Congruence condition on Sylow numbers

Statement
Let $$G$$ be a finite group and $$p$$ a prime. Let $$n_p$$ be the $$p$$- Sylow number of $$G$$, i.e., the number of $$p$$- Sylow subgroups of $$G$$. Then:

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

Since all the $$p$$-Sylow subgroups are conjugate, $$n_p$$ equals the index of any $$p$$-Sylow subgroup. Thus, this is equivalent to the following: if $$P$$ is a $$p$$-Sylow subgroup:

$$[G:N_G(P)] \equiv 1 \mod p$$.

Other parts of Sylow's theorem

 * Sylow's theorem: The whole theorem.
 * Sylow subgroups exist
 * Sylow implies order-conjugate: Any two Sylow subgroups for the same prime are conjugate.
 * Sylow implies order-dominating: Any $$p$$-subgroup is contained in some conjugate of any given $$p$$-Sylow subgroup.
 * Divisibility condition on Sylow numbers: If the order of $$G$$ is $$p^km$$ where $$m$$ is relatively prime to $$p$$, then:

$$n_p | m$$.

This fact is often used along with the congruence condition on Sylow numbers.

Generalizations

 * Congruence condition on number of subgroups of given prime power order: Let $$G$$ be a finite group and $$p^r$$ be any prime power dividing the order of $$G$$. Then, the number of subgroups of order $$p^r$$ in $$G$$ is congruent to $$1$$ modulo $$p$$.
 * Congruence condition on Sylow numbers in terms of maximal Sylow intersection: This states that if the intersection of two $$p$$-Sylow subgroups has index at least $$p^k$$, then $$n_p$$, the number $$p$$-Sylow subgroups, is congruent to $$1$$ modulo $$p^k$$.
 * Congruence condition on number of Sylow subgroups containing a given subgroup of prime power order: This states that if $$G$$ is a finite group and $$A$$ is a $$p$$-subgroup of $$G$$, the number of $$p$$-Sylow subgroups of $$G$$ containing $$A$$ is congruent to $$1$$ modulo $$p$$.
 * Congruence condition on factorization of Hall numbers for a finite solvable group.

Converse
A converse of sorts might be: whenever $$a$$ is a natural number such that $$a \equiv 1 \mod p$$, there exists a finite group $$G$$ such that $$n_p = a$$, i.e., $$a$$ is the number of $$p$$-Sylow subgroups of $$G$$.

This is false. However, some partial converses are true:


 * Converse of congruence condition on Sylow numbers for the prime two: Any odd number occurs as the number of $$2$$-Sylow subgroups in some finite group.
 * Converse of congruence condition for prime power Sylow numbers: If a prime power satisfies the congruence condition, it occurs as a Sylow number.

Applications

 * Prime divisor greater than Sylow index is Sylow-unique
 * Order is product of Mersenne prime and one more implies normal Sylow subgroup

Proof
This proof assumes that we already know that there exist $$p$$-Sylow subgroups of $$G$$.