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Epimorphisms, Closed, and Supersaturated SemigroupsNoor Alam a & Noor Mohammad Khan aa Department of Mathematics , Aligarh Muslim University , Aligarh , IndiaPublished online: 13 Mar 2014.

To cite this article: Noor Alam & Noor Mohammad Khan (2014) Epimorphisms, Closed, and Supersaturated Semigroups,Communications in Algebra, 42:7, 3137-3146, DOI: 10.1080/00927872.2013.781609

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Communications in Algebra®, 42: 3137–3146, 2014Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927872.2013.781609

EPIMORPHISMS, CLOSED, AND SUPERSATURATEDSEMIGROUPS

Noor Alam and Noor Mohammad KhanDepartment of Mathematics, Aligarh Muslim University, Aligarh, India

We show that any almost unitary subsemigroup of a semigroup is closed in thecontaining semigroup and that the class of all left [right] regular semigroups isclosed. Finally, we show that every globally idempotent ideal satisfying a seminormalpermutation identity of a supersaturated semigroup is supersaturated.

Key Words: Almost unitary; Closed and absolutely closed semigroups; Dominion; Epimorphism;Saturated and supersaturated semigroups; Semigroup; Zigzag equations.

2010 AMS Mathematics Subject Classification: 20M10.

1. INTRODUCTION AND PRELIMINARIES

Let U and S be any semigroups with U a subsemigroup of S. FollowingIsbell [8], we say that U dominates an element d of S if for every semigroup T andfor all homomorphisms �� � � S → T , u� = u� for all u ∈ U implies d� = d�. The setof all elements of S dominated by U is called the dominion of U in S, and we denoteit by Dom�U� S�. It is easily seen that Dom�U� S� is a subsemigroup of S containingU . Let � be a class of semigroups. A semigroup U is said to be �-closed if for allS ∈ � such that U is a subsemigroup of S, Dom�U� S� = U . Let � and � be classesof semigroups such that � ⊆ �. Then � is said to be �-closed if every member of� is �-closed. A class � of semigroups is said to be closed if for U� S ∈ � with U asubsemigroup of S, Dom�U� S� = U . A semigroup U is said to be absolutely closedif it is closed in the class of all semigroups. If Dom�U� S� �= S for every properlycontaining semigroup S, then U is said to be saturated (see also [7] for all suchnotions). Following Higgins [4], a semigroup U is said to be supersaturated if everymorphic image of U is saturated. Further, a semigroup U is said to be epimorphicallyembedded in S if Dom�U� S� = S.

A morphism � � S → T is said to be an epimorphism (epi for short) if for allmorphisms �� �; �� = �� implies � = � (where �� � are semigroup morphisms). Itcan be easily checked that � � S → T is epi if and only if i � S� → T is epi and theinclusion map i � U → S is epi if and only if Dom�U� S� = S . Onto morphisms arealways epimorphisms, but the converse is not true in general in the category of all

Received March 3, 2012; Revised February 20, 2013. Communicated by V. Gould.Address correspondence to Dr. Noor Mohammad Khan, Department of Mathematics, Aligarh

Muslim University, Aligarh 202002, India; Email: nm_khan123@yahoo.co.in

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3138 ALAM AND KHAN

semigroups. Infact every epimorphism from a semigroup U is onto is just to say thatU is supersaturated.

Definition 1.1. A band B (a semigroup in which every element is an idempotent)is said to be left[right] regular if it satisfies the identity ax = axa�xa = axa.

A most useful characterization of semigroup dominions is provided by Isbell’sZigzag Theorem.

Result 1.2 ([8, Theorem 2.3] or [6, Theorem VII.2.13]). Let U be a subsemigroupof a semigroup S, and let d ∈ S. Then d ∈ Dom�U� S� if and only if d ∈ U or thereexists a series of factorizations of d as follows:

d = a0y1 = x1a1y1 = x1a2y2 = x2a3y2 = · · · = xma2m−1ym = xma2m� (1)

where m ≥ 1, ai ∈ U , (i = 0� 1� � 2m), xi� yi ∈ S (i = 1� 2� � m), and

a0 = x1a1� a2m−1ym = a2m�

a2i−1yi = a2iyi+1� xia2i = xi+1a2i+1 �1 ≤ i ≤ m− 1�

Such a series of factorization is called a zigzag in S over U with value d, length m,and spine a0� a1� � a2m We refer to the equations in Result 1.2, in whateverfollows, as the zigzag equations.

For any unexplained notations and conventions, one may refer to Clifford andPreston [1, 2] and Howie [6]. An identity of the form

x1x2 · · · xn = xi1xi2 · · · xin �n ≥ 2� (2)

is called a permutation identity, where i is any permutation of the set �1� 2� 3� � n�,and ik, for each k �1 ≤ k ≤ n), is the image of k under the permutation i. Apermutation identity of the form (2) is said to be nontrivial if i is any nontrivialpermutation of the set �1� 2� 3� � n�. The identities

x1x2 = x2x1 �commutativity�

x1x2x3 = x1x3x2 �left normality�

x1x2x3 = x2x1x3 �right normailty�

x1x2x3x4 = x1x3x2x4 �normailty�

are some of the well-known permutation identities.A semigroup U satisfying a nontrivial permutation identity is said to be a

permutative semigroup. Further a nontrivial permutation identity with n ≥ 4 will becalled seminormal if i1 = 1 and in = n.

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EPIMORPHISMS, CLOSED, AND SUPERSATURATED SEMIGROUPS 3139

Definition 1.3 ([6, Exercises II.12]). A subsemigroup U of a semigroup S is calledright unitary [left unitary] in S if

�∀u ∈ U��∀s ∈ S�su ∈ U ⇒ s ∈ U�us ∈ U ⇒ s ∈ U

A subsemigroup U of a semigroup S is said to be unitary in S if U is both left andright unitary in S.

Definition 1.4 ([6, VII.3]). A subsemigroup U of a semigroup S is said to be almostunitary if there exist mappings � S −→ S, � � S −→ S such that

(AU1): 2 = , �2 = �;

(AU2): �st� = � s�t, �st�� = s�t�� ∀s� t ∈ S;

(AU3): �s�� = � s�� for every s ∈ S;

(AU4): s� t� = �s��t ∀s� t ∈ S;

(AU5): U = � U = IU �

(AU6): U is unitary in S�.

A semigroup S is said to be globally idempotent if for all s ∈ S, there exist x� y ∈ Ssuch that s = xy, or equivalently, S2 = S.

Result 1.5 ([9, Theorem 6.4]). Let � be the class of all globally idempotentsemigroups, and let

x1x2 · · · xn = xi1xi2 · · · xin (3)

be any nontrivial permutation identity. Then (3) is equivalent with respect to � tothe following situations:

(i) Commutativity if i1 �= 1 and in �= n;(ii) Left normality if i1 = 1 and in �= n;(iii) Right normality if i1 �= 1 and in = n;(iv) Normality if i1 = 1 and in = n.

In [5], Howie showed that the amalgam �S� T� U is embeddable if U is almostunitary in S and V = �U�� is almost unitary in T which implied that almostunitary subsemigroup of a semigroup is closed in the containing semigroup. We nowprovide, by using zigzag manipulations, new and direct proof of this result and showthat the special semigroup amalgam � = ��S� S′�� U� �i� � U� is embeddable if U isalmost unitary in S.

2. MAIN RESULTS

Theorem 2.1. Let U be a subsemigroup of a semigroup S. If U is almost-unitary inS, then U is closed in S.

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3140 ALAM AND KHAN

Proof. To show that U is closed in S, take any d ∈ Dom�U� S�\U . As d is inDom�U� S�\U , we may let, by Result 1.2, (1) be a zigzag of length m over U withvalue d. Since a0 ∈ U , we have

a0 = �a0� (from (AU5))

= �x1a1� (from the zigzag equations)

= � �x1��a1 (from (AU2))

= � �x1�� �a1� (from (AU5))

= � �x1����a1 (from (AU4))

= � x1��a1 (from (AU3))

Therefore, � x1��a1 ∈ U . Hence, by (AU6), x1� ∈ U .Now

� x1��a2 = �x1�� �a2�� (from (AU4))

= � �x1��a2 (from (AU5))

= �x1a2� (from (AU2))

= �x2a3� (from the zigzag equations)

= � �x2��a3 (from (AU2))

= � �x2�� �a3� (from (AU5))

= � x2��a3 (from (AU4) and (AU3))

⇒ x2� ∈ U (by (AU6)).Continuing this way, we have

xm� ∈ U (4)

Now, we have

d = a0y1

= �a0�y1 (by (AU5))

= �x1a1�y1 (by the zigzag equations)

= � �x1�a1�y1 (by (AU2))

= � �x1� �a1��y1 (by (AU5))

= �� �x1����a1y1 (by (AU4))

= � x1��a1y1 (by (AU3))

= � x1��a2y2 (by the zigzag equations)

= �� x1��a2�y2

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EPIMORPHISMS, CLOSED, AND SUPERSATURATED SEMIGROUPS 3141

= �� x2��a3�y2 (by (AU4), (AU5) and by the zigzag equations)

= � x2��a3y2

= � xm��a2m−1ym

= � xm��a2m

Since by Eq. (4), xm� ∈ U , we have � xm��a2m ∈ U .⇒ d ∈ UHence, Dom�U� S� = U , as required. �

In [11], Scheiblich has shown, by using zigzag manipulations, that the class ofall normal bands is closed. We extend this result for the class of all left[right] regularbands and show that the class of all left[right] regular bands is closed.

Theorem 2.2. The class of all left regular bands is closed.

Proof. Let B be a left regular band and A be a subband of B, and let d be inDom�A�B� \ A. Then, by Result 1.2, we may let (1) be a zigzag in B over A withvalue d of length m and spine a0� a1� a2� � a2m. Now

d = a0y1

= x1a1y1 (by the zigzag equations)

= x1a1a1y1

= x1a1a2y2 (by the zigzag equations)

= x1a1x1a2y2 (as B is a left regular band)

= x1a1x2a3y2 (by the zigzag equations)

= x1a1x2a3a3y2

= x1a1x1a2a3y2 (by the zigzag equations)

= x1a1a2a3y2 (as B is a left regular band)

= a0a2�a3y2� (by the zigzag equations)

=(

1∏i=0

a2i

)�a3y2�

=(

m−2∏i=0

a2i

)�a2m−3ym−1�

= a0a2a4 · · · a2m−4a2m−3ym−1

= x1a1a2a4 · · · a2m−4a2m−2ym (by the zigzag equations)

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3142 ALAM AND KHAN

= x1a1x1a2a4 · · · a2m−4a2m−2ym (as B is a left regular band)

= x1a1x2a3a4 · · · a2m−4a2m−2ym (by the zigzag equations)

= x1a1x2a3x2a4 · · · xm−2a2m−4a2m−2ym (as B is a left regular band)

= x1a1x2a3x2a4 · · · xm−1a2m−3a2m−2ym (by the zigzag equations)

= x1a1x2a3x2a4 · · · xm−1a2m−3xm−1a2m−2ym (as B is a left regular band)

= x1a1x2a3x2a4 · · · xma2m−2xma2m−1ym (by the zigzag equations)

= x1a1x2a3x2a4 · · · xm−1a2m−3xma2m−1a2m−1ym

= x1a1x2a3x2a4 · · · xm−1a2m−3xm−1a2m−2a2m−1ym (by the zizag equations)

= x1a1x2a3x2a4 · · · xm−1a2m−3a2m−2�a2m−1ym� (as B is a leftregular band)

= x1a1x2a3x2a4 · · · xm−2a2m−4a2m−2�a2m� (by the zigzag equations)

= x1a1x2a3a4 · · · a2m−4a2m−2a2m (by the zigzag equations)

= x1a1x1a2a4 · · · a2m−4a2m−2a2m (by the zigzag equations)

= x1a1a2a4 · · · a2m−4a2m−2a2m (as B is a left regular band)

= a0a2a4 · · · a2m−4a2m−2a2m (by the zigzag equations)

=m∏i=0

a2i ∈ A

⇒ d ∈ A. Therefore, Dom�A�B� = A. Hence the theorem is proved. �

Dually, we may prove the following theorem.

Theorem 2.3. The class of all right regular bands is closed.

A semigroup S is said to be right reductive with respect to X if xa = xb forall x in X ⇒ a = b (a� b ∈ S), where X is a subset of S. Dually, we say that asemigroup S is left reductive with respect to X if ax = bx ∀x ∈ X ⇒ a = b (a� b ∈ S).The following result is from Higgins [3].

Result 2.4 ([3, Theorem 8]). A semigroup U is saturated [supersaturated] if theideal Un is saturated [supersaturated] (for some natural number n). In particular,a finite semigroup is saturated [supersaturated] if the ideal generated by theidempotents is saturated [supersaturated].

Whether or not the converse of the above result is true is still an open question.Since �X , the semigroup of all transformations on the set X, is absolutely closed,subsemigroups of absolutely closed or saturated semigroups need not be absolutelyclosed or saturated, in general. But it is not known whether or not an ideal of

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EPIMORPHISMS, CLOSED, AND SUPERSATURATED SEMIGROUPS 3143

a saturated [absolutely closed] semigroup is saturated [absolutely closed]. In [4],Higgins, however, has shown that the converse of the above result holds in somespecial cases and has proved that if S is a supersaturated commutative semigroup,then the same is true for any globally idempotent ideal. He has, in fact, shown thefollowing:

Result 2.5 ([4, Theorem 14]). Let S be a saturated semigroup and suppose that Uis a commutative ideal of S such that Un is globally idempotent for some naturalnumber n. Then U is supersaturated.

Khan and Shah [10] generalized this result by taking U as a permutativeglobally idempotent ideal satisfying a permutation identity x1x2 · · · xn = xi1xi2 · · · xinfor which i1 = 1 and in �= n and thus relaxed the commutativity of U .

Now we further extend this result by taking U as a permutative globallyidempotent ideal satisfying a seminormal permutation identity and, thus, relax theright semicommutativity of U . In fact, we prove the following theorem.

Theorem 2.6. Let S be a supersaturated semigroup, and let U be any ideal of Ssatisfying a seminormal permutation identity. If Un is globally idempotent for somenatural number n, then U is supersaturated.

First, we prove a lemma which is very crucial for the proof of the abovetheorem.

Lemma 2.7. Suppose that a globally idempotent semigroup U is not supersaturated.Then there exists a non-surjective epimorphism � � U → V such that V is right and leftreductive with respect to U�.

Proof. As U is not supersaturated, let � � U → V be a nonsurjective epimorphism.Define a relation � on V as s�t if asu = atu∀a� u ∈ U�. Clearly, � is an equivalencerelation. Now take any d ∈ V \ U� and s�t. Then d = xa for some x ∈ V \ U� anda ∈ U�. Now dsu = �xa�su = x�asu� = x�atu� = �xa�tu = dtu. Hence, for any w ∈V , a� u in U�, wsu = wtu ⇒ awsu = awtu ⇒ ws�wt. So, � is a left congruence.Similarly, we take d = uy. Therefore, asd = as�uy� = �asu�y = �atu�y = at�uy� =atd. Now for any w ∈ V� a� u ∈ U�, asw = atw ⇒ aswu = atwu ⇒ sw�tw.Therefore, � is a right congruence. Hence � is a congruence.

We denote the images of U� and V under �� by U and V , respectively, andshall similarly abbreviate u�, v� by u, v (u ∈ U , v ∈ V ).

From the commutative diagram

one may easily see that the inclusion map i � U → V is epi.

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3144 ALAM AND KHAN

Now, we show that V is right and left reductive with respect to U .

V Is Right Reductive. Let v1 �= v2. Then v1� �= v2� ⇒ �v1� v2� � � ⇒ xv1y �= xv2y

for some x� y ∈ U�. Since U is globally idempotent, so is U�. Therefore, x =x1x2 for some x1� x2 ∈ U�. Then we have x2v1y �= x2v2y. For if x2v1y = x2v2y,then xv1y = xv2y. This is a contradiction as xv1y �= xv2y ⇒ x1�x2v1�y �= x1�x2v2�y ⇒x2v1 �= x2v2 ⇒ x2v1 �= x2v2. Therefore, V is right reductive with respect to U .

V Is Left Reductive. Let v1 �= v2. Then v1� �= v2� ⇒ �v1� v2� � � ⇒ xv1y �= xv2y

for some x� y ∈ U�. Since U is globally idempotent, so is U�. Therefore, y =y1y2 for some y1� y2 ∈ U�. Then, we have xv1y1 �= xv2y1. For if xv1y1 = xv2y1,then xv1y = xv2y. This is a contradiction as xv1y �= xv2y ⇒ x�v1y1�y2 �= x�v2y1�y2 ⇒v1y1 �= v2y1 ⇒ v1y1 �= v2y1. Hence, V is left reductive with respect to U .

Finally, we prove that the inclusion map i: U → V is not surjective. Let d ∈V \ U� and s�t. Then d = uy for some u ∈ U�, y ∈ V \ U�. Let a� v ∈ U�. Thenuav �= uyv (as uav ∈ U�). For if uav = uyv, then ua = uy (as V is right reductivewith respect to U ). As ua ∈ U� and d = uy ∈ V \ U�, we get a contradiction. Henceuav �= uyv implies that �a� y� � �, which implies that a �= y ∀a ∈ U�. Therefore,i�a� = a �= y for some y ∈ V such that y � U . So i is not surjective.

Therefore, � = ��� � U → V is a nonsurjective epimorphism from U in to V ,which is right reductive with respect to the image U� of U . Hence the lemma isproved. �

Proof of the main Theorem. If we prove that Un is supersaturated, then thetheorem follows by Result 2.4. So, we assume that U is globally idempotent idealsatisfying a seminormal permutation identity. Suppose to the contrary that U werenot supersaturated. Then, by Lemma 2.7, there exists a nonsurjective epimorphism� � U −→ V such that V is right and left reductive with respect to U� (which weshall denote by U up to isomorphism).

Let � = � �−1 ∪ 1s. Then, clearly � is an equivalence relation on S. Next, weshow that � is a congruence on S. For this we are required to show that, if u� v ∈ U

and w ∈ S \ U , then u� = v� implies that �uw�� = �vw�� and �wu�� = �wv��.To prove the first equality, suppose that u� v ∈ U and w ∈ S \ U and �uw�� �=

�vw��. Since V is right reductive with respect to U , there exist x ∈ U suchthat x��uw�� �= x��vw��. Then �x�uw��� �= �x�vw���. Since V is left reductivewith respect to U , there exists y ∈ U such that �x�uw���y� �= �x�vw���y�. Then��x�uw��y�� �= ��x�vw��y��. Since U is globally idempotent, x = x1x2 for somex1� x2 ∈ U and y = y1y2 for some y1� y2 ∈ U . Thus, we have

�x1x2�uw�y1y2�� �= �x1x2�vw�y1y2��

⇒ �x1x2u�wy1�y2�� �= �x1x2v�wy1�y2��

Now, by Result 1.5, we have

�x1x2�wy1�uy2�� �= �x1x2�wy1�vy2��

⇒ ��x1x2w�y1uy2�� �= ��x1x2w�y1vy2��

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EPIMORPHISMS, CLOSED, AND SUPERSATURATED SEMIGROUPS 3145

Again, by Result 1.5, we get

��x1x2w�uy1y2�� �= ��x1x2w�vy1y2��

⇒ �x1x2�wu�y1y2�� �= �x1x2�wv�y1y2��

⇒ �x�wu�y�� �= �x�wv�y��

⇒ �x�wu���y� �= �x�wv���y�

⇒ �x�wu��� �= �x�wv���

⇒ ��xw�u��� �= ��xw�v���

⇒ �xw��u� �= �xw��v� ⇒ u� �= v�

Therefore, the statement

u� = v� ⇒ �uw�� = �vw�� ⇒ �

is a right congruence.Next we show that � is a left congruence. Suppose that u� v ∈ U and w ∈ S \

U and �uw�� �= �vw��. Since V is right reductive with respect to U , there existsx ∈ U such that x��wu�� �= x��wv��. Now �x�wu��� �= �x�wv��� ⇒ ��xw�u�� �=��xw�v�� ⇒ �xw��u� �= �xw��v�. Hence u� �= v�.

Again we conclude that the statement u� = v� implies that �wu�� = �wv��.Therefore, � is a left congruence and, hence, a congruence. Denote S/� by S. ThenU�� = U (up to isomorphism).

Now we form the amalgam A of S and V with core U . We extend the partialoperation on A to an associative multiplication. For this, take any a ∈ S\U (= S\U ),v ∈ V\U , and factorize v as v = u1y1 = y′1u

′1 (where u1� u

′1 ∈ U ; y1� y

′1 ∈ V\U ). Now

define av = �au1�y1 and va = y′1�u′1a�. We first show that this is a well-defined binary

operation. For this, suppose that v = u2y2 = y′2u′2 (where u2� u

′2 ∈ U and y2� y

′2 ∈

V\U ). Then for any x ∈ U , as u1y1 = u2y2, we have

�xa�u1y1 = �xa�u2y2

⇒ ��xa�u1�y1 = ��xa�u2�y2 �by associativity of V�

⇒ �x�au1��y1 = �x�au2��y2 �by associativity of S�

⇒ x��au1�y1� = x��au2�y2� �by associativity of V�

As V is right reductive with respect to U , we have that �au1�y1=�au2�y2, andtherefore, the operation is well defined.

Now, we verify the associativity of the above operation. For any a′ ∈ S \ U , wehave a′�av� = a′�au1y1� = �a′au1�y1 = �a′a�u1y1 = �a′a�v (by associativity of V andS, respectively).

Similarly, for any v′ ∈ V , it can be shown that �av�v′ = a�vv′�. The only casethat requires some attention is to show that �av�a′ = a�va′�, where a′ ∈ S� v ∈ V .For this, factorize v as v = a1ya2 (where a1� a2 ∈ V \ U ). Now

�av�a′ = �a�a1ya2��a′ �as v = a1ya2�

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3146 ALAM AND KHAN

= ��aa1�ya2�a′ �as aa1 ∈ U ⊆ S�

= �aa1��ya2�a′ �by associativity of V�

= �aa1�y�a2a′� �as a2a

′ ∈ U ⊆ S�

= �aa1��ya2a′� �by associativity of V�

= a�a1ya2a′� �by associativity of V�

= a�va′��

as required.We now have S �= A = Dom�S�A�. This contradicts the fact that S is

supersaturated. Hence the theorem is proved. �

REFERENCES

[1] Clifford, A. H., Preston, G. B. (1961). The algebraic theory of semigroups. Math.Surveys. No. 7, Vol. I. Providence: Amer. Math. Soc.

[2] Clifford, A. H., Preston, G. B. (1967). The algebraic theory of semigroups. Math.Surveys. No. 7, Vol. II. Providence: Amer. Math. Soc.

[3] Higgins, P. M. (1984). Saturated and epimorphically closed varieties of semigroups.J. Aust. Math. Soc. 36:153–175.

[4] Higgins, P. M. (1985). Epimorphisms, dominions and semigroups. Algebra Univers.21:225–233.

[5] Howie, J. M. (1962). Embedding theorems with amalgamation for semigroups. Proc.London Math. Soc. 12(3):511–534.

[6] Howie, J. M. (1976). An Introduction to Semigroup Theory. London: Academic Press.[7] Howie, J. M., Isbell, J. R. (1967). Epimorphisms and dominions II. Journal of Algebra

6:7–21.[8] Isbell, J. R. (1966). Epimorphisms and dominions. Proc. of the Conference on

Categorical Algebra, LaJolla. Berlin: Lange and Springer, pp. 232–246.[9] Khan, N. M. (1985). On saturated permutative varieties and consequences of

permutation identities. J. Aust. Math. Soc. A 38:186–197.[10] Khan, N. M., Shah, A. H. (2010). Epimorphisms, dominions and permutative

semigroups. Semigroup Forum 80:181–190.[11] Scheiblich, H. E. (1976). On epics and dominions of bands. Semigroup Forum 13:

103–114.

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