Free subgroups of $PSL(2,\mathbb{Z})$ of index 6

There are two "natural" subgroups of $$PSL(2,\mathbb{Z})\cong C_2\ast C_3$$ of index 6. One is the congruence subgroup $$\Gamma_0(2)$$ which is the kernel of the map $$PSL(2,\mathbb{Z})\to PSL(2,\mathbb{Z}/2\mathbb{Z})$$. The other subgroup $$H$$ is the kernel of the map $$C_2\ast C_3\to C_2\times C_3$$. Here are two similarities between these two subgroups:

• Both $$\Gamma_0(2)$$ and $$H$$ are subgroups of $$PSL(2,\mathbb{Z})$$ of index 6.
• Both $$\Gamma_0(2)$$ and $$H$$ are free groups of rank 2.

However, $$PSL(2,\mathbb{Z}/2\mathbb{Z})\cong S_3$$ and $$C_2\times C_3\cong C_6$$ so these are different subgroups of $$PSL(2,\mathbb{Z})$$. Moreover, $$\Gamma_0(2)$$ is freely generated by the matrices $$\begin{bmatrix}1&2\\0&1\end{bmatrix}$$ and $$\begin{bmatrix}1&0\\2&1\end{bmatrix}$$ whereas $$H$$ is freely generated by the matrices $$\begin{bmatrix}2&1\\1&1\end{bmatrix}$$ and $$\begin{bmatrix}1&1\\1&2\end{bmatrix}$$. This last statement can be seen by noting that if $$a=\begin{bmatrix}0&-1\\0&1\end{bmatrix}$$ generates $$C_2$$ and $$b=\begin{bmatrix}-1&-1\\1&0\end{bmatrix}$$ generates $$C_3$$ then $$H$$ is freely generated by $$abab^2$$ and $$ab^2ab$$.

What is going on here? More precisely,

• Are these two subgroups the largest free subgroups of $$PSL(2,\mathbb{Z})$$?
• Are there any other free subgroups of $$PSL(2,\mathbb{Z})$$ of index 6?
• Is there any reason to expect that $$PSL(2,\mathbb{Z})$$ contains two free subgroups of rank 2 and index 6 with different quotients?

Let $$G = \langle x,y \mid x^2,y^3 \rangle \cong {\rm PSL}(2,\mathbb{Z})$$.

Question 1. Yes. Let $$H < G$$, and consider the permutation action of $$G$$ on the (left or right) cosets of $$H$$ in $$G$$. If $$|G:H| < 6$$, then it is not possible for the images of both $$x$$ and $$y$$ to act fixed point freely, and so $$H$$ contains a conjugate of $$x$$ or $$y$$ and hence cannot be free.

Question 2. No, but the two subgroups that you have found are the only two normal subgroups of index $$6$$ in $$G$$. You can see this by observing that there is essentially only one surjective group homomorphism from $$G$$ to each of $$C_6$$ and $$S_3$$ (i.e. up to equivalence under an automorphism of $$C_6$$ or $$S_3$$), so there are only two possible kernels $$H$$.

By a computer calculation, I found that there is also one conjugacy class of non-normal subgroups $$H$$ with $$|G:H| = 6$$ and with $$H$$ free of rank $$2$$, and a representative of this class is $$H=\langle yx, y^{-1}(xy)^3 \rangle$$. The quotient of $$G$$ by the core of $$H$$ is isomorphic to $$S_4$$, and there are three conjugates of $$H$$ in $$G$$.

Question 3. I can only say here that the reason is that we have a proof that this is the case!

Note that the Kurosh Subgroup Theorem says that any subgroup of $$G$$ is a free product of conjugates of $$\langle x \rangle$$, $$\langle y \rangle$$ and a free subgroup of $$G$$. So, for $$H \lhd G$$, if $$H$$ does not contain $$x$$ or $$y$$, then it must be free. I believe that the rank of free subgroups of free products can be calculated using Euler Characteristics, but I don't know the details.

• Very nice. With regards to your comment on Euler characteristics, the answers to mathoverflow.net/questions/43726/… suggest (using Euler Characteristics or tools from Serre's book on Trees) that a free subgroup of $PSL(2,\mathbb{Z})$ of index 6 necessarily has rank 2. – Thomas Browning Jan 27 at 14:28
• . . . nice! . . – janmarqz Jan 27 at 20:57