State all parts of Sylow’s Theorem.
Show that if \(0 < p < q\) are primes and \(p \nmid q - 1\), then any group of order \(pq\) is cyclic.
Solution
1. Sylow’s theorems have the following form.
Let \(H\) be a finite group and let \(\ell\) be prime. Write \(|H|=\ell^am\) with \(\ell\nmid m\). A Sylow \(\ell\)-subgroup is a subgroup of order \(\ell^a\). Such subgroups exist. Every \(\ell\)-subgroup is contained in a Sylow \(\ell\)-subgroup, and any two Sylow \(\ell\)-subgroups are conjugate in \(H\). For a Sylow subgroup \(S\), their number satisfies \[ n_\ell=[H:N_H(S)],\qquad n_\ell\mid m, \qquad n_\ell\equiv1\pmod\ell. \] In particular, \(S\) is normal exactly when \(n_\ell=1\) [@DF04].
2. Under the hypotheses of part (b), both Sylow subgroupsof \(G\) are normal.
For the Sylow \(q\)-subgroups, one has \(n_q\mid p\) and \(n_q\equiv1\pmod q\). Thus \(n_q=1\) or \(p\). Since \(p<q\), the latter is impossible. The Sylow \(q\)-subgroup \(Q\) is unique. Similarly \(n_p=1\) or \(q\) and \(n_p\equiv1\pmod p\). The hypothesis \(p\nmid q-1\) rules out \(n_p=q\), so the Sylow \(p\)-subgroup \(P\) is also unique. Uniqueness implies normality because conjugation preserves subgroup orders.
3. The group \(G\) is cyclic of order \(pq\).
Lagrange’s theorem gives \(P\cap Q=\{1\}\), since its order divides both distinct primes. For \(x\in P\) and \(y\in Q\), normality of both subgroups puts the commutator \(xyx^{-1}y^{-1}\) in \(P\cap Q\). Thus \(x\) and \(y\) commute. The multiplication map \(P\times Q\to G\), \((x,y)\mapsto xy\), is therefore a homomorphism. Its kernel is trivial, since \(xy=1\) forces \(x=y^{-1}\in P\cap Q\). Both its domain and codomain have order \(pq\), so it is an isomorphism.
Each of \(P,Q\) has prime order and is cyclic. If \(x,y\) are generators, then \((x,y)^d=1\) exactly when both \(p\) and \(q\) divide \(d\), that is, exactly when \(pq\mid d\). Hence \((x,y)\) has order \(pq\), and its image generates \(G\).