TITLE: redundant relations
QUESTION [10 upvotes]: The situation is the following: I have a group of matrices (given by generators), and I am trying to find a presentation. Now, this is in general very hard (undecidable?) but here I want to know if people know heuristics for pruning the set of relators -- in other words, by computing further generations of the group, I get a number of words $w_1, \dots, w_k$ which represent the identity. No doubt some of these relations are consequences of others, so is there some nice way to remove the "dependent relations"?

REPLY [11 votes]: This doesn't answer the main question, but it does address the parenthetical question of the decidability of computing presentations in matrix groups. In so doing, I hope it helps to clarify Joel and Benjamin's answers.
Bridson and I showed that the problem of computing presentations in matrix groups over $\mathbb{Z}$ is undecidable, in this paper.  More precisely, we produce a recursive sequence of subsets $S_n\subseteq SL_{m_n}(\mathbb{Z})$ and an r.e. sequence of integers $r_n$ such that:

for each $n$, $\langle S_n\rangle$ is finitely presentable (in principle!), but
the set
$\{n\in\mathbb{N}\mid b_1(\langle S_n\rangle)=r_n\}$
is r.e. but not recursive.

So $b_1(\langle S_n\rangle)$ (the first Betti number) is not computable. Since it's straightforward to compute $b_1$ from a presentation, it follows that presentations are not computable.
Unfortunately, in our examples, $m_n\to\infty$ as $n\to\infty$. It would be great to have examples in which $m_n$ was bounded (and even better to have examples in which $m_n=3$).

REPLY [7 votes]: Let's ignore the source of the group, and concentrate on the question you seem to be asking. Let $G = \langle x_1,x_2,\ldots,x_r \mid r_1,r_2,\ldots,r_s \rangle$ be a finite group presentation, where the $r_i$ are defining relators. Can we decide whether $r_s$ is redundant; that is, whether $G = \langle x_1,x_2,\ldots,x_r \mid r_1,r_2,\ldots,r_{s-1} \rangle$.
As others have pointed out, the theoretical answer is no, but the problem is a semi-decidable probelm in the sense that, if the answer is yes, then you can prove constructively that it is yes. In practice there are two well studied and implemented algorithms that you can use to try and do this.
The first is Todd-Coxeter coset enumeration, which is most useful when the groups are finite, so that is probably not the best choice here.
The second is Knuth-Bendix completion, and that is the most likely to be useful here. You run it on the presentation $\langle x_1,x_2,\ldots,x_r \mid r_1,r_2,\ldots,r_{s-1} \rangle$. It will probably not halt, but that doesn't matter. Just run it for a few minutes or a few days, or whatever. Then interrupt it (or wait until it exceeds some bound - I usually tell it stop when it has produced a million rewrite rules or something). Then you can quickly check whether $r_s$ reduces to the identity under the rewrite rules that you have generated. If $r_s$ really is a consequence of the other relators, then you know that this method will work if you run the program for long eneough but, except in the rare cases when the process completes with a confluent presentation, you cannot use this method to prove the answer is no.
The only practical ways of showing that $r_s$ is not a consequence of the other relators is to compute some of the quotients of $G$ and see if you get different results by leaving out $r_s$. There are algorithms available for computing finite quotients, virtually abelian quotients, virtually nilpotent quoteints, etc.
Knuth-Bendix completion and the quotient algorithms are available in both GAP and Magma.