(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.

  • 2
    $\begingroup$ 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, 2020 at 18:21

2 Answers 2


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.

  • 4
    $\begingroup$ 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. $\endgroup$ Oct 12, 2020 at 19:54
  • 3
    $\begingroup$ @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 $O_p(G)=O_p(\langle R,S\rangle)$. $\endgroup$ Oct 12, 2020 at 20:02
  • $\begingroup$ Title of @GeoffRobinson's reference: Laffey - A remark on minimal Sylow intersections. $\endgroup$
    – LSpice
    Dec 9, 2020 at 20:27

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.