Given two finite presentations of torsion-free groups, is there an algorithm to determine whether the given groups are isomorphic or not?


I have found results for narrower classes (for example, they are briefly reviewed in [The isomorphism problem for finitely generated fully residually free groups](https://www.sciencedirect.com/science/article/pii/S002240490600079X)).



Is the isomorphism problem solvable for torsion-free groups?
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 [Adi1957b] (p. 76 of [my translation](https://arxiv.org/abs/2208.08560), and referenced as in there; he calls the group $F\_0$ in that article), it is not difficult to prove this fact using Novikov's work.


The statement of the Adian-Rabin theorem is today done via Markov properties (as defined by Markov when he proved the corresponding theorem for semigroups a few years earlier). And this is how Adian does it in [Adi1957b], too, but in his main article [Adi1957a], which contains all the detailed proofs, there are no Markov properties, only a slightly smaller family of properties (I call them "pseudo-Markov"). Instead, what Adian does in [Adi1957a] is the following:


Let $A, B$ be any two words in $\mathfrak{A}\_P$. From these two words and $\mathfrak{A}\_P$ he constructs a new finitely presented group, which he calls $\mathfrak{A}\_{q, A, B}$, and which has the following property: $\mathfrak{A}\_{q, A, B}$ is trivial if and only if $A = B$ in $\mathfrak{A}\_P$. Furthermore, the groups $\mathfrak{A}\_{q, A, B}$, for all $A, B$, are all torsion-free (he does not state it directly but it follows from the same type of argument as in his behemoth Main Lemma).


This yields the weaker Adian-Rabin theorem, but it also yields the undecidability you want; the family $\mathfrak{A}\_{q, A, B}$, together with the trivial group, is recursively enumerable (indeed the way one constructs the groups is given explicitly, and in modern terminology it is just taking some HNN-extensions of $\mathfrak{A}\_P$), and so if we could decide the isomorphism problem for torsion-free groups, then we could decide the isomorphism in this class, so we could also decide the word problem for $\mathfrak{A}\_P$, a contradiction.


(Footnote: note that there are Markov properties $\mathcal{P}$ such that the isomorphism problem for all groups with $\mathcal{P}$ is decidable (trivially, we may take the property "being trivial"; but less trivially we may take "being hyperbolic" or "being abelian"). But I suspect that given any Markov property $\mathcal{P}$ such that there are groups with undecidable word problem with $\mathcal{P}$, the isomorphism problem for the class of groups with $\mathcal{P}$ is undecidable; I think I am missing some simple argument for why this is the case.)


[1] *Novikov, P. S.*, [**On the algorithmic insolvability of the word problem in group theory**](https://doi.org/10.1090/trans2/009/01), Am. Math. Soc., Transl., II. Ser. 9, 1-122 (1958). [ZBL0093.01304](https://zbmath.org/?q=an:0093.01304).


