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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?

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  • $\begingroup$ If you assume $S$ to be finite, you should say it. $\endgroup$
    – YCor
    Feb 20, 2019 at 19:18
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    $\begingroup$ @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. $\endgroup$
    – Derek Holt
    Feb 20, 2019 at 19:21
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    $\begingroup$ 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). $\endgroup$
    – YCor
    Feb 20, 2019 at 19:24
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    $\begingroup$ @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. $\endgroup$
    – Ian Agol
    Feb 20, 2019 at 23:58
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    $\begingroup$ @YCor, A RA Coxeter group is virtually free iff the graph is chordal. This is not the case for a 4-gon or higher. $\endgroup$
    – Benjamin Steinberg
    Feb 21, 2019 at 12:11

1 Answer 1

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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.

Maybe there is a generalization of the theory of 1-relator groups (Freiheitzatz and Magnus-Moldovanskii hierarchy) to 1-relator quotients of virtually free groups? If so, the above arguments might carry over to this context. See results of Howie for some progress on this.

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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Feb 21, 2019 at 7:53
  • $\begingroup$ @IanAgol Thanks, this seems like a very interesting approach in general, I just wondered if the community knew of an example before diving into this. $\endgroup$
    – DavidHume
    Feb 21, 2019 at 8:53
  • $\begingroup$ 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. $\endgroup$
    – HJRW
    Feb 21, 2019 at 9:37
  • $\begingroup$ @DerekHolt no need of HNN theory to see that it contains $\mathbf{Z}^2$ (generated by $\{a,a^t\}$): instead just map onto $\mathbf{Z}\wr\mathbf{Z}$. $\endgroup$
    – YCor
    Feb 21, 2019 at 11:29
  • $\begingroup$ I suspect now that such presentations cannot exist. I think at least one should be able to use small cancellation theory to rule out presentations where the power is high, maybe at least 6. $\endgroup$
    – Ian Agol
    Feb 21, 2019 at 19:58

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