Ways to guarantee that a group presentation defines a finite group when randomly generating group presentations

Question

What are some ways to guarantee that a group presentation defines a finite group?

Or, equivalently, is there a convenient sufficient condition for a generators-and-relations presentation of a group defining a finite group that's useful for randomly generating such groups.

If there are no known convenient sufficient conditions, I am okay with a condition that gives me a finite group with high probability.

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I'm interested in ways of generating random finite groups.

I can think of two obvious approaches:

1. generating random instances of some other structure like a graph or a magma and looking at its automorphisms.
2. taking a set of generators and imposing some random relations on it.

I like the generators and relations presentation of groups, since the artifact that it gives me back is easy to compute with.

One problem though is that it seems to hard to tell in general (maybe undecidable?) whether the defined group is finite or not.

For example, $\langle a, b : a^2 = b^2 = 1 \rangle$ is not a finite group because $a \cdot b$ has infinite order.

I can think of a very heavy-handed approach that might allow me to insist that the group in question is finite ... and that is using the equivalent of an axiom schema to set an upper bound (in the sense of the divisibility relation) on the order of each element.

$\langle a, b : \xi^4 = 1 \;\text{and}\; a^2 = b^2 = 1 \rangle$ (where $\xi^4 = 1$ is a relation schema) defines a finite group.

In order to prove this we can write an algorithm for taking any product of generators of length $\ge 8$.

Proof

First, suppose the product is $^{-1}$-free and has length $8$, then the product contains $a \cdot a$, contains $b \cdot b$, or consists of the same segment repeated four times. A rewrite to a strictly shorter product is thus possible.

If the product does contain $^{-1}$, we first normalize it so that only generators are inverted, then we use the rewrites $a^{-1} = a$ and $b^{-1} = b$.

End of Proof

This answer which references the Grigorchuk group, though, makes me think that just insisting that my group is finitely generated and every element has finite order isn't enough.

So, I guess, to sum up the question:

• Is there a nice sufficient condition for group finiteness that I can impose on a generators-and-relations presentation of a group? (Keeping in mind that the intended application is randomly generating small finite groups)
• Are there any conditions that will help me generate a finite group with high probability? (I'm being deliberately vague about the distribution that the relations are drawn from so as not to prematurely rule out a potential result)

Answer 1

Here's the situation I'm referring to (which you can sort of generalize to more than $2$ generators and other "anti-commutativity" relations) :

Say we have $\langle x,y\mid x^n, y^m, yxy^{-2}x^{-3}\rangle $.

Now, whatever group this is, one thing we know is that any element can be written in the form $x^ay^b$, for $0\le a\le n-1,0\le b\le m-1$. Thus, there are at most $n\cdot m$ elements. In particular, $G$ is finite.