This question now has two sequels, [Pointless groups II](https://mathoverflow.net/questions/424456/pointless-groups-ii) (to which @R.vanDobbendeBruyn gave a [counterexample](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091038_424456) for an infinite, imperfect field) and [Pointless groups III](https://mathoverflow.net/questions/424489/pointless-groups-iii), both using revised wording suggested by [@DanielLitt](https://mathoverflow.net/questions/424444/pointless-groups#comment1091016_424453).


Fix a field $k$—not algebraically closed, and, in fact, this question is obviously of interest only if $k$ is finite. (EDIT: @R.vanDobbendeBruyn pointed out in the [comments](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091035_424456) to [Pointless groups II](https://mathoverflow.net/questions/424456/pointless-groups-ii) that my memory that the $k$-points are Zariski-dense in the infinite case is only guaranteed true for reductive groups.) Then "group" means "smooth, connected, affine algebraic group scheme of finite type over $k$".


Is it possible to have a (smooth, connected) group $G$, and a proper (smooth, connected) subgroup $H$, such that $G(k) = H(k)$? If so, then can this even happen for $H$ the trivial subgroup?


I have a small store of counterexamples showing that certain naïve statements about rational points of algebraic groups can fail for small fields $k$. (For example: $C\_G(S)(k) \ne C\_G(S(k))$ for $S$ a maximal torus in $\operatorname{SL}\_2$ when $k$ has $2$ elements.) None of them provides an example of this sort of behaviour, but, then again, it is a very small testbed. If this question is too elementary for MO, then I have no objection to migrating it to MSE.



Pointless groups
Perhaps I'm missing something, but it seems to me that $k=\mathbb{F}\_2$, $G=\mathbb{G}\_{m, k}$ works, where $H$ is the trivial subgroup.


