Sylow $p$-subgroups exist, are conjugate, and their number divides the prime-to-$p$ part while equaling $1$ modulo $p$.Suggest a slogan
Let \(G\) be a finite group and let \(p\) be a prime. Write \(\abs G=p^am\) with \(a\ge0\) and \(p\nmid m\), and let \(n_p\) be the number of Sylow $p$-subgroups of \(G\).
\(G\) has a Sylow \(p\)-subgroup, that is, a subgroup of order \(p^a\).
Any two Sylow \(p\)-subgroups of \(G\) are conjugate in \(G\).
\(n_p\divides m\), \(n_p\equiv1\pmod p\), and \(n_p=[G:N_G(P)]\) for every Sylow \(p\)-subgroup \(P\), where \(N_G(P)\) is the normalizer of \(P\).
State the three Sylow theorems.