# 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).

## 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 $GL_n(k)$.