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

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).

• 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 $R\subset P'$). Oct 11, 2020 at 15:40
• 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. Oct 11, 2020 at 15:42
• 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? Oct 11, 2020 at 17:00
• 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. Oct 12, 2020 at 5:24
• 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. Oct 12, 2020 at 11:05

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 $$3=2+1$$.

• 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 $G$. Oct 10, 2020 at 19:33

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 $$G$$, $$m. Consider the action of $$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. 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 $$N_{S_1}(S_i). Therefore the order of $$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 $$N_i is of order $$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: $$n\ge 2 \& k>1 \& p>2$$ .

• 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 $p^{n-1}$. Oct 11, 2020 at 11:15
• This is maximum of what I can prove so far. Can you prove more? It is not clear even if $G$ is not simple.
– dodd
Oct 11, 2020 at 15:35
• 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 $G$. Oct 12, 2020 at 1:55
• Can we reduce to the case where $\varphi\colon G\to S_{kp+1}$ is injective? Oct 12, 2020 at 1:56
• I am not sure whether the injectivity of $\phi$ will help.
– dodd
Oct 12, 2020 at 2:43