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SEMIDIRECT PRODUCTS AND WREATH PRODUCTS OF

STRONGLY π-INVERSE MONOIDS

YUFEN ZHANG, SHIZHENG LI, AND DESHENG WANG

Abstract. In this paper we determine the necessary and sufficient conditions for the semidirect products and the wreath products of two monoids to be stronglyπ-inverse. Furthermore, we determine

the least group congruence on a stronglyπ-inverse monoid, and we

give some important isomorphism theorems.

1. Introduction

Our terminology and notation will follow [1] and [2].

Let S and T be two monoids, and let End(T) be the endomorphism monoid ofT, and write endomorphisms as exponents to the right of argu-ments. If α: S → End(T) is a homomorphism, and if s ∈ S and t ∈ T, write ts for tα(s), since α(s) End(T) for s S, then for t

1, t2 ∈ T, (t1, t2)s=ts1ts2. Sinceαis a homomorphism, (ts1)s2=ts1s2 for everyt∈T ands1, s2∈S.

The semidirect productS×αTis the monoid with elements{(s, t) :s∈S,

t∈T} and multiplication (s1, t1)(s2, t2) = (s1s2, ts12t2).

In [3], [4] the authors have determined the necessary and sufficient con-ditions forS×αT to be regular, inverse, and orthodox. In this paper we determine the necessary and sufficient conditions forS×αT to be strongly

π-inverse and give their applications to the wreath product.

For a monoidS,E(S) and RegS denote the set of idempotents ofS and the set of regular elements ofS, respectively.

A semigroup isπ-regular if for everys∈S there is anm∈Nsuch that

sm RegS. If S is π-regular and E(S) is a commutative subsemigroup, then we callS a strongly π-inverse semigroup. It is easy to see that RegS

is an inverse subsemigroup of a stronglyπ-inverse semigroupS.

1991Mathematics Subject Classification. 20M18.

Key words and phrases. Wreath product,π-inverse monoid, regular monoid, group

congruence.

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2. Semidirect Products

LetS and T be two monoids and letS×αT be the semidirect product ofS andT, whereα:S →End(T) is a given homomorphism.

Lemma 1. Let S×αT be a stronglyπ-inverse monoid; then (1)both S andT are stronglyπ-inverse monoids;

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Theorem 2. Let S andT be two monoids and let α: S → End(T) be the given homomorphism, and letS×αT be the semidirect product of Sand

T. ThenS×αT is a stronglyπ-inverse monoid iff (1)both S andT are stronglyπ-inverse monoids,

(2)te=t for everytRegT and everyeE(S), and

(3) for every s∈S and t∈T there exists m∈Nsuch thatsmRegS

andts(m)RegT, wherets(m)=tsm−1 tsm−2

· · ·tst.

Proof. The necessity of the assertion is obvious by Lemma 1. We only prove the sufficient part.

ThereforeS×αT is stronglyπ-inverse.

Theorem 3. Let S andT be two monoids and letS×αT be a strongly

Proof. It is a immediate consequence of Lemma 1 and Theorem 2.

Theorem 4. Let S andT be two monoids and letS×αT be a strongly

π-inverse monoid.

(1) (s, t)∈Reg(S×αT)iffs∈RegS andt∈RegT.

(2) For every s ∈ RegS, letα∗(s) be the restriction of α(s) on RegT;

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(3) Defineα∗ : RegS End(RegT) by sα(s); then αis a

homo-morphism fromRegS toEnd(RegT).

(4) Reg(S×αT)∼= Reg(S)×α∗Reg(T).

Proof. (1) Let (s, t)∈Reg(S×αT); then there exists (s1, t1)∈S×αT such that (s, t)(s1, t1)(s, t) = (s, t), and then ss1s = s, ts1sts1t = t. From the latter equation we have (tts

1)s1stts1=tts1, and from Lemma 1 (3), (tts1)s1s=

tts

1, that is,tts1t=t. So thats∈RegS andt∈RegT.

Conversely, let s ∈ RegS and t ∈ RegT; then there exist s1 ∈ S and

t1∈T such thatss1s=s,tt1t=t. Hence

(s, t)(s1, ts11)(s, t) = (ss1s, ts11st s1s

1 t) = (s, tt1t)−(s, t). Therefore (s, t)∈Reg(S×αT).

(2) For every s ∈ RegS and t ∈ RegT, ts RegT; then α(s)

End(RegT).

(3) and (4) are obvious.

3. Least Group Congruence on a Strongly π-Inverse Monoid

Theorem 5. Let S be a stronglyπ-inverse monoid; then the relation

δ=

(s1, s2)∈S×S:s1e=s2e for some e∈E(S)

is the least group congruence onS.

Proof. It is obvious thatδis a left compatible equivalent relation onS. Let

xe=ye, forx, y ∈S ande ∈E(S). For anyz ∈S, sinceS is a strongly

π-inverse monoid, there exist m ∈ N and s S such that zmszm = zm,

szms=s; and we have

xz(zm−1sez) =xezmsez=yezmsez=yz(zm−1sez)

and

(zm−1sez)2=zm−1sezmsez=zm−1szmsez=zm−1szmsez=zm−1sez.

Thus (xz, yz)∈δ.

It is obvious thateδ=f δ= 1 is the identity ofS/δfor everye, f ∈E(S). Now, forsδ∈S/δ there existm∈N,s1S such thatsms

1,s1sm∈E(S). Thus (sms

1)δ = sδ(sm−1s1)δ = 1 and (s1sm)δ = (s1sm−1)δ(sδ) = 1, so thatsδ has an inverse element. This means thatS/δ is a group.

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Theorem 6. Let S andT be two monoids, let S×αT be a strongly π

-inverse monoid, and let δS×αT, δS and δT be the least group congruences

onS×αT,S andT, respectively. Then

(1) for every sδS ∈S/δS define α∗(sδS) :T /δT →T /δT by tδT →tsδT;

thenα∗(

S)∈End(T /δT); (2) define α∗ : S/δ

S → End(T /δT) by sδS → α∗(sδS); then α∗ is a

homomorphism;

(3)S/δS×α∗T /δT = (∼ ×αT)/δS×αT.

Proof. (1) If t1δT = t1δT for t1, t2 ∈ T, then there exists u ∈ E(T) such that t1u=t2uand thents1us=ts2us fors∈S. Since us∈E(T), we have

ts

1δT =ts2δT. Thusα∗(sδS) is well defined for everysδS ∈S/δS. It is easy to see thatα∗(S) is a homomorphism.

(2) If s1δS = s2δS for s1, s2 ∈ S, then s1e = s2e for e ∈ E(S). For arbitrary tδT ∈ T /δT, there exists t1δT ∈ T /δT such that (t1t)δT = 1 ∈

T /δT, and then t1tu = u for some u ∈ E(T). Thus, tut1tu = tu, hence

tu∈RegT. From Lemma 1,teu=tu for everyeE(S) and then teδ T =

tδT, so α∗(eδS) is an identity mapping on T /δT. Thus ts1δT = ts1eδT =

ts2eδ

T =ts2δT, and thenα∗(s1δS) =α∗(s2δS), so thatα∗ is well defined. For arbitrary s1δS, s2δS ∈ S/δS and arbitrary tδT ∈ T /δT, we have

t(s1s2)δ

T = (ts1)s2δT, that is,α∗(s1s2) = α∗(s1δS)α∗(s2δS). Thus α∗ is a homomorphism fromS/δS to End(T /δT).

(3) Define ϕ : (S ×αT)/δS×αT −→ S/δS ×T /δT by (s, t)δS×αT −→ (sδS, tδT). Suppose (s1, t1)δS×αT = (s1, t2)δS×αT. Then there exist (e, u)∈

E(S×αT) such that (s1, t1)(e, u) = (s2, t2)(e, u); thens1e=s2e,te1u=te2u. Sos1δS =s2δS, t1δT =t2δT, and then (s1δS, t1δT) = (s2δS, t2δT). Thusϕ is well defined. ϕis obviously surjective.

If (s1δS, t1δT) = (s2δS, t1δT), thens1δS =s2δS, t1δT =t2δT, and then there existe ∈E(S) and u∈E(T) such that s1e=s2e, t1u=t2u. From Lemma 1 (2), te

1u=te1ue=te2ue=t2eu. Hence (s1, t1)(e, u) = (s2, t2)(e, u), and then (s1, t1)δS×αT = (s2, t2)δS×αT. Thusϕis one-to-one.

It is easy to see that ϕis a homomorphism. Thus (S×αT)/δS×α∗T ∼=

δS×α∗T /δT.

Corollary 7. Let S be a strongly π-inverse monoid. Then for every

s∈S there existe, f∈E(S)such that se, f s∈RegS.

Proof. From the proof of Theorem 6 we know that se ∈ RegS for some

e ∈ E(S). Using a similar way, we can prove that the binary relation defined onS by

σ=

(s1, s2) :f s1=f s2 for some f ∈E(S)

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4. Wreath Products

LetSbe a monoid,Sacts on a setXfrom the left, that is,sx∈X, 1x=x

and s(rx) = (sr)xfor every s, r∈S andx∈X. LetT also be a monoid, then the wreath product SwxT = S×αTX, where TX ={f : X −→ T is a function} is the Cartesian power of T, that is, f g(x) = f(x)g(x) for everyf, g∈TX and everyxX, and where the homomorphismα:S−→ End(TX) is defined by (fs)(x) =f(sx) for everysS,f TX andxX.

Lemma 8. Let T be a monoid and letR={T′T

|T′| ≤ |X|}. Then

TX is a stronglyπ-inverse monoid iff (1)T is a stronglyπ-inverse monoid, and

(2) for every T′ R there exists mN such that(t)m RegT for all

t′ T.

Proof. Suppose thatTX is stronglyπ-inverse andTR. Then there exists g ∈TX such that g(X) = T. Let m Nsuch that gm RegTX; then (t′)m= (g(x))m=gm(x)RegT for alltT.

Now, for each t ∈ T, let T′ = {t}; then there exists m N such that tmRegT. ThusT isπ-regular.

For everyu, v∈E(T), define g:X −→T byg(x) =uand h:X −→T

byh(x) =vfor allx∈X. Then,g, h∈E(TX) and

uv=g(x)h(x) =gh(x) =h(x)g(x) =vu.

ThusT is a stronglyπ-inverse monoid.

Conversely, for anyg∈TX, we haveg(x)R, and then there existsm

Nsuch thatgm(x) = (g(x))mRegT for allxX, so thatgmRegTX.

ThusTX isπ-regular.

For eachg, h∈E(TX) we haveg(x),h(x)E(T) for all xX. Then

gh(x) =g(x)h(x) =h(x)g(x) = (hg)(x)

for allx∈X, and thengh=hg. ThusTX is stronglyπ-inverse.

Lemma 9. Let S and T be two monoids; S acts on a set X from the left. Then the following conditions are equivalent:

(1)for each e∈E(S)andg∈RegTX

,ge=g

;

(2)|T|= 1or ex=xfor eache∈E(S)andx∈X.

Proof. Suppose that (1) holds. If there existe∈E(S) andx∈X such that

ex6=x, then fort′ RegT define g:X−→T by

g(y) = (

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We haveg∈RegTX. Hence ge=g, and thent=g(x) =ge(x) =g(ex) = 1. Thus RegT = {1}. But for each t ∈ T there exists m ∈ N such that

tmRegT. Sotm= 1, and thentRegT. Hencet= 1, therefore|T|= 1. Conversely, assume (2) holds. Lete∈E(S) andg∈RegTX. If|T|= 1, then (1) holds. If |T| 6= 1, then ex= xfor e ∈ E(S) and all x∈ X. So

ge(x) =g(x) for all xX, which means thatge=g.

Theorem 10. LetS andT be two monoids; S acts on a setX from the left. Then the wreath product SwxT is a stronglyπ-inverse monoid iff

(1)S andT are stronglyπ-inverse monoids,

(2) for each subsetT′ of T with |T| ≤ |X| there existsmN such that

(t′)mRegTfor alltT,

(3)|T|= 1or ex=xfor everye∈E(S)and all x∈X, and

(4)for each x∈S andg∈TX there exists mNsuch that smRegS

andgs(m)(x)RegT for all xX, wheregs(m)=gsm−1

· · ·gsgTX.

Proof. It is easy to see that gs(m) RegTX iff gs(m)(x) RegT for all

x∈X. Thus Theorem 10 is an immediate consequence of Lemma 8, Lemma 9, and Theorem 2.

Recall that the standard wreath productSwT of two monoids is formed by the left regular representation ofS on itself, and we have

Theorem 11. The standard wreath productSwT of two monoidsS and

T is a stronglyπ-inverse monoid iff

(1)both S andT are stronglyπ-inverse monoids,

(2) for each subsetT′ of T with |T| ≤ |S| there existsmNsuch that

(t′)mRegT for alltT,

(3)S is a group or|T|= 1, and

(4) for everys∈S andg∈TS there existsmNsuch that gs(m)(x) RegT for allx∈S.

Acknowledgement

The authors would like to thank Professor T. Saito for his careful reading of the paper and his constructive suggestions.

References

1. S. Bogdanowi´c, Semigroups with a system of subsemigroups. Univer-sity of Novy Sad, Institute of Mathematics, Novy Sad, 1985.

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3. W. R. Nico, On the regularity of semidirect products. J. Algebra

80(1983), 29–36.

4. T. Saito, Orthodox semidirect products and wreath products of mo-noids. Semigroup Forum38(1989), 347–354.