See my previous question for the motivation for this question:
Are infinite cyclic normal subgroups "virtually" central?
Let $G$ be a (finitely generated) group and let $K \lhd G$ be a normal subgroup such that $K$ is isomorphic to $\mathbb{Z}^n$ for some $n \ge 2$. When does $G$ have a finite-index subgroup $H$ such that $K \le Z(H)$?
The simple argument to answer my previous question won't work since $\text{Aut}(\mathbb{Z}^n) = GL_n(\mathbb{Z})$ is infinite for $n \ge 2$.
EDIT: I remembered that I do know of one sufficient condition. Namely, if $K$ is quasi-isometrically embedded in $G$, then such an $H$ does exist.
To try to expand and clarify my question: Are there any (other) well-known assumptions or conditions that are sufficient to guarantee there exists a finite-index subgroup $H \le G$ and a finite-index subgroup $K' \le K$ such that $K' \le Z(H)$?