Are commensurable subgroups contained in a finite overgroup? Let $\Gamma$ and $\Lambda$ be commensurable discrete lattices in a semi-simple Lie group $G$, i.e. the intersection $\Gamma\cap\Lambda$ has finite index in both $\Gamma$ and $\Lambda$. Let $\Omega=\langle\Gamma,\Lambda\rangle$. When is $[\Omega:\Gamma]$ and $[\Omega:\Lambda]$ finite? Moreover, can we say anything about either index? If $\Gamma$ and $\Lambda$ are not arithmetic, then by Margulis they are finite index inside their commensurator. So I am only interested in the case when $\Gamma$ and $\Lambda$ are arithmetic.

Let me assume that you implicitly mean $G$ to be semisimple and the discrete subgroups to be (arithmetic) lattices.

Here's an example where they are not contained in a finite index overgroup. Namely $G=\mathrm{SL}_2(\mathbf{R})$ (you can do something similar in $G=\mathrm{SL}_d(\mathbf{R})$ for $d\ge 2$), $\Gamma=\mathrm{SL}_2(\mathbf{Z})$, and $\Lambda=g^{-1}\Gamma g$ with $g=\begin{pmatrix}2 & 0\\\ 0 & 1\end{pmatrix}$. Thus $$\Lambda=\left\\{\begin{pmatrix}a & b\\\ c & d\end{pmatrix}\in\mathrm{SL}_2(\mathbf{Z}[1/2]): a,d\in\mathbf{Z},b\in\frac12\mathbf{Z},c\in2\mathbf{Z}\right\\}.$$ Hence $\Omega$ contains the matrix $\begin{pmatrix}1 & 0\\\ 1/2 & 1\end{pmatrix}$, and hence its transpose as well (using conjugation by some integral matrix) and these generate $\mathrm{SL}_2(\mathbf{Z}[1/2])$, so $\Omega=\mathrm{SL}_2(\mathbf{Z}[1/2])$.