= Shift-invariant graph groups encode arbitrary sets
{title2=$[b_0,t^nb_0t^{-n}]=1\ \Longleftrightarrow\ n\in S$}
For $S\subseteq\mathbb N_{>0}$, form the <Right-angled Artin group> on generators $b_i$ indexed by $i\in\mathbb Z$, imposing $[b_i,b_j]=1$ precisely when $|i-j|\in S$. Translation $b_i\mapsto b_{i+1}$ is an automorphism. Its <semidirect product> with an infinite cyclic group generated by $t$ is countable and generated by $b_0,t$. If $n\notin S$, killing all $b_i$ except $b_0,b_n$ retracts the graph group onto the rank-two <free group>, so their commutator is nonidentity. This proves the displayed equivalence. For non-recursively-enumerable $S$, <enumerable identities of finitely generated subgroups> rules out an embedding of this two-generated group in any finitely presented group.
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