 Solution

The Sylow theorems state that if a finite group has order $|G|=p^{a}m$ with $p$ prime and $p\nmid m$, it has a subgroup of order $p^a$, called a Sylow subgroup. Every p-group subgroup is contained in a Sylow subgroup, and all Sylow subgroups for the same prime are conjugate subgroups. Their number satisfiesMore precisely $n_p=[G:N_G(P)]$ for a Sylow subgroup $P$, where $N_G(P)$ is its normalizer.