# Congruence condition on Sylow numbers in terms of maximal Sylow intersection

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 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 $p$-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 Maximal Sylow intersection (?).

## Facts used

1. Sylow subgroups exist
2. Fundamental theorem of group actions
3. 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 $p^r$.