I'm trying to prove the following proposition:
Let $G$ be a group. Then $G$ is nilpotent iff ${\rm Inn}(G)$ is nilpotent.
I've proven that if $G$ is nilpotent then ${\rm Inn}(G)$ is nilpotent as well by showing that there exists a central series of ${\rm Inn}(G)$ where each element is of the form:
$${\rm Inn}_{G_{i}}(G):=\{f: x\mapsto hxh^{-1}\mid h\in G_i\}.$$
Now for the other direction, I assume that ${\rm Inn}(G)$ is nilpotent which means there exists a central series of it. But I don't how subgroups of ${\rm Inn}(G)$ look like. Are they all of the form I listed above? If not, what could they look like?
Note: We may not use the fact that $G$ is nilpotent $\Rightarrow G/Z(G)$ is nilpotent.
Since the inner automorphism group $\mathrm{Inn}(G)$ is isomorphic to the quotient $G/Z(G)$, the subgroups of $\mathrm{Inn}(G)$ are in bijection with subgroups of $G$ containing $Z(G)$.
Given a central series of $\mathrm{Inn}(G)$ of finite length, we can obtain a central series from $Z(G)$ to $G$, and thus a central series of $G$ by adding $\{1\}\triangleleft Z(G)$ to the left side. This shows that $G$ is also nilpotent.