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May be the discussion is out of date, but I like to present a simple way to show that the subgroup S generated by $\left(\begin{array}{cc}$ \left(\begin{array}{cc} 1 & 2\cr 0 & 1 \end{array}\right)$end{array}\right) $$and \left(\begin{array}{cc} \left(\begin{array}{cc} 1 & 0\cr 2 & 1 \end{array}\right) end{array}\right)$$ has index 12 in$SL(2,Z)$. First of all Mark Sapir noted that by Kargopolov S is a group of matrices $$\left(\begin{array}{cc} 1+4k_1 & 2n_1\cr 2n_2 & 1+4k_2 \end{array}\right)$$ Lemma. Let$G$be a group,$H H$be a subgroup of$G$,$N$be a normal subgroup of$G$, that is subgroup of$H$. Then ind(G:H)=ind(G/N:H/N). Proof. Indeed, let$x\not\in H$. Then$xN\cap HN\subseteq XH\cap H=\emptyset$. So the natural homomorphism$G\to G/N$sends different classes to different classes. Now consider the natural homomorphism$\phi:SL(2,Z)\to SL(2,Z_4)$.$G=SL(2,Z)$,$H=S$and Ker$(\phi)=N$satisfy the Lemma. So, ind$(SL(2,Z):S)$=ind$(SL(2,Z_4):\phi(S))=(2*4^2+2*2*2+4*2):4=12$1 May be the discussion is out of date, but I like to present a simple way to show that the subgroup S generated by$\left(\begin{array}{cc} 1 & 2\cr 0 & 1 \end{array}\right)$and$\left(\begin{array}{cc} 1 & 0\cr 2 & 1 \end{array}\right)$has index 12 in$SL(2,Z)$. First of all Mark Sapir noted that by Kargopolov S is a group of matrices $$\left(\begin{array}{cc} 1+4k_1 & 2n_1\cr 2n_2 & 1+4k_2 \end{array}\right)$$ Lemma. Let$G$be a group,$H
Now consider the natural homomorphism $\phi:SL(2,Z)\to SL(2,Z_4)$. $G=SL(2,Z)$, $H=S$ and Ker$(\phi)=N$ satisfy the Lemma. So, ind$(SL(2,Z):S)$=ind$(SL(2,Z_4):\phi(S))=(2*4^2+2*2*2+4*2):4=12$