# Sufficient Criterion for Word Equivalence in Idempotent Semigroups

I am reading Baader's paper "The Theory of Idempotent Semigroups is of Unification Type Zero" and I have a doubt about a property of idempotent semigroups.

Let $$AI = \{(xy)z = x(yz), x^2 = x \}$$.

Here is the part I didn't understand:

For a word $$s$$ let $$\alpha(s)$$ denote the alphabet of $$s$$, i.e., the set of symbols occuring in $$s$$. The following facts about $$=_{AI}$$ are well-known (see e.g. [1, 4, 5, 6]).

(2.1) - (Not related)

(2.2) - For any word $$s$$ let $$l(s)$$ be the shortest prefix of $$s$$ satisfying $$\alpha(l(s)) = \alpha(s)$$ and $$r(s)$$ be the shortest suffix of $$s$$ satisfying $$\alpha(r(s)) = \alpha(s)$$. Then $$s =_{AI} t \leftrightarrow l(s) =_{AI} l(t) \text{ and } r(s) =_{AI} r(t)$$.

(2.3) - (Not related)

(2.4) - (Not related)

I was able to understand properties (2.1), (2.3) and (2.4) [which I am not showing], but I was not able to prove the $$\leftarrow$$ part of (2.2).

I thought about a counterexample (that I don't know if it's correct) for it:

Let s = (xyz)x(xyz) and t = (xyz)y(xyz). Then: \begin{align} l(s) = xyz = l(t)\\ r(s) = xyz = r(t) \end{align} But I was not able to prove that $$s = (xyz)x(xyz)$$ is equal modulo AI to $$(xyz)y(xyz) = t$$.

All I noticed is that $$s = (xyz)x(xyz) =_{AI} xyz$$.

Can anyone tell me if my counterexample is correct or what is the mistake in it?

If it's wrong, can anyone help me with the proof of (2.2)?

EDIT: The references [1, 4, 5, 6] are shown below. I could not access reference 1 (because it's in German) and 6 (couldn't find). I could not find the information in the references 4 and 5, but I don't know a lot of algebra so there may be something I am missing.

1 - (German reference)

4 - Howie, J.M (1976) "An Introduction to Semigroup Theory", Academic Press

5 - Kimura, N. (1958) "The structure of idempotent semigroups (I)", Pacific Journal of Math

6 - McLean, D. (1954) "Idempotent Semigroups", Am. Math. Mon 61.

EDIT2: My counterexample is not correct:

As mentioned before, $$s =_{AI} xyz$$. But, the chain that follows shows that $$t =_{AI} xyz$$ too. $$t = (xyz)y(xyz) = (xy)zyxyz = (xyxy)zyxyz = x \ (yxyz)(yxyz) = x (yxyz) = (xyxy)z = xyz$$

• And what do those references [1,4,5,6] tell about it? Can you have access to them? Idempotent semigroups are also known as bands. In particular, regular bands are definable by the equation $zxyz=zxzyz$ (all varieties of bands are definable by a single equation). It is not difficult to prove that $s=t$ in regular bands, but of course, you want to know whether or not that is the case for all bands, so I don't have the answer... Oct 8 '20 at 18:30
• @amrsa Thanks anyway, I didn't know about bands and regular bands, and I think this information may be useful. I have edited the question to put the references. Oct 8 '20 at 20:47