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 yes for for $\operatorname{SO}_3$. $ \operatorname{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 $ \operatorname{SO}_3(\mathbb{R})$.

The group of integer points is also maximal in $ \operatorname{SL}_n(\mathbb R) $. See the [answer](https://math.stackexchange.com/a/3390666) by YCor given to [Is $\operatorname{SL}_2(\mathbb Z)$ a maximal discrete subgroup in $\operatorname{SL}_2(\mathbb R)$?](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 $ \operatorname{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.