Congruence condition on Sylow numbers

This article gives the statement, and possibly proof, of a constraint on numerical invariants that can be associated with a finite group

This article describes a congruence condition on an enumeration, or a count. It says that in a finite group and modulo prime number, the number of Sylow subgroups, i.e., the number of subgroups of prime power order whose index is relatively prime to the order satisfies a congruence condition.
View other congruence conditions | View divisor relations

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 subgroup (?)s 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:

$[G : N_{G} (P)] \equiv 1 mod p$ .

Related facts

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 $p^{k} m$ 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

Proof

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