Is there someone who can give me some hints/references to the proof of this fact?



The free group $F_2$ has index 12 in  SL(2,$\mathbb{Z}$) 
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 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 $SL\_2(\mathbb Z)$.


