= Solution
Write the <group order> as $|G|=p^am$, where $p$ is a <prime number> and $p\nmid m$. A <Sylow p-subgroup> is a <subgroup> of <group order> $p^a$, the largest power of $p$ dividing $|G|$.
The three <Sylow theorems> assert:
* \b[Existence:] a <Sylow p-subgroup> exists, and every <subgroup> of <prime power> order a power of $p$ is contained in one.
* \b[Conjugacy:] any two <Sylow p-subgroups> are conjugate in $G$.
* \b[Counting:] their number $n_p$ satisfies $n_p\mid m$ and $n_p\equiv1\pmod p$.
In particular, a unique <Sylow p-subgroup> is a <normal subgroup>, since all its conjugates must equal it.
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