If $G/Z(G)$ is cyclic then $G$ is abelian – what is the point?

The theorem "if $G/Z(G)$ is cyclic then $G$ is abelian" is a popular exercise.

But what is the point of this theorem if $G/Z(G)$ can only be cyclic if it is trivial?

Does "$G/Z(G)$ is cyclic" actually appears in other proofs or is it just a popular exercise?

A nice application is in proving that the center of a non-abelian cannot be too large:

If $G$ is a non-abelian finite group, then $|Z(G)| \leq \frac {1}{4} |G|$.

The contrapositive is

If $|Z(G)| > \frac {1}{4} |G|$, then $G$ be is abelian.

Indeed, $|Z(G)| > \frac {1}{4} |G|$ implies that $G/Z(G)$ has order at most $3$, and so is cyclic.

This other question is using the fact that $G/Z(G)$ being cyclic implies that $G$ is abelian to show that non abelian groups of order $pq$ have a trivial center Verification of Proof that a nonabelian group G of order pq where p and q are primes has a trivial center

Say you want to prove the following.

Every group $G$ of order $p^2$ ($p$ is a prime) is abelian.

You can apply this result.

Since $G$ is a $p$-group, it has a non trivial center $Z(G)$.

Therefore $|G/Z(G)|=1$ or $p$.

Therefore $G/Z(G)$ is cyclic and $G$ is abelian.

There are many other applications.