$\text{Inn}(G)$ cannot be nontrivial cyclic group

From $G/Z$ cyclic you can get something stronger: Let $a,b \in G$, then $a = z_1g^n$ and $b = z_2g^m$ for some $z_1,z_2 \in Z$. Now $ab = z_1g^n z_2g^m = z_2g^mz_1g^n = ba$. Thus $G$ is abelian, therefore $G = Z$. What now?

Hints:

$$\forall x,y\in G\;\exists z_1,z_2\in Z(G)\;,\;n_1,n_2\in\Bbb N\;\;s.t.\;\; x=g^{n_1}z_1\;,\;y=g^{n_2}z_2\implies$$

$$xy=g^{n_1}z_1g^{n_2}z_2=g^{n_1}g^{n_2}z_1z_2=g^{n_2}g^{n_1}z_2z_1=\ldots$$

The above thus proves $\,G\,$ is abelian, but then $\,Z(G)=G\,$ , so$\;\ldots\;$