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.
293
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;
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 everyt∈RegT and everye∈E(S), and
(3) for every s∈S and t∈T there exists m∈Nsuch thatsm∈RegS
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;
(3) Defineα∗ : RegS →End(RegT) by s→α∗(s); then α∗ is 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,s1∈S 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.
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∗(sδ
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∗(sδS) 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 everye∈E(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)
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 everyx∈X, and where the homomorphismα:S−→ End(TX) is defined by (fs)(x) =f(sx) for everys∈S,f ∈TX andx∈X.
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 m∈N such that(t′)m ∈RegT for all
t′ ∈T′.
Proof. Suppose thatTX is stronglyπ-inverse andT′ ∈R. 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 allt′ ∈T′.
Now, for each t ∈ T, let T′ = {t}; then there exists m ∈ N such that tm∈RegT. 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))m∈RegT for allx∈X, so thatgm∈RegTX.
ThusTX isπ-regular.
For eachg, h∈E(TX) we haveg(x),h(x)∈E(T) for all x∈X. 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) = (
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
tm∈RegT. Sotm= 1, and thent∈RegT. 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 x∈X, 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 existsm∈N such that
(t′)m∈RegT′ for allt′∈T′,
(3)|T|= 1or ex=xfor everye∈E(S)and all x∈X, and
(4)for each x∈S andg∈TX there exists m∈Nsuch that sm∈RegS
andgs(m)(x)∈RegT for all x∈X, wheregs(m)=gsm−1
· · ·gsg∈TX.
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 existsm∈Nsuch that
(t′)m∈RegT for allt′ ∈T′,
(3)S is a group or|T|= 1, and
(4) for everys∈S andg∈TS there existsm∈Nsuch 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.
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.
(Received 17.05.1994)
Authors’ address:
