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 subgroups of $G$. Then:

$n_p\equiv 1modp$.

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:

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

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 $a\equiv 1modp$, 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 $G$.