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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).

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  • $\begingroup$ Write $R=P\cap Q$ and $N=N_G(R)$. Let $n_N$ and $n_G$ be the number of Sylow $p$-subgroups of $N$ and $G$ respectively. We know that $p<n_N\le n_G\le p^2$. If we make the extra assumption that $G$ is solvable, then the result cited in this answer gives the relation $n_N\mid n_G$. This leaves $n_N=n_G$ as the only possibility. Consequently $R$ is contained in all the Sylow $p$-subgroups of $G$ (for all Sylows $P'$, if $P'$ normalizes $R$, then $P'R$ is a $p$-group, hence equal to $P'$, hence $R\subset P'$). $\endgroup$ Oct 11, 2020 at 15:40
  • $\begingroup$ At first I thought that we automatically have $n_N\mid n_G$ here (when $P\le N$), but I'm not sure about that, and cannot prove it. Anyway, this suggests to me that finding an eventual counterexample may by a bit taxing, at least for me. Back to preparing material for remote teaching vector calculus. $\endgroup$ Oct 11, 2020 at 15:42
  • $\begingroup$ Interesting idea. If you have any normal subgroup $N$ then $n_N\bigm|n_G$ and $n_G\bigm|n_{G/N}$. Maybe your argument can be used to show that $G$ is simple? $\endgroup$ Oct 11, 2020 at 17:00
  • $\begingroup$ I'm a bit embarrassed to admit that it took me a while to come up with a counterexample to that divisibility result in the non-solvable case. $n_3(S_5)=10$, $n_3(S_4)=4$. I need to start thinking about groups at least a little bit more :-) Of course, that doesn't say anything about your question. $\endgroup$ Oct 12, 2020 at 5:24
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    $\begingroup$ I think you can assume that $N_G(P)$ is maximal in $G$ (since otherwise $N_G(P)<M<G$ with $n_M|n_G$, and then $n_G<p^2$ is impossible), so the conjugation action on the Sylow $p$-subgroups of $N_G(P)$ is primitive. Then you could use the O'Nan-Scott Theorem and probably reduce to the case when $G$ is almost simple, and then try and use CFSG to finish it. But it would be nicer if there was a more elementary approach. You could ask this question on mathoverflow. $\endgroup$
    – Derek Holt
    Oct 12, 2020 at 11:05

2 Answers 2

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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 $3=2+1$.

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  • $\begingroup$ Yeah, in general if $G$ has $p+1$ Sylow $p$-subgroups then you can look at the homomorphism to $S_{p+1}$. It can be shown that the kernel of this homomorphism has a normal Sylow $p$-subgroup of order $p^{n-1}$, contained in every Sylow $p$-subgroup of $G$. $\endgroup$ Oct 10, 2020 at 19:33
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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: $n\ge 2 \& k>1 \& p>2$ .

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    $\begingroup$ But that wasn't the question. The question is are all of these intersections equal or, equivalently, does $G$ have a normal subgroup of order $p^{n-1}$. $\endgroup$
    – Derek Holt
    Oct 11, 2020 at 11:15
  • $\begingroup$ This is maximum of what I can prove so far. Can you prove more? It is not clear even if $G$ is not simple. $\endgroup$
    – dodd
    Oct 11, 2020 at 15:35
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    $\begingroup$ Here's another way to see that we can assume that the Sylow $p$-subgroups of $G$ are elementary abelian: Consider the homomorphism $\varphi\colon G\to S_{kp+1}$. Let $K=\ker\varphi$. It can be shown that $K$ has a normal Sylow $p$-subgroup (it will be the intersection of all Sylow $p$-subgroups of $G$). Since we can assume that $G$ has no normal $p$-subgroups, we can assume that $K$ has order indivisible by $p$. Then $\varphi$ is injective on Sylow $p$-subgroups of $G$. Since Sylow $p$-subgroups of $S_{kp+1}$ are elementary abelian, so are Sylow $p$-subgroups of $G$. $\endgroup$ Oct 12, 2020 at 1:55
  • $\begingroup$ Can we reduce to the case where $\varphi\colon G\to S_{kp+1}$ is injective? $\endgroup$ Oct 12, 2020 at 1:56
  • $\begingroup$ I am not sure whether the injectivity of $\phi$ will help. $\endgroup$
    – dodd
    Oct 12, 2020 at 2:43

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