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A semigroup presentation $\langle A | R\rangle$ is called tree-like if every relation has the form $ab=c$, $a,b,c$ are in $A$ and if two relations $ab=c, a'b'=c'$ belong to $R$, then $c=c'$ if and only if $a=a'$ and $b=b'$. Example: $P=\langle x | xx=x\rangle$. These presentations correspond to certain closed subgroups of the R. Thompson group $F$ (if you are interested, see this paper, but the question is purely semigroup theoretic). For example, the presentation $P$ corresponds to $F$ itself.

Question. Is it decidable that a semigroup given by a finite tree-like presentation contains an idempotent?

The answer is yes if and only if it is decidable if a closed subgroup of $F$ contains a copy of $F$ (by this paper).

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  • $\begingroup$ What do you call "closed" subgroup of $F$? there's no occurrence of the word "closed" in your linked paper with Guba. $\endgroup$
    – YCor
    Nov 23, 2017 at 21:46
  • $\begingroup$ It is in the Golan paper, first link. The closure of a subgroup H is the subgroup of piecewise H functions where the endpoints of the pieces are finite diadic - by a result of Golan. Closed subgroups are those which coincide with their closure. $\endgroup$
    – user6976
    Nov 24, 2017 at 1:40
  • $\begingroup$ The connection between closed subgroups and tree-like semigroup presentations: every closed subgroup of $F$ is a diagram group corresponding to a tree-like semigroup presentation. $\endgroup$
    – user6976
    Nov 24, 2017 at 2:27
  • $\begingroup$ And how do you input a "closed" subgroup? in principle there are uncountably any subgroups, so you can't input a subgroup. You can input a finitely generated subgroup by inputting a finite generating subset, but this does not tell you if it's "closed". Or is the set of "closed" subgroups countable with some reasonable enumeration? $\endgroup$
    – YCor
    Nov 24, 2017 at 6:51
  • $\begingroup$ A closure of a f.g. subgroup corresponds to a finitely generated tree-like semigroup presentation. We do not know yet if the closure is f.g. itself, but it is determined by a finite data. So the input is a finite tree-like semigroup presentation. $\endgroup$
    – user6976
    Nov 24, 2017 at 9:59

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