Sylow Congruence - Knowlpedia

Sylow Congruence: Let $G$ be a finite group and let $p$ be a prime. Write $|G|=p^a m$ with $a\ge 1$ and $p\nmid m$ . Let $n_p$ be the number of Sylow p-subgroups of $G$ . Then

n_p \equiv 1 \pmod p.

Examples

If $|G|=12$ and $p=3$ , then $n_3\mid 4$ and $n_3\equiv 1\pmod 3$ , so $n_3\in\{1,4\}$ .
If $|G|=21=3\cdot 7$ , then $n_7\mid 3$ and $n_7\equiv 1\pmod 7$ , forcing $n_7=1$ . Hence the Sylow $7$ -subgroup is normal (by the n_p=1 normality criterion).

Remarks

This congruence is part of the standard consequences of Sylow's third theorem. A common proof uses a conjugation action and the orbit decomposition of a finite group action.