Let $U$ be a smooth connected unipotent algebraic group over an algebraic closure $k$ of $\mathbf{F}_p$. Let $U' := [U,U]$ be the derived (= commutator) subgroup, and assume it is central in $U$. Let $V$ be a subgroup of the abstract group $U(k)/U'(k) = (U/U')(k)$.

Is there a subgroup $H$ of $U(k)$ such that

(a) $\pi(H) = V$,

(b) $H\cap U' =[H,H]$?

Note that $H\cap U'$ must contain $[H,H]$, which depends only on $V$ due to the centrality of $U'$ in $U$. So if we define the subgroup $\widetilde{V} = \pi^{-1}(V)$ in $U(k)$ then the question is really asking if $\widetilde{V}$ contains a subgroup $H$ that maps onto $V$ and meets $U'$ in precisely the commutator subgroup $[\widetilde{V}, \widetilde{V}]$.

The case of interest is $U = \mathcal{U}/[\mathcal{U},\mathcal{D}(\mathcal{U})]$, where $\mathcal{U}$ is the unipotent radical of a Borel subgroup of a connected semisimple $k$-group.