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Yes. This is given by OEIS sequence A160870. The number of subgroups of index $n$ in $\mathbf{Z}^m$ is there denoted $T(n,m)$. There is a recursive formula in terms of the divisors of $n$ given at thi …

The answer is yes, such a group exists: this is the main result of [1]. In fact, one can take the Baumslag-Solitar group $\operatorname{BS}(p^r, -p^r) = \langle a, b \mid ba^{p^r}b^{-1} = a^{-p^r} \ra …

I am not very familiar with free metabelian groups, so I apologise in advance if this is trivial. A group $G$ is said to be complete if every automorphism of $G$ is inner. In this case, $\operatorname …

Chuck Miller in [Miller, Charles F., III On group-theoretic decision problems and their classification. Annals of Mathematics Studies, No. 68. Princeton University Press, Princeton, N.J.; University o …

sage should have what you need. Check under the documentation for the modular group. Specifically under generators you can find a couple of working examples for finding the generating set for $\Gamm …

It is known that a right-angled Artin group $A(\Gamma)$ embeds in some one-relator group if and only if $\Gamma$ is a forest. In fact, any such $A(\Gamma)$ embeds in the one-relator group $\operatorna …

A few minutes of running GAP shows that the group has order $6751269$. Hence, as the final edit of the accepted answer points out, the group is finite, and no Baumslag-Solitar group embeds in it.

I'll put my comment here, so that the question has an answer. The function $\operatorname{len}_H$ is what is called an actual distortion function (for $H$ in $G$) by Margolis, Meakin & Šuniḱ (see [1]) …

The groups have trivial center, as pointed out by Sam Nead. Another, more combinatorial, way to show this is to apply the algorithm from [1], which decides whether any given one-relator group has a no …

Novikov's centrally-symmetric group $\mathfrak{A}_P$ is a torsion-free group with undecidable word problem, constructed in [1]. Novikov did not prove it is torsion-free but, as Adian points out in [Ad …

Let $d$ be a square-free positive integer, and let $\mathcal{O}_d$ be the ring of integers of the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$. Consider the Bianchi group $\Gamma_d = \oper …

Here's an argument that uses only basic combinatorial group theory (Reidemeister-Schreier). Let $n \geq 2$, and let $G = \langle a_1, b_1, \dots, a_g, b_g | [a_1, b_1]^n [a_2, b_2] \cdots [a_g, b_g] \ …

A preprint by E. Rauzy appeared today on the arXiv, and gives a negative answer to this question. In other words (if the proof is correct), there exists a f.g. residually finite group with decidable w …

In 1965, Murasugi [1] conjectured that any finitely presented group with deficiency at least two has trivial centre. The year before, he had proved it true for one-relator groups, and in [1] he proved …

Burger and Mozes constructed (Burger and Mozes - Lattices in products of trees) infinite, finitely presented, torsion-free simple groups which split as amalgams of two finitely generated free groups o …