The Sylow count divides the prime-to-$p$ part, is $1$ modulo $p$, and equals a normalizer index.Suggest a slogan
Let \(G\) be a finite group and \(p\) a prime, write \(\abs G = p^a m\) with \(p\nmid m\), let \(n_p\) be the number of Sylow $p$-subgroups of \(G\), and let \(S_p\) be one of them. Then:
\(n_p \divides m\), and in particular \(n_p \leq m\).
\(n_p \equiv 1 \pmod p\).
\(n_p = [G : N_G(S_p)]\), where \(N_G(S_p)\) is the normalizer of \(S_p\) in \(G\).