Shoe the group $SL_2(R)$ is can be created by commutators [duplicate]

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.

marked as duplicate by YCor, YuiTo Cheng, user1729 group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); 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.

2 Answers

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 $$\mathrm{SL}_2(\mathbb{R})$$.

• It looks like this works, thanks! – Gabi G Jun 5 at 7:56

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?