groups with abelian inner automorphism group

The inner automorphism group $\mathrm{Inn}(G)$ is isomorphic to the central quotient of $G$, $G/Z(G)$. if $\mathrm{Inn}(G)$ is abelian, one cannot conclude that $G$ itself is abelian. For example, let $G={𝒟}_8$, the dihedral group of symmetries of the square.

$$G=⟨r,s\mid r^4=s^2=1,rs=s{r}^3⟩$$

and $Z(G)=\{1,r^2\}$. Representatives of the cosets of $Z(G)$ are $\{1,r,s,rs\}$; these define a group of order $4$ that is isomorphic to the Klein $4$-group (http://planetmath.org/Klein4Group) $V_4$. Thus the central quotient is abelian, but the group itself is not.

However, if the central quotient is cyclic, it does follow that $G$ is abelian. For, choose a representative $x$ in $G$ of a generator for $G/Z(G)$. Each element of $G$ is thus of the form $x^{a}z$ for $z\in Z(G)$. So given $g,h\in G$,

$$gh=x^{a_1}{z}_1{x}^{a_2}{z}_2=x^{a_1}{x}^{a_2}{z}_1{z}_2=x^{a_1+a_2}{z}_1{z}_2=x^{a_2}{x}^{a_1}{z}_2{z}_1=x^{a_2}{z}_2{x}^{a_1}{z}_1=hg$$

where the various manipulations are justified by the fact that the $z_i\in Z(G)$ and that powers of $x$ commute with themselves.

Finally, note that if $\mathrm{Inn}(G)$ is non-trivial, then $G$ is nonabelian, for $\mathrm{Inn}(G)$ nontrivial implies that for some $g\in G$, conjugation by $g$ is not the identity, so there is some element of $G$ with which $g$ does not commute. So by the above argument, $\mathrm{Inn}(G)$, if non-trivial, cannot be cyclic (else $G$ would be abelian).