TITLE: The free group $F_2$ has index 12 in  SL(2,$\mathbb{Z}$) 
QUESTION [20 upvotes]: Is there someone who can give me some hints/references to the proof of this fact?

REPLY [4 votes]: May be the discussion is out of date, but I like to present a simple way to show that the subgroup S generated by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, 
$H$ 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$(SL(2,Z_4):\phi(S))=(2*4^2+2*2*2+4*2):4=12$