Dear all,


I have a probably rather simple question: Suppose we have a Matrix $ M\in SL\_2(\mathbb{Q}) $. Does the group $ M^{-1} SL\_2(\mathbb{Z}) M \cap SL\_2(\mathbb{Z})$ then always have finite index in $SL\_2(\mathbb{Z})$? Why? Why not?


I really was not able to solve this problem!


All the best
Karl



Has a conjugation of SL2(Z) finite index in SL2(Z)? (Modular group)
I, for one, am less than thrilled with snobbish kibitzing in the comments. Just answer the question already instead of dropping hints and passing judgment.


The answer is yes for $\text{SL}(n,\mathbb{Z})$. Let $d$ be the product of the denominators in the matrices $M$ and $M^{-1}$. Let $\Gamma\_d \subseteq \text{SL}(n,\mathbb{Z})$ be the subgroup of matrices of the form $I+dA$. This subgroup has finite index because it is the kernel of the congruence homomorphism
$$\text{SL}(n,\mathbb{Z}) \longrightarrow \text{SL}(n,\mathbb{Z}/d),$$
whose target is a finite group. On the other hand, $M\Gamma\_dM^{-1} \subseteq \text{SL}(n,\mathbb{Z})$ because $MIM^{-1} = I$ and $dMAM^{-1}$ is an integer matrix. Thus the intersection in question has finite index because it contains $\Gamma\_d$ as a subgroup.


The argument is quite general: You can replace $\text{SL}$ by other algebraic groups defined over $\mathbb{Z}$, and you can replace $\mathbb{Z}$ by any number field ring and $\mathbb{Q}$ by the corresponding number field.


