Group with fewer than $p^2$ Sylow $p$-subgroups UPDATE: This question has been asked and answered on MathOverflow.
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 (see Group with $p+1$ Sylow $p$-subgroups).
 A: This is true for $p=2$. If there are 3 $2$-Sylow subgroups, the group $G$ acts transitively on the set of Sylow 2-subgroups by conjugation. So there is a nontrivial homomorphism into $S_3$. If the image is cyclic of order $3$ then all Sylow 2-subgroups are in the kernel which has fewer elements than $G$ and we conclude by induction on the order of $G$.
Thus the image is of order $6$. Let $S_i$, $i=1,2,3$ be the Sylow 2-subgroups of $G$. Then there exists $g$  in $G$ such that $S_1^g=S_2, S_2^g=S_3$. Hence the pairwise intersections of the Sylow 2-subgroups are all of the same order. This answers the first question.
Just noticed that the poster knows this because $3=2+1$.
A: Here is a proof that the pairwise intersections of Sylow subgroups have the same order $p^{n-1}$.
Let $S_1,...,S_m$ be all Sylow $p$-subgroups of $G$, $m<p^2$. Consider the action of $S_1$ on the set of these subgroups by conjugation. Then the size of every orbit is the index of the normalizer of $S_i$ in $S_1$, is a power of $p$. This power cannot be $1$ if $i\ne 1$. And it cannot be $\ge p^2$ because $m<p^2$. So the size of every orbit except ${S_1}$ is $p$.
Thus $|N_{S_1}(S_i)|=p^{n-1}$. If we consider the product $N_{S_1}(S_i)S_i$ which is a $p$-group containing $S_i$ and remember that $S_i$ is a Sylow subgroup, we conclude that $N_{S_1}(S_i)<S_i$. Therefore the order of $S_1\cap S_i$ is $p^{n-1}$ for every $i\ne 1$.
Since every subgroup of index $p$ in a $p$-group is normal $S_1\cap S_i$ is normal in both $S_1$ and $S_i$.
Edit. A few more facts:
We can assume that $G$ has no normal $p$-subgroups.
Let $O_1=\{S_1\}$, $O_2,...,O_{k+1}$ be the orbits of the action of $S_1$ on the set of Sylow subgroups.  Let $N_i$, $i=2,...,k+1$ be the intersection of the Sylow subgroups in $O_i$. Then $N_i<S_1$ is of order $p^{n-1}$. Therefore for every Sylow $p$-subgroup $S_j, [S_1,S_1]$ is a normal subgroup of $S_j$. Hence $[S_1,S_1]$ is a normal subgroup of $G$. Thus we can assume that all Sylow $p$-subgroups of $G$ are Abelian. Hence all $N_i$ are Abelian also. Similarly, $S_1^p\le N_i$, so $S_1^p$ is normal in $G$, hence we can assume that all Sylow $p$-subgroups of $G$ are elementary Abelian $p$-groups of size $p^n$.
Unknown cases: $n\ge 2 \& k>1 \& p>2$  .
