Analyze groups of order \(pq\) with \(q<p\).
Hint: consider the cases when \(q\) does or does not divide \(p-1\).
Show that if \(q\) does not divide \(p-1\), then \(G\) is cyclic.
Show that \(G\) is never simple.
Solution
Let \(|G|=pq\) with primes \(q<p\).
1. The Sylow \(p\)-subgroup is unique and normal.
Let \(n_p\) be the number of Sylow \(p\)-subgroups. Sylow’s theorem gives \[ n_p\mid q, \qquad n_p\equiv1\pmod p. \] Thus \(n_p\) is either \(1\) or \(q\). Since \(q<p\), the value \(q\) cannot be congruent to \(1\) modulo \(p\). Hence \(n_p=1\).
2. Consequently \(G\) is never simple.
The unique Sylow \(p\)-subgroup has order \(p\), so it is nontrivial and proper. By 1 it is normal. Hence \(G\) has a nontrivial proper normal subgroup.
3. The number \(n_q\) of Sylow \(q\)-subgroups is either \(1\) or \(p\).
Sylow’s theorem gives \[ n_q\mid p, \qquad n_q\equiv1\pmod q. \] Since \(p\) is prime, \(n_q\in\{1,p\}\).
4. If \(q\nmid(p-1)\), then the Sylow \(q\)-subgroup is also normal.
If \(n_q=p\), then the congruence \(n_q\equiv1\pmod q\) would give \[ p\equiv1\pmod q, \] i.e. \(q\mid(p-1)\). Under the stated hypothesis this is impossible, so \(n_q=1\).
5. If \(q\nmid(p-1)\), then \(G\) is cyclic.
Let \(P\) and \(Q\) be the Sylow \(p\)- and \(q\)-subgroups. By 1 and 4, both are normal. Their intersection is trivial because their orders are coprime, and \[ |PQ|=\frac{|P||Q|}{|P\cap Q|}=pq=|G|, \] so \(G=PQ\). Normal subgroups with trivial intersection commute elementwise, hence \[ G\cong P\times Q\cong C_p\times C_q\cong C_{pq}. \]
6. When \(q\mid(p-1)\), there is a nonabelian group \(C_p\rtimes C_q\) of order \(pq\).
Since \(\operatorname{Aut}(C_p)\cong C_{p-1}\) contains an element of order \(q\), there is a nontrivial homomorphism \[ C_q\longrightarrow\operatorname{Aut}(C_p). \] The associated semidirect product \(C_p\rtimes C_q\) has order \(pq\) and is nonabelian. Thus the divisibility condition is exactly the one allowing the nontrivial action.