May be the discussion is out of date, but I like to present a simple way to show that the subgroup S generated by $\left(\begin{array}{cc} $ \left(\begin{array}{cc} 1 & 2\cr 0 & 1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$(SL(2,Z_4):\phi(S))=(2*4^2+2*2*2+4*2):4=12$
