Regularity of the partial Baer-Levi semigroups with restricted range
Let \(Y\) be a fixed nonempty subset of an infinite set \(X\) and let \(q\) be an infinite cardinal such that \(q\leq|X|\). Let \(PS(X,Y,q)\) denote the semigroup of all partial injective transformations from \(X\) into \(Y\) for which the complement of its range has cardinality \(q\). Then \(PS(X,Y...
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Algebra and Discrete Mathematics| _version_ | 1870196963343859712 |
|---|---|
| author | Singha, Boorapa |
| author_facet | Singha, Boorapa |
| author_institution_txt_mv | [
{
"author": "Boorapa Singha",
"institution": "Chiang Mai Rajabhat University"
}
] |
| author_sort | Singha, Boorapa |
| baseUrl_str | https://admjournal.luguniv.edu.ua/index.php/adm/oai |
| collection | OJS |
| datestamp_date | 2026-07-08T07:55:33Z |
| description | Let \(Y\) be a fixed nonempty subset of an infinite set \(X\) and let \(q\) be an infinite cardinal such that \(q\leq|X|\). Let \(PS(X,Y,q)\) denote the semigroup of all partial injective transformations from \(X\) into \(Y\) for which the complement of its range has cardinality \(q\). Then \(PS(X,Y,q)\) is a generalization of the partial Baer-Levi semigroup. In this paper, we study several types of regularity on \(PS(X, Y,q)\). We characterize all regular, left regular, right regular, completely regular, intra-regular and coregular elements and determine the largest regular subsemigroup of this semigroup. Furthermore, when \(Y\) is finite, we present formulas for counting the total number of elements of each type mentioned above. |
| doi_str_mv | 10.12958/adm2440 |
| first_indexed | 2026-07-09T01:00:10Z |
| format | Article |
| fulltext |
© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 41 (2026). Number 2, pp. 244–259
DOI:10.12958/adm2440
Regularity of the partial Baer-Levi semigroups
with restricted range
Boorapa Singha
Communicated by V. Mazorchuk
Abstract. Let Y be a fixed nonempty subset of an infi-
nite set X and let q be an infinite cardinal such that q ≤ |X|.
Let PS(X,Y, q) denote the semigroup of all partial injective trans-
formations from X into Y for which the complement of its range
has cardinality q. Then PS(X,Y, q) is a generalization of the par-
tial Baer-Levi semigroup. In this paper, we study several types of
regularity on PS(X,Y, q). We characterize all regular, left regu-
lar, right regular, completely regular, intra-regular and coregular
elements and determine the largest regular subsemigroup of this
semigroup. Furthermore, when Y is finite, we present formulas
for counting the total number of elements of each type mentioned
above.
1. Introduction
Let X be a nonempty set and let I(X) denote the symmetric inverse
semigroup on X: that is, the semigroup of all injective mappings α whose
domain, dom α, and range, Xα are subsets of X. When X is infinite and
q is a fixed cardinal such that ℵ0 ≤ q ≤ |X|, we write
BL(X, q) = {α ∈ I(X) : dom α = X and d(α) = q},
The author thanks Thailand Science Research and Innovation (TSRI) and Chi-
ang Mai Rajabhat University for funding support (Fundamental Fund: project number
194650)
2020 Mathematics Subject Classification: 20M20.
Key words and phrases: transformation semigroup, regular element, injective
mapping.
https://doi.org/10.12958/adm2440
B. Singha 245
where d(α) = |X\Xα| is called the defect of α. Then BL(X, q) is a
semigroup which is called the Baer-Levi semigroup of type (|X|, q). It
is known that BL(X, q) is a right cancellative, right simple semigroup
without idempotents. The semigroup BL(X, q) plays a crucial role in
semigroup theory, since every semigroup S satisfying these three pro-
perties can be embedded in a Baer-Levi semigroup of type (p, p), where
p = |S| ([2], Section 8.1).
In 1975, Sullivan [11] introduced the following semigroup:
PS(X, q) = {α ∈ I(X) : d(α) = q},
and call this the partial Baer-Levi semigroup on X. It is clear that
BL(X, q) is a subsemigroup of PS(X, q). Later, Pinto and Sullivan [5]
described Green’s equivalences, regular elements and ideals of PS(X, q).
Moreover, they presented an interesting result that PS(X, q) is neither
right simple nor right cancellative and always contains idempotents, this
result is completely opposite to what occurs in BL(X, q). In addition,
partial orders and maximal subsemigroups on PS(X, q) have been stu-
died in [8–10].
Recently, in 2024, Singha [7] introduced and studied the subsemi-
group PS(X,Y, q) of PS(X, q), which consists of all elements of PS(X, q)
whose range is contained in a fixed nonempty subset Y of X. Symboli-
cally, this subsemigroup of PS(X, q) is defined as :
PS(X,Y, q) = {α ∈ I(X) : d(α) = q and Xα ⊆ Y }.
If X = Y , then PS(X,Y, q) = PS(X, q), hence PS(X,Y, q) is a generali-
zation of PS(X, q). The author in [7] characterized the Green’s relations
and described the natural partial order on PS(X,Y, q). He also showed
that PS(X,Y, q) is not a regular semigroup. These findings, along with
those from other research on PS(X, q) are the motivation for this study.
The main objective of this paper is to study many types of regularity
on PS(X,Y, q). We provide necessary and sufficient conditions for each
element of this semigroup to be regular, left regular, right regular, com-
pletely regular, intra-regular and coregular. In addition, we determine
the largest regular subsemigroup of PS(X,Y, q) and we show that the
set of all completely regular elements of PS(X,Y, q) is a disjoint union
of permutation groups on some appropriate sets. In the final part of this
research, we present some combinatorial proofs for the number of these
elements in PS(X,Y, q) when Y is finite.
246 Regularity of the partial Baer-Levi semigroups
2. Preliminary
Throughout this paper, unless otherwise specified, we suppose that X
is an infinite set, q is an infinite cardinal such that q ≤ |X| and Y is a
nonempty subset of X. For each mapping α ∈ PS(X,Y, q), we write
α =
(
ai
yi
)
,
where the subscript i belongs to an index set I, the abbreviation {yi}
denotes {yi : i ∈ I}, Xα = {yi} ⊆ Y , dom α = {ai} and aiα = yi. We
also write
g(α) = |X\dom α| and r(α) = |Xα|,
and refer to these cardinals as the gap and the rank of α, respectively.
Observe that, |X \Y | ≤ |X \Xα| = q for any α ∈ PS(X,Y, q), therefore
PS(X,Y, q) ̸= ∅ only when |X \ Y | ≤ q.
For a subset A of X, we denote by α|A the restriction of α to A and
we write Aα instead of (A∩ dom α)α for convenience and brevity. Also,
we let idA denote the identity mapping on A, and we write A = B ∪̇ C
to denote A is a disjoint union of B and C. As usual, ∅ denotes the
emptyset, but in some contexts, ∅ is used to denote the empty (one-
to-one) transformation, which is the zero element of I(X). We refer to
Lemma 1 of [6] that, if α ∈ I(X) and Y, Z ⊆ X, then (Y \Z)α = Y α\Zα.
This fact will be used throughout this paper without further discussion.
An element a in a semigroup S is said to be regular if a = axa for
some x ∈ S. We also recall that a is called left regular if a = xa2 for
some x ∈ S, right regular if a = a2x for some x ∈ S and completely
regular if a = axa and ax = xa for some x ∈ S. An element a is said to
be intra-regular if a = ya2z for some y, z ∈ S. Moreover, a is said to be
a coregular element in S if a = axa = xax for some x ∈ S.
By the above definitions, it can be verified that every coregular ele-
ment is a regular, left regular, right regular, completely regular and intra-
regular element. In addition, if a3 = a, then it is clear that a is a
coregular element in S. On the other hand, if a is a coregular, i.e.,
a = axa = xax for some x ∈ S, then a = xax = x(axa)x = (xax)ax =
(xax)(axa)x = (xax)a(xax) = a3. This implies that a is coregular if
and only if a3 = a. Next, if a is completely regular, then it is clear
that a is regular. Moreover, a = (ax)a = (xa)a = xa2 and a = a(xa) =
a(ax) = a2x. Thus, every completely regular element is also left and right
regular. Additionally, Petrich and Reilly [4] proved that an element a of a
B. Singha 247
semigroup S is completely regular if and only if a is both a left and a right
regular element of S. In addition, if an element a of S is both left regular
and right regular such that a = ba2 and a = a2c for some b, c ∈ S, then
a = baa = ba(a2c) = (ba)a2(c). Therefore, a is intra-regular. It follows
that every completely regular element is an intra-regular element.
3. Regularity
We recall from [7, Proposition 1] that, when |X| = q, PS(X,Y, q) con-
tains a zero element which is the empty transformation ∅ with an empty
domain. In this case, it can be directly verified by definition that ∅ is
a regular, left regular, right regular, completely regular, intra-regular
and coregular element of PS(X,Y, q). Therefore, for convenience and
brevity in proving the results in this section, we will omit the case where
α = ∅. However, after the proof in this section is completed, we will find
that α = ∅ still meets the criteria for being a regular, left regular, right
regular, completely regular, intra-regular and coregular element, which
will be presented in Theorem 1, Theorem 3, Theorem 4, Corollary 1,
Theorem 5, and Theorem 7, respectively.
From [7, Proposition 3], the author proved that PS(X,Y, q) is not a
regular semigroup. So, we begin by characterizing all regular elements
of PS(X,Y, q).
Theorem 1. Let α ∈ PS(X,Y, q). Then α is regular if and only if
dom α ⊆ Y and g(α) = q. Moreover, in this case, α−1 is also a regular
element of PS(X,Y, q).
Proof. Suppose that α is regular, then there exists β ∈ PS(X,Y, q) such
that α = αβα. Then xα = xαβα for all x ∈ dom α. As α is injective,
we have x = xαβ ∈ Xβ ⊆ Y , whence dom α ⊆ Xβ ⊆ Y . Next, we
claim that |Xβ \ dom α| ≤ q. Suppose that Xβ \ dom α = {ci}. Then
there exists a set {bi} ⊆ dom β and biβ = ci. As α = αβα, we have
that βα|Xα = idXα. Thus, if bi ∈ Xα, then bi = biβα = ciα, this is
a contradiction since ci /∈ dom α. Therefore, {bi} ⊆ X \ Xα, whence
|Xβ \ dom α| = |{bi}| ≤ |X \ Xα| = d(α) = q as required. Now, as
dom α ⊆ Xβ, we have
X \ dom α = (X \Xβ) ∪̇ (Xβ \ dom α),
where |X\Xβ| = q and |Xβ\dom α| ≤ q, whence g(α) = |X\dom α| = q.
248 Regularity of the partial Baer-Levi semigroups
For the converse, we suppose that dom α ⊆ Y and g(α) = q. Then
Xα−1 = dom α ⊆ Y and d(α−1) = g(α) = q, whence α−1 ∈ PS(X,Y, q).
It is clear that α = αα−1α and α−1 = α−1αα−1. Hence, α and α−1 are
regular elements of PS(X,Y, q).
We recall that, an inverse semigroup is a semigroup S such that for
every element a ∈ S, there exists a unique inverse b ∈ S such that
a = aba and b = bab. Equivalently, S is an inverse semigroup if S is a
regular semigroup whose idempotents commute. By [7, Proposition 2],
the set of all idempotents in PS(X,Y, q) is defined as
E(PS(X,Y, q)) = {idA : A ⊆ Y and |X \A| = q}.
Let R(X,Y, q) be the set of all regular elements of PS(X,Y, q), that is,
R(X,Y, q) = {α ∈ PS(X,Y, q) : dom α ⊆ Y and g(α) = q}.
The next result shows that R(X,Y, q) is the largest regular subsemigroup
of PS(X,Y, q).
Theorem 2. The set R(X,Y, q) is the largest regular subsemigroup of
PS(X,Y, q). Moreover, it is also an inverse subsemigroup of PS(X,Y, q).
Proof. Since every element of R(X,Y, q) is regular, it remains to show
R(X,Y, q) is closed. Let α, β ∈ R(X,Y, q), then dom α ⊆ Y , dom β ⊆ Y
and g(α) = q = g(β). Since dom αβ ⊆ dom α, we have
X \ dom αβ = (X \ dom α) ∪̇ (dom α \ dom αβ)
= (X \ dom α) ∪̇ ((Xα)α−1 \ (Xα ∩ dom β)α−1)
= (X \ dom α) ∪̇ (Xα \ (Xα ∩ dom β))α−1
= (X \ dom α) ∪̇ (Xα \ dom β)α−1.
Since |(Xα\dom β)α−1| ≤ |(X \dom β)α−1| ≤ |X \dom β| = q and |X \
dom α| = q, we get g(αβ) = |X \ dom αβ| = q, whence αβ ∈ R(X,Y, q).
Therefore, R(X,Y, q) is the largest regular subsemigroup of PS(X,Y, q).
Next, since all idempotents of PS(X,Y, q) have the form idA for some
A ⊆ Y such that |X \ A| = q and all of these mappings commute.
Therefore, R(X,Y, q) is an inverse subsemigroup of PS(X,Y, q).
Next, we characterize left regular elements of PS(X,Y, q).
Theorem 3. Let α ∈ PS(X,Y, q). Then α is left regular if and only if
dom α ⊆ Xα.
B. Singha 249
Proof. Suppose that α is left regular. Then α = λα2 for some λ ∈
PS(X,Y, q). Thus, Xα ⊆ Xα2. In general, as Xα2 ⊆ Xα, we have
Xα = Xα2 = (Xα ∩ dom α)α. Consequently, as α is injective, we have
Xα ∩ dom α = dom α, whence dom α ⊆ Xα.
Conversely, suppose that dom α ⊆ Xα. Then
q = |X \Xα| ≤ |X \ dom α| = g(α).
As dom α2 ⊆ dom α, we get q ≤ g(α) ≤ g(α2) and we may write
X \ dom α2 = (X \ dom α) ∪̇ (dom α \ dom α2),
which implies
g(α2) = g(α) + |dom α \ dom α2|. (1)
Suppose, for contradiction, that g(α2) = t > g(α). Then substitute
g(α2) = t in (1), we get |dom α \ dom α2| = t. As dom α ⊆ Xα, we may
write
X \ dom α = (X \Xα) ∪̇ (Xα \ dom α),
which implies
g(α) = d(α) + |Xα \ dom α|
= d(α) + |(Xα \ dom α)α−1|
= d(α) + |(Xα)α−1 \ (Xα ∩ dom α)α−1|
= d(α) + |dom α \ dom α2|
= q + t
= t.
which is a contradiction, whence q ≤ g(α) = g(α2). Next, as dom α ⊆
Xα, we get Xα2 = (Xα ∩ dom α)α = (dom α)α = Xα. Thus, we can
write
α =
(
ai
xi
)
and α2 =
(
bi
xi
)
,
where {bi} = dom α2 ⊆ dom α = {ai} ⊆ Xα = {xi} = Xα2 ⊆ Y . We
consider two cases.
Case 1. If g(α2) = q, then d((α2)−1) = g(α2) = q and X(α2)−1 =
{bi} ⊆ Y , whence (α2)−1 ∈ PS(X,Y, q). Let λ = α(α2)−1 ∈ PS(X,Y, q),
we have α = λα2.
Case 2. If g(α2) > q, then we have q < g(α) = g(α2). We see that
X \ dom α2 = (Y \ dom α2) ∪̇ ((X \ dom α2) ∩ (X \ Y )), (2)
250 Regularity of the partial Baer-Levi semigroups
where |X\dom α2| = g(α2) > q and |(X\dom α2)∩(X\Y )| ≤ |X\Y | ≤ q.
Then, from (2), we have |Y \ dom α2| = |X \ dom α2| > q. Now, write
Y \ dom α2 = {vk} ∪̇ U , where |{vk}| = g(α2) and |U | = q. As g(α) =
g(α2), we can suppose X \ dom α = {uk} and define
λ =
(
ai uk
bi vk
)
.
It is clear that λ is injective, Xλ ⊆ Y and α = λα2. Moreover, we have
d(λ) = |X \ ({bi} ∪ {vk})| = |U |+ |X \ Y | = q, whence λ ∈ PS(X,Y, q).
Hence, in both cases, α is a left regular element of PS(X,Y, q) as
required.
Theorem 4. Let α ∈ PS(X,Y, q). Then α is right regular if and only if
Xα ⊆ dom α.
Proof. Suppose that α is right regular. Then α = α2λ for some λ ∈
PS(X,Y, q). Thus, dom α ⊆ dom α2. As dom α2 ⊆ dom α, we get
that dom α = dom α2 = (Xα ∩ dom α)α−1. Hence Xα ∩ dom α = Xα.
Therefore Xα ⊆ dom α.
Conversely, suppose that Xα ⊆ dom α. Then dom α2 = (Xα ∩
dom α)α−1 = (Xα)α−1 = dom α. We may write
α =
(
ai
bi
)
and α2 =
(
ai
ci
)
.
As {bi} = Xα ⊆ Y and d(α) = q, we can define λ =
(
ci
bi
)
∈ PS(X,Y, q).
Then α = α2λ. Therefore α is right regular.
As we discussed earlier, an element a in a semigroup S is completely
regular if and only if it is both left regular and right regular. Hence, the
result below is readily deduced from Theorem 3 and Theorem 4.
Corollary 1. Let α ∈ PS(X,Y, q). Then α is completely regular if and
only if Xα = dom α.
In view of Corollary 1, a completely regular α in PS(X,Y, q) is a
permutation on a set A, i.e., a bijection on a set A, where A = Xα =
dom α. In addition, as Xα ⊆ Y and d(α) = q, we have that A ⊆ Y and
|X \ A| = q. Let SA denote the group of permutations on a set A for
which A ⊆ Y and |X \A| = q. It is clear that all mappings in SA satisfy
the condition in Corollary 1. By using these results, we can describe the
B. Singha 251
set of all completely regular elements of PS(X,Y, q) in terms of a disjoint
union of permutation groups as follows.
Corollary 2. Let T = {A ⊆ Y : |X \A| = q}. The set of all completely
regular elements in PS(X,Y, q) is precisely the set⋃
A ∈ T
SA.
Unlike regular, left regular, right regular and completely regular ele-
ments, a classification of intra-regular elements of PS(X,Y, q) must be
divided into two cases according to the cardinality of X.
Theorem 5. Suppose that |X| = q. Let α ∈ PS(X,Y, q). Then α is
intra-regular if and only if |dom α| = |dom α2 ∩ Y |.
Proof. Suppose that α is intra-regular. Then α = λα2µ for some λ, µ ∈
PS(X,Y, q). Since Xλ ⊆ Y , we have
|dom α| = |dom λα2µ|
= |(Xλα2 ∩ dom µ)(λα2)−1|
= |Xλα2 ∩ dom µ|
≤ |Xλα2|
= |(Xλ ∩ dom α2)α2|
= |Xλ ∩ dom α2|
≤ |Y ∩ dom α2|.
As dom α2 ⊆ dom α, we have |dom α2∩Y | ≤ |dom α2| ≤ |dom α|. Then
the equality follows, whence |dom α| = |dom α2 ∩ Y |.
Conversely, suppose that |dom α| = |dom α2 ∩ Y |. Write dom α2 ∩
Y = {ai} and α =
(
ci
di
)
, i ∈ I. If |I| < q, then we define
λ =
(
ci
ai
)
and µ =
(
aiα
2
di
)
.
Then α=λα2µ. We also see that d(λ)= |X\{ai}|=q. Moreover, Xλ⊆ Y ,
Xµ = Xα ⊆ Y and d(µ) = d(α) = q, whence λ, µ ∈ PS(X,Y, q). On the
other hand, if |I| = q, then we write {ai} = {ui} ∪̇ {vi} and define
λ′ =
(
ci
ui
)
and µ′ =
(
uiα
2
di
)
.
252 Regularity of the partial Baer-Levi semigroups
Then α = λ′α2µ′. Moreover, Xµ′ = Xα ⊆ Y and d(µ′) = d(α) = q.
Hence µ′ ∈ PS(X,Y, q). In addition, Xλ′ ⊆ Y and {vi} ⊆ X \ Xλ′
where |{vi}| = q, so d(λ′) = q. Therefore λ′ ∈ PS(X,Y, q), whence α is
intra-regular.
Theorem 6. Suppose that |X| > q. Let α ∈ PS(X,Y, q). Then α is
intra-regular if and only if g(α2) ≤ q or q < g(α2) = g(α).
Proof. Suppose that α is intra-regular. Then α = λα2µ for some λ, µ ∈
PS(X,Y, q). We also assume that g(α2) ≰ q, then g(α2) = t for some
infinite cardinal t greater than q. Since
|X \ dom α2| = |(X \ dom α2) ∩Xλ|+ |(X \ dom α2) ∩ (X \Xλ)|,
where |(X \dom α2)∩ (X \Xλ)| ≤ |X \Xλ| = q < t, the above equation
implies |(X \dom α2)∩Xλ| = t. Then, for each vi ∈ (X \dom α2)∩Xλ,
vi = uiλ for some ui ∈ dom λ, where uiλ /∈ dom α2. So uiα = (uiλ)α
2µ is
not defined, whence ui /∈ dom α for all i. Therefore, {ui} ⊆ X \ dom α,
where |{ui}| = t. It follows that q < g(α2) = t ≤ g(α). Clearly, as
dom α2 ⊆ dom α, we must have g(α) ≤ g(α2), whence q < g(α2) = g(α)
as required.
Conversely, suppose that g(α2) ≤ q or q < g(α2) = g(α). Let |X| =
p > q. As the defect of each mapping in PS(X,Y, q) is q, we have that
its rank is p. We write
α =
(
ai
bi
)
and α2 =
(
ci
di
)
,
where i ∈ I, |I| = p. Moreover, since
dom α2 = (dom α2 ∩ Y ) ∪̇ (dom α2 \ Y ),
where |dom α2 \ Y | ≤ |X \ Y | ≤ q, we have |dom α2 ∩ Y | = p. Then we
can write dom α2 ∩ Y = {xi} ∪̇ Q, where |Q| = q and define
µ =
(
xiα
2
bi
)
.
Clearly, Xµ = Xα ⊆ Y and d(µ) = d(α) = q, whence µ ∈ PS(X,Y, q).
Now, if g(α2) ≤ q, then we define
λ =
(
ai
xi
)
.
B. Singha 253
Clearly, α = λα2µ and Xλ ⊆ Y . Moreover,
d(λ) = |X \ {xi}| = |X \ dom α2|+ |dom α2 \ Y |+ |Q| = q,
whence, λ ∈ PS(X,Y, q). Finally, suppose that q < g(α2) = g(α). Since
X \ dom α2 = ((X \ dom α2) ∩ Y ) ∪̇ ((X \ dom α2) \ Y ),
where |(X \ dom α2) \ Y | ≤ |X \ Y | ≤ q, we have |(X \ dom α2) ∩ Y | =
g(α2) > q. We write (X \dom α2)∩Y = {uk} ∪̇ R, where |{uk}| = g(α2)
and |R| = q. Since g(α2) = g(α), we can assume that X \ dom α = {vk}
and define
λ′ =
(
ai vk
xi uk
)
.
We see that α = λ′α2µ and Xλ′ = {xi} ∪ {uk} ⊆ Y . Moreover, d(λ′) =
|X \ ({xi} ∪ {uk})| = |X \ Y |+ |Q|+ |R| = q, whence λ′ ∈ PS(X,Y, q).
Hence, α is intra-regular.
In the following theorem, we characterize the coregular elements of
PS(X,Y, q).
Theorem 7. Let α ∈ PS(X,Y, q). Then α is coregular if and only
if α satisfies the following condition: if xα = y, then yα = x for all
x, y ∈ dom α. Equivalently, α2 = iddom α.
Proof. Suppose that α is coregular. Then α3=α as discussed in Section 2.
For each xi∈dom α, we suppose that xiα = yi. Then xiα = xiα
3 = yiα
2.
Consequently, since α is injective, we get that yiα = xi, which satisfies
the desired condition. In addition, we get that xiα
2 = yiα = xi, which
implies α2 = iddom α as required. Conversely, suppose that α2 = iddom α.
Then α3 = α2α = iddom αα = α. Hence, α is a coregular element of
PS(X,Y, q).
By taking Y = X, the results obtained in this research can be ex-
tended to the semigroup PS(X, q) as follows.
Corollary 3. Let α ∈ PS(X, q). Then the following hold:
(1) α is regular if and only if g(α) = q. Moreover, whenever this
occurs, α−1 is also a regular element in PS(X, q).
(2) α is left regular if and only if dom α ⊆ Xα.
(3) α is right regular if and only if Xα ⊆ dom α.
(4) α is completely regular if and only if Xα = dom α.
254 Regularity of the partial Baer-Levi semigroups
(5) Suppose that |X| = q. Then α is intra-regular if and only if
|dom α| = |dom α2|.
(6) Suppose that |X| > q. Then α is intra-regular if and only if
g(α2) ≤ q or q < g(α2) = g(α).
(7) α is coregular if and only if α satisfies the following condition: if
xα = y, then yα = x for all x, y ∈ dom α. Equivalently, α2 = iddom α.
4. Combinatorial results
In this section, Y is assumed to be a nonempty finite subset of X. Then,
as X is infinite, we have |X| = |X \ Y | ≤ q. Therefore, |X| = q and so
PS(X,Y, q) contains a zero element which is the empty transformation ∅.
First, let us reconsider the results obtained in Section 3 under the
assumption that Y is finite. We observe that, since Y is finite, if α ∈
PS(X,Y, q) such that dom α ⊆ Y , then dom α is also finite. It follows
that g(α) = |X \ dom α| = |X| = q. Hence, the following result is a
direct consequence of Theorem 1.
Corollary 4. Let Y be a nonempty finite subset of X and let α ∈
PS(X,Y, q). Let α ∈ PS(X,Y, q). Then α is regular if and only if
dom α ⊆ Y .
For the next result, by using the condition that Y is a finite set, the
results proved in Section 3 appear simpler.
Corollary 5. Let Y be a nonempty finite subset of X and let α ∈
PS(X,Y, q). Then the following statements are equivalent:
(1) α is left regular.
(2) α is right regular.
(3) α is completely regular.
(4) α is intra-regular.
(5) Xα = dom α.
Proof. We first show that (4) implies (2). To do this, we let α be an
intra-regular element. Since |X| = q, Theorem 5 implies that |dom α| =
|dom α2∩Y |. Consequently, since Y is finite, we have dom α2∩Y is finite,
whence dom α is also a finite set. Since dom α2 ∩Y ⊆ dom α2 ⊆ dom α,
it follows that dom α = dom α2 ∩ Y ⊆ Y . Now we have dom α2 ⊆
dom α ⊆ Y , which implies dom α = dom α2 ∩ Y = dom α2. It fol-
lows that dom α = (Xα∩ dom α)α−1, which means Xα∩ dom α = Xα,
whence Xα ⊆ dom α. Therefore, α is right regular by Theorem 4. Next,
B. Singha 255
we show that (2) implies (3). Suppose that α is right regular, we have
Xα ⊆ dom α by Theorem 4. Since Xα ⊆ Y , we have Xα is also a finite
set. Consequently, as |dom α| = |Xα|, the condition Xα ⊆ dom α im-
plies thatXα=dom α.Therefore, α is completely regular by Corollary 1.
As discussed in Section 2 that (3) implies (4), it can be deduced that (2),
(3) and (4) are equivalent. Next, we show that (1) implies (3). Suppose
that α is left regular, then dom α ⊆ Xα by Theorem 3. Since Y is finite
and dom α ⊆ Xα ⊆ Y . Then dom α is a finite subset of a finite set
Xα, where |dom α| = |Xα|. This implies Xα = dom α, whence α is
completely regular by Corollary 1. We now recall what was discussed in
Section 2, namely that (3) implies (1), so (1) and (3) are equivalent. Fi-
nally, since (3) and (5) are equivalent by Corollary 1, it follows that all
five conditions are equivalent when Y is a finite set.
We know from Corollary 5 that the numbers of left regular, right
regular, completely regular, and intra-regular elements of PS(X,Y, q) are
equal, while the numbers of regular and coregular elements may differ.
Recall that if A and B are finite sets with equal cardinality k, then
there are exactly k! bijections from A onto B.
Theorem 8. Let Y be a nonempty finite subset of X such that |Y | = n.
Then
(1) The total number of regular elements of PS(X,Y, q) is
n∑
k=0
k!
(
n
k
)2
.
(2) The total number of left regular (right regular, completely regular,
intra-regular) elements of PS(X,Y, q) is
n∑
k=0
k!
(
n
k
)
.
Proof. To prove (1), let α be a regular element of PS(X,Y, q). By Corol-
lary 4, we have dom α ⊆ Y . Now, dom α and Xα are two finite subsets
of Y with the same size, say |dom α| = |Xα| = k for some integer k such
that 1 ≤ k ≤ n. Then, there are
(
n
k
)
·
(
n
k
)
=
(
n
k
)2
ways to choose the
domain set and the image set from the set Y . Next, since α is a bijection
from dom α onto Xα, there are k!
(
n
k
)2
possible ways to define a regular
element α. In addition, since ∅ is also regular and 0! = 1 =
(
n
0
)
, we will
256 Regularity of the partial Baer-Levi semigroups
also add the number of this mapping to the formula. Hence, the total
number of regular elements of PS(X,Y, q) is
n∑
k=0
k!
(
n
k
)2
.
For the proof of (2), we have dom α = Xα by Corollary 5. So, there
are
(
n
k
)
ways to choose the domain set, which coincides with the image
set. The remaining proof can be continued following the same steps as
in the proof of (1).
In what follows, let α ∈ PS(X,Y, q) and x, y ∈ dom α such that
x ̸= y. We call an unordered pair {x, y} a symmetric pair of α if x
and y satisfy the condition : if xα = y, then yα = x. We also call x a
fixed point of α if xα = x. It follows by Theorem 7 that, α is coregular
if every element of dom α is contained in some symmetric pairs, or is
a fixed point. For example, let X = N denote the set of all positive
integers, Y = {1, 2, 3, ..., 10} and let q = ℵ0. Define
θ =
(
1 2 3 4 5 6 7
7 2 4 3 5 6 1
)
, δ =
(
1 2 3 4 5 6 7 8
2 1 4 5 3 6 8 7
)
,
then θ, δ ∈ PS(X,Y, q). We can verify, according to the condition of
Theorem 7, that θ is coregular while δ is not. We also see that {1, 7}
and {3, 4} are all symmetric pairs of θ and 2, 5, 6 are all fixed points
of θ, which can be seen that every element of dom θ is either in some
symmetric pairs or is a fixed point. For δ, we see that {1, 2} and {7, 8}
are the only two symmetric pairs of δ, but 3, 4, 5 are neither members
of these pairs nor fixed points of δ. We notice that, if α = idA for some
subset A of Y , then α is coregular and all elements in dom α are fixed
points of α. Consequently, the identity mapping on A has no symmetric
pairs.
To count the number of coregular elements of PS(X,Y, q), we need
to refer to the floor function of a real number x that gives the greatest
integer less than or equal to x, and it is denoted by ⌊x⌋. We observe
that there are at most
⌊p
2
⌋
symmetric pairs for a coregular element α
with |dom α| = p. We also refer to the double factorial of an integer n,
where n ≥ −1, which is denoted by n!!. The double factorial is defined
as follows [1, p. 544–547]:
n!! =
{
2× 4× 6× . . .× n if n > 0 and n is even;
1× 3× 5× . . .× n if n > 0 and n is odd;
and 0!! = (−1)!! = 1. To clarify, n!! is not equal to (n!)!. For example,
B. Singha 257
3!! = 1 × 3 = 3, whereas (3!)! = 6! = 720. For the proof of Theorem 9,
the following result which has been discussed in [3] is required.
Lemma 1. Let n be a positive number. The number of ways to partition
2n distinct objects into n unordered pairs, where the order of the pairs
does not matter is (2n− 1)!!.
To prove the following theorem, we also recall that, if α is coregular
in PS(X,Y, q), then α is left (right, completely, intra) regular. So, by
Corollary 5, we have dom α = Xα ⊆ Y .
Theorem 9. Let Y be a nonempty finite subset of X. Then the following
hold:
(1) If |Y | = 1, then there are exactly two coregular elements of
PS(X,Y, q).
(2) If |Y | = n ≥ 2, then the total number of coregular elements of
PS(X,Y, q) is
(n+ 1) +
n∑
p=2
((
n
p
)
·
⌊ p
2⌋∑
k=0
(
p
2k
)
(2k − 1)!!
)
.
Proof. To prove (1), suppose that Y = {a}. Then, by Theorem 7, we
can see that ∅ and the identity mapping on a singleton set {a} are the
only two coregular elements of PS(X,Y, q).
To prove (2), suppose that |Y | = n ≥ 2. We will count the total
number of possible ways to construct a coregular element α ∈ PS(X,Y, q)
by considering the cases according to the size of dom α.
Case 1 : |dom α| = 0. In this case, the only coregular element with
an empty domain is α = ∅.
Case 2 : |dom α| = 1. By Theorem 7, it is clear that a coregular
element with a single domain is precisely the identity mapping on a sing-
leton set {a} for some a ∈ Y . Since |Y | = n, we get n possible ways to
construct such coregular elements.
Case 3 : |dom α| = p, where 2 ≤ p ≤ n. The process for constructing
coregular elements containing p members in a domain is as follows:
Step 1. Form the domain of α. This step can be done in
(
n
p
)
ways since
dom α ⊆ Y , |Y | = n and |dom α| = p.
Step 2. Form disjoint symmetric pairs in the domain of α, where the
number of these pairs can range from 0 (when α is the identity
258 Regularity of the partial Baer-Levi semigroups
mapping) to
⌊p
2
⌋
. To find the number of ways to form exactly
k disjoint symmetric pairs, where 0 ≤ k ≤
⌊p
2
⌋
, we perform the
following steps:
Step 2.1 Select 2k elements out of p elements to be in k symmetric
pairs. This can be done in
(
p
2k
)
ways.
Step 2.2 Arrange these 2k elements into k unordered pairs, and each
pair will create one symmetric pair. By Lemma 1, the num-
ber of ways to do this is equal to (2k − 1)!!.
Step 2.3 The remaining p − 2k elements (if any) will be designated
as fixed points, i.e., xα = x. This step can only be done in
one way.
In Step 2.2, Lemma 1 cannot be applied when k = 0. However, in this
case, the coregular element α has no symmetric pairs, i.e., it is the iden-
tity mapping on its own domain and
(
p
2k
)
(2k − 1)!! =
(
p
0
)
(−1)!! = 1,
which already represents the number of α. Therefore, we can include
the case k = 0 in all the steps above. Then, multiplying these steps
gives the number of coregular elements with exactly k pairs, which is(
n
p
)
·
⌊ p
2⌋∑
k=0
(
p
2k
)
(2k − 1)!!. Next, by summing over all possible values of p
from 2 to n gives the total number of coregular elements in Case 3, which
is
n∑
p=2
((
n
p
)
·
⌊ p
2⌋∑
k=0
(
p
2k
)
(2k − 1)!!
)
.
Finally, the total number of coregular elements is obtained by com-
bining the numbers from all three cases, which is equal to
(n+ 1) +
n∑
p=2
((
n
p
)
·
⌊ p
2⌋∑
k=0
(
p
2k
)
(2k − 1)!!
)
.
References
[1] Arfken, G.: Mathematical Methods for Physicists (3rd edn.). Academic Press,
Orlando, FL (1985). https://doi.org/10.1016/C2013-0-10310-8
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https://doi.org/10.1016/C2013-0-10310-8
https://doi.org/10.1090/surv/007.2
B. Singha 259
[3] Gould, H., Quaintance, J.: Double fun with double factorials. Math. Magazine
85(3), 177–192 (2012). https://doi.org/10.4169/math.mag.85.3.177
[4] Petrich, M., Reilly, N.R.: Completely Regular Semigroups. John Wiley & Sons,
New York (1999)
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Contact information
B. Singha Department of Mathematics and Statistics,
Faculty of Science and Technology, Chiang
Mai Rajabhat University, Chiangmai 50180,
Thailand
E-Mail: boorapa sin@cmru.ac.th
Received by the editors: 12.11.2025
and in final form 27.03.2026.
https://doi.org/10.4169/math.mag.85.3.177
https://doi.org/10.1017/S0004972700034286
https://doi.org/10.1017/S0004972700034286
https://doi.org/10.1017/S0004972708001330
https://doi.org/10.1017/S0004972708001330
http://dx.doi.org/10.2306/scienceasia1513-1874.2024.011
https://doi.org/10.1155/2011/489674
https://doi.org/10.1155/2011/489674
https://doi.org/10.1017/S0004972709001038
https://doi.org/10.1017/S0004972709001038
https://doi.org/10.4134/BKMS.2012.49.1.109
https://doi.org/10.4134/BKMS.2012.49.1.109
https://doi.org/10.1017/S144678870002396X
Boorapa Singha
|
| id | admjournalluguniveduua-article-2440 |
| institution | Algebra and Discrete Mathematics |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-07-09T01:00:10Z |
| publishDate | 2026 |
| publisher | Lugansk National Taras Shevchenko University |
| record_format | ojs |
| resource_txt_mv | admjournalluguniveduua/d8/6c0fee78a8f47550fb8d5830949cf9d8.pdf |
| spelling | admjournalluguniveduua-article-24402026-07-08T07:55:33Z Regularity of the partial Baer-Levi semigroups with restricted range Singha, Boorapa transformation semigroup, regular element, injective mapping 20M20 Let \(Y\) be a fixed nonempty subset of an infinite set \(X\) and let \(q\) be an infinite cardinal such that \(q\leq|X|\). Let \(PS(X,Y,q)\) denote the semigroup of all partial injective transformations from \(X\) into \(Y\) for which the complement of its range has cardinality \(q\). Then \(PS(X,Y,q)\) is a generalization of the partial Baer-Levi semigroup. In this paper, we study several types of regularity on \(PS(X, Y,q)\). We characterize all regular, left regular, right regular, completely regular, intra-regular and coregular elements and determine the largest regular subsemigroup of this semigroup. Furthermore, when \(Y\) is finite, we present formulas for counting the total number of elements of each type mentioned above. Lugansk National Taras Shevchenko University Thailand Science Research and Innovation (TSRI) Chiang Mai Rajabhat University 2026-07-08 Article Article Peer-reviewed Article application/pdf https://admjournal.luguniv.edu.ua/index.php/adm/article/view/2440 10.12958/adm2440 Algebra and Discrete Mathematics; Vol 41, No 2 (2026) 2415-721X 1726-3255 en https://admjournal.luguniv.edu.ua/index.php/adm/article/view/2440/pdf https://admjournal.luguniv.edu.ua/index.php/adm/article/downloadSuppFile/2440/1346 Copyright (c) 2026 Algebra and Discrete Mathematics |
| spellingShingle | transformation semigroup regular element injective mapping 20M20 Singha, Boorapa Regularity of the partial Baer-Levi semigroups with restricted range |
| title | Regularity of the partial Baer-Levi semigroups with restricted range |
| title_full | Regularity of the partial Baer-Levi semigroups with restricted range |
| title_fullStr | Regularity of the partial Baer-Levi semigroups with restricted range |
| title_full_unstemmed | Regularity of the partial Baer-Levi semigroups with restricted range |
| title_short | Regularity of the partial Baer-Levi semigroups with restricted range |
| title_sort | regularity of the partial baer-levi semigroups with restricted range |
| topic | transformation semigroup regular element injective mapping 20M20 |
| topic_facet | transformation semigroup regular element injective mapping 20M20 |
| url | https://admjournal.luguniv.edu.ua/index.php/adm/article/view/2440 |
| work_keys_str_mv | AT singhaboorapa regularityofthepartialbaerlevisemigroupswithrestrictedrange |