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

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.

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

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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 $\varphi^{-1}(O_p(G/O_{p'}(G))) = O_{p',p}(G)$.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $G/O_{p'}(G)$ is $p'$-core-free [SHOW MORE]
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]
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]
4 $\varphi(Q)$ is self-centralizing, i.e., $C_G(\varphi(Q)) \le \varphi(Q) = O_p(G/O_{p'}(G))$ Fact (3) $G$ is $p$-solvable. Steps (1), (3) [SHOW MORE]
5 $\varphi(C_G(Q)) \le C_G(\varphi(Q))$ This follows from the definition of homomorphism: if an element centralizes $Q$, its image centralizes the image of $Q$.
6 $\varphi(C_G(Q)) \le \varphi(Q)$ Steps (4), (5) Step-combination direct
7 $C_G(Q) \le \varphi^{-1}(\varphi(Q)) = O_{p',p}(G)$ Steps (3), (6) Step-combination direct