TITLE: Is $SL_2(\mathbb Z)$ a maximal discrete subgroup in $SL_2(\mathbb R)$?
QUESTION [5 upvotes]: Endow $SL_2(\mathbb R)$ with the usual $L^2$ topology of $\mathbb R^4$. I wonder if $SL_2(\mathbb Z)$ a maximal discrete subgroup in $SL_2(\mathbb R)$.
I am working through some concrete examples. I tried $SL_2(\mathbb Z[\sqrt 2])$, but it seems that it is no longer discrete due to the fact that $m+n\sqrt 2$ is dense in $\mathbb R$. I also tried the group generated by 
$$ A=\left(\begin{matrix}1&\frac{1}{2}\\0&1\end{matrix}\right), B=\left(\begin{matrix}0&-1\\1&0\end{matrix}\right) $$
in matlab, but still I can't find a limit point or prove that is it actually discrete.

REPLY [6 votes]: Yes. Here's a direct proof working in arbitrary dimension:

$\mathrm{SL}_d(\mathbf{Z})$ is maximal among discrete subgroups of $\mathrm{SL}_d(\mathbf{R})$.

Proof: Let $\Gamma\subset\mathrm{SL}_d(\mathbf{R})$ be a discrete subgroup containing $\Lambda=\mathrm{SL}_d(\mathbf{Z})$. Since $\Lambda$ is a lattice, the index $[\Gamma:\Lambda]$ is finite. Since $\mathbf{Z}^d$ is $\Lambda$-invariant, it has finitely many images by $\Gamma$, and hence $E=\bigcap_{\gamma\in\Gamma}\gamma\mathbf{Z}^d$ has finite index in $\Gamma$, and hence $\Gamma\subset\mathrm{SL}(E)$. Let $u\in\mathrm{GL}_d(\mathbf{Q})$ map $\mathbf{Z}^d$ onto $E$. Then $u\Lambda u^{-1}=\mathrm{SL}(E)$, so $$\Gamma\subset u\Lambda u^{-1}\subset u\Gamma u^{-1}.$$ Since the $\mathrm{GL}_d(\mathbf{R})$-action by conjugation on $\mathrm{SL}_d(\mathbf{R})$ preserves some Haar measure, it preserves the covolume of lattices. We deduce that these inclusions are both equalities. Hence $\Lambda=\Gamma$.