Sylow theorems

If $|G|=p^{a}m$ with $p\nmid m$, then $G$ has subgroups of order $p^a$, every $p$-subgroup lies in one, and all such Sylow subgroups are conjugate. Their number satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.