Group with order \(pm\), with \(p > m\) and \(p\) prime (FGT 1A.9)

Suppose \(\left|G\right| = pm\), where \(p > m\) and \(p\) is prime. Show that \(G\) has a unique subgroup of order \(p\).

Assume there are \(k\) distinct order \(p\) subgroups. Since they have trivial intersection, the number of order \(p\) elements is \(k(p-1)\). By 1A.8, we have \(k(p-1) \equiv -1\) mod \(p\), or \(k(p-1) = lp-1\) for some \(l \in \mathbb{Z}^+\).

The sequences \(\{k(p-1)\}\) and \(\{lp-1\}\) form an arithmetic progression with step size \(p-1\) and \(p\), and clearly they meet at \(p-1\) when \(k = l = 1\). Since \(p\) and \(p-1\) are relative prime, the sequences next meet at \(1(p-1)+p(p-1) = (p+1)(p-1)\), so the next smallest solution has \(k = p+1\). But then there are a total of \((p+1)(p-1)+1 = p^2 > pm\) distinct elements, a contradiction.