Amenable groups with special presentations

Question

Is there a group with a presentation

$\left< X \mid r_i, i \in \mathbb{N} \right>$ (where $X$ is finite) with

$\left< X \mid r_i, i \in A \right>$ is amenable if and only if $A\subset \mathbb{N}$ is infinite.

Answer 1

An almost yes.

For $n,m\ge 2$ consider the group

$$H=H(n,m)=\langle t,x,y\mid txt^{-1}=x^n,\;t^{-1}yt=y^m\rangle$$

Remark: $H$ is a semidirect product $\mathbf{Z}\ltimes(\mathbf{Z}[1/n]\ast\mathbf{Z}[1/m])$, where the positive generator of $\mathbf{Z}$ acts by multiplication by $n$ on the first factor, and by $1/m$ on the second factor.

We have an automorphism $\psi$ of $H$, mapping $(t,x,y)\mapsto (t,x^n,y)$. (It is not inner.)

Define $r_i=[t^{-i}xt^i,y]$; note that $r_i=\psi^{-i}(r_0)$. Then in $H(n,m)$, the relator $r_i$ is a consequence of $r_{i+1}$ for each $i$ (i.e., is in the normal subgroup generated by $r_{i+1}$). Define $G_i$ as quotient of $H$ by the $r_j$ for $j\le i$. Then $G_i$ is isomorphic to $G_0$ for all $i$. Note that this is a non-ascending HNN extension, namely of $\mathbf{Z}^2$, using the isomorphism between $\mathbf{Z}\times m\mathbf{Z}$ and $n\mathbf{Z}\times \mathbf{Z}$ given by the diagonal matrix $(n,1/m)$. In particular it has a non-abelian free subgroup.

On the other hand, $$G_\infty=\langle t,x,y\mid txt^{-1}x^{-n},\;t^{-1}yty^{-m},\;\;r_i:i\ge 0\;\;\rangle=H/\langle\!\langle r_i:i\ge 0\rangle\!\rangle$$ is the metabelian semidirect product $\mathbf{Z}\ltimes(\mathbf{Z}[1/n]\times\mathbf{Z}[1/m])$, where the positive generator of $\mathbf{Z}$ acts by the diagonal matrix $(n,1/m)$.

So, $H$ modulo any finite subfamily of the family $(r_i)$ is non-amenable, and modulo any infinite subfamily is metabelian.

This almost answers the question, except that we have these two additional initial relators. I hope this is enough for your purposes.