## If a group mod its center is cyclic, then the group is abelian

Let $G$ be a group. Prove that if $G/Z(G)$ is cyclic then $G$ is abelian.

Let $G$ be a group and suppose $G/Z(G) = \langle x Z(G) \rangle$ is cyclic.

Note that for all $g \in G$, we have $g \in g Z(G) = x^k Z(G)$ for some integer $k$. In particular, $g = x^k z$ for some integer $k$ and some $z \in Z(G)$.

Now let $g,h \in G$, where $g = x^a z$ and $h = x^b w$ and $z,w \in Z(G)$. We have $gh = x^azx^bw = x^{a+b}zw = x^{b+a}wz = x^bwx^az = hg$. Thus $G$ is abelian.

True, but here we are looking at the converse question. Given that $G/Z(G)$ is cyclic, must $G$ be abelian?