TITLE: Infinitely presented group where every finite sub-presentation is virtually free
QUESTION [11 upvotes]: 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?

REPLY [3 votes]: 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.