# The Relationship between Some Regular Subsemigroups ofHyp

Authors: Weerapong Wongpinit; Sorasak Leeratanavalee
Journal: Journal of Mathematics
 (2014)

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## Abstract

The concept of regular subsemigroups plays an important role in the theory of semigroup. In this work, we study the relationship between some regular subsemigroups on the monoid of all generalized hypersubstitutions of typeτ
=
(
2
).


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## Body

## 1. Introduction

LetS be a semigroup. The class of regular semigroups is one of the most important classes of semigroups. Recall that an element a in a semigroup S is said to beregular if there exists b
∈
S such that a
=
a
b
a. A semigroup S is said to beregular if its element is regular. An element a
∈
S is calledidempotent if a
a
=
a. It is well-known that an idempotent element is an obvious example of a regular element. We denote the set of all idempotent elements of a semigroup S by E
(
S
). Next, we recall some definitions of special elements of a semigroup S.Definition 1.
An elementa of a semigroup S is calledleft (right) regular if there exists b
∈
S such that b
a
2
=
a
(
a
2
b
=
a
) and a is calledintraregular if a
∈
S
a
2
S.Definition 2.
An elementa of a semigroup S is calledcoregular if there is an element b
∈
S such that a
=
a
b
a
=
b
a
b
,
b itscoinverse. A semigroup is said to becoregular if each element of S is coregular.Then we have the following proposition.Proposition 3.
LetS be a semigroup and a
∈
S. Then a is coregular element if and only if a
3
=
a.Proof.
(⇒) Let a be a coregular element in S. Then there is an element b
∈
S such that a
=
a
b
a
=
b
a
b. Thus a
=
b
a
b
=
b
(
a
b
a
)
b
=
(
b
a
b
)
a
(
b
a
b
)
=
a
3.
(⇐) Suppose that a
3
=
a. Then a is coregular element.Definition 4.
An elementa of a semigroup S is calledantiregular if there exists b
∈
S such that a
b
a
=
b and b
a
b
=
a. The elements a and b are then calledanti-inverse.Proposition 5.
LetS be a semigroup and a
∈
S. If a is antiregular element, then a
5
=
a.Proof.
Leta be an antiregular element in S. Then there is an element b
∈
S such that a
b
a
=
b and b
a
b
=
a. Thus a
=
b
a
b
=
(
a
b
a
)
(
b
a
b
)
(
a
b
a
)
=
a
(
b
a
b
)
a
(
b
a
b
)
a
=
a
5.Definition 6.
An elementa of a semigroup S is calledcompletely regular if there exists b
∈
S such that a
b
a
=
a and a
b
=
b
a.Proposition 7 (see [1]).
LetS be a semigroup and a
∈
S. Then a is completely regular element if and only if a is both left regular and right regular.Remark 8.
In general, for any semigroupS and a
∈
S, we have the following relationship:
a is coregular ⇒
a is antiregular ⇒
a is completely regular ⇒
a is left regular, right regular, and intraregular.The consequent question is as follows: is there a semigroup such that all completely regular elements, left regular elements, right regular elements, or intraregular elements are coregular elements?As an example, consider the semigroup of all integers with the usual addition. We have that for any nonzero integersa and b, if a
=
a
+
b
+
a and a
+
b
=
b
+
a, then b
=
-
a but a
=
a
+
b
+
a
≠
b
+
a
+
b
=
b. Then a is completely regular but is not coregular.In this paper, we are interested in the semigroup of generalized hypersubstitutions which is a generalization of hypersubstitutions.LetX
≔
{
x
1
,
x
2
,
x
3
,
…
} be a countably infinite set of symbols calledvariables and let X
n
≔
{
x
1
,
x
2
,
…
,
x
n
} be an n-element set. Let (
f
i
)
i
∈
I be an indexed set which is disjoint from X. Each f
i is called an n
i
-ary operation symbol, where n
i
≥
1 is a natural number. Let τ be a function which assigns every f
i to the number n
i as its arity, written as (
n
i
)
i
∈
I and is called atype.Ann
-ary term of type
τ is defined inductively as follows.(i)
The variablesx
1
,
x
2
,
…
,
x
n are n-ary terms of type τ.
(ii)
Ift
1
,
t
2
,
…
,
t
n
i are n-ary terms of type τ, then f
i
(
t
1
,
t
2
,
…
,
t
n
i
) is an n-ary term of type τ.The smallest set which containsx
1
,
x
2
,
…
,
x
n and is closed under finite application of (ii), denoted by W
τ
(
X
n
). Let W
τ
(
X
)
≔
⋃
n
=
1
∞
‍
W
τ
(
X
n
) and is calledthe set of all terms of type
τ.The concept of a hypersubstitution was introduced first by Denecke et al. in 1991; see [2]. A hypersubstitution of type τ is a mapping σ
:
{
f
i
∣
i
∈
I
}
→
W
τ
(
X
) which assigns every n
i-ary operation symbol f
i to an n
i-ary term. The set of all hypersubstitutions of type τ is denoted by Hyp
(
τ
). Every σ
∈
Hyp
(
τ
) induces a mapping σ
^
:
W
τ
(
X
)
→
W
τ
(
X
) by the following steps:(i)
σ
^
[
x
]
≔
x, for any variable x
∈
X,
(ii)
σ
^
[
f
i
(
t
1
,
…
,
t
n
i
)
]
≔
σ
(
f
i
)
(
σ
^
[
t
1
]
,
…
,
σ
^
[
t
n
i
]
), where σ
^
[
t
j
]
,
1
≤
j
≤
n
i are already defined.A binary operation∘
h on Hyp
(
τ
) is defined by σ
1
∘
h
σ
2
≔
σ
^
1
∘
σ
2 for every σ
1
,
σ
2
∈
Hyp
(
τ
) where ∘ is the usual composition of mappings. Let σ
id be the hypersubstitution where σ
id
(
f
i
)
=
f
i
(
x
1
,
…
,
x
n
i
). Then (
Hyp
(
τ
)
;
∘
h
,
σ
id
) is a monoid with σ
id as an identity element.In 2000, Leeratanavalee and Denecke generalized the concepts of a hypersubstitution, a hyperidentity, and a solid variety to a generalized hypersubstitution, a strong hyperidentity, and a strongly solid variety, respectively; see [3]. A generalized hypersubstitution of type τ is a mapping σ
:
{
f
i
∣
i
∈
I
}
→
W
τ
(
X
) from the set of all n
i-ary operation symbols into the set of all terms built up by elements of the alphabet X
≔
{
x
1
,
x
2
,
…
} and operation symbols from {
f
i
∣
i
∈
I
} which does not necessarily preserve the arity.We denote the set of all generalized hypersubstitutions of typeτ by Hy
p
G
(
τ
). To define a binary operation on Hy
p
G
(
τ
), we defined first the concept of a generalized superposition of terms S
m
:
W
τ
(
X
)
m
+
1
→
W
τ
(
X
) by the following steps:for any termt
∈
W
τ
(
X
),(i)
ift
=
x
j, 1
≤
j
≤
m, then S
m
(
x
j
,
t
1
,
…
,
t
m
)
≔
t
j,
(ii)
ift
=
x
j, m
<
j
∈
N, then S
m
(
x
j
,
t
1
,
…
,
t
m
)
≔
x
j,
(iii)
ift
=
f
i
(
s
1
,
…
,
s
n
i
), then
(1)
S
m
(
t
,
t
1
,
…
,
t
m
)
≔
f
i
S
m
s
1
,
t
1
,
…
,
t
m
,
…
,
S
m
s
n
i
,
t
1
,
…
,
t
m
.Every generalized hypersubstitutionσ can be extended to a mapping σ
^
:
W
τ
(
X
)
→
W
τ
(
X
) by the following steps:(i)
σ
^
[
x
]
≔
x
∈
X,
(ii)
σ
^
[
f
i
(
t
1
,
…
,
t
n
i
)
]
≔
S
n
i
(
σ
(
f
i
)
,
σ
^
[
t
1
]
,
…
,
σ
^
[
t
n
i
]
), for any n
i-ary operation symbol f
i where σ
^
[
t
j
], 1
≤
j
≤
n
i are already defined.We define a binary operation∘
G on Hy
p
G
(
τ
) by σ
1
∘
G
σ
2
≔
σ
^
1
∘
σ
2 where ∘ denotes the usual composition of mappings and σ
1
,
σ
2
∈
Hy
p
G
(
τ
). Let σ
id be the hypersubstitution mapping which maps each n
i-ary operation symbol f
i to the term f
i
(
x
1
,
…
,
x
n
i
). Then (
Hy
p
G
(
τ
)
;
∘
G
,
σ
id
) is a monoid and the monoid (
Hyp
(
τ
)
;
∘
h
,
σ
id
) of all arity preserving hypersubstitutions of type τ forms a submonoid of (
Hy
p
G
(
τ
)
;
∘
G
,
σ
id
).Throughout this present paper, we fix the typeτ
=
(
2
). By σ
t, we denote the generalized hypersubstitution which maps f to the term t
∈
W
(
2
)
(
X
).Proposition 9 (see [3]).
For arbitraryt
,
t
1
,
t
2
,
…
,
t
n
∈
W
τ
(
X
) and for arbitrary generalized hypersubstitutions σ
,
σ
1
,
σ
2 one has (i)
S
n
(
σ
^
[
t
]
,
σ
^
[
t
1
]
,
σ
^
[
t
2
]
,
…
,
σ
^
[
t
n
]
)
=
σ
^
[
S
n
(
t
,
t
1
,
t
2
,
…
,
t
n
)
],
(ii)
σ
^
1
∘
σ
2
^
=
σ
^
1
∘
σ
^
2.


## 2. Main Results

Letτ
=
(
2
) be a fixed type with the binary operation symbol f. Let σ
t
∈
Hy
p
G
(
2
), the set of all variables occurring in t is denoted by var
(
t
)
R
1
≔
σ
t
∣
t
=
f
x
2
,
t
′
where
t
′
∈
W
2
X
such
that
x
1
∉
var
t
′
},
R
2
≔
σ
t
∣
t
=
f
t
′
,
x
1
where
t
′
∈
W
2
(
X
)
such
that
x
2
∉
var
(
t
′
)
},
R
3
≔
σ
t
∣
t
=
f
x
1
,
t
′
where
t
′
∈
W
2
X
such
that
x
2
∉
var
(
t
′
)
},
R
4
≔
σ
t
∣
t
=
f
t
′
,
x
2
where
t
′
∈
W
2
(
X
)
such
that
x
1
∉
var
(
t
′
)
},
R
5
≔
σ
t
∣
t
∈
x
1
,
x
2
,
f
x
1
,
x
2
,
f
x
2
,
x
1,
R
6
≔
σ
t
∣
var
t
∩
x
1
,
x
2
=
∅.In [4], Puninagool and Leeratanavalee showed that ∪
i
=
1
6
R
i is a set of all regular elements in Hy
p
G
(
2
) and (
∪
i
=
3
6
R
i
)
∖
{
σ
f
(
x
2
,
x
1
)
}
=
E
(
Hy
p
G
(
2
)
).In [5], Sudsanit et al. proved that the set of all left regular elements and the set of all right regular elements in Hy
p
G
(
2
) are the same; that is, E
(
Hy
p
G
(
2
)
)
∪
{
σ
f
(
x
2
,
x
1
)
}.Furthermore, Boonmee and Leeratanavalee showed that the set of all intraregular elements inHy
p
G
(
2
) is also E
(
Hy
p
G
(
2
)
)
∪
{
σ
f
(
x
2
,
x
1
)
}; see [6].Proposition 10.
Letσ
t
∈
H
y
p
G
(
2
). Then the following conditions are equivalent: (a)
σ
t is coregular,
(b)
σ
t is antiregular,
(c)
σ
t is completely regular,
(d)
σ
t is left regular,
(e)
σ
t is right regular,
(f)
σ
t is intraregular.Proof.
(a)⇒ (b) ⇒ (c) ⇒ (d) are obtained by Remark 8.
(d)⇒ (e) ⇒ (f) is a consequence of the fact that the set of all left regular, right regular, and intraregular elements in Hy
p
G
(
2
) are the same.
(f)⇒ (a) Let σ
t be intraregular; then σ
t
∈
E
(
Hy
p
G
(
2
)
)
∪
{
σ
f
(
x
2
,
x
1
)
}. So σ
t
=
σ
t
3 and then σ
t is a coregular element.Corollary 11.
E
(
H
y
p
G
(
2
)
)
∪
{
σ
f
(
x
2
,
x
1
)
} is the set of all coregular elements, antiregular elements, completely regular elements, left regular elements, right regular elements, and intraregular elements in H
y
p
G
(
2
).Moreover, the next result describes the relationship between some regular subsemigroups of this semigroup.LetS be a semigroup. A nonempty subset T of S is called asubsemigroup of S if T
2
⊆
T. A nonempty subsemigroup T of S is called aregular subsemigroup if, for any element a of T, there exists an element b
∈
T such that a
=
a
b
a.Proposition 12.
LetR be a regular subsemigroup of H
y
p
G
(
2
). Then the following conditions are equivalent: (a)
R is coregular,
(b)
R is antiregular,
(c)
R is completely regular,
(d)
R is left regular,
(e)
R is right regular,
(f)
R is intraregular.Proof.
(a)⇒ (b) Let σ
t
∈
R be such that σ
t
3
=
σ
t. Then σ
t is an antiregular element.
(b)⇒ (c) Let σ
t be an antiregular element in R. By using Proposition 5, we have σ
t
=
σ
t
5
=
σ
t
∘
G
σ
t
3
∘
G
σ
t and σ
t
3
∘
G
σ
t
=
σ
t
∘
G
σ
t
3. Since σ
t
3
=
σ
t for all σ
t
∈
R, so R is completely regular.
(c)⇒ (d) It is obviously by definition.
(d)⇒ (e) Since σ
t
3
=
σ
t
∘
G
σ
t
2
=
σ
t
2
∘
G
σ
t
=
σ
t for all σ
t
∈
R, so R is right regular.
(e)⇒ (f) Let σ
t
∈
R. Since R is right regular, so there exists σ
s
∈
R such that σ
t
2
∘
G
σ
s
=
σ
t. Since σ
t
=
σ
t
2
∘
G
σ
s
=
σ
t
∘
G
σ
t
∘
G
σ
s
=
σ
t
∘
G
σ
t
2
∘
G
σ
s
∘
G
σ
s
=
σ
t
∘
G
σ
t
2
∘
G
σ
s
2
∈
R
∘
G
σ
t
2
∘
G
R, so σ
t is intraregular.
(f)⇒ (a) Let σ
t
∈
R
⊆
E
(
Hy
p
G
(
2
)
)
∪
{
σ
f
(
x
2
,
x
1
)
}. Then σ
t
=
σ
t
3. So R is coregular.



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*Source: 181397-2014-12-08.xml*