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](https://mathoverflow.net/questions/38126/a-balanced-presentation-of-a-cyclic-by-cyclic-group).


**Motivation 2.** Every tree-like semigroup presentation [corresponds](https://arxiv.org/abs/1608.02572) 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 $\mathbb{Z}[1/9]$).



A balanced tree-like presentation of $S_3$
I see that Jeremy has beaten me to it - but here is a solution found by computer by choosing random entries from the group multiplication table.


This one also generates $S\_3$ as a semigroup. The Magma command $\mathtt{RWSMonoid}$ applies the Knuth-Bendix algorithm to the presentation, and regards it as a monoid presentation. It has order $7$ as a monoid with identity equal to the empty word, and so it has order $6$ as a semigroup presentation.


$$\langle a,b,c,d,e,f\mid a=f^2,b=ab,c=df,d=ec,e=db,f=ce\rangle.$$



```
  > G := Group< a,b,c,d,e,f | a = f^2, b = a*b, c = d*f, d=e*c, e=d*b, f=c*e >;
  > Order(G);
  6
  > Homomorphisms(G,Sym(3))[1];
   Homomorphism of GrpFP: G into GrpPerm: X, Degree 3, Order 2 * 3 induced by
      G.1 |--> Id(X)
      G.2 |--> (1, 2, 3)
      G.3 |--> (1, 3, 2)
      G.4 |--> (1, 2)
      G.5 |--> (1, 3)
      G.6 |--> (2, 3)

> M := Monoid< a,b,c,d,e,f | a = f^2, b = a*b, c = d*f, d=e*c, e=d*b, f=c*e >; 
> R := RWSMonoid(M);
> Order(R);
7

```

