Special groups of order pq

Let \(p\), \(q\) be primes such that \(p < q\) such that \(p \nmid q - 1\).
Let \(G\) be a group of order \(pq\). Then, \(G\) is cyclic.

Remark: The case \(p = q\) is covered by this fact.

Proof: By the Sylow Theorems, \(n_p = n_q = 1\) is forced. This is because \(p \not\equiv 1 \mod q\) and \(q \not\equiv 1 \mod p\).

Let \(H\) and \(K\) be the unique Sylow subgroups of order \(p\) and \(q\), respectively. Both of them are normal in \(G\). (As they are the only Sylow subgroups of that order.)
Moreover, \(H \cap K = (1)\), by Lagrange’s theorem. Also, every element of \(H\) commutes with every element of \(K\). To see this, let \(h \in H\) and \(k \in K\). Normality of \(H\) and \(K\) tells us that \(hkh^{-1}k^{-1} \in H \cap K = (1)\). Thus,

\[G \cong H \times K \cong \mathbb{Z}_p \times \mathbb{Z}_q \cong \mathbb{Z}_{pq},\]

where the last isomorphism follows from the fact that \(p\) and \(q\) are relatively prime.