Show that every group of order \(pqr\) with \(p<q<r\) primes contains a normal Sylow subgroup, and hence is never simple.
Solution
Let \(n_r\) be the number of Sylow \(r\)-subgroups. By Sylow, \[ n_r\equiv1\pmod r, \qquad n_r\mid pq. \] Since \(p,q<r\), if \(n_r\ne1\) then necessarily \[ n_r=pq. \] Distinct Sylow \(r\)-subgroups intersect trivially, so they contribute \[ pq(r-1) \] nonidentity elements.
Now let \(n_q\) be the number of Sylow \(q\)-subgroups. Again, \[ n_q\equiv1\pmod q, \qquad n_q\mid pr. \] If \(n_q\ne1\), then \(n_q\ge q+1\). Since the divisors of \(pr\) are \(1,p,r,pr\) and \(p<q\), this forces \[ n_q\in\{r,pr\}, \] so in particular \(n_q\ge r\). Distinct Sylow \(q\)-subgroups also intersect trivially, hence they contribute at least \[ r(q-1) \] nonidentity elements.
If both \(n_r>1\) and \(n_q>1\), the sets of nonidentity elements counted above are disjoint, because an element cannot simultaneously have order \(r\) and order \(q\). Thus \(G\) would contain at least \[ pq(r-1)+r(q-1)+1 \] elements. But \[ pq(r-1)+r(q-1)+1-pqr =r(q-1)-pq+1>0, \] because \(q-1\ge p\) and \(r>q\). This is impossible.
Therefore either \(n_r=1\) or \(n_q=1\). Hence \(G\) has a normal Sylow subgroup and is not simple.