A characterization of translational hulls of a strongly right type B semigroup

Lemma 2.1

[1] Let S be a semigroup and a, b ∈ S. Then the following statements are equivalent:

1. a 𝓛^* b ;

2. for all x, y ∈ S¹, ax = ay if and only if bx = by.

Corollary 2.2

[1] Let S be a semigroup and e² = e, a ∈ S. Then the following statements are true:

1. a 𝓛^* e if and only if ae = a and for all x, y ∈ S¹, ax = ay implies ex = ey;

2. 𝓛^* is a right congruence on S.

Lemma 2.3

[1] Let S be a right adequate semigroup and a, b ∈ S. Define μ_L = {(a, b) ∈ S × S | (ea)^* = (eb)^* for all e ∈ E(S¹)}. Then μ_L is the largest idempotent-separating congruence on S contained in 𝓛^*.

Lemma 2.4

[1] Let S be a right adequate semigroup. Then the following statements are true:

1. if S is a right ample semigroup, then S can be embedded into an inverse semigroup;

2. if S is a right type B semigroup which is right fundamental, then S is right ample.

Lemma 2.5

[15] Let S be a right type B semigroup. Define a relation on S as follows: (a, b) ∈ σ ⟺ ea = eb, for some e ∈ E(S). Then σ is the least left cancellative monoid congruence on S.

Example 2.1

(Due to Fountain [1]) Let ℕ be the set of all non-negative integers and put I = ℕ × ℕ, S = ℕ ∪ I. Define a multiplication “∘” on S as follows:

It is readily verified that “∘” is associative, that the set of idempotents of S is {0, (0, 0)}. It is not difficult to check that the 𝓛^*–classes of S are ℕ and I. In fact, Fountain [2] proved that S is a right type B semigroup. However, S itself is not strongly rpp, for if otherwise, then

But since (0, 0) ∘ (m, n) = (0, m + n), we have (0, 0) ∘ (m, n) ≠ (m, n). This contradiction shows S is not a strongly right type B semigroup.

Example 2.2

Let ℕ be the set of all non-negative integers and S = {(m, n) ∈ ℕ × ℕ | m ≥ n}. Define a multiplication “∙” on S by

where t = max{n, p}. Then, it is readily verified that (S, ∙) is a semigroup and E(S) = {(m, m) ∈ ℕ × ℕ}. In addition, it is easy to check that S is right type B. On the other hand, since for all (m, n) ∈ S, there exists a unique idempotent (n, n) ∈ S such that

We have that S is a strongly rpp semigroup with (m, n)⁰⁰ = (n, n).

Now, we show that S is also proper. For this purpose, let (m, n), (p, q) ∈ S be such that (m, n)[𝓛^* ∩ σ](p, q). Then m ≥ n, p ≥ q, (m, n) 𝓛^* (p, q) and (m, n)σ (p, q). Hence, n = q, and there exists (k, k) ∈ E(S) such that (k, k) ∙ (m, n) = (k, k) ∙ (p, q). That is, (k, k) ∙ (m, n) = (k, k) ∙ (p, n). Hence, (t, n – m + t) = (s, n – p + s), where t = max{k, m} and s = max{k, p}. Thus t = s and n – m + t = n – p + s, this gives m = p. Therefore, (m, n) = (p, q). That is, S is a proper strongly right type B semigroup.

Corollary 2.6

Let S be a proper right type B semigroup. Then S is right ample.

Proof

Let a ∈ S, e ∈ E(S). Then a𝓛^*a^* and ea𝓛^*(ea)^*. Hence a(ea)^*𝓛^*a^*(ea)^* and ea = eaa^*𝓛^*(ea)^*a^* since 𝓛^* is a right congruence on S. Note that E(S) is a semilattice. We have a(ea)^*𝓛^* a^*(ea)^* = (ea)^*a^*𝓛^* ea. On the other hand, it is easy to see that (1, e) ∈ σ and (1, (ea)^*) ∈ σ from Lemma 2.5. Hence (a, ea) ∈ σ and (a, a(ea)^*) ∈ σ since σ is a congruence on S from Lemma 2.5. Therefore, ea [𝓛^* ⋂ σ] a(ea)^*. But S is proper, we have ea = a(ea)^*. This gives that S is right ample.□

Lemma 3.1

Let S be an rpp semigroup, (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S). Then the following statements are true:

1. ρ₁ = ρ₂ ⟺ (∀ e ∈ E(S)) eρ₁ = eρ₂;

2. if S is strongly rpp, then λ₁ = λ₂ ⟺ (∀ e ∈ E(S)) λ₁e = λ₂e;

3. if E(S) is a semilattice, then (λ₁, ρ₁) = (λ₂, ρ₂) ⟺ λ₁ = λ₂;

4. if S is strongly right type B, then

   (λ₁, ρ₁) = (λ₂, ρ₂) ⟺ ρ₁ = ρ₂ ⟺ λ₁ = λ₂;

Proof

(1) and (2) are trivial.

(3) Let λ₁ = λ₂. Then λ₁f = λ₂f for all f ∈ E(S). Hence, for all e ∈ E(S), we have

since E(S) is a semilattice. Choose an idempotent (eρ₁)^* of S to replace the element f of the above formula. We have (eρ₁)^* = (eρ₂)^*(eρ₁)^*. Similarly, we can get (eρ₂)^* = (eρ₁)^*(eρ₂)^*. Hence, (eρ₁)^* = (eρ₂)^* since E(S) is a semilattice. Thus, for all e ∈ E(S), we have

Therefore, by (1), ρ₁ = ρ₂. That is, (λ₁, ρ₁) = (λ₂, ρ₂).

The converse is clear.

(4) By (3), we only need to prove that ρ₁ = ρ₂ implies that (λ₁, ρ₁) = (λ₂, ρ₂). To see it, let ρ₁ = ρ₂. Then, by(1), for all e ∈ E(S), we have

Similarly, λ₂e = (λ₂e)⁰⁰λ₁e. Hence, λ₁e𝓛λ₂e, and so λ₁e𝓛^*λ₂e. Note that S is strongly right type B. We have (λ₁e)⁰⁰ = (λ₂e)⁰⁰. Hence, λ₁e = λ₂e. By (2), λ₁ = λ₂. Therefore, (λ₁, ρ₁) = (λ₂, ρ₂), as required.□

Proposition 3.2

Let S be a strongly right type B semigroup, and (λ, ρ) ∈ Ω(S). Then the following statements are true:

1. eρ⁰⁰ = λ⁰⁰e = (λe)⁰⁰ ∈ E(S), for all e ∈ E(S);

2. (λ⁰⁰, ρ⁰⁰) ∈ E(Ω(S));

3. (λ, ρ)𝓛^*(λ⁰⁰, ρ⁰⁰);

4. eρ⁰⁰ρ𝓛eρ, for all e ∈ E(S);

5. (λ⁰⁰, ρ⁰⁰)(λ, ρ) = (λ, ρ);

6. E(Ω(S)) = {(λ, ρ) ∈ Ω(S) | λE(S) ⋃ E(S)ρ ⊆ E(S)}.

Proof

1. Let e ∈ E(S). Then by the Definitions of λ⁰⁰ and ρ⁰⁰, we have eρ⁰⁰ = e(λe)⁰⁰ = (λe)⁰⁰e = λ⁰⁰e since E(S) is a semilattice. And (λ⁰⁰e)² = (λ⁰⁰e)(λ⁰⁰e) = (λe)⁰⁰e(λe)⁰⁰e = (λe)⁰⁰ee = (λe)⁰⁰e = λ⁰⁰e ∈ E(S). Note that 𝓛^* is a right congruence. We have λ⁰⁰e = (λe)⁰⁰e𝓛^*(λe)e = λe𝓛^* (λe)⁰⁰. Again since S is 𝓛^*–unipotent, we get that λ⁰⁰e = (λe)⁰⁰. That is, eρ⁰⁰ = λ⁰⁰e = (λe)⁰⁰ ∈ E(S).

2. We first prove that (λ⁰⁰, ρ⁰⁰) ∈ Ω(S). To see it, let a, b ∈ S. Then by the Definitions of λ⁰⁰ and ρ⁰⁰ and (1), we have (ab)ρ⁰⁰ = (ab)(λ(ab)⁰⁰)⁰⁰ = (ab)(λ(ab)⁰⁰b⁰⁰)⁰⁰ = (ab)(λb⁰⁰(ab)⁰⁰)⁰⁰ = ab(λb⁰⁰)⁰⁰(ab)⁰⁰ = (ab)(ab)⁰⁰(λb⁰⁰)⁰⁰ = ab(λ b⁰⁰)⁰⁰ = a(bρ⁰⁰). Hence, ρ⁰⁰ is a right translation of S. Similarly, we can prove that λ⁰⁰ is a left translation of S.

   On the other hand, by (1), we have

   a(λ00b)=a(λ00b00)b=aa00(λ00b00)b=aa00(b00ρ00)b=a(a00b00)ρ00b=a(b00a00)ρ00b=ab00(a00ρ00)b=a(a00ρ00)b00b=(aρ00)b,

   this yields that (λ⁰⁰, ρ⁰⁰) ∈ Ω(S).

   Now, we prove that (λ⁰⁰, ρ⁰⁰) ∈ E(Ω(S)). To see it, let e ∈ E(S). Then by (1), e(ρ⁰⁰)² = (eρ⁰⁰)ρ⁰⁰ = (e(eρ⁰⁰))ρ⁰⁰ = (eρ⁰⁰e)ρ⁰⁰ = eρ⁰⁰ eρ⁰⁰ = eρ⁰⁰. By Lemma 3.1(1), (ρ⁰⁰)² = ρ⁰⁰. Thus, by Lemma 3.1(4), (λ⁰⁰, ρ⁰⁰)² = (λ⁰⁰, ρ⁰⁰)(λ⁰⁰, ρ⁰⁰) = ((λ⁰⁰)², (ρ⁰⁰)²) = (λ⁰⁰, ρ⁰⁰) ∈ E(Ω(S)), as required.

3. Let e ∈ E(S). Then by (1), we have λλ⁰⁰e = λ(eρ⁰⁰) = λ(eeρ⁰⁰) = (λe)(eρ⁰⁰) = (λe)(λ⁰⁰e) = (λe)(λe)⁰⁰ = λe. Hence, by Lemma 3.1(2), λλ⁰⁰ = λ. Therefore, by Lemma 3.1(4), (λ, ρ)(λ⁰⁰, ρ⁰⁰) = (λλ⁰⁰, ρρ⁰⁰) = (λ, ρ).

   On the other hand, let (λ₁, ρ₁), (λ₂, ρ₂) ∈ [Ω(S)]¹, and

   (λ,ρ)(λ1,ρ1)=(λ,ρ)(λ2,ρ2).

   Then, by Lemma 3.1(4), ρρ₁ = ρρ₂, and so eρρ₁ = eρρ₂ for all e ∈ E(S). That is, eρ[(eρ)⁰⁰ρ₁] = eρ[(eρ)⁰⁰ρ₂]. Hence, by Corollary 2.2, (eρ)⁰⁰(eρ)⁰⁰ρ₁ = (eρ)⁰⁰(eρ)⁰⁰ρ₂. That is,

   (eρ)00ρ1=(eρ)00ρ2foralle∈E(S).

   Choose an idempotent (λe)⁰⁰ of S to replace the element e of the above formula. We have

   ((λe)00ρ)00ρ1=((λe)00ρ)00ρ2.

   Note that ((λe)⁰⁰ρ)e = (λe)⁰⁰(λe) = λe. We have ((λe)⁰⁰ρ)⁰⁰e = (λe)⁰⁰. Thus,

   eρ00ρ1=(eρ00)ρ1=(λ00e)ρ1=(λe)00ρ1=[((λe)00ρ)00e]ρ1=[e((λe)00ρ)00]ρ1=e((λe)00ρ)00ρ1=e((λe)00ρ)00ρ2=[((λe)00ρ)00e]ρ2=(λe)00ρ2=eρ00ρ2,

   which implies that ρ⁰⁰ρ₁ = ρ⁰⁰ρ₂. By Lemma 3.1(4),

   (λ00λ1,ρ00ρ1)=(λ00λ2,ρ00ρ2).

   That is,

   (λ00,ρ00)(λ1,ρ1)=(λ00,ρ00)(λ2,ρ2).

   This together with the fact (λ, ρ) = (λ, ρ)(λ⁰⁰, ρ⁰⁰), yields that (λ, ρ)𝓛^*(λ⁰⁰, ρ⁰⁰).

4. Obviously, for any a ∈ S, a𝓛^* a⁰⁰ implies that aρ𝓛^* a⁰⁰ρ, where ρ is a right translation of S. Hence, for all e ∈ E(S), we have

   (eρ)00ρ00L∗(eρ)ρ00=eρρ00=eρL∗(eρ)00

   since ρρ⁰⁰ = ρ (i.e (λ, ρ) = (λ, ρ)(λ⁰⁰, ρ⁰⁰)). Thus, (eρ)⁰⁰ = (eρ)⁰⁰ρ⁰⁰ since S is 𝓛^*–unipotent. Note that eρ⁰⁰ eρ = eρ⁰⁰ρ. We have

   eρ=(eρ)00(eρ)=((eρ)00ρ00)(eρ)=(e(eρ)00ρ00)ρ=(e(eρ)00)ρ00ρ=((eρ)00e)ρ00ρ=(eρ)00eρ00ρ

   Therefore, eρ⁰⁰ρ𝓛eρ for all e ∈ E(S), as required.

5. We first prove that eρ⁰⁰ρ = eρ for all e ∈ E(S).

   (eρ00ρ)(eρ)00=[(eρ00)ρ](eρ)00=[(λ00e)ρ](eρ)00=(λ00e)(λ(eρ)00)=(λ00e)[(λ(eρ)00)00](λ(eρ)00)=(λ00e)[(λ(eρ)00)00(eρ)00](λ(eρ)00)=λ00[e(λ(eρ)00)00(eρ)00](λ(eρ)00)=λ00[(eρ)00e(λ(eρ)00)00](λ(eρ)00)=λ00(eρ)00e(λ(eρ)00)00λ(eρ)00=λ00(eρ)00eλ(eρ)00=eλ00(eρ)00λ(eρ)00=e(λ(eρ)00)00λ(eρ)00=e(λ(eρ)00)=(eρ)(eρ)00=eρ.

   On the other hand, by (4), we have eρ⁰⁰ρ𝓛^*eρ and eρ = (eρ)(eρ)⁰⁰. Thus, by Lemma 2.1, we have eρ⁰⁰ρ = eρ⁰⁰ρ(eρ)⁰⁰ = eρ. By Lemma 3.1(1), ρ = ρ⁰⁰ρ. Therefore, by Lemma 3.1(4), (λ⁰⁰, ρ⁰⁰)(λ, ρ) = (λ, ρ).

6. Let (λ, ρ) ∈ Ω(S) be such that λ E(S) ⋃ E(S)ρ ⊆ E(S). Then eρ ∈ E(S) for all e ∈ E(S). Hence, eρ² = (eeρ)ρ = (eρ)eρ = eρ, and so ρ² = ρ. By Lemma 3.1(4), (λ, ρ)² = (λ, ρ) ∈ E(Ω(S)).

Conversely, if (λ, ρ) ∈ E(Ω(S)), then (λ⁰⁰, ρ⁰⁰) = (λ⁰⁰, ρ⁰⁰)(λ, ρ) since (λ⁰⁰, ρ⁰⁰)𝓛^*(Ω(S))(λ, ρ). On the other hand, by (5), (λ, ρ) = (λ⁰⁰, ρ⁰⁰)(λ, ρ). Thus (λ, ρ) = (λ⁰⁰, ρ⁰⁰), and so λ = λ⁰⁰, ρ = ρ⁰⁰, this gives λ E(S) ⋃ E(S)ρ ⊆ E(S). This completes the proof.□

Theorem 3.3

Let S be a strongly right type B semigroup. Then so is Ω(S).

Proof

By Proposition 3.2 (2) and (3), Ω(S) is an rpp semigroup.

Now, we prove that Ω(S) is right adequate. To see it, let (λ₁, ρ₁), (λ₂, ρ₂) ∈ E(Ω(S)). Then, by Proposition 3.2 (6), eρ₁, eρ₂ ∈ E(S) for all e ∈ E(S). Hence, eρ₁ρ₂ = eρ₁eρ₂ = eρ₂eρ₁ = eρ₂ρ₁, which implies that ρ₁ρ₂ = ρ₂ρ₁. By Lemma 3.1(4),

That is,

Therefore, E(Ω(S)) is a semilattice. That is, Ω(S) is a right adequate semigroup. This together with Proposition 3.2(2), (3) and (5), yields that Ω(S) is a strongly rpp semigroup with semilattice of idempotents E(Ω(S)).

Next, we only need to prove that Ω(S) satisfies Conditions (B1) and (B2). To see it, let (λ₁, ρ₁), (λ₂, ρ₂) ∈ E[(Ω(S))¹], (λ, ρ) ∈ Ω(S), and e ∈ E(S). Then

By Lemma 3.1(1), (ρ₁ρ)⁰⁰(ρ₂ρ)⁰⁰ = (ρ₁ρ₂ρ)⁰⁰. On the other hand, we have

and

By Lemma 3.1(4), [(λ₁, ρ₁)(λ₂, ρ₂)(λ, ρ)]^* = [(λ₁, ρ₁)(λ, ρ)]^*[(λ₂, ρ₂)(λ, ρ)]^*. Therefore, Ω(S) satisfies Condition (B1).

Let (λ₁, ρ₁) ∈ E(Ω(S)), (λ, ρ) ∈ Ω(S) be such that (λ₁, ρ₁) ≤ (λ, ρ)^*. Then (λ₁, ρ₁) ≤ (λ, ρ)⁰⁰ = (λ⁰⁰, ρ⁰⁰) since Ω(S) is 𝓛^*–unipotent. Hence (λ₁, ρ₁) = (λ₁, ρ₁)(λ⁰⁰, ρ⁰⁰) = (λ₁λ⁰⁰, ρ₁ρ⁰⁰), and so ρ₁ = ρ₁ρ⁰⁰. By Lemma 3.1(1), eρ₁ = eρ₁ρ⁰⁰ = eρ₁eρ⁰⁰ = eρ⁰⁰eρ₁ for all e ∈ E(S). Hence, eρ ≤ eρ⁰⁰ = (λe)⁰⁰ = (λe)^*. Again since S satisfies Condition (B2), we have

That is,

where λ_f (ρ_f) is the inner left (right) translation on S¹ determined by f ∈ E(S¹). By Lemma 3.1(1), ρ₁ = (ρ_fρ)⁰⁰. Hence, by Lemma 3.1(4),

where (λ_f, ρ_f) ∈ E[(Ω(S))¹]. That is, Ω(S) satisfies Condition (B2).

Summing up the above arguments, we conclude Ω(S) is a strongly right type B semigroup.□

Corollary 3.4

Let S be a strongly right type B semigroup. Then the following statements are true:

1. for all e ∈ E(S), (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S), (λ₁, ρ₁)𝓛^*(Ω(S))(λ₂, ρ₂) if and only if λ₁e𝓛^*(S)λ₂e;

2. for all (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S), (λ1,ρ1)μLΩ(S)(λ2,ρ2) if and only if for all e ∈ E(S), λ1eμLSλ2e;

Proof

1. By Theorem 3.3, Ω(S) is a strongly right type B semigroup. Let (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S) be such that (λ₁, ρ₁)𝓛^*(Ω(S))(λ₂, ρ₂). Then, by Proposition 3.2(3), (λ100,ρ100)L∗(Ω(S))(λ200,ρ200). Hence, (λ100,ρ100)=(λ200,ρ200) since Ω(S) is 𝓛^*–unipotent. Thus λ100=λ200. By Lemma 3.1(2), λ100e=λ200e for all e ∈ E(S)). Hence,

   λ1eL∗(S)(λ1e)00=λ100e=λ200e=(λ2e)00L∗(S)λ2e.

   Conversely, if for all e ∈ E(S), λ₁e𝓛^*(S)λ₂e, then λ100e=λ200e since S is both 𝓛^*–unipotent and strongly rpp. Hence, by Lemma 3.1(2), λ100=λ200. By Lemma 3.1(4), (λ100,ρ100)=(λ200,ρ200). Thus, by Proposition 3.2(3), (λ₁, ρ₁)𝓛^*(Ω(S))(λ₂, ρ₂).

2. Suppose that (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S) and (λ1,ρ1)μLΩ(S)(λ2,ρ2). Then for all f ∈ E(S),

   (λf,ρf)(λ1,ρ1)μLΩ(S)(λf,ρf)(λ2,ρ2),

   where (λ_f, ρ_f) ∈ E(Ω(S)). Hence,

   (λf,ρf)(λ1,ρ1)L∗(Ω(S))(λf,ρf)(λ2,ρ2),

   That is,

   (λfλ1,λfρ1)L∗(Ω(S))(λfλ2,λfρ2).

By (1), λ_fλ₁e𝓛^*(S)λ_fλ₂e for all e ∈ E(S). That is, fλ₁e𝓛^*(S)fλ₂e. Thus, λ1eμLSλ2e.

Conversely, if for all e ∈ E(S), λ1eμLSλ2e, then λ₁e𝓛^*(S)λ₂e. Note S is a strongly right type B semigroup. We have (λ₁e)⁰⁰ = (λ₂e)⁰⁰. Hence, for all (λ, ρ) ∈ Ω(S), we have

That is, λλ1eμLSλλ2e. Hence, λλ₁e 𝓛^*(S)λλ₂e. By (1), we have

That is,

Choose any idempotent (λ^′, ρ^′) of Ω(S) to replace the element (λ, ρ) of the above formula. We have (λ1,ρ1)μLΩ(S)(λ2,ρ2), as required.□

Proposition 4.1

Let S be a strongly right type B semigroup. Then for all e ∈ E(S), (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S),

(λ₁, ρ₁)σ_Ω(S)(λ₂, ρ₂) ⟺ eρ₁σ_Seρ₂ ⟺ λ₁eσ_Sλ₂e.

Proof

We first prove that (λ₁, ρ₁)σ_Ω(S)(λ₂, ρ₂) implies eρ₁σ_Seρ₂ and λ₁eσ_Sλ₂e for all e ∈ E(S). To see it, let (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S) such that (λ₁, ρ₁)σ_Ω(S)(λ₂, ρ₂). Then, by the definition of σ, there exists (λ, ρ) ∈ E(Ω(S)) such that

that is, (λλ₁, ρρ₁) = (λλ₂, ρρ₂). Hence, λλ₁ = λλ₂ and ρρ₁ = ρρ₂. By Lemma 3.1 (1) and (2), λλ₁e = λλ₂e and eρρ₁ = eρρ₂ for all e ∈ E(S). Thus,

Note that λ(λ₁e)⁰⁰, λ(λ₂e)⁰⁰, eρ ∈ E(S). We have eρeρ₁ = eρeρ₂ implies that eρ₁σ_Seρ₂. On the other hand,

where λe ∈ E(S). Thus, λ₁eσ_Sλ₂e.

Now, we prove that for all e ∈ E(S),

To see it, let eρ₁σ_Seρ₂. Then there exists f ∈ E(S) such that f(eρ₁) = f(eρ₂). That is, (fe)ρ₁ = (fe)ρ₂. Hence, (ef)ρ₁ = (ef)ρ₂ since E(S) is a semilattice. That is, eρ_fρ₁ = eρ_fρ₂. By Lemma 3.1(1), ρ_fρ₁ = ρ_fρ₂. Thus, by Lemma 3.1(4), we have (λ_fλ₁, ρ_fρ₁) = (λ_fλ₂, ρ_fρ₂). That is,

Note that (λ_f, ρ_f) ∈ E(Ω(S)). We have (λ₁, ρ₁)σ_Ω(S)(λ₂, ρ₂).

Similarly, we can prove that for all e ∈ E(S),

This completes the proof.□

Theorem 4.2

Let S be a proper strongly right type B semigroup. Then so is Ω(S).

Proof

By Theorem 3.3, Ω(S) is a strongly right type B semigroup. It only remains to show that Ω(S) is proper. To see it, let (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S) be such that (λ₁, ρ₁)[𝓛^*_Ω(S) ⋂ σ_Ω(S)](λ₂, ρ₂). Then, by Corollary 3.4(1) and Proposition 4.1, we have λ₁e𝓛^*(S)λ₂e and λ₁eσ_Sλ₂e for all e ∈ E(S). Hence, λ₁e[𝓛^*_S ⋂ σ_S]λ₂e. Again, since S is proper, we have λ₁e = λ₂e for all e ∈ E(S). By Lemma 3.1(2), λ₁ = λ₂. Therefore, by Lemma 3.1(4), (λ₁, ρ₁) = (λ₂, ρ₂). This completes the proof.□

Corollary 4.3

Let S be a strongly right type B semigroup which is right fundamental. Then so is Ω(S).

Proof

By Theorem 3.3, Ω(S) is strongly right type B. Let (λ₁, ρ₁), (λ₂, ρ₂) ∈ Ω(S) be such that (λ1,ρ1)μLΩ(S)(λ2,ρ2). Then, by Corollary 3.4(2), λ1eμLSλ2e for all e ∈ E(S). Again, since S is right fundamental, we have λ₁e = λ₂e. By Lemma 3.1(2), λ₁ = λ₂. Hence, by Lemma 3.1(4), (λ₁, ρ₁) = (λ₂, ρ₂). That is, μLΩ(S)=1Ω(S). This completes the proof.□

Corollary 4.4

Let S be a proper strongly right type B semigroup. Then Ω(S) is embeddable into an inverse semigroup.

Proof

It follows from Corollary 2.6, Lemma 2.4 and Theorem 4.2.□

Corollary 4.5

Let S be a strongly right type B semigroup which is right fundamental. Then Ω(S) is embeddable into an inverse semigroup.

Proof

It follows from Lemma 2.4 and Corollary 4.3.□