Is finite verbal subgroup equivalent to finite index of marginal subgroup?

Question

There is a well known fact:

If $G$ is a finitely generated group. Then $|G’| < \infty$ iff $[G:Z(G)]<\infty$.

Suppose $\mathfrak{U}$ is a group variety. Let’s denote the corresponding verbal subgroup as a $V_{\mathfrak{U}}(G)$ and the corresponding marginal subgroup as $M_{\mathfrak{U}}(G)$. Note, that for the variety of all abelian groups $\mathfrak{A}$ (defined for the word $[x, y]$) we have $V_{\mathfrak{A}}(G) = G’$ and $M_{\mathfrak{A}}(G) = Z(G)$.

My question is:

Can the aforementioned statement be generalized to the following one:

If $G$ is a finitely generated group and $\mathfrak{U}$ is a variety, defined by one word. Then $|V_{\mathfrak{U}}(G)| < \infty$ iff $[G:M_{\mathfrak{U}}(G)]<\infty$.

?

Answer 1

No. Theorem 3 from "Об абелевых и центральных расширениях асферических групп" by I.S. Ashmanov and A. Yu. Olshanskii states:

For any finitely based subvariety $\mathfrak{U} \subset \mathfrak{B}_n$ where $n \geq 10^{75}$ there exists a noetherian group $G$ such that $|V_\mathfrak{U}(G)| \leq n$ and $[G:M_\mathfrak{U}(G)] = \aleph_0$

Here $\mathfrak{B}_n$ stands for the variety of all groups of exponent dividing $n$.