Sylow intersection technique

Description

The Sylow intersection technique is a technique used to obtain further and tighter conditions on the Sylow numbers of a simple group. The idea is to start with a given integer, obtain constraints on the Sylow numbers of any simple group whose order is that integer, and finally use these constraints to obtain a contradiction.

Motivation

The idea starts off by revisiting the proof that $n_{p}\equiv 1\mod p$. Recall the proof:

Let $P$ be a $p$-Sylow subgroup and $Syl_{p}(G)$ denote the set of all $p$-Sylow subgroups. $G$ acts on $Syl_{p}(G)$ by inner automorphisms. In particular, $P$ acts on $Syl_{p}(G)$ by conjugation.

Under this action, any member $Q\neq P$ cannot be fixed by all elements in $P$, because if it were, then $PQ$ would be a group of prime power order strictly greater than both $P$ and $Q$ (this follows from the product formula).

Thus, every orbit in $Syl_{p}(G)$ under the action of $P$ (other than $P$ itself) is a nontrivial orbit. Since the size of any orbit must divide the size of $P$, any orbit must have size a power of $p$, hence a multiple of $p$, and we are done.

Getting stronger version

For the general version, first observe that $N_{P}(Q)=P\cap N_{G}(Q)=P\cap Q$ for any $P,Q\in Syl_{p}(G)$.

Our aim is to show that every orbit under the conjugation action of $P$ has size a multiple of $p^{l}$ for some given $l$. In order to do that, it suffices to show that for any Sylow subgroup $Q$, $N_{P}(Q)$ has index at least $p^{l}$ in $Q$. This in turn in equivalent to saying that $P\cap Q$ has index at least $p^{l}$ in both $P$ and $Q$.

If we show that, it then follows directly that $n_{p}\equiv 1\mod p^{l}$.

Finding conditions for the stronger version

Here is a special situation where we can apply the stronger version. Suppose $G$ is a simple group of order $p^{r}q$ where $p$ and $q$ are primes. Let $P$ and $Q$ be $p$-Sylow subgroups.

Suppose $P\cap Q$ has order $p^{r-1}$. Then it has index $p$ in both $P$ and $Q$. From the theory of $p$-groups. it follows that $P\cap Q$ is normal in both $P$ and $Q$, hence it is also normal in the subgroup generated by them.

Since $P$ is a maximal subgroup of $G$ (on account of being a subgroup of prime index), the subgroup generated by $P$ and $Q$ is the whole of $G$. Hence $P\cap Q$ is normal in $G$ and this contradicts the simplicity of $G$.

The upshot is that if $G$ has order $p^{l}q$ then $n_{p}\equiv 1\mod p^{2}$.

But since $n_{p}|q$, we have either $n_{p}=1$ or $q\equiv 1\mod p^{2}$. Hence, any simple group of order $p^{r}q$ where $r>1$ must satisfy $p^{2}|q-1$.