(Non-trivial) presentation of general linear and symplectic group over Z/mZ?

Question

I know that there exists a nice presentation (generators and relations) of the general linear group over a finite field (by Steinberg, I think). Is there also a nice presentation of $GL(n,\mathbb{Z}/m\mathbb{Z})$ for an arbitrary integer $m$? And perhaps also for the symplectic group over $\mathbb{Z}/m\mathbb{Z}$?

I want to do some calculations in these groups with a computer and my first problem was to determine these groups. One solution (my current) is of course to determine both by brute force. But already for small m,n this takes too much time.

Answer 1

Sure, you can certainly work something out if you like. It may well be that an off-the-rack answer to your question is in the literature somewhere, but let me say how I would go about getting there from here.

Note that $\textrm{GL}(n,\mathbb{Z}/m\mathbb{Z})$ is the product over all prime powers $p^a$ exactly dividing $m$ of $\textrm{GL}(n,\mathbb{Z}/p^a\mathbb{Z})$, so you're reduced to the case where $m$ is a prime power.

Second, there is an exact sequence $1 \to U_1 \to \textrm{GL}(n,\mathbb{Z}/p^a\mathbb{Z}) \to \textrm{GL}(n,\mathbb{Z}/p\mathbb{Z}) \to 1$ where I'll use $U_i$ to denote all mod $p^a$ matrices that are congruent to the identity modulo $p^i$. Given the Steinberg generators for $\textrm{GL}(n,\mathbb{Z}/p\mathbb{Z})$, you're reduced to giving generators/relations for $U_1$ (as well as working out how lifts of the Steinberg generators act by conjugation on the generators for $U_1$, as well as what elements of $U_1$ you get by applying the Steinberg relations to the lifts of your generators).

Note that $U_1$ is a big $p$-group, and each $U_i/U_{i+1}$ up to $i=a-1$ is just isomorphic to $\text{M}_n(\mathbb{Z}/p\mathbb{Z})$, so again you can produce generators and relations by devissage. In practice it shouldn't require so many generators as you get out of this. For instance for $n=2$, the group $U_1$ is generated by four elements: the upper-triangular unipotent matrix with $p$ in the upper-right, its transpose, and diagonal matrices with diagonal entries $c,1$ and $1,c$, where $c$ is a generator for the multiplicative group $1 + p\mathbb{Z}/p^a\mathbb{Z}$.

Answer 2

Two things. I'll talk about SL_n rather than GL_n, but that's just a technicality.

1. There is a short exact sequence

$1 \rightarrow SL_n(\mathbb{Z},m) \rightarrow SL_n(\mathbb{Z}) \rightarrow SL_n(\mathbb{Z}/m\mathbb{Z}) \rightarrow 1,$

where $SL_n(\mathbb{Z},m)$ is the level $m$ congruence subgroup of $SL_n(\mathbb{Z})$. There is a nice presentation $\langle S|R \rangle$ for $SL_n(\mathbb{Z})$ due essentially to Magnus; it can be found in Milnor's book on algebraic K-theory (look for the chapter that calculates $K_2(\mathbb{Z})$). The generating set $S$ is just the set of elementary matrices. To get from there to $SL_n(\mathbb{Z}/m\mathbb{Z})$, you just need to add a normal generator for $SL_n(\mathbb{Z},m)$ to $R$. By the solution to the congruence subgroup problem for $SL_n(\mathbb{Z})$ (due to Mennicke, but you are better off looking at Bass-Milnor-Serre's paper), this congruence subgroup is normally generated by the mth power of a single elementary matrix.

A similar idea also works for the symplectic group. To find a presentation for $Sp_{2g}(\mathbb{Z})$, look at Theorem 9.2.13 of Hahn and O'Meara's book "The Classical Groups and K-Theory".

1. I also want to make a comment on DLS's answer. There is a related short exact sequence

$1 \rightarrow V \rightarrow SL_n(\mathbb{Z}/p^{k+1}\mathbb{Z}) \rightarrow SL_n(\mathbb{Z}/p^k\mathbb{Z}) \rightarrow 1.$

The group $V$ has a beautiful description, due essentially to Lee and Szczarba. Namely, it is isomorphic to the additive group underlying the special lie algebra over $\mathbb{Z} / p\mathbb{Z}$ (in particular, it is abelian). Moreover, the action of $SL_n(\mathbb{Z}/p^k\mathbb{Z})$ on $V$ factors through the adjoint representation of $SL_n(\mathbb{Z}/p\mathbb{Z})$ on the special lie algebra.

Lee and Szczarba did not write this down in quite this form, but I wrote out a similar result for the symplectic group in Lemma 3.1 of my paper "The Picard group of the moduli space of curves with level structures".