Show ${\rm SL}(n,q)$ is perfect for odd prime power $q>3$. This is Exercise 3.2.10 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to this search and Approach0, it is new to MSE.
The Details:
On page 8, ibid.,

Let $R$ be a commutative ring with identity. Define ${\rm SL}(n, R)$ to be the set of all $n\times n$ matrices over $R$ with determinant equal to $1$. [It is a group under matrix multiplication, called the special linear group of degree $n$ over $R$.]

On page 74, ibid.,

In case $R={\rm GF}(q)$, [we denote ${\rm SL}(n, R)$ by] $${\rm SL}(n, q).$$

On page 28, ibid., the definition of a derived subgroup $G'$ of a group $G$ is the subgroup generated by all commutators $[x,y]=x^{-1}y^{-1}xy$ of $G$.
To explain the title, I use the

Definition: A group $G$ is perfect if $G=G'.$

The Question:

Let $q$ be an odd prime power greater than $3$. Prove that ${\rm SL}(2,q)$ equals its derived subgroup.

Thoughts:
Let $q=p^\ell>3$ for an odd prime $p$.
My first thought was to try out small values of $q$, like $q=7$; here is a GAP experiment:
gap> G:=SL(2,7);
SL(2,7)
gap> Size(G);
336
gap> DG:=DerivedSubgroup(G);
<group of 2x2 matrices over GF(7)>
gap> Size(DG);
336

But nothing came of such considerations, since the groups are seemingly too big for a naïve approach. The next $q$ value is $3^2=9$: the order of ${\rm SL}(2, 9)$ is $720$ according to GAP. The next $q$ value after that is $11$: the order of ${\rm SL}(2, 11)$ is $1320$ according to GAP.
I don't know what I was hoping to see with using GAP.

My linear algebra is a bit rusty.

I guess I should move on like so . . .
Let $S={\rm SL}(2,q)$. Then
$$S=\left\{\begin{pmatrix} a & b \\ c & d\end{pmatrix}\,\middle|\, a,b,c,d\in GF(q), ad-bc=1\right\}.$$
Let $A=\begin{pmatrix} a & b \\ c & d\end{pmatrix}, B=\begin{pmatrix} x & y \\ z & t\end{pmatrix}\in S$. Then $\det(A)=\det(B)=1$. Consider $[A, B]$, a generator of $S'$. We have $[A,B]\in S$ by definition, so, like I used in the GAP code above, we may conclude that $S'\subseteq S.$
What I need to do is show that an arbitrary $\Xi=\begin{pmatrix} \alpha & \beta \\ \gamma & \delta\end{pmatrix}\in S$ is an element of $S'$. This is where I'm stuck.

It is well-known that
$$S'=\displaystyle\bigcap_{T\le S,\\ X\subseteq T}T,$$
where $X=\{[U,V]\in S\mid U,V\in S\}.$ I'm not sure how this helps.

Please help  :)
 A: Let $q>3$. Then there exists a nonzero element $a\in \Bbb F_q$ with $a^2-1\neq 0$. Let $A=\begin{pmatrix} 1 & -x \cr 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} a & 0 \cr 0 & a^{-1} \end{pmatrix}$ in $SL(2,q)$. Then
\begin{align*}
[A,B] & = ABA^{-1}B^{-1} \\
      & =\begin{pmatrix} 1 & (a^2-1)x \cr 0 & 1 \end{pmatrix}
\end{align*}
So every element of the corresponding transvection group
is a commutator. Since the transvections generate
$SL(2,q)$, this is enough to show that $SL(2,q)$ is perfect.
A: I might approach it via Bruhat decomposition.  Let $B$ be the group of upper triangular matrices in $G$.  It is the semidirect product of the group $T$ of diagonal matrices in $G$ and the group $N$ of upper triangular matrices with $1$s on the diagonal.  First, show that every element of $B$ is a product of commutators in $G$.
Next, use the fact that $G$ is the disjoint union of the double cosets $BwB$, where $w$ ranges through a set of permutation matrices in $G$.  This is called the Bruhat decomposition.  For example, $G = \operatorname{SL}_2$ is the disjoint union of $B = B1B$ and $BwB$, where
$$w = \begin{pmatrix} & 1 \\ -1 \end{pmatrix}.$$
Thus you just have to show that $B$ and every such $w$ is a product of commutators.
