Is the derived group of the G(F) perfect 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.
 A: $\newcommand{\ssc}{\text{sc}}
\newcommand{\der}{\text{der}}
\DeclareMathOperator{\im}{im}$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.
