Suppose \(|G| = pq\) with \(p, q \geq 2\) prime, and let \(H \le G\) be a proper subgroup. Prove that \(H\) must be cyclic.
Solution
By Lagrange’s theorem, \[ |H|\mid pq. \] Since \(H\) is proper, \[ |H|\in\{1,p,q\}. \] The trivial group is cyclic. If \(|H|\) is prime, choose \(h\in H\) with \(h\neq e\). Then the order of \(h\) divides \(|H|\) and is greater than \(1\), hence equals \(|H|\). Therefore \[ H=\langle h\rangle. \] Thus every proper subgroup of \(G\) is cyclic.