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

(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.

• 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. Oct 12, 2020 at 18:21

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.

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