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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 …

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.

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 …

For a finitely presented group $G$, generated by a finite set $A$, the commutator problem is the decision problem: given a word $w$ over the alphabet $A \cup A^{-1}$, can one decide if $w$ is a commut …

The following question arose in the comments on this question, and it seems like a reasonable question to ask in its own right. I've added some additional details. The word problem in any fixed finite …

After some digging, I was able to find that the answer to my question exists: the problem can be undecidable. Rattaggi, in an unpublished manuscript (available here), proved that there exists a finite …

Let $G$ be a finitely presented simple group. By Kuznetsov (1958), $G$ has decidable word problem. However, by Scott [1], $G$ may have undecidable conjugacy problem. Is anything known about other deci …

Is the conjugacy problem for two-relator groups known to be undecidable? The word problem for two-relator groups is a famous open problem (appearing e.g. as Question 9.29 in the Kourovka notebook), an …

Is the intersection of any two finitely generated subgroups of $\operatorname{Aut}(F_2)$ (resp. $\operatorname{Aut}(F_3)$) again finitely generated? That is, does $\operatorname{Aut}(F_2)$ (resp. $\o …

As mentioned in the comments, this is still considered an open problem. I thought I'd flesh out a few aspects. A solution was claimed in 1992 by Juhasz, but it seems to have failed to convince experts …

Let $G = \langle A \mid R \rangle$ be a finitely presented group, given by a finite presentation. If $G$ is abelian, then we can verify this fact: simply verify the fact that $[a, b] = 1$ for all gene …

Polycyclic groups are LERF, by Mal'cev 1948. In particular, all nilpotent and all abelian groups are LERF. As mentioned in the comments, as not all one-relator groups are residually finite, not all on …

Let $G$ be a one-relator group $\langle A \mid R = 1 \rangle$. Then clearly $G$ is finite if and only if it is cyclic of finite order, i.e. can be given by a presentation $\langle a \mid a^n = 1 \rang …