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:

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

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

Related facts

Converse

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

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 $Q=P\cap O_{p^{\prime },p}(G)$. Then, $C_G(Q)\le O_{p^{\prime },p}(G)$, where $C_G(Q)$ is the centralizer of $Q$ in $G$.

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

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