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.

(EDIT: Sorry, the above is slightly inaccurate: I'm not sure it is correct that $S_7$ is generated by two elements of order 2 and 3. It is, however, true for $A_n$ for any $n \ge 9$, so this gives lots of non congruence examples. There does exist a noncongruence subgroup of index 7, but it arises as the preimage of the stabilizer of 1 for a homomorphism $PSL(2, \mathbf{Z}) \to S_7$ with image a transitive subgroup of order 42.)

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.

(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 $S_7$).

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.

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.

(EDIT: Sorry, the above is slightly inaccurate: I'm not sure it is correct that $S_7$ is generated by two elements of order 2 and 3. It is, however, true for $A_n$ for any $n \ge 9$, so this gives lots of non congruence examples. There does exist a noncongruence subgroup of index 7, but it arises as the preimage of the stabilizer of 1 for a homomorphism $PSL(2, \mathbf{Z}) \to S_7$ with image a transitive subgroup of order 42.)