Let \(Z(G)\) be the center of \(G\). Show that if \(G/Z(G)\) is cyclic, then \(G\) is abelian.
Note that Hungerford uses the notation \(C(G)\) for the center.
Solution
Assume that \(G/Z(G)\) is cyclic, generated by \(xZ(G)\) for some \(x\in G\).
1. Every element of \(G\) has the form \(x^m z\) with \(m\in\ZZ\) and\(z\in Z(G)\).
Let \(g\in G\). Since \(xZ(G)\) generates \(G/Z(G)\), there is \(m\in\ZZ\) such that \[ gZ(G)=x^mZ(G). \] Hence \(x^{-m}g\in Z(G)\). Setting \(z=x^{-m}g\) gives \(g=x^m z\).
2. Any two elements of \(G\) commute.
1. Since \(z\) and \(w\) commute with every elementof \(G\),\[ gh=x^mzx^nw=x^{m+n}zw=x^{n+m}wz=x^nwx^mz=hg. \]
3. Therefore \(G\) is abelian.
By 2 every pair of elements of \(G\) commutes.