# 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.

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## Statement

Let be a finite group and a prime dividing the order of . Suppose the intersection of any two distinct -Sylow subgroups has index at least in both of them. Then, if denotes the -Sylow number (?) of , i.e., the number of -Sylow subgroups of , we have:

.

In particular, is a multiple of the index of a Maximal Sylow intersection (?).

## Facts used

- Sylow subgroups exist
- Fundamental theorem of group actions
- Intersection of p-subgroup with Sylow subgroup equals intersection with normalizer: If is a -subgroup of and is a -Sylow subgroup of , then .

## Proof

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

**To prove**: If is the number of -Sylow subgroups of , then .

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

.

Suppose and . Then, the size of the orbit of under the action of is given by (using fact (2)):

.

is the set of those elements of that normalize , and is thus given by . By fact (3), this intersection equals . Hence, its index in , by assumption is a multiple of , and thus, from the above expression, we get that the orbit of has size equal to a multiple of .

Thus, the orbit of any element in other than has size a multiple of . itself, on the other hand, has an orbit of size one. Thus, the total number of members of is congruent to modulo .