Decidability of membership problem for rational matrix groups 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?
 A: 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 :-) ).
