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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 $U$.

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.

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.

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 $U(k)/U'(k)$.

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 $V$.) The case of interest is $U = \mathcal{U}/\mathcal{D}^2(\mathcal{U})$, where $\mathcal{U}$ is the unipotent radical of a Borel subgroup of a connected semisimple $k$-group and {$\mathcal{D}^i(\mathcal{U})$} denotes its derived series.

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 $U$. 

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.

Let U be a unipotent group of Lie type over a finite field F_q. Let pi:U->U/U^1 be the natural projection. Assume, for simplicity, that U^2 = {e}. (Here U^1 = [U,U], U^2 = [U,[U,U]], etc.) Say I have a subgroup V of U/U^1.

Is there a subgroup H < U such that

(a) pi(H) = V,

(b) H\cap U^1 = <[H,H]> ?

(Notice that H\cap U^1 must always contain <[H,H]>, which depends only on V.)

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 $U(k)/U'(k)$.

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 $V$.) The case of interest is $U = \mathcal{U}/\mathcal{D}^2(\mathcal{U})$, where $\mathcal{U}$ is the unipotent radical of a Borel subgroup of a connected semisimple $k$-group and {$\mathcal{D}^i(\mathcal{U})$} denotes its derived series.