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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?

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    $\begingroup$ Why must they be abelian? $\endgroup$ – the_fox Nov 9 '18 at 20:56
  • $\begingroup$ or more concisely every group of order $p^{\alpha}, \alpha \geq 1$ is abelian $\endgroup$ – excalibirr Nov 9 '18 at 21:04
  • $\begingroup$ @exodius Is the quaternion group abelian? $\endgroup$ – Hagen von Eitzen Nov 9 '18 at 21:05
  • $\begingroup$ So a group of order 27 for example must be abelian? $\endgroup$ – the_fox Nov 9 '18 at 21:05
  • $\begingroup$ @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 ? $\endgroup$ – excalibirr Nov 9 '18 at 21:10
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You wrote:

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 $\{P_1,P_2\} \subseteq Syl_3(C_G(Q))$.

Another remark:

You also wrote: If any two of these intersect non-trivially then they are both contained within the centraliser of their intersection.

This is not necessarily true: $P_1 \cap P_2$ in general does not lie in the center $Z(P_1)$ or $Z(P_2)$.

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