Let $G$ be a connected, reductive group over an algebraically closed field $k$. Let $B$ be a Borel subgroup with maximal torus $T$ and unipotent radical $U$. Let $\Phi^+ = \Phi(B,T)$ and $\Delta$ the base of $\Phi = \Phi(G,T)$ corresponding to $\Phi^+$.

A subset $\Psi$ of $\Phi^+$ is called **closed** if whenever $\alpha, \beta \in \Psi$, and $\alpha + \beta$ is a root, we have $\alpha + \beta \in \Psi$. If $\Psi$ is closed, then the root subgroups $U_{\alpha} : \alpha \in \Psi$ directly span a closed, connected subgroup $U_{\Psi}$ of $U$ which is normalized by $T$ and whose Lie algebra is $\bigoplus\limits_{\alpha \in \Psi} \mathfrak g_{\alpha}$.

Furthermore, if $\alpha \in \Phi^+$, and $\Psi \subseteq \Phi^+$ is closed, and all roots of the form $a \alpha + b \Psi$ for $a, b \in \mathbb{Z}^+$ lie in $\Psi$, then the root subgroup $U_{\alpha}$ normalizes $U_{\Psi}$. These facts are proved in Chapter 14 of Borel, *Linear Algebraic Groups*.

In particular, let $\Psi = \Phi^+ - \Delta$. It is easy to see that $\Psi$ is closed and normalized by every root subgroup $U_{\alpha} : \alpha \in \Phi^+$, and in particular, $U_{\Phi^+ - \Delta}$ is a normal subgroup of $U$. Moreover, it is a consequence of 8.32 in Springer's *Linear Algebraic Groups* that for any $x \in U_{\alpha}$ and $y \in U_{\beta}$, the commmutator $xyx^{-1}y^{-1}$ lies in $U_{\Phi^+ - \Delta}$. From here one can argue that $U_{\Phi^+ - \Delta}$ contains the derived group of $U$ by producing a homomorphism $U \rightarrow \prod\limits_{\alpha \in \Delta} \mathbf G_a$ with kernel $U_{\Phi^+ - \Delta}$.

My question is, is $U_{\Phi^+ - \Delta}$ exactly the derived group of $U$? For $G = \textrm{GL}_n$, this does seem to be the case.

Reductive Group Schemesin smf4.emath.fr/Publications/PanoramasSyntheses/2014/42-43/html/… (for a simpler proof via dynamic arguments). $\endgroup$ – nfdc23 Jun 21 '17 at 0:45