Sylow and TI implies CDIN

Statement
Suppose $$G$$ is a finite group, $$p$$ is a prime, and $$P$$ is a $$p$$-Sylow subgroup of $$G$$ that is also a fact about::TI-subgroup: any conjugate of $$P$$ is either equal to $$P$$ or intersects it trivially. Then $$P$$ is a CDIN-subgroup of $$G$$: any two elements of $$P$$ that are conjugate in $$G$$ are conjugate in $$N_G(P)$$.

In fact, something stronger is true: the original conjugating element is itself in $$N_G(P)$$.

Related facts

 * Sylow and TI implies SCDIN
 * Sylow implies WNSCDIN
 * Sylow implies MWNSCDIN

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 $$x,y \in P$$ are conjugate by $$g \in G$$, then $$x$$ and $$y$$ are conjugate in $$N_G(P)$$. In fact, $$g \in N_G(P)$$.

Proof: If $$x$$ or $$y$$ is the identity element, so is the other, and in this case we are done. So, we can assume that neither $$x$$ nor $$y$$ is the identity element.

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 $$x^{g_1g_2 \dots g_r} \in P \cap Q_{r+1}$$ 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 $$x^{g_1g_2\dots g_r} \in P \cap Q_{r+1}$$, 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 $$x$$ and $$y$$ are conjugate via an element in $$N_G(P)$$.