# Does the centraliser of the intersection of any two Sylow $p$-subgroups contain all Sylow $p$-subgroups?

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 $$P$$)

My Question:

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?

• Why must they be abelian? Nov 9 '18 at 20:56
• or more concisely every group of order $p^{\alpha}, \alpha \geq 1$ is abelian Nov 9 '18 at 21:04
• @exodius Is the quaternion group abelian? Nov 9 '18 at 21:05
• So a group of order 27 for example must be abelian? Nov 9 '18 at 21:05
• @the_fox I'm trying to find where I read that but now I can only find its true for $p^2$ perhaps I misunderstood and it's neccarily true beyond this ? Nov 9 '18 at 21:10

I read that given the case where there were 4 Sylow $$3$$-subgroups of order $$9$$ that the fact that $$P_1,P_2≤C_G(Q)$$ implies that there are at least $$2$$ Sylow $$3$$-subgroups are contained in it. And so it contains at least $$1+3=4$$.
This follows immediately from Sylow theory in $$C_G(Q)$$, since $$n_{C_G(Q)} \equiv 1$$ mod $$3$$ and $$\{P_1,P_2\} \subseteq Syl_3(C_G(Q))$$.
This is not necessarily true: $$P_1 \cap P_2$$ in general does not lie in the center $$Z(P_1)$$ or $$Z(P_2)$$.