We will keep the notation in here and here . For a group and we let Recall that the commutator subgroup of is the subgroup generated by the set

**Remark 1**. Using the second part of Problem 1, it is easy to show that

We are now ready to prove that the commutator subgroup of the general linear group is the special linear group unless and has at most elements.

**Problem**. Prove that unless and

**Solution**. Clearly for any we have So and hence In order to show that we only need to prove that contains all elementary matrices because, by Remark 2, is generated by elementary matrices. So we consider two cases.

*Case 1* . : For any choose Then, by the above remark,

*Case 2* . and : The equation has at most three solutions in the field and so, since has more than three elements, we can choose a non-zero element such that Thus is invertible in Now given let and Let

A quick calculation shows that

and

**Remark 2**. In the solution of the above problem, we actually showed that every elementary matrix is in the form for some Thus