TITLE: Non-congruence normal subgroups of $SL_2(\mathbb{Z}[1/2])$
QUESTION [14 upvotes]: Let $G=SL_2(\mathbb{Z}[1/2])$, i.e., the modular group (if you wish) over the ring $\mathbb{Z}[1/2]$ consisting of rationals whose denominators are powers of $2$. Unlike $SL_2(\mathbb{R})$, $G$ is neither simple nor almost-simple: for example, it is possible to define congruence subgroups (for odd modulus $N$).
Is there a non-trivial (i.e., neither $\{I\}$ nor $\{\pm I\}$) normal subgroup of $G$ whose intersection with the unipotent subgroup $U = \left(\begin{matrix} 1 & * \\ 0 & 1\end{matrix}\right)$ is trivial? 
(Of course, such a subgroup would have to be of infinite index; does $G$ have non-trivial normal subgroups of infinite index?)

REPLY [6 votes]: More generally the Kazhdan-Margulis normal subgroup theorem (stating that non-central normal subgroups have finite index) applies to every irreducible lattice in a product of connected semisimple groups over real and $p$-adics of total rank $\ge 2$.
Here the ambiant group is $\mathrm{SL}_2(\mathbf{R})\times\mathrm{SL}_2(\mathbf{Q}_2)$, which has (total) rank 2.
(On the other hand this gives no information about finite index subgroups such as congruence subgroup property, but this is disjoint from the question.)