Dummit Foote 3.1.36 (iff)

$\textit{Theorem}:$ $G/Z$ cyclic $\Longleftrightarrow G$ Abelian. ($Z$ is the center of $G$)

Note: $G/Z$ is indeed a group since $Z \unlhd G$

$\textit{Proof}: \quad$ "$\Longleftarrow$" $G$ Abelian $\Longrightarrow Z=G \Longrightarrow G/Z=G/G=<1>$ which is trivially cyclic.
"$\Longrightarrow$"
$G/Z$ cyclic $\Longrightarrow$ $$\forall gZ \in{G/Z}, \quad \exists aZ\in{G/Z} \quad s.t. \quad gZ=(aZ)^n=a^nZ$$ So, $g$ and $a^n$ are in the same fiber of a homomorphism from $G$ to $G/Z$. $\Longrightarrow \quad \exists z_1\in{Z}$ s.t. $ g=a^kz_1$

Let $hZ \in{G/Z}$. So, $hZ=a^kZ$ for some $k\in{\mathbb{Z}}$. Then, $$gh=a^nz_1 \cdot a^kz_2 = a^na^kz_1z_2=a^{n+k}z_1z_2=a^{k+n}z_2z_1=a^ka^nz_2z_1=a^kz_2a^nz_1=hg$$ since $z_1$ and $z_2$ commute with all elements of $G$.

$\therefore G$ is Abelian $\quad \blacksquare$