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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 $G$.
Well, $n\_p$ is the number of $p$-Sylow subgroups. The conjugates of $p$-Sylow subgroup are precisely the other $p$-Sylow sugroups. So if $n\_p=1$ then the $p$-Sylow subgroup is conjugate only to itself, i.e. it is normal.