How can I see that there is no lattice in $G=\mathrm{PSL}\_2( \mathbb{R})$ which contains $\Gamma\_1=\mathrm{PSL}\_2( \mathbb{Z})$ properly, or equivalently, that $X\_1 =\mathrm{PSL}\_2(\mathbb{Z}) \backslash \mathbb{H}$ is not a finite sheated nontrivial cover of a Riemann surface. 


As a motivation, the eigenvalues of the Laplace Beltrami operator on $\mathrm{PSL}\_2(\mathbb{Z}) \backslash \mathbb{H}$ is conjectured to have only eigenvalues of multiplicity one. So now, if there exists $\Gamma \supset \Gamma\_1$, an eigenvalue on $X\_1$ can be associated to an irreducibel representation $\pi$ of $\mathrm{Ind}\_{\Gamma\_1}^{\Gamma} $, since $Ind\_{\Gamma}^{G}$ $ \mathrm{Ind}\_{\Gamma\_1}^{\Gamma} 1 $ and $\mathrm{Ind}\_{G}^{\Gamma}1 $ are isomorphic, and the eigenvalues appear with multiplicty being square of the dimension of $\pi$ (eigenfunctions =matrix coefficients).


Hence, if there would exists such a thing with a higher dimensional representation, the conjecture would be wrong. That's why the intuition that something should be known here.


What about $G$ reductive (see comment of David Loeffler)!



There is no lattice in PSL(2,R) which contains PSL(2,Z) properly?
We can say something stronger.



> 
> **Theorem:** (Helling 1976) Consider the family of subgroups of $SL\_2(\mathbb{C})$ that are commensurable with a conjugate of $SL\_2(\mathbb{Z})$. The maximal elements of this family under inclusion are precisely the conjugates of $\Gamma\_0(N)^+$ for $N$ a squarefree integer.
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Conway has a nice sketch of a proof in his paper "Understanding groups like $\Gamma\_0(N)$" in the book *Groups, difference sets, and the Monster*. You can show using lattices that the commensurable groups act on the product of $p$-adic Bruhat-Tits buildings for primes $p$, and maximal groups are the groups that stabilize cells (i.e., finite products of edges).


I imagine you can do a similar calculation for higher-rank reductive groups, but the analysis may be harder, since the buildings are no longer trees.


