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 non-trivial cyclic.

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

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

Thus:
 * $$\exists t \in G: G / Z \left({G}\right) = \left \langle {t Z \left({G}\right)} \right \rangle$$

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:

$$ $$ $$

Similarly, $$y x = t^{m + n} z_1 z_2 = x y$$.

This holds for all $$x, y \in G$$, and thus $$G$$ is abelian, thereby making $$Z \left({G}\right) = G$$, and thus $$G / Z \left({G}\right)$$ has one element and is therefore trivial.