Congruences on regular semigroups with $Q$-inverse transversals
We give congruences on a regular semigroup with a $Q$-inverse transversal $S^o$ by the congruence pair (abstractly), which consists of congruences on the structural component parts $R$ and $\Lambda$. We prove that the set of all congruences for this kind of semigroups is a complete lattice.
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| author | Wang, A. Wang, L. Ван, А. Ван, Л. |
| author_facet | Wang, A. Wang, L. Ван, А. Ван, Л. |
| author_sort | Wang, A. |
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| description | We give congruences on a regular semigroup with a $Q$-inverse transversal $S^o$ by the congruence pair (abstractly), which consists of congruences on the structural component parts $R$ and $\Lambda$. We prove that the set of all congruences for this kind of semigroups is a complete lattice. |
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К О Р О Т К I П О В I Д О М Л Е Н Н Я
UDC 512.5
A. Wang (Northwest Univ., China and Chongqing Univ. Technology, China),
L. Wang (Chongqing Univ. Technology, China)
CONGRUENCES ON REGULAR SEMIGROUPS
WITH \bfitQ -INVERSE TRANSVERSALS*
КОНГРУЕНЦIЇ НА РЕГУЛЯРНИХ НАПIВГРУПАХ
З \bfitQ -ОБЕРНЕНИМИ ТРАНСВЕРСАЛЯМИ
We give congruences on a regular semigroup with a Q-inverse transversal So by the congruence pair (abstractly), which
consists of congruences on the structural component parts R and \Lambda . We prove that the set of all congruences for this kind
of semigroups is a complete lattice.
Наведено конгруенцiї на регулярних напiвгрупах з Q-оберненою трансверсаллю So щодо пари конгруенцiй (аб-
страктно), сформованої з конгруенцiй на частинах структурних компонент R i \Lambda . Доведено, що множина всiх
конгруенцiй для такого типу напiвгруп є повною ґраткою.
1. Introduction and preliminaries. The multiplicative inverse transversals of a regular semigroup
were first introduced by Blyth and McFadden in 1982 [1]. An inverse subsemigroup S\circ of a regular
semigroup S is an inverse transversal if | V (x) \cap S\circ | = 1 for any x \in S, where V (x) denoted the
set of inverses of x. In this case, the unique element of V (x) \cap S\circ is denoted by x\circ and (x\circ )\circ is
denoted by x\circ \circ . Throughout this paper S denotes a regular semigroup with an inverse transversal
S\circ and E(S\circ ) = E\circ denotes the semilattice of idempotents of S\circ . An inverse transversal S\circ is a
multiplicative inverse transversal if x\circ xyy\circ \in E\circ , and S\circ is a Q-inverse transversal if S\circ SS\circ \subseteq S\circ .
Let S\circ be a Q-inverse transversal, and let
R = \{ x \in S | x\circ x = x\circ x\circ \circ \} , L = \{ a \in S | aa\circ = a\circ \circ a\circ \} ,
and
I = \{ e \in E(S) | ee\circ = e\} , \Lambda = \{ f \in E(S) | f\circ f = f\} ,
where E(S) = \{ x \in S | x2 = x\} which is the set of idempotents of S. It was shown in [6] that
R and L are orthodox subsemigroups of S with transversal S\circ which is a right ideal of R and a
left ideal of L and that E(R) = I, E(L) = \Lambda . Moreover, E\circ is a multiplicative inverse transversal
of I and \Lambda . In [3], McAlister and McFadden show that I and \Lambda are \scrR -unipotent and \scrL -unipotent
subbands respectively of S. Saito gave the structure theory of a regular semigroup with a Q-inverse
transversal in [6]. The congruences on regular semigroups with inverse transversals were studied
using the congruence triple by Wang and Tang (see [5, 9, 10]). In this paper, we give the congruences
on regular semigroups with Q-inverse transversals by the congruence pair and prove that the set of
all congruences on this kind of semigroups is a complete lattice.
We list already obtained results in [3 – 6], which will be used in this paper.
* Research was supported by the Project Foundation of Chongqing Municipal Education Committee (KJ1500925).
c\bigcirc A. WANG, L. WANG, 2016
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 6 853
854 A. WANG, L. WANG
Lemma 1.1. Let S\circ be a Q-inverse transversal. Then, for e \in E\circ , f \in I [resp. g \in \Lambda ],
e\circ f = e\circ f\circ [resp. ge\circ = g\circ e\circ ].
Lemma 1.2. Let S\circ be a Q-inverse transversal. Then (xy)\circ = (x\circ xy)\circ x\circ = y\circ (xyy\circ )\circ for
every x, y \in S.
Lemma 1.3. S\circ is a Q-inverse transversal if and only if for any s, t \in S\circ , x \in S, (sxt)\circ =
= t\circ x\circ s\circ .
Lemma 1.4. Let S be a band with an inverse transversal S\circ . If S\circ is a left ideal of S, then
e\circ e = e for every e \in S. In this case, S is an \scrL -unipotent band.
For a regular semigroup S, Con (S) denotes the complete lattice of congruences on S. For
\rho \in Con (S), let \rho \circ = \rho | S\circ , \rho I = \rho | I , \rho \Lambda = \rho | \Lambda .
Lemma 1.5. Let S\circ be a Q-inverse transversal. For any \rho \in Con (S) and for x, y \in S, x\rho y
implies x\circ \rho \circ y\circ .
Lemma 1.6. Let R be a regular semigroup with a right ideal inverse transversal S\circ . Suppose
that \Lambda is a band with a left ideal inverse transversal E\circ . Let \Lambda \times R - \rightarrow S\circ described by (\lambda , x) - \rightarrow
- \rightarrow \lambda \ast x be a mapping, such that for any x, y \in R and for any \lambda , \mu \in \Lambda :
(Q1) (\lambda \ast x)y = \lambda \ast (xy) and \mu (\lambda \ast x) = (\mu \lambda ) \ast x;
(Q2) if x \in E\circ or \lambda \in E\circ , then \lambda \ast x = \lambda x.
Define a multiplication on the set
\Gamma \equiv R| \times | \Lambda = \{ (x, \lambda ) \in R\times \Lambda : x\circ x = \lambda \circ \}
by
(x, \lambda )(y, \mu ) = (x(\lambda \ast y), (\lambda \ast y)\circ (\lambda \ast y)\mu ).
Then \Gamma is a regular semigroup with a Q-inverse transversal which is isomorphic to S\circ .
Conversely, every regular semigroup with a Q-inverse transversal can be constructed in this way.
2. Main results. In this section, we first establish a characterization of congruences abstractly
by congruence pair. We describe a congruence pair of the form (\rho R, \rho \Lambda ) with \rho R \in Con (R) and
\rho \Lambda \in Con (\Lambda ) satisfying some conditions in order that they produce a congruence on S naturally.
Let S\circ be a Q-inverse transversal. For \rho \in Con (S), let \rho R = \rho | R. The following lemma shows
that \rho R is determined uniquely by its restrictions to S\circ and I.
Lemma 2.1. For \rho , \sigma \in Con (S), \rho R \subseteq \sigma R \leftrightarrow \rho \circ \subseteq \sigma \circ , \rho I \subseteq \sigma I . Therefore,
\rho R = \sigma R \leftrightarrow \rho \circ = \sigma \circ , \rho I = \sigma I .
Proof. Suppose that \rho \circ \subseteq \sigma \circ , \rho I \subseteq \sigma I . Then for any x, y \in S, by Lemma 1.5,
x\rho Ry =\Rightarrow x\circ \rho \circ y\circ , x, y \in R =\Rightarrow
=\Rightarrow x\circ \circ \rho \circ y\circ \circ , x\circ x\circ \circ \rho \circ y\circ y\circ \circ , xx\circ \rho Iyy
\circ , x, y \in R =\Rightarrow
=\Rightarrow x\circ \circ \rho \circ y\circ \circ , x\circ x\rho \circ y\circ y, xx\circ \rho Iyy
\circ , x, y \in R =\Rightarrow
=\Rightarrow x\circ \circ \sigma \circ y\circ \circ , x\circ x\sigma \circ y\circ y, xx\circ \sigma Iyy
\circ , x, y \in R =\Rightarrow
=\Rightarrow x = xx\circ x\circ \circ x\circ x\sigma yy\circ y\circ \circ y\circ y = y, x, y \in R =\Rightarrow x\sigma Ry.
So \rho R \subseteq \sigma R. The reverse implication is obvious.
Lemma 2.1 is proved.
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 6
CONGRUENCES ON REGULAR SEMIGROUPS WITH Q-INVERSE TRANSVERSALS 855
Suppose \rho R and \rho \Lambda are congruences on R and \Lambda , respectively. Then (\rho R, \rho \Lambda ) is called a
congruence pair for \Gamma if the following conditions hold:
(C1) \rho R| E\circ = \rho \Lambda | E\circ ;
(C2) (\forall z \in R)(\forall \lambda , \mu \in \Lambda ) \lambda \rho \Lambda \mu \Rightarrow (\lambda \ast z)\rho R(\mu \ast z);
(C3) (\forall \nu \in \Lambda )(\forall x, y \in R) x\rho Ry \Rightarrow (\nu \ast x)\rho R(\nu \ast y).
Define \rho (\rho
R,\rho \Lambda ) on \Gamma by
(x, \lambda )\rho (\rho
R,\rho \Lambda )(y, \mu ) \leftrightarrow x\rho Ry, \lambda \rho \Lambda \mu .
Theorem 2.1. Let \Gamma be a regular semigroup having a Q-inverse transversal as in Lemma 1.6,
and (\rho R, \rho \Lambda ) be a congruence pair on \Gamma . Then \rho (\rho
R,\rho \Lambda ) is a congruence on \Gamma .
Conversely, every congruence on \Gamma can be constructed in the above manner.
Proof. Let (\rho R, \rho \Lambda ) be a congruence pair on \Gamma . Obviously, \rho (\rho
R,\rho \Lambda ) is an equivalence on \Gamma . For
(x, \lambda ), (y, \mu ) \in \Gamma , with (x, \lambda )\rho (\rho
R,\rho \Lambda )(y, \mu ), we have x\rho Ry, \lambda \rho \Lambda \mu . Let z \in R and \nu \in \Lambda be such
that (z, \nu ) \in \Gamma . By Lemmas 1.2, 1.3 and condition (C2), we have
(x(\lambda \ast z))\circ = (x\circ x(\lambda \ast z))\circ x\circ =
= (\lambda \ast z)\circ (x\circ x(\lambda \ast z)(\lambda \ast z)\circ )\circ x\circ \rho \circ (\mu \ast z)\circ (y\circ y(\mu \ast z)(\mu \ast z)\circ )\circ y\circ =
= (y(\mu \ast z))\circ
and
(x(\lambda \ast z))(x(\lambda \ast z))\circ \rho I(y(\mu \ast z))(y(\mu \ast z))\circ .
Thus, by Lemma 2.1,
x(\lambda \ast z)\rho Ry(\mu \ast z).
From condition (C2, ) we have (\lambda \ast z)\rho (\mu \ast z) and so (\lambda \ast z)\circ \rho \circ (\mu \ast z)\circ . It follows that
(\lambda \ast z)\circ (\lambda \ast z)\rho \Lambda (\mu \ast z)\circ (\mu \ast z).
Hence
(\lambda \ast z)\circ (\lambda \ast z)\nu \rho \Lambda (\mu \ast z)\circ (\mu \ast z)\nu .
Thus
(x, \lambda )(z, \nu )\rho (\rho
R,\rho \Lambda )(y, \mu )(z, \nu ).
Next, by x\rho Ry and condition (C3), we have
(\nu \ast x)\rho R(\nu \ast y).
It follows that
z(\nu \ast x)\rho Rz(\nu \ast y)
and
(\nu \ast x)\circ (\nu \ast x)\lambda \rho \Lambda (\nu \ast y)\circ (\nu \ast y)\mu .
Hence
(z(\nu \ast x), (\nu \ast x)\circ (\nu \ast x)\lambda )\rho (\rho R,\rho \Lambda )(z(\nu \ast y), (\nu \ast y)\circ (\nu \ast y)\mu ).
That is,
(z, \nu )(x, \lambda )\rho (\rho
R,\rho \Lambda )(z, \nu )(y, \mu ).
Therefore \rho (\rho
R,\rho \Lambda ) is a congruence on \Gamma .
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 6
856 A. WANG, L. WANG
Conversely, assume that \rho is a congruence on \Gamma . We define the following equivalences on R and
\Lambda , respectively:
(\forall x, y \in R) x\rho Ry \leftrightarrow (x, x\circ x)\rho (y, y\circ y),
(\forall \lambda , \mu \in \Lambda ) \lambda \rho \Lambda \mu \leftrightarrow (x\circ x, \lambda )\rho (y\circ y, \mu ).
Since \rho is a congruence on \Gamma , we have \rho R and \rho \Lambda are equivalences on R and \Lambda , respectively.
Let (x, \lambda ), (y, \mu ), (x1, \lambda 1), (y1, \mu 1) \in \Gamma . If x\rho Ry and x1\rho Ry1, then
(x, x\circ x)\rho (y, y\circ y) and (x1, x
\circ
1x1)\rho (y1, y
\circ
1y1).
Now we immediately get
(x, x\circ x)(x1, x
\circ
1x1)\rho (y, y
\circ y)(y1, y
\circ
1y1).
And this implies that
(x(x\circ x \ast x1), (x\circ x \ast x1)\circ (x\circ x \ast x1)x\circ 1x1)\rho (y(y\circ y \ast y1), (y\circ y \ast y1)\circ (y\circ y \ast y1)y\circ 1y1).
Then, by Lemma 1.2,
(xx1, (xx1)
\circ xx1)\rho (yy1, (yy1)
\circ yy1).
So we have proved that xx1\rho Ryy1.
Suppose that \lambda \rho \Lambda \mu and \lambda 1\rho \Lambda \mu 1, then we obtain
(x\circ x, \lambda )\rho (y\circ y, \mu ) and (x\circ 1x1, \lambda 1)\rho (y
\circ
1y1, \mu 1).
Hence
(x\circ x, \lambda )(x\circ 1x1, \lambda 1)\rho (y
\circ y, \mu )(y\circ 1y1, \mu 1).
That is,
(x\circ x(\lambda \ast x\circ 1x1), (\lambda \ast x\circ 1x1)\circ (\lambda \ast x\circ 1x1)\lambda 1)\rho (y
\circ y(\mu \ast y\circ 1y1), (\mu \ast y\circ 1y1)\circ (\mu \ast y\circ 1y1)\mu 1).
By Lemma 1.4, we have
(\lambda \lambda \circ
1, \lambda \lambda 1)\rho (\mu \mu
\circ
1, \mu \mu 1).
Thus, by Lemma 1.1,
(\lambda \circ
1\lambda
\circ , \lambda \lambda 1)\rho (\mu
\circ
1\mu
\circ , \mu \mu 1).
So we have proved \lambda \lambda 1\rho \Lambda \mu \mu 1.
And we have the following cases:
(1) \rho R| E\circ = \rho \Lambda | E\circ is obvious. So condition (C1) holds.
(2) Let x, y \in R and x\rho Ry. Then
(x, x\circ x)\rho (y, y\circ y).
Hence
(z\circ z, \nu )(x, x\circ x)\rho (z\circ z, \nu )(y, y\circ y).
That is,
(z\circ z(\nu \ast x), (\nu \ast x)\circ (\nu \ast x)x\circ x)\rho (z\circ z(\nu \ast y), (\nu \ast y)\circ (\nu \ast y)y\circ y).
It follows that
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 6
CONGRUENCES ON REGULAR SEMIGROUPS WITH Q-INVERSE TRANSVERSALS 857
z\circ z(\nu \ast x)\rho Rz\circ z(\nu \ast y).
By condition (Q1) and z\circ z = \nu \circ , we get
(\nu \ast x)\rho R(\nu \ast y).
Now condition (C3) holds.
(3) Let \lambda , \mu \in \Lambda and \lambda \rho \Lambda \mu . Then
(x\circ x, \lambda )\rho (y\circ y, \mu ).
Hence
(x\circ x, \lambda )(z, \nu \circ )\rho (y\circ y, \mu )(z, \nu \circ ).
It follows that
(x\circ x(\lambda \ast z), (\lambda \ast z)\circ (\lambda \ast z)\nu \circ )\rho (y\circ y(\mu \ast z), (\mu \ast z)\circ (\mu \ast z)\nu \circ ).
Thus
x\circ x(\lambda \ast z)\rho Ry\circ y(\mu \ast z).
By condition (Q1) and Lemma 1.4, we have
(\lambda \ast z)\rho R(\mu \ast z).
Now condition (C2) holds.
Now from the above proof, (\rho R, \rho \Lambda ) is a congruence pair on \Gamma .
By the direct part, \rho (\rho R,\rho \Lambda ) is a congruence. If (x, \lambda )\rho (\rho R,\rho \Lambda )(y, \mu ), then we have
x\rho Ry, \lambda \rho \Lambda \mu .
Thus
(x, x\circ x)\rho (y, y\circ y), (x\circ x, \lambda )\rho (y\circ y, \mu ).
Hence
(x(x\circ x \ast x\circ x), (x\circ x \ast x\circ x)\circ (x\circ x \ast x\circ x)\lambda )\rho (y(y\circ y \ast y\circ y), (y\circ y \ast y\circ y)\circ (y\circ y \ast y\circ y)\mu ).
By Lemma 1.4,
(x, \lambda )\rho (y, \mu ).
Thus, \rho (\rho R,\rho \Lambda ) \subseteq \rho . Since \rho \subseteq \rho (\rho R,\rho \Lambda ) is obvious, \rho (\rho R,\rho \Lambda ) = \rho .
Theorem 2.1 is proved.
Example 2.1. Let R = \{ a, b, c\} and \Lambda = \{ e, f, g\} be semigroups whose multiplication tables
given respectively by
. a b c
a a b a
b b b b
c c b c
. e f g
e e f g
f f f f
g e f g
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 6
858 A. WANG, L. WANG
It is clear that R is a regular semigroup with a right ideal inverse transversal S\circ = \{ b, c\} . The
equivalence \rho R on R with class \{ a, c\} , \{ b\} is a nontrivial congruence. \Lambda is a band with a left ideal
inverse transversal \Lambda \circ = \{ f, g\} . The equivalence \rho \Lambda on \Lambda with class \{ e, g\} , \{ f\} is a nontrivial
congruence. Moreover, S\circ = E(S\circ ) \sim = \Lambda \circ .
By Lemma 1.6, we have \Gamma = \{ (a, e), (a, g), (b, f), (c, e), (c, g)\} is a regular semigroup with a
Q-inverse transversal which is isomorphic to S\circ . It is easy to see that (\rho R, \rho \Lambda ) satisfies the conditions
(C1), (C2) and (C3), thus (\rho R, \rho \Lambda ) is a congruence pair. Using Theorem 2.1, we have the equivalence
\rho on \Gamma with class \{ (a, e), (a, g), (c, e), (c, g)\} , \{ (b, f)\} is a nontrivial congruence.
We denote the set of all congruences on \Gamma and the set of all congruence pairs on \Gamma constructed
as in Theorem 2.1 by C(\Gamma ) and CT (\Gamma ).
Lemma 2.2. If (\rho R1 , \rho
\Lambda
1 ), (\rho
R
2 , \rho
\Lambda
2 ) \in CT (\Gamma ), then
\rho (\rho
R
1 ,\rho \Lambda 1 ) \subseteq \rho (\rho
R
2 ,\rho \Lambda 2 ) \leftrightarrow \rho R1 \subseteq \rho R2 , \rho
\Lambda
1 \subseteq \rho \Lambda 2 .
Proof. Suppose \rho (\rho
R
1 ,\rho \Lambda 1 ) \subseteq \rho (\rho
R
2 ,\rho \Lambda 2 ). Let x\rho R1 y. By the proof of Theorem 2.1,
((x, x\circ x), (y, y\circ y)) \in \rho (\rho
R
1 ,\rho \Lambda 1 ) \subseteq \rho (\rho
R
2 ,\rho \Lambda 2 ).
Hence x\rho R2 y, and immediately we get \rho R1 \subseteq \rho R2 . Similarly, we have \rho \Lambda 1 \subseteq \rho \Lambda 2 .
The reverse implication is obvious.
Lemma 2.2 is proved.
Define \leq on CT (\Gamma ) by
(\rho R1 , \rho
\Lambda
1 ) \leq (\rho R2 , \rho
\Lambda
2 ) \leftrightarrow \rho R1 \subseteq \rho R2 , \rho
\Lambda
1 \subseteq \rho \Lambda 2 .
Then CT (\Gamma ) is a partial ordered set with respect to \leq . By Theorem 2.1 and Lemma 2.2, we can
easily see that C(\Gamma ) and CT (\Gamma ) are isomorphic as partial ordered set.
Proposition 2.1. Let \Omega \subseteq C(\Gamma ) and T\rho = (\rho R, \rho \Lambda ) where \rho \in \Omega . Then
T(\cap \rho \in \Omega \rho ) =
\left( \bigcap
\rho \in \Omega
\rho R,
\bigcap
\rho \in \Omega
\rho \Lambda
\right)
and
T(
\bigvee
\rho \in \Omega \rho ) =
\left( \bigvee
\rho \in \Omega
\rho R,
\bigvee
\rho \in \Omega
\rho \Lambda
\right) .
Proof. The first equality is obvious, we only need to prove the second equality. Let x, y \in R be
such that x(
\bigvee
\rho \in \Omega \rho )Ry. Then
i = (x, x\circ x)
\bigvee
\rho \in \Omega
\rho (y, y\circ y) = j.
Hence, there exist \rho i \in \Omega and ai = (xi, xi
\circ xi) \in \Gamma such that
i\rho 1a1\rho 2a2 . . . an - 1\rho nj.
This implies that
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 6
CONGRUENCES ON REGULAR SEMIGROUPS WITH Q-INVERSE TRANSVERSALS 859
x\rho R1 x1\rho
R
2 x2 . . . xn - 1\rho
R
n y.
We have proved that \left( \bigvee
\rho \in \Omega
\rho
\right) R
\subseteq
\bigvee
\rho \in \Omega
\rho R.
\bigvee
\rho \in \Omega \rho R \subseteq (
\bigvee
\rho \in \Omega \rho )R is obvious. The dually equality can be proved similarly.
Proposition 2.1 is proved.
Now, by summing up the above results, we obtain the following theorem.
Theorem 2.2. Let \Gamma be constructed in Lemma 1.6. Then CT (\Gamma ) forms a complete lattice with
respect to \leq and C(\Gamma ) is isomorphic to CT (\Gamma ) as a complete lattice.
For a semigroup S, the equality relations on S are denoted by \epsilon S . The following theorem
describes the idempotent-separating congruences.
Theorem 2.3. Let \pi be an idempotent-separating congruence on R. Then \rho (\pi ,\epsilon \Lambda ) is an
idempotent-separating congruence on \Gamma , and every such congruence may be obtained in this way.
Proof. Since \pi is an idempotent-separating congruence, \pi | E\circ = \epsilon E\circ . It is easy to see that (\pi , \epsilon \Lambda )
is a congruence pair. For any (x, \lambda ), (y, \mu ) \in E(\Gamma ) with (x, \lambda )\rho (\pi ,\epsilon \Lambda )(y, \mu ), we have \lambda = \mu ,
xx\circ = yy\circ , x\circ x = y\circ y. So x\circ = x\circ x(\lambda \ast x)x\circ = y\circ y(\mu \ast y)y\circ = y\circ . Hence x = xx\circ x\circ \circ x\circ x =
= yy\circ y\circ \circ y\circ y = y. Therefore \rho (\pi ,\epsilon \Lambda ) is an idempotent-separating congruence. It is easy to show the
reverse implication.
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Received 22.04.13,
after revision — 06.03.16
ISSN 1027-3190. Укр. мат. журн., 2016, т. 68, № 6
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| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:14:32Z |
| publishDate | 2016 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/2a/e7b35731f6431e7088626b51578b142a.pdf |
| spelling | umjimathkievua-article-18842019-12-05T09:30:37Z Congruences on regular semigroups with $Q$-inverse transversals конгруенцiї на регулярних напiвгрупах з $Q$-оберненими трансверсалями Wang, A. Wang, L. Ван, А. Ван, Л. We give congruences on a regular semigroup with a $Q$-inverse transversal $S^o$ by the congruence pair (abstractly), which consists of congruences on the structural component parts $R$ and $\Lambda$. We prove that the set of all congruences for this kind of semigroups is a complete lattice. Наведено конгруенцiї на регулярних напiвгрупах з $Q$-оберненою трансверсаллю $S^o$ щодо пари конгруенцiй (абстрактно), сформованої з конгруенцiй на частинах структурних компонент $R$ i $\Lambda$. Доведено, що множина всiх конгруенцiй для такого типу напiвгруп є повною ґраткою. Institute of Mathematics, NAS of Ukraine 2016-06-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1884 Ukrains’kyi Matematychnyi Zhurnal; Vol. 68 No. 6 (2016); 853-859 Український математичний журнал; Том 68 № 6 (2016); 853-859 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1884/866 Copyright (c) 2016 Wang A.; Wang L. |
| spellingShingle | Wang, A. Wang, L. Ван, А. Ван, Л. Congruences on regular semigroups with $Q$-inverse transversals |
| title | Congruences on regular semigroups with $Q$-inverse transversals |
| title_alt | конгруенцiї на регулярних напiвгрупах з $Q$-оберненими трансверсалями |
| title_full | Congruences on regular semigroups with $Q$-inverse transversals |
| title_fullStr | Congruences on regular semigroups with $Q$-inverse transversals |
| title_full_unstemmed | Congruences on regular semigroups with $Q$-inverse transversals |
| title_short | Congruences on regular semigroups with $Q$-inverse transversals |
| title_sort | congruences on regular semigroups with $q$-inverse transversals |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1884 |
| work_keys_str_mv | AT wanga congruencesonregularsemigroupswithqinversetransversals AT wangl congruencesonregularsemigroupswithqinversetransversals AT vana congruencesonregularsemigroupswithqinversetransversals AT vanl congruencesonregularsemigroupswithqinversetransversals AT wanga kongruenciínaregulârnihnapivgrupahzqobernenimitransversalâmi AT wangl kongruenciínaregulârnihnapivgrupahzqobernenimitransversalâmi AT vana kongruenciínaregulârnihnapivgrupahzqobernenimitransversalâmi AT vanl kongruenciínaregulârnihnapivgrupahzqobernenimitransversalâmi |