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This question already has an answer here:

I want to show that $SL_2(R) = [SL_2(R),SL_2(R)]$.

I reduced the problem only show that the matrices $A=\{(1,x),(0,1)\}, B=\{(1,0),(x,1)\}, C=\{(x,0),(0,\frac {1}{x})\}$

are products of commutators.

How can I show this?

Help would be appreciated.

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marked as duplicate by YCor, YuiTo Cheng, user1729 group-theory Jun 5 at 16:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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For any $S\in\mathrm{SL}_2(\mathbb{R})$, we have \begin{equation*} S = \begin{pmatrix} 1&\ \\ a&1 \end{pmatrix} \begin{pmatrix} x&\ \\ \ &x^{-1} \end{pmatrix} \begin{pmatrix} 1&b\\ \ &1 \end{pmatrix} \end{equation*} or \begin{equation*} \begin{aligned} S &= \begin{pmatrix} \ &1\\ 1&\ \end{pmatrix} \begin{pmatrix} 1&\ \\ a&1 \end{pmatrix} \begin{pmatrix} x&\ \\ \ &-x^{-1} \end{pmatrix} \begin{pmatrix} 1&b\\ \ &1 \end{pmatrix}\\ &=\begin{pmatrix} 1&a\\ \ &1 \end{pmatrix} \begin{pmatrix} \ &-1\\ 1&\ \end{pmatrix} \begin{pmatrix} x&\ \\ \ &x^{-1} \end{pmatrix} \begin{pmatrix} 1&b\ \\ \ &1 \end{pmatrix} \end{aligned} \end{equation*} for some $a,b\in \mathbb{R}$ and $x\in \mathbb{R}^*$. Moreover, we have \begin{equation*} \begin{pmatrix} \ &-1\\ 1&\ \end{pmatrix} = \begin{pmatrix} 1&-2\\ \ &1 \end{pmatrix} \begin{pmatrix} 1&1\\ \ &1 \end{pmatrix} \begin{pmatrix} 1&\ \\ 1&1 \end{pmatrix} \begin{pmatrix} 1&-1\\ \ & 1 \end{pmatrix} \end{equation*} and \begin{equation*} \begin{pmatrix} x&\ \\ \ &x^{-1} \end{pmatrix} = \begin{pmatrix} 1&1\\ \ &1 \end{pmatrix} \begin{pmatrix} 1&\ \\ x-1&1 \end{pmatrix} \begin{pmatrix} 1&-x^{-1}\\ \ &1 \end{pmatrix} \begin{pmatrix} 1&\ \\ x-x^2&1 \end{pmatrix}. \end{equation*} Hence $\mathrm{SL}_2({\mathbb{R}})$ is generated by transvections (matrices of form \begin{pmatrix}1&y\\ 0&1\end{pmatrix} or \begin{pmatrix}1&0\\ y&1\end{pmatrix} for some $y\in\mathbb{R}$).

Now it suffices to prove that all transvections are commutators. Note that \begin{equation*} \begin{pmatrix} 1&-x\\ \ &1 \end{pmatrix} \begin{pmatrix} a&\ \\ \ &a^{-1} \end{pmatrix} \begin{pmatrix} 1&x\\ \ &1 \end{pmatrix} \begin{pmatrix} a^{-1}&\ \\ \ &a \end{pmatrix} = \begin{pmatrix} 1&(a^2-1)x\\ \ &1 \end{pmatrix} \end{equation*} for any $x\in\mathbb{R}$ and $a\in\mathbb{R}^*$. This implies every transvection is a commutator (similar for lower triangular transvections). Thus the commutator subgroup is indeed the whole group, because all transvections generate $\mathrm{SL}_2(\mathbb{R})$.

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    $\begingroup$ It looks like this works, thanks! $\endgroup$ – Gabi G Jun 5 at 7:56
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But that is not true, since $SL_2(R)$ is not closed under the bracket. Do you mean the Lie algebra $\mathfrak{sl}_2(R)=\{A \in M_2(R) | tr(A)=0\}$? Or is your first equation wrong?

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