Algebra in superextensions of groups, I: zeros and commutativity
aa r X i v : . [ m a t h . GN ] F e b ALGEBRA IN SUPEREXTENSIONS OF GROUPS, I:ZEROS AND COMMUTATIVITY
T.BANAKH, V.GAVRYLKIV, O.NYKYFORCHYN
Abstract.
Given a group X we study the algebraic structure of its superex-tension λ ( X ). This is a right-topological semigroup consisting of all maximallinked systems on X endowed with the operation A ◦ B = { C ⊂ X : { x ∈ X : x − C ∈ B} ∈ A} that extends the group operation of X . We characterize right zeros of λ ( X )as invariant maximal linked systems on X and prove that λ ( X ) has a rightzero if and only if each element of X has odd order. On the other hand, thesemigroup λ ( X ) contains a left zero if and only if it contains a zero if and onlyif X has odd order | X | ≤
5. The semigroup λ ( X ) is commutative if and onlyif | X | ≤
4. We finish the paper with a complete description of the algebraicstructure of the semigroups λ ( X ) for all groups X of cardinality | X | ≤ Contents
Introduction 11. Self-linked sets in groups 42. Maximal invariant linked systems 93. Right zeros in λ ( X ) 164. (Left) zeros of the semigroup λ ( X ) 185. The commutativity of λ ( X ) 196. The superextensions of finite groups 20References 26 Introduction
After the topological proof of the Hindman theorem [H1] given by Galvin andGlazer , topological methods become a standard tool in the modern combinatoricsof numbers, see [HS], [P]. The crucial point is that any semigroup operation ∗ defined on a discrete space X can be extended to a right-topological semigroup Mathematics Subject Classification. Unpublished, see [HS, p.102], [H2] operation on β ( X ), the Stone- ˇCech compactification of X . The extension of theoperation from X to β ( X ) can be defined by the simple formula:(1) U ◦ V = { A ⊂ X : { x ∈ X : x − A ∈ V} ∈ U} , where U , V are ultrafilters on X and x − A = { y ∈ X : xy ∈ A } . Endowedwith the so-extended operation, the Stone- ˇCech compactification β ( X ) becomes acompact right-topological semigroup. The algebraic properties of this semigroup(for example, the existence of idempotents or minimal left ideals) have importantconsequences in combinatorics of numbers, see [HS], [P].The Stone- ˇCech compactification β ( X ) of X is the subspace of the double power-set P ( P ( X )), which is a complete lattice with respect to the operations of unionand intersection. In [G2] it was observed that the semigroup operation extends notonly to β ( X ) but also to the complete sublattice G ( X ) of P ( P ( X )) generated by β ( X ). This complete sublattice consists of all inclusion hyperspaces over X .By definition, a family F of non-empty subsets of a discrete space X is calledan inclusion hyperspace if F is monotone in the sense that a subset A ⊂ X belongsto F provided A contains some set B ∈ F . On the set G ( X ) there is an impor-tant transversality operation assigning to each inclusion hyperspace F ∈ G ( X ) theinclusion hyperspace F ⊥ = { A ⊂ X : ∀ F ∈ F ( A ∩ F = ∅ ) } . This operation is involutive in the sense that ( F ⊥ ) ⊥ = F .It is known that the family G ( X ) of inclusion hyperspaces on X is closed inthe double power-set P ( P ( X )) = { , } P ( X ) endowed with the natural producttopology. The induced topology on G ( X ) can be described directly: it is generatedby the sub-base consisting of the sets U + = {F ∈ G ( X ) : U ∈ F} and U − = {F ∈ G ( X ) : U ∈ F ⊥ } where U runs over subsets of X . Endowed with this topology, G ( X ) becomes aHausdorff supercompact space. The latter means that each cover of G ( X ) by thesub-basic sets has a 2-element subcover.The extension of a binary operation ∗ from X to G ( X ) can be defined in thesame way as for ultrafilters, i.e., by the formula (1) applied to any two inclusionhyperspaces U , V ∈ G ( X ). In [G2] it was shown that for an associative binaryoperation ∗ on X the space G ( X ) endowed with the extended operation becomes acompact right-topological semigroup. The algebraic properties of this semigroupswere studied in details in [G2]. LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 3
Besides the Stone- ˇCech compactification β ( X ), the semigroup G ( X ) containsmany important spaces as closed subsemigroups. In particular, the space λ ( X ) = {F ∈ G ( X ) : F = F ⊥ } of maximal linked systems on X is a closed subsemigroup of G ( X ). The space λ ( X )is well-known in General and Categorial Topology as the superextension of X , see[vM], [TZ]. Endowed with the extended binary operation, the superextension λ ( X )of a semigroup X is a supercompact right-topological semigroup containing β ( X )as a subsemigroup.The space λ ( X ) consists of maximal linked systems on X . We recall that asystem of subsets L of X is linked if A ∩ B = ∅ for all A, B ∈ L . An inclusionhyperspace
A ∈ G ( X ) is linked if and only if A ⊂ A ⊥ . The family of all linkedinclusion hyperspace on X is denoted by N ( X ). It is a closed subset in G ( X ).Moreover, if X is a semigroup, then N ( X ) is a closed subsemigroup of G ( X ). Thesuperextension λ ( X ) consists of all maximal elements of N ( X ), see [G1], [G2].In this paper we start a systematic investigation of the algebraic structure of thesemigroup λ ( X ). This program will be continued in the forthcoming papers [BG2]and [BG3]. The interest to studying the semigroup λ ( X ) was motivated by thefact that for each maximal linked system L on X and each partition X = A ∪ B of X into two sets A, B either A or B belongs to L . This makes possible to applymaximal linked systems to Combinatorics and Ramsey Theory.In this paper we concentrate on describing zeros and commutativity of the semi-group λ ( X ). In Proposition 3.1 we shall show that a maximal linked system L ∈ λ ( X ) is a right zero of λ ( X ) if and only if L is invariant in the sense that xL ∈ L for all L ∈ L and all x ∈ X . In Theorem 3.2 we shall prove that a group X admits an invariant maximal linked system (equivalently, λ ( X ) contains a rightzero) if and only if each element of X has odd order. The situation with (left) zerosis a bit different: a maximal linked system L ∈ λ ( X ) is a left zero in λ ( X ) if andonly if L is a zero in λ ( X ) if and only if L is a unique invariant maximal linkedsystem on X . The semigroup λ ( X ) has a (left) zero if and only if X is a finitegroup of odd order | X | ≤ X is isomorphic to the cyclic group C , C or C ). The semigroup λ ( X ) rarely is commutative: this holds if and only if thegroup X has finite order | X | ≤ A of a group X is called self-linked if A ∩ xA = ∅ for all x ∈ X . In Proposition 1.1 weshall give lower and upper bounds for the smallest cardinality sl ( X ) of a self-linkedsubset of X . We use those bounds to characterize groups X with sl ( X ) ≥ | X | / T.BANAKH, V.GAVRYLKIV, O.NYKYFORCHYN
In Section 2 we apply self-linked sets to evaluating the cardinality of the (rect-angular) semigroup ↔ λ ( X ) of maximal invariant linked systems on a group X . InTheorem 2.2 we show that for an infinite group X the cardinality of ↔ λ ( X ) equals2 | X | . In Proposition 2.3 and Theorem 2.6 we calculate the cardinality of ↔ λ ( X ) forall finite groups X of order | X | ≤ X with | ↔ λ ( X ) | = 1.In Sections 4 and 5 these results are applied for characterizing groups X whosesuperextensions have zeros or are commutative.We finish the paper with a description of the algebraic structure of the superex-tensions of groups X of order | X | ≤ C n we denote the cyclic group of order n and by D n the dihedral group of cardinality2 n , that is, the isometry group of the regular n -gon. For a group X by e we denotethe neutral element of X . For a real number x we put ⌈ x ⌉ = min { n ∈ Z : n ≥ x } and ⌊ x ⌋ = max { n ∈ Z : n ≤ x } . Self-linked sets in groups
In this section we study self-linked subsets in groups. By definition, a subset A of a group G is self-linked if A ∩ xA = ∅ for each x ∈ G . In fact, this notion can bedefined in the more general context of G -spaces.By a G -space we understand a set X endowed with a left action G × X → X of a group G . Each group G will be considered as a G -space endowed with theleft action of G . An important example of a G -space is the homogeneous space G/H = { xH : x ∈ G } of a group G by a subgroup H ⊂ G .A subset A ⊂ X of a G -space X defined to be self-linked if A ∩ gA = ∅ for all g ∈ G . Let us observe that a subset A ⊂ G of a group G is self-linked if and onlyif AA − = G .For a G -space X by sl ( X ) we denote the smallest cardinality | A | of a self-linkedsubset A ⊂ X . Some lower and upper bounds for sl ( G ) are established in thefollowing proposition. Proposition 1.1.
Let G be a finite group and H be a subgroup of G . Then (1) sl ( G ) ≥ (1 + p | G | − / ; (2) sl ( G ) ≤ sl ( H ) · sl ( G/H ) ≤ sl ( H ) · ⌈ ( | G/H | + 1) / ⌉ . (3) sl ( G ) < | H | + | G/H | .Proof.
1. Take any self-linked set A ⊂ G of cardinality | A | = sl ( G ) and considerthe surjective map f : A × A → G , f : ( x, y ) xy − . Since f ( x, y ) = xy − = e forall ( x, y ) ∈ ∆ A = { ( x, y ) ∈ A : x = y } , we get | G | = | G \ { e }| + 1 ≤ | A \ ∆ A | + 1 = sl ( G ) − sl ( G ) + 1, which just implies that sl ( G ) ≥ (1 + p | G | − / LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 5 H be a subgroup of G . Take self-linked sets A ⊂ H and B ⊂
G/H = { xH : x ∈ G } having sizes | A | = sl ( H ) and |B| = sl ( G/H ). Fix any subset B ⊂ G such that | B | = |B| and { xH : x ∈ B } = B . We claim that the set C = BA isself-linked. Given arbitrary x ∈ G we should prove that the intersection C ∩ xC is not empty. Since B is self-linked, the intersection B ∩ x B contains the coset bH = xb ′ H for some b, b ′ ∈ B . It follows that b − xb ′ ∈ H = AA − . The latterequality follows from the fact that the set A ⊂ H is self-linked in H . Consequently, b − xb ′ = a ′ a − for some a, a ′ ∈ A . Then xC ∋ xb ′ a = ba ′ ∈ C and thus C ∩ xC = ∅ .The self-linkedness of C implies the desired upper bound sl ( G ) ≤ | C | ≤ | A | · | B | = sl ( H ) · sl ( G/H ) . sl ( G/H ) ≤ ⌈ ( | G/H | + 1) / ⌉ . Take any subset A ⊂ G/H of size | A | = ⌈ ( | G/H | + 1) / ⌉ and note that | A | > | G/H | /
2. Then for each x ∈ G the shift xA has size | xA | = | A | > | G/H | /
2. Since | A | + | xA | > | G/H | , the sets A and xA meet each other. Consequently, A is self-linked and sl ( G/H ) ≤ | A | = ⌈ ( | G/H | + 1) / ⌉ .3. Pick a subset B ⊂ G of size | B | = | G/H | such that BH = G and observe thatthe set A = H ∪ B is self-linked and has size | A | ≤ | H | + | B | − B ∩ H is a singleton). (cid:3) Theorem 1.2.
For a finite group G (i) sl ( G ) = ⌈ ( | G | + 1) / ⌉ > | G | / if and only if G is isomorphic to one of thegroups: C , C , C , C , C × C , C , D , ( C ) ; (ii) sl ( G ) = | G | / if and only if G is isomorphic to one of the groups: C , C , C × C , D , Q .Proof. I. First we establish the inequality sl ( G ) < | G | / G not iso-morphic to the groups appearing in the items (i), (ii). Given such a group G weshould find a self-linked subset A ⊂ G with | A | < | G | / G contains a subgroup H of order | H | = 3 and index | G/H | = 3. Then sl ( H ) = 2 and we can apply Proposition 1.1(2) to conclude that sl ( G ) ≤ sl ( H ) · sl ( G/H ) ≤ · < / | G | / . | G | / ∈ { , , } and G contains a subgroup H of order n = | H | ≥ m = | G/H | ≥
3. It this case n + m − < nm/ sl ( G ) ≤ | H | + | G/H | − n + m − < nm/ G is cyclic of order n = | G | ≥
9. Given a generator a of G , construct asequence ( x i ) ≤ i ≤ n/ letting x = a , x = a , x = a , x = a , and x i = x i − a i T.BANAKH, V.GAVRYLKIV, O.NYKYFORCHYN for 5 < i ≤ n/
2. Then the set A = { x i : 2 ≤ i ≤ n/ } has size | A | < n/ G is cyclic of order | G | = 7. Given a generator a of G observe that A = { e, a, a } is a 3-element self-linked subset and thus sl ( G ) ≤ < | G | / G contains a cyclic subgroup H ⊂ G of prime order | H | ≥
7. By the precedingtwo cases, sl ( H ) < | H | / sl ( G ) ≤ sl ( H ) · sl ( G/H ) < | H | · | G || H | = | G | / . | G | > | G | / ∈ { , , } . If | G | is prime or | G | = 15, then G is cyclicof order | G | ≥ sl ( G ) < | G | / | G | = 2 p forsome prime number p , then G contains a cyclic subgroup of order p ≥ sl ( G ) < | G | / | G | = 4 n for some n ≥
4, then by Sylow’sTheorem (see [OA, p.74]), G contains a subgroup H ⊂ G of order | H | = 4 andindex | G/H | ≥
4. Then sl ( G ) < | G | / | G | = nm = 15 for some odd numbers n, m ≥ sl ( G ) < | G | / | G | = 8, then G is isomorphic to one of the groups: C , C × C , ( C ) , D , Q . All those groups appear in the items (i), (ii) and thus are excluded fromour consideration.8) If | G | = 10, then G is isomorphic to C or D . If G is isomorphic to C ,then sl ( G ) < | G | / G is isomorphic to D , then G containsan element a of order 5 and an element b of order 2 such that bab − = a − . Nowit is easy to check that the 4-element set A = { e, a, b, ba } is self-linked and hence sl ( G ) ≤ < | G | / G with | G | = 12. It is well-known that thereare five non-isomorphic groups of order 12: the cyclic group C , the direct sum oftwo cyclic groups C ⊕ C , the dihedral group D , the alternating group A , andthe semidirect product C ⋊ C with presentation h a, b | a = b = 1 , aba − = b − i .If G is isomorphic to C , C ⊕ C or A , then G contains a normal 4-elementsubgroup H . By Sylow’s Theorem, G contains also an element a of order 3. Takinginto account that a / ∈ H and Ha − = a − H , we conclude that the 5-element set A = { a } ∪ H is self-linked and hence sl ( G ) ≤ < | G | / G is isomorphic to C ⋊ C , then G contains a normal subgroup H of order3 and an element a ∈ G such that a / ∈ H . Observe that the 5-element set A = H ∪{ a, a } is self-linked. Indeed, AA − ⊃ H ∪ aH ∪ a H ∪ Ha − = G . Consequently, sl ( G ) ≤ < | G | / D . It contains an element a generating a cyclic subgroup of order 6 and an element b of order 2 such that bab − = a − . Consider the 5-element set A = { e, a, a , b, ba } and note that AA − = { e, a, a , b, ba } · { e, a , a , b, ba } = G . This yields the desired inequality sl ( G ) ≤ < | G | / LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 7
Therefore we have completed the proof of the inequality sl ( G ) < | G | / sl ( G ) = ⌈ ( | G | + 1) / ⌉ > | G | / G with | G | ≤ sl ( G ) > | G | / G is isomorphic to D or C . First weconsider the case G = D . In this case G contains a normal 3-element subgroup T .Assuming that sl ( G ) ≤ | G | / A . Withoutloss of generality we can assume that the neutral element e of G belongs to A (otherwise replace A by a suitable shift xA ). Taking into account that AA − = G ,we conclude that A T and thus we can find an element a ∈ A \ T . This elementhas order 2. Then AA − = { e, a, b } · { e, a, b − } = { e, a, b, a, e, ba, b − , ba, e } 6 = G, which is a contradiction.Now assume that G is isomorphic to C . In this case G is the 3-dimensionallinear space over the field C . Assuming that sl ( A ) ≤ | G | /
2, find a 4-elementself-linked subset A ⊂ G . Replacing A by a suitable shift, we can assume that A contains a neutral element e of G . Since AA − = G , the set A contains threelinearly independent points a, b, c . Then AA − = { e, a, b, c } · { e, a, b, c } = { e, a, b, c, ab, ac, bc } 6 = G, which contradicts the choice of A .III. Finally, we prove the equality sl ( G ) = | G | / G = C , then sl ( G ) ≥ a of G the 3-element subset A = { e, a, a } is self-linkedin G , which yields sl ( G ) = 3 = | G | / | G | = 8, then sl ( G ) ≥ G is cyclic of order 8 and a is a generator of G , then the set A = { e, a, a , a } is self-linked and thus sl ( C ) = 4.If G is isomorphic to C ⊕ C , then G has two commuting generators a, b suchthat a = b = 1. One can check that the set A = { e, a, a , b } is self-linked andthus sl ( C ⊕ C ) = 4.If G is isomorphic to the dihedral group D , then G has two generators a, b connected by the relations a = b = 1 and bab − = a − . One can check that the4-element subset A = { e, a, b, ba } is self-linked. T.BANAKH, V.GAVRYLKIV, O.NYKYFORCHYN If G is isomorphic to the group Q = {± , ± i, ± j, ± k } of quaternion units, thenwe can check that the 4-element subset A = {− , , i, j } is self-linked and thus sl ( Q ) = 4. (cid:3) In the following proposition we complete Theorem 1.2 calculating the values ofthe cardinal sl ( G ) for all groups G of cardinality | G | ≤ Proposition 1.3.
The number sl ( G ) for a group G of size | G | ≤ can be foundfrom the table: G C C C C C ⊕ C C D C C ⊕ C D Q C sl ( G ) 2 2 3 3 3 3 4 4 4 4 4 5 G C C C C C ⊕ C C D C C ⊕ C D A C ⋊ C sl ( G ) 3 4 4 4 4 4 4 4 5 5 5 5 Proof.
For groups G of order | G | ≤
10 the value of sl ( G ) is uniquely determined bythe lower bound sl ( G ) ≥ √ | G |− from Proposition 1.1(1) and the upper boundfrom Theorem 1.2. It remains to consider the groups G of order 11 ≤ | G | ≤ G is cyclic of order 11 or 13, then take a generator a of G and check thatthe 4-element set A = { e, a , a , a } is self-linked, witnessing that sl ( C ) = 4.2. If G is cyclic of order 12, then take a generator a for G and check that the4-element subset A = { e, a, a , a } is self-linked witnessing that sl ( G ) = 4.It remains to consider all other groups of order 12. Theorem 1.2 gives us anupper bound sl ( G ) ≤
5. So, we need to show that sl ( G ) > G with | G | = 12.3. If G is isomorphic to C ⊕ C or A , then G contains a normal subgroup H isomorphic to C ⊕ C . Assuming that sl ( G ) = 4, we can find a 4-element self-linkedsubset A ⊂ G . Since AA − = G , we can find a suitable shift xA such that xA ∩ H contains the neutral element e of G and some other element a of H . Replacing A by xA , we can assume that e, a ∈ A . Since A H , there is a point b ∈ A \ H .Since the quotient group G/H has order 3, bH ∩ Hb − = ∅ .Concerning the forth element c ∈ A \ { e, a, b } there are three possibilities: c ∈ H , c ∈ b − H , and c ∈ bH . If c ∈ H , then bH = bH ∩ AA − = b ( A ∩ H ) − consists of3 elements which is a contradiction. If c ∈ b − H , then H = H ∩ AA − = { e, a } ,which is absurd. So, c ∈ bH and thus c = bh for some h ∈ H . Since h = h − , weget cb − = bhb − = bh − b − = bc − . Then H = H ∩ AA − = { e, a, cb − , bc − } hascardinality | H | = |{ e, a, cb − = bc − }| ≤
3, which is not true. This contradictioncompletes the proof of the inequality sl ( G ) > C ⊕ C and A .4. Assume that G is isomorphic to the dihedral group D . Then G containsa normal cyclic subgroup H of order 6, and for each b ∈ G \ H and a ∈ H we LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 9 get b = e and bab − = a − . Assuming that sl ( D ) = 4, we can find a 4-element self-linked subset A ⊂ G . Let a be a generator of the group H . Since a ∈ AA − = G , we can find two element x, y ∈ A such that a = xy − . Then theshift Ay − contains e and a . Replacing A by Ay − , if necessary, we can assumethat e, a ∈ A . Since A H , there is an element b ∈ A \ H . Concerning theforth element c ∈ A \ { e, a, b } there are two possibilities: c ∈ H and c / ∈ H . If c ∈ H , then the set A H = A ∩ H = { e, a, c } contains three elements and is equalto bA − H b − , which implies bA − H = A H b − = bA − H ∪ A H b − = AA − ∩ bH = bH .This is a contradiction, because | H | = 4 > | bA − H | . Then c ∈ bH and hence H = H ∩ AA − = { e, a, a − , bc − , cb − } which is not true because | H | = 6 > G is isomorphic to the semidirect product C ⋊ C and hencehas a presentation h a, b | a = b = 1 , aba − = b − i . Then the cyclic subgroup H generated by b is normal in G and the quotient G/H is cyclic of order 4. Assumingthat sl ( G ) = 4, take any 4-element self-linked subset A ⊂ G .After a suitable shift of A , we can assume that e, b ∈ A . Since A H , there isan element c ∈ A \ H . We claim that the fourth element d ∈ A \ { e, b, c } does notbelong to H ∪ cH ∪ c − H . Otherwise, AA − ⊂ H ∪ cH ∪ c − H = G . This impliesthat one of the elements, say c belongs to the coset a H and the other to aH or a − H . We lose no generality assuming that d ∈ aH . Then c = a b i , d = ab j forsome i, j ∈ {− , , } . It follows that aH = aH ∩ AA − = { d, db − , cd − } = { ab j , ab j − , a b i − j a − } = { ab j , ab j − , ab j − i } which implies that i = − c = a b − . In this case we arrive to a contra-diction looking at a H ∩ AA − = { c, cb − , c − , bc − } = { a b − , a b − , ba , b a } 6∋ a . (cid:3) Problem 1.4.
What is the value of sl ( G ) for other groups G of small cardinality?Is sl ( G ) = ⌈ (1 + p | G | − / ⌉ for all finite cyclic groups G ? Maximal invariant linked systems
In this section we study (maximal) invariant linked systems on groups. Aninclusion hyperspace A on a group X is called invariant if x A = A for all x ∈ X .The set of all invariant inclusion hyperspaces on X is denoted by ↔ G ( X ). By [G2], ↔ G ( X ) is a closed rectangular subsemigroup of G ( X ) coinciding with the minimalideal of G ( X ). The rectangularity of ↔ G ( X ) means that A ◦ B = B for all A , B ∈ ↔ G ( X ).Let ↔ N ( X ) = N ( X ) ∩ ↔ G ( X ) denote the set of all invariant linked systems on X and ↔ λ ( X ) = max ↔ N ( X ) be the family of all maximal elements of ↔ N ( X ). Elements of ↔ λ ( X ) are called maximal invariant linked systems . The reader shouldbe concisions of the fact that maximal invariant linked systems need not be maximallinked! Theorem 2.1.
For every group X the set ↔ λ ( X ) is a non-empty closed rectangularsubsemigroup of G ( X ) .Proof. The rectangularity of ↔ λ ( X ) implies from the rectangularity of ↔ G ( X ) estab-lished in [G2, §
5] and the inclusion ↔ λ ( X ) ⊂ ↔ G ( X ).The Zorn Lemma implies that each invariant linked system on X (in particular, { X } ) can be enlarged to a maximal invariant linked system on X . This observationimplies the set ↔ λ ( X ) is not empty. Next, we show that the subsemigroup ↔ λ ( X )is closed in G ( X ). Since the set ↔ N ( X ) = N ( X ) ∩ ↔ G ( X ) is closed in G ( X ), itsuffices to show that ↔ λ ( X ) is closed in ↔ N ( X ). Take any invariant linked system L ∈ ↔ N ( X ) \ ↔ λ ( X ). Being not maximal invariant, the linked system L can beenlarged to a maximal invariant linked system M that contains a subset B ∈ M\L .Since M ∋ B is invariant, the system { xB : x ∈ X } ⊂ M is linked. Observe that B / ∈ L and B ∈ M ⊃ L implies X \ B ∈ L ⊥ and B ∈ L ⊥ . We claim that O ( L ) = B − ∩ ( X \ B ) − ∩ ↔ N ( X ) is a neighborhood of L in ↔ N ( X ) that misses theset ↔ λ ( X ). Indeed, for any A ∈ O ( L ), we get that A is an invariant linked systemsuch that B ∈ A ⊥ . Observe that for every x ∈ X and A ∈ A we get x − A ∈ A bythe invariantness of A and hence the set B ∩ x − A and its shift xB ∩ A both are notempty. This witnesses that xB ∈ A ⊥ for every x ∈ X . Then the maximal invariantlinked system generated by A ∪ { xB : x ∈ X } is an invariant linked enlargement of A , which shows that A is not maximal invariant linked. (cid:3) Next, we shall evaluate the cardinality of ↔ λ ( X ). Theorem 2.2.
For any infinite group X the semigroup ↔ λ ( X ) has cardinality | ↔ λ ( X ) | = 2 | X | .Proof. The upper bound | ↔ λ ( X ) | ≤ | X | follows from the chain of inclusions: ↔ λ ( X ) ⊂ G ( X ) ⊂ P ( P ( X )) . Now we prove that | ↔ λ ( X ) | ≥ | X | . Let | X | = κ and X = { x α : α < κ } be aninjective enumeration of X by ordinals < κ such that x is the neutral element of X . For every α < κ let B α = { x β , x − β : β < α } . By transfinite induction, choose atransfinite sequence ( a α ) α<κ such that a = x and a α / ∈ B − α B α A <α where A <α = { a β : β < α } . LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 11
Consider the set A = { a α : α < κ } . By [HS, 3.58], the set U κ ( A ) of κ -uniformultrafilters on A has cardinality | U κ ( A ) | = 2 κ . We recall that an ultrafilter U is κ -uniform if for every set U ∈ U and any subset K ⊂ U of size | K | < κ the set U \ K still belongs to U .To each κ -uniform ultrafilter U ∈ U κ ( A ) assign the invariant filter F U = T x ∈ X x U .This filter can be extended to a maximal invariant linked system L U . We claimthat L U = L V for two different κ -uniform ultrafilters U , V on A . Indeed, U 6 = V yields a subset U ⊂ A such that U ∈ U and U / ∈ V . Let V = A \ U . Since U , V are κ -uniform, | U | = | V | = κ .For every α < κ consider the sets U α = { a β ∈ U : β > α } ∈ U and V α = { a β ∈ V : β > α } ∈ V .It is clear that F U = [ α<κ x α U α ∈ F U and F V = [ α<κ x α V α ∈ F V . Let us show that F U ∩ F V = ∅ . Otherwise there would exist two ordinals α, β and points u ∈ U α , v ∈ V β such that x α u = x β v . It follows from u = v that α = β .Write the points u, v as u = a γ and v = a δ for some γ > α and δ > β . Then wehave the equality x α a γ = x β a δ . The inequality u = v implies that γ = δ . We loseno generality assuming that δ > γ . Then a δ = x − β x α a γ ∈ B − δ B δ A <δ which contradicts the choice of a δ .Therefore, F U ∩ F V = ∅ . Taking into account that the linked systems L U ⊃ F U ∋ F U and L V ⊃ F V ∋ F V contain disjoint sets F U , F V , we conclude that L U = L V .Consequently, | ↔ λ ( X ) | ≥ |{L U : U ∈ U κ ( A ) }| = | U κ ( A ) | = 2 κ . (cid:3) The preceding theorem implies that | ↔ λ ( G ) | = 2 c for any countable group G .Next, we evaluate the cardinality of ↔ λ ( G ) for finite groups G .Given a finite group G consider the invariant linked system L = { A ⊂ X : 2 | A | > | G |} and the subset ↑L = {A ∈ ↔ λ ( G ) : A ⊃ L } of ↔ λ ( G ). Proposition 2.3.
Let G be a finite group. If sl ( G ) ≥ | G | / , then ↔ λ ( G ) = ↑L . Proof.
We should prove that each maximal invariant linked system
A ∈ ↔ λ ( G )contains L . Take any set L ∈ L . Taking into account that sl ( G ) ≥ | G | / A ∈ A is self-linked, we conclude that | A | ≥ | G | / A intersectseach shift xL of L (because | A | + | xL | > | G | ). Since the set L is self-linked, we getthat the invariant linked system A ∪ { xL : x ∈ G } is equal to A by the maximalityof A . Consequently, L ∈ A and hence L ⊂ A . (cid:3) In light of Proposition 2.3 it is important to evaluate the cardinality of the set ↑L . In | G | is odd, then the invariant linked system L is maximal linked and thus ↑L is a singleton. The case of even | G | is less trivial.Given an group G of finite even order | G | , consider the family S = { A ⊂ G : AA − = G, | A | = | G | / } of self-linked subsets A ⊂ G of cardinality | A | = | G | /
2. On the family S considerthe equivalence relation ∼ letting A ∼ B for A, B ∈ S if there is x ∈ G such that A = xB or X \ A = xB . Let S / ∼ the quotient set of S by this equivalence relationand s = |S / ∼ | stand for the cardinality of S/ ∼ . Proposition 2.4. | ↔ λ ( G ) | ≥ |↑L | = 2 s .Proof. First we show that ∼ indeed is an equivalence relation on S . So, assumethat S 6 = ∅ . Let us show that G \ A ∈ S for every A ∈ S . Let B = G \ A . Assumingthat B / ∈ S , we conclude that B ∩ xB = ∅ for some x ∈ G . Since | B | = | A | = | G | / xB = A and G \ A = B = x − A . The equality A ∩ x − A = ∅ implies x − / ∈ AA − = G , which is a contradiction.Taking into account that A = eA for every A ∈ S , we conclude that ∼ is areflexive relation on S . If A ∼ B , then there is x ∈ X such that A = xB or G \ A = xB . This implies that B = x − A or X \ B = x − A , that is B ∼ A and ∼ is symmetric. It remains to prove that the relation ∼ is transitive on S . So let A ∼ B ∼ C . This means that there exist x, y ∈ G such that A = xB or G \ A = xB and B = yC or G \ B = yC . It is easy to check that in these cases A = xyC or X \ A = xyC .Choose a subset T of S intersecting each equivalence class of ∼ at a single point.Observe that |T | = |S / ∼ | = s . Now for every function f : T → { , } considerthe maximal invariant linked system L f = L ∪ { xT : x ∈ G, T ∈ f − (0) } ∪ { x ( G \ T ) : x ∈ G, T ∈ f − (1) } . It can be shown that |↑L | = |{L f : f ∈ T }| = 2 |T | = 2 s . (cid:3) LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 13
This proposition will help us to calculate the cardinality of the set ↔ λ ( G ) for allfinite groups G of order | G | ≤ Theorem 2.5.
The cardinality of ↔ λ ( G ) for a group G of size | G | ≤ can be foundfrom the table: G C C C C ⊕ C C D C C C D C ⊕ C C Q sl ( G ) 2 2 3 3 3 4 3 3 5 4 4 4 4 ↔ λ ( X ) 1 1 1 1 1 1 2 3 1 2 4 8 8 Proof.
We divide the proof into 5 cases.1. If sl ( G ) > | G | /
2, then L is a unique maximal invariant linked system andthus | ↔ λ ( X ) | = 1. By Theorem 1.2, sl ( G ) > | G | / | G | ≤ G isisomorphic to D or C .2. If sl ( G ) = | G | /
2, then | ↔ λ ( G ) | = 2 s where s = |S / ∼ | . So it remains to calculatethe number s for the groups C , D , C ⊕ C , C , and Q .2a. If G is cyclic of order 6, then we can take any generator a on G and byroutine calculations, check that S = { xT, x ( G \ T ) : x ∈ G } where T = { e, a, a } . It follows that s = |S / ∼ | = 1 and thus | ↔ λ ( G ) | = |↑L | = 2 s = 2 . G is cyclic of order 8, then we can take any generator a on G and byroutine verification check that S = { xA, G \ xA, xB, G \ xB, C, G \ xC : x ∈ G } where A = { e, a, a , a } , B = { e, a, a , a } , and C = { e, a, a , a } . It follows that s = |S / ∼ | = 3 and thus | ↔ λ ( G ) | = |↑L | = 2 s = 8 . G is isomorphic to C ⊕ C and let G = { x ∈ G : xx = e } be the Boolean subgroup of G . We claim that a 4-element subset A ⊂ G is self-linked if and only if | A ∩ G | is odd.To prove the “if” part of this claim, assume that | A ∩ G | = 3. We claimthat A is self-linked. Let A = A ∩ G and note that G = A A − ⊂ AA − because | A | = 3 > | G | /
2. Now take any element a ∈ A \ G and note that AA − ⊃ aA − ∪ A a − . Observe that both aA − = aA and A a − = a − A are3-element subsets in the 4-element coset aG . Those 3-element sets are different.Indeed, assuming that aA − = A a − we would obtain that a A = A whichimplies that | A | = 3 is even. Consequently, aG = aA − ∪ A a − ⊂ AA − andfinally G = AA − . If | A ∩ G | = 1, then we can take any a ∈ A \ G and consider the shift Aa − which has | Aa − ∩ G | = 3. Then the preceding case implies that Aa − is self-linkedand so is A .To prove the “only if” part of the claim assume that | A ∩ G | is even. If | A ∩ G | =4, then A = G and AA − = G G − = G = G . If | A ∩ G | = 0, then A = G a for any a ∈ A and hence AA − = G aa − G − = G = G . If | A ∩ G | = 2, then | G ∩ AA − | ≤ AA − = G .Thus S = { A ⊂ G : | A | = 4 and | A ∩ G | is odd } . Each set A ∈ S has a unique shift aA with aA ∩ G = { e } . There are exactly foursubsets A ∈ S with A ∩ G = { e } forming two equivalence classes with respect tothe relation ∼ . Therefore s = 2 and | ↔ λ ( G ) | = |↑L | = 2 s = 4 . G is isomorphic to the dihedral group D of isometries ofthe square. Then G contains an element a of order 4 generating a normal cyclicsubgroup H . The element a commutes with all the elements of the group G .We claim that for each self-linked 4-element subset A ⊂ G we get | A ∩ H | = 2.Indeed, if | A ∩ H | equals 0 or 4, then A = Hb for some b ∈ G and then AA − = Abb − A − = H = G . If | A ∩ H | equals 1 or 3, then replacing A by a suitableshift, we can assume that A ∩ H = { e } and hence A = { e } ∪ B for some 3-elementsubset B ⊂ G \ H . It follows that G \ H = AA − \ H = ( B ∪ B − ) = B = G \ H .This contradiction shows that | A ∩ H | = 2. Without loss of generality, we canassume that A ∩ H = { e, a } (if it is not the case, replace A by its shift Ax − where x, y ∈ A are such that yx − = a ). Now take any element b ∈ A \ H . Since G is notcommutative, we get ab = ba . Observe that ba / ∈ A (otherwise A = { e, b, a , ba } would be a subgroup of G with AA − = A = G ). Consequently, the 4-th element c ∈ A \ { e, a , b } of A should be of the form c = ba or c = ba = ab . Observe thatboth the sets A = { e, a , b, ba } and A = { e, a , b, ab } are self-linked. Observe alsothat a ( G \ A ) = a · { a, a , ba , ba } = { e, a , ab, b } = A . Consequently, s = |S / ∼ | = 1 and | ↔ λ ( G ) | = 2 s = 2.2e. Finally assume that G is isomorphic to the group Q = {± , ± i, ± j, ± k } ofquaternion units. The two-element subset H = {− , } is a normal subgroup in X .Let S ± = { A ∈ S : H ⊂ A } and observe that each set A ∈ S has a left shift in S . Take any set A ∈ S ± and pick a point a ∈ A \ { , − } . Observe that the 4-thelement b ∈ A \ { , − , a } of A is not equal to − a (otherwise, A is a subgroup of G ). LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 15
Conversely, one can easily check that each set A = { , − , a, b } with a, b ∈ G \ H and a = − b is self-linked. This means that S ± = {{− , , a, b } : a = − b and a, b ∈ G \ H } and thus |S ± | = C − A ∈ S the set − A ∈ S and there axactly two shifts of X \ A that belong to S . This means that theequivalence class [ A ] ∼ of any set A ∈ S intersects S in four sets. Consequently, s = |S / ∼ | = |S ± | / / | ↔ λ ( G ) | = |↑L | = 2 s = 8 .
3. If | G | = 7, then L is one of three elements of ↔ λ ( G ). The other two elementscan be found as follows. Consider the invariant linked system L = { A ⊂ G : | A | ≥ } and observe that L ⊂ A for each A ∈ ↔ λ ( G ). Indeed, assuming that some A ∈ L does not belong to A , we would conclude that B = G \ A ∈ A by the maximalityof A . Since | G \ B | ≤ x ∈ G \ BB − . It follows that B, xB are twodisjoint sets in A which is not possible. Thus L ⊂ A .Observe that L ⊂ A ⊂ L ∪ L , where L = { A ⊂ G : | A | = 3 , AA − = G } . Given a generator a of the cyclic group G , consider the 3-element set T = { a, a , a } and note that T T − = G and T − ∩ T = ∅ . By a routine calculation, one can checkthat L = { xT, xT − : x ∈ G } . Since T and T − are disjoint, the invariant linked system A cannot contain boththe sets T and T − . If A contains none of the sets T, T − , then A = L . If A contains T , then A = ( L ∪ { xT : x ∈ G } ) \ { y ( G \ T ) : y ∈ G } . If T − ∈ A , then A = ( L ∪ { xT − : x ∈ G } ) \ { y ( G \ T − ) : y ∈ G } . And those are the unique 3 maximal invariant systems in ↔ λ ( G ). (cid:3) In the following theorem we characterize groups possessing a unique maximalinvariant linked system.
Theorem 2.6.
For a finite group G the following conditions are equivalent: (1) | ↔ λ ( G ) | = 1 ; (2) sl ( G ) > | G | / ; (3) | G | ≤ or else G is isomorphic to D or C .Proof. (2) ⇒ (1). If sl ( G ) > | G | /
2, then L = { A ⊂ G : | A | > | G | / } is a uniquemaximal invariant linked system on G (because invariant linked systems composeof self-linked sets).(1) ⇒ (2) Assume that sl ( G ) ≤ | G | / A ⊂ G with | A | ≤ | G | /
2. If | G | is odd, then L is maximal linked and then any maximal invari-ant linked system A containing the self-linked set A is distinct from L , witnessingthat | ↔ λ ( G ) | > G is even, then we can enlarge A , if necessary, and assume that | A | = | G | / B = G \ A of A is self-linked too. Assuming theconverse, we would find some x / ∈ BB − and conclude that B ∩ xB = ∅ , whichimplies that A = G \ B = xB and hence x − A = B . Then the sets A and x − A aredisjoint which contradicts x − ∈ AA − = G . Thus BB − = G which implies that { xB : x ∈ G } is an invariant linked system. Since | G | = 2 | A | is even, the unions A = { xA : x ∈ G } ∪ L and B = { xB : x ∈ G } ∪ L are invariant linked systemsthat can be enlarged to maximal linked systems ˜ A and ˜ B , respectively. Since thesets A ∈ A ⊂ ˜ A and B ∈ B ⊂ ˜ B are disjoint, ˜ A 6 = ˜ B are two distinct maximalinvariant systems on G and thus | ↔ λ ( G ) | ≥ ⇔ (3) follows from Theorem 1.2(i). (cid:3) Right zeros in λ ( X )In this section we return to studying the superextensions of groups and shalldetect groups X whose superextensions λ ( X ) have right zeros. We shall show thatfor every group X the right zeros of λ ( X ) coincide with invariant maximal linkedsystems.We recall that an element z of a semigroup S is called a right (resp. left ) zero in S if xz = z (resp. zx = z ) for every x ∈ S . This is equivalent to saying that thesingleton { x } is a left (resp. right) ideal of S .By [G2, 5.1] an inclusion hyperspace A ∈ G ( X ) is a right zero in G ( X ) if andonly if A is invariant. This implies that the minimal ideal of the semigroup G ( X )coincides with the set ↔ G ( X ) of invariant inclusion hyperspaces and is a compactrectangular topological semigroup. We recall that a semigroup S is called rectan-gular if xy = y for all x, y ∈ S .A similar characterization of right zeros holds also for the semigroup λ ( X ). Proposition 3.1.
A maximal linked system L is a right zero of the semigroup λ ( X ) if and only if L is invariant.Proof. If L is invariant, then by proposition 5.1 of [G2], L is a right zero in G ( X )and consequently, a right zero in λ ( X ). LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 17
Assume conversely that L is a right zero in λ ( X ). Then for every x ∈ X we get x L = L , which means that L is invariant. (cid:3) Unlike the semigroup G ( X ) which always contains right zeros, the semigroup λ ( X ) contains right zeros only for so-called odd groups. We define a group X to be odd if each element x ∈ X has odd order. We recall that the order of an element x is the smallest integer number n ≥ x n coincides with the neutralelement e of X . Theorem 3.2.
For a group X the following conditions are equivalent: (1) the semigroup λ ( X ) has a right zero; (2) some maximal invariant linked system on X is maximal linked (which canbe written as ↔ λ ( X ) ∩ λ ( X ) = ∅ ); (3) each maximal invariant linked system is maximal linked (which can be writ-ten as ↔ λ ( X ) ⊂ λ ( X ) ); (4) for any partition X = A ∪ B either AA − = X or BB − = X ; (5) each element of X has odd order.Proof. The equivalence (1) ⇔ (2) follows from Proposition 3.1.(2) ⇒ (4) Assume that λ ( X ) contains an invariant maximal linked system A .Given any partition X = A ∪ A , use the maximality of A to find i ∈ { , } with A i ∈ A . We claim that A i A − i = X . Indeed, for every x ∈ X the invariantness of A implies that xA i ∈ A and hence A i ∩ xA i = ∅ , which implies x ∈ A i A − i .(4) ⇒ (3) Assume that for every partition X = A ∪ B either AA − = X or BB − = X . We need to check that each maximal invariant linked system L ismaximal linked. In the other case, there would exist a set A ∈ L ⊥ \ L . Since L 6∋ A is maximal invariant linked system, some shift xA of A does not intersect A and thus x / ∈ AA − . Then our assumption implies that B = X \ A has property BB − = X , which means that the family { xB : x ∈ X } is linked. We claim that B ∈ L ⊥ . Assuming the converse, we would find a set L ∈ L with L ∩ B = ∅ andconclude that A ∈ L because L ⊂ X \ B = A . But this contradicts the choice of A ∈ L ⊥ \ L . Therefore B ∈ L ⊥ and L ∪ { L ⊂ X : ∃ x ∈ X ( xB ⊂ L ) } is an invariant linked system that enlarges L . Since L is a maximal invariant linkedsystem, we conclude that B ∈ L , which is not possible because B does not intersect A ∈ L ⊥ . The obtained contradiction shows that L ⊥ \ L = ∅ , which means that L belongs to λ ( X ) and thus is an invariant maximal linked system.The implication (3) ⇒ (2) is trivial. ¬ (5) ⇒ ¬ (4) Assume that X \ { e } contains a point a whose order is even orinfinity. Then the cyclic subgroup H = { a n : n ∈ Z } generated by a decomposesinto two disjoint sets H = { a n : n ∈ Z + 1 } and H = { a n : n ∈ Z } suchthat aH = H . Take a subset S ⊂ X meeting each coset Hx , x ∈ X , in asingle point. Consider the disjoint sets A = H S and A = H S and note that aA = A = X \ A and aA = X \ A , which implies that a / ∈ A i A − i for i ∈ { , } .Since A ∪ A = X , we get a negation of (4).(5) ⇒ (4) Assume that each element of X has odd order and assume that X admits a partition X = A ⊔ B such that a / ∈ AA − and b / ∈ BB − for some a, b ∈ X .Then aA ⊂ X \ A = B and bB ⊂ X \ B = A . Observe that baA ⊂ bB ⊂ A and by induction, ( ba ) i A ⊂ A for all i >
0. Since all elements of X have finiteorder, ( ba ) n = e for some n ∈ N . Then ( ba ) n − A ⊂ A implies A = ( ba ) n A ⊂ baA ⊂ bB ⊂ A and hence bB = A . It follows from X = bA ⊔ bB = bA ⊔ A = B ⊔ A that bA = B . Thus x ∈ A if and only if bx ∈ B .Let H = { b n : n ∈ Z } ⊂ X be the cyclic subgroup generated by b . By ourassumption it is of odd order. On the other hand, the equality bB = A = b − B implies that the intersections H ∩ A and H ∩ B have the same cardinality because b ( B ∩ H ) = A ∩ H . But this is not possible because of the odd cardinality of H . (cid:3) (Left) zeros of the semigroup λ ( X )An element z of a semigroup S is called a zero in S if xz = z = zx for all x ∈ S .This is equivalent to saying that z is both a left and right zero in S . Proposition 4.1.
Let X be a group. For a maximal linked system L ∈ λ ( X ) thefollowing conditions are equivalent: (1) L is a left zero in λ ( X ) ; (2) L is a zero in λ ( X ) ; (3) L is a unique invariant maximal linked system on X .Proof. (1) ⇒ (3) Assume that Z is a left zero in λ ( X ). Then Z x = Z for all x ∈ X and thus Z − = { Z − : Z ∈ Z} LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 19 is an invariant maximal linked system on X , which implies that the group X is oddaccording to Theorem 3.2. Note that for every right zero A of λ ( X ) we get Z = Z ◦ A = A which implies that Z is a unique right zero in λ ( X ) and by Proposition 3.1 a uniqueinvariant maximal linked system on X .(3) ⇒ (2) Assume that Z is a unique invariant maximal linked system on X .We claim that Z is a left zero of λ ( X ). Indeed, for every A ∈ A and x ∈ X we get x Z ◦ A = Z ◦ A , which means that
Z ◦ A is an invariant maximal linked system. ByProposition 3.1,
Z ◦ A is a right zero and hence
Z ◦ A = Z because Z is a uniqueright zero. This means that Z is a left zero, and being a right zero, a zero in λ ( X ).(2) ⇒ (1) is trivial. (cid:3) Theorem 4.2.
The superextension λ ( X ) of a group X has a zero if and only if X is isomorphic to C , C or C .Proof. If X is a group of odd order | X | ≤
5, then ↔ λ ( X ) ⊂ λ ( X ) because X is oddand | ↔ λ ( X ) | = 1 by Theorem 2.6. This means that λ ( X ) contains a unique invariantmaximal linked system, which is the zero of λ ( X ) by Proposition 4.1.Now assume conversely that the semigroup λ ( X ) has a zero element Z . ByProposition 3.1 and Theorem 3.2, X is odd and thus ↔ λ ( X ) ⊂ λ ( X ). Since the zero Z of λ ( X ) is a unique invariant maximal linked system on X , we get | ↔ λ ( X ) | ≤ X has order | X | ≤ D or C . Since X isodd, X must be isomorphic to C , C or C . (cid:3) The commutativity of λ ( X )In this section we detect groups X with commutative superextension. Theorem 5.1.
The superextension λ ( X ) of a group X is commutative if and onlyif | X | ≤ .Proof. The commutativity of the superextensions λ ( X ) of groups X of order | X | ≤ X has commutative superextension λ ( X ). Then X iscommutative. We need to show that | X | ≤
4. First we show that | ↔ λ ( X ) | = 1.Assume that ↔ λ ( X ) contains two distinct maximal invariant linked systems A and B . Taking into account that A , B ∈ ↔ λ ( X ) ⊂ ↔ G ( X ) and each element of ↔ G ( X )is a right zero in G ( X ) (see [G2, 5.1]) we conclude that A ◦ B = B 6 = A = B ◦ A . Extend the linked systems systems A , B to maximal linked systems ˜ A ⊃ A and˜
B ⊃ B . Because of the commutativity of λ ( X ), we get A = B ◦ A ⊂ ˜ B ◦ ˜ A = ˜ A ◦ ˜ B ⊃ A ◦ B = B . This implies that the union
A ∪ B 6 = A is an invariant linked system extending A ,which is not possible because of the maximality of A . This contradiction shows that | ↔ λ ( X ) | = 1. Applying Theorem 2.6, we conclude that | X | ≤ X is isomorphicto C .It remains to show that the semigroups λ ( C ) and λ ( C ) are not commutative.The non-commutativity of λ ( C ) will be shown in Section 6.To see that λ ( C ) is not commutative, take any 3 generators a, b, c of C andconsider the sets A = { e, a, b, abc } , H = { e, a, b, ab } , H = { e, a, bc, abc } . Observethat H , H are subgroups in C . For every i ∈ { , } consider the linked system A i = h{ H , H } ∪ { xA : x ∈ H i }i and extend it to a maximal linked system ˜ A i on C .We claim that the maximal linked systems ˜ A and ˜ A do not commute. Indeed,˜ A ◦ ˜ A ∋ [ x ∈ H x ∗ ( x − bA ) = bA = { e, b, ba, ac } , ˜ A ◦ ˜ A ∋ [ x ∈ H x ∗ ( x − bcA ) = bcA = { a, c, bc, abc } . It follows from bA ∩ bcA = ∅ that ˜ A ◦ ˜ A = ˜ A ◦ ˜ A . (cid:3) The superextensions of finite groups
In this section we shall describe the structure of the superextensions λ ( G ) offinite groups G of small cardinality (more precisely, of cardinality | G | ≤ λ ( G ) growth very quickly as | G | tends to infinity.The calculation of the cardinality of | λ ( G ) | seems to be a difficult combinatorialproblem related to the still unsolved Dedekind’s problem of calculation of the num-ber M ( n ) of inclusion hyperpspaces on an n -element subset, see [De]. We were ableto calculate the cardinalities of λ ( G ) only for groups G of cardinality | G | ≤
6. Theresults of (computer) calculations are presented in the following table: | G | | λ ( G ) | | λ ( G ) /G | S containing a group G . Inthis case S can be thought as a G -space endowed with the left action of the group G . So we can consider the orbit space S/G = { Gs : s ∈ S } and the projection LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 21 π : S → S/G . If G lies in the center of the semigroup S (which means that theelements of G commute with all the elements of S ), then the orbit space S/G admitsa unique semigroup operation turning
S/G into a semigroup and the orbit projection π : S → S/G into a semigroup homomorphism. A subsemigroup T ⊂ S will becalled a transversal semigroup if the restriction π : T → S/G is an isomorphism ofthe semigroups. If S admits a transversal semigroup T , then it is a homomoprhicimage of the product G × T under the semigroup homomorphism h : G × T → S, h : ( g, t ) gt. This helps to recover the algebraic structure of S from the structure of a transversalsemigroup.For a system B of subsets of a set X by hBi = { A ⊂ X : ∃ B ∈ B ( B ⊂ A ) } we denote the inclusion hyperspace generated by B .Now we shall analyse the entries of the above table. First note that each group G of size | G | ≤ C , C , C , C , C ⊕ C , C . It will be convenient to think of the cyclic group C n as themultiplicative subgroups { z ∈ C : z n = 1 } of the complex plane.6.1. The semigroups λ ( C ) and λ ( C ) . For the groups C n with n ∈ { , } thesemigroup λ ( C n ) coincides with C n while the orbit semigroup λ ( C n ) /C n is trivial.6.2. The semigroup λ ( C ) . For the group C the semigroup λ ( C ) contains thethree principal ultrafilters 1 , z, − z where z = e πi/ and the maximal linked in-clusion hyperspace ⊲ = h{ , z } , { , − z } , { z, − z }i which is the zero in λ ( C ). Thesuperextension λ ( C ) is isomorphic to the multiplicative semigroup C = { z ∈ C : z = z } of the complex plane. The latter semigroup has zero 0 and unit 1 whichare the unique idempotents.The transversal semigroup λ ( C ) /C is isomorphic to the semilattice 2 = { , } endowed with the min-operation.6.3. The semigroups λ ( C ) and λ ( C ⊕ C ) . The semigroup λ ( C ) contains 12elements while the orbit semigroup λ ( C ) /C contains 3 elements. The semigroup λ ( C ) contains a transversal semigroup λ T ( G ) = { , △ , (cid:3) } where 1 is the neutral element of C = { , − , i, − i } , △ = h{ , i } , { , − i } , { i, − i }i and (cid:3) = h{ , i } , { , − i } , { , − } , { i, − i, − }i . The transversal semigroup is isomorphic to the extension C = C ∪ { e } of thecyclic group C by an external unit e / ∈ C (such that ex = x = xe for all x ∈ C ).The action of the group C on λ ( C ) is free so, λ ( C ) is isomorphic to λ T ( C ) ⊕ C .The semigroup λ ( C ⊕ C ) has a similar algebraic structure. It contains atransversal semigroup λ T ( C ⊕ C ) = { e, △ , (cid:3) } ⊂ λ ( C ⊕ C )where e is the principal ultrafilter supported by the neutral element (1 ,
1) of C ⊕ C and the maximal linked inclusion hyperspaces △ and (cid:3) are defined by analogy withthe case of the group C : △ = h{ (1 , , (1 , − } , { (1 , , ( − , } , { (1 , − , ( − , }i and (cid:3) = h{ (1 , , (1 , − } , { (1 , , ( − , } , { (1 , , ( − , − } , { (1 , − , ( − , , ( − , − }i . The transversal semigroup λ T ( C ⊕ C ) is isomorphic to C and λ ( C ⊕ C ) isisomorphic to C ⊕ C ⊕ C .We summarize the obtained results on the algebraic structure of the semigroups λ ( C ) and λ ( C ⊕ C ) in the following proposition. Proposition 6.1.
Let G be a group of cardinality | G | = 4 . (1) The semigroup λ ( G ) is isomorphic to C ⊕ G and thus is commutative; (2) λ ( G ) contains two idempotents; (3) λ ( G ) has a unique proper ideal λ ( G ) \ G isomorphic to the group C ⊕ G . The semigroup λ ( C ) . Unlike to λ ( C ), the semigroup λ ( C ) has compli-cated algebraic structure. It contains 81 elements. One of them is zero Z = { A ⊂ C : | A | ≥ } , which is invariant under any bijection of C . All the other 80 elements have 5-element orbits under the action of C , which implies that the orbit semigroup λ ( C ) /C consists of 17 elements. Let π : λ ( C ) → λ ( C ) /C denote the orbitprojection.It will be convenient to think of C as the field { , , , , } with the multi-plicative subgroup C ∗ = { , − , , − } of invertible elements (here − − x, y, z ∈ C we shall write xyz instead of { x, y, z } .The semigroup λ ( C ) contains 5 idempotents: U = h i , Z , Λ = h , , , , i , Λ = h , , , , i ,
2Λ = h , , , , i , LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 23 which commute and thus form an abelian subsemigroup E ( λ ( C )). Being a semi-lattice, E ( λ ( C )) carries a natural partial order: e ≤ f iff e ◦ f = e . The partialorder Z ≤ Λ , ≤ Λ ≤ U on the set E ( λ ( C )) is designed at the picture: Z r (cid:0)(cid:0)❅❅ r Λ r r ❅❅ ❅❅(cid:0)(cid:0) r U The other distinguished subset of λ ( C ) is p E ( λ ( C )) = {L ∈ λ ( C ) : L ◦ L ∈ E ( λ ( C )) } == {L ∈ λ ( C ) : L ◦ L ◦ L ◦ L = L ◦ L} . We shall show that this set contains a point from each C -orbit in λ ( C ).First we show that this set has at most one-point intersection with each orbit.Indeed, if L ∈ p E ( λ ( C )) and L ◦ L 6 = Z , then for every a ∈ C \ { } , we get( L + a ) ◦ ( L + a ) ◦ ( L + a ) ◦ ( L + a ) = L ◦ L ◦ L ◦ L + 4 a == L ◦ L + 4 a = L ◦ L + 2 a = ( L + a ) ◦ ( L + a ) . witnessing that L + a / ∈ p λ T ( C ).By a direct calculation one can check that the set λ T ( C ) contains the followingfour maximal linked systems:∆ = h , , i , Λ = h , , , i , Θ = h , , , , , , i , Γ = h , , , , i . For those systems we get∆ ◦ ∆ = ∆ ◦ ∆ ◦ ∆ = Λ , Λ ◦ Λ = Λ ◦ Λ ◦ Λ = Λ , F ◦
Θ =
F ◦
Γ = Z for every F ∈ λ ( C ) \ C . All the other elements of λ ( C ) can be found as images of ∆ , Θ , Γ , Λ under theaffine transformations of the field C . Those are maps of the form f a,b : x ax + b mod 5 , where a ∈ { , − , , − } = C ∗ and b ∈ C . The image of a maximal linked system L ∈ λ ( C ) under such a transformation will be denoted by a L + b .One can check that a Λ = Λ for each a ∈ C ∗ while Λ = − Λ, and Θ = − Θ.Since the linear transformations of the form f a, : C → C , a ∈ C ∗ , are autho-morphisms of the group C the induced transformations λf a, : λ ( C ) → λ ( C )are authomorphisms of the semigroup λ ( C ). This implies that those transforma-tions do not move the subsets E ( λ ( C )) and p E ( λ ( C )). Consequently, the set p E ( λ ( C ) contains the maximal linked systems: a ∆ , a Θ , a Λ , a Γ , a ∈ Z ∗ , which together with the idempotents form a 17-element subset T = E ( λ ( C )) ∪ (cid:8) a ∆ , a Θ : a ∈ { , } (cid:9) ∪ { a Λ , a Γ : a ∈ Z ∗ } that projects bijectively onto the orbit semigroup λ ( C ) /C . The set T looks asfollows (we connect an element x ∈ T with an idempotent e ∈ T by an arrow if x ◦ x = e ): Z (cid:0)(cid:0)❅❅ Λ 2ΛΛ ❅❅ ❅❅(cid:0)(cid:0) U− ΓΘ ✲ ✛(cid:0)(cid:0)✒ Γ ✂✂✂✍ ❇❇❇▼ − ❅❅■ − Λ ✑✑✸ ∆ ✲ Λ ◗◗s − ◗◗❦ ✛ ✑✑✰ The set p E ( λ ( C )) includes 24 elements more and coincides with the union T ∪ √Z where √Z = { a Θ + b, a
Γ + b : a ∈ Z ∗ , b ∈ C } . Since each element of λ ( C ) can be uniquely written as the sum L + b for some L ∈ T and b ∈ C , the multiplication table for the semigroup λ ( C ) can berecovered from the Cayley table for multiplication of the elements from T : LGEBRA IN SUPEREXTENSIONS OF GROUPS, I 25 ◦ Λ Λ ∆ Λ − Λ
2Λ 2∆ 2Λ − a Θ , a ΓΛ Λ Λ Λ Λ Λ 2Λ 2Λ 2Λ 2Λ Z Λ Λ Λ Λ Λ Λ
Z Z Z Z Z ∆ ∆ Λ Λ Λ Λ 2Θ 2Θ 2Θ 2Θ Z Λ Λ Λ Λ Λ Λ 2Θ + 2 2Θ + 2 2Θ + 2 2Θ + 2 Z− Λ − Λ Λ Λ Λ Λ 2Θ − − − − Z
2Λ 2Λ
Z Z Z Z
2Λ 2Λ 2Λ 2Λ Z
2∆ 2∆ Θ Θ Θ Θ 2Λ 2Λ 2Λ 2Λ Z Θ − − − − Z− − Θ + 1 Θ + 1 Θ + 1 Θ + 1 2Λ 2Λ 2Λ 2Λ Z Θ Θ Θ Θ Θ Θ
Z Z Z Z Z
2Θ 2Θ
Z Z Z Z
2Θ 2Θ 2Θ 2Θ Z Γ Γ Θ + 1 Θ + 1 Θ + 1 Θ + 1 2Θ + 2 2Θ + 2 2Θ + 2 2Θ + 2 Z− Γ − Γ Θ − − − − − − − − Z
2Γ 2Γ Θ − − − − Z− −
2Γ Θ + 1 Θ + 1 Θ + 1 Θ + 1 2Θ − − − − Z Looking at this table we can see that T is not a subsemigroup of λ ( C ) andhence is not a transversal semigroup for λ ( C ). This is not occasional. Proposition 6.2.
The semigroup λ ( C ) contains no transversal semigroup.Proof. Assume conversely that λ ( C ) contains a subsemigroup T that projects bi-jectively onto the orbit semigroup λ ( C ) /C . Then T must include the set E ( λ ( C ))of idempotents and also the subset p E ( λ ( C )) \ √Z . Consequently, T ⊃ {U , Z , Λ , − Λ , ∆ , , Λ , − Λ , , − } . Since 2Λ ◦ Λ = Θ − = Θ = 2∆ ◦ Λ, then there are two different points in theintersection T ∩ (Θ + C ) which should be a singleton. This contradiction completesthe proof. (cid:3) Analysing the Cayley table for the set T we can establish the following prop-erties of the semigroup λ ( C ). Proposition 6.3. (1)
The maximal linked system Z is the zero of λ ( Z ) . (2) λ ( C ) contains 5 idempotents: U , Z , Λ , Λ , , which commute. (3) The set of central elements of λ ( C ) coincides with C ∪ {Z} . (4) All non-trivial subgroups of λ ( C ) are isomorphic to C . Summary table.
The obtained results on the superextensions of groups G with | G | ≤ K ( λ ( G )) stands forthe minimal ideal of λ ( G ). | G | | λ ( G ) | λ ( G ) | E ( λ ( G )) | K ( λ ( G )) maximal group2 2 C C C C ∪ { ⊲ } { ⊲ } C C × G C × G C × G T · C {Z} C References [BG2] T. Banakh, V. Gavrylkiv. Algebra in superextension of groups, II: cancelativity and centers,preprint.[BG3] T. Banakh, V. Gavrylkiv. Algebra in superextension of groups, III: the minimal ideal of λ ( G ), preprint.[De] R. Dedekind, ¨Uber Zerlegungen von Zahlen durch ihre gr¨ussten gemeinsammen Teiler // InGesammelte Werke, Bd. (1897), 103–148.[G1] V. Gavrylkiv. The spaces of inclusion hyperspaces over noncompact spaces , Matem. Studii. (2007), 92–110.[G2] V. Gavrylkiv,
Right-topological semigroup operations on inclusion hyperspaces , Matem.Studii. (to appear)[H1] N. Hindman,
Finite sums from sequences within cells of partition of N // J. Combin. TheorySer. A (1974), 1–11.[H2] N. Hindman, Ultrafilters and combinatorial number theory // Lecture Notes in Math. (1979), 49–184.[HS] N. Hindman, D. Strauss, Algebra in the Stone- ˇCech compactification, de Gruyter, Berlin,New York, 1998.[vM] J. van Mill, Supercompactness and Wallman spaces, Math. Centre Tracts. . Amsterdam:Math. Centrum., 1977.[P] I. Protasov. Combinatorics of Numbers, VNTL, Lviv, 1997.[OA] L.A. Skorniakov et al., General Algebra, Nauka, Moscow, 1990 (in Russian).[TZ] A. Teleiko, M. Zarichnyi. Categorical Topology of Compact Hausdofff Spaces, VNTL, Lviv,1999. Ivan Franko National University of Lviv, Ukraine
E-mail address : [email protected] Vasyl Stefanyk Precarpathian National University, Ivano-Frankivsk, Ukraine
E-mail address ::