## Commutator subgroup of GL(n,k)

Posted: January 29, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $SL(n,k)' = SL(n,k).$