Let \(p\), \(q\) be two distinct primes. Prove that there is at most one non-abelian group of order \(pq\) and describe the pairs \((p,q)\) such that there is no non-abelian group of order \(pq\).
Solution
1. Without loss of generality, suppose \(p<q\). Then every group \(G\) of order \(pq\) has a unique Sylow \(q\)-subgroup \(Q\), hence\[ Q\trianglelefteq G, \qquad Q\cong C_q. \]
The number \(n_q\) of Sylow \(q\)-subgroups satisfies \[ n_q\mid p, \qquad n_q\equiv1\pmod q. \] Since \(p<q\), the only possibility is \(n_q=1\). A group of prime order is cyclic, so \(Q\cong C_q\).
2. If \(P\) is a Sylow \(p\)-subgroup, then\[ G\cong C_q\rtimes_\varphi C_p \]for some homomorphism\[ \varphi:C_p\to\operatorname{Aut}(C_q). \]
We have \(Q\cap P=1\) and \[ |QP|=|Q||P|=pq=|G|, \] so \(G=QP\). Since \(Q\trianglelefteq G\), this is a semidirect product. Conjugation by \(P\) on \(Q\) gives the action \(\varphi\).
3. The semidirect product is nonabelian exactly when \(\varphi\) is nontrivial.
If \(\varphi\) is trivial, then \(P\) centralizes \(Q\), so \[ G\cong C_q\times C_p\cong C_{pq}, \] which is abelian. Conversely, a nontrivial action means some element of \(P\) acts nontrivially by conjugation on \(Q\), so \(G\) is nonabelian.
4. A nontrivial action exists if and only if\[ p\mid(q-1). \]
Since \(Q\cong C_q\), \[ \operatorname{Aut}(Q)\cong(\mathbb Z/q\mathbb Z)^\times\cong C_{q-1}. \] A nontrivial homomorphism \(C_p\to C_{q-1}\) is injective because \(p\) is prime. Such an injection exists exactly when \(C_{q-1}\) has an element of order \(p\), equivalently when \(p\mid(q-1)\).
5. When \(p\mid(q-1)\), there is exactly one nonabelian group of order \(pq\) up to isomorphism.
The cyclic group \(\operatorname{Aut}(C_q)\cong C_{q-1}\) has a unique subgroup of order \(p\). Therefore all injective homomorphisms \[ C_p\hookrightarrow\operatorname{Aut}(C_q) \] have the same image. Any two differ only by an automorphism of the source \(C_p\), and precomposing the action by an automorphism of the complement does not change the semidirect product up to isomorphism. Hence there is a unique nontrivial semidirect-product isomorphism type.
6. Therefore, for distinct primes \(p,q\), there is at most one nonabelian group of order \(pq\). Such a group exists precisely when the smaller prime divides one less than the larger prime.
If \(r=\min\{p,q\}\) and \(s=\max\{p,q\}\), apply <1>1–<1>5 with \(r<s\). Thus no nonabelian group exists exactly when \[ \boxed{r\nmid(s-1)}. \]