P-solvable implies p-constrained

Verbal statement
Any p-solvable group is a  p-constrained group.

Statement with symbols
Suppose $$G$$ is a finite group and $$p$$ is a prime number. Suppose further that $$G$$ is $$p$$-solvable. Then, if $$P$$ is a $$p$$-Sylow subgroup, we have:

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

In other words, $$G$$ is $$p$$-constrained.

Converse

 * p-constrained not implies p-solvable
 * The converse is partially true: p-constrained implies not simple non-abelian

Facts used

 * 1) uses::Equivalence of definitions of Sylow subgroup of normal subgroup: This states that the intersection of a Sylow subgroup and a normal subgroup is a Sylow subgroup of the normal subgroup.
 * 2) uses::Sylow satisfies image condition
 * 3) uses::Pi-separable and pi'-core-free implies pi-core is self-centralizing

Proof
Given: A finite group $$G$$ that is $$p$$-solvable for some prime $$p$$. $$P$$ is a $$p$$-Sylow subgroup.

To prove: Let $$Q = P \cap O_{p',p}(G)$$. Then, $$C_G(Q) \le O_{p',p}(G)$$, where $$C_G(Q)$$ is the centralizer of $$Q$$ in $$G$$.

Proof: Let $$\varphi:G \to G/O_{p'}(G)$$ be the natural quotient map. Note that $$\varphi^{-1}(O_p(G/O_{p'}(G))) = O_{p',p}(G)$$.