Let $F$ be a local field of characteristic $0$ and $G$ a connected split reductive group over $F$.


Let's look at the derived groups. We have $(G(F),G(F)) \subset (G,G)(F)$ and this inclusion is of finite index according to [this MO question](https://mathoverflow.net/questions/133072/on-the-f-rational-points-of-the-derived-group-of-a-connected-reductive-algebra).


My question is : do we have (maybe under stronger assumptions) $U(F) \subset (G(F),G(F))$ for any unipotent subgroup $U \subset G$ ?


If no, same question for a given unipotent subgroup $U \subset G$.



When does the derived subgroup of $G(F)$ contains the $F$-points of unipotent subgroups of $G$
Yes, since in characteristic 0 all unipotent groups are connected and split. More generally, for any field $k \ne \mathbf{F}\_2$ and any connected reductive $k$-group $G$ and split unipotent smooth connected closed $k$-subgroup $U \subset G$, necessarily $U(k) \subset (G(k), G(k))$. (This is false for ${\rm{SL}}\_2(\mathbf{F}\_2) = S\_3$, whose commutator subgroup $A\_3$ has order 3.) 


In effect, this will reduce to the analogue for the non-reductive $H := \mathbf{G}\_m \ltimes \mathbf{G}\_{\rm{a}}$ with semi-direct product via either the usual scaling action or its precomposition with squaring on $\mathbf{G}\_m$, taking $U$ to be $\mathbf{G}\_{\rm{a}} = \mathscr{R}\_u(H)$. In such cases, $(t,0)(1,x)(t,0)^{-1}(1,x)^{-1} = (1, (t^e-1)x)$ with $e = 1, 2$. But as $x$ varies through $k$ and $t$ varies through $k^{\times}$, $(t^e-1)x$ varies through $k$ when we can choose $t \in k^{\times}$ with $t^e - 1 \ne 0$; i.e., $k \ne \mathbf{F}\_2$ when $e = 1$ and $k \ne \mathbf{F}\_2, \mathbf{F}\_3$ when $e = 2$. To deal with $\mathbf{F}\_3$ below, we'll need to do a bit more work. (Note that ${\rm{SL}}\_2(\mathbf{F}\_3) = S\_4$ does have commutator subgroup which contains the 3-Sylow subgroups.)


In the initial setup, every $U$ lies in $\mathscr{R}\_u(P)$ for a minimal parabolic $k$-subgroup $P$ of $G$, so it suffices to treat such unipotent radicals (which are $k$-split). Of course, if $G$ has no proper parabolic $k$-subgroup (i.e., if $P = G$) then the only such $U$ is 1 and there is nothing to do, so we may assume $P \ne G$ and hence $G$ has a non-central split $k$-torus.


For a split maximal $k$-torus $S$ in $P$, the unipotent radical of $P$ is the direct product variety (under multiplication in $G$) given by $\prod\_{c \in \Phi^+\_{\rm{nd}}} U\_c$ with the root groups associated to non-divisible elements in a positive system of roots in the relative root system $\Phi(G,S)$ (here ${\rm{Lie}}(U\_c)$ is the $c$-weight space in ${\rm{Lie}}(G)$ unless $c$ is multipliable, in which case it is the span of the weight spaces for $c$ and $2c$). The hypothesis that $P \ne G$ ensures that the root system is non-empty (otherwise there is nothing to do).
It suffices to treat the $U\_c$'s separately, so by replacing $G$ with $\mathscr{D}(Z\_G(S\_c))$ for the codimension-1 subtorus $S\_c = (\ker c)^0\_{\rm{red}}$ killed by $c$ we may focus on the case of semisimple $G$ with $k$-rank 1. 


For the purpose of handling $k = \mathbf{F}\_3$, the following additional reduction steps are also convenient and make sense with arbitrary $k$. We may replace $G$ with its simply connected central cover $\widetilde{G}$ (as the preimage of $P$ in $\widetilde{G}$ is a minimal parabolic $k$-subgroup whose unipotent radical maps isomorphically onto $U$), so $G$ is simply connected. Hence, $G = \prod\_i {\rm{R}}\_{k\_i/k}(G\_i)$ for a finite separable extensions $k\_i/k$ and *absolutely simple* connected semisimple $k\_i$-groups $G\_i$ that are simply connected. The $k$-rank 1 conditions forces all but one $G\_i$ to be $k\_i$-anisotropic, so we can drop those factors and thereby get to the case where there's a single $i = i\_0$. Then $P = {\rm{R}}\_{k\_{i\_0}/k}(P\_{i\_0})$ and $U = {\rm{R}}\_{k\_{i\_0}/k}(U\_{i\_0})$, so on $k$-points we can pass to $(G\_{i\_0}, U\_{i\_0})$; i.e., we can assume $G$ is also *absolutely simple* over $k$.


If the relative root system is A$\_1$ then $U\_c$ is a direct sum of copies of the 1-dimensional representation $\mathbf{G}\_{\rm{a}}(c)$ of $S = \mathbf{G}\_m$ via a nontrivial character $c$ that is at worst 2-divisible in the character group (since $\langle c, c^{\vee} \rangle = 2$). If the relative root system is BC$\_1$ then $U\_c$ is an extension of a direct sum of copies $\mathbf{G}\_{\rm{a}}(c)$ by a direct sum of copies of $\mathbf{G}\_{\rm{a}}(2c)$ where $c$ is a basis of the character group, and this extension of $k$-groups remains short exact on $k$-points due to additive Hilbert 90. 


Hence, it now suffices to show that if $H := \mathbf{G}\_m \ltimes \mathbf{G}\_{\rm{a}}(\chi)$ for the character $\chi$ of $\mathbf{G}\_m$ given by $\chi(t) = t$ or $\chi(t) = t^2$ then $(H(k), H(k))$ contains $U(k)$ where $U = \mathscr{R}(H)$ is $\mathbf{G}\_{\rm{a}}(\chi)$. The computation near the start handles this when $|k| > 3$. What if $k = \mathbf{F}\_3$? This is where the above passage to the absolutely simple and simply connected case (not yet used) is useful. The only such cases over a finite field are ${\rm{SL}}\_2$ and ${\rm{SU}}\_3$. For the first case, by inspection ${\rm{SL}}\_2(\mathbf{F}\_3) = S\_4$, whose commutator subgroup $A\_4$ contains the 3-Sylow subgroups. In the other case, a direct argument shows that $G(k)$ is perfect whenever $|k| > 2$.


