# Non-congruence normal subgroups of $SL_2(\mathbb{Z}[1/2])$

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?)

• Nice question with a nice answer! – GH from MO Dec 11 '15 at 18:14

The answer is no. Every normal non-central subgroup of $G=SL_2({\mathbb Z}[1/2])$ has finite index and is a congruence subgroup. This is proved in greater generality in a paper by J-P.Serre, in Annals of math (see .http://www.ams.org/mathscinet-getitem?mr=272790 for a reference)
• I should have said that the specific result for $SL_2({\mathbb Z}[1/p])$ is due to Mennicke. – Venkataramana Dec 12 '15 at 2:54
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.