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?)

  • $\begingroup$ Nice question with a nice answer! $\endgroup$
    – GH from MO
    Dec 11, 2015 at 18:14

2 Answers 2


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)

  • 2
    $\begingroup$ I should have said that the specific result for $SL_2({\mathbb Z}[1/p])$ is due to Mennicke. $\endgroup$ Dec 12, 2015 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.

(On the other hand this gives no information about finite index subgroups such as congruence subgroup property, but this is disjoint from the question.)

  • 1
    $\begingroup$ The title of the question mentions non-congruence subgroups. That is why I mentioned the congruence subgroup property $\endgroup$ Dec 12, 2015 at 1:47
  • $\begingroup$ Yes, the title could rather have been "Infinite index normal subgroups..." $\endgroup$
    – YCor
    Dec 12, 2015 at 10:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.