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

Question

(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.

Answer 1

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.

Answer 2

Now that I understand things better, let me also give a direct proof (using essentially the same idea as Brodkey's theorem).

Let $P,Q,R$ be Sylow $p$-subgroups of $G$, let $N=N_G(P\cap Q)$. We know that $P\cap Q$ has order $p^{n-1}$, so $P\leq N$ and $Q\leq N$. In other words, $P$ and $Q$ are Sylow $p$-subgroups of $N$. The intersection $R\cap N$ is a $p$-subgroup of $N$, so $R\cap N\leq P^g$ for some $g\in N$. We have $$R\cap Q^g=R\cap(N\cap Q^g)=(R\cap N)\cap Q^g\leq P^g\cap Q^g=(P\cap Q)^g=P\cap Q.$$ Now suppose that $P\neq Q$. Then $P\cap Q$ has order $p^{n-1}$, so $R\cap Q^g$ must also have order $p^{n-1}$, and in fact we must have $R\cap Q^g=P\cap Q$. Thus, $P\cap Q\leq R$. But $R$ is any arbitrary Sylow $p$-subgroup of $G$, so the conjecture is proven.