Does the 6-element group $S_3$ have a finite (balanced) **semigroup** presentation of the form $$\langle a_1,...,a_n\mid a_1=u_1, a_2=u_2,...,a_n=u_n\rangle$$ where $u_1,u_2,...,u_n$ are semigroup words? Let us call such a semigroup presentation {\it tree-like}.

**Edit:** By a semigroup word, I mean a word without inverses of letters. By a semigroup presentation, I mean a presentation in the class of semigroups (not groups). For example, $\langle a,b\mid a^3=a, b=ab^2a\rangle$ defines $S_3$ in the class of groups but a "bigger" semigroup in the class of semigroups. In that semigroup, $b$ does not divide $a$, so it is not a group.

**Motivation 1.** The cyclic group of order $n$ has a tree-like semigroup presentation $\langle a\mid a=a^{n+1}\rangle$. Matt Brin noted that the 8-element quaternion group has the tree-like presentation $\langle a,b,c \mid a=bc, b=ca, c=ab\rangle$. The groups $S_m, m>3$ do not have tree-like presentations because they do not even have balanced (same number of relations and generators) presentations at all since their Schur multipliers are non-trivial. The group $S_3$ has a balanced presentation.

**Motivation 2.** Every tree-like semigroup presentation corresponds to a (closed) subgroup of the R. Thompson group $F$. For example as shown by Guba the Brin tree-like presentation of the quaternion group corresponds to a copy of the Thompson group $F_9$ (the group of piecewise linear homeomorphisms of $[0,1]$ with slopes of the form $9^k$ and break points of the derivative from $\mathbb{Z}[1/9]$).

balancedpresentation (number of relators = number of generators) is a strong restriction. More generally, see the notion of deficiency en.wikipedia.org/wiki/Presentation_of_a_group#Deficiency $\endgroup$5more comments