## Return to Answer

3 1 typo

To elaborate on Qiaochu's answer. The subgroup generated by the two matrices

$$\left[ \begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array} \right]$$ and

$$\left[ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right]$$ is the Sanov subgroup. It consists, by an exercise in Kargapolov-Merzlyakov, of matices matrices of the form $$\left[ \begin{array}{cc} 4k+1 & 2l \\ 2m & 4n+1 \end{array} \right]$$ and det=1. The congruence subgroup $\Gamma(2)$ consists of matrices of the form $$\left[ \begin{array}{cc} 2k+1 & 2l \\ 2m & 2n+1 \end{array} \right]$$ and det=1. Those matrices from $\Gamma(2)$ and not in the Sanov subgroup have the form $$\left[ \begin{array}{cc} 4k+3 & 2l \\ 2m & 4n+3 \end{array} \right].$$ Taking the product of any two such matrices gives us a matrix from the Sanov subgroup. So the Sanov subgroup has index 2 in $\Gamma(2)$, and index 12 in $SL_2(\mathbb Z)$.

2 added 18 characters in body; added 5 characters in body

To elaborate on Qiaochu's answer. The subgroup generated by the two matrices

$\left[\begin{array}{cc}$\left[ \begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}\right]$end{array} \right]$$and \left[ \left[ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right] right]$$ is the Sanov subgroup. It consists, by an exercise in Kargapolov-Merzlyakov, of matices of the form$\left[ $\left[ \begin{array}{cc} 4k+1 & 2l \\ 2m & 4n+1 \end{array} \right]$ right]$$and det=1. The congruence subgroup \Gamma(2) consists of matrices of the form \left[ \left[ \begin{array}{cc} 2k+1 & 2l \\ 2m & 2n+1 \end{array} \right] right]$$ and det=1. Those matrices from $\Gamma(2)$ and not in the Sanov subgroup have the form $\left[$\left[ \begin{array}{cc} 4k+3 & 2l \\ 2m & 4n+3 \end{array} \right]$. right].$$Taking the product of any two such matrices gives us a matrix from the Sanov subgroup. So the Sanov subgroup has index 2 in$\Gamma(2)$, and index 12 in$SL_2(\mathbb Z)$. 1 To elaborate on Qiaochu's answer. The subgroup generated by the two matrices$\left[\begin{array}{cc} 1 & 2 \ 0 & 1 \end{array}\right]$and$\left[ \begin{array}{cc} 1 & 0 \ 2 & 1 \end{array} \right]$is the Sanov subgroup. It consists, by an exercise in Kargapolov-Merzlyakov, of matices of the form$\left[ \begin{array}{cc} 4k+1 & 2l \ 2m & 4n+1 \end{array} \right]$and det=1. The congruence subgroup$\Gamma(2)$consists of matrices of the form$\left[ \begin{array}{cc} 2k+1 & 2l \ 2m & 2n+1 \end{array} \right]$and det=1. Those matrices from$\Gamma(2)$and not in the Sanov subgroup have the form$\left[ \begin{array}{cc} 4k+3 & 2l \ 2m & 4n+3 \end{array} \right]$. Taking the product of any two such matrices gives us a matrix from the Sanov subgroup. So the Sanov subgroup has index 2 in$\Gamma(2)$, and index 12 in$SL_2(\mathbb Z)\$.