Prove that if \(G/Z(G)\) is cyclic, then \(G\) is abelian.
Solution
Write \(Z=Z(G)\) and suppose \[ G/Z=\langle gZ\rangle. \]
1. Every element of \(G\) can be written as \(g^m z\) for some \(m\in\ZZ\) and \(z\in Z\).
If \(x\in G\), then its coset lies in the cyclic quotient, so \[ xZ=(gZ)^m=g^mZ \] for some \(m\). Hence \[ g^{-m}x\in Z, \] so \(x=g^m z\) for some \(z\in Z\).
2. Any two elements of \(G\) commute.
1. Since the \(z_i\) are central,\[ xy=g^m z_1g^n z_2=g^{m+n}z_1z_2 =g^{m+n}z_2z_1=g^n z_2g^m z_1=yx. \]
3. Therefore \(G\) is abelian.
By 2, every pair of elements of \(G\) commutes.