Discreteness of a particular matrix group

Question

I want to understand the structure of some matrix groups. Unfortunately I know very little group theory, so I need some help.

First some motivation: I know that in the group $(\mathbb R_{>0},\times)$, the subgroup $\langle2\rangle$ generated by $\{2\}$ is discrete, but the subgroup $\langle2,3\rangle$ is not. Furthermore, for any $x\in\mathbb R_{>0}$, one can obtain the best approximation (under whatever metric) of $x$ in $\langle2\rangle$ by a simple "descent" algorithm: starting from any $a\in\langle2\rangle$, check which of the three values $2^{-1}a$, $a$, and $2a$ is closest to $x$, go there, and repeat. This algorithm does not work for $\langle2,3\rangle$, which is not discrete. However, this is not simply a consequence of discreteness: descent does not work for the discrete subgroup $\langle4,8\rangle$ either.

I am interested in answering these questions for the matrix group $$G = \left\langle \begin{bmatrix}2&0 \\ 0&1\end{bmatrix}, \begin{bmatrix}1&0 \\ 0&2\end{bmatrix}, \begin{bmatrix}1&1 \\ 0&1\end{bmatrix}, \begin{bmatrix}1&0 \\ 1&1\end{bmatrix} \right\rangle,$$ and its generalizations to higher dimensions. Namely,

1. Is $G$ discrete?
2. Is there a simpler way to characterize whether a given matrix is in $G$?
3. Does it permit a descent algorithm for matrix approximation (say under the Frobenius norm $\|M\|_F = \sum |m_{ij}|^2$)?

It is clear that the subgroup generated by the first two matrices is discrete, since it is simply the set of diagonal matrices with entries from $\langle2\rangle$. The subgroup generated by the last two matrices is a subset of the set of all integer matrices with determinant $1$; I believe it is equal to that set, but I don't even know how to prove that.

Answer 1

No, $G$ is not discrete in $\mathrm{GL}_2(\mathbf{R})$. Indeed, write your 4 generators as $d_1(2),d_2(2),e_{12}(1),e_{21}(1)$. Then $d_1(2)^{-n}e_{12}(1)d_1(2)^{n}=e_{12}(2^{-n})$ accumulates when $n\to\infty$ at the identity matrix $e_{12}(0)$.

Actually, $G$ has index 2 in $\mathrm{GL}_2(\mathbf{Z}[1/2])$ and consists exactly of those matrices of positive determinant in $\mathrm{GL}_2(\mathbf{Z}[1/2])$.

Indeed those matrices $e_{12}(2^{-n})$ and $e_{21}(2^{-n})$ generate $\mathrm{SL}_2(\mathbf{Z}[1/2])$ ($e_{12}(1/2)$ and $e_{21}(1)$ are actually enough), and using $d_1(2)$ one gets all matrices in $\mathrm{GL}_2(\mathbf{Z}[1/2])$ of positive determinant. Since all given generators have positive determinant, this is all that can be done.

Note: the standard way to view such a group as discrete is to view it diagonally embedded into $\mathrm{GL}_2(\mathbf{R})\times\mathrm{GL}_2(\mathbf{Q}_2)$.