Quotient of Group by Center Cyclic implies Abelian

Theorem
Let $G$ be a group.

Let $Z \left({G}\right)$ be the center of $G$.

Let $G / Z \left({G}\right)$ be the quotient group of $G$ by $Z \left({G}\right)$.

Let $G / Z \left({G}\right)$ be cyclic.

Then $G$ is abelian, so $G = Z \left({G}\right)$.

That is, the group $G / Z \left({G}\right)$ can not be a cyclic group which is non-trivial.

Proof
Suppose $G / Z \left({G}\right)$ is cyclic.

Then by definition:
 * $\exists \tau \in G / Z \left({G}\right): G / Z \left({G}\right) = \left \langle {\tau} \right \rangle$

Since $\tau$ is a coset by $Z \left({G}\right)$, Thus:
 * $\exists t \in G: \tau = t Z \left({G}\right)$

Thus each coset of $Z \left({G}\right)$ in $G$ is equal to $\left({t Z \left({G}\right)}\right)^i = t^i Z \left({G}\right)$ for some $i \in \Z$.

Now let $x, y \in G$.

Suppose $x \in t^m Z \left({G}\right), y \in t^n Z \left({G}\right)$.

Then $x = t^m z_1, y = t^n z_2$ for some $z_1, z_2 \in Z \left({G}\right)$.

Thus:

This holds for all $x, y \in G$, and thus $G$ is abelian.

Thus by Group is Abelian iff Center Equals Group $Z \left({G}\right) = G$.

Therefore Quotient of Group by Itself it follows that $G / Z \left({G}\right)$ is the trivial group.