5
$\begingroup$

Does every index $p$ subgroup of $SL(2,\mathbb{Z}_p)$ contain the principal congruence subgroup $\Gamma(p)$?

Equivalently, must it be the preimage of an index $p$ subgroup of $SL(2,\mathbb{Z}/p\mathbb{Z})$?

$\endgroup$
6
  • $\begingroup$ It was known to Galois that ${\rm PSL}(2,\mathbb{Z}/p\mathbb{Z})$ has a subgroup of index $p$ if and only if $p \leq 11$. This turns the problem into a finite computation. $\endgroup$ Nov 12 '18 at 14:00
  • $\begingroup$ @JeremyRouse Your remark seems to reduce the problem to checking that $SL(2,\mathbb{Z}_p)$ doesn't have an index $p$ subgroup for $p > 11$. I can't see how that's a finite computation... Also, do you have a reference for the nonexistence of index $p$ subgroups for $p > 11$? $\endgroup$ Nov 13 '18 at 5:29
  • $\begingroup$ Sorry. I don't think my first comment was useful. I can give you a reference though. Serre proves that for $p \geq 5$, there is no proper closed subgroup of $SL(2,\mathbb{Z}_{p})$ that surjects onto $SL(2,\mathbb{Z}/p\mathbb{Z})$. (The reference is to Lemma 3 of Section 3.4 of Chapter IV of the book "Abelian $\ell$-adic representations and elliptic curves".) From this, one gets a positive answer to your question for $p \geq 5$, and a computation is necessary for $p = 2$ and $p = 3$. $\endgroup$ Nov 13 '18 at 13:03
  • $\begingroup$ Curiously enough, this conversation thread (between @JeremyRouse and OP) took place long after my answer was posted. Perhaps reading the answer would have been more useful. For those interested in mathematics rather than references, it might be useful to point out that Serre's argument (even if he doesn't use this language) consists of two main points. First, one exploits the irreducibility of the adjoint representation (which is exactly what I use, and holds for $p \ge 3$), and second... $\endgroup$ Nov 13 '18 at 23:29
  • $\begingroup$ (continued) the non-existence of any section $\mathrm{SL}_2(\mathbf{Z}/p \mathbf{Z}) \rightarrow \mathrm{SL}_2(\mathbf{Z}/p^2 \mathbf{Z})$, which holds for $p \ge 5$). The latter is not relevant for the question at hand, since the existence of a section plus irreducibility of the adjoint would lead to a subgroup of index $p^3$. $\endgroup$ Nov 13 '18 at 23:30
9
$\begingroup$

Yes.

Let $H$ be the subgroup, and let $N$ be the normal closure. The index of $N$ in $\Gamma = \mathrm{SL}(2,\mathbb{Z}_p)$ has index dividing $p!$ which is not divisible by $p^2$. Hence either:

  1. $N$ contains the principal congruence subgroup $\Gamma(p)$ and you win,
  2. $N \cap \Gamma(p)$ has index $p$ inside $\Gamma(p)$.

If $p > 2$, one can now apply the following observations:

  1. The abelianization of $\Gamma(p)$ is $V = \Gamma(p)/\Gamma(p^2)$. (It's easy to write down topological generators of $\Gamma(p^2)$ using commutators when $p > 2$.)

  2. As a module for $\Gamma/\Gamma(p) = \mathrm{SL}(2,\mathbb{F}_p)$ under conjugation, $V$ is the adjoint representation and is irreducible.

This gives a contradiction, since, if $N$ is normal, then $\Gamma(p)/(N \cap \Gamma(p))$ will be a proper quotient of $V$ under the action of $\mathrm{SL}(2,\mathbb{F}_p)$.

If $p = 2$, then you can work a little harder with explicit computations and use a similar argument (now the abelianization of $\Gamma(2)$ is something close to $\Gamma(2)/\Gamma(8)$) or instead simply make the stupid observation that $\mathrm{SL}(2,\mathbf{Z})$ is dense in $\Gamma$ and thus there is an injection from subgroups of $\Gamma$ (of any index) to subgroups of $\mathrm{SL}(2,\mathbf{Z})$ of the same index (not a bijection because of the failure of the congruence subgroup property). However, the latter is well known to have abelianization $\mathbf{Z}/12 \mathbf{Z}$ and so has a unique index $2$ subgroup.

$\endgroup$
5
  • $\begingroup$ Firstly, let me apologize for not responding sooner. I was working on two projects A and B. I got stuck on project A, which prompted me to ask this question, at which point I switched to project B. When I saw your answer, I liked the argument and upvoted, but due to my shameful lack of familiarity with Lie theory, I knew it would take significant effort to understand the argument, which is why I opted instead to make a little more progress on project B before I committed time to understanding your argument. I responded to Jeremy's comments because it was easy for me. $\endgroup$ Nov 14 '18 at 20:11
  • $\begingroup$ I've just spent the last hour thinking about your argument and trying to recall my meagre knowledge of Lie theory (which mostly come from a self-study of Serre's Harvard lecture notes). Unfortunately, I'm still confused on both of your points (1) and (2). For (1) - You say it's easy to write down topological generators of $\Gamma(p^2)$ using commutators when $p > 2$. How do you do this? (and show that they actually generate $\Gamma(p^2)$?) For (2) - the way I see it, since $\Gamma(p)$ is a standard subgroup of $SL(2,\mathbb{Z}_p)$ constructed from giving the unit disc a group law from the ... $\endgroup$ Nov 14 '18 at 20:21
  • $\begingroup$ ...formal group law on $SL(2,\mathbb{Z}_p)$, we know that $\Gamma(p)/\Gamma(p^2)$ is abelian, and is isomorphic to the additive group $\mathbb{F}_p^3$, and hence the conjugation action of $SL(2,\mathbb{Z}_p)$ on $\Gamma(p)$ descends to an action of $SL(2,\mathbb{F}_p)$ on $\Gamma(p)/\Gamma(p^2)$. Since $\Gamma(p)$ is basically the Lie algebra of $SL(2,\mathbb{Z}_p)$, the former action is (essentially by definition) the adjoint representation of $SL(2,\mathbb{Z}_p)$. Unfortunately, I don't see why this implies that $\Gamma(p)/\Gamma(p^2)$ is the adjoint representation of $SL(2,\mathbb{F}_p)$... $\endgroup$ Nov 14 '18 at 20:27
  • $\begingroup$ ... Of course it makes sense to me that $V$ should be the adjoint representation of $SL(2,\mathbb{F}_p)$, but I wouldn't be able to write a proof down if I had to. I suspect the passage between the adjoint representation of $SL(2,\mathbb{Z}_p)$ and $SL(2,\mathbb{F}_p)$ is best expressed through the language of algebraic groups, which I will try to study now (probably via Milne's book for lack of a better idea), but this will take some time, and I do have an unrelated deadline on Friday which I need to meet. This is my reason for interacting with Jeremy Rouse's comments before yours. $\endgroup$ Nov 14 '18 at 20:35
  • $\begingroup$ By the way, I like your name. $\endgroup$ Nov 14 '18 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.