Let \(G\) be a group with center \(Z(G)\). Show that if \(G/Z(G)\) is cyclic, then \(G\) is abelian.
Solution
Suppose \(G/Z(G)=\langle gZ(G)\rangle\) for some \(g\in G\).
1. Every \(x\in G\) can be written \(x=g^iz\) with \(i\in\ZZ\) and \(z\in Z(G)\).
The coset \(xZ(G)\) lies in \(\langle gZ(G)\rangle\), so \(xZ(G)=g^iZ(G)\) for some \(i\in\ZZ\), and then \(z=g^{-i}x\in Z(G)\).
2. Any \(x,y\in G\) commute.
By step 1, write \(x=g^iz_1\) and \(y=g^jz_2\) with \(z_1,z_2\in Z(G)\). Since \(z_1,z_2\) are central and powers of \(g\) commute with each other, \[ xy=g^iz_1g^jz_2=g^{i+j}z_1z_2=g^jz_2g^iz_1=yx. \]
3. Q.E.D.
Step 2 shows that \(G\) is abelian.