Consider a collection of Sylow $p$-subgroups. If any two of these intersect non-trivially then they are both contained within the centraliser of their intersection. Now assume that $P_1$ and $P_2$ are abelian I believe if $P_1, P_2$ are abelian then obviously every one of the elements of this group must commute with those in $Q$. ($Q$ denotes the intersection of these $P$)
I read that given the case where there were $4$ Sylow $3$-subgroups of order $9$ that the fact that $P_1,P_2 \leq C_G(Q)$ implies that there are at least $2$ Sylow $3$-subgroups are contained in it. And so it contains at least $1+3=4$.
So in other words it contains all of them. so my question is can we assume if $P_1,P_2,P_3,P_4$ are all abelian and of the same order that then the centraliser of the intersection of any two of then a subgroup which contains every $P$ a $p$-Sylow subgroup?