On an equivalence between regular ordered Γ-semigroups and regular ordered semigroups

Lemma 2.1

Every element of Ω γ 0 can be represented by an irreducible word which has the form ( γ , x , γ ′ ) , ( γ , x ) , ( x , γ ) , γ or x, where x ∈ S and γ , γ ′ ∈ Γ .

Proof

First, we have to prove that the reduction system arising from the given presentation is Noetherian and confluent, and therefore any element of Ω γ 0 is given by a unique irreducible word from S ∪ Γ . Second, we have to prove that the irreducible words have one of these five forms. So if ω is a word of the form ω = ( u , x , γ , y , v ) for γ ∈ Γ , x , y ∈ S and u , v possibly empty words, then ω reduces to ω ′ = ( u , x γ y , v ) . And if ω = ( u , x , y , v ) , then it reduces to ω ′ = ( u , x γ 0 y , v ) . In this way, we obtain a reduction system which is length reducing and therefore it is Noetherian. To prove that this system is confluent, from Newman’s lemma, it is sufficient to prove that it is locally confluent. For this, we need to see only the overlapping pairs.

1. ( x , y , z ) → ( x γ 0 y , z ) and ( x , y , z ) → ( x , y γ 0 z ) which both reduce to ( x γ 0 y γ 0 z ) ;

2. ( x , γ , y , z ) → ( x γ y , z ) and ( x , γ , y , z ) → ( x , γ , y γ 0 z ) which both reduce to ( x γ y γ 0 z ) ;

3. ( x , y , γ , z ) → ( x γ 0 y , γ , z ) and ( x , y , γ , z ) → ( x , y γ z ) which both reduce to ( x γ 0 y γ z ) ;

4. ( x , γ , y , γ ′ , z ) → ( x γ y , γ ′ , z ) and ( x , γ , y , γ ′ , z ) → ( x , γ , y γ ′ z ) which both reduce to ( x γ y γ ′ z ) ;

5. ( γ 1 , γ 2 , γ 3 ) → ( γ 1 • γ 2 , γ 3 ) and ( γ 1 , γ 2 , γ 3 ) → ( γ 1 , γ 2 • γ 3 ) which both reduce to γ 1 • γ 2 • γ 3 .

To complete the proof, we need to show that the irreducible word representing the element of Ω γ 0 has one of the five forms stated. If the word which has neither a prefix nor a suffix made entirely of letters from Γ , then it reduces to an element of S by performing the appropriate reductions. If the word has the form ( α , ω , α ′ ) , ( α , ω ) or ( ω , α ′ ) , where ω is a word which has neither a prefix nor a suffix made entirely of letters from Γ , and α , α ′ have only letters from Γ , then it reduces to an element of one of the first three forms.□

Definition 2.1

We define an order relation ≤ Ω γ 0 in terms of ≤ S as follows:

1. For every x , y ∈ S , we let x ≤ Ω γ 0 y ⇔ x ≤ S y ;

2. For every x , y ∈ S , and γ ∈ Γ we let γ x ≤ Ω γ 0 γ y ⇔ x ≤ S y ;

3. For every x , y ∈ S , and γ ∈ Γ we let x γ ≤ Ω γ 0 y γ ⇔ x ≤ S y ;

4. For every x , y ∈ S , and γ ∈ Γ we let, γ x γ ′ ≤ Ω γ 0 γ y γ ′ ⇔ x ≤ S y ;

5. The restriction of the relation in Γ is taken to be the equality.

Proposition 2.1

The triple ( Ω γ 0 , ⋅ , ≤ Ω γ 0 ) is an ordered semigroup.

Definition 2.2

For C ⊆ S , we define L S ( C ) = { x ∈ S : ∃ c ∈ C such that x ≤ S c } , and for every D ⊆ Ω γ 0 , L Ω γ 0 ( D ) = { w ∈ Ω γ 0 : ∃ d ∈ D such that w ≤ Ω γ 0 d } .

Lemma 2.2

L Ω γ 0 ( C ) = L S ( C ) for C ⊆ S .

Proof

First, we prove that L S ( C ) ⊆ L Ω γ 0 ( C ) . If y ∈ L S ( C ) , then y ≤ S c for c ∈ C . Since y , c ∈ S , then y ≤ S c by Definition 2.1 is equivalent to y ≤ Ω γ 0 c , hence y ∈ L Ω γ 0 ( C ) .

Conversely, if ω ∈ L Ω γ 0 ( C ) , then w ≤ Ω γ 0 c for c ∈ C and by Definition 2.1 we must have that w ∈ S , and that ω ≤ S c , proving that ω ∈ L S .□

Lemma 2.3

Let x ∈ S by an arbitrary element. The following hold true.

1. The principal left ordered ideal of Ω γ 0 generated by x is the set ( x ) l ≤ Ω γ 0 = ( x ) l ≤ S ∪ Γ ( x ) l ≤ S , where ( x ) l ≤ S = L S ( x ∪ S Γ x ) is the left ordered ideal of S generated by x and Γ ( x ) l ≤ S = { γ y | γ ∈ Γ , y ∈ L S ( S Γ x ∪ x ) } .

2. The principal right ordered ideal of Ω γ 0 generated by x is the set ( x ) r ≤ Ω γ 0 = ( x ) r ≤ S ∪ ( x ) r ≤ S Γ , where ( x ) r ≤ S = L S ( x Γ S ∪ x ) is the right ordered ideal of S generated by x and ( x ) r ≤ S Γ = { y γ | γ ∈ Γ , y ∈ L S ( x Γ S ∪ x ) } .

Proof

We obtain the proof for (i) since the proof for (ii) is dual to that of (i). So we have to prove that A ∈ ( x ) l ≤ Ω γ 0 if and only if A ∈ ( x ) l ≤ S ∪ Γ ( x ) l ≤ S . Indeed, if A ∈ ( x ) l ≤ Ω γ 0 , then A ≤ Ω γ 0 B , where B ∈ L Ω γ 0 ( Ω γ 0 x ∪ x ) . If B = x , we have A ≤ γ 0 x and by Definition 2.1 A ≤ S x , so A ∈ ( x ) l ≤ S . If B ∈ Ω γ 0 x , then B may have these forms:

1. B = ( α y β ) x = α ( y β x ) . In this case, A ≤ Ω γ 0 α ( y β x ) which forces A = α z , where z ≤ S y β x , hence z ∈ ( x ) l ≤ S and then A ∈ Γ ( x ) l ≤ S .

2. B = ( α y ) x = α ( y γ 0 x ) . In this case, A ≤ Ω γ 0 α ( y γ 0 x ) . One can see that in the same way as above, A ∈ Γ ( x ) l ≤ S .

3. B = ( y α ) x = y α x . In this case, A ≤ Ω γ 0 y α x and by definition 2.1 we have that A ≤ S y α x , therefore, A ∈ ( x ) l ≤ S .

4. B = ( α ) x = α x . In this case, A ≤ Ω γ 0 α x , then A = α z , where z ≤ S x , hence z ∈ ( x ) l ≤ S and then A ∈ Γ ( x ) l ≤ S .

Conversely, if A ∈ ( x ) l ≤ S ∪ Γ ( x ) l ≤ S , then either A ≤ S B where B ∈ ( x ) l ≤ S or A ≤ Ω γ 0 B , where B = α C with α ∈ Γ and C ∈ ( x ) l ≤ S . In the first case, it follows at once that A ∈ ( x ) l ≤ Ω γ 0 . In the second case, when B = α C and C ∈ ( x ) l ≤ S , the inequality A ≤ Ω γ 0 α C implies that A = α A ′ with A ′ ≤ S C , and as a consequence A ′ ∈ ( x ) l ≤ S . Thus, A ∈ Γ ( x ) l ≤ S .□

Proposition 3.1

S is a regular ordered Γ -semigroup if and only if Ω γ 0 is a regular ordered semigroup.

Proof

If S is a regular ordered Γ -semigroup, then for all a ∈ S , ∃ x ∈ S and γ 1 , γ 2 ∈ Γ , such that a ≤ S a γ 1 x γ 2 a . To prove Ω γ 0 is a regular ordered semigroup, we have to prove that every element of Ω γ 0 have an ordered inverse in Ω γ 0 . By Lemma 2.1, we have that the elements of Ω γ 0 can be represented by an irreducible word which has only five forms. We prove regularity for elements of each of these five forms. So let first α 1 a α 2 ∈ Ω γ 0 . To find its ordered inverse we take a ≤ S a γ 1 x γ 2 a and then by Definition 2.1 we have α 1 a α 2 ≤ Ω γ 0 α 1 ( a γ 1 x γ 2 a ) α 2 = ( α 1 a α 2 ) ( α 2 − 1 γ 1 x γ 2 α 1 − 1 ) ( α 1 a α 2 ) which tells us that α 1 a α 2 is regular in Ω γ 0 and α 2 − 1 γ 1 x γ 2 α 1 − 1 ∈ Ω γ 0 is its ordered inverse. Second, for showing that α a ∈ Ω γ 0 is regular, we take a ≤ S a γ 1 x γ 2 a and then by Definition 2.1 we can write α a ≤ Ω γ 0 α ( a γ 1 x γ 2 a ) = ( α a ) ( γ 1 x γ 2 α − 1 ) ( α a ) . This tells us that α a is regular and as its inverse we can take γ 1 x γ 2 α − 1 ∈ Ω γ 0 . In the same way, one may prove that a α ∈ Ω γ 0 is regular with inverse α − 1 γ 1 x γ 2 ∈ Ω γ 0 . Furthermore, we prove that a ∈ Ω γ 0 is regular with inverse γ 1 x γ 2 ∈ Ω γ 0 , since by Definition 2.1, a ≤ S a γ 1 x γ 2 a and then a ≤ Ω γ 0 a γ 1 x γ 2 a . And finally, γ ∈ Ω γ 0 is regular since γ ≤ Ω γ 0 γ γ − 1 γ . Hence, we showed that Ω γ 0 is a regular ordered semigroup. Conversely, if Ω γ 0 is regular ordered semigroup, then every a ∈ S has an inverse in Ω γ 0 . To show that S is a regular ordered Γ -semigroup, we show that very a ∈ S has an inverse in S. For this, we distinguish between the five following forms. First, if the inverse of a in Ω γ 0 has the form α x β ∈ Ω γ 0 , for x ∈ S , then a ≤ Ω γ 0 a α x β a by Definition 2.1 implies that a ≤ S a α x β a , showing that a is regular in S. Second, if the inverse of a in Ω γ 0 has the form α x , then a ≤ Ω γ 0 a ( α x ) a = a α x γ 0 a and then Definition 2.1 implies that a ≤ S a α x γ 0 a , which means that a is regular in S. Third, the inverse of a in Ω γ 0 has the form x α , this case is similar to the second case. Fourth, the inverse of a in S is x, then a ≤ Ω γ 0 a x a = a γ 0 x γ 0 a and by Definition 2.1 a ≤ S a γ 0 x γ 0 a , which means that a is regular in S. Finally, if the inverse have the form α ∈ Γ , then a ≤ Ω γ 0 a γ a ≤ Ω γ 0 a γ a γ a and by Definition 2.1 we have that a ≤ S a γ a γ a and a is regular in the ordered Γ semigroup S.□

Lemma 3.1

If ( S , Γ , ≤ S ) is an ordered Γ -semigroup such that one-sided ideals of ( S , Γ , ≤ S ) are idempotent, and for every right ideal R and every left ideal L of ( S , Γ , ≤ S ) , L S ( R Γ L ) is a quasi ideal of ( S , Γ , ≤ S ) , then for every a ∈ S , ( a ) r ≤ S ∩ ( a ) l ≤ S = L S ( ( a ) r ≤ S Γ ( a ) l ≤ S ) .

Proof

Observe first that

from which we derive that ( a ) r ≤ S = L S ( a Γ S ) . In a similar fashion, one can get that ( a ) l ≤ S = L S ( S Γ a ) . Furthermore, since

is a quasi ideal, we have

From this and the previous assumptions, we obtain

proving thus the nonobvious inclusion ( a ) r ≤ S ∩ ( a ) l ≤ S ⊆ L S ( ( a ) r ≤ S Γ ( a ) l ≤ S ) .□

Theorem 3.1

The following are logically equivalent.

1. An ordered Γ -semigroup ( S , Γ , ≤ S ) is regular if and only if, one-sided ideals of ( S , Γ , ≤ S ) are idempotent, and for every right ideal R and every left ideal L of ( S , Γ , ≤ S ) , L S ( R Γ L ) is a quasi ideal of ( S , Γ , ≤ S ) .

2. An ordered semigroup ( S , ⋅ , ≤ S ) is regular if and only if, one-sided ideals of ( S , ⋅ , ≤ S ) are idempotent, and for every right ideal R and every left ideal L of ( S , Γ , ≤ S ) , L S ( R Γ L ) is a quasi ideal of ( S , ⋅ , ≤ S ) .

Proof

( i ) ⇒ ( i i ) is trivial since any regular ordered semigroup can be regarded as an regular ordered Γ -semigroup, where Γ is a singleton. Also, one-sided ideals and quasi ideals in ordered semigroups are the same as those in ordered Γ -semigroups when Γ is a singleton.

( i i ) ⇒ ( i ) . Assume first that the ordered Γ -semigroup S is regular. Then by Proposition 3.1, Ω γ 0 is a regular ordered semigroup. Theorem 3.2 of [4] implies that for every right ideal R and every left ideal L of Ω γ 0 , R ∩ L = L Ω γ 0 ( R L ) , L Ω γ 0 ( R L ) is a quasi-ideal of Ω γ 0 , and also every right and left ideal of the semigroup Ω γ 0 is idempotent. Let A now be an ordered right ideal of S and consider the subset R = L Ω γ 0 ( A ∪ A Γ ) of Ω γ 0 . We show that R = L Ω γ 0 ( A ∪ A Γ ) is a right ideal of the ordered semigroup Ω γ 0 . To this end, we have to prove that it satisfies the two conditions of right ideals: (1) L Ω γ 0 ( A ∪ A Γ ) Ω γ 0 ⊆ L Ω γ 0 ( A ∪ A Γ ) , (2) if a ∈ L Ω γ 0 ( A ∪ A Γ ) and b ≤ Ω γ 0 a , then b ∈ L Ω γ 0 ( A ∪ A Γ ) . To show the first we have to show that for every b ∈ L Ω γ 0 ( A ∪ A Γ ) and every C ∈ Ω γ 0 , b C ∈ L Ω γ 0 ( A ∪ A Γ ) . Since b ∈ L Ω γ 0 ( A ∪ A Γ ) , either b ≤ S a where a ∈ A , or b ≤ Ω γ 0 a γ , in which case b = a ′ γ and a ′ ≤ S a . In the first case, when a , b ∈ S and b ≤ S a , it follows immediately that b C ≤ Ω γ 0 a C whatever the value of C is, since ≤ Ω γ 0 is a compatible relation. But still we have to prove that a C ∈ L Ω γ 0 ( A ∪ A Γ ) and this depends on the value of C. Since

then

where the last inclusion comes from the fact that A is an ordered right ideal of ( S , Γ , ≤ S ) . It remains to prove that the same holds true in the second case when b = a ′ γ with a ′ ∈ S and a ′ ≤ S a , and γ ∈ Γ . Depending on the value of C we have to prove that a γ C ∈ L Ω γ 0 ( A ∪ A Γ ) . Indeed,

All the above verifications prove the first condition, while the second condition is obvious. So R = L Ω γ 0 ( A ∪ A Γ ) is a right ideal of the ordered semigroup Ω γ 0 and from [4] it follows that R = L Ω γ 0 ( A ∪ A Γ ) is an idempotent. Passing now from the ordered semigroup Ω γ 0 to the ordered Γ semigroup S, we show that if every right ideal of the ordered semigroup Ω γ 0 is idempotent, then so is every right ideal of the ordered Γ -semigroup S. Let A be a right ideal of the ordered Γ semigroup S, we have to prove A is an idempotent in S, that is, L S ( A Γ A ) = A . Since A is a right ideal, L S ( A Γ A ) ⊆ L S ( A Γ S ) ⊆ A . To prove the converse, we utilize the fact that L Ω γ 0 ( A ∪ A Γ ) is an idempotent in Ω γ 0 , thus

This implies that every a ∈ A is lower with respect to ≤ Ω γ 0 than some element of A or some element of A Γ . The second case is impossible from the way we have defined ≤ Ω γ 0 , so it remains that there is some γ ∈ Γ , and a ′ , a ″ ∈ A such that a ≤ Ω γ 0 a ′ γ a ″ . But this is the same as to say that a ≤ S a ′ γ a ″ , so a ∈ L S ( A Γ A ) , and as a result A ⊆ L S ( A Γ A ) . One can show that left ideals of S too are idempotent by first proving in a similar fashion to above that for every left ideal B of S, the set L Ω γ 0 ( B ∪ Γ B ) is a left ideal of Ω γ 0 . Finally, if A is a right ideal and B a left ideal of the ordered Γ semigroup S, we have to prove that L S ( A Γ B ) is a quasi ideal of ( S , Γ , ≤ S ) , which means that: