_{1}

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In [1], a new consequence of the (restricted) wreath product for arbitrary monoids A and B with an underlying set
_{}. Let us denote it by
. Actually, in the same reference, it has been also defined the generating and relator sets for
, and then proved some finite and infinite cases about it. In this paper, by considering the product, we show Green’s relations L and R as well as we present the conditions for this product to be left cancellative, orthodox and finally left (right) inverse(s).

Let A and B be arbitrary monoids. In [

We recall the fundamentals of the construction of A ⋇ B which will be needed to form our results. We note that this product is based on the wreath product and we may refer to ( [

( f , b ) ( g , b ′ ) = ( f g b , b b ′ ) , (1)

where, g b : B → A is given by

( y ) b g = ( y b ) g ( y ∈ B ) . (2)

Dually the restricted wreath product of the monoid B by the monoid A is the set B ⊕ A × A with the multiplication defined by

( f , a ) ( g , a ′ ) = ( f g a , a a ′ ) , (3)

where, g a : A → B is given by

( x ) g a = ( a x ) g ( x ∈ A ) . (4)

Now for P 1 = ( a 1 , b 1 ) , P 2 = ( a 2 , b 2 ) ∈ A × B , let us define

P 1 P 2 = ( a 1 a 2 , b 1 b 2 ) .

After that the new derivation for the wreath product of A and B, denoted by A ⋇ B , is the set A ⊕ B × ( A × B ) × B ⊕ A with the multiplication

( f , P 1 , g ) ( h , P 2 , k ) = ( f h b 1 , P 1 P 2 , g a 2 k ) , (5)

where, h b 1 : B → A and g a 2 : A → B are defined by

( y ) b 1 h = ( y b 1 ) h ( y ∈ B ) (6)

and

( x ) g a 2 = ( a 2 x ) g ( x ∈ A ) . (7)

In fact, A ⋇ B is a monoid with the identity ( 1 ¯ , ( 1 A ,1 B ) , 1 ˜ ) , where 1 ¯ and 1 ˜ are defined by

( b ) 1 ¯ = 1 A , ( a ) 1 ˜ = 1 B , (8)

respectively, for all b ∈ B and a ∈ A .

In the light of Green’s relations, it is well known that one may prove some computational results (for example, the minimal number of generators etc.) on the monoid structure (which will be kept for a future work and so not investigated in here). Hence, in this section, we only characterize Green’s relations L and R (cf. [

Proposition 2.1 Let A ⋇ B be the new derivation of wreath product of a monoid A by a monoid B. Then

1) ( f 1 , P 1 , g 1 ) L ( h 1 , P 2 , k 1 ) in A ⋇ B implies that g 1 L k 1 in B ⊕ A , and P 1 L P 2 in A × B ,

2) ( f 1 , P 1 , g 1 ) R ( h 1 , P 2 , k 1 ) in A ⋇ B implies that f 1 R h 1 in A ⊕ B , and P 1 R P 2 in A × B .

Proof. 1) Suppose that ( f 1 , P 1 , g 1 ) L ( h 1 , P 2 , k 1 ) in A ⋇ B . So there exist ( f 2 , P 3 , g 2 ) , ( h 2 , P 4 , k 2 ) ∈ A ⋇ B such that

( f 2 , P 3 , g 2 ) ( f 1 , P 1 , g 1 ) = ( h 1 , P 2 , k 1 ) , (9)

( h 2 , P 4 , k 2 ) ( h 1 , P 2 , k 1 ) = ( f 1 , P 1 , g 1 ) . (10)

These two equations can also be written as

( f 2 ( f b 3 1 ) , P 3 P 1 , g 2 a 1 g 1 ) = ( h 1 , P 2 , k 1 ) , (11)

( h 2 ( h b 4 1 ) , P 4 P 2 , k 2 a 2 k 1 ) = ( f 1 , P 1 , g 1 ) . (12)

Hence, by the equality of components, we obtain

f 2 ( f b 3 1 ) = h 1 , : a 3 a 1 = a 2 , : b 3 b 1 = b 2 and g 2 a 1 g 1 = k 1 , (13)

h 2 ( h b 4 1 ) = f 1 , : a 4 a 2 = a 1 , : b 4 b 2 = b 1 and k 2 a 2 k 1 = g 1 . (14)

It follows that g 1 L k 1 in B ⊕ A while P 1 L P 2 in A × B .

Similar proof can be applied for 2). Hence the result. +

Theorem 2.2 Assume that the product A ⋇ B is obtained by a monoid A and a group B. Then

( f 1 , P 1 , g 1 ) R ( h 1 , P 2 , k 1 ) ∈ A ⋇ B ⇔ f 1 R h 1 ∈ A ⊕ B and P 1 R P 2 ∈ A × B .

Proof. By Proposition 2.1, ( f 1 , P 1 , g 1 ) R ( h 1 , P 2 , k 1 ) ∈ A ⋇ B implies the existence of f 1 R h 1 ∈ A ⊕ B and P 1 R P 2 ∈ A × B .

To prove the converse, let us suppose that f 1 R h 1 in A ⊕ B and P 1 R P 2 in A × B . In fact, f 1 R h 1 in A ⊕ B gives that there exist l 1 and l 2 in A ⊕ B such that

f 1 l 1 = h 1 and h 1 l 2 = f 1 .

Also, P 1 R P 2 in A × B implies that there exist ( c , d ) , ( c ′ , d ′ ) in A × B such that

( a 1 , b 1 ) ( c , d ) = ( a 2 , b 2 ) and ( a 2 , b 2 ) ( c ′ , d ′ ) = ( a 1 , b 1 ) .

To show that ( f 1 , P 1 , g 1 ) R ( h 1 , P 2 , k 1 ) in A ⋇ B , we have to find two elements ( f 2 , P 3 , g 2 ) and ( h 2 , P 4 , k 2 ) ∈ A ⋇ B such that these must satisfy

( f 1 , P 1 , g 1 ) ( f 2 , P 3 , g 2 ) = ( h 1 , P 2 , k 1 ) , (15)

( h 1 , P 2 , k 1 ) ( h 2 , P 4 , k 2 ) = ( f 1 , P 1 , g 1 ) . (16)

From these above, we obtain

f 1 ( f b 1 1 ) = h 1 , P 1 P 3 = P 2 , g 1 a 3 g 2 = k 1 , (17)

h 1 ( h b 2 2 ) = f 1 , P 2 P 4 = P 1 , k 1 a 4 k 2 = g 1 . (18)

Since A ⊕ B is a group (because B is a group), we have

g 1 a 3 g 2 = k 1 ⇒ g 2 = ( g 1 a 3 ) − 1 k 1 , (19)

k 1 a 4 k 2 = g 1 ⇒ k 2 = ( k 1 a 4 ) − 1 g 1 . (20)

Therefore, we set f 2 = l ( b 1 ) − 1 1 and h 2 = l ( b 2 ) − 1 2 . Hence

( f 1 , P 1 , g 1 ) ( f 2 , P 3 , g 2 ) = ( f 1 , P 1 , g 1 ) ( l ( b 1 ) − 1 1 , P 3 , ( g 1 a 3 ) − 1 k 1 ) = ( f 1 ( l ( b 1 ) − 1 1 ) b 1 , P 1 P 3 , g 1 a 3 ( g 1 a 3 ) − 1 k 1 ) = ( f 1 l 1 B 1 , P 1 P 3 , k 1 ) = ( h 1 , P 2 , k 1 ) .

With a similar way, one can also show that

( h 1 , P 2 , k 1 ) ( h 2 , P 4 , k 2 ) = ( f 1 , P 1 , g 1 ) .

Hence, ( f 1 , P 1 , g 1 ) R ( h 1 , P 2 , k 1 ) , as required. +

In this section, we will illustrate some algebraic properties of the new wreath product A ⋇ B in terms of the algebraic properties of the monoids A and B themselves. The following Theorem characterize when new wreath product M = A ⋇ B is a group.

Theorem 3.1 The new derivation of wreath product M = A ⋇ B of monoids A and B is a group if and only if both A and B are groups.

Proof. Suppose A and B are both, groups, then M = A ⋇ B is a monoid with identity ( 1 ¯ , ( 1 A ,1 B ) , 1 ˜ ) where 1 ¯ and 1 ˜ are defined by

( b ) 1 ¯ = 1 A , ( a ) 1 ˜ = 1 B . (21)

Now, let ( f , ( a , b ) , g ) ∈ M . Define

f ′ : B → A by ( y ) f ′ = ( ( y b − 1 ) f ) − 1 such that f ′ ∈ A ⊕ B , and y , b ∈ B ,

g ′ : A → B by ( x ) g ′ = ( ( a − 1 x ) g ) − 1 such that g ′ ∈ B ⊕ A , and x , a ∈ A .

Then

( f , ( a , b ) , g ) ( f ′ , ( a − 1 , b − 1 ) , g ′ ) = ( f b f ′ , ( a a − 1 , b b − 1 ) , g a − 1 g ′ ) = ( 1 ¯ , ( 1 A ,1 B ) , 1 ˜ )

Since ( y ) f f ′ b = ( y ) f ( y b ) f ′ = ( y ) f ( ( y b b − 1 ) f ) − 1 = ( y ) f ( ( y ) f ) − 1 = 1 A , and ( x ) g a − 1 g ′ = ( x ) g a − 1 ( x ) g ′ = ( a − 1 x ) g ( ( a − 1 x ) g ) − 1 = 1 B . Hence ( f ′ , ( a − 1 , b − 1 ) , g ′ ) is a right inverse for ( f , ( a , b ) , g ) . Also

( f ′ , ( a − 1 , b − 1 ) , g ′ ) ( f , ( a , b ) , g ) = ( f ′ f b , ( a − 1 a , b − 1 b ) , g ′ a g ) = ( 1 ¯ , ( 1 A , 1 B ) , 1 ˜ )

Since ( y ) f ′ f b − 1 = ( y ) f ′ ( y b − 1 ) f = ( ( y b − 1 ) f ) − 1 ( y b − 1 ) f = 1 A , and ( x ) g ′ a g = ( x ) g ′ a ( x ) g = ( a x ) g ′ ( ( x ) g ) − 1 = ( ( a − 1 a x ) g ) − 1 ( x ) g = ( ( x ) g ) − 1 ( x ) g = 1 B . Hence, ( f ′ , ( a − 1 , b − 1 ) , g ′ ) is a left inverse for ( f , ( a , b ) , g ) , therefore, M is a group.

Conversely, assume that M = A ⋇ B is a group, let ( f , ( 1 A , 1 B ) , 1 ˜ ) − 1 = ( k , ( 1 A , 1 B ) , l ) so

( f , ( 1 A , 1 B ) , 1 ˜ ) ( k , ( 1 A , 1 B ) , l ) = ( f k 1 B , ( 1 A , 1 B ) , 1 ˜ a l ) = ( 1 ¯ , ( 1 A , 1 B ) , 1 ˜ )

and

( k , ( 1 A , 1 B ) , l ) ( f , ( 1 A , 1 B ) , 1 ˜ ) = ( k f 1 B , ( 1 A , 1 B ) , l a 1 ˜ ) = ( 1 ¯ , ( 1 A , 1 B ) , 1 ˜ )

Then f k 1 B = 1 ¯ , 1 ˜ a l = 1 ˜ , k f 1 B = 1 ¯ , and l a 1 ˜ = 1 ˜ . Since

( b ) f k 1 B = ( b ) 1 ¯ ⇒ ( b ) f ( b ) k 1 B = 1 A ⇒ ( b ) f ( b 1 B ) k = 1 A ⇒ ( b ) f k = 1 A and

( b ) k f 1 B = ( b ) 1 ¯ ⇒ ( b ) k ( b ) f 1 B = ( b ) 1 ¯ ⇒ ( b ) k ( b 1 B ) f = 1 A ⇒ ( b ) k f = 1 A .

Hence, k = f − 1 ∈ A ⊕ B , therefore, A ⊕ B is a group and hence B is a group. Similarly we get l = g − 1 ∈ B ⊕ A , if we suppose that ( 1 ¯ , ( 1 A , 1 B ) , g ) − 1 = ( k , ( 1 A , 1 B ) , l ) , therefore B ⊕ A is a group and hence A is a group. +

We first recall that a semigroup S is called left-cancellative if c a = c b ⇒ a = b and right-cancellative if a c = b c ⇒ a = b , for all a , b , c ∈ S (cf. [

Theorem 3.2 A and B are cancellative monoids if and only if M = A ⋇ B is cancellative monoid.

Proof. Assume that A and B are left cancellative monoids. Suppose

( f , P 1 , g ) ( k , P 2 , l ) ( h , P 3 , j ) ∈ A ⋇ B ,

where f , k , h ∈ A ⊕ B , g , l , j ∈ B ⊕ A and P 1 , P 2 , P 3 ∈ A × B . Therefore

( f , P 1 , g ) ( h , P 3 , j ) = ( k , P 2 , l ) ( h , P 3 , j )

⇒ ( f h b 1 , P 1 P 3 , g a 3 j ) = ( k h b 2 , P 2 P 3 , l a 3 j )

⇒ f h b 1 = k h b 2 , P 1 P 3 = P 2 P 3 , g a 3 j = l a 3 j

⇒ ( ∀ b ∈ B ) ( b ) f h b 1 = ( b ) k h b 2

⇒ ( ∀ b ∈ B ) ( b ) f ( b b 1 ) h = ( b ) k ( b b 2 ) h ∧ b 1 = b 2

[Since B is left cancellative]

⇒ f = k ∧ b 1 = b 2 .

Also

P 1 P 3 = P 2 P 3 and g a 3 j = l a 3 j

⇒ ( ∀ a ∈ A ) ( a ) g a 3 j = ( a ) l a 3 j

⇒ ( ∀ a ∈ A ) ( a ) g a 3 ( a ) j = ( a ) l a 3 ( a ) j

⇒ ( ∀ a ∈ A ) ( a 3 a ) g ( a ) j = ( a 3 a ) l ( a ) j

[Since A is left cancellative]

⇒ ( ∀ a ∈ A ) ( a ) g = ( a ) l

[Since B is left cancellative]

⇒ g = l ∧ a 1 = a 2 .

As a result, M = A ⋇ B is actually a right cancellative monoid. In fact, one may prove with a similar way for left cancellative. Hence M = A ⋇ B is cancellative.

On the other hand, the converse part of the proof is clear.

Hence the result. +

In [

Recall that the semigroup S is called orthodox if the set of idempotents E ( S ) is a subsemigroup of S. An orthodox semigroup S is left (respectively, right) inverse if e g e = g e (respectively, e g e = e g ) for every e , g ∈ E ( S ) . For more details reader refer [

Theorem 3.3 If A ⋇ B is an orthodox monoid or left (right) inverse, then each of A and B has the same property.

Proof. Applying [ [^{2} = k. Since A ⋇ B is an orthodox monoid, then for an element ( f , ( a , b ) , h ) , ( g , ( c , d ) , k ) ∈ E ( A ⋇ B ) , we have

( f , ( a , b ) , h ) , ( g , ( c , d ) , k ) = ( f g b , ( a c , b d ) , h c k ) ∈ E ( A ⋇ B )

⇒ ( f g b , ( a c , b d ) , h c k ) ( f g b , ( a c , b d ) , h c k ) = ( f g b , ( a c , b d ) , h c k )

⇒ ( ( f g b ) f b d g b , ( ( a c ) 2 , ( b d ) 2 ) , ( h c k ) a c ( h c k ) ) = ( f g b , ( a c , b d ) , h c k )

⇒ ( f g b ) f b d g b = f g b , ( h c k ) a c ( h c k ) = h c k , ( a c ) 2 = a c , ( b d ) 2 = b d

⇒ ( f g ) 2 = f g in A ⊕ B and ( h k ) 2 = h k in B ⊕ A .

Furthermore

( x ) ( f g b ) b d f g b = ( x ) f g b

⇔ ( b d x ) ( f g b ) ( x ) f ( x ) b g = ( x ) f ( x ) b g

⇔ ( b d x ) f ( b d x ) b g ( x ) f ( x b ) g = ( x ) f ( x b ) g

⇔ ( b d x ) f ( b d x b ) g ( x ) f ( x b ) g = ( x ) f ( x b ) g

⇔ ( f g ) 2 = f g for every x ∈ B .

Similar, calculation shows that ( h k ) 2 = h k , for every y ∈ A . Hence the set of idempotents of A ⊕ B and B ⊕ A are subsemigroups. The result follows from [ [

Now let us suppose that A ⋇ B is the left inverse. Then, for any element ( f , ( a , b ) , h ) , ( g , ( c , d ) , k ) ∈ E ( A ⋇ B ) , where f , g ∈ E ( A ⊕ B ) and h , k ∈ E ( B ⊕ A ) , we certainly have

( f , ( a , b ) , h ) ( g , ( c , d ) , k ) ( f , ( a , b ) , h ) = ( g , ( c , d ) , k ) ( f , ( a , b ) , h )

⇒ ( f g b f b d , ( a c a , b d b ) , h c a k a h ) = ( g f d , ( c a , d b ) , k a h )

⇒ f g b f b d = g f d , h c a k a h = k a h , a c a = c a , b d b = d b

⇒ f g f = g h , h k h = k h

and

( x ) f g b f b d = ( x ) g f d

⇔ ( x ) f ( x ) b g ( x ) b d f = ( x ) g ( x ) d f

⇔ ( x ) f ( x b ) g ( x b d ) f = ( x ) g ( x d ) f .

Hence, f g f = g h for every x ∈ B . Similar, calculation shows that h k h = k h for every y ∈ A . We thus conclude that A and B are actually left inverses.

The same proof can be applied to show right inverse case as well. +

Note 3.4 1) The other inclusion of Theorem 3.3 is left for future work. Following Caito [

2) There is also a particular class of regular monoids, namely coregular monoids [

In this paper, the author investigated some specific theories such as Green’s relations, left cancellative, orthodox, left (right) inverse etc. over new type of wreath products over monoids. Of course, there are still so many different properties that can be checked on this important product. On the other hand, in Note 3.4, we indicated some problems for future studies.

The authors declare no conflicts of interest regarding the publication of this paper.

Wazzan, S.A. (2019) New Properties over a New Type of Wreath Products on Monoids. Advances in Pure Mathematics, 9, 629-636. https://doi.org/10.4236/apm.2019.98032