My set up is the following: I have an affine algebraic group $G$ over a $p$adic field $F$, we assume that $G$ is semisimple and simply connected. I have an abstract subgroup $H\leq G(F)$ of the group of points of $G$. I know that $H$ is open and that $[G(F):H]<\infty$. I would like to conclude that $H=G(F)$, is this true in general?

$\begingroup$ Yes, I think it's in Margulis' book ($G(F)$ is generated by unipotents). $\endgroup$– YCorCommented Apr 4, 2018 at 19:12

1$\begingroup$ Do you happen to have a precise reference? The one I've found is in Margulis (Discrete Subgroups of Semisimple Lie Groups) Theorem 5.1. But that only applies to isotropic $k$almost simple groups. My hypothesis are stronger than in the theorem, which leads me to believe it should be true. $\endgroup$– Ian GleasonCommented Apr 4, 2018 at 19:30

$\begingroup$ You're right and more precisely Margulis applies for isotropic $k$simple simply connected groups, and products thereof. For simply connected semisimple groups, actually this is optimal: if $G$ is anisotropic, then $G(k)$ is profinite and hence has open subgroup of arbitrary large finite index, and more generally the latter holds if $G$ has an anisotropic factor. To summary, the answer to your question is yes iff every $F$simple factor of $G$ is $F$isotropic. $\endgroup$– YCorCommented Apr 4, 2018 at 19:54
1 Answer
It's true if and only if every $F$simple factor of $G$ is $F$isotropic (i.e., has nonzero $F$rank).
Indeed, if this is the case, then we can reduce to the $F$simple case. By Theorem 5.1 in Margulis' book (which uses the simple connectedness assumption), $G(F)$ is generated by its 1dimensional unipotent subgroups, which are isomorphic to $F$ as groups and hence have no proper subgroup of finite index. Hence $G(F)$ has no proper of finite index.
Conversely, if $G$ has a simple factor $H$ that is $K$anisotropic, then $H(F)$ is an infinite profinite group and is a direct factor of $G$, so this yields open subgroups in $G$ of arbitrary large index.