I don't seem to have seen any explicit generators for the principal congruence subgroups of $SL(n, \mathbb{Z}),$ for $n>2,$ although it is known (Sury+Venkataramana) is that the number of generators is bounded independently of the modulus. It is certainly possible to write a program to generate the generators (since the subgroup is given as a kernel of a map to a finite group, if we fix a generating set, the kernel is generated by the loops in the induced Cayley graph of the image group), but this seems painful (and will probably produce really awful generating sets without much further work).

For the self-proclaimed obtuse The question is: is there a known way to produce, given $n, p,$ a generating set of the level $p$ principal congruence subgroup in $SL(n, \mathbb{Z}).$ Particularly a nice, small generating set.

EDIT The Sury and Venkataramana paper is actually available for free. and having overcome my laziness somewhat, I see that they do actually construct a generating set using an idea somewhat related to @Andy's, but the generating set is quite large. Since $SL(n, \mathbb{Z})$ itself can be generated by three elements, one is tempted to make a rash conjecture that the minimal size of a generating set of $\Gamma_n(p)$ (with the obvious notation) is bounded independently of $n$ and $p.$ BUT (EDIT), as pointed out by @Andy Putman in his answer and his comment, this would be rash indeed, since the rank of the abelianization of $\Gamma_n(p)$ is $n^2-1.$

  • $\begingroup$ Why on earth did someone vote to close? This is a perfectly good question... $\endgroup$ Sep 25 '11 at 21:35
  • $\begingroup$ By the way, you might be interested in the following question I asked a while ago : mathoverflow.net/questions/2757/… $\endgroup$ Sep 25 '11 at 22:15
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    $\begingroup$ @Andy: I think someone had put a close vote, because previously the question did not have a "for the self-proclaimed obtuse" response that actually states what the question is! $\endgroup$
    – Suvrit
    Sep 26 '11 at 7:27
  • $\begingroup$ As I discussed in my answer, the rank of the abelianization of $\Gamma_n(p)$ is $n^2-1$, so any generating set must have size at least $n^2-1$. $\endgroup$ Sep 26 '11 at 14:00
  • $\begingroup$ @Andy: Oops, did not process that part. $\endgroup$
    – Igor Rivin
    Sep 26 '11 at 14:10

I'm pretty sure that no such explicit generating set exists in the literature. However, at some point I spent some time thinking about this and I have a conjectural generating set.

Normal generators. Let $\Gamma_n$ denote $SL_n(\mathbb{Z})$ and let $\Gamma_n(p)$ denote the level $p$ congruence subgroup of $\Gamma_n$. For $1 \leq i,j \leq n$ such that $i \neq j$, let $e_{ij}$ be the $(i,j)$-elementary matrix. Define $S = \{e_{ij}\ |\ 1 \leq i,j \leq n, i \neq j\}$ and $S(p) = \{e_{ij}^p\ |\ 1 \leq i,j \leq n, i \neq j\}$. Clearly we have $S(p) \subset \Gamma_n(p)$. Moreover, the proof of the congruence subgroup theorem (by Mennicke and Serre) shows that $\Gamma_n(p)$ is the normal closure in $\Gamma_n$ of the subgroup generated by $S(p)$.

Abelianization of congruence subgroup. However, $S(p)$ does not generate $\Gamma_n(p)$. This follows from the computation of the abelianization of $\Gamma_n(p)$ by Lee and Szczarba. Briefly, they defined a homomorphism $$\phi : \Gamma_n(p) \rightarrow \mathfrak{sl}_n(\mathbb{Z}/p).$$ Here $\mathfrak{sl}_n(\mathbb{Z}/p)$ is the abelian group of $n \times n$ matrices of trace $0$ with entries in $\mathbb{Z}/p$. The definition of $\phi$ is as follows. An element of $\Gamma_n(p)$ is of the form $1 + p A$ for some matrix $A$. Define $\phi(1+pA) = A$ modulo $p$. It is easy to see that the trace of $A$ mod $p$ is $0$. Moreover, this map is a homomorphism due to the identity $$(1+pA)(1+pB) = 1+p(A+B)+p^2 AB.$$ Lee and Szczarba proved two things about $\phi$.

  1. They proved that $\phi$ is surjective. I'll say more about this below.
  2. They proved that the kernel of $\phi$ (which, by the way, is clearly $\Gamma_n(p^2)$) is the whole commutator subgroup of $\Gamma_n(p)$ for $n \geq 3$.

This implies that the abelianization of $\Gamma_n(p)$ is $\mathfrak{sl}_n(\mathbb{Z}/p)$.

They only normally generate. Now, above I claimed that $S(p)$ does not generate $\Gamma_n(p)$. If $G$ is the subgroup generated by $S(p)$, it is easy to see that $\phi(G)$ is exactly the group of matrices whose diagonals are $0$. The problem thus is that you don't get the diagonal matrices.

Hitting the diagonal. We thus need generators that hit the diagonal matrices. For $1 \leq i < n$, let $f_i$ be the result of inserting the $2 \times 2$ matrix $\left( \begin{array}{ll} 1+p & -p \\ p & 1-p\\ \end{array} \right)$ into the $n \times n$ identity matrix with the upper left at position $(i,i)$. Define $T = \{f_i\ |\ 1 \leq i < n\}$. I conjecture that $S(p) \cup T$ generates $\Gamma_n(p)$. Here is my evidence for this.

  1. Let $H$ be the subgroup generated by $S(p) \cup T$. Then $\phi(H)$ is all of $\mathfrak{sl}_n(\mathbb{Z}/p)$.

  2. I can prove this by hand for $p=2$ and $p=3$.

Conjectural proof sketch. My proof for $p=2$ and $p=3$ should work in general, so let me briefly describe it. By the congruence subgroup property business discussed above, the normal closure in $\Gamma_n$ of $H$ is $\Gamma_n(p)$. To prove that $H = \Gamma_n(p)$, therefore, it is enough to prove that $H$ is normal. To do this, it is enough prove the following. Let $e_{ij} \in S$ be an elementary matrix (a generator for $\Gamma_n$). Then for $s \in S(p) \cup T$, it is enough to prove that $e_{ij} s e_{ij}^{-1}$ and $e_{ij}^{-1} s e_{ij}$ can be written as words in $S(p) \cup T$. For $p=2$ and $p=3$, this is fairly easy, but I haven't managed to do the computations for higher $p$.


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