Let $G$ be a group with generators and relations.

I know that in general it is difficult to determine what a group is from its generators and relations. I am interested in learning about techniques for figuring out the order of a group from the given information.

For example, I know that if the number of generators exceeds the number of relations then the group has infinite order. If the number of generators equals the number of relations then the group is cyclic or has infinite order.

Let $G= <x, y|x^2 = y^3 = (xy)^4 = 1>$. My hunch is that G has finite order because $(xy)^4$ is somehow independent of $x^2$ and $y^3$. But if the exponent on $xy$ were bigger, say $(xy)^6=1$ that relation becomes redundant.

My question is: is this sort of thinking correct? Furthermore: my method will only tell me if $G$, or its modification, is finite (or infinite). If $G$ is finite how can I figure out the order of the group? I know that the orders divide the order of the group, but I am looking for a specific number.

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    $\begingroup$ It's not just difficult in general; it's actually impossible, in the sense that there does not exist an algorithm that given two presentations of two groups decides if they are isomorphic. One can't even tell if a presentation determines the trivial group. As for what can be said, are you familiar with the Todd-Coxeter algorithm? en.wikipedia.org/wiki/Todd%E2%80%93Coxeter_algorithm $\endgroup$ – Qiaochu Yuan Oct 13 '10 at 11:10
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    $\begingroup$ Also, (xy)^6 = 1 is not redundant. x and y don't commute. $\endgroup$ – Qiaochu Yuan Oct 13 '10 at 11:14

The group generated by two elements $a,b$ subject only to the relations $a^2=1$, $aba=bab$ has the same number of generators and relations, but is neither cyclic nor infinite (it is the non-abelian group of order 6).

Be careful that a relation like $a^6 = 1$ does not imply $a$ has order 6, only that it has order dividing 6.

There is no practical and general method of determining the order of a group given by a presentation. You may want to look into "quotient methods", where you can determine the order of nice quotients of your group. If that order is small, then you can find a finite presentation for the kernel, and try again. For instance, finding the abelian invariants of a finitely presented group is easy, and if those invariants are small, then finding a presentation of the derived subgroup is easy. Repeat, until you hit the perfect residuum, the largest perfect subgroup. Now apply simple group recognition methods to determine the simple quotients of your group. Find the entire top layer, and then hope beyond hope you can find the kernel. Now repeat the derived subgroup method on the kernel.

You can often use GAP to compute abelian invariants and derived subgroups fairly easily, as long as the index is kept under a thousand or so.


Intuition in these cases is treacherous. Your example $G$ is an von Dyck group, a subgroup of index 2 in the more well-known triangle groups. The general von Dyck group has presentation $$\langle x, y\mid x^m=y^n=(xy)^p=1\rangle$$ and is finite iff $1/m+1/n+1/p>1$.

  • $\begingroup$ And triangle groups are Coxeter groups: en.wikipedia.org/wiki/Coxeter_group . It's easy to find the order of a Coxeter group; either it's on the (well-understood) list of finite Coxeter groups, or it's infinite. $\endgroup$ – Qiaochu Yuan Oct 13 '10 at 9:27
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    $\begingroup$ @Qiaochu, that's an interesting definition of easy :) $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '10 at 13:43

While somewhat "dated" at this point you might want to look at the book of Coxeter and Moser, Generators and Relations, for work in this area.



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