Lobachevskii Journal of Mathematics
Vol. 13, 2003, 51 – 56
©Niovi Kehayopulu and Michael Tsingelis
Niovi Kehayopulu and Michael Tsingelis
A NOTE ON SEMI-PSEUDOORDERS IN SEMIGROUPS
(submitted by M. M. Arslanov)
ABSTRACT. An important problem for studying the structure of an ordered
semigroup S
is to know conditions under which for a given congruence
ρ on
S the set
S∕ρ is an
ordered semigroup. In [1] we introduced the concept of pseudoorder in ordered
semigroups and we proved that each pseudoorder on an ordered semigroup
S induces a
congruence σ
on S such
that S∕σ
is an ordered semigroup. In [3] we introduced the concept of
semi-pseudoorder (also called pseudocongruence) in semigroups
and we proved that each semi-pseudoorder on a semigroup
S induces a
congruence σ
on S such
that S∕σ
is an ordered semigroup. In this note we prove that the converse
of the last statement also holds. That is each congruence
σ on a
semigroup (S,.)
such that S∕σ
is an ordered semigroup induces a semi-pseudoorder on
S.
________________
Key words and phrases. Pseudoorder, pseudocongruence, semi-pseudoorder.
2000 Mathematical Subject Classification. 06F05, 20M10.
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For a given ordered semigroup (S,.,≤)
is essential to know if there exists a congruence
ρ on
S such that
S∕ρ be an
ordered semigroup. This plays an important role for studying the structure of ordered
semigroups. If S
is a semigroup (resp. an ordered semigroup), by a congruence on
S we mean an
equivalence relation σ
on S such
that (a,b) ∈ σ
implies (ac,bc) ∈ σ
and (ca,cb) ∈ σ for
all c ∈ S. If
S is a semigroup
and σ a
congruence on S,
then the set S∕σ := {(a)σ ∣a ∈ S}
( (a)σ is the
σ-class
of S
containing a
(a ∈ S)) is a semigroup and the
operation on S∕σ is defined via the
operation on S. The following
question is natural: If (S,.,≤) is an
ordered semigroup and σ a
congruence on S, then is the
set S∕σ an ordered semigroup
? A probable order on S∕σ
could be the relation ” ≼ ”
on S∕σ defined by
means of the order ” ≤ ”
on S,
that is
≼: = {(t,z) ∈ S∕σ × S∕σ ∣∃(a,b) ∈≤ such that t = (a)σ,z = (b)σ}
= {((x)σ, (y)σ) ∣∃a ∈ (x)σ,b = (y)σ such that (a,b) ∈≤}.
But this relation is not an order, in general. An example can be
found in [1]. The following question arises: Is there a congruence
σ on
S for which
S∕σ is an ordered
semigroup ? This let to the concept of pseudoorder introduced by the same authors in [1]. Let
(S,.,≤) be an ordered
semigroup. A relation ρ
on S is
called pseudoorder if
- ≤⊆ ρ
- (a,b) ∈ ρ and (b,c) ∈ ρ imply (a,c) ∈ ρ.
- (a,b) ∈ ρ implies (ac,bc) ∈ ρ and (ca,cb) ∈ ρ for each c ∈ S.
According to Lemma 1 in [1], if (S,.,≤)
is an ordered semigroup and σ
a pseudoorder on S,
then the relation σ̄
on S
defined by
σ̄ := {(a,b) ∈ S × S ∣(a,b) ∈ σ and (b,a) ∈ σ}
is a congruence on S
and the set
S∕σ̄ is
an ordered semigroup. So according to [1],
each pseudoorder on an ordered semigroup
S induces a
congruence σ̄
on S such
that S∕σ̄
is an ordered semigroup. For a further study of pseudoorders in ordered
semigroups we refer to [2]. On the other hand, the concept of pseudocongruences
in semigroups has been introduced by the same authors in [3]. If
(S,.) is a semigroup, by a
pseudocongruence on S
we mean a relation ρ
on S
such that
- (a,a) ∈ ρ∀a ∈ S
- (a,b) ∈ ρ and (b,c) ∈ ρ imply (a,c) ∈ ρ.
- (a,b) ∈ ρ implies (ac,bc) ∈ ρ and (ca,cb) ∈ ρ for each c ∈ S.
If (S,.,≤)
is an ordered semigroup, then each pseudoorder on
S is a
pseudocongruence on S.
Indeed, if ρ
is a
pseudoorder on S
and a ∈ S,
then (a,a) ∈≤⊆ ρ.
Pseudocongruences can be also called semi-pseudoorders, and
from now on we will keep that terminology of semi-pseudoorders.
We have seen in [3], that each semi-pseudoorder on a semigroup
S induces a
congruence ρ
on S such that
S∕ρ is an ordered
semigroup. In this paper we prove that the converse of this statement also holds. For a semigroup
(S,.) we define a
multiplication ” ∗ ”
on on S∕ρ
defined by (a)ρ ∗ (b)ρ := (ab)ρ. If
(S,.) is a semigroup and
ρ a congruence on
S and if there exists
an order relation ”∇”
on S∕ρ such
that the (S∕ρ,∗,∇)
is an ordered semigroup, then there exists a semi-pseudoorder
σ on
S such that
ρ = σ̄. So each
congruence ρ on
a semigroup (S,.)
such that S∕ρ
is an ordered semigroup induces a semi-pseudoorder on
S.
If (S,.) is a semigroup
and σ a
semi-pseudoorder on S,
we define
σ̄ := σ ∩ σ−1.
The relation σ̄ is
a congruence on S.
Indeed: If a ∈ S,
then (a,a) ∈ σ,
then (a,a) ∈ σ−1,
so (a,a) ∈ σ ∩ σ−1 := σ̄. If
(a, b) ∈σ̄, then
(a, b) ∈ σ and
(a, b) ∈ σ−1, then
(b, a) ∈ σ−1 and
(b, a) ∈ σ, so
(b, a) ∈ σ−1 ∩ σ := σ̄. If
(a, b) ∈σ̄ and
(b, c) ∈σ̄, then
(a, b) ∈ σ,
(a, b) ∈ σ−1,
(b, c) ∈ σ,
(b, c) ∈ σ−1, then
(a, c) ∈ σ and
(a, c) ∈ σ−1, thus
(a, c) ∈ σ ∩ σ−1 := σ̄. Let
(a, b) ∈σ̄ and
c ∈ S. We
have (a,b) ∈ σ
and (a,b) ∈ σ−1.
Since (a,b) ∈ σ,
c ∈ S, we
have (ac,bc) ∈ σ,
(ca,cb) ∈ σ. Since
(a, b) ∈ σ−1, we
have (b,a) ∈ σ,
then (bc,ac) ∈ σ,
(ca,cb) ∈ σ, hence
(ac,bc) ∈ σ−1,
(ca,cb) ∈ σ−1. Then
we have (ac,bc) ∈ σ ∩ σ−1 := σ̄
and (ca,cb) ∈ σ ∩ σ−1 := σ̄.
[It might be also noted that σ̄ = {(a,b) ∈ S × S ∣(a,b) ∈ σ and (b,a) ∈ σ}.
Hence σ̄ is a
congruence on S (cf.
[3])]. Since σ̄ is a
congruence on S,
the set S∕σ̄ with
the operation ” ∗ ”
on S∕σ̄ defined
by (a)σ̄ ∗ (b)σ̄ := (ab)σ̄ is
a semigroup (It is known).
If (S,.) is a
semigroup and σ
a
semi-pseudoorder on S,
we define a relation ”∇”
on S∕σ̄ as follows:
(a)σ̄∇(b)σ̄ if and only
if (a,b) ∈ σ. The
relation ”∇” on
S∕σ̄ is well defined.
Indeed: Let (a)σ̄ = (c)σ̄,
(b)σ̄ = (d)σ̄ and
(a)σ̄ ≼ (b)σ̄. Since
(a)σ̄∇(b)σ̄, we have
(a, b) ∈ σ. Since
(a)σ̄ = (c)σ̄, we have
(a, c) ∈σ̄ := σ ∩ σ−1 ⊆ σ−1, then
(c, a) ∈ σ. Since
(b)σ̄ = (d)σ̄, we
have (b,d) ∈σ̄ := σ ∩ σ−1 ⊆ σ,
then (b,d) ∈ σ.
Then (c,d) ∈ σ,
and (c)σ̄∇(d)σ̄.
(Cf. also [3]).
Theorem. Let (S, .) be a semigroup. If
σ is a semi-pseudoorder
on S, then the set
(S∕σ̄,∗,∇) is an ordered semigroup.
Let ρ be a congruence
on S and suppose there
exists an order relation ” ≼ ”
on S∕ρ such
that (S∕ρ̄,∗,∇)
be an ordered semigroup. Then there exists a semi-pseudoorder
σ on
S such
that
ρ = σ̄ and ≼ = ∇.
Proof. For the first part of the Theorem we refer to the Theorem in [3]. Let now
ρ be a
congruence on S
and ” ≼ ” an order
on S∕ρ such that
(S∕ρ,∗,≼) be an ordered
semigroup. Let σ
be the relation on S
defined by
σ := {(a,b) ∈ S × S ∣(a)ρ ≼ (b)ρ}.
1) σ is a
semi-pseudoorder on S.
In fact:
Let a ∈ S.
Since (a)ρ ≼ (a)ρ,
we have (a,a) ∈ σ.
Let (a,b) ∈ σ,
(b, c) ∈ σ. Then
(a)ρ ≼ (b)ρ,
(b)ρ ≼ (c)ρ, then
(a)ρ ≼ (c)ρ, and
(a, c) ∈ σ. Let
(a, b) ∈ σ and
c ∈ S. Then
(a)ρ ≼ (b)ρ and
(c)ρ ∈ S∕ρ. Since
(S∕ρ,∗,≼) is an ordered
semigroup, we have (a)ρ ∗ (c)ρ ≼ (b)ρ ∗ (c)ρ,
then (ac)ρ ≼ (bc)ρ,
and (ac,bc) ∈ σ.
Similarly (a,b) ∈ σ
and c ∈ S,
imply (ca,cb) ∈ σ.
2) ρ = σ̄.
Indeed: We have
(a,b) ∈ ρ ⇔ (a)ρ = (b)ρ
⇔ (a)ρ ≼ (b)ρ and (b)ρ ≼ (a)ρ
⇔ (a,b) ∈ σ and (b,a) ∈ σ
⇔ (a,b) ∈σ̄.
3) ≼ = ∇.
Indeed:
Let (a)ρ ≼ (b)ρ.
Since (a,b) ∈ σ,
we have (a)σ̄∇(b)σ̄.
By 2), ρ = σ̄.
So (a)σ̄ = (a)ρ
and (b)σ̄ = (b)ρ.
Then (a)ρ∇(b)ρ.
Let (a)ρ∇(b)ρ.
Since ρ = σ̄,
we have (a)ρ = (a)σ̄
and (b)ρ = (b)σ̄.
Then (a)σ̄∇(b)σ̄,
hence (a,b) ∈ σ,
and (a)ρ ≼ (b)ρ.
Remark 1. If (S,.,≤) is an
ordered semigroup and ρ
a pseudoorder on S,
then the mapping
f(S,.,≤) →(S∕ρ̄,∗,∇) ∣a →(a)ρ̄
is a homomorphism. In fact, if a,b ∈ S,
then
f(ab) := (ab)ρ̄ := (a)ρ̄ ∗ (b)ρ̄ = f(a) ∗ f(b).
Let now a ≤ b.
Since (a,b) ∈≤⊆ ρ, we
have (a,b) ∈ ρ. Then,
since ρ is a
semipseudoorder on S,
we have (a)ρ̄∇f(b),
and f(a)∇f(b).
□
For a semigroup S, we
denote by SP(S) the set of
semi-pseudoorders on S
and by C(S) the set of
congruences on S. Let
” ≈ ” be the equivalence
relation on S
defined as follows:
ρ ≈ σ if and only if ρ̄ = σ̄.
Remark 2. If S is a
semigroup and ρ a
semi-pseudoorder on S,
then the mapping
f : SP(S)∕≈→C(S) ∣(ρ)≈ →ρ̄
is (1-1) and onto. In fact: The mapping
f is well
defined: If ρ is a
semi-pseudoorder on S,
then ρ̄ is a
congruence on S.
Let ρ,σ ∈SP(S) and
(ρ)≈ = (σ)≈. Then we
have ρ ≈ σ,
and ρ̄ = σ̄.
f is (1-1):
Let ρ,σ ∈SP(S) such
that ρ̄ = σ̄.
Then ρ ≈ σ,
and (ρ)≈ = (σ)≈.
f is onto:
Let ρ ∈C(S).
Then ρ = ρ−1
and ρ is a
semi-pseudoorder on S.
Thus ρ ∈SP(S)
and
f((ρ)≈) := ρ̄ := ρ ∩ ρ−1 = ρ ∩ ρ = ρ.
□
For a semigroup S, we
denote by OC(S) the set of
all congruences ρ on
S for which there exists
an order relation ”∇”
on S∕ρ such
that (S∕ρ,∗,∇)
is an ordered semigroup.
Remark 3. If S
is a semigroup, then the mapping
f : SP(S)∕≈→OC(S) ∣(ρ)≈ →ρ̄
is (1-1) and onto. In fact: The mapping
f is well defined: If
ρ is a semi-pseudoorder
on S, then
ρ ̄ is a congruence on
S. Then, by the Theorem, the
set (S∕ρ̄,∗,∇) is an ordered semigroup.
Which means that ρ̄ ∈OC(S).
Let ρ,σ ∈SP(S) and
(ρ)≈ = (σ)≈. Then we
have ρ ≈ σ,
and ρ̄ = σ̄.
f is (1-1):
Let ρ,σ ∈SP(S)
and ρ̄ = σ̄.
Then ρ ≈ σ,
and (ρ)≈ = (σ)≈.
f is onto:
Let ρ ∈OC(S).
By the Theorem, there exists a semi-pseudoorder
σ on
S such
that ρ = σ̄.
Then σ ∈SP(S),
and f((σ))≈ := σ̄ = ρ.
Acknowledgment. This research was supported by the Special Research Account
of the University of Athens (Grant No. 70/4/5630).
References
[1] N. Kehayopulu, M. Tsingelis, On subdirectly irreducible ordered semigroups, Semigroup
Forum 50 (1995), 161-177.
[2] N. Kehayopulu, M. Tsingelis, Pseudoorder in ordered semigroups, Semigroup Forum 50
(1995), 389-392.
[3] N. Kehayopulu, M. Tsingelis A note on pseudocongruences in semigroups, Lobachevskii
J. Math. 11 (2002), 19-21.
UNIV. OF ATHENS, DEPT. OF MATHEMATICS HOME: NIOVI KEHAYOPULU, NIKOMIDIAS
18, 161 22 KESARIANI, GREECE
E-mail address: nkehayop@cc.uoa.gr
ReceivedSeptember 30, 2003