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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 $[G:M_{\mathfrak{U}}(G)]<\infty$.

?

Chain Markov
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    This doesn't address your question, but it is just to observe that $|G:Z(G)|$ finite implies $|G'|$ finite even when $G$ is not finitely generated. But the reverse implication can fail to hold in groups that are not finitely generated. – Derek Holt Nov 21 '19 at 09:00
  • It might be easy to prove for fg groups that if the verbal subgroup is finite then the marginal subgroup has finite index. Is this correct? – YCor Nov 21 '19 at 22:12
  • For the converse (for arbitrary groups) it might be wary to assume the variety finitely based. Indeed assuming the marginal subgroup has finite index then maybe implies that he defining words of the varieties take finitely many values. – YCor Nov 21 '19 at 22:14

1 Answers1

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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 $n$.

Chain Markov
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