Derived subgroup of general linear group is special linear group

Statement

Let $k$ be a field and $n$ be a natural number. The derived subgroup (i.e., commutator subgroup) of the general linear group $G L_{n} (k)$ (the group of invertible $n \times n$ matrices) is the special linear group $S L_{n} (k)$ (the group of $n \times n$ matrices of determinant $1$ ), under either of these conditions:

$n \geq 3$ .
$k$ has at least three elements.

In other words, the only case where the result does not hold is when $n = 2$ and $k$ is the field of two elements. (In the case $n = 1$ , the result holds vacuously).

Related facts

Special linear group is perfect: This is not true for all fields, but is always true when $n \geq 3$ , or when the field has more than three elements. Note that this in particular means that the derived series of the general linear group stabilizes at the special linear group.
Lower central series of general linear group stabilizes at special linear group: This is true when $n \geq 3$ , or when the field has more than two elements.

Facts used

Proof

Observe that:

$S L_{n} (k)$ is the kernel of the determinant homomorphism from $G L_{n} (k)$ to the multiplicative group of nonzero elements of $k$ , which is Abelian. Hence, $S L_{n} (k)$ contains the commutator subgroup of $G L_{n} (k)$ .
By facts (1) and (2), $S L_{n} (k)$ is contained in the commutator subgroup of $G L_{n} (k)$ .
Thus, $S L_{n} (k)$ equals the commutator subgroup of $G L_{n} (k)$ .