7
$\begingroup$

Let $\Gamma(N)$ be the principal congruence subgroup of level $N$ in $\mathrm{SL}_n(\mathbf{Z})$, where $n\geq 3$. Then $\Gamma(N)$ is residually $p$-finite for all primes $p$ dividing $N$.

Can $\Gamma(N)$ be residually $p$-finite for any prime $p$ that does not divide $N$ ?

On a related note: $\Gamma(N)$ is residually $p$-finite for only finitely many primes $p$. The proof I know is somewhat indirect: 1) (Rhemtulla) if a group is residually $p$-finite for infinitely many primes $p$, then it is orderable. 2) (Witte) no finite index subgroup of $\mathrm{SL}_n(\mathbf{Z})$, where $n\geq 3$, is orderable. Is there a more direct / hands-on proof?

$\endgroup$
1
  • $\begingroup$ Since it's not said explicitly, let me emphasize that both answers show in particular that that if $p$ does not divide $N$ then $\Gamma(N)$ does not admit the cyclic group $C_p$ as a quotient (and a fortiori is not residually-$p$). $\endgroup$
    – YCor
    Feb 16, 2020 at 18:09

2 Answers 2

12
$\begingroup$

No. In fact, I claim that if $G$ is any solvable group and $\phi : \Gamma(N) \rightarrow G$ is a surjection, then $G$ is a finite group and all primes that divide $|G|$ also divide $N$. The key is the following beautiful theorem of Lee and Szczarba.

Theorem: If $n \geq 3$ and $\Gamma(N)$ is the level $N$ principal congruence subgroup of $\mathrm{SL}_n(\mathbb{Z})$, then $[\Gamma(N),\Gamma(N)] = \Gamma(N^2)$.

See their paper

MR0422498 (54 #10485) Lee, Ronnie; Szczarba, R. H. On the homology and cohomology of congruence subgroups. Invent. Math. 33 (1976), no. 1, 15–53. DOI link

Anyway, this implies that the derived series of $\Gamma(N)$ is $$\Gamma(N) > \Gamma(N^2) > \Gamma(N^4) > \cdots.$$ Any surjection to a solvable group thus contains $\Gamma(N^{2^k})$ in its kernel for some $k$. But it also follows from Lee-Szczarba's work that $\Gamma(M)/\Gamma(M^2)$ is an abelian group all of whose elements have order $M$. This implies that all the primes which divide the order of $\Gamma(N) / \Gamma(N^{2^k})$ also divide $N$. The desired result follows.

$\endgroup$
10
$\begingroup$

No – since $\mathrm{SL}_n(\mathbb{Z})$ has the congruence subgroup property the profinite completion of $\Gamma(N)$ is the same as the congruence subgroup of level $N$ in $\mathrm{SL}_n(\widehat{ \mathbb{Z}})$. Since $\mathrm{SL}_n(\mathbb{Z}_q)$ does not have any quotient which is a $p$-group, one sees that $\Gamma(N)$ does not have $p$-quotients unless $p$ divides $N$.

$\endgroup$

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.