Maximal Sylow intersection conjugate in the whole group to another subgroup of the Sylow subgroup is conjugate in the normalizer

Statement
Suppose $$P, Q$$ are $$p$$-Sylow subgroups of a finite group $$G$$, and $$P \cap Q$$ is a fact about::maximal Sylow intersection. In other words, $$P \cap Q$$ is of maximal order among all possible intersections of distinct Sylow subgroups. Then, any conjugate of $$P \cap Q$$ in $$G$$ that is contained in $$P$$ is conjugate to $$P \cap Q$$ in $$N_G(P)$$.

Facts used

 * 1) uses::Alperin's fusion theorem in terms of tame intersections

Proof
(This proof uses the right-action convention)

Given: A maximal Sylow intersection $$H = P \cap Q$$, a conjugate $$H^g$$ contained in $$P$$ for some $$g \in G$$.

To prove: $$H$$ and $$H^g$$ are conjugate in $$N_G(P)$$.

Proof: By fact (1) applied to the subsets $$H, H^g$$ of $$P$$, there exist $$p$$-Sylow subgroups $$Q_1, Q_2, Q_3, \dots, Q_n$$ of $$G$$ such that $$P \cap Q_i$$ is tame, and elements $$g_i \in N_G(P \cap Q_i)$$ such that $$g = g_1g_2 \dots g_n$$, $$H \le P \cap Q_1$$ and $$H^{g_1g_2 \dots g_i} \le P \cap Q_{i+1}$$ for all <$$1 \le i \le n - 1$$.

We prove the claim by induction on $$n$$, i.e., we assume that the claim is true for all $$m < n$$.

We consider the following two cases:


 * 1) $$P \cap Q_i = P \cap Q$$ for all $$1 \le i \le n$$: In this case, $$g_i \in N_G(P \cap Q)$$ for all $$i$$, so $$g \in N_G(P \cap Q)$$, so $$H^g = H$$. In particular, $$H$$ and $$H^g$$ are conjugate by the identity element, so they are conjugate in $$N_G(P)$$.
 * 2) There exists some $$i$$ such that $$P \cap Q_i \ne P \cap Q$$:
 * 3) Consider the smallest $$i$$ for which $$P \cap Q_i \ne P \cap Q$$. Then, $$g_k \in N_G(P \cap Q)$$ for $$k < i$$, so $$H^{g_1g_2 \dots g_{i-1}} = H$$. Thus, $$H \le P \cap Q_i$$. By maximality of the Sylow intersection, $$P \cap Q_i = P$$, so $$N_G(P \cap Q_i) = N_G(P)$$. Thus, $$H$$ is conjugate to $$H^{g_1g_2 \dots g_i}$$ in $$N_G(P)$$.
 * 4) Note that $$H^{g_1g_2 \dots g_i} = H^{g_i}$$ is a maximal Sylow intersection involving $$P$$: it equals $$P \cap Q^{g_i}$$. Replacing $$H$$ by $$H^{g_i}$$, and $$g$$ by $$g_{i+1} \dots g_n$$, we conclude that $$H^{g_1g_2 \dots g_i}$$ is conjugate in $$N_G(P)$$ to $$H^{g_1g_2 \dots g_n}$$ by induction on $$n$$. Thus, $$H$$ and $$H^g$$ are conjugate in $$N_G(P)$$.