Theorem: is a semigroup if and only if for any

Further Results in the Theory of Generalised Inflations of Semigroups R.A.R. Monzo Abstract We determine the structure of semigroups that satisfy xyzw ∈ {xy, xw, zy, zw} . These semigroups are precisely those whose power semigroup is a generalised inflation of a band. The structure of generalised inflations of the following types of semigroups is determined: the direct product of a group and a band, a completely simple semigroup and a free semigroup F(X) on a set X. In the latter case the semigroup must be an inflation of F(X) . We also prove that in any semigroup that equals its square, the power semigroup is a generalised inflation of a band if and only if it is an inflation of a band.

INTRODUCTION In this paper we obtain further results in the theory of generalised inflations of semigroups. In Section 1 we give preliminary definitions and notation. We also list some results regarding the "gross" structure of bands and, in particular, the "fine" structure of royal bands and almost commutative bands. This notation and these results will be used throughout the paper. For a given semilattice Y and a given (disjoint) collection {Sα : α ∈Y } of rectangular bands, Result 1.1 provides an explicit construction for all associative products ∗ on the disjoint union S = ∪ Sα

(α ∈Y )

such that {S , ∗} is a semilattice Y of rectangular bands Sα , (α ∈Y ) . Such products are in 1-1 correspondence with “compatible” collections of (idempotent) mappings on S and each mapping idempotent homomorphisms

θ a ( a ∈ Sα ) in the collection is a union of

θ a / Sβ : S β → Sαβ ( β ∈Y ) . Since the form of all homomorphisms from one rectangular

band into another is known and since every band is a semilattice of rectangular bands, Result 1.1 provides a method of constructing all bands. In Section 2 we extend the results of Clarke and Monzo [1] to the broader version of generalised inflations, as defined in this paper. Theorem 2.19 gives insight into the structure of generalised inflations of an arbitrary semigroup. Conditions for a semigroup U to have a symmetric generalised inflation that is not an inflation of U follow. These conditions are necessary and sufficient when the semigroup is a generalised inflation of a union of groups. Theorem 2.19 also aids in the construction of an example of a point inflation of a semigroup U that is not an inflation of U . It is shown in Proposition 2.10 that a point inflation S of a generalised inflation U of a union of groups is an inflation of U and, in Proposition 2.11, that a point inflation of a Brandt semigroup U is an inflation of U . We also note the results of Wang and Wismath [15] that a symmetric generalised inflation of a Brandt semigroup U is an inflation of U and that any semigroup whose square is a Brandt semigroup is a point generalised inflation (of its square). Section 3 concerns the construction of semigroups from the (known) structure of their square. We give a new proof of Tamura's Theorem, which states that if the square of a semigroup is a semilattice of groups then the semigroup is an inflation (of its square). In the remainder of the section we determine the structure of semigroups whose square is any of the following: a direct product of a group and a band, a completely simple semigroup or a free semigroup F(X) on a set X. In the latter case the semigroup is an inflation of F(X). In Section 4 we explore the concept of an inclusion class of semigroups, a concept that generalises the concept of a variety of semigroups. We review and extend the results of Redei [13], Giraldes and Howie [3], Pondelicek [12] and Pelikan [10]. In this section it is shown that the collection of all semigroups whose power semigroup is a generalised inflation of a band is an inclusion class. The same is true for the collection of all semigroups whose power semigroup is an inflation of a band and the collection of all semigroups whose power semigroup is a band. In all three cases a construction for such semigroups is determined. We consider the concept of inclusion class a fruitful notion and, in fact, Theorem 4.9 links the concepts of inclusion class, power semigroup, generalised inflation and almost commutative band. Other examples of inclusion classes that determine the following interesting collections of semigroups are given: the union of all left-zero and all right-zero semigroups; all semigroups in which every subsemigroup is a left ideal; all semigroups that are an inflation of a tree; and all chains C of semigroups Sα (α ∈ C) , where each Sα is a rectangular band or a group of order two, with xy = yx = x whenever x ∈ Sα , y ∈ S β and trivial unless

α ≥ β , ∀β ∈ C.

α≺β

and with any group Sα

2 1. Preliminary definitions and results

S is a semigroup in which x = x 2 for every x ∈ S (that is, every element of S is idempotent). We write x ≤ y if x = xy = yx and we write x ≺ y if x ≤ y and x ≠ y. A chain is a semigroup S in which x ≤ y or y ≤ x for every { x, y} ⊆ S . A semilattice is a band S in which xy = yx for every { x, y} ⊆ S . A semigroup S is a band B of semigroups

A band

Sα (α ∈ B ) if S is a disjoint union of the semigroups Sα , indexed by the band B , and Sα S β ⊆ Sαβ for every {α , β } ⊆ B .

B = BS .

In such a case we will write

A semigroup

S is a left-zero [right-zero] semigroup if xy = x [ xy = y ] for every

{ x, y} ⊆ S . We define S ∈ L0 [R0 ] to mean that S is a left-zero [right-zero] semigroup. A semigroup S is called a rectangular band if S = L × R ( L ∈ L0 , R ∈ R0 ) , in which case we will write x = ( x1 , x2 ) ∈ S ∈RB . Note that a band S ∈RB if and only if xyz = xz for every { x, y, z} ⊆ S . Note that S ∈RB if and only if for every x ∈ S , xyx = x. Let n ∈ N = {1, 2, ...} . Then L n [ R n ] denotes the left-zero [right-zero] semigroup on the set {1, 2,..., n} . A Brandt semigroup S = {0} ∪ { gij : g ∈ G, and {i, j} ⊆ I } where G is a group and I a set and where multiplication is gij hkl = ( gh)il if j = k and 0 otherwise. A completely simple semigroup (or a Rees I × Λ matrix semigroup S over G with sandwich matrix P )

(

consists of S = G × I × Λ , where G is a group and I and Λ are sets, with multiplication ( g , i, λ )( h, j , σ ) = gPλ j h, i,

σ)

where P= {Pλi : λ ∈ Λ, i ∈ I} is a Λ × I matrix of the elements Pλi ∈ G . It is well-known that a semigroup that is a union of groups is a semilattice of completely simple semigroups (cf. [2], Theorem 4.6). Henceforth, if S is a union of groups we may •

∪ Cα ,

therefore write S =

{α , β } ⊆ Y

where

Y

α ∈Y

is a semilattice, Cα is a completely simple semigroup and Cα Cβ ⊆ Cαβ for all

.

It is also well-known that any band

S is a semilattice Y S of rectangular bands Sα (α ∈Y S ) . (See, for example, McLean [6]). For a

a to be Sa = Sα where a ∈ Sα and α ∈Y S . Note that if Sa = Sb and ab = ba then a = b . For any semigroup S with a ∈ S we define Ka to be the constant map x a (for any x ∈ S ) and Γa to be the map x axa (for any x ∈ S ). All mappings will be written on the left, unless otherwise stated. We will write ≅ [ ≅ ] to denote a semigroup or groupoid isomorphism [anti-isomorphism]. The straightforward proof of Result 1.1 is omitted.

band S , a ∈ S , we define the rectangular band component Sa of

Result 1.1 Let S =

•

∪ Sα (α ∈Y ) be a union of the pairwise disjoint rectangular bands Sα , indexed by a given semilattice Y .

Suppose that for every

α ∈Y

(b) for every β ∈Y ,

θ a/S β : S β → Sαβ

and every a ∈ Sα there exists a mapping

θa : S → S

is a homomorphism and (c) for every

such that : (a)

{β , γ } ⊆ Y

θ a/Sα = Ka

and every

( b,c ) ∈ Sβ × Sγ ,

θ aθ b c = ⎡⎣θ(θ b )(θ a ) c ⎤⎦ ⎡⎣θ(θ a )(θ b ) c ⎤⎦ . a

Then define 2)

b

b

a

S is a semilattice Y of the rectangular bands Sα (α ∈Y ) with multiplication *K S , as follows. For a ∈ Sα and b ∈ S β we

θ aθ a c = (θ a a c )(θ aa c ) ,

5) if

(

)

a *K S b ≡ (θ a b ) (θ b a ) . Then for any {α , β , γ } ⊆ Y and any a,a, b,c ∈ Sα × Sα × S β × Sγ ,

αβγ = αβ then θ a b = θ aθ b c ,

Also, any semigroup (Note: We call

3) 6)

θ aθ b c = θθ bθθ a c , a

4)

b

a *K S b *K S c = (θaθ b c )(θcθ b a )

and

1) θ a = (θ a ) ,

θ aθ b = (θ aθ b )

2

2

,

7) θ a b = a *K S b *K S a .

S that is a semilattice Y = Y S of the rectangular bands Sα (α ∈Y ) can be expressed in the above manner.

K S = {θ a : a ∈ S } a compatible collection of mappings on S with respect to Y .)

We define

M ( S ,Y

) as the collection of all associative multiplications * on S such that {S , ∗} is a semilattice YS = Y Sα (α ∈Y ) . We define Ω ( S ,Y ) to be the collection of all compatible collections of mappings on

rectangular bands

3 of the

S with

respect to Y . Using Result 1.1 it is straightforward to prove the following. Result 1.2 Define a mapping Φ :

Ω ( S ,Y ) → M ( S ,Y

)

as follows: Φ

( K S ) = *K

S

{α , β } ⊆ Y and every

, where for every

a ∈ Sα , b ∈ S β we define a *K S b ≡ (θa b ) (θ b a ) . Then Φ is well-defined and is a bijection. Theorem 1.3 Let

K S ∈ Ω ( S ,Y ) . Then {K S ,

Proof: ( ⇒ ) Suppose that such that

{K S , } is a semigroup. Then for any {α , β } ⊆ Y

θ a θb = θ x . Suppose that x ∈ Sσ

αβδ = σδ . If δ = β

then

αβ = σβ

and x = θ x ⎡⎣(θ a b )(θ b a ) ⎤⎦ = (θ a

and if

for some σ ∈Y . Then for any

δ =σ

then

αβσ = σ

and any {a, b, c} ∈ Sα × S β × Sγ , (θ a

Corollary 1.4 Proof:

( ⇒ ) If

δ ∈Y , c ∈ Sδ , (θ a θb ) c = θ x c σ = αβσ = ααβ = αβ

and so

. Hence x ∈ Sσ = Sαβ

θb ) θ c = θθ b θ c = θθθ a

ab

c

(θb θ c ) .

= θ(θa θb )c = θθa (θb c ) = θ a θθb c = θ a

■

{K S , } is a semigroup then θ a (θb θ a ) = θ a θθ a = θθ θ a = θθ b = θ a θb . b

θa

a b

a

(θb θ a ) = θ a θb .Then for any c ∈ S , θ a (θbθ a c ) 1 = θ a (θ(θ a )(θ b)c ) 1 = θθ ⎡⎣(θ a )(θ b)⎤⎦ c 1 = θθ b c 1 . b

b

a

and any {a, b} ⊆ Sα × S β there exists x ∈ S

a

(θ aθb )θ a c 2 = θ(θ a )(θ b)θ a c 2 = θθ ⎣⎡(θ a )(θ b)⎦⎤ ⎡θ(

θθ b = θ a

a

{K S , } is a semigroup if and only if for any {a, b} ⊆ S , θ a θb = θ a θb θ a .

( ⇐ ) Assume that Also,

θb = θθ b .

θ a θb = θθ b . Then for any {α , β , γ } ⊆ Y

{K S , } is a semigroup.

Thus,

and so

for any {a, b} ⊆ S , θ a

θb ) ⎡⎣(θ a b )(θb a ) ⎤⎦ = θ a ⎡⎣(θbθ a b ) (θb2 a ) ⎤⎦ = θ aθb a = θ a b.

( ⇐ ) Suppose that for any {a, b} ⊆ S ,

So

} is a semigroup if and only if

a

a

b

a

⎣

⎤

θba )(θ ab ) a ⎦

a

a

b

a

a

c = θθab(θaθb a ) c = θθab c . 2

2

2

(θb θ a ) = θ a θb and therefore {K S , } is a semigroup.

Corollary 1.5 Let

K S ∈ Ω ( S ,Y )

{

. Then S , ∗K S

■

} ≅ {K , } [ {S , ∗ } ≅ {K , } ] implies {S , ∗ } is a semilattice Y KS

S

←

left-zero [right- zero] semigroups. Conversely, for any semigroup {S , •} that is a semilattice semigroups there exists ∗K S ∈ M ( S ,Y

) such that {S , •} ≅ {S , ∗K

S

KS

S

Y

of

of left- zero [right- zero]

} ≅ {K , } [ {S , •} ≅ {S , ∗ } ≅ {K , } ]. S

KS

←

S

Lemma 1.6 A band is regular if and only if it satisfies the identity xyzyx = xyxzxyx . Proof:

( ⇒ ) Suppose that a band satisfies the identity xyzx = xyxzx . Then,

xyzyx = x[ y ( zy )]x = xyx( zy ) x = x[( yxz ) y ] x = x( yxz ) xyx = ( xyx) z ( xyx) . ( ⇐ ) Suppose that a band satisfies the identity xyzyx = xyxzxyx. Then,

xyxzx = x( yxz )3 x = x( yxz ) x( yxz ) x( yxz ) x = ( xyxzxyx)( xzxyxzx) = ( xyzyx )( xzyzx ) = ( xyz ) 2 yxxz ( yzx) 2 = = ( xyzx) yzyxzyz ( xyzx). But ( xyzx) and ( yzyxzyz ) are in the same rectangular band component and so xyxzx = xyzx.

■

Corollary 1.7

{K S , } is a semigroup if and only if

Proof: Define ∗ ≡ ∗K s . For any

{α , β , γ } ⊆ Y

{S , ∗ } is a regular band if and only if {K , } is a full groupoid. KS

4

S

and any {a, b, c} ⊆ Sα × S β × Sγ , by 7 ) of Result 1.1 we have

θ a b = a ∗ b ∗ a . Clearly, θ a θb = θ a θb θ a if and only if a ∗ b ∗ c ∗ b ∗ a = a ∗ b ∗ a ∗ c ∗ a ∗ b ∗ a . Therefore by Lemma 1.6 ■ {K S , } is a semigroup iff {S , ∗} is regular. The remainder of the proof follows from Theorem 1.3. The following types of bands will arise later in this paper. An almost commutative band is a band in which any two elements are either in the same rectangular band component or they commute. A royal band is a band that is not generated by any proper subset of

S if α ≠ αβ ≠ β for some {α , β } ⊆ Y S then Sαβ = 1 . For if g ∈ Sαβ then

itself. Note that in any almost commutative band

g commutes with any e ∈ Sα and any f ∈ S β . But ef ∈ Sαβ and so g = gefg = efgef = ef . Result 1.8 (Hall [4], Theorem 1.) Let semilattice Y and such that Sα = 1 if

{Sα : α ∈Y }

α = βγ

be a collection of pairwise disjoint rectangular bands indexed by the

for some

β , γ ∈Y

and

β ≠α ≠γ;

only element of Sα . Let multiplication in each Sα be denoted by juxtaposition. Put S =

if this is the case then we let eα denote the •

∪ Sα (α ∈Y ) and define a multiplication

∗ on S as follows: for any α , β ∈Y and any a ∈ Sα , b ∈ Sβ , define a ∗ b = a , if α ≺ β ; a ∗ b = ab , if α = β ; a ∗ b = b , if

β ≺α

and a ∗ b = eαβ , if

α ≠ αβ ≠ β

.

Then ( S , ∗) is an almost commutative band. Conversely any almost commutative band can be obtained in this way. Result 1.9 (Giraldes and Howie [3], Theorem 2.2) A semigroup S is a royal band if and only if ab ∈ {a,b} for all

(α ∈Y ) , where each α ≺ β (α , β ∈Y ) .

if and only if S is a chain Y of semigroups Sα

ab = ba = a whenever a ∈ Sα , b ∈ Sβ and Result 1.10 (Monzo [9], Lemma 1) If S

⎛

t

⎝

i =1

⎞

n

{a,b} ⊆ S

Sα is either a left-zero or right-zero semigroup and where

is a band for some positive integer

n then for any positive integer t and for any

n

∈ ( S n ) t where ( a i ) ∈ ( S n ) . αi ∏αi ⎠ i =1

{a1 , a 2 ,..., a t } ⊆ S , ⎜ ∏ a i ⎟

n

2. Generalised inflations and their properties

Definition 2.1. A semigroup

S is called an inflation of its subsemigroup U if S is a disjoint union of the sets Sa ( a ∈U ) , where:

1) a ∈ U implies a ∈ Sa

and

2) x ∈ Sa and y ∈ S b imply xy = ab .

S is an inflation of its subsemigroup U if and only if S 2 ⊆ U and there is an idempotent homomorphism θ which maps S onto U .

Definition 2.2. Equivalently,

S is an inflation of (its subsemigroup) U if and only if there is a mapping θ of S onto U such that θ /U = idU and such that xy = ( xθ )( yθ ) , for any x , y ∈ S . Definition 2.3. Alternatively,

With this definition of inflation in mind we proceed to define a generalised inflation preliminaries.

S of (its subsemigroup) U . First, here are some

5

Notation. Let U be a semigroup. Then FR (U ) ⎡⎣ FL (U ) ⎤⎦ denotes the collection of all mappings on U written on the right [left]. For a ∈ U we consider the constant mapping

Ka

on U to be an element of both

FR (U ) and FL (U ) .

Definition 2.4. A semigroup

S is said to be a generalised inflation of (its subsemigroup) U if S is a disjoint union of the sets 1) a ∈ U implies a ∈ Sa , 2) for any x ∈ S there exists an α x ∈ FR (U ) and β x ∈ FL (U )

Sa ( a ∈U ) where

such that x ∈ Sa and y ∈ S b imply xy = bα x β y a

and

3) a ∈ U implies

α a = Ka = β a .

α x and β x are called the left and right substitution functions for x respectively. If for every x ∈ S , and every a ∈ U , aα x = β x a then S is called a symmetric generalised inflation of U . In this case α x is called the substitution function for x . A symmetric generalised inflation S of U in which aα x = bα x for any x ∈ S and any a,b ∈ U is called a constant The mappings

S 2 = U 2 and U ∪ {x} is an inflation [ generalised inflation ; symmetric generalised inflation ]of U for every x ∈ S then S is called a point inflation [ point generalised inflation ; symmetric point generalised inflation] of U .

generalised inflation of U . When

The following Result is well-known and will be applied in Section 4. Result 2.5 A semigroup is an inflation of a band if and only if it satisfies the equation xy = x y if and only if it satisfies the 2

2

equations xy = xy = x y . 2

2

Note: The definition of generalised inflation given above is more general still than that given in [1]. In fact, Definition 1 of [1] fits the definition of a symmetric generalised inflation given above. The proofs of Propositions 2.6-2.9 are straightforward and are omitted. Proposition 2.6 If Proposition 2.7

S 2 has an identity element then S is an inflation of S 2 .

S is an inflation of U if and only if S is a constant generalised inflation of U .

S is a generalised inflation of U then S n +1 is isomorphic to U n +1 for every positive integer n . In fact for every positive integer n .

Proposition 2.8 If

S n +1 = U n +1

Proposition 2.9 If

S is a [generalised] inflation of U and if U ⊆ T ⊆ S then T is a [generalised] inflation of U .

Notice that Proposition 2.9 implies that an inflation of U is a point inflation of U . The converse is not true, as will be shown below. However, we do have the following result. Proposition 2.10 Let U be a generalised inflation of its subsemigroup V , where V is a union of groups. Suppose that inflation of U . Then S is an inflation of U . Proof: For any

S is a point

x ∈ S − U we let a x denote the element of U which satisfies xa = a x a and

ax = aa x , for every a ∈ U . ( Such an element exists because S is a point inflation of U .) If x ∈ U then we let a x denote x . Then for any { x, y} ⊆ S , xy and

a x a y are in S 2 = U 2 = V 2 = V . But V is a union of groups and so ( xy ) and ( a x a y )

exist in V . Therefore, xy = xyxy

( xy )

a x a y ( a x a y ) xy xy ( xy ) −1

−1

−1

−1

−1

= a x a y xy ( xy ) = a x a y ( a x a y ) a x a y xy ( xy ) = −1

−1

−1

= a x a y ( a x a y ) xy = a x a y ( a x a y ) a x a y = a x a y . Now S is the disjoint union −1

−1

∪ Su ( u ∈ U ) , where Su = { x ∈ S : a x = u} and u ∈ Su ( u ∈U ) . Also, if x ∈ Su and y ∈ Sv then xy = a x a y = uv , so S is ■ an inflation of U .

6 Proposition 2.11 A point inflation S of a Brandt semigroup U is an inflation of U .

{ }

Proof: Let { x, y} ⊆ S − U . Then there exists x, y ⊆ U such that , for any u ∈ U , xu = xu , ux = u x , yu = yu and

uy = u y. We need to show that xy = x y. If xy = 0 then x y = gij 1 jj = x y1 jj = x y1 jj = xy1 jj = 0. Therefore xy = 0 implies x y = 0. If xy = gij ≠ 0 then xy = gij = gij 1 jj = xy1 jj = x y1 jj = x y1 jj. Since xy ≠ 0 , y = hlj for some {l , j} ⊆ I . Then

■

xy = xhlj 1 jj = xhlj = x y . The proofs of Theorems 2.12 to 2.15 are similar to the proofs of Theorems 1,2,3 and 5 in [1] and are therefore omitted. Theorem 2.12 If

S is a generalised inflation of U and T is an inflation of S then T is a generalised inflation of U .

Theorem 2.13 If S is a symmetric generalised inflation of a commutative semigroup U then S is commutative. Theorem 2.14 A generalised inflation of a cancellative semigroup U is an inflation of U . 2

Theorem 2.15 (Clarke & Monzo [1], 5) If S is a union of groups then

S is a symmetric generalised inflation of S 2 .

Theorem 2.16 (Wang & Wismath [15], 3.1) A symmetric generalised inflation of a Brandt semigroup U is an inflation of U . 2

2

Theorem 2.17 (Wang & Wismath [15], p.19) If S is a Brandt semigroup then S is a point generalised inflation of S . 2

In [15] Wang and Wismath give an example of a semigroup T in which T is a Brandt semigroup but T is not an inflation of

T 2 . It follows that T is not a symmetric generalised inflation of T 2 .

Ψ ∈ FL (U ) ⎡⎣ FR (U )⎤⎦ is said to be a left [right] translation of U if for any u, v ∈U , Ψ ( uv ) = ( Ψu ) v ⎡⎣( uv ) Ψ = u ( vΨ )⎤⎦ . Then LT (U ) ⎡⎣RT (U ) ⎤⎦ will denote the collection of all left [right] translations of U . The proofs of Theorem 2.19 and Corollary 2.20 are omitted.

Definition 2.18 A mapping

Theorem 2.19 Let

S = ∪ Sa ( a ∈U ) be a disjoint union, where U is a semigroup and a ∈ U implies a ∈ Sa .

Suppose that for each

x ∈ S there exist α x ∈ FR (U ) and β x ∈ FL (U ) such that

1) the mappings Ψ x : a

aα x a and Φ x : a

aβ x a satisfy Ψ x ∈ LT (U ) and Φ x ∈ RT (U ) ,

x ∈ S , and for every a,b ∈ U , aΨ x b = aΦ x b , 3) a ∈ U implies α a = Ka = β a , and 4) for any a,b,c ∈ U with x ∈ Sa , y ∈ S b and z ∈ Sc we have

2) for each

Ψ x Ψ y = Ψ bα x β ya , Φ x Φ y = Φ bα x β y a

and

bα x ⎡⎣( β y a ) Φ z ⎤⎦ = ⎡⎣ Ψ x ( cα y )⎤⎦ β z b .

S as follows: for a,b ∈ U with x ∈ Sa , y ∈ Sb define x * y = bα x β y a . Then {S ,*} is a semigroup and, therefore, S is a generalised inflation of U . Conversely, any generalised inflation of U can be constructed in this

Define a multiplication * on manner.

x ∈ S and any c ∈ U , cα x = β x c then {S ,*} is a symmetric generalised inflation of U . Conversely, any symmetric generalised inflation of U can be constructed in this manner. Also, if for any

Note: In Theorem 2.19, if

U 3 = {0} then the conditions 1) , 2) and 4) are satisfied for any mappings α x

and β x , ( x ∈ S ) . This fact aids in the construction of the semigroups in the Examples 2.21 and 2.22 below.

7 Corollary 2.20: A semigroup U has a symmetric generalised inflation that is not a point inflation of U if there exists

α ∈ FR (U )

( ab )α ab = aα ab , ab ⎡⎣( ab )α ⎤⎦ = ab ( bα ) and a ( aα ) b = a ( bα ) b ; and 2 2 2 2 2) for any a ∈ U , a ( aα ) ( aα ) = a ( cα ) and ( aα ) ( aα ) a = ( cα ) a , 3) 2 2 4) for any a ∈ U such that a = ( cα ) there exists an element v ∈ U such that

such that for any a,b ∈ U 1)

there exists an element c ∈ U such that

( cα ) either

2

( cα ) α v ( vα ) ≠ va 2

and

commute and or

( vα ) v

≠ av . If U is a generalised inflation of a union of groups and S is a symmetric generalised

inflation of U that is not a point inflation of U then there exists Example 2.21 Let Then

α ∈ FR (U )

that satisfies 1) through 4).

S = {a, b, c, 0, x} be a groupoid with all products equal to 0 except ca = ac = ax = b . Let U = {a, b, c, 0} .

S is a generalised inflation of U that is not a symmetric generalised inflation of U .

S = U ∪ { x, y} where U is as in Example 1, and where all products in S are equal to 0, except ca = ac = xy = yx = b . Then S is a point inflation of U , but S is not an inflation of U .

Example 2.22 Let

3. Constructing semigroups from their square

U is a semigroup of a particular known structure, then how can we construct all semigroups S for which S = U ? Tamura [14] showed that if U is a semilattice of groups then any semigroup S for 2 which S = U is an inflation of U . We give a new and direct proof of Tamura's result. First we give some definitions.

In this section we explore the following question: given that 2

Notation 3.1 Suppose that U is a band element of Gα and u will denote

B

of groups. If u ∈ Gα

α . If U = G × B

for some

α∈B

then 1u or 1α will denote the identity

where G is a group and B is a band, then for any u ∈ U we let G u and Bu

denote the group and band components of u in G and B respectively. In this case B ≅ B and we can identify Bu with u . This notation, used in the proof of the following well-known Lemma, is also used throughout the remainder of this Section. Lemma 3.2 In a semilattice Y of groups idempotents commute. Proof: For any so

α , β ∈Y

,

{1 1

α β

, 1β 1α } ⊆ Gαβ and so 1α1β = 1α1β (1β1α )

−1

(1 1 ) = 1 1 1 β α

α β α

. Similarly,

1α1β = 1β1α1β and ■

1α1β = 1β1α1β = 1β1α .( Note: This implies 1α1β = 1αβ .)

Theorem 3.3 (Tamura, [14]) If

S 2 is a semilattice of groups then S is an inflation of S 2 .

•

Proof : Clearly,

{

}

S = ∪ Sa ( a ∈ S 2 ) , where for any a ∈ S 2 , Sa = s ∈ S : s1s2 = a . We’ll show that for any a,b ∈ S 2 , given

s ∈ Sa and t ∈ Sb , st = ab . Let a ∈ ( S 2 ) and b ∈ ( S 2 ) . Then a = s1s2 = 1s2 s = a1s2 and so γ = γα , where 1s2 ∈ ( S 2 ) . γ σ α Then

s2 = (1s2 s )( s1s2 ) = a 2 ∈ ( S 2 ) and so α = γ and {a, 1s2 , s 2 } ⊆ ( S 2 ) . Similarly, {b, 1t 2 , t 2 } ⊆ ( S 2 ) . Now α γ σ

ab = 1s2 st1t 2 ∈ ( S 2 )

γσ

because

and by Lemma 3.2,

ab = 1s2 1t 2 (1s2 st1t 2 )1s2 1t 2 = 1s2 1t 2 st1s2 1t 2 .We need only show that st ∈ ( S 2 ) , γσ

1s21t 2 is the identity of the group ( S 2 ) . Assume that st ∈ ( S 2 ) . Then ( st ) = ( st ) = sts tst = ( sts ) γσ ξ 3

2

( tst )

2

=

= st s 2 ts ts t 2 st and so ξ = ξγσ . Then s2 t 2 = (1s s2 )( t 21t ) = (1s s )( st )( t1t ) and so γσ = γξσ = ξ . So ab = st

■

8

Definition 3.4 A semigroup S is idempotent inversive if for each a ∈ S there exists x ∈ S such that ax is idempotent.

Definition 3.5 A semigroup S is rectangular if for any {a ,b,x,y} ⊆ S , whenever three elements of the set {ax,bx,ay ,by} are equal, all four are equal. Yamada [16] showed that a semigroup is idempotent inversive and rectangular if and only if it is an inflation of a rectangular group. In [1], Theorem 19, a method is given to construct all (symmetric) generalised inflations of a given rectangular group U = G × E , where G is a given group and E is a given rectangular band. We proceed to generalise Theorem 19 of [1], in Theorem 3.10 below. Definition 3.6 A mapping H ∈ FL (U ) is called a- fixed if an almost a- endomorphism on

Ha = a , transitive if for any {a,b} ⊆ U , a ( Ha ) b = a ( Hb ) b ,

U if H is a- fixed and for any {b,c} ⊆ U , H ( bc ) = H ( b ) bcabc H ( c ) , and an a- mapping if

H is a transitive, almost a- endomorphism on U . In the latter case we will write Η = H a is an a- mapping. Definition 3.7 A collection H is called a 1- compatible collection of mappings on

U if every H ∈ H is an a- mapping, for some

a ∈U , and if for any {H a , H b } ⊆ H and any c ∈ U , H a H b c = ( H a b )( H b a ) c ( H b a )( H a b ) .

Lemma 3.8 Suppose that S = 2

•

∪ Cα α Y

is a union of groups,

x ∈ S , x 2 ∈ Cα and u ∈ Cβ . Then (1): { xu, ux} ⊆ Cαβ .

∈

u and v are in the same subgroup of C β then (2): x ( ux )

Also, if

Proof: We have

( xu )

−1

( ux ) and x ( vx ) ( vx ) are in the same subgroup of −1

Cαβ .

ux 2u = ( ux )( xu ) ∈ Cαβ and so ( xux ) = ( xu ) ( x 2 ) ( ux ) ∈ Cαβ . Therefore xux ∈ Cαβ . Then 2

= ( xux ) u ∈ Cαβ and so xu ∈ Cαβ . Also, ( ux ) = u ( xux ) ∈ Cαβ and so ux ∈ Cαβ . So { xu, ux} ⊆ Cαβ and (1) holds.

2

2

It follows that

{x (ux )

−1

( ux ) , x ( vx ) ( vx )} ⊆ Cαβ . Note that x ( vx ) ( vx ) = x ( vx )( vx ) −1

−1

= ( xv )

−1

( xv ) x ( vx ) ( vx −1 ) =

−1 −1 −2 −1 uv −1vx = x ( ux ) uv −1v ⎡ x ( vx ) ( vx ) ⎤ = x ( ux )( ux ) u ⎡ x ( vx ) ( vx ) ⎤ = ⎣ ⎦ ⎣ ⎦ −2 −1 −1 −1 −1 −1 − 1 = xvv ux ( ux ) u ⎡ x ( vx ) ( vx ) ⎤ = ⎡( xv ) ( xv ) x ⎤ ⎡vv ( ux ) u ⎤ ⎡ x ( vx ) ( vx ) ⎤ . It then follows from (1) that the last ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦

= ( xv )

−1

( xv ) x . So x ( ux ) ( ux ) = x ( ux )

−1

−1

−1

three terms in square brackets are all in Cαβ . Then it follows from the definition of multiplication in Cαβ that x ( ux )

x ( vx )

−1

( vx ) are in the same subgroup of

( ux )

and

■

Cαβ and so (2) holds.

Lemma 3.9 Let S be a generalised inflation of

U , where U is a band B of groups. For each a ∈U and each

x ∈ Sa = { x ∈ S : 1x2 x = a = x1x2 } define Φa,x : B → B as follows: for any e ∈ B , Φ a,x e ≡ x ( ux )

element of

−1

−1

( ux ) , where

u is any

U such that u = e . Then each Φ a,x is a well-defined a − mapping and each pair Φ a,x and Φ b,y are

1- compatible. Proof:

S 2 is a union of groups it follows from Lemma 3.8 that each Φ a,x is well-defined.

Since

Φ a,x a = x ( ax )

−1

( ax ) = x1x

= a , so Φ a,x is a - fixed. Now for any {b, c} ⊆ U and any x ∈ Sa , b ( xbx ) c = ( bx ) c = 2

2

= ( bx ) c = ( bx ) c = bxc = b ( xc ) = b ( xc ) = b ( xcx ) c . Therefore (a) b ( xbx ) c = b ( xcx ) c . 2

2

(

Now xbx = ( xbx ) = ( xb ) x bx 3

x ( bx )

−1

2

2

) ( bx ) = xb x bx

( bx ) = x ( bx )( bx ) ( bx ) = xb x ( bx )

xcx = x ( cx )

−2

−1

2

−2

2

bx and it follows from Lemma 3.8 that xbx = xb bx . Also,

bx and again by Lemma 3.8, x ( bx )

−1

( bx ) = xb bx . Similarly,

( cx ) . It then follows from (a) that b ( Φ a,x b ) c = b ( Φ a,x c ) c and so Φ a,x is transitive.

Then

( )

( )

Φ a,x bc = Φ a,x bc = xbcx = xb cx = ( xb )

Now

Proposition 7 of [9] that Φ a,x is an

( cx )

2

2

(

) (

= xbx bc xcx = Φ a,x b bc Φ a,x c

)

9 and it follows by

a -mapping.

Finally, suppose that y ∈ S b . In the proof of Theorem 5, [1], it was shown that xy ∈ xy S ∩ Sx y and it follows that 2

2

xy 2 xy x 2 y = xy . Then, Φ a,x Φ b,y c = xycyx = xy c yx = xy 2 xy x 2 y c yx 2 yx y 2 x = xbx yay c yay xbx and so

(

)(

)(

)(

Φ a,x Φ b,y c = Φ a,x b Φ b,y a c Φ b,y a Φ a,x b

)

■

which implies that Φ a,x and Φ b,y are 1-compatible.

Theorem 3.10 Let U = G × B where G is a given group and B is a given band. Let

X = {S a }a∈U be a collection of mutually disjoint sets such that

2) for any

a ∈U , if there is no h ∈ G such that h 2 = Ga , then Sa = {a} . Let Y = {Φa,x : a ∈U , x ∈ Sa } be a collection

of 1-compatible mappings on B , where each Let

1) a ∈ U implies a ∈ Sa and

Z = { f a,x : U

Φ a,x is a Ba − mapping on B and such that

} ⊆ F (U ) satisfy

U such that a ∈U , x ∈ Sa

(

4) for any a ∈ U ,

L

5) for any a ∈ U and any x ∈ S a − {a} , G a = G fa,x a

) and 2

6) for any

3) for any a ∈ U ,

{a,b} ⊆ U and any

Φ a,a = Γ Ba .

f a,a = Ka ,

(

)

x ∈ Sa , f a,x b = G fa,x a , Φ a,x Bb .

•

Define

* on S = ∪ Sa ( a ∈U ) as follows: for any {a,b} ⊆ U and any ( x, y ) ∈ Sa × S b , x * y = ( f a,x b ) ( f b,y a ) . Then and for any a ∈ U , S a = {a}

{S ,*} is a generalised inflation of U

∪ { x ∈ S : x*x = a} .

U can be formed in this manner.

Conversely, every generalised inflation of

Proof of Theorem 3.10 ( ⇒ ) We show that

* is associative. Let a,b ∈ U with x ∈ Sa and y ∈ Sb . First note that if x = a then

x*y = a*y = a f b,y a

y = b then

and

similarly

if

x*y = x*b = ( f a,x b ) b . It follows that ∀ c ∈U

{

}( f c) ( f

z ∈ Sc , ( x*y ) *z = ( f a,x b ) ( f b,y a ) f c,z ⎡⎣( f a,x b ) ( f b, y a ) ⎤⎦ and x* ( y*z ) = f a,x ⎡⎣( f b,y c ) ( f c,z b ) ⎤⎦ definition of the f 's and the definition of multiplication in U , it is clear that G x *( y *z ) = G ( x *y )*z .

b,y

c,z

and

any

b ) . From the

By Proposition 3, [9], (a): Φ a,x B b ∈ Bαβ where Ba ∈ Bα and Bb ∈ B β . Then

{

}

B x*( y*z ) = Φ a,x ⎡⎣( Φ b,y Bc ) ( Φ c,z Bb ) ⎤⎦ ( Φ b,y Bc ) ( Φ c,z Bb ) . Using the facts that Φ a,x is an almost a -endomorphism, that the

Φ a,x ’s are 1-compatible and that (a) holds, it follows that B x*( y*z ) = ⎡⎣( Φ a,x Bb ) ( Φ b,y Ba ) Bc ⎤⎦ ⎡⎣ Bc ( Φ a,x Bc ) ( Φ b,y Bc ) ( Φ c,z Bb ) ⎤⎦ where each factor in square brackets is in Bαβγ ∈RB , where Bc ∈ Bγ . Let σ = ( Φ a,x Bb ) ( Φ b,y Ba ) Bc ∈ Bαβγ .

) ( Φ B ) B = ( Φ B ) ( Φ B ) ( Φ B ) B = ( Φ B ) ( Φ B ) ⎡⎣Φ ( Φ B )⎤⎦ B and σ = ( Φ B ) ( Φ B ) ( Φ B ) . Similarly, B ( Φ B ) ( Φ B ) ( Φ B ) = ( Φ B ) ( Φ B ) ( Φ Similar calculations show that B( ) = ( Φ B ) ( Φ B ) {Φ ⎡( Φ B ) ( Φ B ) ⎤} = ⎣ ⎦ = ⎡⎣( Φ B ) ( Φ B ) ( Φ B ) B ⎤⎦ ⎡⎣ B ( Φ B ) ( Φ B ) ⎤⎦ = = ⎡⎣( Φ B ) ( Φ B ) ( Φ B ) ⎤⎦ ⎡⎣( Φ B ) ( Φ B ) ( Φ B ) ⎤⎦ and so B( ) = B ( ) . (

Then Φ a,x Bb a,x

1

b,y

b

a

b,y

c

a

a,x

c,z

b

b

b,y

b

b,y

a

c,z

a

a,x

b

b,y

a

c,z

b

c,z

c

c

1

a,x

x *y *z

a,x

a

a

a

a,x

b

b,y

b

b,y

c

b,y

a,x

a

c,z

c

c

c

a,x

b,y

c,z

b

c

a,x

b,y

c,z

b

b

b,y

a

c,z

a,x

2

a

b

c,z

b

x *y *z

b,y

x * y *z

b

a

c

b,y

c

c,z

Bb ) . 2

Therefore inflation of

* is associative. Since, clearly, S *S =U and U is a union of groups, by Theorem 2.15 {S ,*} is a generalised

10

U . Then x*x = ( f a,x a ) = ( G a , Φ a,x Ba ) = ( G a , Ba ) = a . If y*y = a then a = y*y = b and so y ∈ Sa . It follows

that S a = {a}

2

∪ { x ∈ S : x*x = a} .

(⇐)

If S is a generalised inflation of U = G × B then , by Proposition 2.8,

S 2 = U 2 = U . For a ∈U we define

f a,a = Ka and Φ a,a = Γ Ba . We also define Sa = {a} ∪{ x ∈ S : 1x2 x = a} . Let X ≡ {S a : a ∈ U } . Then X is a collection of

mutually disjoint sets satisfying 1) and 2) of Theorem 3.10. For every x ∈ S a − {a} we define

f a,x b ≡ x ( bx )

follows. For b ∈ U ,

−1

( bx ) and for e ∈ B , Φ a,xe ≡ Bx(cx )

follows from Lemma 3.9 that the Φ 's are well-defined. If we define is a collection of 1-compatible mappings on B , where each 3.10. Then define

−1

( cx )

f a,x : U → U and Φ a,x : B → B as

, where c is any element of

Y ≡ {Φa,x : a ∈U , x ∈ Sa }

U such that Bc = e . It

then , by Lemma 3.9 again,

Φ a,x is a Ba − mapping on B and where Y satisfies 3) of Theorem

Z ≡ { f a,x : a ∈U , X ∈ Sa } . The fact that Z

satisfies 4), 5) and 6) of Theorem 3.10 follows from Proposition

2 [1] and the proof of Theorem 5 [1] . Furthermore, it also follows from the proof of Theorem 5 [1] that for every

x ∈ Sa , y ∈ S b we have xy = ( f a,x b ) ( f b,y a ) .

{a,b} ⊆ U and any ■

Theorem 3.11 Suppose that S = F ( X ) , the free semigroup on a non-empty set X . Then S is an inflation of 2

( )

Proof. Let x ∈ S . Then x

t

s

i=1

j=1

S2 .

= ( x 3 ) and so we can write x 2 = m = ∏ mi and x 3 = n = ∏ n j , 2

m = t . But m = n and so 2s = 3t . That is , t = 2q and s = 3q for some positive integer q . We then 3

where n = s and

⎣(

2 3

Y

...m2q )⎤⎦ = ⎡⎣ ( n1...nq )( nq+1...n2q )( n2q+1...n3q )⎤⎦ . It follows that m1...mq = mq+1...m2q = n1...nq = nq+1... n2q = n2q+1... n3q . So m = w2 and n = w3 , where w = m1...mq . We write

have ⎡ m1...mq

)( m

2

3

2

q+1

•

w = wx . Then S =∪ Sa ( a ∈ S 2 ) , where Sa = { x ∈ S : wx = a} . Then for x in Sa and y in Sb , wx wy = x 3 y 3 = x 2 ( xy ) y 2 = wx 3

3

2

( xy ) wy 2

and since F ( X ) is cancellative,

xy = wx wy = ab .

■

Ι × Λ matrix semigroup without zero over the group G , with sandwich matrix U = {( a, i, λ ) : a ∈ G, i ∈Ι, λ ∈ Λ} with multiplication consists of the set

Definition 3.12. Recall that a Rees

P = {Pλ i ∈ G : λ ∈ Λ , i ∈ Ι} ,

( a,i,λ )( b, j,μ ) = ( aPλ jb, i, μ ) and that U

is also called a completely simple semigroup.

Theorem 3.13 below determines all semigroups S in which

S 2 = U , where U is as in Definition 3.12.

Ι × Λ matrix semigroup without zero over the group G with sandwich matrix P . Let X = S( a ,i ,λ ) : ( a, i, λ ) ∈ U be a collection of mutually disjoint sets that satisfy the following. 1) ( a, i, λ ) ∈ U implies

Theorem 3.13 Let U = G × Ι × Λ be the Rees

{

( a , i , λ ) ∈ S ( a ,i , λ ) Let

}

and 2) if

{

∃ h ∈ G such that hPλi h = a then S( a ,i ,λ ) = {( a, i, λ )} .

Y = Φ (xa ,i ,λ ) : ( a, i, λ ) ∈ U, x ∈ S( a ,i ,λ )

} be a collection of pairwise 1- compatible endomorphisms on the rectangular band

Ι × Λ . We will write Φ (xa ,i ,λ ) ( j , μ ) = ( jx , μ x ) . Assume that Y satisfies the following. 3) ( a, i, λ ) ∈ U implies

Φ(( a ,i ,λ )) = K( i ,λ ) , 4) each Φ (xa ,i ,λ ) is ( i, λ ) − fixed and a ,i , λ

( j , μ ) ∈ Ι × Λ , P Pμi a P P = P Pλi a P P and (5B) if x = ( a , i, λ ) then ax = a and otherwise ax Pλi ax = a . x ∈ S( a ,i ,λ ) } be a collection of mappings of U into itself, where for any ( b, j , μ ) ∈ U ,

5) for any x ∈ S ( a ,i ,λ ) , ∃ a x ∈ G such that (5A)

Let

{

Z = f x( a ,i ,λ ) : ( a, i, λ ) ∈ U, f x(

6)

a ,i , λ )

−1 μ jx

for any

−1 x λi μ xi

−1 λ jx

11

−1 x λ j μx j

•

( b, j, μ ) = ( Pλ−1j Pλi ax Pλ j Pμ−1j , jx , μx ) , where ax is as in 5) above. Let S = ∪ S( a ,i ,λ ) ( ( a, i, λ ) ∈U ) . x

x

Define a product

* on S as follows. For x ∈ S( a ,i ,λ ) and y ∈ S( b , j , μ ) , 7) x*y ≡ f x( a ,i ,λ ) ( b, j , μ ) f y(b , j , μ ) ( a, i, λ ) .

Then {S ,*} is a semigroup and is a generalised inflation of

U . Conversely, any generalised inflation of U has the above form.

( ⇒ ) Let x ∈ S ( a ,i ,λ ) , y ∈ S ( b , j , μ ) , z ∈ S ( c , k ,α ) . We must show firstly that

Proof of Theorem 3.13

Z

( x*y ) *z = x* ( y*z ) . It

* and the fact that the endomorphisms in Y are pairwise 1-compatible that the Ι × Λ components of ( x*y ) *z = x* ( y*z ) are equal. Furthermore, the G component of x*y is

follows from the definition of the mappings in

(

(

)

, the definition of

) (

G x*y = Pλ−1jx Pλi ax Pλ j Pμ−x1j Pμx iy Pμ−i1y Pμ j by Pμi Pλ−y1i = Pλ−1jx Pλi ax Pλ j Pμ−x1j By (5A),

(

G x*y = Pλ−1jx Pλi ax Pλ j Pμ−x1j

)(P

b P P(−μ1x )

μx j y μ j

y

j

) = (P

−1

)(P

P −1 P b P P(−μ1x )

μ x i y μ iy μ j y μi

) (

y

i

).

)

P a Pλ j by Pμ j Pλ−y1j .

λ jx λ i x

(

) and similarly if y = ( b, j, μ ) then = G P c P P( ) = ⎡( P P a ) P ( b P P ) ⎤ P c P ⎣ ⎦ ) = (P P a ) (P P P b P ) (c P P ) =

Notice that if x = ( a , i, λ ) then by 3) and (5B), G x*y = a Pλ j by Pμ j Pλy j

(

)

G x*y = Pλ−1jx Pλi ax Pλ j b . Therefore G ( x*y )*z

(

= ( P Pλi ax ) P b P P P −1 λ jx

−1 λ j y μ j λy j λy k

(

)

) (c P

z αk

−1 μz k

P

-1

x *y λ y k z α k −1 λ jx

−1

-1

λ jx λ i x

λy k z

λi x

λky

−1

μky

μ j y μk

λj

z αk

-1 y μ j λy j

P -1 =

λy k z α k μz k

−1

μz k

= Pλ−1jx Pλi a x Pλ k y ⎡ Pμ−k1y Pμ j by Pμ k ( cz Pα k Pμ−z1k ) ⎤ = Pλ−1jx Pλi a x Pλ k y G y *z = G x*( y *z ) ⎣ ⎦ and so * is an associative multiplication and {S ,*} is a semigroup. Since S *S = U it follows from Theorem 2.14 that S is a

(

)

generalised inflation of

(

U.

( ⇐ ) Suppose that S is a generalised inflation of

)

U . For ( a, i, λ ) ∈ U define

{

}

S( a ,i ,λ ) ≡ {( a, i, λ )} ∪{ x ∈ S − U : x 2 = ( a, i, λ )} and define Φ (( aa ,,ii ,,λλ )) ≡ K( i ,λ ) . Define X ≡ S( a ,i ,λ ) : ( a, i, λ ) ∈ U . For a ,i , λ x ∈ S( a ,i ,λ ) define Φ (x ) : Ι × Λ → Ι × Λ as follows: ( j , μ )

( jx , μ x ) , where jx is the Ι − component of x ( a, j , λ ) and μ x is the Λ − component of ( a, i, μ ) x . Define Y ≡ {Φ (xa ,i ,λ ) : ( a, i, λ ) ∈ U , x ∈ S( a ,i ,λ ) } . Now define ax ≡ the −1

{

G − component of x ( a, i, λ ) ⎡⎣ x ( a, i, λ ) ⎤⎦ x . Define Z ≡ f x( a ,i ,λ ) : ( a, i, λ ) ∈ U , x ∈ S( a ,i ,λ ) defined as follows. For ( b, j , μ ) ∈ U , sets

X ,Y ,Z

f x(

a ,i , λ )

} where

f x(

a ,i , λ )

: U → U is

( b, j, μ ) ≡ ( Pλ−1j Pλi ax Pλ j Pμ−1j , jx , μx ) . Then, using the proof of Theorem 5 [1], the x

x

■

can be shown to satisfy all the conditions 1) to 7) of Theorem 3.13.

Definition 3.14. A semigroup S is categorical if every ideal Ι of S satisfies the following property. For any { x, y , z} ⊆ S ,

xyz ∈ Ι implies xy ∈ Ι or yz ∈ Ι . Lemma 3.15 (McMorris and Satyanarayana [7], Theorem 3.6) If S is a commutative, categorical semigroup then semilattice of groups. Definition 3.16. A semilattice S is a tree if for any {e,f,g} ⊆ S , e ≤ g and

S 2 is a

f ≤ g implies e ≤ f or f ≤ e .

Theorem 3.17 (Monzo, [8], Theorem 3.) A semigroup S is a commutative, categorical semigroup if and only if abelian groups if and only if S is an inflation of a tree of abelian groups.

S 2 is a tree of

12 4. Power semigroups and inclusion classes Definition 4.1. The power semigroup of S is denoted by P ( S ) ; that is, P ( S ) = {A ⊆ S : A ≠ ∅} with multiplication defined as AB = {ab : a ∈ A , b ∈ B } . When the structure of

P ( S ) is known, the structure of S may be able to be determined, as in the following cases.

Theorem 4.2 (Redei, [13] and Giraldes and Howie, [3])

The following conditions on a semigroup S are equivalent.

1)P ( S ) is a band ,

2) for any { x, y} ⊆ S , xy ∈ { x, y} , •

3) S = ∪ Sα (α ∈Y ) where Y is a chain, each Sα ∈ L0 ∪R0 and if x ∈ Sα and y ∈ Sβ , with α ≺ β , then xy = yx = x . 4) S is not generated by any proper subset and 5) Every non-empty subset of S is a subsemigroup . Theorem 4.3 (Pondelicek, [12]) The following conditions on a semigroup S are equivalent.

1)P ( S ) is a regular semigroup. 2) A = A3 for every A ∈ P ( S ) ,

3) every element A ∈ P ( S ) with at most three elements is either regular or right regular or left regular ,

4) for any { x, y, z} ⊆ S , xyz ∈ { x, y , z} and 5) S = K ∪ L , where K satisfies xy ∈ { x, y} for every { x, y} ⊆ K . So K has the structure as in Theorem 12, (3). Also,

L ∈ {∅, {q}} . If L = {q} then Y is a chain with a maximal element δ , Kδ = {i} and 5a) i = q 2 and ix = x = xi for all x ∈ S

and

5b) if i ≠ e ≠ q , e ∈ S , then qe = e = eq .

Theorem 4.4 (Pelikan, [10]) Consider the following property. For a given integer n

0,

( An ) : for any { x1 , x2 ,..., xn } ⊆ S , x1 x2 ...xn ∈ { x1 , x2 ,..., xn } . Then, ( An ) implies ( A2 ) and ( An ) implies ( A3 ) for any n = 2 k + 1 ( k ≥ 2 ) .

for any n = 2 k

(k ≥ 2)

Definition 4.5 An inclusion class of semigroups is a collection of all semigroups that satisfy a given set of k number of inclusions as follows:

(1) W1 = {w1,1 , w1,2 ,..., w1,n } ⊆ {t1,1 , t1,2 ,..., t1,m } = T1 ; ( 2 ) W2 = {w2,1 , w2,2 ,..., w2,n } ⊆ {t2,1 , t2,2 ,..., t2,m } = T2 … ( k ) Wk = {wk ,1 , wk ,2 ,..., wk ,n } ⊆ {tk ,1 , tk ,2 ,..., tk ,m } = Tk , where the w ' s and the t ' s are words over some alphabet. 1

1

2

k

[

2

k

Notation: We write W1 ⊆ T1 ;W2 ⊆ T2 ;...;Wk ⊆ Tk

]

to denote the collection of all semigroups that satisfy the k inclusions

W1 ⊆ T1 ,...,Wk ⊆ Tk .

{

}

{

}

Examples. 1) ⎡⎣ xyz ∈ { x, z}⎤⎦ = R0 ∪ L0 ; 2) ⎡⎣ xy ∈ x, y ⎤⎦ = { S :P ( S ) is a band} and 3) ⎡⎣ xyz ∈ x, y, z ⎤⎦ = { S :P ( S ) is regular}. In examples 1-3 above the structure of S can be determined. Now, it follows from Definition 2.4, Proposition 2.8 and Theorem 2.15 that a semigroup is a generalised inflation of a band if and only if its square is a band. Therefore, it is easy to see that 4) { S :

(P ( S ) )

2

is a band } = ⎡⎣ xyzw ∈ { xy , xw, zy , zw}⎤⎦

and, using 2.5, that

13

5) { S :P ( S ) is an inflation of a band } = ⎡⎣ xyzw ∈ { xz , xw, yz , yw}⎤⎦ . Also,

{

6) S ∈ ⎡ xy ∈ y , y

⎣

2

3

}⎤⎦ if and only if every subsemigroup of S is a left ideal of S (c.f. [11] ).

The following two Examples are given without proof. 7) S ∈ ⎡⎣ xyx ∈ { x, y}⎤⎦ if and only if S is a chain Y of semigroups Sα order two, xy = yx = x whenever x ∈ Sα , y ∈ S β and

α≺β

(α ∈Y ) where Sα

is a rectangular band or a group of

and where any group Sα is trivial unless

α ≥ β , ∀β ∈Y

.

8) S ∈ ⎡⎣ xyz ∈ { xy , yz} ; xy = yx ⎤⎦ if and only if S is a tree if and only if S is an inflation of a tree. 2

We proceed to find a construction of the semigroups in Examples 4) and 5). Lemma 4.6 A semilattice Y satisfies the inclusion xyzw ∈ { xy , xw, zy , zw} if and only if Y is a tree. Proof. ( ⇒ ) Suppose that e ≤ g and

f ≤ g . Then ef = egfg ∈ {eg , fg } = {e, f } . So either ef = e or ef = f . That is, either

e ≤ f or f ≤ e . So Y is a tree. ( ⇐ ) Consider any efgh ∈Y . Now efh ≤ f and fgh ≤ f and so efgh ∈ {efh , fgh} . Also ef ≤ e and eh ≤ e and so efh ∈ {ef , eh} . Similarly, fgh ∈ { fg , gh} and so efgh ∈ {efh, fgh} ⊆ {ef , eh, gf , gh} . Lemma 4.7 If

■

S = S 2 then P ( S ) is a generalised inflation of a band if and only if it is an inflation of a band . In other words S

satisfies xyzw ∈ { xy , xw, zy , zw} if and only if it satisfies xyzw ∈ { xz , xw, yz , yw} . Proof. ( ⇒ ) Let S satisfy xyzw ∈ { xz , xw, yz , yw} . Then x y = xy . So 2

Therefore,

2

x2 = x4 and x 3 = x 6 = x 4 x 2 = x 2 x 2 = x 2 .

xy = x 2 y 2 = x ( x 2 y 2 ) = x 2 y and similarly, xy = xy 2 . Then

xy = xy 2 = ( xy ) y = ( xy ) y = ( xy ) ( xy 2 ) = ( xy ) . So S 2 = S is a band. We want to show that AB = ( AB ) for any non2

2

2

empty sets A, B ⊆ S , which is equivalent to xyzw ∈ { xy , xw, zy , zw} . Now

AB ⊆ ( AB) because S is a band. Let 2

{a,a} ⊆ A and {b,b} ⊆ B . We want to show that aba b ∈ AB . CASE 1. abab = aa = aabb ∈ AB . CASE 2. abab = ab ∈ AB . CASE 3. abab = ba = aaba ∈ {ab,aa} . CASE 4. abab = bb = aabb ∈ AB . But abab = aa

implies abab ∈ AB , as in CASE 1 and so abab ∈ AB . So

( AB)

2

⊆ AB and so AB = ( AB ) . 2

( ⇐ ) Assume that xyzw ∈ { xy , xw, zy , zw} . Then xyxy ∈ { xy} and so S = S 2 is a band. So AB ⊆ A 2 B 2 for any

{ }

{ }

non-empty sets A, B ⊆ S . We wish to show that A B ⊆ AB . Let a,a ⊆ A and b,b ⊆ B . Consider aabb . Let 2

2

a ∈ Sα , a ∈ S β , b ∈ Sγ , b ∈ Sσ for some {α , β , γ , σ } ⊆ Y S . Now for any { x, y} ⊆ S xyx = xyxx = xxyx ∈ { x , xy} ∩ { x , yx} and, similarly, yxy ∈ { y , xy} ∩ { y , yx} . So either xyx = x and yxy = y or

xy = yx . That is, either α = β or xy = yx . So S is an almost commutative band. Now if β ≠ γ then a and b commute and so aabb = ab ab ∈ AB . Then, if

β =γ =σ ≠α

β =γ ≠σ

then b commutes with both a and b and then aabb = abab ∈ AB . Similarly, if

{

}

[

]

then aabb = abab ∈ AB . If a,a,b,b ⊆ Sα ∈RB ⊆ xyz = xz then aabb = ab ∈ AB . So

A 2 B 2 ⊆ AB . Therefore, AB = A 2 B 2 which implies xyzw ∈ { xz , xw, yz , yw} .

■

14 Lemma 4.8 The following are equivalent in any semigroup (1)

P ( S ) is an inflation of a band ;

(2)

P ( S ) is a generalised inflation of a band;

S=S . 2

(3) S satisfies xyzw ∈ { xz , xw, yz , yw} ; (4) S satisfies xyzw ∈ { xy , xw, zy , zw} and (5) S is an almost commutative band, Y S is a tree and Sα ∈ L0 ∪ R0 whenever Proof.

α ≺ β (α , β ∈YS

).

The equivalence of (1), (2), (3), and (4) has already been proved in Lemma 4.7.

Assume that (1) through (4) hold. We have shown in the proof of Lemma 4.7 that S is an almost commutative band. By Lemma 4.6,

YS

is a tree. Now suppose that there exists

{α , β } ⊆ Y S

such that

α ≺ β . If Sα ∉ L0 ∪R0

then there exists {e, f } ⊆ Sα with

e = ( e1 , e2 ) , f = ( f1 , f 2 ) and such that e1 ≠ f1 and e2 ≠ f 2 . Let g ∈ S β . Then ( e1 , f 2 ) = ef = eefg ∈ {e, fe, f } , which is a contradiction. Therefore Sα ∈ L0 ∪R0 and we have shown that (5) holds. Suppose that (5) holds. We will show that (4) holds. Let

αβ ≠ αβγσ ≠ γσ

x ∈ Sα , y ∈ Sβ , z ∈ Sγ , w ∈ Sσ for some {α , β , γ , σ } ⊆ Y S . If

then, since S is an almost commutative band, by Result 1.8,

4.6, (4) holds in YS and, using Result 1.10, (4) holds in

Sαβγσ = 1 . Then, since YS is a tree, by Lemma

S . If αβγσ = αβ ≺ γσ then xyzw = xy . If αβγσ = γσ ≺ αβ then

xyzw = zw . If αβ = αβγσ = γσ and αβγσ ≺ ξ for some ξ ∈ {α , β , γ , σ } then Sαβγσ = Sαβ = Sγσ ∈ L0 ∪ R0 and so xyzw ∈ { xy , zw} . If αβγσ = α = β = γ = σ then xyzw = xw . So we have shown that (4) holds.

■

Theorem 4.9 The following conditions on a semigroup S are equivalent.

( 4.9.1)P ( S ) is a generalised inflation of a band, 2 ( 4.9.2 ) ⎡⎣P ( S ) ⎤⎦ is a band, ( 4.9.3) for any { x, y, z , w} ⊆ S , xyzw ∈ { xy,xw,zy,zw} ( 4.9.4 ) a) S 2 is an almost commutative band, b) YS 2 is a tree, c) for any d)

( 4.9.5)

{x ∈ S : x ∈ ( S ) } is an inflation of ( S ) ∈ L ∪R , ( S ) ∈ L ∪ R or {x ∈ S : x ∈ ( S ) } is an inflation of ( S ) .

{α , β } ⊆ Y S

for any α ∈YS 2

,

2

with

α ≺ β,

2

2

2

2

α

0

2

α

2

0

0

α

0

and

2

α

α

the semigroup S has the following form: a)

S=

•

∪ Tα where Y

α∈Y

b) ∀

is a tree, each Tα is a semigroup and

α ∈Y , either ( Tα ) ∈ L0 ∪R0 2

or Tα is an inflation of a rectangular band

{α , β } ⊆ Y where α ≺ β , Tα is an inflation of ( Tα ) ∈ L0 ∪R0 , 2 if ∃{α , β , γ } ⊆ Y such that α = βγ and β ≠ α ≠ γ then ( Tα ) = {eα } and 2

c) ∀ d)

e) S has multiplication

∗ defined as follows: for any xα ∈ Tα and any yβ ∈ Tβ ,

xα ∗ yβ = ( xα ) if α ≺ β , 2

xα ∗ yβ = ( yβ ) if β ≺ α 2

and

xα ∗ yβ = eαβ if α ≠ αβ ≠ β ,

xα ∗ yβ = xα yα if α = β .

15 Proof of Theorem 4.9 We have already shown the equivalence of 4.9.1, 4.9.2 and 4.9.3. (4.9.3) ⇒ (4.9.4) xyxy ∈ { xy} and so that

S 2 = S 4 and S 2 satisfies { xyzw} ⊆ { xy , xw, zy , zw} . It then follows from Lemma 4.8

S 2 is an almost commutative band,YS 2 is a tree and for α ≺ β (α , β ∈YS 2 ) , ( S 2 ) ∈ L0 ∪ R0 and so 4.9.4, a) and b) hold. α

Now suppose that

{α , β } ⊆ Y S

2

with

( )

and ez = z with { x , z} ⊆ S , h ∈ S

α ≺ β . Let ex = x 2

2

2

β

( )

and {ex , ez } ⊆ S

2

α

.

Using Result 1.10, xh = ( xh ) = ( xhx ) h = ( xhex hx ) h = xh ( ex x ) hxh = x ( hex )( xh ) = xex ( xh ) = x h = ex h = ex . 2

2

2

Similarly, hz = ez . Then ( xh ) z = x ( hz ) and so ex z = xez . Using Corollary 5 [9], we get

(S ) 2

α

{

(

x 2 ∈ L0 then ex = H xx 2 ez and then x z = H x2 z

X = x ∈ S : x2 ∈ ( S 2 )

} is an inflation of ( S ) 2

α

α

)(H

z z2

4

ex ( Hezz ex ) = Hexx ez ez . If

(

x 2 ) = Hexx ez = ex = ex ez = x 2 z 2 and so

( )

. Similarly, if S

)

∈R0 then X is an inflation of ( S 2 ) and so 4.9.4 c)

2

α

α

holds.

α ∈YS

Now suppose that

2

{

( )

and that S

α

( ) 2

α

. We show that for any

α

) ( H x ) where the H ’s are compatible f ∈ ( S ) , H f = x . Let x = ( x , x ) . Let

(

is a (rectangular) band. By Corollary 5 [9], for any { x , y} ⊆ X , x y = H almost endomorphisms on S

( ) } is a semigroup whose square

∉ L0 ∪ R0 . By Result 1.10, X = x ∈ S : x 2 ∈ S 2

2

x ∈ X and any

x x2

y

y y2

2

2

2

x x2

α

g = ( g1 , g 2 ) ∈ ( S 2 ) where x1 ≠ g1 and x2 ≠ g 2 .Then

2

2

1

2

α

( x1 , g2 ) ( g1 , x2 ) x = ( x1 , x2 ) x = x 3 = ( x 3 ) = x 2 ∈ {( x1 , g2 ) , ( x1 , g2 ) x, g , ( g1 , x2 ) x} . Thus, x 2 = ( x1 , x2 ) = ( x1 , g 2 ) x = ( x1 , g 2 ) ( H xx ( x1 , g2 ) ) . By Result 1 [9, p.357,] then, H xx maps any element of ( S 2 ) α 2

2

2

2

that has

x

second component distinct from x 2 to an element with second component equal to x 2 . Similarly, H x 2 maps any element with first x

component distinct from x1 to an element with first component equal to x1 . Also by Corollary 5 [9], H x 2 fixes

( )

therefore shown that H x 2 = K x 2 . Hence, xy = x y and by Result 2.5 , X is an inflation of S 2

x

2

2

α

x2 . We have

. Thus 4.9.4 d) holds.

Suppose that { x, y , z , w} ⊆ S with

(4.9.4) ⇒ (4.9.3)

x 2 = e ∈ ( S 2 ) , y 2 = f ∈ ( S 2 ) , z 2 = g ∈ ( S 2 ) , w2 = h ∈ ( S 2 ) α

β

αβ = γσ , αβ ≺ γσ , γσ ≺ αβ

or

γ

(S ) 2

αβγσ

σ

= 1 . If ( S 2 )

αβγσ

{α , β , γ , σ } ⊆ Y S

for some

2

. Now either

= 1 then (using Result 1.10 ) we see that

( )

xyzw = x 2 y 2 z 2 w 2 = efgh ∈ {ef , eh, gf , gh} . If xyzw = ef then αβγσ = αβ and so xy ∈ S 2

αβ

= { xyzw} . Similarly,

xyzw = eh implies xyzw = xw , xyzw = gf implies xyzw = zy and xyzw = gh implies xyzw = zw . If

(S ) 2

αβγσ

Similarly, if

Now if

≠ 1 and αβ ≺ γσ then by Result 1.10, xy ∈ ( S 2 )

αβ

(S ) 2

αβγσ

(S ) 2

αβγσ

( )

and yz ∈ S

2

γσ

. Using Result 1.8 we have

xyzw = xy.

≠ 1 and γσ ≺ αβ then xyzw = zw.

≠ 1 , αβ = γσ and αβγσ ≺ ξ ∈ {α , β , γ , σ } then ( S 2 )

xyzw ∈ { xy , zw} . If αβγσ = α = β = γ = σ and ( S

2

)

α

αβ

= (S2 )

αβγσ

= ( S 2 ) ∈ L0 ∪R0 and therefore γσ

∉ L0 ∪ R0 then, by 4.9.4 (d) , xyzw = xw . So 4.9.3 holds.

Suppose that S satisfies the conditions of 4.9.4. We define Y

(4.9.4) ⇒ (4.9.5)

{

Tα ≡ xα ∈ S : xα ∈ ( S 2 ) 2

α

16

≡ YS 2 ,

} and * ≡ multiplication in S . Then , using Result 1.10, the conditions a), b), c), and d) in 4.9. 5 hold. ( )

Using Result 1.10 again, xα * yβ = xα yβ ∈ S

= {eαβ } when α ≠ αβ ≠ β . If α ≺ β then xα = xα yβ = 2

2

αβ

2

2

2 2 2 2 = yβ xα and xα *yβ = xα yβ = ( xα yβ )( xα yβ xα )( yβ xα yβ ) = xα ( yβ ) xα yβ = ⎡⎢ xα ( yβ ) xα ⎤⎥ yβ = ⎣ ⎦ 2 2 2 2 2 4 2 2 2 = xα ⎡⎢( yβ ) ( xα ) ( yβ ) ⎤⎥ xα yβ = ( xα ) yβ = ⎡⎢( xα ) ( yβ ) ⎤⎥ yβ = ( xα ) ( yβ ) = ( xα ) . Similarly, if β ≺ α then ⎣ ⎦ ⎣ ⎦ 2

xα * yβ = ( yβ ) and, finally, if α = β then xα *yβ = xα yβ in Tα . Thus, 4.9.5 holds. 2

(4.9.5) ⇒ (4.9.4) Suppose that 4.9.5 holds. Let each of the cases where the order of the set

{α , β , γ } ⊆ Y

. Then it is fairly straightforward to show that ∗ is associative in

{α , β , γ } equals d ∈ {1,2,3} . Now by 4.9.5 (e),

•

S *S = ∪ ( Tα ) , (α ∈Y ) , and so 2

S *S is a band and Y S *S = Y . Thus, since Y is a tree, 4.9.4 (b) holds. By 4.9.5 (e), any two elements of S *S are either in the same rectangular band component of S *S , or they commute. Thus, S *S is an almost commutative band and 4.9.4 (a) holds. By 4.9.5 (b) and (c),4.9.4 (c) and (d) hold. So 4.9.4 holds. We have therefore shown that 4.9.4 and 4.9.5 are equivalent and so Theorem 4.9 is ■ proved. Corollary 4.10

P ( S ) is an inflation of a band if and only if S ∈ ⎡⎣ xyzw ∈ { xz, xw, yz, yw}⎤⎦ if and only if S 2 is a tree Y of

inflations of rectangular bands Tα

(α ∈Y ) , where Y

satisfies 4.9.5 c) and d) and where S has multiplication as in 4.9 .5 e).

Open Questions (1) (2) (3)

U then is T a generalised inflation of U ? Find a construction for semigroups in which P ( S ) ∈ ⎡ xy = ( xy ) ⎤ for some integer n ≥ 2 . If T is a generalised inflation of S and S is a generalised inflation of n

⎣

} and {T , ∗ } are n − chains of n copies of L × R ( n ∈ N − { 1 } ) and {K , } and {K , } are isomorphic as partial groupoids then are {S , ∗ } and {T , ∗ } isomorphic? {

If S , ∗K S

KT

n

KS

(4)

⎦

n

S

T

KT

Find necessary and sufficient conditions on a semigroup U such that every semigroup inflation [generalised inflation] of U .

S with S 2 = U is an

(5)

Give necessary and sufficient conditions on a partial groupoid G such that G ≅ {K S ,

(6)

For {m, n, p} ⊆ N , find a formula for the number of non-isomorphic p − chains of p copies of L m × R n .

(7)

In which semigroups is

(8)

It is known that

(9)

Find a construction of semigroups in the inclusion class ⎡⎣ xyx ∈ { y , yx}⎤⎦ . ( Note: The square of each such semigroup is a semilattice of semigroups that are either right-zero semigroups or groups of order two.) 2

P (S )

(P ( S ) )

2

} for some band S .

a regular semigroup? Find a construction for such semigroups.

is regular if and only if P ( S ) ∈ ⎡⎣ x = x ⎤⎦ . Is the same true for 3

2

(P ( S ) )

2

?

(10) If S is a Brandt semigroup then is S a generalised inflation of S ? ( This question was posed by Wang and Wismath in [15], where they call a generalised inflation a “left-right generalised inflation”. In a sense , an affirmative answer to this question would “extend” Theorem 2.15.)

17 Acknowledgments The author is most grateful to the referee for the detailed scrutiny of and comprehensive feedback on the first version of this paper and to Ralph and Ida Monzo for their enduring and sustaining love. References

[1] Clarke, G. T. and Monzo, R. A. R., A Generalisation of the Concept of an Inflation of a Semigroup, Semigroup Forum 60 (2000) 172-186. [2] Clifford, A. H. and Preston G. B., "The Algebraic Theory of Semigroups", Volume I, Math. Surveys of the American Math. Soc. 7, Providence, R. I., 1961. [3] Giraldes, E. and Howie, J. M., Semigroups Of High Rank, Proc. Of the Edin. Math. Soc.28 (1985) 13-34. [4] Hall, T. E., Almost Commutative Bands, Glasgow Math. Journal 13 (1972) 176-178. [5] Howie, J. M., "An Introduction to Semigroup Theory", Academic Press, London, New York, San Francisco, 1976. [6] McLean, D., Idempotent Semigroups, Amer. Math. Monthly 61 (1954), 110-113. [7] McMorris, F. R. and Satyanarayana, M., Categorical Semigroups, Proc. Amer. Math. Soc. (1972) 271-277. [8] Monzo, R. A. R., Categorical Semigroups, Semigroup Forum 6 (1973) 59-68. [9] Monzo, R. A. R., Pre-compatible Almost Endomorphisms and Semigroups Whose Cube is a Band, Semigroup Forum 67 (2003) 355-372. [10] Pelikan, J., On Semigroups In Which Products Are Equal To One Of The Factors, Periodica. Mathematica Hungarica 4 (1973) 103-106. [11] Petrich,M., A Simple Construction Of Semigroups All Of Whose Subsemigroups Are Left Ideals, Semigroup Forum 4 (1972) 262-266. [12] Pondelicek, B., On Semigroups Having Regular Globals, Colloq. Mathem. Soc. Janos Bolyai (1976). [13] Redei, L., "Algebra I", Permagon Press, Oxford, 1967. [14] Tamura, T., Semigroups Satisfying Identity

xy = f ( x, y ) , Pacific J. Math. 31 (1962) 513-521.

[15] Wang, Q. and Wismath, S. L., Null Extensions and Generalised Inflations of Brandt Semigroups, Semigroup Forum 74 (2007) 274-292. [16] Yamada, M., A Note on Middle Unitary Semigroups, Kodai Math. Sem. Rep. 7 (1955) 49-52.

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