Prove that any group of size 15 is cyclic.
Solution
Let \(G\) be a group of order \(15=3\cdot5\).
1. \(G\) has a unique Sylow \(3\)-subgroup \(P\cong\ZZ/3\ZZ\) and a unique Sylow \(5\)-subgroup \(Q\cong\ZZ/5\ZZ\), and both are normal in \(G\).
By the Sylow theorems, the number \(n_3\) of Sylow \(3\)-subgroups satisfies \(n_3\equiv1\pmod3\) and \(n_3\mid5\), so \(n_3=1\). Likewise \(n_5\equiv1\pmod5\) and \(n_5\mid3\), so \(n_5=1\). A unique Sylow \(p\)-subgroup is normal, since conjugation permutes the Sylow \(p\)-subgroups. Groups of prime order are cyclic.
2. \(G\cong P\times Q\).
By Lagrange’s theorem, \(|P\cap Q|\) divides \(\gcd(3,5)=1\), so \(P\cap Q=\{e\}\). For \(p\in P\) and \(q\in Q\), the commutator \([p,q]=pqp^{-1}q^{-1}\) equals \((pqp^{-1})q^{-1}\in Q\) because \(Q\lhd G\), and equals \(p(qp^{-1}q^{-1})\in P\) because \(P\lhd G\). Hence \([p,q]\in P\cap Q=\{e\}\), so \(pq=qp\). Therefore \(\varphi\colon P\times Q\to G\), \(\varphi(p,q)=pq\), is a group homomorphism. Its kernel is \(\{(p,p^{-1}) : p\in P\cap Q\}=\{(e,e)\}\), so \(\varphi\) is injective, and \(\abs{P\times Q}=15=\abs{G}\) makes it an isomorphism.
3. \(G\) is cyclic.
By step 2 and the Chinese remainder theorem, \(G\cong\ZZ/3\ZZ\times\ZZ/5\ZZ\cong\ZZ/15\ZZ\). Equivalently, if \(P=\langle a\rangle\) and \(Q=\langle b\rangle\), then \(a\) and \(b\) commute and have coprime orders, so \(ab\) has order \(\operatorname{lcm}(3,5)=15\) and \(G=\langle ab\rangle\).
4. Q.E.D.
Step 3 proves that every group of order \(15\) is cyclic.