We will keep the notation in here and here . For a group G and a,b \in G we let [a,b]=aba^{-1}b^{-1}. Recall that G', the commutator subgroup of G, is the subgroup generated by the set \{[a,b]: \ a,b \in G \}.

Remark 1. Using the second part of Problem 1, it is easy to show that E_{ij}(\alpha \beta)=[E_{ir}(\alpha), E_{rj}(\beta)].

We are now ready to prove that the commutator subgroup of the general linear group GL(n,k) is the special linear group SL(n,k) unless n=2 and k has at most 3 elements.

Problem. Prove that GL(n,k)'=SL(n,k) unless n=2 and |k| \leq 3.

Solution. Clearly for any A,B \in GL(n,k) we have \det (ABA^{-1}B^{-1})=1. So [A,B] \in SL(n,k) and hence GL(n,k)' \subseteq SL(n,k). In order to show that SL(n,k) \subseteq GL(n,k)', we only need to prove that GL(n,k)' contains all elementary matrices because, by Remark 2, SL(n,k) is generated by elementary matrices. So we consider two cases.

Case 1 . n \geq 3 : For any i \neq j choose r \notin \{i,j \}. Then, by the above remark, E_{ij}(\alpha)=[E_{ir}(\alpha), E_{rj}(1)].

Case 2 .  n=2 and |k| > 3 : The equation x(x^2-1)=0 has at most three solutions in the field k and so, since k has more than three elements, we can choose a non-zero element \gamma \in k such that \gamma^2 \neq 1. Thus \gamma^2 - 1 is invertible in k. Now given \alpha \in k, let \beta_1=\alpha (\gamma^2 - 1)^{-1} and \beta_2 = \alpha \gamma^2 (1-\gamma^2)^{-1}. Let

A = \begin{pmatrix} \gamma & 0 \\ 0 & \gamma^{-1} \end{pmatrix}.

A quick calculation shows that

E_{12}(\alpha)=[A, E_{12}(\beta_1)] and E_{21}(\alpha)=[A, E_{21}(\beta_2)]. \Box

Remark 2. In the solution of the above problem, we actually showed that every elementary matrix is in the form [a,b] for some a,b \in SL(n,k). Thus SL(n,k)' = SL(n,k).


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