A subgroup of $SL\_2(\mathbb{R})$ is called **arithmetic** if it is commensurable with $SL\_2(\mathbb{Z})$.


An arithmetic subgroup is called **congruence** if it contains a subgroup of type $\Gamma(N)$ for some $N\in \mathbb{N}$.



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> Question: What are concrete examples of subgroups of $SL\_2(\mathbb{R})$, which are arithmetic, but not congruence?
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I have heard the Belyi theorem produces some examples, but I have never seen a concrete one.



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> Further Question: Can such things exist in higher rank Lie groups (real rank $\geq 2$)?
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Concrete examples of noncongruence, arithmetic subgroups of SL(2,R)
I would quibble with your definition of "arithmetic". There are discrete subgroups of $SL(2, \mathbf{R})$ coming from quaternion algebras, which are not commensurable with $SL(2, \mathbf{Z})$, but are commensurable with $G(\mathbf{Z})$ for some twisted form $G$ of $SL(2)$ which becomes isomorphic to the usual form over $\mathbf{R}$. These are arithmetic subgroups of $G$, but not of the usual $SL(2)$. So the notion of "arithmetic subgroup" depends on a choice of model over a number field.


With that quibble out of the way, examples of noncongruence arithmetic subgroups of $SL\_2 / \mathbf{Q}$ are not too hard to come by. The easiest construction I know goes like this. We know $G = PSL(2, \mathbf{Z})$ has a rather simple presentation; it's the free product of $C\_2$ and $C\_3$. So if we have any other group $H$ containing elements $a,b$ of orders 2 and 3, there is a unique homomorphism $G \to H$ sending the generators to these elements.


In particular, $S\_7$ is generated by two such elements, so there is a normal subgroup of $G$ with quotient isomorphic to $S\_7$. But $S\_7$ is not a quotient of $SL(2, \mathbf{Z} / m)$ for any $m$ (easy check once you know that $PSL(2, \mathbf{Z}/p)$ is simple for $p > 2$) hence this subgroup cannot be congruence.


(If you look not at the kernel of this homomorphism but the preimage of the stabilizer of 1, then you get a noncongruence subgroup of index 7, which is, if I remember correctly, the smallest possible.)


The congruence subgroup problem for more general reductive groups is a difficult problem and one that's far from solved. As I mentioned above, it depends on a choice of model over a number field. The group $SL(n) / \mathbf{Q}$ has no non-congruence arithmetic subgroups for $n \ge 3$ (Bass-Milnor-Serre), while $SL(2)$ over an imaginary quadratic field has lots of them (it behaves like $SL(2) / \mathbf{Q}$). There are many more recent results; googling brings up a [survey by Raghunathan](http://www.ias.ac.in/mathsci/vol114/nov2004/Pm2396.pdf).


(EDIT: I suddenly realized I wasn't sure if the bit about $S\_7$ being a quotient of $PSL(2, \mathbf{Z})$ was true or not. So I wrote a computer program and it is indeed true that the elements (2,3,4)(5,6,7) and (1,2)(3,5)(4,6) generate $S\_7$).


