Sylow and TI implies SCDIN

Statement
Suppose $$G$$ is a finite group, $$p$$ is a prime, and $$P$$ is a $$p$$-Sylow subgroup of $$G$$, such that $$P$$ is a fact about::TI-subgroup of $$G$$: any conjugate of $$P$$ is either equal to $$P$$ or intersects $$P$$ trivially. Then $$P$$ is a SCDIN-subgroup of $$G$$: any two elements $$A,B$$ of $$P$$ that are conjugate by $$g$$ in $$G$$ are conjugate by some element $$h$$ in $$N_G(P)$$, and moreover, $$h$$ and $$g$$ have the same element-wise action on $$A$$.

In fact, something stronger is true: $$g$$ itself is in $$N_G(P)$$.

Similar facts

 * Sylow and TI implies CDIN
 * Sylow implies WNSCDIN: Combines Sylow implies pronormal and pronormal implies WNSCDIN
 * Sylow implies MWNSCDIN: Combines Sylow implies pronormal and pronormal implies MWNSCDIN
 * Abelian and Sylow implies SCDIN: Combines Sylow implies pronormal and abelian and pronormal implies SCDIN

Opposite facts

 * Sylow not implies CDIN
 * Hall not implies WNSCDIN

Facts used

 * 1) uses::Alperin's fusion theorem in terms of tame intersections
 * 2) uses::Sylow implies order-conjugate

Proof
Given: A finite group $$G$$, a prime $$p$$, a $$p$$-Sylow subgroup $$P$$ of $$G$$ such that the intersection of $$P$$ with any distinct conjugate of itself is trivial.

To prove: If $$A,B \subseteq P$$ are conjugate by $$g \in G$$, then $$A$$ and $$B$$ are conjugate in $$N_G(P)$$. In fact, $$g \in N_G(P)$$.

Proof: If $$A$$ or $$B$$ is empty or has only one element, the same is true for the other, and in this case we are done. So, we can assume that neither $$A$$ nor $$B$$ is contained in the trivial group.

By fact (1), there exist $$p$$-Sylow subgroups $$Q_1, Q_2, \dots, Q_n$$ such that $$P \cap Q_i$$ is a tame intersection, and elements $$g_i \in N_G(P \cap Q_i)$$ such that $$g = g_1g_2 \dots g_n$$ and $$A^{g_1g_2 \dots g_r} \subseteq P \cap Q_{r+1}$$ for $$1 \le r \le n - 1$$.

However, since $$P$$ is TI, its intersection with any other Sylow subgroup (note also fact (2)) is trivial. Since $$A^{g_1g_2\dots g_r} \subseteq P \cap Q_{r+1}$$, we are forced to have that all the $$P \cap Q_i$$ are nontrivial, and hence, equal to $$P$$. This forces $$Q_i = P$$ for all $$i$$, hence $$g_i \in N_G(P)$$. Hence, $$g$$, which is the product of the $$g_i$$s, is in $$N_G(P)$$, and $$A$$ and $$B$$ are conjugate via an element in $$N_G(P)$$.