On the Separator of Subsets of Semigroups

Semigroup Forum (2011) 83:289–303DOI 10.1007/s00233-011-9306-x

R E S E A R C H A RT I C L E

On the separator of subsets of semigroups

Attila Nagy

Received: 24 November 2010 / Accepted: 6 April 2011 / Published online: 1 May 2011© Springer Science+Business Media, LLC 2011

Abstract By the separator SepA of a subset A of a semigroup S we mean the set ofall elements x of S which satisfy conditions xA ⊆ A, Ax ⊆ A, x(S − A) ⊆ (S − A),(S−A)x ⊆ (S−A). In this paper we deal with the separator of subsets of semigroups.

In Sect. 2, we investigate the separator of subsets of special types of semigroups.We prove that, in the multiplicative semigroup S of all n × n matrices over a fieldF and in the semigroup S of all transformations of a set X with |X| = n < ∞, theseparator of an ideal I �= S is the unit group of S. We show that the separator of asubset of a group G is a subgroup of G. Moreover, the separator of a subset A ofa completely regular semigroup [a Clifford semigroup; a completely 0-simple semi-group] S is either empty or a completely regular semigroup [a Clifford semigroup, acompletely simple semigroup (supposing ∅ �= A �= S)] of S.

In Sect. 3 we characterize semigroups which satisfy certain conditions concerningthe separator of their subsets. We prove that every subset A of a semigroup S with ∅ ⊂A ⊂ S has an empty separator if and only if S is an ideal extension of a rectangularband by a nil semigroup. We also prove that every subsemigroup of a semigroup S isthe separator of some subset of S if and only if ∅ ⊂ SepA ⊆ A is satisfied for everysubsemigroup A of S if and only if S is a periodic group if and only if A = SepA forevery subsemigroup A of S.

In Sect. 4 we apply the results of Sect. 3 for permutative semigroups. We show thatevery permutative semigroup without idempotent elements has a non-trivial group ora group with a zero homomorphic image. Moreover, if a finitely generated permuta-tive semigroup S has no neither a non-trivial group homomorphic image nor a groupwith a zero homomorphic image then S is finite.

Communicated by Mohan S. Putcha.

Research supported by the Hungarian NFSR grant No K77476.

A. Nagy (�)Department of Algebra, Institute of Mathematics, Budapest University of Technology andEconomics, Pf. 91, 1521 Budapest, Hungarye-mail: [email protected]

mailto:[email protected]

290 A. Nagy

Keywords Semigroup · Separator of a subset of a semigroup · Permutativesemigroups

1 Introduction

By the idealizator Id A of a subset A of a semigroup S we mean the set of all elementsx of S which satisfy conditions xA ⊆ A, Ax ⊆ A (see [12]).

By the separator SepA of a subset A of a semigroup S we mean the meet of Id A

and Id(S −A). With other words, SepA is the set of all elements x of S which satisfyall of the conditions xA ⊆ A, Ax ⊆ A, x(S − A) ⊆ (S − A), (S − A)x ⊆ (S − A)

(see [3]).We note that, for every subset A of a semigroup S, SepA is either empty or a

subsemigroup of S. Moreover, SepA = Sep(S − A). Specially, Sep∅ = SepS = S. IfS is a monoid then the identity element belongs to SepA for every subset A of S. Thebasic properties of the separator of subsets of semigroups were published in [3]. Herewe present some of them which will be used in our investigations.

Lemma 1 (Theorem 3 of [3]) If A is a subset of a semigroup S such that SepA �= ∅then either SepA ⊆ A or SepA ⊆ S − A.

A subset A of a semigroup S is said to be separator including (excluding) ifSepA ⊆ A (SepA ⊆ (S − A)). In the case of SepA = ∅, A is considered separatorincluding as well as separator excluding (see [3]).

We note that if A is a right ideal or a left ideal of a semigroup S with A �= S thenA is separator excluding.

Lemma 2 (Theorem 7 of [3]) If SepA �= ∅ for a subset A of a semigroup S thenSepA is separator including, that is, Sep SepA ⊆ SepA.

We say that a subset A of a semigroup S is unitary in S if ab ∈ A implies eithera, b ∈ A or a, b /∈ A for every elements a, b ∈ S.

Lemma 3 (Theorem 8 of [3]) A subsemigroup A of a semigroup S is unitary in S ifand only if A = A.

In this paper we deal with further properties of the separator of subsets of semi-groups. For notations and notion not defined here, we refer to [1, 2] or [4].

2 Separator of subsets of special types of semigroups

First of all we deal with the separator of ideals of semigroups. Let I be an ideal ofa semigroup S. The congruence ρ on S defined by (a, b) ∈ ρ if and only if a = b ora, b ∈ I is called the Rees congruence of S modulo I . We shall write S/I instead ofS/ρ. We may describe S/I as the result of collapsing I into a single (zero) element,while the elements of S outside of I retain their identity.

On the separator of subsets of semigroups 291

Theorem 1 For an ideal I of a semigroup S, Sep I = SepS/I ({0}), where SepS/I ({0})denotes the separator of the zero element in the Rees factor semigroup S/I .

Proof It is obvious. �

An element a of a semigroup S with zero 0 is called a left divisor of zero if thereis an element c ∈ S such that c �= 0 and bc = 0. The notion of the right divisor of zerois the dual of the notion of the left side case. An element will be called a divisor ofzero if it is a left divisor or a right divisor of zero.

Theorem 2 In an arbitrary semigroup S with a zero 0, Sep({0}) is the set of allnon-divisors of zero.

Proof It is obvious. �

In the next we investigate the separator of ideals of the multiplicative semigroupMn(F) of all n×n matrices over a field F (n is a positive integer) and of the semigroupTX of all transformations of a finite set X.

First of all, introduce the following notations: let Ek [E∗k ] denote the n×n diagonal

matrix, such that the first [last] k elements of the diagonal are the identity element ofF and the others are the zero of F.

Let I be an ideal of Mn(F). It is clear that I contains a matrix A with rank(A) = k

if and only if I contains the matrix Ek if and only if I contains all of the matrix ofMn(F) whose rank is k. It is easy to see that Ei Ej = Emin{i,j}. From this and theabove mentioned facts we have the following.

Lemma 4 For every ideal I of the multiplicative semigroup Mn(F) of all n × n

matrices over a field F, there is an integer kI with 0 ≤ kI ≤ n such that I = {A ∈Mn(F) : rank(A) ≤ kI }.

An ideal I of a semigroup S will be called a proper ideal of S if I �= S.

Theorem 3 If I is a proper ideal of the multiplicative semigroup Mn(F) of all n × n

matrices over a field F then the separator of I is the unit group of Mn(F).

Proof Let I be a proper ideal of Mn(F). By Lemma 4, I = {A ∈ Mn(F) : rank(A) ≤k}. As rank(AB) = rank(BA) = rank(A) for a regular matrix B and an arbitrary ma-trix A of Mn(F), we have that the unit group of Mn(F) is contained by Sep I . Asthe ideals of semigroups are separator excluding, Sep I is the unit group of Mn(F) ifkI = n − 1.

Consider the case when kI < n − 1. Let A be an arbitrary matrix of Mn(F) suchthat kI < rank(A) ≤ n−1. Let i denote the rank of A. Then there are regular matricesB and C of Mn(F) such that

BAC = Ei .

It is a matter of checking to see that EiE∗kI +1 is either the zero matrix or a diagonal

matrix in which the (kI + 1)th, . . . , ith positions of the diagonal contain the identity

292 A. Nagy

element of F and the other elements are the zero of F. Thus rank(EiE∗kI +1) ≤ kI

(because i < n) and so EiE∗kI +1 ∈ I . Thus

BACE∗kI +1 ∈ I.

As B is regular, we have

ACE∗kI +1 ∈ I.

As rank(CE∗kI +1) = kI + 1, we have

CE∗kI +1 /∈ I.

Thus A /∈ Sep I . Consequently Sep I is the unit group of Mn(F). Thus the theorem isproved. �

Theorem 4 If I is a proper ideal of the semigroup TX of all transformations of a setX with |X| = n < ∞ then Sep I = Sn, where Sn denotes the group of all permutationsof X.

Proof Let I be a proper ideal of TX (|X| = n < ∞). By Theorem 2.9 and Exercise3(a) for §2.2 of [1, 2], there is an integer kI (0 ≤ kI ≤ n) such that I is the set of alltransformations of X whose rank is less then or equal to kI . It is clear that Sn ⊆ Sep I .In case kI = n − 1, Sep I = Sn.

Consider the case when kI < n − 1. Let α be a transformation of X whose rank i

satisfies the condition kI < i < n. Let Xα = {j1, . . . , ji}. As i < n, X − Xα �= ∅. Letx0 ∈ X − Xα be an arbitrary fixed element. Let α∗ be a transformation of X whichsatisfies the following conditions: Xαα∗ = {j1, . . . , jkI

} and xα∗ = x0 for every x /∈Xα. As the rank of α∗ is kI + 1, we have α∗ /∈ I . As the rank of αα∗ is kI , we haveαα∗ ∈ I . Thus α /∈ Sep I . Consequently, Sep I = Sn. �

In the next we deal with the separator of subsets of groups, completely regularsemigroups, Clifford semigroups and completely 0-simple semigroups.

Theorem 5 Let S be a semigroup and G be a subgroup of S. If A is a subset of S

such that SepA ∩ G �= ∅ then SepA ∩ G is a group.

Proof Let e denote the identity element of G. Let a ∈ A be an arbitrary element.Assume ea /∈ A. Then, for an arbitrary t ∈ SepA ∩ G, ta = (te)a = t (ea) /∈ A whichis a contradiction, because ta ∈ A. Thus eA ⊆ A. We can prove, in a similar way, thatAe ⊆ A and e(S −A) ⊆ S −A, (S −A)e ⊆ S −A. Thus e ∈ SepA. Let t ∈ SepA∩G

be an arbitrary element. Let t−1 denote the inverse of t in the group G. Then, for everya ∈ A, t t−1a = ea ∈ A and so t−1a ∈ A (because t ∈ SepA). Similarly, at−1 ∈ A.Thus t−1A ⊆ A and At−1 ⊆ A. We can prove, in a similar way, that t−1(S − A) ⊆S − A and (S − A)t−1 ⊆ S − A. Consequently t−1 ∈ SepA. Thus SepA ∩ G is agroup. �

Theorem 6 For every subset A of a group G, SepA is a subgroup of G.


Proof Let A be a subset of a group G. As the identity element of G is in SepA, wehave SepA �= ∅. Applying Theorem 5 for A and G (as a subgroup of G), we canconclude that SepA is a subgroup of G. �

An element a of a semigroup S is called completely regular, if there is an elementx in S such that axa = a and ax = xa. A semigroup S is called a completely regularsemigroup if every element of S is completely regular. It is known that a semigroupis completely regular if and only if it is a (disjoint) union of groups.

Theorem 7 For every subset A of a completely regular semigroup S, SepA is eitherempty or a completely regular subsemigroup of S.

Proof Let S be a completely regular semigroup. Then it is a union of disjoint sub-groups Gα , α ∈ Y . Let A be a subset of S such that SepA �= ∅. Then SepA is asubsemigroup of S. Let α be an arbitrary element of Y for which SepA ∩ Gα �= ∅.By Theorem 5, SepA∩Gα is a subgroup of Gα . Let Y ′ = {α ∈ Y : SepA∩Gα �= ∅}.Then SepA is a union of all subgroups SepA ∩ Gα where α ∈ Y ′. Thus SepA iscompletely regular. �

A semigroup is called a Clifford semigroup if S is regular (for every element a ∈ S

there is an element x ∈ S such that axa = a) and the idempotent elements of S are inthe centre of S. It is known that a semigroup is a Clifford semigroup if and only if itis a semilattice of groups.

Theorem 8 The separator of a subset of a Clifford semigroup S is either empty or aClifford subsemigroup of S.

Proof Let S be a Clifford semigroup. Then it is a semilattice Y of groups Gα (α ∈ Y ).Let A be a subset of S such that SepA �= ∅. Let Y ′ = {α ∈ Y : SepA∩Gα �= ∅}. Then,by Theorem 7, SepA is the union of all subgroups SepA ∩ Gα where α ∈ Y ′. It isclear that Y ′ is a subsemigroup of the semilattice of Y and so SepA is a semilatticeY ′ of subgroups SepA ∩ Gα , (α ∈ Y ′). Thus SepA is a Clifford semigroup. �

It is known (see, for example, Lemma 1 of [11]) that if A is a subsemigroup ofa finite completely 0-simple semigroup S = M0(G; I, J ;P) such that A does notcontain the zero of S then there are subsets I ′ ⊆ I and J ′ ⊆ J and a subgroup G′ ofG such that A = {(i, h, j); i ∈ I ′, j ∈ J ′, h ∈ G′} and pj,i ∈ G′ for any j ∈ J ′ andi ∈ I ′. In the proof of Lemma 1 of [11], the finiteness of the completely 0-simplesemigroup is used for only proving that A ∩ H is a group for every H-class H of S

for which A ∩ H �= ∅. It is clear that if A is a completely 0-simple semigroup S suchthat A does not contain the zero of S then every H-class H of S with A ∩ H �= ∅is a subgroup of S (see also Corollary 2.52b of [1, 2]). Thus we can formulate thefollowing result.

Lemma 5 If A is a subsemigroup of a completely 0-simple semigroup S = M0(G :I, J ;P) such that A does not contain the zero of S and, for every H-class H of S

294 A. Nagy

with A ∩ H �= ∅, A ∩ H is a subgroup of the group H then there are subset I ′ ⊆ I

and J ′ ⊆ J and a subgroup G′ of G such that A = {(i, h, j); i ∈ I ′, j ∈ J ′, h ∈ G′}and pj,i ∈ G′ for any j ∈ J ′ and i ∈ I ′.

Proof See the above remark and the proof of Lemma 1 of [11]. �

A subset A of a semigroup S will be called a proper subset, if A �= ∅ and A �= S.

Theorem 9 If A is a proper subset of a completely 0-simple semigroup S =M0(G; I, J ;P) then SepA = ∅ or there are subsets I ′ ⊆ I and J ′ ⊆ J and a sub-group G′ of G such that SepA = {(i, h, j); i ∈ I ′, j ∈ J ′, h ∈ G′} and pj,i ∈ G′ forany j ∈ J ′ and i ∈ I ′.

Proof Let S be a completely 0-simple semigroup. Let A be a proper subset of S

such that SepA �= ∅. It is easy to see that 0 /∈ SepA. From this it follows that ifSepA ∩ H �= ∅ for some H-class H of S then H is a (maximal) subgroup of S

(Corollary 2.52b of [1, 2]). By Theorem 5, SepA ∩ H is a group. Thus our assertionfollows from Lemma 5. �

Theorem 10 The separator of a subset of a completely simple semigroup S is eitherempty or a completely simple subsemigroup of S.

Proof By Theorem 9 it is obvious. �

3 On the separator of subsets of arbitrary semigroups

Let S be a semigroup. A single valued mapping of S into itself is called a transfor-mation of S. A transformation λ(·) of S is called a left translation of S if, for everyx, y ∈ S, λ(xy) = (λ(x))y. A transformation (·)ρ of S is called a right translationof S if, for every x, y ∈ S, (xy)ρ = x((y)ρ). From a left translation λ and a righttranslation ρ of S, we can define a single valued mapping of S into the direct productS × S. For every x ∈ S, let (λ,ρ)(x) = (λ(x), (x)ρ). The mappings (λ,ρ) of S intoS × S defined above are called bitranslations of S. A left translation λ and a righttranslation ρ of S are said to be linked if, for every x, y ∈ S, x(λ(y)) = ((x)ρ)y

(in this case we shall say that (λ,ρ) is a linked bitranslation of S). It is knownthat the set �(S) of all linked bitranslations of S is a monoid under the operation(λ1, ρ1)(λ2, ρ2) = (λ1 ◦ λ2, ρ1 ◦ ρ2), where λ1 ◦ λ2 and ρ1 ◦ ρ2 are defined as fol-lows: for every x ∈ S, (λ1 ◦ λ2)(x) = λ1(λ2(x)) and (x)(ρ1 ◦ ρ2) = ((x)ρ1)ρ2. Theidentity element of �(S) is (ι, ι), where ι is the identity transformation of S. Themonoid �(S) is called the translational hull of S.

It is easy to see that λa and ρa defined by λa(x) = ax, (x)ρa = xa are left andright translations of S, respectively, such that they are linked. The set

�0(S) = {(λa, ρa) : a ∈ S}


is a subsemigroup of �(S); it is called the inner part of �(S). It is easy to check that,for every (λ,ρ) ∈ �(S) and (λa, ρa) ∈ �0(S),

(λ,ρ)(λa, ρa) = (λλ(a), ρλ(a), (λa, ρa)(λ,ρ) = (λ(a)ρ, ρ(a)ρ).

Thus �0(S) is an ideal of �(S).It is known that �� : a → (λa, ρa) is a homomorphism of S onto the inner part

of �(S); this homomorphism is injective if and only if S is weakly reductive (that is,for every a, b ∈ S, the assumption “ax = bx and xa = xb are satisfied for all x ∈ S”implies a = b).

For a subset A of a semigroup S, let

�SepA(S) = {(λ,ρ) ∈ �(S) : λ(A), (A)ρ ⊆ A,λ(S − A), (S − A)ρ ⊆ S − A}.As �(S) is a monoid, �SepA(S) is not empty for every subset A of S. It is easy tosee that �SepA(S) is a submonoid of �(S). It is clear that �SepA(S) = �Sep(S−A)(S)

for every subset A of S. We note that �Sep∅(S) = �SepS(S) = �(S).It is easy to see that, for a subset A of S,

SepA = {x ∈ S : ��(x) = (λx, ρx) ∈ �SepA(S)}.In this section we concentrate our attention to semigroups S which satisfies one of

the following conditions.

Condition 1 �0(S) ∩ �SepA(S) = ∅ for every proper subset A of S.

Condition 2 ��(A) ∩ �SepA(S) �= ∅ for every subsemigroup A of S.

It is easy to see that Condition 1 is equivalent to the condition: SepA = ∅ for everyproper subset A of S. Using also Lemma 1, it is clear that Condition 2 is equivalentthe condition: ∅ ⊂ SepA ⊆ A is satisfied for every subsemigroup A of S.

A semigroup S is called a rectangular band if every element of S is idempotentand S satisfies the identity axa = a.

A semigroup S with a zero 0 is called a nil semigroup if, for every a ∈ S, there isa positive integer n such that an = 0.

We show that, in a semigroup S, every proper subset A has an empty separator ifand only if S is an ideal extension of a rectangular band by a nil semigroup. We alsoshow that ∅ ⊂ SepA ⊆ A is satisfied for every subsemigroup A of a semigroup S ifand only if S is a periodic group.

It is known that every semigroup is a semilattice of semilattice indecomposablesemigroups. A special subclass of semilattice indecomposable semigroups is the classof Archimedean semigroups. A semigroup S is called an Archimedean semigroupif, for every elements a and b of S, there are positive integers n and m such thatan ∈ S1bS1 and bm ∈ S1aS1. A semigroup is called a Putcha semigroup if, for everyelements a, b ∈ S, the assumption b ∈ S1aS1 implies bm ∈ S1a2S1 for some positiveinteger m. The Putcha semigroups play an important role in the theory of semigroups.

296 A. Nagy

It is known that a semigroup S is a semilattice of Archimedean semigroups if andonly if it is a Putcha semigroup (see Theorem 2.1 of [8] or Theorem 2.1 of [4]). In theinvestigation of some types of Archimedean semigroups S not containing idempotentelements, the subset

Sa = {b ∈ S : (∃i, j, k ∈ N+) aibaj = ak}(N+ is the set of all positive integers) defined for elements a ∈ S plays an importantrole (see, for example, Theorem 1.42 [4]). The subset Sa will play an important rolein our investigation, too. The next lemma contains a basic result about the subsets Sa

of an arbitrary semigroup S (a ∈ S).

Lemma 6 For every element a of an arbitrary semigroup S, a ∈ SepSa .

Proof Let a be an arbitrary element of a semigroup S. It is clear that a ∈ Sa . Letb ∈ Sa be arbitrary. Assume aibaj = ak . We can suppose that i, j > 1. Then

ak = a(i−1)(ab)aj = ai(ba)a(j−1)

and so ab, ba ∈ Sa . Thus

a ∈ Id Sa.

Let y ∈ (S −Sa) be an arbitrary element. If ay was in Sa then there would be positiveintegers k,m,n such that

an = ak(ay)am = a(k+1)yam

and so y would be in Sa which would be a contradiction. Thus ay /∈ Sa . Similarly,ya /∈ Sa . Hence

a ∈ Id(S − Sa).

Consequently,

a ∈ SepSa.�

3.1 Semigroups with Condition 1

Theorem 11 For an arbitrary semigroup S, the following conditions are equivalent.

(1) �0(S) ∩ �SepA(S) = ∅ for every proper subset A of S.(2) SepA = ∅ for every proper subset A of S.(3) Every proper subset of S is separator excluding.(4) For every couple (a, b) ∈ S × S, there are i, j, k ∈ N+ such that aibaj = ak .(5) S is an ideal extension of a rectangular band by a nil semigroup.

Proof It is clear that conditions (1) and (2) are equivalent (see above).(2) implies (3). Assume (2) for a semigroup S. Let A be a proper subset of S such

that A is not separator excluding. Then SepA �= ∅ which contradicts the assump-tion (2).


(3) implies (4): Assume (3) for a semigroup S. Let a be an arbitrary element of S.By Lemmas 6 and 1, ∅ ⊂ Sa is separator including. Then Sa = S. Thus (4) is satisfied.

(4) implies (5): Assume (4) for a semigroup S. Then, for every a ∈ S, there arepositive integers i, j, k such that

a2k = (a2)k = (a2)ia(a2)j = a2(i+j)+1.

The power on the left side is even, on the right side is odd. Thus the order of a is finiteand so S is a periodic semigroup. Thus S has at least one idempotent element. LetE denote set of all idempotent elements of S. Let e ∈ E be arbitrary. By condition(4), for every b ∈ S, ebe = e from which we get (eb)2 = eb and (be)2 = be. ThusES,SE ⊆ E which means that E is an ideal of S. As S is periodic, the Rees factorsemigroup S/E defined by the ideal E of S is a nil semigroup. By condition (4),ef e = e for every e, f ∈ E and so E is a rectangular band. Thus (5) is satisfied.

(5) implies (2): Assume (5) for a semigroup S. Then the set E of all idempotentelements of S is a rectangular band, E is an ideal of S and, for every element a of S,there is a positive integer n such that an ∈ E. Suppose (in an indirect way) that thereis a proper subset A of S such that SepA �= ∅. Then, by Lemma 1, either A or S −A isseparator including. Let B denote one of A and S − A which is separator including.Then B is a proper subset of S and ∅ �= SepB ⊆ B . As SepB is a subsemigroupof S, for every x ∈ SepB , there is a positive integer such that xn ∈ SepB ∩ E. ThusSepB ∩E �= ∅. As E is a rectangular band, for arbitrary f ∈ E, ef e = e ∈ SepB ⊆ B

which implies f ∈ B . Hence E ⊆ B . Let s ∈ S be an arbitrary element. Then es ∈E ⊆ B which implies s ∈ B , because e ∈ SepB . Thus B = S which is a contradiction.Thus S satisfies (2). �

Corollary 1 The separator of every proper subset of a rectangular band is empty.

Proof By Theorem 11, it is obvious. �

Corollary 1 shows that a rectangular band is an example for a completely regu-lar semigroup [a Clifford semigroup, a completely simple semigroup] in which theseparator of every proper subset is empty.

3.2 Semigroups with Condition 2

Theorem 12 For an arbitrary semigroup S, the following conditions are equivalent.

(1) ��(A) ∩ �SepA(S) �= ∅ for every subsemigroup A of S.(2) For every subsemigroup A of S, ∅ ⊂ SepA ⊆ A.(3) S is a periodic group.(4) Every subsemigroup of S is unitary.(5) A = SepA for every subsemigroup A of S.(6) For every subsemigroup A of S there is a subset K of S such that A = SepK .

Proof Conditions (1) and (2) are equivalent (see above).(2) implies (3): Assume (2) for S. Let R be a right ideal of a semigroup S. Then

there is a subset K of S such that R = SepK . Then RK ⊆ R and RK ⊆ K . Thus K ∩

298 A. Nagy

SepK �= ∅. By Lemma 1, R ⊆ K which implies K = S. Hence R = SepK = SepS =S. Consequently S is right simple. We can prove, in a similar way, that S is leftsimple. Thus S is a group. Let a ∈ S be arbitrary. Consider the cyclic subsemigroup〈a〉 generated by a. As

A = {a2, a3, . . .}is a subsemigroup of S and so ∅ ⊂ SepA ⊆ A, there is an integer n ≥ 2 such thatan ∈ SepA. Then

〈a〉 � an+1 /∈ A

and so an+1 = a. Thus a has a finite order. Consequently, S is a periodic group andso (3) is satisfied.

(3) implies (4). As every subsemigroup of a periodic group G is a subgroup andso is unitary in G, (3) implies (4).

(4) implies (5). See Lemma 3.(5) implies (6). It is obvious.(6) implies (2). Assume (6) for a semigroup S. Let A be a proper subsemigroup

of S. Then there is a subset K of S such that SepK = A. By Lemma 2, Sep SepK ⊆SepK , and so SepA = A. Thus (2) is satisfied. �

4 On the separator of subsets of permutative semigroups

In this section we concentrate our attention to the separator of subsets of permutativesemigroups. We show that a permutative semigroup S satisfies Condition 1 if and onlyif every subsemigroup of S has an empty separator. We show that every permutativesemigroup without idempotent elements has a non-trivial group or a group with azero homomorphic image. Moreover, if a finitely generated permutative semigroupS has no neither a non-trivial group homomorphic image nor a group with a zerohomomorphic image then S is finite.

A semigroup S is called a permutative semigroup ([7]) if it satisfies a non-trivialpermutation identity, that is, there is a non-identity permutation σ of {1,2, . . . , n}such that S satisfies the identity

x1x2 · · ·xn = xσ(1)xσ(2) · · ·xσ(n).

A subset A of a semigroup S is said to be reflexive in S if, for every a, b ∈ S,ab ∈ A if and only if ba ∈ A (see [1, 2]).

Lemma 7 If A is a subset of a permutative semigroup S such that SepA �= ∅ then A

is reflexive.

Proof Let S be a permutative semigroup. Then, by Theorem 1 of [9], there existsa positive integer k such that, for every u,v ∈ Sk and x, y ∈ S, we have uxyv =uyxv. Let A be an arbitrary subset of S with SepA �= ∅. Assume xy ∈ A for somex, y ∈ S. Then, for arbitrary u,v ∈ SepA, we get ukxyvk ∈ A, because SepA is a


subsemigroup of S (and so uk, vk ∈ SepA). As ukyxvk = ukxyvk , we get yx ∈ A.Hence A is reflexive in S. �

Lemma 8 If S is a permutative semigroup and A is a subset of S such that SepA �= ∅then Sep SepA = SepA.

Proof If SepA �= ∅ for a subset A of an arbitrary semigroup S then, by Lemma 2,

Sep SepA ⊆ SepA.

We prove that SepA ⊆ Sep SepA also holds if S is permutative. Let t ∈ SepA bearbitrary. As SepA is a subsemigroup of S, we have

t (SepA) ⊆ SepA

and

(SepA)t ⊆ SepA.

Thus

t ∈ Id(SepA).

We show that t ∈ Id(S − SepA). Indirectly, assume that t /∈ Id(S − SepA). Thenthere is an element c ∈ S − SepA such that tc ∈ SepA or ct ∈ SepA. We show thatboth inclusions are impossible. Consider the case when tc ∈ SepA. Then tca ∈ A forevery a ∈ A and so

ca ∈ A,

because t ∈ SepA. By Lemma 7,

ac ∈ A

also holds. Hence

c ∈ Id A.

We show that c ∈ Id(S − A) also holds. As tc ∈ SepA, for every b ∈ S − A, tcb ∈S − A and so

cb ∈ S − A,

because t ∈ SepA. By Lemma 7,

bc ∈ S − A.

Hence

c ∈ Id(S − A),

indeed. Then

c ∈ Id A ∩ Id(S − A) = SepA

300 A. Nagy

which contradict c ∈ S − SepA.We can prove, in a similar way, that the inclusion ct ∈ SepA is impossible. Thus

tc, ct /∈ SepA.

Hence

t ∈ Id(S − SepA).

Consequently

t ∈ Id(SepA) ∩ Id(S − SepA) = Sep SepA.

Thus

SepA ⊆ Sep SepA,

indeed. Hence,

Sep SepA = SepA. �

Corollary 2 If H is a subsemigroup of a permutative semigroup S such that H =SepA for some subset A of S then H is reflexive and unitary in S.

Proof By Lemma 7 and Lemma 8, it is obvious. �

The subset Sa of a semigroup S (a ∈ S) was defined in Sect. 2. The next lemma isabout Sa when S is a permutative semigroup.

Lemma 9 Sa is a reflexive unitary subsemigroup of a permutative semigroup S forevery a ∈ S.

Proof See Lemma 1 of [6]. �

The reflexive unitary subsemigroups of a semigroup S play an important role inthe description of the group or a group with a zero congruences of S. It is known (see,for example, Theorem 1.41 (and the remark before Theorem 1.41) of [4]) that if H isa reflexive unitary subsemigroup of a semigroup S then the factor semigroup S/PH

is either a group or a group with a zero adjoined, where

PH = {(a, b) ∈ S × S : (∀x, y ∈ S) xat ∈ H ⇔ xby ∈ H }is the principal congruence of S defined by H . Conversely, every group or group witha zero adjoined congruence of a semigroup can be so constructed.

In [5] it was shown how can we construct the monoid congruences of permutativesemigroups (see Theorem 2 of [5]). We proved the following.

Lemma 10 (Theorem 2 of [5]) Let {Ai, i ∈ I } be a family of subsets of a permutativesemigroup S such that A = ∩i∈I Sep(Ai) is not empty. Then

P{Ai, i∈I } = {(a, b) ∈ S × S : (∀x, y ∈ S)(∀i ∈ I ) xay ∈ Ai ⇔ xby ∈ Ai}


is a monoid congruence of S such that A is the identity element of S/P{Ai, i∈I }. Con-versely, every monoid congruence of a permutative semigroup can be so constructed.

For subsets Ai, i ∈ I considered in Lemma 10, SepAi �= ∅. Thus, by Corollary 2,SepAi is a reflexive unitary subsemigroup of S. Thus the principal congruence PA

of S is a group or a group with a zero congruence.In the next corollary, we characterize permutative semigroups which has no neither

a non-trivial group homomorphic image nor a group with a zero adjoined homomor-phic image.

Theorem 13 On a permutative semigroup S, the following conditions are equiva-lent.

(1) S has no neither a non-trivial group homomorphic image nor a group with a zeroadjoined homomorphic image.

(2) Every proper subset of S has an empty separator (that is S is an ideal extensionof a rectangular band by a nil semigroup (see Theorem 11)).

(3) Every proper subsemigroup of S has an empty separator.(4) S has no a non-trivial monoid homomorphic image.

Proof (1) implies (2). If A is a proper subset of a permutative semigroup S such thatSepA �= ∅ then, by Corollary 2, SepA is a reflexive unitary subsemigroup of S. AsSepA ⊆ A or SepA ⊆ (S − A) (see Lemma 1), we get S �= SepA. Thus the principalcongruence PSepA of S is either a non-trivial group or a group with a zero adjoinedcongruence. Thus (1) implies (2).

(2) implies (3). It is evident.(3) implies (4). Let S be a permutative semigroup satisfying (3). If S had a

non-trivial monoid homomorphic image then, by Lemma 10 and the remark afterLemma 10, S would have a subsemigroup A �= S such that A = SepA which wouldbe a contradiction. Thus (3) implies (4).

(4) implies (1). It is obvious. �

It is known that every permutative semigroup is a Putcha semigroup, that is, asemilattice of Archimedean semigroups (see [7]). It is easy to see that if S is anArchimedean semigroup then it has no homomorphic image which is a group with azero adjoined, because, for every reflexive unitary subsemigroup A of S,

WA = {a ∈ S : (∀x, y ∈ S1) xay /∈ A}is empty (see Theorem 1.41 of [4] and the fact that the right residue of A equalsWA, because A is reflexive and unitary). Thus, by Theorem 13, an Archimedeanpermutative semigroup S has no a non-trivial group homomorphic image if and onlyif S is an ideal extension of a rectangular band by a nil semigroup and so it has anidempotent element. Thus we get the following corollary.

Corollary 3 (Theorem 3 of [6]) An Archimedean permutative semigroup withoutidempotent element has a non-trivial group homomorphic image.

302 A. Nagy

The following theorem is a generalization of Corollary 3.

Theorem 14 Every permutative semigroup without idempotent elements has a non-trivial group or a group with a zero adjoined homomorphic image.

Proof Let S be a permutative semigroup which has no idempotent elements. Let a

be an arbitrary element of S such that a ∈ Sa2 . Then there are positive integers i, j, k

such that a2iaa2j = a2k , that is, a2(i+j)+1 = a2k . The power on the left is odd, on theright is even. Thus a has a finite order and so S has an idempotent element. This isa contradiction. Thus a /∈ Sa2 and so Sa2 �= S for every a ∈ S. By Lemma 9, Sa2 is aproper reflexive unitary subsemigroup of S. Thus the factor semigroup S/Sa2 of S isa non-trivial group or a group with a zero adjoined. The theorem is proved. �

Theorem 15 If S is a finitely generated permutative semigroup then S is either finiteor S has a non-trivial group or a group with a zero adjoined homomorphic image.Thus every infinite finitely generated permutable semigroup has a non-trivial groupor a group with a zero adjoined homomorphic image.

Proof Let S be a finitely generated permutative semigroup. Assume that Sa = S forevery element a of S. Then a ∈ Sa2 and so a is periodic (a has a finite order) for everya ∈ S. Then S is a permutative finitely generated periodic semigroup. By the solutionof the Burnside problem for semigroups (see [10] or Theorem 1.1 of [4]), S is finite.

Consider the case when Sa �= S for some a ∈ S. Then, by Lemma 9, Sa is a properreflexive unitary subsemigroup of S and so the factor semigroup S/Sa of S is eithera non-trivial group or a group with a zero adjoined. �

Corollary 4 If a finitely generated permutative semigroup S has no neither a trivialgroup homomorphic image nor a group with a zero adjoined homomorphic image (orequivalently, SepA = ∅ for every proper subset A of S) then S is finite.

Proof By Theorem 15, it is obvious. �

Corollary 5 If a finitely generated permutative semigroup S has no a proper unitarysubsemigroup then S is finite.

Proof Using also Corollary 2, a permutative semigroup S has no neither a non-trivialgroup homomorphic image nor a group with a zero adjoined homomorphic image ifand only if S has no proper unitary subsemigroup. Thus the assertion follows fromCorollary 4. �

Acknowledgement I would like to express my hearty thanks to the referee for his useful suggestions.

References

1. Clifford, A.H., Preston, G.B.: The Algebraic Theory of Semigroups, vol. I. AMS, Providence (1961)2. Clifford, A.H., Preston, G.B.: The Algebraic Theory of Semigroups, vol. II. AMS, Providence (1967)


3. Nagy, A.: The separator of a subset of a semigroup. Publ. Math. (Debr.) 27(1–2), 25–30 (1980)4. Nagy, A.: Special Classes of Semigroups. Kluwer Academic, Dordrecht (2001)5. Nagy, A.: On commutative monoid congruences of semigroups. Pure Math. Appl. 13(3), 389–392

(2002)6. Nagy, A., Jones, P.R.: Permutative semigroups whose congruences form a chain. Semigroup Forum

69, 446–456 (2004)7. Nordahl, T.E.: On permutative semigroup algebras. Algebra Univers. 25, 322–333 (1988)8. Putcha, M.S.: Semilattice decomposition of semigroups. Semigroup Forum 6, 12–34 (1973)9. Putcha, M.S., Yaqub, A.: Semigroups satisfying permutation identities. Semigroup Forum 3, 68–73

(1971)10. Restivo, A., Reutenauer, C.: On the Burnside problem for semigroups. J. Algebra 89, 102–104 (1984)11. Reutenauer, C.: Semisimplicity of the algebra associated to a biprefix code. Semigroup Forum 23,

327–342 (1981)12. Shevrin, L.N.: Completely simple semigroups without zero and idealizators of subsemigroups. Izv.

Vysš. Ucebn. Zaved., Mat. 55, 157–160 (1966) (Russian)

On the Separator of Subsets of Semigroups

Documents

Transcript of On the Separator of Subsets of Semigroups