Congruence condition on Sylow numbers in terms of maximal Sylow intersection

Statement
Let $$G$$ be a finite group and $$p$$ a prime dividing the order of $$G$$. Suppose the intersection of any two distinct $$p$$-Sylow subgroups has index at least $$p^r$$ in both of them. Then, if $$n_p$$ denotes the $$p$$-fact about::Sylow number of $$G$$, i.e., the number of $$p$$-Sylow subgroups of $$G$$, we have:

$$n_p \equiv 1 \mod p^r$$.

In particular, $$n_p - 1$$ is a multiple of the index of a fact about::maximal Sylow intersection.

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Fundamental theorem of group actions
 * 3) uses::Intersection of p-subgroup with Sylow subgroup equals intersection with normalizer: If $$P$$ is a $$p$$-subgroup of $$G$$ and $$Q$$ is a $$p$$-Sylow subgroup of $$G$$, then $$P \cap Q = P \cap N_G(Q)$$.

Proof
Given: A finite group $$G$$, a prime $$p$$ dividing the order of $$G$$. The intersection of any two distinct $$p$$-Sylow subgroups has index at least $$p^r$$ in both of them.

To prove: If $$n_p$$ is the number of $$p$$-Sylow subgroups of $$G$$, then $$n_p \equiv 1 \mod p^r$$.

Proof: There exists at least one $$p$$-Sylow subgroup by fact (1). Let $$P$$ be a $$p$$-Sylow subgroup. Let $$S$$ be the set of all $$p$$-Sylow subgroups. $$P$$ acts on $$S$$ by conjugation: in other words, given $$g \in P$$ and $$Q \in S$$, we define:

$$g \cdot Q = gQg^{-1}$$.

Suppose $$Q \in S$$ and $$Q \ne P$$. Then, the size of the orbit of $$Q$$ under the action of $$P$$ is given by (using fact (2)):

$$\frac{|Q|}{\operatorname{Stab}_P(Q)}$$.

$$\operatorname{Stab}_P(Q)$$ is the set of those elements of $$P$$ that normalize $$Q$$, and is thus given by $$P \cap N_G(Q)$$. By fact (3), this intersection equals $$P \cap Q$$. Hence, its index in $$Q$$, by assumption is a multiple of $$p^r$$, and thus, from the above expression, we get that the orbit of $$Q$$ has size equal to a multiple of $$p^r$$.

Thus, the orbit of any element in $$S$$ other than $$P$$ has size a multiple of $$p^r$$. $$P$$ itself, on the other hand, has an orbit of size one. Thus, the total number of members of $$S$$ is congruent to $$1$$ modulo $$p^r$$.