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.
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.