Has a conjugation of SL2(Z) finite index in SL2(Z)? (Modular group) 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
 A: There is a second proof which Tony Scholl hints at in the comments. This is probably secretly equivalent to the argument Greg writes up, but I find it easier to think about.
$SL_2(\mathbb{Z})$ is the subgroup of $SL_2(\mathbb{Q})$ preserving the lattice $L_1:=\mathbb{Z}^2$ inside $\mathbb{Q}^2$. Similarly, $M SL_2(\mathbb{Z}) M^{-1}$ is the gorup of matrices preserving $L_2 := M \cdot L_1$. So the group we are interested in is the group of matrices sending $L_1$ and $L_2$ to themselves. 
Choose an integer $N$ such that $L_1 \cap L_2 \supseteq N L_1$ and $L_1 + L_2 \subseteq N^{-1} L_1$. Let $\Gamma$ be the subgroup of $SL_2(\mathbb{Z})$ which acts trivially on $L_1/ N^2 L_1$. The subgroup $\Gamma$ has finite index as it is the kernel of $SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/N^2)$. 
Now, $\Gamma$ stabilizes $N L_1$ and $N^{-1} L_1$, and acts trivially on $(N^{-1} L_1)/(N L_1)$. In particular, any lattice $K$ with $N^{-1} L_1 \supseteq K \supseteq N L_1$ will be taken to itself by $\Gamma$. We chose $N$ so that $L_2$ lies between $N^{-1} L_1$ and $N L_1$. So $\Gamma$ takes $L_2$ to itself, and we deduce that $\Gamma$ is contained in the group we are interested in. So the group we are interested in has index $\leq [SL_2(\mathbb{Z}) : \Gamma]$ in $SL_2(\mathbb{Z})$.
A: 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.
