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.

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    $\begingroup$ Hint: actually, one can even show that the $r$-Sylow subgroup is normal. $\endgroup$ – Alexander Gruber Mar 22 '14 at 4:42

Suppose $n_r \ne 1$; then by Sylow's Theorems, we have that $n_r \mid pq$ and $n_r \equiv 1 \mod r$. This implies that

$$n_r \ge r + 1 > r > p, q$$ so that $n_r = pq$. Each $r$-Sylow subgroup contributes $r - 1$ elements, plus the identity; thus, we've accounted for $$pq (r - 1) + 1 = |G| - pq + 1$$

distinct elements in the group.

Now suppose in addition that $n_q \ne 1$; then $n_q | pr$ and $n_q \ge q + 1$ imply that $n_q \ge r$. Again counting elements, we've accounted for

$$r \cdot (q - 1) > p(q - 1) = pq - p$$

new elements, bringing our total to $|G| - p + 1$. Since a $p$-Sylow subgroup contributes $p - 1$ more elements, and we only have $p - 1$ slots left, it follows that $n_p = 1$.


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