Maximal Sylow intersection is tame

Definition
Suppose $$G$$ is a finite group and $$p$$ is a prime number. Suppose $$H = P \cap Q$$ is a fact about::maximal Sylow intersection: in other words, $$H$$ is an intersection of two $$p$$-Sylow subgroups $$P,Q$$ such that no intersection of distinct Sylow subgroups has order bigger than the order of $$H$$. Then, $$H$$ is a tame Sylow intersection: $$N_P(H)$$ and $$N_Q(H)$$ are both Sylow subgroups of $$N_G(H)$$.

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Sylow implies order-dominating
 * 3) uses::Prime power order implies nilpotent
 * 4) uses::Nilpotent implies normalizer condition

Proof
Given: A finite group $$G$$, a maximal $$p$$-Sylow intersection $$H = P \cap Q$$.

To prove: $$H$$ is a tame Sylow intersection.

Proof: Suppose $$N_P(H)$$ is not $$p$$-Sylow in $$N_G(H)$$. We derive a contradiction.


 * 1) Let $$S$$ be a $$p$$-Sylow subgroup of $$N_G(H)$$ containing $$N_P(H)$$: $$S$$ exists by facts (1) and (2). Note that $$S$$ is not contained in $$P$$.
 * 2) Let $$R$$ be a $$p$$-Sylow subgroup of $$G$$ containing $$S$$. $$R$$ is distinct from $$P$$: Such an $$R$$ exists by fact (2). Distinctness follows because, by assumption, $$S$$ is not contained in $$P$$.
 * 3) $$N_P(H) \le P \cap R$$: The intersection $$P \cap R$$ is a $$p$$-subgroup of $$G$$ containing $$P \cap S$$, which in turn equals $$N_P(H)$$.
 * 4) $$H$$ is a proper subgroup of $$P \cap R$$: By facts (3) and (4) applied to the group $$P$$, $$H$$ is a proper subgroup of $$P$$, so $$H$$ is properly contained in $$N_P(H)$$.
 * 5) $$H$$ is not a maximal Sylow intersection: Indeed, $$P \cap R$$ is an intersection of distinct Sylow subgroups of order bigger than that of $$H$$.

Thus, we have achieved the desired contradiction.