Sylow Congruence

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

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 .

Proof sketch (via an action): Fix a Sylow $p$-subgroup $P\le G$ and let $P$ act on the set $\mathrm{Syl}_p(G)$ of Sylow $p$-subgroups by conjugation: $x\cdot Q = xQx^{-1}$. Orbit sizes are powers of $p$, so every orbit has size $1$ or a multiple of $p$. The subgroup $P$ itself is fixed (it is a fixed point), so there is at least one orbit of size $1$. Therefore $|\mathrm{Syl}_p(G)|\equiv 1 \pmod p$, i.e. $n_p\equiv 1\pmod p$.

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 ).