Generating matrices for $SL_2(\mathbb{Z})/\{\pm I_2\}$

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 $G$.

• What is $I_{2}$ – user9413 Apr 28 '12 at 19:38
• This group is called the the modular group, denoted $PSL_2(\mathbb{Z})$ Induct on the sum of the absolute values of the entries. – Brett Frankel Apr 28 '12 at 19:39
• @Chandrasekhar I'm pretty sure $I_2$ is the $2 \times 2$ identity matrix. – Brett Frankel Apr 28 '12 at 19:40
• @Brett: I thought they would be using the notation $I$ for that :) – user9413 Apr 28 '12 at 19:41
• 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. – Michael Hardy Apr 28 '12 at 19:58