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 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. Sylow subgroups exist
2. Sylow implies order-dominating
3. Prime power order implies nilpotent
4. 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. $N_{P}(H)\leq P\cap R$: 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.