Applied Mathematics, 2015, 6, 1047-1068 Published Online June 2015 in SciRes. http://www.scirp.org/journal/am http://dx.doi.org/10.4236/am.2015.66096
Zappa-Szép Products of Semigroups
Suha Wazzan Mathematics Department, King Abdulaziz University, Jeddah, Saudi Arabia Email: [email protected]
Received 12 May 2015; accepted 6 June 2015; published 9 June 2015
Copyright © 2015 by author and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/
Abstract The internal Zappa-Szép products emerge when a semigroup has the property that every element has a unique decomposition as a product of elements from two given subsemigroups. The external version constructed from actions of two semigroups on one another satisfying axiom derived by G. Zappa. We illustrate the correspondence between the two versions internal and the external of Zappa-Szép products of semigroups. We consider the structure of the internal Zappa-Szép product as an enlargement. We show how rectangular band can be described as the Zappa-Szép product of a left-zero semigroup and a right-zero semigroup. We find necessary and sufficient conditions for the Zappa-Szép product of regular semigroups to again be regular, and necessary conditions for the Zappa-Szép product of inverse semigroups to again be inverse. We generalize the Billhardt λ-semidirect product to the Zappa-Szép product of a semilattice E and a group G by constructing an inductive groupoid.
Keywords Inverse Semigroups, Groups, Semilattice, Rectangular Band, Semidiret, Regular, Enlargement, Inductive Groupoid
1. Introduction The Zappa-Szép product of semigroups has two versions internal and external. In the internal one we suppose that S is a semigroup with two subsemigroups A and B such that each sS∈ can be written uniquely as s= ab with aA∈ and bB∈ . Then since ba∈ S, we have ba= a′′ b with aA′∈ and bB′∈ determined uniquely by a and b. We write a′ = ba ⋅ and bb′ = a . Associativity in S implies that the functions (ab, ) b⋅ a and (ab, ) ba satisfy axioms first formulated by Zappa [1]. In the external version we assume that we have semigroups A and B and assume that we have maps AB×→ A, defined by (ab, ) b⋅ a and a map AB×→ B defined by (ab, ) ba which satisfy Zappa axioms [1]. For groups, the two versions are equal, but as we show in this paper for semigroups this is true for only some
How to cite this paper: Wazzan, S. (2015) Zappa-Szép Products of Semigroups. Applied Mathematics, 6, 1047-1068. http://dx.doi.org/10.4236/am.2015.66096 S. Wazzan
special kinds of semigroups. Zappa-Szép products of semigroups provide a rich class of examples of semigroups that include the self- similar group actions [2]. Recently, [3] uses Li’s construction of semigroup C*-algebras to associate a C*-algebra to Zappa-Szép products and gives an explicit presentation of the algebra. They define a quotient C*-algebra that generalises the Cuntz-Pimsner algebras for self-similar actions. They specifically discuss the Baumslag-Solitar groups, the binary adding machine, the semigroup NN × , and the ax+ b -semigroup ZZ × . In [4] they study semigroups possessing E-regular elements, where an element a of a semigroup S is E-regular if a has an inverse a such that aa, a a lie in E⊆ ES( ) . They also obtain results concerning the extension of (one-sided) congruences, which they apply to (one-sided) congruences on maximal subgroups of regular semigroups. They show that a reasonably wide class of DE -simple monoids can be decomposed as Zappa-Szép products. In [5] we look at Zappa-Szép products derived from group actions on classes of semigroups. A semidirect product of semigroups is an example of a Zappa-Szép product in which one of the actions is taken to be trivial, and semidirect products of semilattices and groups play an important role in the structure theory of inverse semigroups. Therefore Zappa-Szép products of semilattices and groups should be of particular interest. We show that they are always orthodox and ℑ-unipotent, but are inverse if and only if the semilattic acts trivially on the group, that is when we have the semidirect product. In [5] we relate the construction (via automata theory) to the λ -semidirect product of inverse semigroups devised by Billhardt. In this paper we give general definitions of the Zappa-Szép product and include results about the Zappa-Szép product of groups and a special Zappa-Szép product for a nilpotent group. We illustrate the correspondence between the internal and external versions of the Zappa-Szép product. In addition, we give several examples of both kinds. We consider the structure of the internal Zappa-Szép product as an enlargement. We show how a rectangular band can be described as the Zappa-Szép product of a left-zero semigroup and a right-zero semigroup. We characterize Green’s relations ( and ) of the Zappa-Szép product MG of a monoid M and a group G. We prove some results about regular and inverse Zappa-Szép product of semigroups. We construct from the Zappa-Szép product of a semilattice E and a group G, an inverse semigroup by constructing an inductive groupoid. We rely on basic notions from semigroup theory. Our references for this are [6] and [7].
2. Internal Zappa-Szép Products Let S be a semigroup with subsemigroups A and B such that each element sS∈ is uniquely expressible in the form s= ab with aA∈ and bB∈ . We say that S is the “internal” Zappa-Szép product of A and B, and write SA= B. Since ba∈ S with bB∈ and aA∈ , we must have unique elements aA′∈ and bB′∈ so that ba= a′′ b . This defines two functions (ab, ) b⋅ a and (ab,.) ba Since ba∈ S with bB∈ and aA∈ , we must have unique elements aA′∈ and bB′∈ so that ba= a′ b. Write a′ = ba ⋅ and bb′ = a . This defines two function AB×→ A, (ab, ) b⋅ a and AB×→ B, (ab,.) ba Thus ba=( b ⋅ a) ba . Using these definitions, we have for all aa, ′∈ A and bb, ′∈ B that
(ab)( a′′ b) = a( b ⋅ a ′) ba′ b ′.
Thus the product in S can be described in terms of the two functions. Using the associativity of the semigroup S and the uniqueness property, we deduce the following axioms for the two functions. By the associativity of S, we have b( aa′′) = ( ba) a . Now aa′ b( aa′′) = b ⋅( aa) b and
a′ (baa) ′′=⋅=⋅⋅( baba) a( ba)( b aa a ′)( b) . Thus, by uniqueness property, we have the following two properties
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(ZS2) b⋅=⋅⋅( aa′′) ( b a)( ba a ). a′ (ZS3) (bba) = aa′. Similarly by the associativity of S, we have (bb′′) a= b( b a). Now a (bb′) a=( bb ′′ ⋅ a)( bb ) and bba( ′) = b(( b ′′ ⋅ ab) a) =⋅⋅ b( b ′ ab) ba′⋅ b ′ a. Thus, by uniqueness property, we have the following two properties (ZS1) bbab′′⋅=⋅( ba ⋅). a (ZS4) (bb′′) = bba′⋅ b a. In the following we illustrate which subsemigroups may be involved in the internal Zappa-Szép product. Lemma 1. If the semigroup S is the internal Zappa-Szép product of A and B then A B= RI( A) LI( B). Proof. Consider xAB∈ . Then since (ax) b= a( xb) we have ax= a for all a∈= A, xb b for all bB∈ . Thus x is a right identity for A and left identity for B, whereupon A B⊆ RI( A) LI( B) . Observe that RI( A) LI( B) ⊆ A B, thus A B= RI( A) LI( B).
Of course, if S is a monoid and A and B are submonoids then RI( A) LI( B) = {1.S } Proposition 1. If SA= B the internal Zappa-Szép product of A and B, then RI( A) LI( B) ≠∅. Proof. We use Brin’s ideas in [8] Lemma 3.4. If aA∈ then aa= ′β ( a) for unique aA′∈ and β (aB)∈ , giving us a function β :,AB→ and likewise, if bB∈ then b= α ( bb) ′ for some unique bB′∈ and some function α :.BA→ But for all bB∈ we have ab= a′β ( a) b, and therefore aa= ′ and β (ab) = b: that is β (a) is a left identity for B. Similarly, bb= ′ and α (b) is a right identity for A. In particular, α (b) and β (a) are idempotents. Now
a2 aa12= aa 12β( aa 12) = aa 12 ββ( a 2) = a 1( a 1) a 2 = a 1( ββ( a 1) ⋅ a 2) ( a 1) . Therefore
aa12= a 1(β ( a 1) ⋅ a 2) (1) a2 β(aaaa12) = ββ( 2) = ( 1) Similarly
===α αα⋅=α(b2 ) α bb12( bb 12) bb 12 b 1( b 2) b 2( b 1( b 2))( b 1 b 2) ( b 1) bb 12. Therefore
αα(bb12) =( b 1) = b 1 ⋅ α( b 2) (2) α(b2 ) bb12= b 1 b 2.
Set aa1 = and for any b∈= Ba, 2 α ( b) in (1): β(a) = β( ab α( )) = βα( ( b)).
Hence β is constant: β (a) = fB ∈ for all aA∈ . Similarly, setting ba1 = β ( ) and bb2 = in (2): α(baba) = αβ( ( ) ) = αβ( ( )) Hence α is constant: α (b) = eA ∈ for all bB∈ . But now we have that for all aA∈ and b∈== B, ae a af and fb= b = eb. But then putting ae= and bf= we have eeefff22= = = = and in particular ef= .
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Lemma 2. Let SA= B, the internal Zappa-Szép product of A and B and e∈ ES( ) be a right identity for A and a left identity for B. Then ea = e =( be ⋅ ) and ea=⋅= e a,. be be a Proof. We have e∈ A Ba,,, ∈∈ Ab B then ea=( eae ⋅ ) , but ae= a, thus ea= e( ae) = ( ea) e and by uniqueness we have ea= e ⋅ a, ee= a . Similarly, since eb= b we have be=( b ⋅ e) be . Also be=( eb) e = e( be). Thus ebebeb=⋅=,.e Hence ea = e =( be ⋅ ) and ea=⋅= e a,. be be In an internal Zappa-Szép product SA= B, we find an idempotent e∈ RI( A) LI( B). This shows (for example) that a free semigroup cannot be a Zappa-Szép product. But in a monoid Zappa-Szép product
MAB= of submonoids A and B the special idempotent e∈ RI( A) LI( B) must be 1M , since we have 1M = ab uniquely. Then for all x∈=== Ax,1 xe xM xab( ) =( xab) and thus x= xa and be= . Similarly 2 ae= and 1.M =ab = e = e In the following we give a definition of the enlargement of a semigroup introduced in [8] for regular semigroups, and in [9] this concept is generalized to non-regular semigroups by describing a condition (enlarge- ment) under which a semigroup T is covered by a Rees matrix semigroup over a subsemigroup. We describe the enlargement concept for internal Zappa-Szép products. Definition 1. A semigroup T is an enlargement of a subsemigroup S if STS= S and TST= T . Example 1. [9] Let S be any semigroup and let I be a set of idempotents in S such that S= SIS . Then S is an enlargement of ISI because S( ISI) S=( SIS) IS = SIS = S, and (ISI) S( ISI) = I( SIS)( ISI) = IS( ISI) = I( SIS) I = ISI. If Ie= { } and S= SeS, then S is an enlargement of the local submonoid eSe . Proposition 2. Let S be the internal Zappa-Szép product of subsemigroups A and B. Then S is an enlargement of a local submonoid eSe for some e∈ RI( A) LI( B), and eSe is the internal Zappa-Szép product of the sub- monoids A and B where A==∈==∈{ xxeaaAB:,} ,{ yybebB :,.} Proof. We have e∈ ES( ) such that e is a right identity for A and a left identity for B. Then S== AB AeeB = AeB ⊆⊆ SeS S so S= SeS for e∈ ES( ). So S is an enlargement of the local submonoid eSe ( eSe is a monoid with identity e). It is clear that A and B are submonoids of eSe . We must show that each element z∈ eSe is uniquely expressible as z= xy with x∈∈ Ay,. B If z∈ eSe then z= ese,. s ∈ S But s= ab for unique a∈∈ Ab,, B and so z= eabe( ) =( ea)( be) = xy, where xeaAybeB=∈=∈,. Since AA⊆ and BB⊆ this expression is unique, because zS∈= A B. There- fore each element z∈ eSe is uniquely expressible as z= xy with x∈∈ Ay,. B We note that if TA= B such that T is an enlargement of S= eTe = A B , where AA⊆ and BB⊆ , if AB, are regular with the assumption that if ea= ea′ then aa= ′ and if be= b′ e then bb= ′. Then A, B
are regular, since if A is regular monoid, then for each xA1 ∈ there exists xA2 ∈ such that
xxx121= x 1,. xxx 212 = x 2 Now xxx121= x 1 which implies ea1211211 ea ea= ea a a = ea . Since A has a right identity
e, then aaa121= a 1. Similarly we get aaa212= a 2. Thus A is regular. Similarly, we get B is regular. Following [9] we describe the Rees Matrix cover for the Zappa-Szép product TA= B such that T = TeT and is an enlargement of S= eTe for some idempotent eT∈ where SS2 = such that Ae = A and eB = B and S is the Zappa-Szép product of A and B, SA= B with A={ x:, x = ea a ∈⊆ A} T and B={ y:, y = be b ∈⊆ B} T . Then by Corollary 4 in [9] the Rees matrix semigroup is given by M= MSUVP( ;,;) = MeTeABP( ;,;) such that T= xT = Ty since T= TST. For each zAB∈ , we xA∈∈ yB
can find rz ∈ TS and rz′∈ ST such that z= rrzz′. So if xA∈ , then rx = e ∈ TS and rx′ = a ∈ ST such that x= ea = eeae ∈ TSST. Similarly, for yB∈ , yberr= = bb′, therefore rbb = and reb′ = . Now for each ′ zAB∈ , fix elements rrzz,,′∈ T and define BA× matrix P by putting pyx=∈ r y r x ( ST)( TS) ⊆= STS S. Thus M= M( eTe;,; A B P) is the Rees matrix cover for TA= B where the map θ :M( SABP ;,;) → T
defined by θθ( x,, s y) = rxy sr′ =( ea,, s be) = easbe is the covering map (θ is a strict local isomorphism from M to T along which idempotents can be lifted).
3. Green’s Relations and on Zappa-Szép Products In this Section we give some general properties of the Zappa-Szép product. We characterize Green’s relations ( and ) of the Zappa-Szép product MG of a monoid M and a group G.
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Proposition 3. [10] Let AB be a Zappa-Szép product of semigroups A and B. Then
(i) (ab11,,)( ab 2 2) ⇒ b 1 b 2 in B;
(ii) (ab11,,)( ab 2 2) ⇒ a 1 a 2 in A.
Proof. Suppose (ab11,,)( ab 2 2) in AB , then there exist ( xy11,,,) ( xy 2 2)∈ A B such that ( xy1,, 1)( ab 11) = ( ab 2 , 2) and ( x2,, y 2)( ab 2 2) = ( ab 11 ,) . Then
a1 ( x1( y 1⋅= a 1),, yb 11) ( a 2 b 2) and
a2 ( x2( y 2⋅= a 2), yb 22) ( ab 11 ,.) Hence
aa12 x111( y⋅= a) a 211,, yb = b 2222 x( y ⋅= a) a 122 ,. yb = b 1
It follows that bb12 in B. Similar proof for (ii). Proposition 4. In the Zappa-Szép product MG of a monoid M and a group G. Then (mg,)( nh ,) ⇔ m n in M . Proof. By Proposition 3 we have (mg,,) ( nh) implies that mn in M. To prove the converse suppose that mn in M then there exist z1 and z2 in M such that mz1 = n and nz2 = m. To show that (mg,,) ( nh) we have to find ( xy11, ) and ( xy22, ) in MG such that
(mg,,)( x11 y) = ( nh ,)
(nh,)( x22 , y) = ( mg ,.)
x1 x2 Then m( g⋅= x11) ng,, y = hn( h ⋅= x 2) m and hy2 = g. Hence −−11 xx12 yg12= ( ) hand yh= ( ) g.
−−11 Therefore we set x1=⋅=⋅ g zx 12,. h z 2 Hence −− −111 −−11 =−−11 ⋅gz⋅1 = ⋅⋅ gz⋅⋅11 gz (mg,,)( x11 y) ( mg ,) g z1 ,( g) h m( g g z1), g( g) h =−1 ⋅= = (mggzhmzhnh( 11),) ( ,,) ( )
Similarly (nh,)( x22 , y) = ( mg ,.) Hence (mg,) ( nh ,.) But from the following example we conclude that the action of the group G is a group action is a necessary condition. Example 2. Let A= { et,, f , b} be a Clifford semigroup with the following multiplication table. Note that
{et, } ≅ C2 and { fb,,} ≅ C2
e t f b
e e t f b
t t e b f
f f b f b
b b f b f
Let (,+) , the group of integers. Suppose that the action of on A for each m∈ is as follows: me⋅= fmt,, ⋅= bm ⋅ f = fmb , ⋅ = b .
Observe that m⋅= a fa for all aA∈ . The action of A on as follows:
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meft= mm,, ==−=− mm mm , b m .
Thus Zappa-Szép axioms are satisfied, since define φ :,A→≅{ fb} C2 by eφ= f φ = ft, φφ = b = b φ = ψ → is a morphism (this is easy to see from the fact that (a) af ). Now define :,{ fb} S (where S is the group of permutations on ) by ψ= ψα = f Ib , where mmα = − , for all m∈ . Clearly ψ is a morphism (of groups). Now for aA∈ and m∈ we de- fine the action arises from the composition φψ as follows (am,) maa by m= ma( φψ ) and (am,) ma⋅ by ma ⋅= fa . We therefore have (ZS1,) ( ZS 3,) ( ZS 2) and (ZS4) as following. For (ZS1:) m⋅( m′′ ⋅=⋅ a) m( fa) = f2 a = fa =( m + m) ⋅ a, and for (ZS3:)
a′ mmaamaamamaa′ = (( ′)φψ) = ( φψ)( ′′ φψ) = ( a )( φψ ) = ( a ) , For (ZS2:) m⋅=( aa′′) faa , and using af= fa for all aA∈ we have (m⋅ a)( ma ⋅= a′) fafa ′′ = faa For (ZS4:) ′ a m+∈ mif a{ ef ,} , (mm+=+′′) ( mm)( aφψ ) = −−m m′ if a ∈{ tb ,} . and +∈′ ′⋅ m mif a{ ef ,} , mmaafaa+=+= mmm′′ −−m m′ if a ∈{ tb ,} . Thus M= A ={( am,:) a ∈∈ Am , B} the Zappa-Szép product of A and B. The set EM( ) = {( f,0)} , since ( f,0)( f ,0) =⋅=( ff( 0) ,0f 0) ( f ,0) and (am, )∈ E( M) if and only if a= ama( ⋅ ) and m= mma , since mma = or −m for all aA∈ so mma +=2 m or 0.≠ m
Now, we note that 1 acts non-trivially. We have et and bf in A but (em, ) not -related to (tn, ) where mn, ∈ since if we suppose (em,,) ( tn) then there exist ( xy11, ) and ( xy22, ) in A such that
(em,,)( x11 y) = ( tn ,)
x1 (em( ⋅= x11),, m y) ( tn)
But emx( ⋅∈1 ) { fb, } so (em, ) not -related to (tn,.) To calculate the -class of (em,,) if (em,,) ( an) then ea and so ae= or at= we prove at= is impossible so ae= . If (em,,) ( en) for some n∈ then we have
(em,,)( x11 y) = ( en ,)
x1 (e( m⋅= x11),, m y) ( en)
But emx( ⋅∈1 ) { fb, } so (em, ) -related only to itself. Similarly (tn, ) -related only to itself. To calculate the -class of ( fm, ) suppose ( f,, m) ( an) then af= or ab= . Let af= so
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( fm,,) ( fn) then
( fm,,)( x11 y) = ( fn ,)
( fn,,)( x22 y) = ( fm ,)
x1 x1 and so (mx⋅11,, m += y) ( fn, ) which implies xe1 = or xf1 = so mm= . Thus ( fm,,) ( fn) for all ∈ mn,. By similar calculation we have ( f, m) ( bn ,.) So if 1 acts non-trivially we have a different structure for the -classes of A and A.
Proposition 5. Let MG be the semidirect product of a monoid M and a group G. Then (mg,,)( nh) ⇔⋅( g−−11 m) ( h ⋅ n) in M .
Proof. Suppose that (mg,,)( nh) then there exist ( xy11,,,) ( xy 2 2) in MG such that
( x11,, y)( mg) = ( nh ,)
( x22,, y)( nh) = ( mg ,)
−1 −−11 Then ( x11( y⋅= m),, yg 1) ( nh) and ( x22( y⋅= n), yh 2) ( mg ,.) Hence y1 = hg , y21= gh = y there- fore −1 x11( y⋅= m) x 1( hg ⋅= m) n which implies −−11 − 1 (1) (hxgm⋅1 )( ⋅=⋅) hn and −1 x22( y⋅= n) x 2( gh ⋅= n) m which implies
−−11 − 1 (2) ( gxhn⋅2 )( ⋅=) gm ⋅ Thus by (1) and (2) we have ( gm−−11⋅⋅)( hn) in M. −−11 −−11 Now suppose ( gm⋅⋅)( hn) in G then there exist z1 and z2 in G such that zg1 ( ⋅=⋅ m) h n and −−11 −1 zh2 ( ⋅= n) g ⋅ m. Therefore (h⋅ z1 )( hg ⋅= m) n. Hence =⋅⋅=⋅−−11 (n, h) (( h z11)( hg m), h) ( h z ,, hg)( m g )
−1 −1 Therefore we set ( x11,, y) =( h ⋅ z 1 hg ) in MG other formula by symmetry ( x22,, y) =( g ⋅ z 2 gh ) so (mg,)( nh ,.) m n Proposition 6. If ( gm−−11⋅⋅)( hn) such that ( gg−−11) = and (hh−−11) = in G, then (mg,,)( nh) in MG . −−11 −−11 Proof. Suppose ( gm⋅⋅)( hn) then there exist z1 and z2 in M such that zg1 ( ⋅=⋅ m) h n which implies that
z1 −1 −− 11 nhzhgm=⋅( 12)( ⋅) and zhn( ⋅=) gm ⋅.
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z1 −1 z2 −1 We set ( x11,, y) =( hzhg ⋅ 1 ) and ( x22,, y) =( gz ⋅ 2 gh ) then
m x,, y mg=⋅=⋅⋅ x y m , ygm hz hgzz11−−11 m, hg g ( 11)( ) ( 1( 1) 1 ) (( 1)( ) ( ) )
− −1 1⋅ mn− zg1( m) −1hn1⋅− 11− = nh, ( g) g= ( nh, g g) = n,( h) = ( nh,.) ( )
Similarly ( x22, y)( nh ,) = ( mg ,.) Hence (mg,,)( nh) in MG . 4. Regular and Inverse Zappa-Szép Products The main goal of this Section is to determine some of the algebraic properties of Zappa-Szép products of semigroups in terms of the algebraic properties of the semigroups themselves. The (internal) Zappa-Szép product MAB= , of the regular subsemigroups A and B need not to be regular in general. A special case of the Zappa-Szép product is the semidirect product for which one of the actions is trivial. We use Theorem 2.1 [11] to construct an example of regular submonoids such that their semidirect product is not regular. Example 3. Let T= {1, ef , , 0} be a commutative monoid with 0, each of whose elements is idempotent and such that ef= f . Let S= {1, ab , } be a monoid with two left zeros a and b. Then both S and T are regular semigroups. Let 1 acts trivially, ea= e b = ef, a = 0 a = 0, f b = 0 b = f ,1 ab = 1 = 1. There is no e2 = eS ∈ such that Ss= Se, t ∈ te T for all s∈∈ St,. T Thus the semidirect product RST= is not regular. For example, (af, ) is not a regular element of R. Example 4. Take A= {1, ef , } such that eeffefffe22=,, = = = , and Bb= {1, } such that bb2 = . Let 1∈ A act trivially on B, 1ef= 1 = 1,bb e = f = 1, and 1∈ B act trivially on A, b⋅=1 1, be ⋅= fb , ⋅ f = f . Then A and B are regular monoids but their Zappa-Szép product MAB= is not regular. However, there are criteria we can prove that the internal Zappa-Szép product MAB= of regular A and B is regular as the following Propositions illustrated. a Proposition 7. If A is a regular monoid, B is a group, 1B⋅=aa ,1( BB) = 1 for all aA∈ , then MAB= is regular. Proof. Let (ab, )∈= M A B where aA∈ and bB∈ , we have to find (cd, )∈= M A B such that a (ab,,)( cd)( ab ,) = ( ab ,) . Now (ab,,)( cd)( ab ,) =⋅⋅( a( b c)(( bdcc) a),( bd) b) and so we choose
− −1 a −⋅1 ba1 ′ c (cd,) = b ⋅ a′ ,, b where a′∈ Va( ). Since we must have (bd) b= b but B is a group, so ( ) a −1 a c c c a (bd) =1.B Suppose we are given c, and choose db= ( ) since B is a group, so (bd) =(1BB) = 1, and ⋅c ⋅= ⋅ ⋅= ⋅ ′∈ then abc( )(( bd) a) abc( )(1.B a) abca( ) Since A is regular, we choose any a Va( ) and set
− −1 −1 ′ −−11′ ′ ′′ −⋅1 ′ ba1 ′ cb= ⋅ a, thus b⋅=⋅ c b( b ⋅ a) =( bb) ⋅=⋅= a1.B a a Thus (ab,) b⋅= a , b( ab ,) ( ab ,,) ( ) whereupon MAB= is regular. Proposition 8. Let A be a left zero semigroup and B be a regular semigroup. Suppose that for all bB∈ , there x exists some aA∈ such that bba = , and for all xA∈ there exists some b′∈ Vb( ) such that (bb′′) = . Then MAB= is regular. Proof. Let (ab, )∈= M A B where aA∈ and bB∈ . We have to find (cd, )∈= M A B such that cda( ⋅ ) a (ab,,)( cd)( ab ,) = ( ab ,) . Now (ab,,)( cd)( ab ,) =⋅⋅( a( b c( d a) ), b db) and we choose (cd,) = ( cb ,,′) where b′∈ Vb( ). Now since A is a left zero semigroup cda( ⋅ ) a ca c (a( b⋅⋅ c( d a) ), b db) =( abdb ,,) and by our assumptions we can choose cA∈ such that bb= , and bd′ = that is fixed by a. Then (ab,)( cd ,)( ab ,) =( abb ,ca′′ b) =( abbb ,) = ( ab ,.) Thus MAB= is regular. Theorem 1. [12] For any arbitrary semigroup S, Reg( S ) is a subsemigroup of S if and only if the product
1054 S. Wazzan of any pair of idempotents in S is regular. We now give a general necessary and sufficient condition for Zappa-Szép products of regular semigroups to be regular. Consider the internal Zappa-Szép product MAB= of regular semigroups A and B. Then each mM∈ is uniquely a product of regular elements: m= ab where a,. b∈ Reg( M ) Hence M is regular if and only if Reg( M ) is a subsemigroup of M and so by Hall’s Theorem 1, M is regular if and only if the product of any two idempotents is regular. But in fact we need only consider products of idempotents eA∈ and fB∈ , as our next theorem shows. Theorem 2. Let A and B be regular subsemigroups and e∈ EA( ) and f∈ EB( ). Then ef∈ Reg( M ) if and only if MAB= is regular. Proof. Given mM∈ with m= ab (uniquely) where a∈∈ Ab,, B since A and B are regular subsemi- groups, then there exist a′∈ Va( ) and b′∈ Vb( ). Then aa′ ∈ E( A) and bb′∈ E( B). Set e= aa′ and f= bb′. Then by the assumption (a′′ a)( bb)∈ Reg( M ) and the set M( a′′ a,, bb) == M( e f) { g∈= V( ef) E( M) :. ge fg= g} is not empty, see [14]. Because ef∈ Reg( M ) let x∈ V( ef ) , and let g= fxe. Then (ef) g( ef) = ( ef) fxe( ef) = ef22 xe f = efxef = ef and g( ef) g= fxe( ef) fxe= fxe22 f xe = f( xefx) e = fxe = g and so g∈ V( ef ). Also g2 = f( xefxe) = fxe = g, and so g∈ EM( ). Also ge= fxee = g, fg= ffxe = g. Thus g∈ Mef( ,.) Also b′′ ga∈ V( m) because (ab)( b′′ ga)( ab) = afgeb = ageb = agb = aa′′ agbb b = a( egf) b we have egf= ef ,
egf= efgef( ) =( ef) gef( ) = ef . Then (ab)( b′′ ga)( ab) = a( ef) b = ( aa′′ a)( bb b) = ab, and (b′′ ga) ab( b ′′ ga) = b ′ gefga ′ = b ′ g2 a ′ = b ′′ ga . and so b′′ ga∈ V( m) . Then m= ab is a regular element which implies that MAB= is regular. Conversely, if MAB= is the regular internal Zappa-Szép product of the regular subsemigroups A and B, each element m of M is uniquely written in the form m= ab where aA∈ and bB∈ . Thus if eA∈ and fB∈ this implies that ef∈ M , then ef∈ Reg( M ). Corollary 1. If A and B are regular and EA( ), EB( ) act trivially, then MAB= is regular. Proof. If we take e∈ EA( ) and f∈ EB( ) , then ef is an idempotent in M. Because (ef)( ef) =⋅== e( f e)( fe ) f eeff ef , since EA( ), EB( ) act trivially. Therefore ef∈ Reg( M ). Hence MAB= is regular. In this case: if m= ab ∈ M , we can find m′∈ Vm( ). First find g∈== M( aa′′, bb) M( e ,, f) e aa′ and f= bb′, g= fefe = fe since idempotents of A commute with those of B. Then m′′= b( ef) a ′ = xy for some ′ ′′( fa⋅ ) x∈∈ Ay,. B Thus m′′=⋅( beb) ′e( fa ⋅ ′) f a =⋅( be ′)( b ′ e ⋅⋅( fa ′′))( b ea) f where x= ba′′ ⋅ and
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a′ a′ yb= ( ′) . Then m′= xy =( b ′′ ⋅ a)( b ′) . Now we discuss inverse Zappa-Szép products. Let S and T be inverse semigroups/monoids with θ : S→ End( T ) and let PST= be the semidirect product of S and T. We can see from the following example that P need not be inverse. Example 5. [11] Let Sa= {1, } be the commutative monoid with one non zero-identity idempotent a. Let Te= {1, , 0} be the commutative monoid with zero an with ee= 2. Then S and T are both inverse monoids, and there is a homomorphism θ : S→ End( T ) given by 11a = and eeaa=0. = Then PST= is regular. However, the element (ae, )∈ P has two inverses, namely (ae, ) and (a,0) , and hence P cannot be an inverse monoid. A complete characterization of semidirect products of monoids which are inverse monoids is given in Nico [11]. Theorem 3. [11] A semidirect product PST= of two inverse semigroups S and T will be inverse if and only if ES( ) acts trivially. In the general case of the Zappa-Szép product of inverse semigroups PS= T we have also P need not be inverse semigroup as we can see from the following example. Example 6. Let A= E = { e,, f ef } where e22= e,, f = f ef = fe and B= G = {1, a , b , ab} —Klein 4-group where a22=1, b = 1, ab = ba . Suppose that the action of G on E is defined by: aefbefaefefafebfe⋅=,, ⋅= ⋅ = ,,, ⋅ = ⋅ =
and bef⋅= efabe,, ⋅= eab ⋅= f fabef , ⋅= ef . and 1∈G acts trivially (that is each of ab, permutes {ef, } non-trivially). The action of E on G is defined by:
ae= bb,, e = ba f = bb , f = ba , ef = b , and
e f ef bef = b,( ab) = 1,( ab) = 1,( ab) = 1.
We check Zappa-Szép axioms by the following: define φ :GC→=2 { 1, t} by φφφ(a) = t,( b) = t ,( ab) = 1.
This is a homomorphism of groups since C2 is the automorphism group of E. We have an action of G on E using φ : if gG∈ and xE∈ define gx⋅=φ ( g)( x). We have a homomorphism ψ :GG→ given by abψ = and bbψ = . The action of E on G is given by: gge= f = g ef = gψ We note that φψ( gg) = φ( ), for all gG∈ . For (ZS1) we have for gh, ∈ G ghx⋅⋅=( ) g(φ( hx)( )) = φφ( g) ( hx)( ) = φ( ghxghx)( ) = ⋅. Since ψψ2 = it is clear that (ZS3) holds. For (ZS2) we have for xy, ∈ E g⋅=( xy) φ( g)( xy) = φφ( g)( x) ( g)( y) and ( gxgy⋅)( x ⋅=) φφψφφ( gx)( ) ( g)( y) =( gx)( ) ( gy)( ) So (ZS2) holds and for (ZS4) x ( gh) =( gh)ψ = ( g ψψ)( h )
1056 S. Wazzan and ghhx⋅ x = ( gψψ)( h). Thus (ZS4) holds. Then ME= G. Since every element of M is regular, then M is regular. EM( ) is a closed subsemigroup of M, so ME= G is orthodox, but since EM( ) is not commutative for example ( f, a)( f ,1) = ( ef , b) while ( f,1)( fa ,) = ( fa , ) , then M is not inverse. The achievement of necessary and sufficient conditions was difficult; so we try to find an inverse subset of the Zappa-Szép product of inverse semigroups. This achieved and described in Section 9. We have given the necessary conditions for Zappa-Szép products of inverse semigroups to be inverse in the following theorem. Theorem 4. PS= T is an inverse semigroup if (i) S and T are inverse semigroups; (ii) ES( ) and ET( ) act trivially; (iii) For each p=( st, ) ∈ E( P) where sS∈ and tT∈ , then s and t act trivially on each other. Proof. We know that PS= T is regular. Since a regular semigroup is inverse if and only if its idem- potents commutes, it suffices to show that idempotents of PS= T commute. If (at,,,) ( bu) are idem- potents of PS= T, then (at,,)( at) =( at ,) =( a( t ⋅ a) , tta ) Thus a=⋅= at( a) and t tta , and b=⋅= bu( b) and u uub , By (iii) a and t act trivially on each other, b and u act trivially on each other, then a= ab2, = bt 22 ,, = tu = u 2 . But since S and T are inverse semigroup, then idempotents commutes that is ab= ba∈∈ S ,= tu ut T , Then (at,,)( bu) =( a( t ⋅ b) ,, tub ) but t and c are idempotents they are act trivially then (a, t)( b , u) ===⋅=( ab , tu) ( ba , ut) ( b( u a) , ua t) ( b , u)( a ,. t) Thus PS= T is inverse. 5. External Zappa-Szép Products Let A and B be semigroups, and suppose that we are given functions AB×→ A, (ab, ) b⋅∈ a A and AB×→ B, (ab, ) ba ∈ B where aA∈ and bB∈ . satisfying the Zappa-Szép rules (ZS1,) ( ZS 2,) ( ZS 3) and (ZS4.) Then the set AB× with the product defined by: (ab,,)( cd) =( a( b ⋅ c) , bdc ) is a semigroup, the external Zappa-Szép product of A and B, which is written as AB .
If A and B are semigroups that both have zero elements ( 0A and 0B respectively), and we have in addition to (ZS1,) ( ZS 2,) ( ZS 3) and (ZS4) for all aA∈ and bB∈ the following rule: 0A a (ZS5) bb⋅0AA = 0,( ) = 0,0BB ⋅= a 0,0 A( B) = 0. B
Then by Proposition [10] we have that S is a semigroup with zero (0AB ,0) . But from the following example we deduce that (ZS9) is not a necessary condition:
Example 7. If A and B are semigroups with 0A and 0B respectively, acting trivially on each other. Then (ZS5) is not satisfied. However, in the Zappa-Szép product S= AB × we have a (0AB ,0)(ab ,) =⋅=( 0A( 0 B a) ,0 B b) ( 0 A a ,0 B b) =( 0 AB ,0) , and
0A (ab,)( 0,0AB) =⋅==( a( b 0A) , b 0 B) ( a 0,0 A b B) ( 0,0. AB) Thus (0AB ,0 ) is zero for AB .
From the following example we deduce that the zeros 0A and 0B of A and B respectively do not necessarily
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give a zero for the external Zappa-Szép product AB .
Example 8. If A is a monoid with identity 1A and zero 0A and B is a semigroup with zero 0,B such that a the action of A on B is trivial action (ab, ) b= b and the action of A on B is (ab,1) b⋅= a A for all a∈∈ Ab,. B Then Zappa-Szép rules (ZS1-) ( ZS 4) are satisfied. But (ZS5) is not satisfied since
b ⋅=≠0AA 1 0, A and (0AB ,0 ) is not a zero for AB . The Zappa-Szép rules can be demonstrated using a geometric picture: draw elements from A as horizontal arrows and elements from B as vertical arrows. The rule ba=( b ⋅ a)( ba ) completes the square
From the horizontal composition we get (ZS2) and (ZS3) as follows:
From the vertical composition we get (ZS1) and (ZS4) as follows:
These pictures show that a Zappa-Szép product can be interpreted as a special kind of double category. This viewpoint on Zappa-Szép products underlies the work of Fiedorowicz and Loday [13]. In the theory of quantum groups Zappa-Szép product known as the bicrossed (bismash) product see [14].
6. Internal and External Zappa-Szép Products In general, there is not a perfect correspondence between the internal and external Zappa-Szép product of semi- groups. For one thing, embedding of the factors might not be possible in an external product as the following example demonstrates. Example 9. Consider the external Zappa-Szép product P = , where for all m∈ and n∈ we have nm⋅=+ n m and nm = 0, so that the multiplication in P is ′′ ′m′ ′ ′′ (mn,,)( mn) =( mnm( ⋅), nn) =( m ++ n mn,.) Then for each k ∈ the subset k ={(mk,:) m ∈ } is a subgroup of P isomorphic to (with identity (−kk, ) ). However P cannot be an internal Zappa-Szép pro- duct of subsemigroups Z, N isomorphic to and respectively: If ( pq, ) generates N, then the second
1058 S. Wazzan coordinate of every non-identity element of N is q, and so the second coordinate of any product xy with xZ∈ and yN∈ is equal to q. However, under some extra hypotheses, the external product can be made to correspond to an internal product for example: • if we assume the two factors A and B involving in the external Zappa-Szép product have an identities ele-
ments 1A and 1B respectively such that the following is satisfied 1A a (ZS6) b⋅1AA = 1 , b = b ,1 B ⋅= aa ,1 BB = 1 , for all aA∈ and bB∈ . So if MAB= , the external Zappa-Szép product of A and B, then each A and B are embedded in
MAB= Define τ M : AAB→ by τ MB(aa) = ( ,1 ) . We claim τ M is an injective homomorphism since a2 τ MB(aa12) = ( aa 12,1 ) and ττMM(a1) ( a 2) =( a 1,1B)( a 2 ,1 B) =⋅=( a1( 1 B a 2) ,1 BB 1) ( aa12 ,1B) . Thus τM is a homo- morphism, also τ M is injective since ττMM(a1) =( a 2) ⇔( a 1,1 B) =( a 2 ,1 B) ⇔= aa 12 . Denote its image by A . Define ψ M : B→→ AAB by ψ MA(ab) = (1, ) , then ψ M is also an injective homomorphism. Denote its image by B . Observe that (ab,) = ( a ,1BA)( 1 , b) . Thus M= A B = AB . This decomposition is evidently unique. Thus MA= B is the internal Zappa-Szép product of A and B . • If A is a left zero semigroup and B is a right zero semigroup, then the external Zappa-Szép product of A and
B is a rectangular band and it is the internal Zappa-Szép product of A={( ab,: ) a ∈ A} and
B={( ab ,:) b ∈ B} where ab, are fixed elements of A and B respectively. Note that in a left-zero semi- group A, RI( A) = A and in a right-zero semigroup B, LI( B) = B, and we have the following Theorem: Theorem 5. M is the internal Zappa-Szép product of a left-zero semigroup A and a right-zero semigroup B if and only if M is a rectangular band. Proof. Let A be a left-zero semigroup and B a right-zero semigroup. In the rectangular band M= AB × , let
A={( ab,: ) a ∈ A} and B={( ab ,:) b ∈ B} , where ab, are fixed elements. Then ′ ′ (ab,) =( ab ,,)( a b) ∈× A B uniquely and (ab,,)( a b) =( ab , ) ∈ A , and (ab,,)( ab) =( ab ,) ∈ B . So M= AB × is the internal Zappa-Szép product of A and B , MA= B, where AA≅ (as left-zero semi- group) and BB≅ (as right-zero semigroup).
Conversely, Let MAB= where A is left-zero semigroup and B is right-zero semigroup. Then aa12= a 1 for all aa12, ∈ A and bb12= b 2 for all bb12,.∈ B M is a rectangular band if for all xy, ∈ M then xyx= x, where x= ab11 and y= ab22 for unique aa12, ∈ A and bb12,.∈ B Now
aa21 a 2 xyxababababab=11221111 =⋅⋅=⋅⋅( 21) ( babb 2 121) abab 11( 21) ( bab 2 11)
(ba21⋅⋅) (ba21) aa22 a 2 =aba11( ⋅ 2)( b 1 ⋅⋅( ba 2 1))( b 1) b 1= ab 11( ) b 1 = abx 11 = Thus M is a rectangular band.
7. Examples
1) Let A= { et,, f , b} be a Clifford semigroup. Note that {et, } ≅ C2 and { fb,.} ≅ C2 Let B =(,, +) the group of integers. Suppose that the action of on A for each m∈ is as follows: eft b me⋅= fmt,, ⋅= bm ⋅ f = fmb , ⋅ = b . The action of A on as follows: m= mm,, ==−=− mm mm , m . Then the Zappa-Szép multiplication is associative. Thus M= A B ={( xm,:) x ∈∈ Am , B} the Zappa-Szép product of A and B. The set EM( ) of idempotents of M is the empty set, since ( xm, )∈ E( M) if and only if x= xmx( ⋅ ) and m= mmx , since mmx = or −m for all xA∈ so mmx = 2 m or 0.≠ m 2) Suppose that A is a band. Then the left and right regular actions of A on itself allows us to form the Zappa- Szép product MA= A, since if we define b⋅= a ba and ba = ba with ab,,∈ A we obtain the multi- a2 plication (ab1,, 1)( ab 2 2) =⋅=( aba 1( 1 2) , bb 1 2) ( ababab 112 , 122) . Then MA= A is the external Zappa-Szép product of A and A. Moreover M is a band if and only if A is a rectangular band, in which case M is a rectangular band. 3) Let A= E = { e,, f ef } where e22= e,, f = f ef = fe and B= G = {1, a , b , ab} —Klein 4-group, where a22=1, b = 1, ab = ba . Suppose that the action of G on E is defined by: aefbefaefefafebfe⋅=,, ⋅= ⋅ = ,,, ⋅ = ⋅ = and bef⋅= efabe,, ⋅= eab ⋅= f fabef , ⋅= ef . and 1∈G acts trivially. The action of E on G is defined by: ae= bb,, e = ba f = bb , f = ba , ef = b , and
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e f ef bef = b,( ab) = 1,( ab) = 1,( ab) = 1. Then ME= G is the Zappa-Szép product of E and G. Since every element of M is regular, then M is regular. EM( ) is a closed subsemigroup of M, so ME= G is orthodox, but since EM( ) is not commutative for example ( f, a)( f ,1) = ( ef , b) while ( f,1)( fa ,) = ( fa , ) , then M is not inverse. 4) For groups, MAB= is the Zappa-Szép product of subgroups A and B if and only if, MB= A since for any mM∈ we have m−1 = ab for unique aA∈ and bB∈ . This implies that m= ba−−11 for unique bB−1 ∈ and aA−1 ∈ . Thus MB= A. But this is not true in general for semigroups or monoids. Let Aa= {1, } be a commutative monoid with one non-identity idempotent a. Let B= {1, ef , } be a com- mutative monoid with two idempotents e and f and ef= fe = e. Let B act trivially on A and 1∈ A act trivially on B and 1a= 1,efe aa = = . Then A= {(1,1) ,( aA ,1)} ≅ and B= {(1,1) ,( 1, efB) ,( 1,)} ≅ . Then MA= B is the internal Zappa-Szép product of A and B . But MB≠ A, since (1,b)( a ,1) =⋅=( 1( b a) , baa 1) ( ab ,) , so (af, ) can not be written as ba′′. Moreover, (a,1)( 1, f) = ( af , ) so BA is not a submonoid of M.
8. Zappa-Szép Products and Nilpotent Groups In this section we consider a particular Zappa-Szép product for nilpotent groups. Note that G being nilpotent of class at most 2 is equivalent to the commutator subgroup G′ being contained in the center ZG( ) of G. Now, let G be a group and let G act on itself by left and right conjugation as follows: b⋅= a bab−−11and ba = a ba . In the following we show that these actions let us form a Zappa-Szép product PG= G if and only if G is nilpotent group of class at most 2. Proposition 9. Let G be nilpotent group of class at most 2. Then the left and right conjugation actions of G on it self can be used to form the Zappa-Szép product PG= G. Proof. Let G act on itself by left and right conjugation as follows: b⋅= a bab−−11and ba = a ba where ab,.∈ G Thus the multiplication is given by:
−−11 (a1,, b 1)( a 2 b 2) = ( abab 1121 , a 2 bab 122) . We prove that the Zappa-Szép rules are satisfied if G is a nilpotent group of class less than or equal 2, which implies that [ab,,] c = 1 for all abc,,∈ G . For (ZS1) and (ZS3) clear they are hold. a1 (ZS2) b⋅=⋅⋅( aa12) ( b a 1)( b a 2). −1 LHS. ..: b⋅=( aa12) baab 12 , RHS. ..:
a1 −1 − 1 −−11 −− 11 (b⋅ a1)( b ⋅= a 2) ( ba 1 b)( a 112 ba ⋅= a) ba 1 b a 11211 ba a a b a since G is nilpotent of class ≤ 2,then
a1 −−11 −− 11 −1 −− 11 − 1 −1 (b⋅ a1)( b ⋅= a 2) ba 1 b a 1 ba 121 a a b a 1= ba 121 a a a 1 b a 1 bb a 1= ba 12 a b = R. H .. S Thus (ZS2) holds. a ba2⋅ a (ZS4) (bb12) = b 1 b 2. a −1 RHS. ..: (bb12) = a bb12 a, LHS. ..: −1 b2⋅ aaa b22 ab bb12= b 1 b 2 since G is nilpotent of class 2, then
ba2⋅ a −−11 −− 11 −−11 −− 11 − 1 b1 b 2= ba 2 b 2 bbab 12 2 a ba 2= ba 2 b 2 bab 2 2 a bba 12= a bba 12 = RHS. ..
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Thus (ZS4) holds. Hence PG= G is the Zappa-Szép product. Proposition 10. If the left and right conjugation actions of G on itself satisfy the Zappa-Szép rules, then G is nilpotent of class at most 2. Proof. Suppose the Zappa-Szép rules satisfied, we prove that G is nilpotent of class ≤ 2. If (ZS2) holds,
a1 then for all aab12,,∈ G we have b⋅=⋅⋅( aa12) ( b a 1)( b a 2) . Thus
ba a b−1= ba b −− 11 a ba a a −− 11 b a 12 1 1 121 1 −1 −− 11 −− 11 ab2 = babaaaba1 121 1
−−11 −−11 −−11 Therefore a112211 b aba= a a b ab. Hence a11 b ab is central in G. Similarly if (ZS4) holds. Combining Propositions 9 and 10 we prove the following: Proposition 11. P is the Zappa-Szép product of the group G and G with left and right conjugation actions of G on itself if and only if G is nilpotent of class at most 2. Next we prove the following: Lemma 3. The center of PG= G is ZG( )× ZG( ). Proof. Suppose (ab,.)∈ Z( P) Since for all (cd, )∈ P we have (a,, b)( c d) = ( abcb−−11 , c bcd ) . and (c,, d)( a b) = ( cdad−−11 , a dab) . Then (1) abcb−−11= cdad . and (2) c−−11 bcd= a dab. Put in (1) c =1: then a= dad −1 for all dG∈ . Therefore ad= da. So a∈ ZG( ). Put in (2) d =1: then b= c−1 bc for all cG∈ . Therefore cb= bc. So b∈ ZG( ). So ZP( ) ={( ab,) :, ab ∈ ZG( )} ≅× ZG( ) ZG( ), since if aa,′′ ,, bb∈ Z( G) . Then (ab,,)( ab′′) = ( ab ′′ ,,)( ab) =( aabb ′ , ′) ∈ P . Lemma 4. If PG= G then P is abelian if and only if G is abelian. Proof. If G is abelian then G is nilpotent of class 1 if and only if ZG( ) = G. This implies ZP( ) =×= G G P and so P is abelian. If P is abelian then ZP( ) = P, but ZP( ) = ZG( ) × ZG( ) . Thus P= G G = ZG( ) × ZG( ) if and only if G= ZG( ). Hence G is abelian group. In which case P= GG × . Proposition 12. If P is the non-abelian Zappa-Szép product PG= G and G is nilpotent group of class at most 2, then P is nilpotent of class 2. Proof. We have G is nilpotent group of class ≤ 2 if and only if for all abc,,∈ G we have −−11 −− 11 a b abc= ca b ab, that is [abc,,] = cab[ ] the commutator elements are central. Let (ab,) ,,( cd) be a commutator in P. We prove it is central in P. We have
−−11 −1 (ab,,,) ( cd) = ( ab ,) ( cd ,) ( ab ,,)( cd) = (( cd ,)( ab ,)) ( ab ,,.)( cd) Now −−11 −1 −− 11 − 1 −1 − 1 (a,,, b)( c d) = ( abcb c bcd) = ( acc bcb, c bcb bd) = ( ac c, b ,, c b bd )
−1 − 1 −−11 = (accb,, bdcb ,) = ( acbd ,)( cb ,,,. cb ) and −−11 −1 − 1 (c,,, d)( a b) = ( cdad a dab) = ( ca a, d ,, a d db)
−−11 − 1 − 1 −− 11 = (acc a ca a, d ,, a d bdd b db)
−−11 = (ac[ c, a] a , d ,, a d bd[ d , b])
−−11 = (ac, bd)([ c ,,, a] [ d b])( a , d ,, a d ) .
−−11 Write xadycbucavdb=, , = , , =[ ,,] = [ ,.] Then
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−1 −−11 − 1 (a, b) ,,( c d) = (( c , d)( a , b)) ( a , b)( c , d) = ( x , x) ( u , v) ( ac , bd) ( ac , bd)( y , y)
−−11 Since cb, , ad , ,,[ ca] ,[ db ,] are commutators, then −1 −1 − 1 −1 −−11 (ab,) ,,( cd) = ( ad , , ad , ) ([ ca ,,,] [ db]) ( cb , ,, cb ) .
This implies that (ab,) ,,( cd) ∈× G′′ G . Thus P′⊆×⊆ G ′′ G ZG( ) × ZG( ) = ZP( ). So commutators in P are in the center ( and P is not abelian) so P is nilpotent of class 2. Combining Propositions 11, 12 and Lemma 4 we have the following. Theorem 6. Let G be a group that is nilpotent of class at most 2, and let PG= G with left and right con- jugation action of G on it self. Then: 1) P is abelian if and only if G is abelian, in which case P= GG × ; 2) If G is non-abelian and hence nilpotent of class 2, then P is also nilpotent of class 2.
9. Zappa-Szép Products of Semilattices and Groups The Zappa-Szép product of inverse semigroups need not in general be an inverse semigroup. This is even the case for the semidirect product as we see (Nico [11] for example) However, Bernd Billhardt [15] showed how to get around this difficulty in the semidirect product of two inverse semigroups by modifying the definition of semidirect products in the inverse case to obtain what he termed λ -semidirect products. The λ -semidirect product of inverse semigroups is again inverse. In this Section, we construct from the Zappa-Szép product P of a semilattice E and a group G, an inverse semigroup by constructing an inductive groupoid. We assume the additional axiom for the identity element 1∈G we have 1.⋅=ee Note that if 1⋅=ee for all eE∈ , then (ZS81) e = 1 holds, since by cancellation in the group G. e gge=( 1) = g1⋅ee 11 = g ee
1= 1.e
We consider the following where E a semilattice and G a group, and subset BP ( ) of the Zappa-Szép product PE= G: e B P=∈= eg,: P g−−11 g . ( ) {( ) ( ) } We form a groupoid from the action of the group G on the set E which has the following features:
• vertex set: EB( ( P)) ={( e,1) : e ∈≅ E} E;
• arrow set: BEG ( ) ; −1 −1 • an arrow (eg, ) starts at (eg,)( eg ,) = ( e ,1) , finishes at (eg,) ( eg ,) =( g−1 ⋅ e ,1) ;
−−11 • the inverse of the arrow (eg, )∈ B ( P) is ( g⋅ eg,;) • the identity arrow at e is (e,1) ; and • an arrows (eg, ) and ( fh, ) are composable if and only if gef−1 ⋅= , in which case the composite arrow is (eg( ⋅ f),. ghf )
ge−1⋅ Lemma 5. If (eg,,)∈= B ( P E G) then gg= .
−⋅−⋅−ee−−11 −1 Proof. we have 11= e for all eE∈ , then 11=e =( gggggg1) = ge( 11) = ge . Then gg= ge⋅ . −−11 Lemma 6. Suppose that (eg,,)∈ B then (eg, ) is a regular element and ( g⋅∈ eg,) V( eg ,.) Proof. We have
−1 (eggeg,,,)( −11⋅ −)( eg) =( eggeg( −1 ⋅),ge ⋅− g 1)( eg ,) =( e ,1)( eg ,) =⋅=( ee( 1) ,1e g) ( eg , ) and
−1 ( geg−−11⋅,,,,,,)( eggeg)( −− 11 ⋅=⋅) ( gegggeg −− 1e)( −− 11 ⋅=⋅) ( ge − 11ge ⋅−−− g 1) =⋅( geg 11 ,)
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Thus (eg, ) is a regular. Proposition 13. If PE= G, where E is a semilattice and G is a group then e B P=∈= eg,: P g−−11 g ( ) {( ) ( ) } with composition defined by (eg,,)( f h) =( e( g ⋅ f) , gf h) if ( ge−1 ⋅=,1) ( f ,1) and 1∈G acts trivially on E is a groupoid. Proof.
We have to prove f (eg( ⋅∈ f), gh) B ( P) ⋅ −−11egf( ) eg( ⋅ f), ghf ∈⇔ B( P) gh ff = gh ( ) ( ) ( ) Now
−1 egf( ⋅⋅) egf( ) f egf( ⋅) −−11( g) ⋅⋅ egf( ) −1 ghff = h−−11 g = h gf ( ) ( ) ( ) Since gef−1 ⋅= implies that gg⋅−1 ⋅= e g ⋅ f, this implies that e= gf ⋅ , therefore
−1 gf ⋅⋅ gf e ( gf ) = ( g−1) ( g − 1) =( gg −− 11) = . Then −− egf( ⋅ ) ff11ee( ) e −1( gee) ⋅⋅( ) −− 11( ge) f = −−11 ff= ( gh) h( g) h ( g) −1⋅ −−11 −1ge −− 11 e ff = h g= h( g) = ( gh) .
f (eg( ⋅ f), gh) starts at (e,1) But (eg( ⋅ f), ghf ) starts at (eg( ⋅ f),1) and gf⋅= e, so eg( ⋅= f) e,
f −1 (eg( ⋅ f), gh) ends at (hf⋅ ,1) −1 But (eg( ⋅ f), ghf ) ends at (h−1 ( gf ) ⋅⋅ eg( f),1)
−1 hg−1( f ) ⋅ egfhgehgehf( ⋅) =−−112 ⋅ = − 1 ⋅ − 1 ⋅= − 1 ⋅.
Thus BP ( ) is a groupoid. Now we introduce an ordering on BP ( ) . The ordering is giving as follows:
−1 (eg,) ≤( f , h) ⇔≤ e f and g = hhe⋅ .
Lemma 7. The ordering on BP ( ) defined by −1 (eg,) ≤( f , h) ⇔≤ e f and g = hhe⋅ .
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is transitive. Proof. We have to prove that if (eg,) ≤≤( f , h) ( lk ,,) then (eg,) ≤ ( lk ,.) We have −1 −1 (eg,,) ≤( f h) ⇔≤ e f and gh= he⋅ and ( f,, h) ≤( lk) ⇔≤ f l and hk= kf⋅ . Since
e ge⋅ −⋅1 ge −1 −1 11=ee =( gg−1 ) =( g −1) g . Thus ( gge ) = ( −1 ) . We conclude gh=he⋅− = ( h e) and
−1 −1 hk=kf⋅− = ( k f) . Now, efl≤≤ implies that el≤ and −1 −1 e −−11−1 gh=( −e ) =(( k −f ) ) =( k −−fe ) =( ke) = k k ⋅ e. Thus ≤ is transitive.
Lemma 8. The ordering on BP ( ) defined by he−1⋅ B ( P)( eg,) ≤( f , h) ⇔≤ e f and g = h . is antisymmetric. Proof. We have to prove if (eg,,) ≤ ( f h) and ( f,, h) ≤ ( eg) , then (eg,) = ( f ,. h) Now −1 −1 (eg,,) ≤( f h) ⇔≤ e f and gh= he⋅ and ( f,, h) ≤( eg) ⇔≤ f e and hg= gf⋅ . Thus ef= and −1 −−11e −1 gh=( −−e) =(( gf) ) =( g −f) = g gf ⋅ = h. Thus ≤ is antisymmetric. e Proposition 14. B P=∈= eg,: P g−−11 g with ordering defined by ( ) {( ) ( ) }
−1 (eg,) ≤( f , h) ⇔≤ e f and g = hhe⋅ . is a partial order set. Proof. Clear from the definition of the ordering that ≤ is reflexive. By Lemma 7 and Lemma 8 ≤ is transitive
and antisymmetric. Thus (BP ( ),≤) is a partial order set.
Next we prove that (BP ( ),≤) is an ordered groupoid. −−11 Lemma 9. If (eg,,) ≤ ( f h) , then (eg,,) ≤ ( f h) for all (eg,,,) ( f h)∈ B ( P) . −1 Proof. Suppose that (eg,) ≤ ( f ,, h) so that ef≤ and gh= he⋅ . Now, we have
−1 −1 g−11⋅= e( hhe ⋅) ⋅= eh −−e ⋅= eh ⋅ e. Thus ( gehfhehfhfhe−−11⋅)( ⋅=) ( −− 11 ⋅)( ⋅=) ( − 1 ⋅)( − 1 ⋅)
f =(hfh−1 ⋅)(( − 1) ⋅ ehfehege) = − 1 ⋅ = −− 11 ⋅= ⋅ and f (hfge−−−−−−111111⋅)( ⋅=) ( hfhe ⋅)( ⋅=) ( hfh ⋅)(( ) ⋅= ehfehege) −−− 111 ⋅= ⋅= ⋅. Therefore g−−11⋅≤ eh ⋅ f. Also
hge⋅⋅−1 hhe ⋅⋅ −− 11 hhe⋅ 1. e e (h−1) =( h − 1) =( h − 1) =( h − 1) =( hg −− 11) = .
−−11 and hence (eg,,) ≤ ( f h) as required. Lemma 10. If ( ps,,) ≤ ( eg) and (qt,,) ≤ ( f h) such that the composition ( ps,,)( qt) and (eg,,)( f h) are defined, then ( ps,,)( qt) ≤ ( eg ,)( f , h)
for all ( ps,,,) ( eg) ,,,( qt) ( f , h)∈ B ( P) . Proof. Suppose that ( ps,,)( qt) = ( pu ,) and (eg,,)( f h) = ( ek ,) are defined
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Then we have −1 pe≤=and sggp⋅ . and −1 q≤= fand thhq⋅ . and we have the following
−1 where uststst=q = sp⋅ =
−1 where kghghgh=f = ge⋅ = . Now
−−11 −−− 111 ustghgh= = ( gp⋅⋅)( hq) = ( gp ⋅⋅⋅)( hsp)
−−−−−−111111 = ( ghgp⋅)( hgp ⋅⋅) = ( gh gp ⋅)( hgp ⋅) and
−1 −−11 −1⋅gh⋅ p −−11⋅hh⋅⋅( g p) −−11⋅⋅−1 −− 11 ⋅ kkp= gh( ) = ghhgp = g hhgp= ( g gp)( h hgp). −1 Then uk= kp⋅ . Moreover, pe≤ . Thus ( pu,,) ≤ ( ek) as required. −1 ∈ ≤ gf⋅ Lemma 11. If (eg, ) B ( P) and ( f ,1) is an identity such that ( fe,1) ( ,1) , then ( fg, ) is the restriction of (eg, ) to (e,1) .
Proof. Suppose (eg, )∈ B ( P) and ( f ,1) is an identity such that ( fe,1) ≤ ( ,1) , since −1 −1 f −1 gf⋅= −− f = f gf⋅ ∈ ( g) gg( ) , then ( fg,.) B ( P) Also
−−11−1 − 1 ( fg,,gf⋅)( fg gf ⋅) = ( fg ,gf⋅−)( g f⋅ fg , − f)
− f ⋅ −−11gf = −−ff ⋅⋅ − f − f= fg( ) g f, ( g) g( f,1) .
−1 Moreover, ( f,, ggf⋅ ) ≤ ( eg) and unique by definition.
−1 Thus ( fg, gf⋅ ) is the restriction of (eg, ) to (e,1) .
Proposition 15. (BP ( ),≤) is an inductive groupoid. Proof. We prove that (OG1,) ( OG 2) and (OG3) hold. By Lemma 9 we have (OG1,) by Lemma 10
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(OG2) holds and by Lemma 11 (OG3) holds. Since the partially ordered set of identities forms a meet semi-
lattice EB( ( P)) ≅ E. Thus (BP ( ),≤) is an inductive groupoid. Theorem 7. If PE= G, where E is a semilattice and G is a group, then
e B P=∈= eg,: P g−−11 g ( ) {( ) ( ) } is an inverse semigroup with multiplication defined by
−−11 (eg,,)( f h) =( e( g ⋅ f) , gf h hg⋅ e) .
−1 −1 −1 Proof. Let (eg,,,) ( f h)∈ B ( P) since (eg,) ( eg ,) =( g ⋅ e ,1) and ( fh,)( fh ,) = ( f ,1) , we form the pseudoproduct (eg,,) ⊗ ( f h) using the greatest lower pound. −−11 =(egegfhfh,,,,) ( ) ∧( )( ) =( ge−1 ⋅∧,1) ( f ,1) =( ge −− 11 ⋅ ,1)( f ,1) =(( gef ⋅) ,1) and − ge1⋅ hg−−11⋅⋅ e h − 1 f ( )( ) −−11 ( ( fh, )) =⋅=⋅((( g ef) ,1) ( fh , )) ( g efh) , and −1 −1 gg⋅⋅−1 ef −−11 − 1 − 1( ) ((eg,,) ) =⋅=⋅( ( g eg)) ( g e) f ,( g ) −1 −1⋅ − ( ge) −1 gg⋅⋅1 e g ⋅ f eg( ⋅ f) −−11( ) −−11 =g⋅= ef,, g g⋅ ef g ( ) ( ) (( ) ( ) ) −−11 egf( ⋅⋅) egf( ) − −− =g1 ⋅⋅ g 11 ef,. g ( ) ( ) ( )
−−11 egf( ⋅⋅) ( gf) −−e 1 −−11 Now, since (eg, )∈ B ( P) then gg= and so ( gg) = ( ) . We have −1 ( gf⋅ ) − ( gg1 ) = f . Then ((eg,) ) =⋅⋅( gff( g−1 e) f,, g ) Therefore (eg,,,) ⊗=( f h) (( eg) )( ( f , h))
− ge1⋅ hg−−11⋅⋅ e h − 1 f ( )( ) ff−−11 =⋅⋅( g( g efg) ,,)( g ⋅ efh) − ge1⋅ hg−−11⋅⋅ e h − 1 f g−1⋅ ef ( )( ) ff−1 ( ) =( g ⋅⋅( g ef) ),( g) h
h−−11⋅⋅ g ef ff−1 ( ) =⋅⋅g( g e) fgh,.
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−−11 −1 Now, we have in the ordering defined on BP ( ), ( gefge⋅) ≤⋅ and ge⋅ acts trivially on g this ( g−1⋅ ef) implies that gg= f . Then we have
−1 −( g⋅ ef) −− f ⋅⋅=1 ⋅⋅=⋅⋅ 11 g( g ef) g( g ef) g( g ef)
−1 =⋅( gg−⋅11 ⋅ e)(( gge ⋅ f)) =( gge − ⋅)( gf ⋅=) egf( ⋅) and f −1 −− 11 h−−11⋅⋅( gef) h−1⋅ fge( − 1 ⋅) (h ⋅ f)( h) ⋅⋅( ge) ghfff= gh = gh
−−11⋅⋅ −1 h( ge) −−11 = gf( h h⋅⋅ f) = ghf hg e. Therefore
−−11 (eg,,)( f h) =( e( g ⋅ f) , gf h hg⋅ e) .
Thus BP ( ) is an inverse semigroup. We summarize the main results of this paper in the following: 1) We characterize Green’s relations ( and ) of the Zappa-Szép product MG of a monoid M and a m group G we prove that (mg,,)( nh) ⇔ m n in M. And If ( gm−−11⋅⋅)( hn) such that ( gg−−11) = and −−n (hh11) = in G, then (mg,,)( nh) in MG . 2) We prove that the internal Zappa-Szép product S of subsemigroups A and B is an enlargement of a local submonoid eSe for some e∈ RI( A) LI( B), and eSe is the internal Zappa-Szép product of the submonoids A and B where A==∈==∈{ xxeaaAB:,} ,{ yybebB :,.} And M is the internal Zappa-Szép product of a left-zero semigroup A and a right-zero semigroup B if and only if M is a rectangular band. 3) We give the necessary and sufficient conditions for the internal Zappa-Szép product MAB= of re- gular subsemigroups A and B to again be regular. We prove that MAB= is regular if and only if ef∈ Reg( M ) where e∈ EA( ) and f∈ EB( ). 4) The Zappa-szep products PG= G of the nilpotent group G with left and right conjugation action of G on it self is abelian if and only if G is abelian, in which case P= GG × ; and if G is non-abelian and hence nilpotent of class 2, then P is also nilpotent of class 2. 5) The Zappa-Szép product of inverse semigroups need not in general be an inverse semigroup. In this paper we give the necessary conditions for their existence and we modified the definition of semidirect products in the inverse case to obtain what we termed λ -semidirect products. The λ -semidirect product of inverse semi- groups is again inverse. We construct from the Zappa-Szép product P of a semilattice E and a group G, an inverse semigroup by constructing an inductive groupoid.
References [1] Zappa, G. (1940) Sulla construzione dei gruppi prodotto di due dadi sottogruppi permutabili tra loro. Atti del secondo congresso dell’Unione Matematica Italiana, Cremonese, Roma, 119-125. [2] Lawson, M. (2008) A Correspondence between a Class of Monoids and Self-Similar Group Actions I. Semigroup Fo- rum, 76, 489-517. http://dx.doi.org/10.1007/s00233-008-9052-x [3] Brownlowe, N., Ramagge, J., Robertson, D. and Whittaker, M. (2014) Zappa-Szép Products of Semigroups and Their C*-Algebras. Journal of Functional Analysis, 266, 3937-3967. http://dx.doi.org/10.1016/j.jfa.2013.12.025 [4] Gould, V. and Zenab, R. (2013) Semigroups with Inverse Skeletons and Zappa-Szép Products. CGASA, 1, 59-89. [5] Gilbert, N.D. and Wazzan, S. (2008) Zappa-Szép Products of Bands and Groups. Semigroup Forum, 77, 438-455. http://dx.doi.org/10.1007/s00233-008-9065-5 [6] Wazzan, S.A. (2008) The Zappa-Szép Product of Semigroups. PhD Thesis, Heriot-Watt University, Edinburgh. [7] Lawson, M.V. (1998) Inverse Semigroups. World Scientific, Singapore City. [8] Lawson, M.V. (1996) Enlargements of Regular Semigroups. Proceedings of the Edinburgh Mathematical Society (Se-
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ries 2), 39, 425-460. http://dx.doi.org/10.1017/S001309150002321X [9] Lawson, M.V. and Marki, L. (2000) Enlargement and Covering by Rees Matrix Semigroups. Springer-Verlag, Berlin, 191-195. [10] Kunze, M. (1983) Zappa Products. Acta Mathematica Hungarica, 41, 225-239. http://dx.doi.org/10.1007/BF01961311 [11] Nico, W.R. (1983) On the Regularity of Semidirect Products. Journal of Algebra, 80, 29-36. http://dx.doi.org/10.1016/0021-8693(83)90015-7 [12] Hall, T.E. (1982) Some Properties of local Subsemigroups Inherited by Larger Subsemigroups. Semigroup Forum, 25, 35-49. http://dx.doi.org/10.1007/BF02573586 [13] Fiedorowicz, Z. and Loday, J.L. (1991) Crossed Simplicial Groups and Their Associated Homology. Transactions of the American Mathematical Society, 326, 57-87. http://dx.doi.org/10.1090/S0002-9947-1991-0998125-4 [14] Kassel, C. (1995) Quantum Groups. Graduate Texts in Mathematics, No. 155, Springer, New York. [15] Billhardt, B. (1992) On a Wreath Product Embedding and Idempotent Pure Congruences on Inverse Semigroups. Se- migroup Forum, 45, 45-54. http://dx.doi.org/10.1007/BF03025748
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