Let \(G\) have order \(pqr\) with primes \(p<q<r\). Show that \(G\) has a normal Sylow subgroup. In particular, show that \(G\) is not simple.
Solution
Let \(n_r\) be the number of Sylow \(r\)-subgroups. Sylow gives \[ n_r\mid pq, \qquad n_r\equiv1\pmod r. \] Since \(p,q<r\), if \(n_r\ne1\) then necessarily \[ n_r=pq. \] If \(n_r=1\), the Sylow \(r\)-subgroup is normal and we are done.
Assume therefore that \(n_r=pq\). Distinct Sylow \(r\)-subgroups have trivial intersection, so they contribute \[ pq(r-1) \] nonidentity elements. Thus only \[ pqr-pq(r-1)=pq \] elements remain.
Now let \(n_q\) be the number of Sylow \(q\)-subgroups. If \(n_q>1\), then \[ n_q\equiv1\pmod q, \qquad n_q\mid pr. \] Because \(p<q\), this forces \[ n_q\ge r. \] Distinct Sylow \(q\)-subgroups also intersect trivially, so they would contribute at least \[ r(q-1) \] nonidentity elements among the \(pq-1\) nonidentity elements not already lying in Sylow \(r\)-subgroups. But \[ r(q-1)>q(q-1)>pq-1 \] because \(r>q>p\). This is impossible.
Hence \[ n_q=1, \] so the Sylow \(q\)-subgroup is normal.
Thus every group of order \(pqr\) has a nontrivial proper normal Sylow subgroup and is therefore not simple.