Let $ G $ be a simple linear algebraic group. Let $ G_\mathbb{R} $ be the real points of $ G $. Let $ G_\mathbb{Z} $ be the integer points of $ G $. Is $ G_\mathbb{Z} $ a maximal closed subgroup? In other words, are the only closed subgroups $ H $ of $ G_\mathbb{R} $ such that
$$
G_\mathbb{Z} \subset H \subset G_\mathbb{R}
$$
just $ G_\mathbb{R} $ and $ G_{\mathbb{Z}} $?

The answer is no for $ SO_3 $. $ SO_3(\mathbb{Z}) $ is the 24 element octohedral group ( isomorphic to the symmetric group $ S_4 $ ) and this group is a maximal finite subgroup of rotations and indeed it turns out that adding any other rotation and taking the closure will generate all of $ SO_3(\mathbb{R})$.

The answer also appears to be no for $ SL_n $. See second answer by Ycor given here:

https://math.stackexchange.com/questions/3388661/is-sl-2-mathbb-z-a-maximal-discrete-subgroup-in-sl-2-mathbb-r

Although this only states that $ SL_n(\mathbb{Z}) $ is maximal among discrete subgroups I'm guessing that it is in fact maximal among closed subgroups as well in this case.