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 $$Q \ne P$$ 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 $$p^rq$$ 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 $$p^lq$$ 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 $$p^rq$$ where $$r > 1$$ must satisfy $$p^2|q-1$$.