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? 



Open subgroups of finite index of p-adic semisimple groups
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 1-dimensional 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.


