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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?)

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

2 Answers 2

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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)

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    $\begingroup$ 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
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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.)

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    $\begingroup$ 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

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