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(This question is originally from Math.SE where it was suggested that I ask the question here)

Let $G$ be a finite group with fewer than $p^2$ Sylow $p$-subgroups, and let $p^n$ be the power of $p$ dividing $\lvert G\rvert$. I can show that if $P$ and $Q$ are any two distinct Sylow $p$-subgroups of $G$ then $\lvert P\cap Q\rvert=p^{n-1}$. I was wondering if this intersection is necessarily the same across all Sylow $p$-subgroups of $G$.

Is the intersection $P\cap Q$ the same for any two distinct Sylow $p$-subgroups $P$ and $Q$?

We might as well assume that $G$ has more than one Sylow $p$-subgroup, in which case here are two equivalent formulations:

Does the intersection of all Sylow $p$-subgroups of $G$ necessarily have order $p^{n-1}$?

Must there exist a normal subgroup of $G$ of order $p^{n-1}$?

I'm looking for a proof or counterexample of this conjecture.

I know that the conjecture holds in the case where $G$ has $p+1$ Sylow $p$-subgroups.

There is some good partial progress in the comments and answers of the Math.SE link.

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    $\begingroup$ As I commented in the MSE post, I believe that $N_G(P)$ must be a maximal subgroup of $G$, since $N_G(P) < M < G$ would make it impossible for $G$ to have less than $p^2$ Sylow $p$-subgroups. So the conjugation action of $G$ on the Sylow $p$-subgroups is primitive, and we can apply the O'Nan-Scott Theorem. I haven't thought it through, but I am guessing that we can reduce to the almost simple case, and then of course one could resort to the classification. $\endgroup$ – Derek Holt Oct 12 at 18:21
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The conjecture follows quickly from Brodkey's Theorem: Let $G$ be a finite group and $p$ a prime. Suppose that Sylow $p$-subgroups of $G$ are abelian. If $O_p(G)=1$, then there exist Sylow $p$-subgroups $P$ and $Q$ of $G$ such that $P\cap Q=1$.

Here $O_p(G)$ is the intersection of all Sylow $p$-subgroups of $G$, or equivalently the largest normal $p$-subgroup of $G$. (Note that $O_p(G/O_p(G))=1$.) Brodkey's theorem can be found several places on the web. It is an exercise in section 1E of Isaacs's Finite Group Theory.

Now, your assumption implies that $\Phi(P)\le P\cap Q\le Q$ for all Sylow $p$-subgroups $P,Q$ of $G$, so $\Phi(P)\le O_p(G)$. Pass to $\bar G=G/O_p(G)$. Then $\bar P$ is an (elementary) abelian Sylow $p$-subgroup of $\bar G$, and $O_p(\bar G)=1$. (This much was already noted on Math.SE.) Now Brodkey's Theorem gives you $\bar P\cap \bar Q=\bar 1$ for some Sylow subgroups $P,Q$ of $G$, so $P\cap Q=O_p(G)$, as you conjectured.

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    $\begingroup$ londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/blms/… gives a nice generalization by T. Laffey of the Theorem of J. Brodkey. Laffey proves that if $O_{p}(G) = 1$ and $R,S$ are Sylow $p$-subgroups of $G$ with $R \cap S$ minimal, then $Z(R) \cap Z(S) = 1$. But Brodkey's Theorem is very elegant and enough for this problem. $\endgroup$ – Geoff Robinson Oct 12 at 19:54
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    $\begingroup$ @GeoffRobinson: Thanks for that. Another generalization is Theorem 1.38 in Isaacs's book: If $R$ and $S$ are Sylow $p$-subgroups of $G$ with $R\cap S$ minimal, then $O_p(G)=O_p(\langle R,S\rangle)$. $\endgroup$ – Richard Lyons Oct 12 at 20:02

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