If \(|G| = pq\) (distinct primes), is \(G\) necessarily nilpotent?
Solution
No. For example, \(S_3\) has order \(6=2\cdot3\) but is not nilpotent: its center is trivial, so its upper central series cannot reach \(S_3\).
More generally, for a finite group, nilpotence is equivalent to normality of every Sylow subgroup. If \(|G|=pq\) with primes \(p<q\), the Sylow \(q\)-subgroup is always normal. A nonabelian group of order \(pq\) exists exactly when \(p\mid(q-1)\); it is a semidirect product \[ C_q\rtimes C_p \] with nontrivial action, and its Sylow \(p\)-subgroups are not normal. Hence such a group is not nilpotent. The case \(p=2,q=3\) is \(S_3\).