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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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    What is $I_{2}$2012-04-28
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    This group is called the the modular group, denoted $PSL_2(\mathbb{Z})$ Induct on the sum of the absolute values of the entries.2012-04-28
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    @Chandrasekhar I'm pretty sure $I_2$ is the $2 \times 2$ identity matrix.2012-04-28
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    @Brett: I thought they would be using the notation $I$ for that :)2012-04-28
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    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.2012-04-28
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    Read here http://math.stackexchange.com/questions/5333/on-the-generators-of-the-modular-group , including links there. Very clear exposition.2012-05-18

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