P-constrained and p-stable implies abelian normal subgroup of Sylow subgroup is contained in (p',p)-core

Statement
Suppose $$G$$ is a finite group and $$p$$ is a prime number such that $$G$$ is both a fact about::p-constrained group and a fact about::p-stable group. Then, if $$A$$ is an fact about::abelian normal subgroup of $$P$$ (in the sense of fact about::abelian normal subgroup of group of prime power order), $$A$$ is contained in $$O_{p',p}(G)$$.

Facts used

 * 1) uses::Frattini's argument: If $$H$$ is a normal subgroup of $$G$$ and $$Q$$ is a $$p$$-Sylow subgroup of $$H$$, then $$HN_G(P) = G$$.
 * 2) uses::Sylow satisfies permuting transfer condition

Proof
Given: A $$p$$-constrained and $$p$$-stable group $$G$$. An abelian normal subgroup $$P$$ of a $$p$$-Sylow subgroup $$P$$ of $$G$$.

To prove: $$A \le O_{p',p}(G)$$.

Proof:


 * 1) Let $$Q = P \cap O_{p',p}(G)$$. Then $$Q$$ is $$p$$-Sylow in $$O_{p',p}(G)$$ and $$O_{p'}(G)Q = O_{p',p}(G)$$ which is normal in $$G$$: $$Q$$ is $$p$$-Sylow in $$O_{p',p}(G)$$ by fact (2), and its image modulo $$O_{p'}(G)$$ is therefore $$p$$-Sylow in $$O_{p',p}(G)/O_{p'}(G)$$. The latter is a $$p$$-group, so the image of $$Q$$ is the whole group. Thus, $$O_{p'}(G)Q = O_{p',p}(G)$$.
 * 2) $$A \le N_G(Q)$$ and $$[[Q,A],A]$$ is trivial: By definition, $$Q$$ is the intersection of $$P$$ and a normal subgroup of $$G$$, so $$Q$$ is normal in $$P$$. Thus, $$P \le N_G(Q)$$, and so, $$A \le N_G(Q)$$. Further, since $$Q \le P$$, $$[[Q,A],A] \le [[P,A],A]$$. The latter is trivial because $$[P,A] \le A$$ by normality of $$A$$ and $$[A,A]$$ is trivial by abelianness of $$A$$.
 * 3) $$AC_G(Q)/C_G(Q) = O_p(N_G(Q)/C_G(Q))$$: This follows from the previous step and the definition of $$p$$-stability.
 * 4) $$C_G(Q) \le O_{p',p}(G) = O_{p'}(G)Q$$: This follows from the previous step and the definition of $$p$$-constraint.
 * 5) $$G = O_{p'}(G)N_G(Q)$$: Note that by fact (1), setting $$H = O_{p',p}(G)$$, we get that $$O_{p',p}(G)N_G(Q) = G$$. By step (1), $$O_{p',p}(G) = O_{p'}(G)Q$$, we get get $$O_{p'}(G)(QN_G(Q)) = G$$. Since $$Q \le N_G(Q)$$, we obtain $$O_{p'}(G)N_G(Q) = G$$.
 * 6) Let $$\varphi:G \to G/O_{p'}(G)$$ be the quotient map. Then $$\varphi(AQ)/\varphi(Q) \le O_p(\varphi(G)/\varphi(Q))$$:
 * 7) By step (4), $$\varphi(Q) = \varphi(C_G(Q))$$. Thus, $$\varphi(AC_G(Q)/C_G(Q)) = \varphi(AQ)/\varphi(Q)$$.
 * 8) By step (5), $$\varphi(N_G(Q)) = \varphi(G)$$. Thus, $$\varphi(N_G(Q)/C_G(Q)) = \varphi(G)/\varphi(C_G(Q))$$. Using step (4) again, $$\varphi(N_G(Q)/C_G(Q)) = \varphi(G)/\varphi(Q)$$.
 * 9) Plugging these into step (3) yields the required: $$\varphi(AQ)/\varphi(Q) \le O_p(\varphi(G)/\varphi(Q))$$.
 * 10) $$A \le Q$$:
 * 11) The right side of the previous step is the $$p$$-core of $$\varphi(G)/\varphi(Q)$$. But step (1) says that $$O_{p'}(G)Q = O_{p',p}(G)$$, so $$\varphi(Q) = O_p(\varphi(G))$$. Thus, $$O_p(\varphi(G/Q))$$ is trivial.
 * 12) Thus, the right side is trivial, and we obtain that the left side is also trivial. This yields $$\varphi(AQ) = \varphi(Q)$$, forcing $$AQ \le O_{p'}(G)Q$$.
 * 13) On the other hand, both $$A$$ and $$Q$$ are subgroups of $$P$$, so $$AQ \le P$$. Thus, $$AQ \le P \cap O_{p'}(G)Q$$. Since $$Q \le P$$ and $$P \cap O_{p'}(G)$$ is trivial, we get $$AQ \le Q$$, forcing $$A \le Q$$.