Let $|G|=pqr$ where $p, q$ and $r$ are prime and $p < q < r$. Prove that $G$ has a normal Sylow subgroup for either $p, q$ or $r$.
Let $n_p, n_q, n_r$ denote the number of Sylow subgroups for $p,q$ and $r$ respectively. I want to show that one of these equals 1. I have gone through and looked at all the cases and have come to the conclusion that $n_p=n_q=1$. I think I may have made a mistake. And anyways, even if I haven't it's a terribly long horrible way of proving this. I think there has to be a smarter way.
Just to be more clear, what I did was use the Sylow Theorems and the fact that $p,q$ and $r$ are prime. So there are four possibilities for each "$n$". I also got that either $n_r=q$ or $n_r=1$, which seems crazy but I'm not sure.