Lobachevskii Journal of Mathematics Vol. 13, 2003, 45 – 50 \begin{matrix}c \\ ◯ \end{matrix}Niovi Kehayopulu and Michael Tsingelis

Niovi Kehayopulu and Michael Tsingelis

THE EMBEDDING OF AN ORDERED SEMIGROUP INTO AN LE-SEMIGROUP

(submitted by M. M. Arslanov)

Abstract. In this paper we prove the following: If S is an ordered semigroup, then the set P(S) of all subsets of S with the multiplication ” ∘ ” on P(S) defined by ”A ∘ B := (AB] if A, B ∈ P(S), A ≠ ∅, B ≠ ∅ and A ∘ B := ∅ if A = ∅ or B = ∅ is an le-semigroup having a zero element and S is embedded in P(S).

If (S, ., ≤) is an ordered semigroup, for A ⊆ S, we define (A] := {t ∈ S | t ≤ a for some a ∈ A}. For A = {a}, we write (a] instead of ({a}]. An element 0 of S is called the zero element of S if 0 ≤ x and 0x = x0 = 0 for all x ∈ S [1]. Let (S, ., ≤), (T, ∘, ⪯) be ordered semigroups, f : S → T a mapping from S into T.

The mapping f is called isotone if x, y ∈ S, x ≤ y implies f(x) ≤ f(y). f is called reverse isotone if x, y ∈ S, f(x) ⪯ f(y) implies x ≤ y. [Each reverse isotone mapping is (1-1): Let x, y ∈ S, f(x) = f(y). Since f(x) ⪯ f(y), we have x ≤ y. Since f(y) ⪯ f(x), we have y ≤ x.] f is called a homomorphism if it is isotone and satisfies f(xy) = f(x) ∘ f(y) for all x, y ∈ S. f is called an isomorphism if it is onto, homomorphism and reverse isotone. S and T are called isomorphic if there exists an isomorphism between them [3]. S is embedded in T if, by definition, S is isomorphic to a subset of T, i.e., if there exists a mapping f : S → T which is homomorphism and reverse isotone [4]. An l-semigroup (: lattice ordered semigroup) is a semigroup S at the same time a lattice satisfying the conditions a(b ∨ c) = ab ∨ ac and (a ∨ b)c = ac ∨ bc for all a, b, c ∈ S [1]. By an le-semigroup we mean an l-semigroup having a greatest element ”e” (i.e. e ≥ a for all a ∈ S) [2]. We denote by P(S) the set of all subsets of S.

Theorem. Let (S, ., ≤) be an ordered semigroup. We define a multiplication ” ∘ ” on P(S) as follows:

2000 Mathematical Subject Classification. 06F05.

∘ : P(S) × P(S) → P(S) | (A, B) → A ∘ B

where

A ∘ B : = \{ \begin{matrix}(AB] & if & A, B ∈ P(S)\\{∅\} \\ ∅ & if & A = ∅ or B = ∅ \end{matrix}

Then (P(S), ∘, ⊆) is an le-semigroup having a zero element and (S, ., ≤) is embedded in (P(S), ∘, ⊆).

Proof. First of all, the set P(S) in non-empty. The multiplication ” ∘ ” on P(S) is well defined. Moreover, we have the following:

1) The multiplication ” ∘ ” on P(S) is associative. In fact:

Let A, B, C ∈ P(S). If A = ∅ or B = ∅ or C = ∅, then (A ∘ B) ∘ C = ∅ and A ∘ (B ∘ C) = ∅, so (A ∘ B) ∘ C = A ∘ (B ∘ C).

Let A, B, C ∈ P(S)\{∅}. We have A ∘ B, B ∘ C ∈ P(S)\{∅}. Let now x ∈ (A ∘ B) ∘ C := ((A ∘ B)C]. Then x ≤ yc for some y ∈ A ∘ B, c ∈ C. Since y ∈ A ∘ B := (AB], we have y ≤ ab for some a ∈ A, b ∈ B. Then

x ≤ (ab)c = a(bc); a ∈ A, bc ∈ BC ⊆ (BC] := B ∘ C,

so x ∈ (A(B ∘ C)] := A ∘ (B ∘ C). Similarly, A ∘ (B ∘ C)] ⊆ (A ∘ B) ∘ C.

2) (P(S), ∘, ⊆) is an le-semigroup: Let A, B, C ∈ P(S). Then A ∘ (B ∪ C) = (A ∘ B) ∪ (A ∘ C). Indeed: If A = ∅, then A ∘ (B ∪ C) = ∅, A ∘ B = ∅, A ∘ C = ∅.

If B = ∅, then A ∘ (B ∪ C) = A ∘ C, (A ∘ B) ∪ (A ∘ C) = A ∘ C.

If C = ∅, then A ∘ (B ∪ C) = A ∘ B, (A ∘ B) ∪ (A ∘ C) = A ∘ B.

Let A, B, C ∈ P(S)\{∅}. We have

A ∘ (B ∪ C) := (A(B ∪ C)], A ∘ B := (AB], A ∘ C := (AC].

Since (AB], (AC] ⊆ (A(B ∪ C)], we have (AB] ∪ (AC] ⊆ (A(B ∪ C)]. Let now t ∈ (A(B ∪ C)]. Then t ≤ ax for some a ∈ A, x ∈ B ∪ C. If x ∈ B, then t ∈ (AB] ⊆ (AB] ∪ (AC]. If x ∈ C, then t ∈ (AC] ⊆ (AB] ∪ (AC]. Similarly, for any A, B, C ∈ P(S)\{∅}, we have (A∪B)∘C = (A∘C)∪(B ∘C). Finally, S is the greatest element and ∅ the zero element of P(S).

3) We consider the mapping

f : (S, ., ≤) → (P(S), ∘, ⊆) | a → f(a) := (a].

The mapping f is well defined. Moreover,

EMBEDDING OF AN ORDERED SEMIGROUP 47

A) The mapping f is a homomorphism. Indeed: Let a, b ∈ S. We have (a], (b] ∈ P(S)\{∅} (since a ∈ (a], b ∈ (b]). Thus we have

f(a) ∘ f(b) = (a] ∘ (b] := ((a](b]] = (ab] := f(ab).

Let a, b ∈ S, a ≤ b. Then f(a) := (a] ⊆ (b] := f(b). B) The mapping f is reverse isotone: Let a, b ∈ S, f(a) ⊆ f(b). Then a ∈ (a] ⊆ (b], and a ≤ b. Remark. More generally, we have the following: If A, B_{i} ∈ P(S), i ∈ I, then

A ∘ (⋃_{i∈I} B_{i}) = ⋃_{i∈I}(A ∘ B_{i}) and (⋃_{i∈I} B_{i}) ∘ A = ⋃_{i∈I}(B_{i} ∘ A).

In fact, A) If A = ∅, then A∘(⋃_{i∈I} B_{i}) = ∅, and A∘B_{i} = ∅ for all i ∈ I, so ⋃_{i∈I}(A ∘ B_{i}) = ∅.

Thus A ∘ (⋃_{i∈I} B_{i}) = ⋃_{i∈I}(A ∘ B_{i}).

B) If A ≠ ∅, then

I) If ⋃_{i∈I} B_{i} = ∅, then A ∘ (⋃_{i∈I} B_{i}) = ∅. Since ⋃_{i∈I} B_{i} = ∅, we have B_{i} = ∅ for all i ∈ I, then A ∘ B_{i} = ∅ for all i ∈ I, and ⋃_{i∈I}(A ∘ B_{i}) = ∅. Then

A ∘ (⋃_{i∈I} B_{i}) = ⋃_{i∈I}(A ∘ B_{i}).

II) Let ⋃_{i∈I} B_{i} ≠ ∅. We put J := {i ∈ I | B_{i} = ∅}, K := {i ∈ I | B_{i} ≠ ∅}.

Clearly I = J ∪ K and J ∩ K = ∅. If K = ∅, then I = J, B_{i} = ∅ for all i ∈ I, and ⋃_{i∈I} B_{i} = ∅. Impossible. Thus K ≠ ∅.

α) Let J = ∅. Then I = K, B_{i} ≠ ∅ for all i ∈ I. Since A ≠ ∅ and B_{i} ≠ ∅ for all i ∈ I, we have A ∘ B_{i} := (AB_{i}] for all i ∈ I. Then

⋃_{i∈I}(A ∘ B_{i}) = ⋃_{i∈I}(AB_{i}].

Besides, ⋃_{i∈I}(AB_{i}] = (A(⋃_{i∈I} B_{i})]. Thus we have

⋃_{i∈I}(A ∘ B_{i}) = (A(⋃_{i∈I} B_{i})]..........(∗)

Since A ≠ ∅ and (⋃_{i∈I} B_{i}) ≠ ∅, we have A ∘ (⋃_{i∈I} B_{i}) = (A(⋃_{i∈I} B_{i})]. Then, by (*),

A ∘ (⋃_{i∈I} B_{i}) = ⋃_{i∈I}(A ∘ B_{i}).

β) Let J ≠ ∅. Then B_{i} = ∅ for all i ∈ I, ⋃_{i∈I} B_{i} = ∅, and

∅ ≠ ⋃_{i∈I} B_{i} = (⋃_{i∈J} B_{i}) ∪ (⋃_{i∈K} B_{i}) = ⋃_{i∈K} B_{i}.

Since A ≠ ∅ and ⋃_{i∈I} B_{i} ≠ ∅, we have

A ∘ (⋃_{i∈I} B_{i}) = (A(⋃_{i∈I} B_{i})] = (A(⋃_{i∈K} B_{i})] = ⋃_{i∈K}(AB_{i}].

Since A ≠ ∅ and B_{i} ≠ ∅ for all i ∈ K, we have A ∘ B_{i} := (AB_{i}] for all i ∈ K, and ⋃_{i∈K}(A ∘ B_{i}) = ⋃_{i∈K}(AB_{i}]. Thus we have

A ∘ (⋃_{i∈I} B_{i}) = ⋃_{i∈K}(A ∘ B_{i})..........(∗∗)

Since B_{i} = ∅ for all i ∈ J, we have A∘B_{i} = ∅ for all i ∈ J, then ⋃_{i∈J}(A ∘ B_{i}) =

∅. Then

⋃_{i∈K}(A ∘ B_{i}) ∪ ⋃_{i∈J}(A ∘ B_{i}) = ⋃_{i∈K}(A ∘ B_{i})..........(∗ ∗ ∗)

By (**) and (***), we have

A ∘ (⋃_{i∈I} B_{i}) = ⋃_{i∈K}(A ∘ B_{i}) ∪ ⋃_{i∈J}(A ∘ B_{i}) = ⋃_{i∈K∪J}(A ∘ B_{i}) = ⋃_{i∈J}(A ∘ B_{i}). □

Example. We consider the ordered semigroup

S = {x, y, z} defined by the multiplication and the figure below:

. x y z
x x x z
y x y z
z x z z

Applying the Theorem of this note, the ordered semigroup (S, ., ≤) is embedded into the le-semigroup L = {a, b, c, d, e, f, g, h}, defined by the multiplication ”.”and the order ” ≤_{L} ” below:

. a b c d e f g h
a a a g a g g g h
b a b g d g g g h
c a g g g g g g h
d a d g d g g g h
e a g g g g g g h
f a g g g g g g h
q a g g g g g g h
h h h h h h h h h
≤_{L}: = {(a, a), (a, d), (a, e), (a, g), (b, b), (b, d), (b, f), (b, g), (c, d), (c, e),

(c, f), (c, g), (d, d), (d, g), (e, e), (e, g), (f, f), (f, g), (g, g), (h, a), (h, b), (h, c), (h, d), (h, e), (h, f), (h, g), (h, h)}.

We give the covering relation ”≺’ and the figure of S.

\begin{aligned}≺ & = & \{(a, d), (a, e), (b, d), (b, f), (c, e), (c, f), (d, g), (e, g), (f, g), \\ & & (h, a), (h, b), (h, c)\}. \end{aligned}

The embedding is given by the mapping:

f : (S, ., ≤) → (L, ., ≤_{L}) | \begin{matrix}x → a \\ y → b \\ z → g \end{matrix}

This research was supported by the Special Research Account of the University of Athens (Grant No. 5630).

50 NIOVI KEHAYOPULU AND MICHAEL TSINGELIS

References

[1] Birkhoff G., ”Lattice Theory”, Amer. Math. Soc. Coll. Publ. Vol. XXV, Providence, Rh. Island, 1967.

[2] Kehayopulu N., On intra-regular ∨e-semigroups, Semigroup Forum 19 (1980), 111-121.

[3] Kehayopulu N. and M. Tsingelis, On subdirectly irreducible ordered semigroups, Semigroup Forum 50 (1995), 161-177.

[4] Kehayopulu N. and M. Tsingelis, The embedding of an ordered semigroup in a simple one with identity, Semigroup Forum 53 (1996), 346-350.

Univ. of Athens, Dept. of Mathematics; Home address: Niovi Kehayopulu, Nikomidias 18, 161 22 KESARIANI, GREECE E-mail address: nkehayop@cc.uoa.gr

Received September 30, 2003