I am interested in a reference and proof for some version of the following (folklore?) statement:


``Let $G$ be a (semi)simple Lie group (with no compact factors and trivial centre) and let $\Gamma$ be an (irreducible) arithmetic lattice. Then $\mathrm{Comm}\_G(\Gamma)$ is a simple group."


I have yet been unable to locate a precise statement or proof of this result. However, it is briefly alluded to on page 2 of [this](https://pdfs.semanticscholar.org/74b1/6c6829bb3355486df4ce8e3002e7b1969b93.pdf) open problem list.



Reference request: The commensurator of an arithmetic lattice is a simple group
Hmm, that's not exactly true. For example, $Comm\_{SL\_2(\mathbb{R})}(SL\_{2}(\mathbb{Z}))$ contains the normal subgroup $\pm I$. 


For a special case describing the commensurator (when the complexified algebraic group has trivial center), see Ex. 4, §5.2. <https://arxiv.org/abs/math/0106063> For the general case, map to the product of the projective groups of the simple factors, which will have trivial center when complexified, and hence satisfy the conditions of the exercise. Then pull back to get the full description. I think then that sometimes the commensurator will simple modulo its center.


Under certain hypotheses this is known. See [Tits' Simplicity Theorem](https://en.wikipedia.org/wiki/Reductive_group#Structure_of_semisimple_groups_as_abstract_groups). But as indicated in Yves de Cornulier's comment about, the commensurator will not always be simple. 


