P-constrained not implies p-solvable

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.

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:


 * 1) $$G$$ 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 $$C_G(P \cap O_{p',p}(G)) \le O_{p,p}(G)$$.
 * 2) $$G$$ 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.