PROCEDIMIENTOS DE INSTRUMENTACIÓN.-

In the previous chapter we gave an example of the Cayley digraph of a right-zero semigroup with three elements generated by a three element set (see Figure 3.1 on

page 26). As indicated in Figure 3.1, the Cayley digraph of a right-zero semigroup system (R, Σ, f ) has the following form: for every element a ∈ R and every r ∈ Σ, the digraph Cay(R; Σ) contains an r-labeled edge from a to rf . The Cayley digraph also has the useful property that the left action of S on Cay(R; Σ) (corresponding to left translation by elements of S) is trivial. In other words, if c ∈ S and P ⊆ Cay(R; Σ), then cP = P . This enables us to simplify the product for graph expansion elements. Lemma 4.4.1. Let (R, Σ, f ) be a semigroup system of a right-zero semigroup R. Then (r, P, c)(s, Q, d) = (r, P ∪ Q ∪ {(c, s)}, d).

Proof: We derive that:

(r, P, c)(s, Q, d) = (r, P ∪ c(Q1_s), cd)

= (r, P ∪ cQ ∪ {c(1, s)}, d)

= (r, P ∪ Q ∪ {(c, s)}, d). _

This simplification will be useful as we describe many of the properties of right-zero semigroup graph expansions.

Proposition 4.4.2. Let (R, Σ, f ) be a semigroup system of a right-zero semigroup R. The following are true about M(R; Σ):

(a) an element (r, P, c) is idempotent if and only if (c, r) ∈ E(P ); (b) E M(R; Σ)
is a subsemigroup if and only if |R| = 1;

(c) elements (r, P, c) and (s, Q, d) commute if and only if s = r and c = d; (d) an element (r, P, c) is regular if and only if there is a non-empty c −→ rf

path in P ;

(e) Reg M(R; Σ)
is a subsemigroup if and only if |R| = 1; moreover, if |R| = 1, then Reg M(R; Σ)
= E M(R; Σ)
;

(f ) if (r, P, c) is regular, then (s, Q, d) is an inverse if and only if P = Q and (c, s), (d, r) ∈ E(P );

(g) M(R; Σ) is locally a semilattice.

Proof: To prove part (a), suppose (r, P, c) is idempotent. Then from Theorem 3.2.1 (b), cP_r1 ∈ P , which gives that (c, r) ∈ E(P ). Conversely, if (c, r) ∈ E(P ), then cP1

r = cP ∪ {(c, r)} = P . Again appealing to Theorem 3.2.1 (b), (r, P, c) is

idempotent.

Moving on to part (b), suppose E M(R; Σ)
is a subsemigroup. Let r, s ∈ Σ. We wish to show that rf = sf . Let P be the digraph consisting of the r-labeled loop at rf . Similarly, let Q be the digraph consisting of the s-labeled loop at sf . By part (a), both (r, P, rf ) and (s, Q, sf ) are idempotent elements of M(R; Σ). Moreover, the element (r, P, rf )(s, Q, sf ), which we show below, is idempotent.

Thus from part (a), the digraph P ∪(rf )Q1_scontains the edge (sf, r). This implies that rf = sf . Hence |R| = 1. For the converse, assume |R| = 1. Let (r, P, c) and (s, Q, c) be idempotents. From (a), we know that (c, r) ∈ E(P ). Hence (c, r) ∈ E(P ∪ Q), whereupon we see that the product is idempotent.

In order to show (c), suppose elements (r, P, c) and (s, Q, d) commute. Then (r, P ∪ cQ1

s, d) = (s, Q ∪ dPr1, c). Immediately we see that r = s and c = d. To

prove the converse, assume that r = s and c = d. Let (r, P, c), (r, Q, c) ∈ M(R; Σ). Recall that left-translation corresponds to the identity map for subdigraphs, i.e. that

cP = P and cQ = Q. Therefore,

(r, P, c)(r, Q, c) = (r, P ∪ cQ1_r, c)

= (r, cP ∪ Q ∪ {(c, r)}, c) = (r, Q ∪ cP_r1, c)

= (r, Q, c)(r, P, c).

We now wish to prove (d). Suppose (r, P, c) is regular. Let (s, Q, d) be an inverse. Then it follows that

(r, P, c) = (r, P, c)(s, Q, d)(r, P, c) = r, P ∪ Q ∪ {(c, s), (d, r)}, c
.

The subdigraph Q, by virtue of being sf -rooted, contains an sf −→ d path. Affixing the edges (c, s) and (d, r) to this path produces the desired non-empty c −→ rf path in P . For the converse, suppose that (r, P, c) is such that P contains a non-empty c −→ rf path. Thus, there exists some vertex x ∈ V (P ) such that (x, r) ∈ E(P ). From the properties of right zero semigroups, xc = c and xP1

r = P . Moreover, if we

combine the c −→ rf path with a rf −→ x path, we obtain a c −→ x path in P . Using Theorem 3.2.1(c), we conclude that (r, P, c) is regular.

We turn to proving (e). The same argument as used in part (b) shows that if Reg M(R; Σ)
is a subsemigroup, then |R| = 1. Conversely, if |R| = 1, then (a) and (d) imply that all regular elements are idempotent, i.e. we obtain that Reg M(R; Σ)
= E M(R; Σ)
. Thus (b) implies that Reg M(R; Σ)
is a subsemi- group.

In order to show (f), suppose that (r, P, c) and (s, Q, d) are inverses. Then (r, P, c) = (r, P, c)(s, Q, d)(r, P, c) = (r, P ∪ Q ∪ {(c, s), (d, r)}, c).

with (s, Q, d) on the left hand side shows that P ⊆ Q. Hence P = Q. On the other hand, if we assume that P = Q and (c, s), (d, r) ∈ E(P ), then the converse is evident upon examining the relevant products.

Finally, we turn to the proof of (g). Assume (r, P, c) is idempotent and let (s, Q, d) ∈ M(S; Σ). Consider the element

(r, P, c)(s, Q, d)(r, P, c) = r, P ∪ Q ∪ {(c, s), (d, r)}, c
.

Since (c, r) ∈ E(P ) ⊆ E P ∪ Q ∪ {(c, s), (d, r)}
, we know by (a) that the element (r, P, c)(s, Q, d)(r, P, c) is idempotent. Appealing to (c), all elements of

(r, P, c)M(S; Σ)(r, P, c) commute. Thus M(S; Σ) is locally a semilattice. _ The Path Expansion for Right-Zero Semigroups

The properties of the graph expansion of right-zero semigroup systems carry over to the path expansion. We give these results in Proposition 4.4.3. As before, when there is no difference between the result or proof for the path expansion with that for the graph expansion in Proposition 4.4.2, we indicate this with the comment “no change” and omit the proof.

Proposition 4.4.3. Let (R, Σ, f ) be a semigroup system of a right-zero semigroup R. The following are true about Path(R; Σ):

(a0) an element (r, P, c) is idempotent if and only if (c, r) ∈ E(P ) (no change); (b0) E Path(R; Σ)
is a subsemigroup if and only if |R| = 1;

(c0) elements (r, P, c) and (s, Q, d) commute if and only if s = r and c = d (no change);

(d0) an element (r, P, c) is regular if and only if P is strongly connected; (e0) Reg Path(R; Σ)
is a subsemigroup if and only if |R| = 1; moreover, if

(f0) if (r, P, c) ∈ Path(S; Σ) is regular, then (s, Q, d) is an inverse if and only if P = Q and (c, s), (d, r) ∈ E(P );

(g0) Path(R; Σ) is locally a semilattice.

Proof: We start with a proof of (b0). If E Path(R; Σ)
is a subsemigroup, then the same argument as given for Proposition 4.4.2(b) shows that |R| = 1. Conversely, if |R| = 1, then R is also a left zero semigroup and by Proposition 4.3.3, we know that Path(R; Σ) = M(R; Σ). From Proposition 4.3.2(b), we have that E Path(R; Σ)
is a subsemigroup.

In order to show (d0), we suppose (r, P, c) is regular. Let x, y ∈ V (P ). Due to the properties of the path expansion, the digraph P contains the following paths: rf −→ x, rf −→ y, x −→ c, and y −→ c. Moreover, from the regularity of (r, P, c), we know there is an element (s, Q, d) such that

(r, P, c) = (r, P, c)(s, Q, d)(r, P, c) = r, P ∪ Q ∪ {(c, s), (d, r)}, c
.

The digraph Q contains a path sf −→ d. If we attach the edges (c, s) and (d, r) to this path, we see that P contains a c −→ rf path. Combining this path with the earlier paths, we have paths from x to y and vice versa. Thus P is strongly connected. We now wish to prove (e0). If Reg Path(R; Σ)
is a subsemigroup, then we can again rely on the argument used to prove Proposition 4.4.2(b) to show that |R| = 1. If |R| = 1, then as we noted before, R is a left zero semigroup and hence we have that Path(R; Σ) = M(R; Σ). Thus applying Proposition 4.3.2(d) yields that

E Path(R; Σ)
=Reg Path(R; Σ)
. Then (b0) implies that Reg Path(R; Σ)
is a subsemigroup.

One direction of (f0) follows from Proposition 4.4.2(f). For the converse, assume that (r, P, c) is regular and that (c, s), (d, r) ∈ E(P ). From Proposition 4.4.2(f), we know that (s, P, d) is an inverse of (r, P, c) in M(S; Σ). We wish to show that (s, P, d) is in the path expansion. Since (r, P, c) ∈ Path(S; Σ), there is a path rf −→ c in P that traverses every edge of P . Similarly, as (s, P, d) ∈ M(S; Σ) there is a sf −→ d in

P . Combining these two paths via the edge (c, s), which we assumed to be in E(P ), we obtain a path sf −→ d which traverses every edge of P . Thus (s, P, d) ∈ Path(S; Σ), as desired.

Finally, if we replace E M(R; Σ)
by E Path(R; Σ)
in the proof of Proposition 4.4.2(g), we obtain a proof of (g0). _