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Subgroup membership problem for $BS(1,2)$ is decidable, as answered in this link: https://math.stackexchange.com/a/3220976/465070

$BS(1,2)$ is isomorphic to a subgroup of $GL(2,\mathbb{Q})$ (the group of $2 \times 2$ rational matrices with determinant $\pm 1$).

Are there any other known results about the subgroup membership problem (or semigroup membership problem) for other rational matrix groups?

quid
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usercs
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    It sounds plausible that the subgroup membership problem is decidable in each finitely generated virtually solvable subgroup of $\mathrm{GL}_d(\mathbf{Q})$ — these are the same as finitely generated, virtually (solvable torsion-free) groups. (This more addresses the question before being edited, i.e., not only dimension 2.) – YCor May 10 '19 at 21:01
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    I think this is unknown (the $SL(2,Q)$ question). We really do not know much about existence of "exotic" subgroups of this group. You can even replace ${\mathbb Q}$ with ${\mathbb R}$ or with some totally real number field; the situation is still quite unclear. – Moishe Kohan May 10 '19 at 21:09

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The subgroup membership problem is not decidable for groups which are linear over $\mathbb{Z}$ (that is, group of integer matrices).

The Mihailova subgroup of direct products. Note that $F_2\times F_2$ is linear over $\mathbb{Z}$, where $F_2$ is the free group of rank two. A striking example of a finitely generated subgroup with undecidable membership problem is the "Mihailova subgroup" of $F_2\times F_2$ (the reference is K. A. Mihailova, The occurrence problem for direct products of groups Dokl. Acad. Nauk SSRR 119 (1958), 1103-1105.). The idea is as follows: take a surjective homomorphism $\phi:F_2\rightarrow G$ where $G=\langle \mathbf{x}\mid\mathbf{r}\rangle$ has insoluble word problem and consider the diagonal subgroup $$\Delta=\{(g, g)\in G\times G\mid g\in G\}.$$ Then the subgroup membership problem for $\Delta$ is not decidable in $G\times G$, and hence the membership problem for $Q=(\phi\times\phi)^{-1}(\Delta)$ is not decidable in $F_2\times F_2$. Moreover, $Q$ is finitely generated; it is generated by the set $$\{(x, x)\mid x\in\mathbf{x}\}\cup \{(R, 1)\mid R\in\mathbf{r}\}\cup \{(1, R)\mid R\in\mathbf{r}\}.$$

Rips' construction. A different kind of example can be found using Rips' construction. I wrote a length post about this construction here, which I don't want to repeat. The idea is that for every finitely presentable group $Q$ there exists a "small cancellation" group $H$ and a two-generated subgroup $N$ such that $H/N\cong Q$. So if $Q$ has insoluble word problem then the membership problem for $N$ is undecidable. The issue here is that it is not obvious that $H$ is linear over $\mathbb{Z}$; this was a major result of Dani Wise and his co-authors (although some years before he made his fame and fortune, before he proved that all small cancellation groups are linear over $\mathbb{Z}$ and a million other amazing things, Wise wrote a paper called "a residually finite version of Rips' construction" which is all you need here :-) ).

user1729
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  • The undecidability of the membership for $F_2 \times F_2$ shows that membership problem for $SL(4,\mathbb{Z})$ and therefore for $SL(4,\mathbb{Q})$ is undecidable. What I wanted to ask was the decidability of $2\times 2$ rational matrix groups. I am editing the question. – usercs May 10 '19 at 21:01
  • Could you possibly ask the 2x2 question separately? It us completely different, and as the comments to the question suggest, admit have a completely different answer. Also, this answer took quite a while to type up! – user1729 May 11 '19 at 09:14
  • Sorry but I couldn't as it was marked as duplicate when I tried to post it. May be you can post a question about the decidability of membership in linear groups and post your answer. – usercs May 11 '19 at 10:32
  • @cssstyler You probably need to rollback the edits here first, and also then edit the new question to look different enough from this one. (Probably linking to this one would be enough.) – user1729 May 12 '19 at 19:30