Question

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

Answer 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 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 $SL_2(\mathbb Z)$.

Answer 2

There have already been a lot of answers to this question and I'm not going to go over the details of a proof that it is true. But I want to say something about the real strength of any of these standard arguments. The arguments generally have one of three starting points: (1) That $\text{PSL}(2,\mathbb{Z})$ is isomorphic to $C_2 * C_3$, a free product of cyclic groups; (2) The group action of $\text{SL}(2,\mathbb{Z})$ on the hyperbolic plane or upper half plane; or (3) A group action of $\text{SL}(2,\mathbb{Z})$ on trees. The first starting point arguably dodges the question, because obtaining that the whole group is a free product is closely related to finding a free subgroup, The other two starting points are "geometric". But geometry is ultimately a human description of some other, more formal mode of reasoning. So what are the geometric arguments really saying?

It is important to realize that the statement that a group $G$ is free is ultimately a uniqueness statement for words in $G$ in its claimed free generators. More precisely, if $G$ is freely generated by $a$ and $b$, that says that every element of $G$ has unique reduced word in $a$ and $b$ and their inverses. In the case of this subgroup of $\text{SL}(2,\mathbb{Z})$ or $\text{PSL}(2,\mathbb{Z})$, the uniqueness property ultimately comes from and generalizes another, more familiar uniqueness property: The uniqueness of the Euclidean algorithm with positive remainders.

Suppose that you have two coprime positive integers $a$ and $b$ and you are allowed to subtract $a$ from $b$, or $b$ from $a$, but you are not allowed to create negative integers. Then there is a unique path from $(a,b)$ to $(1,1)$. The same statement in reverse, in linear algebra form, says that there is a unique word in the matrices $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$ which, when applied to the vector $(1,1)$, gives you the vector $(a,b)$. Thus, these two matrices generate a free semigroup.

The uniqueness property is established by induction, and the obvious fact that if $a < b$ then you can only subtract $a$ from $b$, and vice versa. Obvious though it may be, this type of fact is what lets you build the tree on which $\text{SL}(2,\mathbb{Z})$ acts. Or, in the hyperbolic geometry version of the argument, the inequality $a < b$ is associated with the sides of the ideal polygons that eventually yield the same tree. In the upper half plane model, the rational numbers $a/b$ with $a < b$ are a ray in the real numbers, and their convex hull in the hyperbolic plane is a half hyperbolic plane whose boundary is the geodesic connecting $\infty$ to 1. Uniqueness says that the orbit of this geodesic under the above semigroup consists of disjoint geodesics.

Answer 3

Well $SL_2(\mathbb{Z})\cong \mathbb{Z}/6*_{\mathbb{Z}/2} \mathbb{Z}/4$. $SL_2(\mathbb{Z})$ acts on a tree $T$. Then you can consider its abelianization. It is $\mathbb{Z}/12$. So the commutator subgroup $\Gamma$ has index 12. It is free, as it acts freely on the same tree $T$. See for example Serre, trees for details.

EDIT: One can also investigae the rank of the free group. Thatfor one has to consider the quotient graphs $SL_2(\mathbb{Z})\backslash T$ and $\Gamma\backslash T$. The first Graph is a interval, where the stabilizers of representatives of one endpoint are isomorphic to $\mathbb{Z}/6$ (resp. $\mathbb{Z}/4$ for the other endpoint). The stabilizer of a representative of the inverval is $\mathbb{Z}/2$.

Now $\Gamma$ acts freely on $T$ and $[SL_2(\mathbb{Z}):\Gamma]=12$. So the first endpoint (which is a $SL_2(\mathbb{Z})$-orbit) consists of $12/6=2$ $\Gamma$-orbits and the second endpoint consists of $12/4=3$ $\Gamma$-orbits. Analogously the edge consists of $6$ $\Gamma$-orbits.

So the Euler Characteristic of $\Gamma\backslash T$ is $2+3-6=-1$, so this graph is homotopy equivalent to a wedge of $2$ circles and hence its fundamental group (which is isomorphic to $\Gamma$ by Serre, trees or covering theory) is $F_2$.

Answer 4

Start with the formula for $SL_2(\mathbb Z)$ in HenrikRueping's answer. Divide by the central $\mathbb Z/2$ to conclude that $PSL_2(\mathbb Z)$ is $\mathbb Z/3*\mathbb Z/2$. The kernel of abelianization for $SL_2(\mathbb Z)$ is the same as the kernel of abelianization for $PSL_2(\mathbb Z)$. The kernel of $G*H \to G\times H$ is always a free group with basis $ghg^{-1}h^{-1}$, $g\ne 1$, $h\ne 1$.

Answer 5

Maybe this works and maybe it doesn't: the index of the congruence subgroup $\Gamma(2)$ in $\text{SL}_2(\mathbb{Z})$ is $|\text{SL}_2(\mathbb{F}_2)| = 6$, and $\Gamma(2)$ contains a free subgroup of index $2$ generated by $\left[ \begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array} \right]$ and $\left[ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right]$, which follows from the fact that its action on the hyperbolic plane is free.

Answer 6

Thanks for the question!; I asked this myself several times, without definite answer. But now I got it. We know from Henrik Rüpings answer that the kernel $K$ of $Sl_2 (Z) \to Z/12$ is a free group. Thus it remains to determine the rank. There is a notion of Euler characteristic for groups (it has rational values), see Brown's book, cohomology of groups. Theorem IX 6.3 (or better IX 7.3 b) of that book states that $\chi(Sl_2 (Z))= \chi (Z/12) \chi(K)$ in our case. The Euler number of $Z/12$ is $\frac{1}{12}$, the Euler number of $Sl_2 (Z)$ is $-\frac{1}{12}$, so the Euler number of $K$ is $-1$. But the Euler number of $F_n$ is $1-n$, so $n=2$, and you are done.

I would like to see a more elementary answer, though; one should be able to see this using the action of the upper half plane; then glue $12$ fundamental domains together in the correct way and see that the result is $C \setminus {0,1}$. But I never managed to do this properly.

Answer 7

I'm late to the party, but here's how I think about this fact. The approach I will present is perhaps slightly higher tech than some, but has the advantage that it's a completely general way of computing a free finite-index subgroup of a virtually free group.

To keep things simple, I'll do $PSL_2(\mathbb{Z})$, and then explain how to do $SL_2(\mathbb{Z})$ at the end.

As you presumably know, $\Gamma=PSL_2(\mathbb{Z})\cong \mathbb{Z}/2*\mathbb{Z}/3$. You should think of it as the fundamental group of a graph of groups $G$, with one edge and two vertices, one of which, $u$, is labelled by $\mathbb{Z}/2$ and one, $v$, labelled by $\mathbb{Z}/3$.

Just as in the case of ordinary topology, any (conjugacy class of a) subgroup $H\subseteq\Gamma$ corresponds to a covering map $\widehat{G}\to G$. (This is a trivial consequence of Bass--Serre theory: take $\widehat{G}=T/H$ where $T$ is the Bass--Serre tree.) Coverings of graphs of groups are just like coverings of graphs, except that sometimes a neighbourhood looks like a quotient by a non-free action of a group, in which case you have to remember point stabilisers (cf. orbifolds). The subgroup $H$ is free if and only if every vertex of $\widehat{G}$ is labelled by the trivial group, in which case we can think of $\widehat{G}$ as a genuine graph and $H$ as its fundamental group.

To prove that $PSL_2(\mathbb{Z})$ has a free subgroup of index $6$, we now just have to construct a covering map $\widehat{G}\to G$ of degree $6$, where $\widehat{G}$ is a genuine graph. But this is easy. To do this, take three vertices $u_1,u_2,u_3$ of valence two and two vertices $v_1,v_2$ of valence three. Now glue them together however you like, such that each edge adjoins one $u_i$ and one $v_j$. (I think there are exactly two ways of doing this, up to permuting the $i$'s and $j$'s.) The covering map $\widehat{G}\to G$ sends $u_i\mapsto u$ and $v_j\mapsto v$, and each edge goes to the unique edge of $G$. Whichever way you do the gluing, the graph $\widehat{G}$ has rank two, so the subgroup is isomorphic to $F_2$.

It's also easy to see that these subgroups are the minimal-index torsion-free subgroups: any covering of $G$ by a genuine graph $\widehat{G}$ must have even degree, so that the vertices above $u$ have trivial label; and it must have degree divisible by three, so that the vertices above $v$ have trivial label.

Exactly the same argument can be made to work for $SL_2(\mathbb{Z})$, which you of course should think of as $\mathbb{Z}/4*_{\mathbb{Z}/2}\mathbb{Z}/6$. Indeed, if I have it right, these techniques should prove the following theorem, which is presumably standard, although I don't know a reference. (Warning: I haven't thought about this very hard.)

Theorem: If $\Gamma=A*_CB$ for $A,B$ finite, then a minimal-index free subgroup is of index $d=\mathrm{lcm}(|A|,|B|)$, and of rank

$1-d\left(\frac{1}{|A|}+\frac{1}{|B|}-\frac{1}{|C|}\right)$.

Note that this technique is powerful enough to let you compute all the conjugacy classes of all the subgroups of a given index, if you want. (Update: though there are some subtleties if you really want to do this; see below.)

(At the moment what I'm seeing doesn't quite square with some of the other answers/comments: I get one normal free subgroup of $PSL_2(\mathbb{Z})$ of index six, and one conjugacy class of non-normal free subgroups of index six, corresponding to a regular covering of $G$ and an irregular covering, respectively; this seems at odds with Mark's remark about different maps to $\mathbb{Z}/12$. But it's late, so I'm not going to figure out what's going on now.)

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Update

I had forgotten an important subtlety. A covering map of graphs of groups $\widehat{G}\to G$ includes an extra piece of information, viz:

For each vertex $\hat{v}$ of $\widehat{G}$ with image $v$ in $G$, and for each edge $e$ incident at $v$, a covering map specifies a bijection between the set of edges of $\widehat{G}$ incident at $\hat{v}$ that map to $e$ and the set of double cosets

$\widehat{G}_{\hat{v}}\backslash G_v/G_e$

(where $G_v$ is the group labelling $v$, etc). As such bijections always exist, this doesn't arise if you want to prove the existence of a subgroup. But the correct notion of equivalence for covering maps takes these bijections into account, so two different covering maps $\widehat{G}\to G$ may have the same underlying graph-map. This happens for the commutator subgroup and the Sanov subgroup: in both cases, the underlying graph is the first subdivision of the theta graph.

Answer 8

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$(SL(2,Z_4):\phi(S))=(2*4^2+2*2*2+4*2):4=12$

Answer 9

A remark on the original question: it follows from either the results of Gilman/Miasnikov/Osin or of Richard Aoun (for linear groups) that the subgroup generated by a random (in an appropriate sense) pair of elements in $SL(2, \mathbb{Z})$ is free. It can also be shown that such a subgroup will (with very high probability) be of infinite index in the modular group.