Is the derived group of the G(F) perfect

Question

Let $G$ be a connected reductive group defined over a finite field $F$ of characteristic $> 3$. Is it true that the commutator group of $G(F)$ is perfect? This is true if $G$ is assumed to be semisimple.

Answer 1

$\newcommand{\ssc}{\text{sc}} \newcommand{\der}{\text{der}} \DeclareMathOperator{\im}{im}$Yes, this is true for a reductive $F$-group $G$, if for the simply connected semisimple $F$-group $G^\ssc$, see below, the group of $F$-points $G^\ssc(F)$ is perfect, that is, the derived group $G^\ssc(F)^\der$ coincides with $G^\ssc(F)$.

Theorem. Let $G$ be a (connected) reductive group over a field $F$. Let $G^\ssc$ denote the universal cover of the commutator subgroup $[G,G]$ of $G$. Consider Deligne's homomorphism $$\rho\colon\, G^\ssc\twoheadrightarrow [G,G]\hookrightarrow G.$$ If the group $G^\ssc(F)$ is perfect, then $G(F)^\der=\rho\bigl(G^\ssc(F)\bigr)$, and hence $G(F)^\der$ is perfect as well.

Proof. Consider the regular maps of $F$-varieties \begin{align*} \varkappa&\colon G\times G\to G,\qquad (x,y)\mapsto xyx^{-1}y^{-1}\\ \varkappa^\ssc&\colon G^\ssc\times G^\ssc\to G^\ssc,\quad (x,y)\mapsto xyx^{-1}y^{-1}. \end{align*} By definition, \begin{align*} G(F)^\der&=\bigl\langle\varkappa\bigl(G(F)\times G(F)\bigr)\bigr\rangle,\\ G^\ssc(F)^\der&=\bigl\langle\varkappa^\ssc\bigl(G^\ssc(F)\times G^\ssc(F)\bigr)\bigr\rangle, \end{align*} where $\langle S \rangle$ denotes the subgroup generated by a set $S$. Deligne observed that $\varkappa$ factors via $G^\ssc$, that is, there exists a regular map $$\widetilde\varkappa\colon G\times G\to G^\ssc$$ such that $$\varkappa=\rho\circ\widetilde\varkappa\colon\, G\times G\,\to\, G^\ssc\,\to\, G.$$ It follows that $\im\varkappa_F\subseteq \im\rho_F$, where we use the subscript $_F$ to denote maps on $F$-points, for instance, $\rho_F\colon G^\ssc(F)\to G(F)$. Therefore, \begin{align}\tag{1}\label{1} G(F)^\der=\langle\im \varkappa_F\rangle\subseteq\langle\im\rho_F\rangle=\rho\bigl(G^\ssc(F)\bigr). \end{align}

Moreover, we have $$\widetilde\varkappa\circ(\rho\times\rho)=\varkappa^\ssc\colon\ G^\ssc\times G^\ssc\ \to\ G\times G\ \to\ G^\ssc$$ and $$\rho\circ\widetilde\varkappa\circ(\rho\times \rho)=\rho\circ\varkappa^\ssc\colon\, G^\ssc\times G^\ssc\to G\times G\to G^\ssc\to G.$$ We see that \begin{align*} \im\varkappa_F=\im(\rho\circ\widetilde\varkappa)_F\supseteq\im&\bigl(\rho\circ\widetilde\varkappa\circ(\rho\times \rho)\bigr)_F\\ &=\im(\rho\circ\varkappa^\ssc)_F=\rho(\im \varkappa^\ssc_F), \end{align*} whence \begin{align}\tag{2}\label{2} G(F)^\der=\langle\im\varkappa_F\rangle\supseteq\rho\langle\im\varkappa^\ssc_F\rangle= \rho \bigl(G^\ssc(F)^\der\bigr). \end{align}

From \eqref{1} and \eqref{2} we obtain that \begin{align}\tag{3}\label{3} \rho\bigl(G^\ssc(F)^\der\bigr)\subseteq G(F)^\der\subseteq \rho\bigl(G^\ssc(F)\bigr). \end{align}

Now assume that $G^\ssc(F)$ is perfect. Thus means that $$G^\ssc(F)^\der=G^\ssc(F),\quad\text{whence}\quad \rho\bigl(G^\ssc(F)^\der\bigr)=\rho\bigl(G^\ssc(F)\bigr),$$ and we obtain from \eqref{3} that $G(F)^\der=\rho\bigl(G^\ssc(F)\bigr)$. Since $G^\ssc(F)$ is perfect, so is $\rho\bigl(G^\ssc(F)\bigr)$. Thus $G(F)^\der$ is perfect, as required.