If we initially take into account the fact that $G/Z$ is cyclic, then I know how to prove that $G$ is abelian. However I wanted an alternate proof which starts by assuming that $G/Z$ is abelian(which is true since it is cyclic) and then it proceeds to prove this result.  We would eventually need to consider that $G/Z$ is cyclic (because being only abelian is not sufficient to arrive at result) but I could not figure out how to bring this in.  Here's what I did:  Since $G/Z$ is abelian,  $$Za Zb=Zb Za$$ for some $a,b \in G.$ Since $Z$ is a normal subgroup as well so left cosets are same as right cosets. $$\therefore ZZab=ZZba$$ Thus we have finally, $$Zab=Zba$$  Please keep the proof same till here and show that $ab=ba$.
Continuing my work through Dummit &amp; Foote's "Abstract Algebra", 3.1.36 asks the following (which is exactly the same as exercise 5 in ):     Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian. [If $G/Z(G)$ is cyclic with generator $xZ(G)$, show that every element of $G$ can be written in the form $x^az$ for some $a \in \mathbb{Z}$ and some element $z \in Z(G)$]      The hint is actually the hardest part for me, as the quotient groups are somewhat abstract. But once I have the hint, I can write:  $g, h \in G$ implies that $g = x^{a_1}z_1$ and $h = x^{a_2}z_2$, so \begin{align*}gh &amp;= (x^{a_1}z_1)(x^{a_2}z_2)\\\  &amp;= x^{a_1}x^{a_2}z_1z_2\\\ &amp; = x^{a_1 + a_2}z_2z_1\\\ &amp;= \ldots = (x^{a_2}z_2)(x^{a_1}z_1) = hg. \end{align*} Therefore, $G$ is abelian.  1) Is this right so far?  2) How can I prove the "hint"?
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