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 $$GL_n(k)$$ (the group of invertible $$n \times n$$ matrices) is the special linear group $$SL_n(k)$$ (the group of $$n \times n$$ matrices of determinant $$1$$), under either of these conditions:


 * $$n \ge 3$$.
 * $$k$$ 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 $$n \ge 3$$, 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 $$n \ge 3$$, or when the field has more than two elements.

Facts used

 * 1) uses::Every elementary matrix is a commutator of invertible matrices
 * 2) uses::Elementary matrices generate the special linear group

Proof
Observe that:


 * $$SL_n(k)$$ is the kernel of the determinant homomorphism from $$GL_n(k)$$ to the multiplicative group of nonzero elements of $$k$$, which is Abelian. Hence, $$SL_n(k)$$ contains the commutator subgroup of $$GL_n(k)$$.
 * By facts (1) and (2), $$SL_n(k)$$ is contained in the commutator subgroup of $$GL_n(k)$$.
 * Thus, $$SL_n(k)$$ equals the commutator subgroup of $$GL_n(k)$$.