International Journal of Pure and Applied Mathematics
Volume 101 No. 1 2015, 21-31
ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)
url: http://www.ijpam.eu
doi: http://dx.doi.org/10.12732/ijpam.v101i1.3
AP
ijpam.eu
SOME REGULAR EQUIVALENCE RELATION ON
THE SEMIHYPERGROUP OF THE PARTIAL
TRANSFORMATION SEMIGROUP ON
A SET AND LOCAL SUBSEMIHYPERGROUPS
WITH THAT REGULAR EQUIVALENCE RELATION
Ruangvarin Intarawong Sararnrakskul1 § , Sajee Pianskool2
1 Department
of Mathematics
Faculty of Science
Srinakharinwirot University
Bangkok, 10110, THAILAND
2 Department of Mathematics and Computer Science
Faculty of Science
Chulalongkorn University
Bangkok, 10330, THAILAND
Abstract: A hyperoperation ◦ on a nonempty set H is a function from H × H
into P ∗ (H) where P ∗ (H) is the set of all nonempty subset of H and (H, ◦)
is call a hypergroupoid. A hypergroupoid (H, ◦) is called a semihypergroup if
the hyperoperation ◦ is associative. Thus, semihypergroups generalize semigroups. Moreover, if S is a semigroup; we can define a hyperoperation ◦ on S
in order to make (S, ◦) a semihypergroup. In 2013, R.I. Sararnrakskul defined
a hyperoperation ◦ on the partial transformation semigroup P (X) to make
a semihypergroup. In this paper, we define a regular equivalence relation ρ
on (P (X), ◦) so that P (X)/ρ is a semihypergroup and then we studies some
subsemihypergroup of P (X)/ρ.
AMS Subject Classification: 20M20, 20N20
Received:
November 10, 2014
§ Correspondence
author
c 2015 Academic Publications, Ltd.
url: www.acadpubl.eu
22
R.I. Sararnrakskul, S. Pianskool
Key Words: partial transformation semigroup, local subsemigroup, semihypergroup, regular equivalence relation
1. Introduction
Hyperstructures theory was first initiated by F. Marty in 1934 at the 8th
Congress of Scandinavian Mathematicians when he defined hypergroups as a
generalization of group. Since then this theory has enjoyed a rapid development
and many researchers have contributed to study of hypergroup with classical algebraic structure, modern algebra and its application. Elementary concepts and
results on hypergroups can be found in [2]. In 2003, Corsini and Leoreanu presented numerous application of hyperstructures theory in their book [3]. These
applications can be used in various areas such as geometry, topology, combinatorics, theory of binary relations, theory of fuzzy sets, probability theory, codes
theory, automata theory especially the theories of group and semigroups.
First of all, we recall some basic definitions and examples of hypergroup
theory from [2]. Let H be a nonempty set and P ∗ (H) the set of all nonempty
subsets of H. A hyperoperation on H is a function ◦ : H × H → P ∗ (H), the
image of (x, y) ∈ H × H under ◦ is denoted by x ◦ y and called the hyperproduct
of x and y, and (H, ◦) is[called a hypergroupoid. For nonempty subsets A and
B of H, let A ◦ B =
a ◦ b, A ◦ b = A ◦ {b} and a ◦ B = {a} ◦ B where
a∈A,b∈B
a, b ∈ H.
A hypergroupoid (H, ◦) is called a semihypergroup if the hyperoperation ◦
is associative, that is, (x ◦ y) ◦ z = x ◦ (y ◦ z) for all x, y, z ∈ H.
A semihypergroup (H, ◦) is called a hypergroup if x ◦ H = H ◦ x = H for
all x ∈ H.
A hypergroupoid (H, ◦) is called commutative if x◦y = y ◦x for all x, y ∈ H.
If (H, ◦) is a hypergroupoid [semihypergroup], then for a nonempty subset
K of H, K is called a subhypergroupoid [subsemihypergroupoid] if K ◦ K ⊆ K,
that is, x ◦ y ⊆ K for all x, y ∈ K.
One can see that semihypergroups are generalization of hypergroups. Besides, we can define a hyperoperation ◦ on a semigroup S which makes (S, ◦) a
semihypergroup.
The partial transformation semigroup, the full transformation semigroup,
the 1-1 partial transformation semigroup or the symmetric inverse semigroup
and the symmetric group on a nonempty set X are denoted P (X), T (X), I(X)
and G(X), respectively.
SOME REGULAR EQUIVALENCE RELATION ON...
23
Example 1.1. (see [10]) Let S be a semigroup and P be a nonempty
subset of S and define x ◦ y = xP y(= {xty | t ∈ P }) for all x, y ∈ S. Then (S, ◦)
is a semihypergroup. In particular, if S is a group, then (S, ◦) is a hypergroup.
Then by Example 1.1, the semigroup P (X) along with I(X) is also a semihypergroup under the hyperoperation ◦, i.e.,
α ◦ β = αI(X)β = {αγβ | γ ∈ I(X)}
for all α, β ∈ P (X).
Our aim is to study another hyperoperation which is related to (P (X), ◦)
and regular equivalet relation ρ on P (X) which make P (X)/ρ is a semihypergroup. And we also study some subsemihypergroup on its.
Next, we recall some basic definitions and theorems of hypergroup theory
from [2] which make help to investigate P (X)/ρ is a semihypergroup.
Let ρ be an equivalence relation on a hypergroupoid (H, ◦) and A, B ⊆ H,
define A ρ B if and only if:
(i) ∀ a ∈ A, ∃ b ∈ B, aρb and ∀ b ∈ B, ∃ a ∈ A, aρb or
(ii) ∀ a ∈ A, ρ(a) ∩ B 6= ∅ and ∀ b ∈ B, ρ(b) ∩ A 6= ∅, where ρ(x) = {t ∈
H | tρx} ⊆ H.
An equivalence relation ρ is call regular if xρy then x ◦ aρy ◦ a and a ◦ xρa ◦ y
for all x, y, a ∈ H.
Theorem 1.2. (see [2]) Let (H, ◦) be a hypergroupoid ρ an equivalence
on H. Define ⊗ on H/ρ by
aρ ⊗ bρ = {xρ | x ∈ a ◦ b}
for all a, b ∈ H where xρ is the ρ − class of H containing x. Then:
(i) ρ is regular if and only if ⊗ is well-defined.
(ii) If (H, ◦) is a semihypergroup and ρ is regular then (H/ρ, ⊗) is a semihypergroup.
2. Main Result
First, we note that for any mapping α, α can be written as
−1 xα
.
α=
x
x∈ ran α
24
R.I. Sararnrakskul, S. Pianskool
Proposition 2.1. Define a relation on (P (X), ◦) by
αρβ
if and only if
| ran α| = | ran β|
for all α, β ∈ P (X). Then ρ is an equivalence relation.
Lemma 2.2. If α, β ∈ P (X) such that αρβ, then αI(X)γρβI(X)γ, that
is, α ◦ γρβ ◦ γ for any γ ∈ P (X).
Proof. Let α, β ∈ P (X) and assume that αρβ. Then | ran α| = | ran β|.
So, there is an bijective mapping θ : ran α → ran β. Let f ∈ I(X). Since
θ is a bijection from ran α onto ran β, ran α = (ran β)θ −1 = ran βθ −1 and
θ −1 f ∈ I(X). Therefore
ran(αf γ) = [ran α ∩ dom(f γ)](f γ)
= [ran βθ −1 ∩ dom(f γ)](f γ)
= ran(βθ −1 · f γ)
= ran(βθ −1 f γ).
Hence, | ran(αf γ)| = | ran(βθ −1 f γ)|, that is, αf γρβ(θ −1 f )γ.
Similarly, θf ∈ I(X) and ran β = (ran α)θ = ran αθ Therefore
ran(βf γ) = [ran β ∩ dom(f γ)](f γ)
= [ran αθ ∩ dom(f γ)](f γ)
= ran(αθ · f γ)
= ran(αθf γ).
So, | ran(βf γ)| = | ran(αθf γ)|, that is, α(θf )ρβf γ.
Lemma 2.3. Let α, β ∈ P (X) and γ ∈ I(X) such that αρβ. If f ∈ I(X)
then there exists h ∈ I(X) such that | ran γf α| = | ran γhβ|.
Proof. There is an bijective mapping θ : ran α → ran β because of αρβ. Let
f ∈ I(X). Since θ is a bijection from ran α onto ran β and ran γf α ⊆ ran α,
there exist B ⊆ ran β such that θ| ran γf α : ran γf α → B is a bijection. We
define α∗ ∈ I(X) by
for each y ∈ ran γf α, choose xy ∈ yα−1 ∩ ran f , that is,
xy
∗
.
α =
y∈ran γf α
y
xy ∈yα−1 ∩ran f
SOME REGULAR EQUIVALENCE RELATION ON...
25
So,
ran γf α = ran γf α∗ .
Next, we define β ∗ ∈ I(X) by
for each y ∈ B, choose xy ∈ yβ −1 , that is,
xy
β∗ =
y∈B
y
(1)
.
xy ∈yβ −1
Therefore f α∗ θ(β ∗ )−1 ∈ I(X) and ran β ∗ = ran β. Let h = f α∗ θ(β ∗ )−1 .
Since ran γf α∗ ⊆ ran α = dom θ and θ is injection,
| ran γf α∗ | = |(ran γf α∗ )θ| = | ran(γf α∗ θ)|.
(2)
Since ran γf α∗ ⊆ ran α = dom θ, (ran γf α∗ )θ = ran(γf α∗ θ).
Similarly, ran(γf α∗ θ) = (ran γf α∗ )θ ⊆ B = ran β ∗ = dom(β ∗ )−1 then
[ran(γf α∗ θ)](β ∗ )−1 = ran(γf α∗ θ(β ∗ )−1 ).
Because ran(γf α∗ θ(β ∗ )−1 ) ⊆ dom β then
[ran γf α∗ θ(β ∗ )−1 ]β = ran(γf α∗ θ(β ∗ )−1 β).
(3)
(4)
Since ran(γf α∗ θ) ⊆ B = dom(β ∗ )−1 and β| dom β ∗ = β ∗ is a bijection,
[ran(γf α∗ θ)](β ∗ )−1 β = ran(γf α∗ θ).
(5)
Therefore,
ran(γf α∗ θ(β ∗ )−1 β) = [ran γf α∗ θ(β ∗ )−1 ]β
∗
∗ −1
= [(ran(γf α θ))](β )
β
from (4)
from (3)
∗
= ran(γf α θ)
from (5).
Thus
ran(γf α∗ θ(β ∗ )−1 β) = ran(γf α∗ θ).
Consequently,
| ran(γhβ)| = | ran(γ(f α∗ θ(β ∗ )−1 )β)|
= | ran(γf α∗ θ)|
∗
Hence the proof is complete.
from (6)
= | ran(γf α )|
from (2)
= | ran γf α|
from (1).
(6)
26
R.I. Sararnrakskul, S. Pianskool
Lemma 2.4. Let α, β ∈ P (X) and γ ∈ I(X) such that αρβ. If f ∈ I(X)
then there exists k ∈ I(X) such that | ran γkα| = | ran γf β|.
Proof. Let θ : ran α → ran β be a bijection. Since θ −1 is a bijection from
ran β onto ran α and ran γf β ⊆ ran β, there exist A ⊆ ran α such that θ|−1
ran γf β :
ran γf β → A is a bijection. We define β ∗ ∈ I(X) by
for each y ∈ ran γf β, choose xy ∈ yβ −1 ∩ ran f , that is,
xy
.
β∗ =
y∈ran γf β
y
xy ∈yβ −1 ∩ran f
Thus
ran γf β = ran γf β ∗ .
Next, we define α∗ ∈ I(X) by
for each y ∈ A, choose xy ∈ yα−1 , that is,
xy
∗
α =
y∈A
y
xy
(7)
.
∈yα−1
Therefore f β ∗ θ −1 (α∗ )−1 ∈ I(X) and ran α∗ = ran α. Let k = f β ∗ θ −1 (α∗ )−1 .
It is enough to show that | ran γkα| = | ran γf β|.
Since ran γf β ∗ ⊆ ran β = dom θ −1 and θ −1 is injection,
| ran γf β ∗ | = |(ran γf β ∗ )θ −1 | = | ran(γf β ∗ θ −1 )|.
(8)
Since ran γf β ∗ ⊆ ran β = dom θ −1 , (ran γf β ∗ )θ −1 = ran(γf β ∗ θ −1 ). Similarly,
ran(γf β ∗ θ −1 ) = (ran γf β ∗ )θ −1 ⊆ A = ran α∗ = dom(α∗ )−1 then
[ran(γf β ∗ θ −1 )](α∗ )−1 = ran(γf β ∗ θ −1 (α∗ )−1 ).
Because ran(γf β ∗ θ −1 (α∗ )−1 ) ⊆ dom α then
[ran γf β ∗ θ −1 (α∗ )−1 ]α = ran(γf β ∗ θ −1 (α∗ )−1 α).
(9)
(10)
Since ran(γf β ∗ θ −1 ) ⊆ A = dom(α∗ )−1 and α| dom α∗ = α∗ is a bijection,
[ran(γf β ∗ θ −1 )](α∗ )−1 α = ran(γf β ∗ θ −1 ).
(11)
Therefore,
ran(γf β ∗ θ −1 (α∗ )−1 α) = [ran γf β ∗ θ −1 (α∗ )−1 ]α
from (10)
SOME REGULAR EQUIVALENCE RELATION ON...
= [ ran(γf β ∗ θ −1 ) ](α∗ )−1 α
∗ −1
= ran(γf β θ
)
27
from (9)
from (11).
Thus
ran(γf β ∗ θ −1 (α∗ )−1 α) = ran(γf β ∗ θ −1 ).
(12)
Consequently,
| ran(γkα)| = | ran(γ(f β ∗ θ −1 (α∗ )−1 )α)|
= | ran(γf β ∗ θ −1 )|
∗
from (12)
= | ran(γf β )|
from (8)
= | ran γf β|
from (7).
Lemma 2.5. If α, β ∈ P (X) such that αρβ, then γI(X)αργI(X)β, that
is, γ ◦ αργ ◦ β for any γ ∈ P (X).
Proof. Let α, β, γ ∈ P (X) such that αρβ. Then | ran α| = | ran β|. So, there
is an bijective mapping θ : ran α → ran β. From Lemma 2.3 and Lemma 2.4
there exists h ∈ I(X) such that | ran γf α| = | ran γhβ| and k ∈ I(X) such that
| ran γkα| = | ran γf β|, respectively. Therefore γI(X)αργI(X)β.
Theorem 2.6. Define the equivalence relation on (P (X), ◦) by
αρβ
if and only if
| ran α| = | ran β|
for all α, β ∈ P (X). Then ρ is regular.
Proof. It obtained directly from Lemma 2.2 and Lemma 2.5.
Corollary 2.7. Let ρ be an equivalence relation on (P (X), ◦) defined by
αρβ
if and only if
| ran α| = | ran β|
for all α, β ∈ P (X) and define ⊗ on P (X)/ρ by
αρ ⊗ βρ = {γρ | γ ∈ α ◦ β}
where γρ is the ρ − class of P (X) containing γ. Then (P (X)/ρ, ⊗) is a semihypergroup. Moreover, (G(X)/ρ, ⊗) is a subsemihypergroup of (P (X)/ρ, ⊗).
28
R.I. Sararnrakskul, S. Pianskool
Proof. Its follows from Theorem 1.2 and Theorem 2.6.
In [4], [6], [7] and [8], the authors use the word a “ local subsemigroup ”of
a semigroup S to mean a subsemigroup of S of the form eSe where e is an
idempotent of S. In 2008, [9] the authors were motivated by this definition and
defined a “ local subset ”and a “ local subsemigroup ”of a semigroup S in more
general sense as follows : a local subset of a semigroup S is a subset of S of
the form eAe where e is an idempotent of S and A is a subsemigroup of S.
Note that a local subset of a semigroup S need not be a subsemigroup of S.
Then they were interested in finding a necessary and sufficient condition for an
idempotent e of S which guarantees that eAe becomes a subsemigroup of S for
a given subsemigroup A of S. They called a local subset eAe of a semigroup S
a local subsemigroup of S if eAe is a subsemigroup of S.
Denote by E(S) the set of all idempotents of a semigroup S, that is,
E(S) = {x ∈ S | x2 = x}.
The cardinality of a set X is denoted by |X|. The domain and range of
a mapping α are denoted by dom α and ran α, respectively, and the value
of α at x ∈ dom α is written by xα. For A ⊆ dom α, α|A denotes the
restriction of α to A.
We also have that
E(P (X)) = {α ∈ P (x) | ran α ⊆ dom α and xα = x for all x ∈ ran α}
= {α ∈ P (x) | ran α ⊆ dom α and α|ran α = 1ran α }
([1], page 12). In [9],the authors provided a neccessary and sufficient condition
of α ∈ E(T (X)), where X is finite, so that αG(X)α is a local subsemigroup of
T (X) as follows:
Theorem 2.8 ([9]). Let X be a finite nonempty set and α ∈ E(T (X)).
Then αG(X)α is a local subsemigroup of T (X) if and only if either:
(i) α = 1X , the identity mapping on X, or
(ii) for every a ∈ ran α, |aα−1 | ≥ | ran α|.
In the second case, αG(X)α = αT (ran α) ∼
= T (ran α).
From Example 1.1, if S is a semigroup and ◦ is a hyperoperation on S which
makes (S, ◦) a semihypergroup. For this reason in 2013, R.I. Sararnrakskul definea local subset and a local subsemihypergroup on (S, ◦) in the similar way as
SOME REGULAR EQUIVALENCE RELATION ON...
29
follows : a local subset of a semihypergroup (S, ◦) is a subset of S of the form
e ◦ A ◦ e where e is an idempotent of the semigroup S and A is a subsemihypergroup of S; moreover, if a local subset e ◦ A ◦ e is a subsemihypergroup of
S, then it is called a local subsemihypergroup of S.
Furthermore, she provided a neccessary and sufficient condition of α ∈
E(P (X)), where X is finite, so that α ◦ G(X) ◦ α is a local subsemihypergroup
of P (X) as follows: α ◦ G(X) ◦ α is a local subsemihypergroup of P (X) if and
only if either:
(i) α = 1A , for some nonempty subset A of X, or
(ii) for every a ∈ ran α, |aα−1 | ≥ | ran α|.
If α satisfies (ii) then α ◦ G(X) ◦ α = αP (ran α) ∼
= P (ran α) under good
isomorphism.
Lemma 2.9. (see [10]) Let X be a nonempty set and α ∈ E(P (X)) \{0}.
Then
α ◦ G(X) ◦ α = αI(X)α.
Later on, we will find some subsemihypergroup on P (X)/ρ which is related
to Theorem 2.8 and we a call local subsemihypergroup of P (X)/ρ.
Corollary 2.10. Let X be a nonempty set and α ∈ E(P (X)) \{0}. If
α = 1A for some nonempty subset A of X then
α ◦ G(X) ◦ α = I(A).
Lemma 2.11. (see [10]) Let X be a nonempty set and α ∈ E(P (X)) \{0}.
Then αP (ran α) and P (ran α) are subsemihypergroups of P (X) and
αP (ran α) ∼
= P (ran α)
under good isomorphism of semihypergroup.
Lemma 2.12. (see [10]) Let X be a nonempty set and α ∈ E(P (X))\{0}.
If |aα−1 | ≥ | ran α| for every a ∈ ran α, then αI(X)α = αP (ran α).
Theorem 2.13. Let X be a nonempty set and α ∈ E(P (X)) \ {0}. Then
the local subset αρ ⊗ G(X)/ρ ⊗ αρ of P (X)/ρ is a local subsemihypergroup of
P (X)/ρ if:
30
R.I. Sararnrakskul, S. Pianskool
(i) α = 1A for some nonempty subset A of X, or
(ii) |aα−1 | ≥ | ran α| for every a ∈ ran α.
Proof. If α satisfies (i) then by Corollary 2.10, α ◦ G(X) ◦ α is a subsemihypergroup of P (X).
Assume that α satisfies (ii). Then by Lemma 2.9 and Lemma 2.10 we
have α ◦ G(X) ◦ α = αI(X)α = αP (ran α) and then by Lemma 2.11 we have
αP (ran α) is a subsemihypergroup of P (X). Therefore α ◦ G(X) ◦ α is a subsemihypergroup of P (X).
Next, we let γ, γ ′ ∈ G(X). Then
(αρ ⊗ γρ) ⊗ αρ = {xρ | x ∈ α ◦ γ} ⊗ αρ
[
=
xρ ⊗ αρ
x∈α◦γ
=
[
{yρ | y ∈ x ◦ α}
x∈α◦γ
= {yρ | y ∈ α ◦ γ ◦ α}.
Similarly, (αρ ⊗ γ ′ ρ) ⊗ αρ = {yρ | y ∈ α ◦ γ ′ ◦ α}. Therefore
(αρ ⊗ γρ ⊗ αρ) ⊗ (αρ ⊗ γ ′ ρ ⊗ αρ) = {xρ | x ∈ α ◦ γ ◦ α} ⊗ {yρ | y ∈ α ◦ γ ′ ◦ α}
[
=
xρ ⊗ yρ
x∈α◦γ◦α
y∈α◦γ ′ ◦α
=
[
{zρ | z ∈ x ◦ y}
x∈α◦γ◦α
y∈α◦γ ′ ◦α
= {zρ | z ∈ (α ◦ γ ◦ α) ◦ (α ◦ γ ′ ◦ α)} (13)
Since α ◦ G(X) ◦ α is a subsemihypergroup of P (X), there exist β ∈ G(X) such
that
(α ◦ γ ◦ α) ◦ (α ◦ γ ′ ◦ α) = α ◦ β ◦ α.
(14)
Consequently,
(αρ ⊗ γρ ⊗ αρ) ⊗ (αρ ⊗ γ ′ ρ ⊗ αρ) = {zρ | z ∈ (α ◦ γ ◦ α) ◦ (α ◦ γ ′ ◦ α)}
from (13)
= {zρ | z ∈ α ◦ β ◦ α}
from (14)
SOME REGULAR EQUIVALENCE RELATION ON...
31
= αρ ⊗ βρ ⊗ αρ
⊆ αρ ⊗ G(X)/ρ ⊗ αρ.
Hence αρ ⊗ G(X)/ρ ⊗ αρ is a subsemihypergroup of P (X)/ρ.
Acknowledgments
This research was supported by Government Fund Fiscal Year 2014, under
contract number 051/2557, Srinakharinwirot University.
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