# Residual $p$-finiteness of principal congruence subgroups

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?

• 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-$p$). – YCor Feb 16 at 18:09

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 $$[\Gamma(N),\Gamma(N)] = \Gamma(N^2)$$.
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.
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 $$N$$.