Prove that every group of order 15 is cyclic.
Prove that every group of order \(255 = 3 \cdot 5 \cdot 17\) is cyclic.
Solution
1. In part (a), every group \(G\) of order \(15\) is cyclic.
By the Sylow theorems, \(n_3\equiv1\pmod3\) and \(n_3\mid5\), so \(n_3=1\); and \(n_5\equiv1\pmod5\) and \(n_5\mid3\), so \(n_5=1\). Thus the Sylow subgroups \(P\cong\ZZ/3\) and \(Q\cong\ZZ/5\) are normal, and \(P\cap Q=1\) because their orders are coprime. Hence \(G=PQ\cong P\times Q\cong\ZZ/3\times\ZZ/5\cong\ZZ/15\) by the Chinese remainder theorem.
Let \(G\) be a group of order \(255=3\cdot5\cdot17\).
2. \(G\) has a normal Sylow \(17\)-subgroup \(N\cong\ZZ/17\), and \(N\subseteq Z(G)\).
By the Sylow theorems, \(n_{17}\equiv1\pmod{17}\) and \(n_{17}\mid15\); the divisors \(1,3,5,15\) of \(15\) leave \(n_{17}=1\), so \(N\) is normal. Conjugation gives a homomorphism \(G\to\Aut(N)\cong(\ZZ/17)^\times\), a group of order \(16\). Its image has order dividing \(\gcd(255,16)=1\), so \(G\) centralizes \(N\).
3. \(G\) is abelian.
1.By step 2, \(N\subseteq Z(G)\), so \(G/Z(G)\) is a quotient of \(G/N\) and is cyclic.A group whose quotient by its center is cyclic is abelian: if \(G/Z(G)=\langle gZ(G)\rangle\), then any two elements have the form \(g^iz\) and \(g^jz'\) with \(z,z'\in Z(G)\), and these commute.
4. In part (b), \(G\) is cyclic.
By step 3 and the structure theorem for finite abelian groups, \(G\) is a product of cyclic groups of prime-power order whose orders multiply to \(255=3\cdot5\cdot17\); since \(255\) is squarefree, \(G\cong\ZZ/3\times\ZZ/5\times\ZZ/17\cong\ZZ/255\) by the Chinese remainder theorem.
5. Q.E.D.
Steps 1 and 4 answer parts (a) and (b).