# Infinitely presented group where every finite sub-presentation is virtually free

Does there exist a finitely generated non-finitely presentable group $$G=\langle S \mid R\rangle$$ with $$S$$ finite, where we can enumerate $$R=\{r_1,r_2,\ldots\}$$ such that for every finite subset $$I\subset \mathbb{N}$$, the group $$G_I=\langle S \mid \{r_i\mid i\in I\}\rangle$$ is virtually free?

One potential candidate is the lamplighter $$\mathbb{Z}_2\wr\mathbb{Z}$$ with the presentation $$\langle a,t \mid a^2, [a,t^kat^{-k}] \textrm{ for all } k\in\mathbb{Z}\rangle.$$

(edit) Let me ask a more specific question related to lamplighters. Is it true that if $$G_I=\langle a,t \mid a^2, [a,t^kat^{-k}] \textrm{ for all } k\in I\rangle$$ is virtually free for some finite set $$I\subset\mathbb N$$, then for every large enough $$k$$, $$G_{I\cup\{k\}}$$ is also virtually free?

• If you assume $S$ to be finite, you should say it. – YCor Feb 20 '19 at 19:18
• @YCor I have to confess that I have been guilty of writing "Let $G = \langle X \rangle$ be a finitely generated group'' without explicitly saying that $X$ is assumed to be finite. – Derek Holt Feb 20 '19 at 19:21
• The truncated presentations $\Gamma_J=\langle a,t\mid a^2,[a,t^kat^{-k}]:k\in J\rangle$ are HNN extensions of finite groups, hence virtually free, when $J$ is a segment (i.e. equal to $\{1,\dots,n\}$ -- we can ignore those relators for $k\le 0$ since they are redundant). – YCor Feb 20 '19 at 19:24
• @YCor: I think the group $G_{\{1,3\}}$ is not virtually free. The kernel of the map to $\mathbb{Z}$ given by $a\mapsto 0, t\mapsto 1$ is a right-angled Coxeter group with vertices labelled by $a_k, k\in\mathbb{Z}$ ($a_k=t^kat^{-k}$ in $G_I$). Then I think $\langle a_0, a_1, a_2, a_3\rangle \cong (\mathbb{Z}_2 \ast \mathbb{Z}_2) \times (\mathbb{Z}_2\ast \mathbb{Z}_2)$, so is not virtually free. – Ian Agol Feb 20 '19 at 23:58
• @YCor, A RA Coxeter group is virtually free iff the graph is chordal. This is not the case for a 4-gon or higher. – Benjamin Steinberg Feb 21 '19 at 12:11

Here are some observations on your question. Firstly, consider the case that we take the presentation $$\langle S | r_i\rangle$$ for some $$i$$. This is a 1-relator group, and it will be virtually free iff $$r_i$$ is a power of a primitive element of $$S$$ (see Theorem 3 of this paper in the torsion case; and otherwise this answer and induction in the torsion-free case).
As a special case, suppose there is a relator $$r_i$$ which is not a proper power. Then it must be a primitive element in $$S$$, so $$\langle S | r_i\rangle$$ must be a free group on 1 fewer generator. Thus by induction on the rank, we may assume that all of the $$r_i$$ are proper powers of primitive elements.
• But the easiest argument is that $\langle a,t \mid [a,a^t] \rangle \cong \langle a,b,t \mid [a,b] = 1, b = a^t \rangle$ is an HNN extension of ${\mathbb Z}^2$ and hence has ${\mathbb Z^2}$ as a subgroup and cannot be virtually free. Of course that still requires the theory of HNN extensions. – Derek Holt Feb 21 '19 at 7:53
• Perhaps it's worth pointing out that it's not terribly difficult to ensure that all the one-relator subpresentations are free. Just add an extra generator $t$ to $S$, add the relation $t$, and change every $r_i$ to $r_it$. Of course, it's much harder to do something similar for larger subsets of relators, but then one gets out of the theory of one-relator groups. – HJRW Feb 21 '19 at 9:37
• @DerekHolt no need of HNN theory to see that it contains $\mathbf{Z}^2$ (generated by $\{a,a^t\}$): instead just map onto $\mathbf{Z}\wr\mathbf{Z}$. – YCor Feb 21 '19 at 11:29