(March 2007) Let \(p,q\ge 2\) be prime numbers, and let \(G\) be a group of order \(|G|=pq\). Let \(H\subsetneq G\) be a proper subgroup (i.e., \(H\) is a subgroup but is not the full group \(G\)). Prove that \(H\) is cyclic.
Solution
Proof. Since|G|<∞, Lagrange’s Theorem says that |H| ⏐⏐|G|. Thus, |H| is one of the four numbers 1,p,q,pq . However, since H is proper, we must have|H|<|G| =pq. Three cases remain: If|H| = 1, then H ={e} is the trivial group, which is generated by e and so is cyclic. Otherwise,H has prime orderp orq, and therefore it is cyclic by (a corollary of) Lagrange’s Theorem. QED