Let \(G\) be a group of order \(|G| = p q\), where \(p\) and \(q\) are distinct primes with \(p > q\). Prove that \(G\) possesses a unique (and hence normal) Sylow \(p\)-subgroup of order \(p\).
Solution
Let \(n_p\) be the number of Sylow \(p\)-subgroups. Sylow’s theorem gives \[ n_p\mid q,\qquad n_p\equiv1\pmod p. \] Since \(q\) is prime, \(n_p\) is either \(1\) or \(q\). But \(p>q\), so \(q\not\equiv1\pmod p\). Hence \(n_p=1\).
Thus the Sylow \(p\)-subgroup is unique and therefore normal. Its order is \(p\), so it is nontrivial and proper in a group of order \(pq\).