---
tag: 0Q8DGAZSJ4
kind: statement
role: theorem
name: Third Sylow theorem
msc: [20D20]
uses: [0JWZX6FKSH, 098D2E7DEH]
sources:
  - url: https://en.wikipedia.org/wiki/Sylow_theorems
    locator: Statement
---

## Statement

Let $G$ be a finite [[group|G5S1M9]] and let $p$ be a prime number.
Then the number $n_p$ of [[Sylow p-subgroups|0053CQE9Y2]] of $G$ satisfies $n_p \equiv 1 \pmod p$.

## Remarks

The proof lets a Sylow $p$-subgroup $P$ act by conjugation on $\operatorname{Syl}_p(G)$, and $P$ is the only fixed point.
Together with [[the divisibility condition|0M3PYJ3TSE]], this often forces $n_p = 1$.
In $S_4$, $n_2 = 3$ and $n_3 = 4$, and both numbers are $\equiv 1$ modulo the prime.

## Explanations

### L1 human

The number of Sylow $p$-subgroups is one more than a multiple of $p$.
This simple fact, together with a divisibility condition, leaves only a few possible values.
Often the only possible value is $1$, and then the Sylow subgroup is normal.