= Solution
The <Sylow theorems> state that if a <finite group> has order $|G|=p^am$ 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 satisfies
$$
\boxed{n_p\equiv1\pmod p,\qquad n_p\mid m}.
$$
More precisely $n_p=[G:N_G(P)]$ for a <Sylow subgroup> $P$, where $N_G(P)$ is its <normalizer>.
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