Does a lattice in $SL_n(\mathbb R)$ which is contained in $SL_n(\mathbb Z)$ have finite index in $SL_n(\mathbb Z)$? A lattice $H$ in a locally compact group $G$ is a discrete subgroup such that the coset space $G/H$ admits a finite $G$-invariant measure.
I have read several places that any lattice H in $\operatorname{SL}_n(\mathbb{R})$ which is contained in $\operatorname{SL}_n(\mathbb{Z})$ must have finite index in $\operatorname{SL}_n(\mathbb{Z})$. But I have been unable to prove this.
I have tried using the correspondence between the Haar measure on $\operatorname{SL}_n(\mathbb{R})$ and the counting measure on $\operatorname{SL}_n(\mathbb{Z})$, where we can partition $\operatorname{SL}_n(\mathbb{R})$ into sets each containing one element of $\operatorname{SL}_n(\mathbb{Z})$, and then normalizing s.t. each of these has measure one. But this seemed to lead nowhere. Also just restricting the measure on $\operatorname{SL}_n(\mathbb{R})/H$ to $\operatorname{SL}_n(\mathbb{Z})/H$ does not work either since the latter has measure zero.
Thanks a lot to the ones who will answer.
 A: First, for $H\subset \Gamma\subset G$ with $G$ unimodular, $\Gamma$ discrete, fixing a Haar measure on $G$, there is a unique $G$-invariant measure on $\Gamma\backslash G$ such that
$$
\int_{\Gamma\backslash G} \sum_{\gamma\in \Gamma} \varphi(\gamma\cdot g)\;dg \;=;\ \int_G \varphi(g)\;dg
$$
for all $\varphi\in C^o_c(G)$. Suppose $\Gamma\backslash G$ has finite measure. Similarly, by the same general uniqueness results, there is a unique measure on $H\backslash G$ such that
$$
\int_{\Gamma\backslash G} \sum_{\gamma\in H\backslash \Gamma} \varphi(hg)\;dg
\;=\; \int_{H\backslash G} \varphi(g)\;dg
$$
for all $\varphi\in C^o_c(H\backslash G)$. This set-up answers most questions about the trio $H\subset \Gamma \subset G$. For example, yes, if $H\backslash G$ has finite volume, then $H$ must be of finite index in $\Gamma$, or else the sum over $H\backslash \Gamma$ is infinite...
A: If $F\subset\operatorname{SL}_n(\mathbb{R})$ is a (measurable) fundamental domain for the left-action of $\operatorname{SL}_n(\mathbb{Z})$ on $\operatorname{SL}_n(\mathbb{R})$, and $\{k_i\}_{i\in H\backslash\operatorname{SL}_n(\mathbb{Z})}$ is a collection of representatives from the cosets of $H$ in $\operatorname{SL}_n(\mathbb{Z})$, then $$\cup_ik_iF$$ is a fundamental domain for the action of $H$ on $\operatorname{SL}_n(\mathbb{R})$.  There must therefore be only finitely many $k_i$ since $H$ is a lattice.
