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Let $\Gamma(N)$ be the principal congruence subgroup of level $N$ in $SL_n(\mathbf{Z})$, 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 a prime $p$ which 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 $SL_n(\mathbf{Z})$, where $n\geq 3$, is orderable. Is there a more direct / hands-on proof?

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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.

THM : If $n \geq 3$ and $\Gamma(N)$ is the level $N$ principal congruence subgroup of $SL_n(\mathbb{Z})$, 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.

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.

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No -- since $SL_n(\mathbb{Z})$ has congruence subgroup property the profinte completionh of $\Gamma(N)$ is the same as the congruence subgroup of level $N$ in $SL_n(\widehat{ \mathbb{Z}})$. Since $SL_n(\mathbb{Z}_q)$ does not have any quotient which is a $p$ group. One sees that $\Gamma(N)$ does not have $p$ quotients unless $p$ divides $N$.

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