Sylow subgroups of groups with order \(pqr\), \(p < q < r\) (D&F 4.5.16)

Let \(\left|G\right| = pqr\), where \(p,q,r\) are primes with \(p < q < r\). Prove that \(G\) has a normal Sylow subgroup for either \(p\), \(q\) or \(r\).

Suppose \(G\) has no normal Sylow subgroups. Then \(n_p, n_q, n_r > 1\). This combined with the fact that \(n_p \mid qr\), \(n_q \mid pr\) and \(n_r \mid pq\) implies the possibilities for \(n_p\), \(n_q\) and \(n_r\) are \(\{q,r,qr\}\), \(\{p,r,pr\}\) and \(\{p,q,pq\}\) respectively. Furthermore, using the fact that \(p < q < r\) and \(n_p = 1+kp\) (likewise for \(n_q\) and \(n_r\)), we can narrow the possibilities for \(n_q\) and \(n_r\) to \(\{r,pr\}\) and \(\{pq\}\) respectively.

Now we shall choose the smallest possibilities for \(n_p\), \(n_q\) and \(n_r\) (that is, \(q\), \(r\) and \(pq\) respectively). Clearly all the Sylow subgroups have trivial intersection, so the total number of distinct nonidentity elements is \(q(p-1)+r(q-1)+pq(r-1) = pqr+qr-q-r > pqr\), which is a contradiction. Since these were the smallest possibilities, all other possibilities lead to the same contradiction. This completes the proof.