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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 $ab=ba$.

NotNotLogic
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    You have to use the assumption that $G/Z$ is cyclic (you are still trying to get away with assuming only that $G/Z$ is abelian). This assumption implies that there exists $g\in G$ such that $G/Z=\langle gZ\rangle$. In particular, every coset can be written as $g^kZ$. You have to use this to relate $a$ to $b$. – David Hill Sep 07 '17 at 20:58
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    $Zab=Zba$ is not enough to conclude that $ab=ba$, for it leaves open the possibility that $ab=zba$ for some $z\in Z, z\neq1$. On the other hand, when $G/Z$ is cyclic, we can write $a=g^kz_1$ and $b=g^\ell z_2$ for some integers $k,\ell$ and some $z_1,z_2\in Z$, $gZ$ a generator of $G/Z$, and this makes a huge difference. – Jyrki Lahtonen Sep 07 '17 at 21:01

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