P-solvable implies p-constrained

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., p-solvable group) must also satisfy the second group property (i.e., p-constrained group)
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Statement

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:

$C_{G}(P\cap O_{p',p}(G))\leq O_{p',p}(G)$.

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

Facts used

1. 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. Sylow satisfies image condition
3. 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, $C_{G}(Q)\leq O_{p',p}(G)$, 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 $\varphi ^{-1}(O_{p}(G/O_{p'}(G)))=O_{p',p}(G)$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$G/O_{p'}(G)$ is $p'$-core-free				[SHOW MORE] The inverse image of a normal $p'$-subgroup in $G/O_{p'}(G)$ is a normal $p'$-subgroup of $G$ containing $O_{p'}(G)$. But $O_{p'}(G)$ is the unique largest normal $p'$-subgroup, so there is no nontrivial normal $p'$-subgroup in $G/O_{p'}(G)$.
2	$Q$ is a $p$-Sylow subgroup of $O_{p',p}(G)$	Fact (1)	$P$ is $p$-Sylow in $G$, and $Q=P\cap O_{p',p}(G)$.		[SHOW MORE] We combine the fact ,given, and the observation that $O_{p',p}(G)$ is normal in $G$.
3	$\varphi (Q)=O_{p}(G/O_{p'}(G))$, or equivalently $\varphi ^{-1}(\varphi (Q))=O_{p',p}(G)$.	Fact (2)		Step (2)	[SHOW MORE] By combining Step (2) and Fact (2), $\varphi (Q)$ is a $p$-Sylow subgroup of $\varphi (O_{p',p}(G))=O_{p',p}(G)/O_{p'}(G)=O_{p}(G/O_{p'}(G))$. But the latter is a $p$-group, so $\varphi (Q)$ is the whole group.
4	$\varphi (Q)$ is self-centralizing, i.e., $C_{G}(\varphi (Q))\leq \varphi (Q)=O_{p}(G/O_{p'}(G))$	Fact (3)	$G$ is $p$-solvable.	Steps (1), (3)	[SHOW MORE] By step (1), $G/O_{p'}(G)$ is $p'$-core-free. Also, $G$ is $p$-solvable, i.e., $\{p\}$-separable, hence $G/O_{p'}(G)$ is also $\{p\}$-separable. Thus, by Fact (3), the $p$-core of $G/O_{p'}(G)$ is self-centralizing. Step (3) yields that the $p$-core of $G/O_{p'}(G)$ is $\varphi (Q)$, hence $\varphi (Q)$ is self-centralizing.
5	$\varphi (C_{G}(Q))\leq C_{G}(\varphi (Q))$				This follows from the definition of homomorphism: if an element centralizes $Q$, its image centralizes the image of $Q$.
6	$\varphi (C_{G}(Q))\leq \varphi (Q)$			Steps (4), (5)	Step-combination direct
7	$C_{G}(Q)\leq \varphi ^{-1}(\varphi (Q))=O_{p',p}(G)$			Steps (3), (6)	Step-combination direct