(This question is [originally from Math.SE](https://math.stackexchange.com/questions/3859476) 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 $G$.



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> Is the intersection $P\cap Q$ the same for any two distinct Sylow $p$-subgroups $P$ and $Q$?
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We might as well assume that $G$ has more than one Sylow $p$-subgroup, in which case here are two equivalent formulations:



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> Does the intersection of all Sylow $p$-subgroups of $G$ necessarily have order $p^{n-1}$?
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> Must there exist a normal subgroup of $G$ of order $p^{n-1}$?
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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.



Group with fewer than $p^2$ Sylow p-subgroups
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.


