Problem

Prove that if \(G/Z(G)\) is cyclic then \(G\) is abelian.

Solution

Suppose \(G/Z(G)\) is cyclic.

By definition of center, \(Z(G) = \{g \in G \mid gx = xg \text{ for all } x \in G\}\).

\(Z(G)\) is a subgroup of \(G\). By Theorem 3.3, the cosets of \(Z(G)\) form \(G/Z(G)\) with operation defined by \(uZ(G) \circ vZ(G) = (uv)Z(G)\). Since \(G/Z(G)\) is cyclic, \(G/Z(G) = \langle xZ(G) \rangle\) for some \(x \in G\).

It follows that for every element \(g \in G\), \(g \in x^aZ(G)\), and thus \(g = x^az\) for some \(a \in \mathbb{Z}\) and \(z \in Z(G)\).

For \(g_1, g_2 \in G\), we can let \(g_1 = x^az_1\) and \(g_2 = x^bz_2\) where \(a,b \in \mathbb{Z}\) and \(z_1,z_2 \in Z(G)\). Then,

$$ g_1g_2 = x^az_1x^bz_2 = x^ax^bz_1z_2 = x^ax^bz_2z_1 = x^bx^az_2z_1 = x^bz_2x^az_1 = g_2g_1. $$

Therefore, \(G\) is abelian.

$$\tag*{$\blacksquare$}$$