Let $k$ be a nonarchimedean local field and $G$ a reductive $k$-group,
which we assume to be semisimple and simply-connected. Recall that an abstract group $H$ is *perfect* if it is generated by commutators, that is, equals its derived subgroup.

**Question**: Is $G(k)$ perfect?

When $G$ is isotropic, $G(k)$ is known to be perfect. This is a consequence of the solution to the Kneser–Tits problem for $k$: the group $G(k)^+$ generated by (the rational points of) additive subgroups of $G$ is the same as the group generated by root subgroups of $G$ for a fixed maximal split torus, and the latter group is generated by commutators. It is known that $G(k)=G(k)^+$ when $G$ is simply-connected.

So the main thrust of my question is when $G$ is anistropic, or equivalently (hence the title), when $G(k)$ is compact in the analytic topology. In type $A$, Platonov and Jančevskiĭ have proved (On a conjecture of Harder) by computations with division rings that $G(k)$ is perfect. (**Clarification**: This result is for $D^\times$, not $\text{SL}_1(D)$.)

**Related question**: Are there any anisotropic simple $k$-groups not of type $A$?

~~I realize that if the answer to the related question is negative then this work of Platonov and Jančevskiĭ answers my first question.~~ I once looked through the tables in Tits's Boulder notes Classification of semisimple algebraic groups and it seemed like there were compact simple reductive groups only in type $A$. But this result was so surprising, and the tables so hard for me to follow, that I was not confident in the correctness of my understanding.

Bonus points for questions that work with a more general $k$, say, a complete discretely valued field with residue field of dimension $\leq1$.

**Clarification**: I was too blithe about what happens in type $A$. As krl pointed out in the comments, $\operatorname{SL}_1(D)$ is not perfect, and in fact, its commutator subgroup is its pro-unipotent radical. Riehm's paper "The norm 1 group of a $\mathfrak P$-adic division algebra" carefully analyzes the situation. So in the end, the answer to my question is "no" overall, but "yes" in many cases.

Related questionis "no", there are no other such groups, but I know of no better reference than Tits's Boulder notes. $\endgroup$3more comments