On Maximal Subsemigroups of Partial Baer-Levi Semigroups

Lemma 3.1. Let S be a semigroup and suppose that S is a disjoint union of a subsemigroup T and an ideal I of S. Then,

• (a)

  for any maximal subsemigroup M of T, M ∪ I is a maximal subsemigroup of S;

• (b)

  for any maximal subsemigroup N of S such that T∖N ≠ ∅ and T∩N ≠ ∅, the set T∩N is a maximal subsemigroup of T.

Proof. To see that (a) holds, let M be a maximal subsemigroup of T. Since I is an ideal, we have M ∪ I is a subsemigroup of S. Clearly, M ∪ I⊊T ∪ I = S. If a ∈ S∖(M ∪ I), then a ∈ T∖M and thus T = 〈M ∪ {a}〉⊆〈M ∪ I ∪ {a}〉. Since 〈M ∪ I ∪ {a}〉 contains I, we have S = T ∪ I =   〈M ∪ I ∪ {a}〉 and so M ∪ I is maximal in S as required.

To prove (b), let N be a maximal subsemigroup of S, where T∖N ≠ ∅ and T∩N ≠ ∅, and let a ∈ T∖N. Since N is maximal in S, we have 〈N ∪ {a}〉 = S. Thus, for each b ∈ T∖N, b = c₁c₂ ⋯ c_n for some natural n and some c_i ∈ N ∪ {a} for all i = 1,2, …, n. Since b ∉ N, we have c_i = a for some i. Moreover, since b ∉ I, we have c_j ∈ T∩N for all j ≠ i. It follows that T∖N⊆〈(T∩N)∪{a}〉, therefore

Corollary 3.2. Suppose that p > r > q ≥ ℵ₀. If M is a maximal subsemigroup of S_r, then $$ is a maximal subsemigroup of PS(q).

Lemma 3.3. Let p > q ≥ ℵ₀ and suppose that M is a maximal subsemigroup of PS(q). Then,

• (a)

  S_r∩M ≠ ∅ for all q ≤ r < p;

• (b)

  if there exists α ∉ M with g(α) < p, then S_k∖M ≠ ∅ for some q ≤ k < p.

Proof. To show that (a) holds, we first note that S_q is contained in S_r for all q ≤ r < p. If S_q∩M = ∅, then $$ and thus $$ by the maximality of M. But $$ where $$ is a subsemigroup of PS(q) (since $$ is an ideal), so we get a contradiction. Therefore, ∅ ≠ S_q∩M⊆S_r∩M for all q ≤ r < p.

To show that (b) holds, suppose there is α ∉ M with g(α) = k < p. If k < q, then α ∈ S_r∖M for all q ≤ r ≤ p. Otherwise, if q ≤ k, then α ∈ S_k∖M. Hence (b) holds.

Lemma 3.4. Suppose that p ≥ r > q ≥ ℵ₀. Then G_r = BL(q) · α · BL(q) for each α ∈ G_r.

Proof. Let α ∈ G_r and β, γ ∈ BL(q). Since

For the converse, if α, β ∈ G_r, then |X∖dom α | = r = |X∖dom β|. Since p > q, every element in PS(q) has rank p, so we write

Theorem 3.5. Suppose that p > q ≥ ℵ₀. Then M is a maximal subsemigroup of PS(q) if and only if M equals one of the following sets:

• (a)

  PS(q)∖G_p = {α ∈ PS(q) : g(α) < p};

• (b)

  $$, where q ≤ r < p and N is a maximal subsemigroup of S_r.

Proof. Let α, β ∈ PS(q) be such that g(α) < p and g(β) < p. Clearly |Xα∖dom β | ≤|X∖dom β | = g(β) < p. Then

We now suppose that M is a maximal subsemigroup of PS(q) such that M ≠ PS(q)∖G_p. Then there exists α ∉ M with g(α) < p. Thus, Lemma 3.3 implies that S_k∖M ≠ ∅ and S_k ∩ M ≠ ∅ for some k, where q ≤ k < p. Since $$, Lemma 3.1(b) implies that S_k∩M is maximal in S_k. We also see that

Lemma 3.6. Suppose that p ≥ q ≥ ℵ₀, and let A be a nonempty subset of X such that |X∖A | ≥ q. Then,

• (a)

  for any cardinal k such that 0 ≤ k ≤ p, there exist α, β ∈ PS(q) such that g(α) = k = g(β) and $$;

• (b)

  for each $$, |dom γ∖Aγ⁻¹ | = |X∖A | = |Xγ∖A| and |Aγ⁻¹ | = |A|.

Proof. To show that (a) holds, let |X∖A| = r ≥ q, and let k be a cardinal such that 0 ≤ k ≤ p. We write $$ where |R | = r and |Q | = q. If r = p, then |A ∪ R | ≥ r = p; if not, then |X∖A | < p, and this implies |A | = p, and so |A ∪ R | = p. Fix a ∈ A and let B = (A∖{a}) ∪ R. Then, |B | = p and |X∖B | = |Q ∪ {a}| = q. We write $$ where |K | = k and |L | = p. Then there exists a bijection α : L → B and so g(α) = k, d(α) = q. Also, since A ⊈ B = Xα, we have $$.

To find $$ with g(β) = k, we consider two cases. First, if r = p, we write $$ where |P| = p,   |Q| = q,    | K | = k. Fix a ∈ A and define

To see that (b) holds, suppose that there is $$, then A⊆Xγ and |Xγ∖A | ≥ q. So |Aγ⁻¹ | = |A| since γ is injective. Also,

Lemma 3.7. Suppose that p ≥ q ≥ ℵ₀, and let A be a nonempty subset of X such that |X∖A | ≥ q. Then $$ is a proper subsemigroup of PS(q).

Proof. Let $$. If A ⊈ Xαβ, then $$. Now we suppose that A⊆Xαβ. Then, A⊆Xβ and since $$, we either have Aβ⊆A⊆dom β, or |Xβ∖A | < q. If |Xβ∖A | < q, then

Remark 3.8. For any cardinal r such that q ≤ r ≤ p, $$ is a proper subsemigroup of S_r but it is not maximal when q < r. To see this, suppose $$ is maximal and choose $$ such that g(α) = r and g(β) = 0 (possible by Lemma 3.6(a)). Then $$ where dom β = X. Moreover $$, and so

Since M_A is maximal in BL(q), a subsemigroup of PS(q), it is natural to think that $$ is maximal in PS(q). But when p > q, by taking r = p, the above observation shows that this claim is false since S_p = PS(q). Thus, $$ is not always a maximal subsemigroup of PS(q).

Theorem 3.9. Suppose that p ≥ r ≥ q ≥ ℵ₀, and let A be a nonempty subset of X such that |X∖A | ≥ q. Then $$ is a maximal subsemigroup of S_r precisely when r = q.

Proof. In Remark 3.8, we have shown that $$ is not maximal in S_r when r > q. It remains to show $$ is maximal in S_q. Let $$. Then g(α), g(β) ≤ q and Lemma 3.6(b) implies that

Now define γ ∈ P(X) by

On the other hand, if A ⊈ dom β, then there exists w ∈ A∩(X∖dom β). In this case, we rewrite $$ and $$ where |J | = s, |K | = q. Like before, we write A = {a_i} and $$ where {b_j} = dom α∖Aα⁻¹, then

Lemma 3.10. Suppose that p ≥ r ≥ q ≥ ℵ₀. Let k be a cardinal such that k = 0 or q ≤ k ≤ r. Then

Proof. Since k ≤ r, we have S_r∖G_k⊊S_r. If k = 0, then S_r∖G₀ = S_r∖BL(q), and this is a subsemigroup of S_r since, for α, β ∈ S_r∖BL(q), dom (αβ)⊆dom α⊊X, and this implies αβ ∈ S_r∖BL(q). Now suppose q ≤ k ≤ r and let α, β ∈ S_r be such that g(αβ) = k. We claim that g(α) = k or g(β) = k. To see this, assume that g(α) ≠ k. Since

Note that

Remark 3.11. Observe that, if 0 < k < q then S_r∖G_k is not a semigroup for all q ≤ r ≤ p. To see this, let α ∈ BL(q) and β = id_Xα∖K for some subset K of Xα such that |K | = k (possible since |Xα | = p > k), then α, β ∈ PS(q) since d(β) = d(α) + k = q. Moreover, since g(α) = 0 and g(β) = q ≠ k, we have α, β ∈ S_r∖G_k. But

Theorem 3.12. Suppose that p ≥ r ≥ q ≥ ℵ₀. Then the following statements hold:

• (a)

  S_r∖G₀ is a maximal subsemigroup of S_r;

• (b)

  if p > q, then for each cardinal k such that q ≤ k ≤ r, S_r∖G_k is a maximal subsemigroup of S_r.

Proof. By Lemma 3.10, S_r∖G₀ is a subsemigroup of S_r. To see that it is maximal, let α, β ∈ G₀ = BL(q)⊆S_q. By [3, Theorem  5], S_q = β · R(q), and this implies that α = βγ for some γ ∈ R(q)⊆S_r∖G₀. Hence S_r∖G₀ is maximal in S_r.

Now suppose that p > q and let q ≤ k ≤ r. Let α, β ∈ G_k. If k = q, then G_k = R(q)⊆S_q and, by [3, Theorem  6], S_q = BL(q) · β · BL(q). If k > q, then G_k = BL(q) · β · BL(q) (by Lemma 3.4). Therefore, α = γβμ for some γ, μ ∈ BL(q)⊆S_r∖G_k, and so S_r∖G_k is maximal in S_r.

Corollary 3.13. Suppose that p > q ≥ ℵ₀ and let A be a nonempty subset of X such that |X∖A | ≥ q. Then the following sets are maximal subsemigroups of PS(q):

• (a)

  $$;

• (b)

  N_k = {α ∈ PS(q) : g(α) ≠ k} where k = 0 or q ≤ k ≤ p.

Proof. By Theorem 3.9, $$ is maximal in S_q. Then Corollary 3.2 implies that $$ is maximal in PS(q). But

Theorem 3.14. Suppose that p > q ≥ ℵ₀ and k equals 0 or q. Let A be a nonempty subset of X such that |X∖A | ≥ q. Then the two classes of maximal subsemigroups $$ and S_q∖G_k of S_q are always disjoint.

Proof. By Theorems 3.9 and 3.12, $$ and S_q∖G_k are maximal subsemigroups of S_q. By Lemma 3.6(a), there exists $$ with g(α) = k. Then $$ but α ∉ S_q∖G_k, that is, $$. Also, $$ by the maximality of $$ and S_q∖G_k. Therefore, $$ is not equal to S_q∖G_k.

Lemma 4.1. Suppose that p = q ≥ ℵ₀ and M is a maximal subsemigroup of PS(q). Then the following statements hold:

• (a)

  M contains all α ∈ PS(q) with r(α) < p,

• (b)

  if R(q)⊆M, then M∩BL(q) = ∅.

Proof. Suppose that there exists α ∉ M with r(α) = k < p. Then g(α) = p, and we write in the usual way

To show that (b) holds, suppose that R(q)⊆M. If there exists α ∈ M∩BL(q), then [3, Theorem  5] implies that PS(q) = α · R(q)⊆M (note that S_q = PS(q) when p = q), so M = PS(q), contrary to the maximality of M. Thus M∩BL(q) = ∅.

Remark 4.2. If p > q, then every α ∈ PS(q) has rank p. This contrasts with Lemma 4.1(a). Also, by Corollary 3.13, if p > q and q < k ≤ p, N_k is a maximal subsemigroup of PS(q) containing R(q) ∪ BL(q), this contrasts with Lemma 4.1(b).

Theorem 4.3. Suppose that p = q ≥ ℵ₀, and let A be a nonempty subset of X such that |X∖A | ≥ q. The following are maximal subsemigroups of PS(q):

• (a)

  $$;

• (b)

  N₀;

• (c)

  N_p ∪ J_p.

Proof. If p = q, then S_q = PS(q), and so (a) holds by Theorem 3.9. Also, by taking r = p in Theorem 3.12(a), we see that (b) holds. To show that (c) holds, take r = p = k in Lemma 3.10, we have N_p = S_p∖G_p is a subsemigroup of PS(q). Moreover, N_p ∪ J_p is also a subsemigroup of PS(q) since J_p is an ideal. To show the maximality of N_p ∪ J_p, let α, β ∈ PS(q)∖(N_p ∪ J_p). Then g(α) = g(β) = p = r(α) = r(β). Write in the usual way

Remark 4.4. When p = q, if M is a maximal subsemigroup containing R(q), then

Remark 4.5. As we showed in Section 3, to see all maximal subsemigroups of PS(q) when p > q, it is necessary to describe all maximal subsemigroups of S_r where q ≤ r < p. So we leave this as a direction for future research.