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«0-TIGHT COMPLETELY 0-SIMPLE SEMIGROUPS BY HSING Y. WU Abstract A semigroup is 0-tight if each of its congruences is uniquely determined by each of ...»

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Bulletin of the Institute of Mathematics

Academia Sinica (New Series)

Vol. 1 (2006), No. 4, pp. 475-484

0-TIGHT COMPLETELY 0-SIMPLE SEMIGROUPS

BY

HSING Y. WU

Abstract

A semigroup is 0-tight if each of its congruences is uniquely

determined by each of the congruence classes which do not contain

zero. We classify finite 0-tight rectangular 0-bands, and characterize 0-tight completely 0-simple semigroups. Finally, we obtain

correspoding results about tight completely simple semigroups.

1. Introduction Throughout this paper we shall use the terminology and notation of Howie [2]. We recall several definitions: a semigroup S is called 0-simple if, for any a, b ∈ S \ {0}, there exist x, y ∈ S such that xay = b. A completely 0simple semigroup S is a 0-simple semigroup such that every idempotent z of S has the property that zf = f z = f = 0 implies z = f. The following result is due to Rees [6]. Every completely 0-simple semigroup is isomorphic to a certain Rees matrix semigroup. This construction has led to an extensive study of congruences on completely 0-simple semigroups (see [3], [5]).

We begin with an analysis of congruences on a particular finite completely 0-simple semigroup. The result turns out to be the structure theorem for finite 0-tight rectangular 0-bands (see Theorem 2.3 in [8]). We recall that a rectangular 0-band is a semigroup (I × Λ) ∪ {0} whose product is given in Received September 19, 2005 and in revised form January 3, 2006.

AMS Subject Classification: Primary 20M20; Secondary 08A35.

Key words and phrases: Completely 0-simple, semigroups, tight.

The results of this paper are due to [8].

476 HSING Y. WU [December

terms of a Λ × I matrix P = (pλi ) with entries in {0, 1} as follows:

(i, µ) if pλj = 1 (i, λ)(j, µ) = 0 if pλj = 0 (i, λ)0 = 0(i, λ) = 00 = 0, where P is regular, in the sense that no row or column of P consists entirely of zeros. The term tight was introduced by Schein [7]. A semigroup is called 0-tight if each of its congruences is uniquely determined by each of the congruence classes which do not contain zero. Using this structure theorem we then describe a characterization of 0-tight rectangular 0-bands.

The main aim of Section 3 is to characterize 0-tight completely 0-simple semigroups. A completely 0-simple semigroup M0 [G; I, Λ; P ] is 0-tight if and only if the rectangular 0-band (I × Λ) ∪ {0} is 0-tight. For the notation M0 [G; I, Λ; P ], see p. 88 in [1].

By a similar approach to the study of 0-tight completely 0-simple semigroups, we investigate further tight completely simple semigroups. We recall that a semigroup S is called completely simple if S has no proper ideals and every idempotent z of S has the property that zf = f z = f implies z = f.

A semigroup is called tight if each of its congruences is uniquely determined by each of the congruence classes (see [7]). Our last result of Section 3 is a characterization of tight completely simple semigroups.

2. 0-Tight Rectangular 0-Bands

Every semigroup S with zero has exactly one of the following properties:

(1) S is 0-tight.

(2) No congruence on S except S × S is uniquely determined by each of its congruence classes which do not contain zero.

(3) There exists a congruence ρ = S × S on S such that ρ is uniquely determined by each of its congruence classes which do not contain zero.

Also, there exist two congruences on S which have the same congruence class that does not contain zero.

2006] 0-TIGHT COMPLETELY 0-SIMPLE SEMIGROUPS 477 A semigroup with the property in (2) is called 0-tight-free. A semigroup with the property in (3) is of the third type.

Now we proceed to the classification of finite rectangular 0-bands. Suppose S is a rectangular 0-band with I = {1,..., m} and Λ = {1,..., n}. In [2] an equivalence relation EI on I is defined by the rule that

–  –  –

The relation EI is related to a unique partition of m in the following way:

there are r EI -equivalence classes and mi -element EI -classes for 0 ≤ i ≤ r ≤ |I| if

–  –  –

Also, the relation EΛ is related to a unique partition of n in the following:

there are s EΛ -equivalence classes and nj -element EΛ -classes for 0 ≤ j ≤ s ≤ |Λ| if

–  –  –

where n1,..., ns ∈ N and n1 ≥ · · · ≥ ns.

We recall that a proper congruence ρ on a completely 0-simple semigroup is defined by 0 ρ = {0}. Supppose ρ is a relation on S \ {0}. According to Lemma 3.5.6 in [2], every proper congruence ρ ∪ {(0, 0)} on S is defined by the rule that

–  –  –

proper congruences are not uniquely determined by each of their congruence classes which do not contain zero.

Case 1. Let P = (pλi ).

Suppose the equivalence relations EI and EΛ

correspond to partitions of m and n, respectively, as follows:

–  –  –

where r, s ≥ 2.

We look at the case where |I| 2 or |Λ| 2. For |I| 2, suppose m1,..., mr are not all equal to 1. Clearly, for every equivalence S1 = I × I, there exists an equivalence S2 = S1 such that iS1 = iS2 for some i ∈ I according to the selection rules for the counting problems.





Similarly, for |Λ| 2, suppose n1,..., ns are not all equal to 1. For every equivalence T1 = Λ × Λ, there exists an equivalence T2 = T1, and λT1 = λT2 for some λ ∈ Λ. Let ρt correspond to (St, Tt ) for 1 ≤ t ≤ 2. We have (i, λ)ρ1 = (i, λ)ρ2, for some (i, λ) ∈ I × Λ.

When iS1 = {i} and λT1 = {λ}, a one-element S1 -class and a oneelement T1 -class give the existence of a one-element ρ1 -class (see Example 2.2).

Case 2. Suppose r = 1 = s in case 1.

Then there exists a proper congruence ρ which corresponds to (I × I, 1Λ ) and ρ is uniquely determined by each of its congruence classes which do not contain zero. However, there exists a congruence which has a one-element class that is not equal to {0}.

Hence such a finite rectangular 0-bands is of the third type.

Remark. Suppose n ≥ |Λ| ≥ 2 and m ≥ |I| ≥ 2. When r = 1 and s ≥ 2, or r ≥ 2 and s = 1, we cannot find a regular matrix P. In other words, in each of these cases we cannot find a rectangular 0-band whose multiplication is in terms some regular matrix P. Now suppose n ≥ |Λ| ≥ 2 and m ≥ |I| ≥ 2|Λ|. When m1 = · · · = mr = 1 and n1 = · · · = ns = 1, again we cannot find a regular matrix P.

So far we have discussed the main part of the following classification.

(1) If |Λ| = 1 and |I| = 1 or 2, then S is 0-tight.

2006] 0-TIGHT COMPLETELY 0-SIMPLE SEMIGROUPS 479 (2) If |Λ| = 1 and m ≥ |I| ≥ 3, then S is of the third type.

(3) If |Λ| = 2 and |I| = 2, then S is 0-tight.

(4) If |Λ| = 2 and |I| = 3, then S can be 0-tight or 0-tight-free or of the third type.

(5) If |Λ| = 2 and m ≥ |I| ≥ 4, then S is either 0-tight-free or of the third type.

(6) If |Λ| = n ≥ 3 and |I| = n,..., 2n −1, then S can be 0-tight or 0-tight-free or of the third type.

(7) If |Λ| = n ≥ 3 and m ≥ |I| ≥ 2n, then S is either 0-tight-free or of the third type.

Note that there is a duality between |Λ| and |I|. For example, in order to know in which category a rectangular 0-band with |Λ| = 4 and |I| = 3 is, we simply check the rectangular 0-band with |Λ| = 3 and |I| = 4. Here we point out the number 2n − 1 coincides with that in Corollary 2.2 in [4].

Next, we recall the Green’s equivalence H.

–  –  –

Also, a semigroup S is called congruence-free if S has no congruences other than 1S and S ×S. [2] is a reference for the classification of finite congruencefree semigroups. The following theorem is the structure theorem for finite 0-tight rectangular 0-bands.

Theorem 2.1.

Suppose I = {1,..., m} and Λ = {1,..., n} are finite sets. If |I| ≤ 2 and |Λ| ≤ 2, then a rectangular 0-band is 0-tight. In other cases a rectangular 0-band is 0-tight if and only if it is congruence-free with zero. Conversely, every finite 0-tight semigroup S with H = 1S is isomorphic to one of this kind.

Proof. First every congruence-free semigroup with zero is 0-tight. We show other cases apart from |I| ≤ 2 and |Λ| ≤ 2 by verifying the following

equivalent statements on a finite rectangular 0-band S:

–  –  –

We shall show that (1) ⇔ (2) ⇔ (3).

(1) ⇒ (2). Suppose EI = 1I or EΛ = 1Λ. From (1) and (2) on page 3, it follows that m1,..., mr are not all equal to 1 or n1,..., ns are not all equal to 1. By case 1 on page 4, there exists a proper congruence which is not uniquely determined by each of its congruence classes that do not contain zero. It is a contradiction.

(2) ⇒ (1). If EI = 1I, then S = 1I. Also, EΛ = 1Λ implies that T = 1Λ.

By (3) on page 3, S has only one proper congruence, and hence S is 0-tight.

(2) ⇒ (3). Suppose column i and column j of the regular matrix P are identical. Then {λ ∈ Λ : pλi = 0} = {λ ∈ Λ : pλj = 0}.

This gives (i, j) ∈ EI. It is a contradiction. Similarly, if EΛ = 1Λ, then no two rows of the matrix P are identical.

(3) ⇒ (2). Suppose no two columns of the matrix P are identical. Since P has entries in {0, 1}, we have EI = 1I. Similarly, Suppose no two rows of the matrix P are identical. Since P has entries in {0, 1}, we have EΛ = 1Λ.

We are done. Conversely, first every 0-tight semigroup is 0-simple. Also, it is known that every finite 0-simple semigroup S with H = 1S is isomorphic to a finite rectangular 0-band. The remaining part of the proof follows the classification of finite 0-tight rectangular 0-bands.

Now we apply Theorem 2.1 to arbitrary rectangular 0-bands.

Theorem 2.2.

Suppose S is a rectangular 0-band. If |I| ≤ 2 and |Λ| ≤ 2, then S is 0-tight. If |I| ≥ 3 or |Λ| ≥ 3, then the following statements

are equivalent:

(1) S is 0-tight.

(2) 1S is uniquely determined by each of its congruence classes which do not contain zero.

(3) EI = 1I and EΛ = 1Λ.

(4) S is congruence-free with zero.

Proof. It suffices to show that (2) implies (3). Let us assume that EI = 1I and EΛ = 1Λ. Let i and λ be fixed elements in I and Λ, respectively. We consider T -classes {λ} and Λ \ {λ}. Suppose the congruence ρ corresponds 2006] 0-TIGHT COMPLETELY 0-SIMPLE SEMIGROUPS 481 to (1I, T ). Then ρ and 1S have the same congruence class {(i, λ)}. We are done.

Example 2.2.

Let |I| = 2 = |Λ|. We find out the reason why partitions of m and n do not always give the equivalences EI and EΛ, respectively.

Since P is regular, there are only 7 possible P ’s. Let P1 be a 2 × 2 matrix which consists of all 1’s. Suppose each of P2 to P5 is a 2 × 2 matrix which consists of one 0 and three 1’s. Suppose each of P6 to P7 is a 2 × 2 matrix which consists of two 0’s and two 1’s with either 0’s or 1’s on the main diagonal.

We check the rectangular 0-band whose multiplication is in terms of P1.

Now EI = I × I = {1, 2} × {1, 2} and EΛ = Λ × Λ = {1, 2} × {1, 2}. In this case, m = 2 = n. Let S1 = 1I, and S2 = I × I, and T1 = 1Λ, and T2 = Λ × Λ.

Suppose ρkt corresponds to (Sk, Tt ), where 1 ≤ k ≤ 2 and 1 ≤ t ≤ 2. Then we obtain the following proper congruences which are uniquely determined by each of their congruence classes that do not contain zero.

(1) ρ11 ∪ {(0, 0)}, the identity congruence.

(2) ρ12 ∪ {(0, 0)} has classes {0}, {(1, 1), (1, 2)}, {(2, 1), (2, 2)}.

(3) ρ21 ∪ {(0, 0)} has classes {0}, {(1, 1), (2, 1)}, {(1, 2), (2, 2)}.

(4) ρ22 ∪ {(0, 0)} has classes {0}, {(1, 1), (1, 2), (2, 1), (2, 2)}, Next, we look at the rectangular 0-band whose multiplication is in terms of some Pf, where 2 ≤ f ≤ 7. Then EI = 1I and EΛ = 1Λ. Now m = 1 + 1 = n. For each 2 ≤ f ≤ 7, we obtain one proper congruence which is uniquely determined by each of its congruence classes that do not contain zero. When m = 1 + 1 and n = 2, we cannot say EI = 1I and EΛ = Λ × Λ. In other words, if EI = 1I and EΛ = Λ × Λ, then we cannot find a regular matrix equal to some Pf for 1 ≤ f ≤ 7.

Example 2.2.

Let m = 2 + 2 = n. We give an example of a 0tight-free rectangular 0-band. Now there are 2 two-element EI -classes and 2 two-element EΛ -classes. Here we only discuss one possible arrangement of elements of each class selected from the sets I and Λ.

Suppose S1 -classes are {1, 2}, {3, 4}, and S2 -classes are {1, 2}, {3}, {4}, and S3 -classes are {1}, {2}, {3}, {4}. Suppose T1 -classes are {1, 3}, {2, 4}, and T2 -classes are {1, 3}, {2}, {4}, and T3 -classes are {1}, {2}, {3}, {4}.

482 HSING Y. WU [December Also, let ρkt correspond to (Sk, Tt ), where 1 ≤ k ≤ 3 and 1 ≤ t ≤ 3. We

obtain the following result:

(1) (1, 3) ρ11 = (1, 3) ρ12 = (1, 3) ρ21 = (1, 3) ρ22 = {(1, 1), (1, 3), (2, 1), (2, 3)}.

(2) (1, 1) ρ13 = (1, 1) ρ23 = {(1, 1), (2, 1)}.

(3) (1, 1) ρ31 = (1, 1) ρ32 = {(1, 1), (1, 3)}.

(4) (4, 4) ρ23 = (4, 4) ρ33 = {(4, 4)}.

–  –  –

The Green’s equivalence H (see (4) on page 5) plays a major role in analyzing 0-tight completely 0-simple semigroups. Notice that H is a congruence on a completely 0-simple semigroup. We consider a morphism which maps a rectangular 0-band Y into the quotient semigroup S/H by sending (i, λ) into H-class Hiλ, where Hiλ is either a group or Hiλ 2 = 0 We can show that Y is isomorphic to S/H. In other words, a completely 0-simple semigroup S is a rectangular 0-band Y of H-classes, where an Hclass H is either a group or H 2 = 0. Conversely, suppose S is a rectangular 0-band Y of sets Hα with α ∈ Y. Here Hα is either a group or Hα 2 = 0. We can show that S is a completely 0-simple semigroup.

Theorem 3.1.



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