) and hence, by Fact (1), is not a $p$-constrained group.