Examples of finitely presented subgroups of $\operatorname{GL}(n,\mathbb{Z})$ with unsolvable decision problems

Question

1. $\DeclareMathOperator\GL{GL}\DeclareMathOperator\Aut{Aut}$Does there exist a finitely presented subgroup of $\GL(n,\mathbb{Z})$ for which it is known that the conjugacy problem is unsolvable (if yes, please give a reference and some examples)? Note that it is known that $\GL(n,\mathbb{Z})$ itself has solvable conjugacy problem [The conjugacy problem in $\GL(n, \mathbb{Z})$ by B. Eick, T. Hofmann and E.A. O’Brien].

2. For a different/related decision problem: In Orbit decidability and the conjugacy problem for some extensions of groups, O. Bogopolski, A. Martino, and E. Ventura provide a construction of finitely generated subgroups of $\GL(n, \mathbb{Z})$, for $n$ large enough, for which the orbit problem with respect to $\mathbb{Z}^n$ is undecidable. As part of their construction, they use Mihailova’s construction. However, all the groups provided by Mihailova's construction which satisfy the conditions stated by Bogopolski et al. so that the resulting group has undecidable orbit problem will be finitely generated, but not finitely presented. Is there a way to construct finitely presented subgroups of $\GL(n, \mathbb{Z})$ with unsolvable orbit problem?

By the orbit problem I mean: Let $G$ be a group and $H \leq \Aut(G)$. Then we define the orbit problem for $H$ with respect to $G$ to be the problem of determining, given any two $u, v\in G$ whether or not there exists a $z\in H$ such that $u^z=v$.

Answer 1

Yes. If there is a f.g. subgroup $\Gamma$ of $\operatorname{GL}_n(\mathbf{Z})$ whose orbit problem on $\mathbf{Z}^n$ is indecidable, then there is a f.p. subgroup $G$ of $\operatorname{GL}_{n+3}(\mathbf{Z})$ with undecidable conjugacy problem.

Namely this group is as follows: pick a f.g. free group $F$ surjecting onto $\Gamma$ and consider the semidirect product $G=F\ltimes\mathbf{Z}^n$ (through the $\Gamma$-action).

That $G$ is $\mathbf{Z}$-linear in dimension $n+3$ can be seen as follows: let $g_1,\dots,g_k$ form a generating subset for $\Gamma$, $x_1,\dots,x_k$ freely generate a subgroup in $\operatorname{GL}_2(\mathbf{Z})$. Then $G$ can be viewed as the subgroup generated by the $(1+n+2)$-block matrices $\operatorname{diag}(I_1,g_j,x_j)$ and the matrices $e_{1j}=I_{n+3}+E_{1j}$ for $2\le j\le n+1$.

The orbit condition implies that there is no algorithm to determine whether two products $\prod_{j=2}^{n+1}e_{1j}e_{1j}^{k_j}$, $\prod_{j=2}^{n+1}e_{1j}e_{1j}^{k'_j}$ are conjugate ($(k_2,\dots,k_{n+1})$, $(k'_2,\dots,k'_{n+1})\in\mathbf{Z}^n$ being the input).

Answer 2

Finitely presented $\mathbb{Z}$-linear groups with unsolvable conjugacy problem are known to exist, although writing them down explicitly will be extremely painful! I'm not sure (or have perhaps forgotten…) if there is a published reference, but here are the ingredients you need.

Start with a finitely presented group $Q$ with unsolvable word problem, of type $F_3$. (That is, $Q$ has a classifying space with finitely many cells of dimension $\leq 3$.)

Next, apply Haglund–Wise's version of the Rips construction. This gives a short exact sequence

$$1\to K\to \Gamma\to Q\to 1$$

where $K$ is finitely generated and $\Gamma$ is torsion-free, hyperbolic and linear over $\mathbb{Z}$.

The example is now the fibre product $P\leq \Gamma\times \Gamma$ of the epimorphism $\Gamma\to Q$. This is clearly linear over $\mathbb{Z}$, and is finitely presented by the 1-2-3 Theorem of Bridson–Howie–Miller–Short. Finally, $P$ has unsolvable conjugacy problem, say by Proposition 7.10 in Conjugacy in normal subgroups of hyperbolic groups by Martino–Minasyan.

I think the linked paper of Martino–Minasyan should have most of the references needed to fill in the details in this answer.

More generally, the technique I describe here is the standard way to upgrade the finitely generated examples provided by Mihailova's construction to finitely presented examples — it can surely also be used to answer your Question 2.