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. Every elementary matrix is a commutator of invertible matrices
2. 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 $GL_n(k)$.