7
$\begingroup$

The answer to the original question is no, see JSE's answer!

Are the $\Gamma(N)$ the only normal congruence subgroups of $\mathrm{PSL}_2(\mathbb{Z})$ with no finite subgroups (elliptic elements)? What about the normal subgroups of $\Gamma(4)$, which does not contain torsion elements? Here, $\gamma \in \Gamma(N)$, if $\gamma \cong 1 \mod N$.

$\newcommand\o{\mathfrak o}$I state the question in the local picture for general $G$: Let $\o$ be a local ring and $G \subset GL(n)$ a group. Then it makes sense to speak about $G( p^r)$, which consists of matrices which are congruent to $1$ modulo $p^r$, where $p$ is the maximal ideal in $\o$. From $\o \rightarrow \o/p^r$, we get an exact sequence $$ G(p^r) \rightarrow G(\o) \rightarrow G( \o / p^r).$$ Is it surjective? I guess the $G(p^r)$ form a basis of neighborhoods for $1$ and are normal open compact subgroups. Under which conditions are these all normal subgroups in $G(\o)$ with no finite subgroups?

$\endgroup$
2
  • 1
    $\begingroup$ You need your local ring to be compact in the $p$-adic topology to say that the congruence subgroups are compact. This is not true for local rings in general, but it is true for complete dvrs. $\endgroup$
    – S. Carnahan
    Apr 1, 2011 at 5:27
  • $\begingroup$ I edited to link to JSE's answer while this was on the front page, and on the way thought that it might be acceptable to change $o$ for a local ring to the more visually distinct $\mathfrak o$. I hope that that was OK. If not, then I apologise, and please feel free to change back, which you can do all at once just by changing \newcommand\o{\mathfrak o} to \newcommand\o{o}. $\endgroup$
    – LSpice
    Sep 23, 2023 at 23:00

2 Answers 2

14
$\begingroup$

Almost but not quite. A congruence subgroup has to contain some $\Gamma(N)$. So if it is normal its image in $\text{SL}_2(\mathbb{Z})/\Gamma(N)$ is a normal subgroup of $\text{SL}_2(\mathbb{Z}/N\mathbb{Z})$. That group is almost simple but not quite. A proper normal subgroup need not be trivial; it could be $\pm1$ for instance. Or if $N$ is very small you have a few more choices; for instance, the commutator subgroup of $\text{SL}_2(\mathbb{Z})$ is a finite-index congruence subgroup of (if I recall correctly) index $12$, containing $\Gamma(6)$.

$\endgroup$
3
  • $\begingroup$ Sorry, I meant of course $PSL$ instead of $SL$. What for subgroup of $\Gamma(4)$ instead of $\Gamma(1)$? Do the elipptic elements/finite subgroups give the trouble? $\endgroup$
    – Marc Palm
    Mar 31, 2011 at 15:47
  • $\begingroup$ Please forgive me that I changed the questions accordingly.... $\endgroup$
    – Marc Palm
    Mar 31, 2011 at 16:08
  • 1
    $\begingroup$ Gamma(4) will have new normal subgroups; Gamma(4)/Gamma(8) is (Z/2Z)^3 so you have a whole bunch of normal subgroups in between those two. $\endgroup$
    – JSE
    Mar 31, 2011 at 17:07
6
$\begingroup$

(i) Normal congruence subgroups of PSL(2, Z) have been classified by D. L. McQuillan, in Classification of Normal Congruence Subgroups of the Modular Group, Amer. J. Math. 87 (1965), 285–296. (ii) The only torsion normal subgroups of PSL(2, Z) are the powers subgroups P_n= < x^n : x\in PSL(2,Z)>, where n = 2 or 3.

One may now list all normal congruence subgroups of level N.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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