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> If $n\_p=1$, then the $p$-Sylow subgroup is normal.
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I've used this fact several times, and it's about time I knew why it's true.
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I don't want to add something different here, but good to know that $n\_p=[G:N\_G(P)]$ when $P$ is a $p$- sylow of $G$. So if $n\_p=1$ then $N\_G(P)=G$ so $P$ is normal in $G$.
Let $N$ be the single Sylow $p$-group. Suppose it is not normal. Then conjugation by some element $g$ gives a group $gNg^{-1}$ different from $N$ with the same number of elements as $N$. This is another Sylow $p$-group, distinct from the first, a contradiction.