MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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 $\gamma \cong 1 \mod N$.

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?

share|cite|improve this question
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. – S. Carnahan Apr 1 '11 at 5:27
up vote 11 down vote accepted

Almost but not quite. A congruence subgroup has to contain some Gamma(N). So if it is normal its image in SL_2(Z) / Gamma(N) is a normal subgroup of SL_2(Z/NZ). That group is almost simple but not quite. A proper normal subgroup need not be trivial; it could be +-1 for instance. Or if N is very small you have a few more choices; for instance, the commutator subgroup of SL_2(Z) is a finite-index congruence subgroup of (if I recall correctly) index 12, containing Gamma(6).

share|cite|improve this answer
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? – Marc Palm Mar 31 '11 at 15:47
Please forgive me that I changed the questions accordingly.... – Marc Palm Mar 31 '11 at 16:08
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. – JSE Mar 31 '11 at 17:07

(i) Normal congruence subgroups of PSL(2, Z) has been classified by D. L. McQuillan, 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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