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Does there exist generators $S$ and $T$ for the modular group $\Gamma=PSL(2,\mathbb{Z})$ with the following property:

$$S+S^{-1}+T+T^{-1}=0$$

Here is a candidate:

$$S=\left[\begin{array}{cc} -1 & 0 \\ 1 & -1 \\ \end{array}\right], \,T=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}\right]$$

Just not sure if these two generate $\Gamma$.

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Yes, your choice for $S$ and $T$ generate the modular group $\Gamma=PSL_2(\mathbb{Z})$. Other choices would also work.

The standard choice of generators is $$ \Gamma = \left \langle \left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right], \left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right] \right \rangle.$$

You already have $T=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right]$. Notice that $TST=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] = - \left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$. Since we can obtain the standard generators for $\Gamma$ as products of $S$, $T$, and their inverses, we conclude that $S$ and $T$ generate $\Gamma$.


Just to clarify for anyone not familiar with all the matrix groups, $PSL_2(\mathbb{Z})=SL_2(\mathbb{Z})/\left\{1,-1\right\}$ is the group of 2x2 matrices over the integers with determinant 1, modulo the subgroup of scalar transformations, which in this case is just $\{1,-1\}$. That's why it doesn't matter if I obtain $\left[ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right]$ or $\left[ \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix} \right]$; they represent the same element in $PSL_2(\mathbb{Z})$.

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    Ah, $TST$ does it. Thanks!2012-05-13
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    Also, $\left( TST \right)^{2}=-I_{2}$, so $S$ and $T$ generate the full $SL(2,\mathbb{Z})$.2012-05-15
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    @JacksonWalters: That's true, but you actually get the identity for free whenever you construct a group from generators. You even have the identity if you use the empty set as a generating set, since the empty product is the identity. Another way of looking at it is $S^0 = T^0 = I_2$.2012-05-16
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    No, no. I understand that $I_{2}=-I_{2}$ in $PSL(2,\mathbb{Z})$. I was trying to make the additional point that since $S$ and $T$ generate every $2 \times 2$ matrix with determinant 1 and integer entries modulo a sign, and since they also generate $-I_{2}$, you can actually get any matrix with determinant 1 and integer entries.2012-05-19
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    @JacksonWalters: OK, I understand what you mean now. That's nice to know. I suppose any choice of (representatives of) generators for $PSL(2,\mathbb{Z})$ will be generators of $SL(2,\mathbb{Z})$, since $(\pm \left[ \begin{smallmatrix} 0 & 1 \\-1 & 0 \end{smallmatrix} \right])^2 = -I_2$.2012-05-19