# Reference request: The commensurator of an arithmetic lattice is a simple group

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 open problem list.

• I doubt this. Let $G$ be an $\mathbf{R}$-split $\mathbf{Q}$-form of $\mathrm{SL}_n$ that is anisotropic over $\mathbf{Q}_p$ for some prime $p$. So $G(\mathbf{Z})$ is an arithmetic cocompact lattice in $\mathrm{SL}_n(\mathbf{R})$. Then I'd expect that ($\ast$) the commensurator in $G(\mathbf{R})$ of $G(\mathbf{Z})$ to be contained in $G(\mathbf{Q})$ . But the latter is residually finite, since it is contained in $G(\mathbf{Q}_p)$ which is profinite. Still ($\ast$) has to be double checked. – YCor Oct 29 '19 at 17:47

Hmm, that's not exactly true. For example, $$Comm_{SL_2(\mathbb{R})}(SL_{2}(\mathbb{Z}))$$ contains the normal subgroup $$\pm I$$.
• Yes of course we need to take $\mathrm{Comm}_G(\Gamma)/Z(\mathrm{Comm}_G(\Gamma))$. ls it obvious why there should be no other quotients of $\textrm{Comm}_G(\Gamma)$? – Sam Hughes Oct 29 '19 at 16:40