# Proving a cyclically presented group is infinite using GAP.

I apologise if this is too broad or is otherwise off topic.

## The Problem:

I'm looking for ways to prove that a cyclically presented group, given in terms of generators and relators, is infinite using GAP.

In particular, I'm looking for ways to prove that $H(r, n, s)$ is infinite using GAP for certain $r,n,s\in\Bbb N$. I'm looking at $1\le s<r\le10$ and $2\le n\le10$.

The group $H(6,8,3)$ is infinite for example, whereas $\lvert H(10,7,3)\rvert=9,997,813$. The order of the group $H(4,8,3)$ is unknown.

## What I have so far.

Let G be the cyclically presented group.

• There's Size(G);, of course, but that can be inconclusive if G is sufficiently large. It does sometimes say that G is infinite.

• There's AbelianInvariants(DerivedSubgroup(G)); (and taking AbelianInvariants of the derived subgroup of the derived subgroup of G and so on); if this has 0 in its output it means that $\Bbb Z$ is a component of the abelianisation of G, so that G is infinite. This is also inconclusive if G is sufficiently large, but so far, it seems to handle more than Size(G);.

The group $I=$$I(r,n,s)$ is a degree $n$ extension of $H(r,n,s)$, so applying the above to $I$ can be effective.

• Sure, @ahulpke: I'm looking at $1\le s<r\le 10$ and $2\le n\le 10$. The group $H(6, 8, 3)$ is infinite for example, whereas $\lvert H(10, 7, 3)\rvert=9,997,813$. – Shaun Oct 24 '17 at 8:02