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Here are 3 basic observations regarding $SL_2(\mathbb{Z})$:

  1. The abelianization of $SL_2(\mathbb{Z})$ is isomorphic to $\mathbb{Z}/12\mathbb{Z}$ and so $SL_2(\mathbb{Z})$ has a finite index subgroup for each $d|12$.

  2. We have a surjection $SL_2(\mathbb{Z})\rightarrow PSL_2(\mathbb{Z})\simeq C_2*C_3$. In 1901, Miller proved that for $n\geq 9$, $A_n$ is generated by a pair of elements of order $2$ and $3$. Since $[A_n:A_{n-1}]=n$ it follows that for $n\geq 9$, $SL_2(\mathbb{Z})$ has a subgroup of index $n$.

  3. We have a surjective map $\pi:SL_2(\mathbb{Z})\rightarrow PSL_2(\mathbb{F}_5)\simeq A_5$ and since $[A_5:A_4]=5$ this gives a finite index subgroup of index $5$ in $SL_2(\mathbb{Z})$

What about the indices 7 and 8 ?

We do have a surjective map $\pi:SL_2(\mathbb{Z})\rightarrow PSL_2(\mathbb{F}_7)=:G$ where $|G|=7\cdot 24$. However, the $7$-Sylow subgroup of $G$ cannot have a group complement otherwise $G$ would be solvable...no luck

added: The reasoning above regarding $PSL_2(\mathbb{F}_7)$ is wrong: $24$ is divisible by two distinct primes... After checking, $PSL_2(\mathbb{F}_7)$ does contain a subgroup of order 24 and therefore $7$ is a realized index! For an explicit description of such a subgroup under the description $GL_3(\mathbb{F}_2)\simeq PSL_2(\mathbb{F}_7)$ see link

Original Question: Do there exist subgroups of index $7$ and $8$ in $SL_2(\mathbb{Z})$ ?

Updated Question: Does there exist a subgroup of index $8$ in $SL_2(\mathbb{Z})$ ?

The answer is YES as was pointed out by @kabenyuk in the comments below. The normalizer of a 7-Sylow sugroup of $PSL_2(F_7)$ is a group of order 21. Therefore $8$ is a realized index as well. So all integers $n\geq 1$ are realized indices of $SL_2(\mathbb{Z})$.

Hugo Chapdelaine
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  • Not at a computer to double-check, but $(12)(34)$ and $(135)(467)$ might generate $A_7$? Their product is a 7-cycle at least, and their commutator is of shape $(abcd)(ef)$. – Steve D Jul 13 '24 at 13:10
  • The group PSL$(2,7)$ has a subgroup of order $24$ and a subgroup of order $21$. – kabenyuk Jul 13 '24 at 13:13
  • Sorry, the generators I gave give $PSL(2,7)$. But changing $(12)(34)$ to $(12)(34)(56)$ gives all of $S_7$< which gives the index $7$ subgroup. – Steve D Jul 13 '24 at 14:01
  • Thanks for pointing that out and my apologies for being too careless in my writing of the original post. – Hugo Chapdelaine Jul 13 '24 at 14:18
  • @HugoChapdelaine As I have already noted in my first comment the group PSL$(2,7)$ contains a subgroup of order $21$. You can easily find it. – kabenyuk Jul 13 '24 at 14:26
  • The original question has been above. I wanted to try and get all the $A_n$ as quotients of $SL(2,\mathbb{Z})$. The case $n<4$ is trivial, and for $n\ge4$ $A_n$ is centerless, so we can focus on surjections from $PSL(2,\mathbb{Z})$. So we need generators of orders $2$ and $3$.

    Miller's paper handles $n\ge9$. We also have \begin{align} A_4 &=\langle (12)(34), (123)\rangle\ A_4 &= \langle (12)(34), (135)\rangle \end{align}

     – Steve D Jul 13 '24 at 15:13
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    $A_6$ doesn't work: you can only get subgroups that look like $S_3$, $S_4$, $A_4$, or $A_5$.

    $A_7$ only adds $C_6$ and $PSL(2,7)$ to this list.

    $A_8$ adds some direct products, as well as $GL(2,4)$.

     – Steve D Jul 13 '24 at 15:13
  • Thanks @Steve D. This is a nice complement to the original question. – Hugo Chapdelaine Jul 13 '24 at 15:20

1 Answers1

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Actually we can give a straightforward uniform argument that $\Gamma \cong PSL_2(\mathbb{Z})$, and hence $SL_2(\mathbb{Z})$, has a subgroup of index $n$ for every positive integer $n$. The sequence of counts of such subgroups is A005133 on the OEIS, and the sequence of counts of conjugacy classes of subgroups is A121350 on the OEIS.

Since $\Gamma \cong C_2 \ast C_3$, the problem is equivalent to writing down a permutation $a$ of order $2$ and a permutation $b$ of order $3$ which generate a group acting transitively on a set of size $n$ (and note that in this approach it is not necessary to know anything about what group this is). These can be visualized concretely in terms of their Schreier graphs, which are graphs on $n$ vertices containing two types of edges, namely

  • undirected edges describing the action of $a$ (which switches the two vertices any $a$-edges connect), and
  • directed $3$-cycles of edges describing the action of $b$ (which cyclically permute the three vertices these $b$-cycles connect).

The resulting group action is transitive iff this graph is connected, and the only rule for how to construct them is that each vertex can be connected to at most one $a$-edge and contained in at most one $b$-cycle. Now it's not hard to construct such a graph for every $n$ by induction on $n$, possibly with a little casework depending on the value of $n \bmod 3$ or at worst $n \bmod 6$. Here are some explicit examples of permutations for $n = 7, 8$:

  • For $n = 7$ we can take $a = (12)(34)(56), b = (135)(467)$.
  • For $n = 8$ we can take $a = (12)(34)(56)(78), b = (135)(468)$.

And here are pictures of their Schreier graphs, drawn in quiver, with $a$-edges in red and $b$-cycles in blue:

enter image description here

enter image description here

For more discussion of this see my blog posts Finite index subgroups of the modular group and Drawing subgroups of the modular group. The first post also explains how to write down a generating function counting the number of subgroups of index $n$, while the second presents a Poincare dual visualization of the subgroups coming from some version of Bass-Serre theory, or some version of the theory of dessins, which even lets you read off a presentation of the subgroup. The idea is to think of $B \Gamma$ as a wedge sum of $B C_2$ and $B C_3$ and to try to visualize its covering spaces as such wedge sums.

Qiaochu Yuan
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  • Thanks a lot @Qiaochu. My feeling of awkwardness about my naive question is fully rewarded by this wonderful and magic generating series identity which begins your blog! – Hugo Chapdelaine Jul 13 '24 at 20:15
  • @Hugo: yes, it's one of my favorite generating function identities! I've only ever seen it buried in an exercise in Stanley's Enumerative Combinatorics (Part II, I think) but it really deserves to be better-known. – Qiaochu Yuan Jul 13 '24 at 20:22
  • The "source" looks great. I Would not be surprised if there are other gold nuggets buried in Stanley two-volume set. – Hugo Chapdelaine Jul 13 '24 at 20:47