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Let $G = \operatorname{SL}_2(\mathbb{Z})/\{\pm I_2\}$. Can anyone help me in proving that $T =\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right] / \{\pm I_2\}$ and $S = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] / \{\pm I_2\}$ generates $G$.

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  • $\begingroup$ What is $I_{2}$ $\endgroup$ – user9413 Apr 28 '12 at 19:38
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    $\begingroup$ This group is called the the modular group, denoted $PSL_2(\mathbb{Z})$ Induct on the sum of the absolute values of the entries. $\endgroup$ – Brett Frankel Apr 28 '12 at 19:39
  • $\begingroup$ @Chandrasekhar I'm pretty sure $I_2$ is the $2 \times 2$ identity matrix. $\endgroup$ – Brett Frankel Apr 28 '12 at 19:40
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    $\begingroup$ @Brett: I thought they would be using the notation $I$ for that :) $\endgroup$ – user9413 Apr 28 '12 at 19:41
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    $\begingroup$ I changed \{{\pm {I_{2}}}\I_2\} to \{\pm I_2\}. All these extra curly braces that serve no purpose set a bad example for anyone trying to learn $\TeX$ or $\LaTeX$ by looking at these postings. Braces are needed for things like I_{24}, but writing {{{ {x}^{{2}} }}} instead of x^2 obfuscates things. $\endgroup$ – Michael Hardy Apr 28 '12 at 19:58

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