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?
  • $\begingroup$ your a has a typo $\endgroup$
    – the_lar
    Commented Jun 3 at 14:09

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Jan 27, 2019 at 14:28
  • $\begingroup$ . . . nice! . . $\endgroup$
    – janmarqz
    Commented Jan 27, 2019 at 20:57

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