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 can be written uniquely as with and Then since we have with and determined uniquely by a and b. We write and Associativity in S implies that the functions and 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 defined by and a map defined by 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 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, and the -semigroup.

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 such that lie in. 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 -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 -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 -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 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] .

Let S be a semigroup with subsemigroups A and B such that each element is uniquely expressible in the form with and We say that S is the “internal” Zappa-Szép product of A and B, and write Since with and, we must have unique elements and so that This defines two functions and Since with and, we must have unique elements and so that Write and This defines two function and Thus Using these definitions, we have for all and that

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

Now

and

Thus, by uniqueness property, we have the following two properties

Similarly by the associativity of S, we have

Now

and

Thus, by uniqueness property, we have the following two properties

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

Proof. Consider Then since we have for all for all Thus x is a right identity for A and left identity for B, whereupon. Observe that thus W

Of course, if S is a monoid and A and B are submonoids then

Proposition 1. If the internal Zappa-Szép product of A and B, then

Proof. We use Brin’s ideas in [8] Lemma 3.4. If then for unique and giving us a function and likewise, if then for some unique and some function But for all we have and therefore and: that is is a left identity for B. Similarly, and is a right identity for A. In particular, and are idempotents. Now

Therefore

Similarly

Therefore

Set and for any in (1):

Hence is constant: for all Similarly, setting and in (2):

Hence is constant: for all But now we have that for all and

and But then putting and we have and in particular W

Lemma 2. Let the internal Zappa-Szép product of A and B and be a right identity for A and a left identity for B. Then and

Proof. We have then, but thus and by uniqueness we have

Similarly, since we have. Also Thus Hence and W

In an internal Zappa-Szép product we find an idempotent This shows (for example) that a free semigroup cannot be a Zappa-Szép product. But in a monoid Zappa-Szép product of submonoids A and B the special idempotent must be, since we have uniquely. Then for all and thus and Similarly and

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 and.

Example 1. [9] Let S be any semigroup and let I be a set of idempotents in S such that. Then S is an enlargement of ISI because and If and then S is an enlargement of the local submonoid.

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 and eSe is the internal Zappa-Szép product of the sub-

monoids and where

Proof. We have such that e is a right identity for A and a left identity for B. Then so for So S is an enlargement of the local submonoid (is a monoid with identity e). It is clear that and are submonoids of. We must show that each element is uniquely expressible as with If then But for unique and so where

Since and this expression is unique, because There- fore each element is uniquely expressible as with W

We note that if such that T is an enlargement of, where and if are regular with the assumption that if then and if then Then A, B are regular, since if is regular monoid, then for each there exists such that Now which implies Since A has a right identity e, then. Similarly we get. Thus A is regular. Similarly, we get B is regular.

Following [9] we describe the Rees Matrix cover for the Zappa-Szép product such that T = TeT and is an enlargement of for some idempotent where such that Ae = A and eB = B and S is the Zappa-Szép product of and with and Then by Corollary 4 in [9] the Rees matrix semigroup is given by

such that since For each, we

can find and such that So if then and such that Similarly, for therefore and Now for each

fix elements and define matrix P by putting

Thus is the Rees matrix cover for where the map

defined by is the covering map (is a strict local isomorphism from M to T along which idempotents can be lifted).

In this Section we give some general properties of the Zappa-Szép product. We characterize Green’s relations (and) of the Zappa-Szép product of a monoid M and a group G.

Proposition 3. [10] Let be a Zappa-Szép product of semigroups A and B. Then

(i) in B;

(ii) in A.

Proof. Suppose in then there exist such that

and. Then

and

Hence

It follows that in B. Similar proof for (ii). W

Proposition 4. In the Zappa-Szép product of a monoid M and a group G. Then

Proof. By Proposition 3 we have implies that in M. To prove the converse suppose that in M then there exist z₁ and z₂ in M such that and To show that we have to find and in such that

Then and Hence

Therefore we set Hence

Similarly Hence W

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 be a Clifford semigroup with the following multiplication table. Note that and

Let, the group of integers. Suppose that the action of on A for each is as follows:

Observe that for all. The action of A on as follows:

Thus Zappa-Szép axioms are satisfied, since define by

is a morphism (this is easy to see from the fact that). Now define (where is the group of permutations on) by

where for all. Clearly is a morphism (of groups). Now for and we de- fine the action arises from the composition as follows

and

We therefore have and as following.

For and for

For and using for all we have

For

and

Thus the Zappa-Szép product of A and B. The set

since and if and only if and

since or for all so or

Now, we note that acts non-trivially. We have and in A but not -related to where since if we suppose then there exist and in such that

But so not -related to To calculate the -class of if

then and so or we prove is impossible so If

for some then we have

But so -related only to itself. Similarly -related only to itself. To calculate the -class of suppose then or Let so then

and so which implies or so Thus for all By similar calculation we have So if acts non-trivially we have a different structure for the -classes of and A.

Proposition 5. Let be the semidirect product of a monoid M and a group G. Then

Proof. Suppose that then there exist in such that

Then and Hence there- fore

which implies

and

which implies

Thus by (1) and (2) we have in M.

Now suppose in G then there exist and in G such that and

. Therefore Hence

Therefore we set in other formula by symmetry so W

Proposition 6. If such that and in G, then

in

Proof. Suppose then there exist and in M such that which implies that

We set and then

Similarly Hence in W

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 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 be a commutative monoid with 0, each of whose elements is idempotent and such that Let be a monoid with two left zeros a and b. Then both S and T are regular semigroups. Let 1 acts trivially, There is no such that for all Thus the semidirect product is not regular. For example, is not a regular element of R.

Example 4. Take such that and such that Let act trivially on B, and act trivially on A, Then A and B are regular monoids but their Zappa-Szép product is not regular.

However, there are criteria we can prove that the internal Zappa-Szép product of regular A and B is regular as the following Propositions illustrated.

Proposition 7. If A is a regular monoid, B is a group, for all then is regular. W

Proof. Let where and we have to find such that

. Now and so we choose

where Since we must have but B is a group, so

Suppose we are given c, and choose since B is a group, so and

then Since A is regular, we choose any and set

thus Thus

whereupon is regular. W

Proposition 8. Let A be a left zero semigroup and B be a regular semigroup. Suppose that for all, there exists some such that and for all there exists some such that Then is regular.

Proof. Let where and We have to find such that

. Now and we choose

where Now since A is a left zero semigroup

and by our assumptions we can choose such that

and that is fixed by a. Then Thus is regular. W

Theorem 1. [12] For any arbitrary semigroup S, is a subsemigroup of S if and only if the product 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 of regular semigroups A and B. Then each is uniquely a product of regular elements: where Hence M is regular if and only if 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 and, as our next theorem shows.

Theorem 2. Let A and B be regular subsemigroups and and Then if and only if is regular.

Proof. Given with (uniquely) where since A and B are regular subsemi- groups, then there exist and Then and Set and Then by the assumption and the set

is not empty, see [14] . Because let, and let Then

and

and so Also

and so Also

Thus Also because

we have

Then

and

and so. Then is a regular element which implies that is regular.

Conversely, if 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 where and Thus if and this implies that then W

Corollary 1. If A and B are regular and act trivially, then is regular.

Proof. If we take and, then is an idempotent in M. Because

since act trivially. Therefore Hence

is regular. W

In this case: if, we can find First find and

since idempotents of A commute with those of B. Then for some

Thus where and

Then

Now we discuss inverse Zappa-Szép products. Let S and T be inverse semigroups/monoids with and let 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 be the commutative monoid with one non zero-identity idempotent a. Let be the commutative monoid with zero an with Then S and T are both inverse monoids, and there is a homomorphism given by and Then is regular. However, the element has two inverses, namely and 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 of two inverse semigroups S and T will be inverse if and only if acts trivially.

In the general case of the Zappa-Szép product of inverse semigroups we have also P need not be inverse semigroup as we can see from the following example.

Example 6. Let where and―Klein 4-group where Suppose that the action of G on E is defined by:

and

and acts trivially (that is each of permutes non-trivially). The action of E on G is defined by:

and

We check Zappa-Szép axioms by the following: define by

This is a homomorphism of groups since is the automorphism group of E. We have an action of G on E using: if and define

We have a homomorphism given by and The action of E on G is given by:

We note that for all For we have for

Since it is clear that holds. For we have for

and

So holds and for

and

Thus holds. Then. Since every element of M is regular, then M is regular. is a closed subsemigroup of M, so is orthodox, but since is not commutative for example while, 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. is an inverse semigroup if

(i) S and T are inverse semigroups;

(ii) and act trivially;

(iii) For each where and, then s and t act trivially on each other.

Proof. We know that is regular. Since a regular semigroup is inverse if and only if its idem- potents commutes, it suffices to show that idempotents of commute. If are idem- potents of, then

Thus

and

By (iii) a and t act trivially on each other, b and u act trivially on each other, then

But since S and T are inverse semigroup, then idempotents commutes that is

Then but t and c are idempotents they are act trivially then

Thus is inverse. W

Let A and B be semigroups, and suppose that we are given functions and where and satisfying the Zappa-Szép rules and Then the set with the product defined by: is a semigroup, the external Zappa-Szép product of A and B, which is written as

If A and B are semigroups that both have zero elements (and respectively), and we have in addition to and for all and the following rule:

Then by Proposition [10] we have that S is a semigroup with zero But from the following example we deduce that is not a necessary condition:

Example 7. If A and B are semigroups with and respectively, acting trivially on each other. Then (ZS5) is not satisfied. However, in the Zappa-Szép product we have

and

Thus is zero for

From the following example we deduce that the zeros 0_A and 0_B of A and B respectively do not necessarily give a zero for the external Zappa-Szép product

Example 8. If A is a monoid with identity and zero and B is a semigroup with zero such that the action of A on B is trivial action and the action of A on B is for all Then Zappa-Szép rules are satisfied. But (ZS5) is not satisfied since and is not a zero for

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 completes the square

From the horizontal composition we get and 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] .

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 where for all and we have and so that the multiplication in P is

Then for each the subset is

a subgroup of P isomorphic to (with identity). However P cannot be an internal Zappa-Szép pro- duct of subsemigroups Z, N isomorphic to and respectively: If generates N, then the second coordinate of every non-identity element of N is q, and so the second coordinate of any product with and 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 and respectively such that the following is satisfied

for all and

So if, the external Zappa-Szép product of A and B, then each A and B are embedded in

Define by. We claim is an injective homomorphism since

and. Thus τ_M is a homo-

morphism, also is injective since Denote its image by. Define by, then is also an injective homomorphism. Denote its image by. Observe that Thus. This decomposition is evidently unique. Thus is the internal Zappa-Szép product of and.

・ 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 and where are fixed elements of A and B respectively. Note that in a left-zero semi- group A, and in a right-zero semigroup 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, let

and where are fixed elements. Then

uniquely and and So is the internal Zappa-Szép product of and, , where (as left-zero semi- group) and (as right-zero semigroup).

Conversely, Let where A is left-zero semigroup and B is right-zero semigroup. Then for all and for all M is a rectangular band if for all then where and for unique and Now

Thus M is a rectangular band. W

1) Let be a Clifford semigroup. Note that and Let the group of integers. Suppose that the action of on A for each is as follows: The action of A on as follows: Then the Zappa-Szép multiplication is associative. Thus the Zappa-Szép product of A and B. The set of idempotents of M is the empty set, since if and only if and since or for all so or

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 since if we define and with we obtain the multi- plication Then 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 where and―Klein 4-group, where Suppose that the action of G on E is defined by: and and acts trivially. The action of on is defined by: and Then is the Zappa-Szép product of E and G. Since every element of M is regular, then M is regular. is a closed subsemigroup of M, so is orthodox, but since is not commutative for example while, then M is not inverse.

4) For groups, is the Zappa-Szép product of subgroups A and B if and only if, since for any we have for unique and This implies that for unique and Thus But this is not true in general for semigroups or monoids. Let be a commutative monoid with one non-identity idempotent a. Let be a com- mutative monoid with two idempotents e and f and Let B act trivially on A and act trivially

on B and Then and Then

is the internal Zappa-Szép product of and. But, since

so can not be written as. Moreover, so

is not a submonoid of M.

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 being contained in the center of G. Now, let G be a group and let G act on itself by left and right conjugation as follows:

In the following we show that these actions let us form a Zappa-Szép product 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.

Proof. Let G act on itself by left and right conjugation as follows:

where Thus the multiplication is given by:

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 for all For and clear they are hold.

:

:

since G is nilpotent of class £ 2,then

Thus holds.

:

:

since G is nilpotent of class 2, then

Thus holds. Hence is the Zappa-Szép product. W

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 holds,

then for all we have. Thus

Therefore Hence is central in G. Similarly if holds. W

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 is

Proof. Suppose Since for all we have and

Then

and

Put in (1): then for all Therefore So Put in (2): then for all Therefore So So

since if Then

W

Lemma 4. If 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 This implies and so P is abelian.

If P is abelian then but. Thus if and only if Hence G is abelian group. In which case W

Proposition 12. If P is the non-abelian Zappa-Szép product 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 we have that is the commutator elements are central. Let be a commutator in P. We prove it is central in P. We have

Now

and

Write Then

Since are commutators, then

This implies that Thus So commutators in P are in the center ( and P is not abelian) so P is nilpotent of class 2. W

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 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

2) If G is non-abelian and hence nilpotent of class 2, then P is also nilpotent of class 2.

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 -semidirect products. The -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 we have

Note that if for all then holds, since by cancellation in the group G.

We consider the following where E a semilattice and G a group, and subset of the Zappa-Szép product:

We form a groupoid from the action of the group G on the set E which has the following features:

・ vertex set:;

・ arrow set:;

・ an arrow starts at finishes at

・ the inverse of the arrow is

・ the identity arrow at e is and

・ an arrows and are composable if and only if, in which case the composite arrow

is

Lemma 5. If then

Proof. we have for all then Then W

Lemma 6. Suppose that then is a regular element and

Proof. We have

and

Thus is a regular. W

Proposition 13. If where E is a semilattice and G is a group then

with composition defined by

if and acts trivially on E is a groupoid.

Proof.

We have to prove

&

Now

Since implies that this implies that therefore

Then

& starts at

But starts at and so

& ends at

But ends at

Thus is a groupoid. W

Now we introduce an ordering on. The ordering is giving as follows:

Lemma 7. The ordering on defined by

is transitive.

Proof. We have to prove that if then We have

and and and Since

Thus We conclude and

Now, implies that and

Thus £ is transitive. W

Lemma 8. The ordering on defined by

is antisymmetric.

Proof. We have to prove if and, then Now

and and and Thus and

Thus £ is antisymmetric. W

Proposition 14. with ordering defined by

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 is a partial order set. W

Next we prove that is an ordered groupoid.

Lemma 9. If, then for all

Proof. Suppose that so that and Now, we have

Thus

and

Therefore Also

and hence as required. W

Lemma 10. If and such that the composition and are defined, then

for all

Proof. Suppose that and are defined

Then we have

and

and we have the following

where

where Now

and

Then Moreover, Thus as required. W

Lemma 11. If and is an identity such that then is the

restriction of to

Proof. Suppose and is an identity such that since

then Also

Moreover, and unique by definition.