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Algebra in superextensions of groups, I: zeros and commutativity


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Journal Algebra Discrete Math.

c Journal “Algebra and Discrete Mathematics”

Algebra in superextensions of groups, I:

zeros and commutativity

T. Banakh, V. Gavrylkiv, O. Nykyforchyn

Communicated by M. Ya. Komarnytskyi

Abstract. Given a groupX we study the algebraic structure of its superextension λ(X). This is a right-topological semigroup consisting of all maximal linked systems on X endowed with the operation

A ◦ B={CX:{xX :x−1C∈ B} ∈ A}

that extends the group operation ofX. We characterize right zeros ofλ(X)as invariant maximal linked systems onX and prove that λ(X)has a right zero if and only if each element ofXhas odd order.

On the other hand, the semigroupλ(X)contains a left zero if and only if it contains a zero if and only ifXhas odd order|X| ≤5. The semigroupλ(X)is commutative if and only if |X| ≤4. We finish the paper with a complete description of the algebraic structure of the semigroupsλ(X)for all groupsX of cardinality|X| ≤5.


After the topological proof of the Hindman theorem [H1] given by Galvin and Glazer1, topological methods become a standard tool in the modern combinatorics of numbers, see [HS], [P]. The crucial point is that any semigroup operation∗ defined on a discrete spaceX can be extended to a right-topological semigroup operation on β(X), the Stone- ˇCech com- pactification ofX. The extension of the operation fromX to β(X) can

2000 Mathematics Subject Classification: 20M99, 54B20.

Key words and phrases: Superextension, right-topological semigroup.

1Unpublished, see [HS, p.102], [H2]


Journal Algebra Discrete Math.

be defined by the simple formula:

U ◦ V ={A⊂X:{x∈X:x−1A∈ V} ∈ U}, (1) whereU,Vare ultrafilters onXandx−1A={y∈X:xy ∈A}. Endowed with the so-extended operation, the Stone- ˇCech compactification β(X) becomes a compact right-topological semigroup. The algebraic properties of this semigroup (for example, the existence of idempotents or minimal left ideals) have important consequences 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 union and intersection. In [G2] it was observed that the semigroup operation extends not only to β(X) but also to the com- plete sublattice G(X) of P(P(X)) generated by β(X). This complete sublattice consists of all inclusion hyperspaces overX.

By definition, a family F of non-empty subsets of a discrete space X is called aninclusion hyperspace if F is monotone in the sense that a subset A ⊂ X belongs to F provided A contains some set B ∈ F. On the setG(X) there is an important transversality operation assigning to each inclusion hyperspaceF ∈G(X) the inclusion hyperspace

F={A⊂X:∀F ∈ F (A∩F 6=∅)}. 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 in the double power-set P(P(X)) ={0,1}P(X) endowed with the natural product topology. The induced topology on G(X) can be de- scribed directly: it is generated by the sub-base consisting of the sets

U+={F ∈G(X) :U ∈ F} andU ={F ∈G(X) :U ∈ F} where U runs over subsets of X. Endowed with this topology, G(X) becomes a Hausdorff supercompact space. The latter means that each cover of G(X)by the sub-basic sets has a 2-element subcover.

The extension of a binary operation∗fromXtoG(X)can be defined in the same way as for ultrafilters, i.e., by the formula (1) applied to any two inclusion hyperspacesU,V ∈G(X). In [G2] it was shown that for an associative binary operation ∗ on X the space G(X) endowed with the extended operation becomes a compact right-topological semigroup. The algebraic properties of this semigroups were studied in details in [G2].

Besides the Stone- ˇCech compactification β(X), the semigroup G(X) contains many important spaces as closed subsemigroups. In particular, the space

λ(X) ={F ∈G(X) :F =F}


Journal Algebra Discrete Math.

of maximal linked systems onX is a closed subsemigroup ofG(X). The space λ(X) is well-known in General and Categorial Topology as the superextension ofX, see [vM], [TZ]. Endowed with the extended binary operation, the superextensionλ(X)of a semigroup X is a supercompact right-topological semigroup containing β(X) as a subsemigroup.

The spaceλ(X) consists of maximal linked systems onX. We recall that a system of subsetsL ofX is linkedif A∩B 6=∅ for all A, B ∈ L. An inclusion hyperspaceA ∈G(X) is linked if and only ifA ⊂ A. The family of all linked inclusion hyperspace on X is denoted by N2(X). It is a closed subset in G(X). Moreover, if X is a semigroup, then N2(X) is a closed subsemigroup ofG(X). The superextensionλ(X) consists of all maximal elements ofN2(X), see [G1], [G2].

In this paper we start a systematic investigation of the algebraic struc- ture of the semigroupλ(X). This program will be continued in the forth- coming papers [BG2] – [BG4]. The interest to studying the semigroup λ(X) was motivated by the fact that for each maximal linked system L onX and each partitionX =A∪B ofX into two setsA, B eitherA or B belongs toL. This makes possible to apply maximal linked systems to Combinatorics and Ramsey Theory.

In this paper we concentrate on describing zeros and commutativity of the semigroupλ(X). In Proposition 3.1 we shall show that a maximal linked systemL ∈λ(X)is a right zero ofλ(X)if and only ifLis invariant in the sense thatxL∈ Lfor allL∈ Land allx∈X. In Theorem 3.2 we shall prove that a group X admits an invariant maximal linked system (equivalently, λ(X) contains a right zero) if and only if each element of X has odd order. The situation with (left) zeros is a bit different: a maximal linked system L ∈λ(X) is a left zero in λ(X) if and only if L is a zero in λ(X) if and only if L is a unique invariant maximal linked system on X. The semigroup λ(X) has a (left) zero if and only if X is a finite group of odd order|X| ≤5(equivalently, X is isomorphic to the cyclic group C1,C3 or C5). The semigroupλ(X) rarely is commutative:

this holds if and only if the groupX has finite order|X| ≤4.

We start the paper studying self-linked subsets of groups. By defi- nition, a subset A of a group X is called self-linked if A∩xA 6= ∅ for all x ∈X. In Proposition 1.1 we shall give lower and upper bounds for the smallest cardinality sl(X) of a self-linked subset ofX. We use those bounds to characterize groupsX withsl(X)≥ |X|/2 in Theorem 1.2.

In Section 2 we apply self-linked sets to evaluating the cardinality of the (rectangular) semigroup λ(X) of maximal invariant linked systems on a groupX. In Theorem 2.2 we show that for an infinite groupX the cardinality ofλ(X)equals22|X|. In Proposition 2.3 and Theorem 2.6 we


Journal Algebra Discrete Math.

calculate the cardinality of λ(X) for all finite groupsX of order |X| ≤8 and also detect groups X with |λ(X)| = 1. In Sections 4 and 5 these results are applied for characterizing groups X whose superextensions have zeros or are commutative.

We finish the paper with a description of the algebraic structure of the superextensions of groupsX of order|X| ≤5.

Now a couple of words about notations. Following the algebraic tradi- tion, byCnwe denote the cyclic group of ordernand byD2nthe dihedral group of cardinality2n, that is, the isometry group of the regularn-gon.

For a group X by e we denote the neutral element of X. For a real numberx we put

⌈x⌉= min{n∈Z:n≥x}and ⌊x⌋= max{n∈Z:n≤x}. 1. Self-linked sets in groups

In this section we study self-linked subsets in groups. By definition, a subsetAof a groupGisself-linkedifA∩xA6=∅for eachx∈G. In fact, this notion can be defined in the more general context of G-spaces.

By aG-space we understand a setX endowed with a left action G× X → X of a group G. Each group G will be considered as a G-space endowed with the left 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 subsetA⊂Xof aG-spaceXdefined to beself-linkedifA∩gA6=∅ for all g ∈G. Let us observe that a subsetA ⊂G of a group G is self- linked if and only ifAA−1=G.

For a G-space X by sl(X) we denote the smallest cardinality |A|of a self-linked subsetA ⊂X. Some lower and upper bounds for sl(G) are established in the following proposition.

Proposition 1.1. Let G be a finite group and H be a subgroup of G.


1) sl(G)≥(1 +p

4|G| −3)/2;

2) sl(G)≤sl(H)·sl(G/H)≤sl(H)· ⌈(|G/H|+ 1)/2⌉. 3) sl(G)<|H|+|G/H|.

Proof. 1. Take any self-linked set A ⊂G of cardinality |A|=sl(G) and consider the surjective map f : A×A → G, f : (x, y) 7→ xy−1. Since f(x, y) = xy−1 =e for all (x, y) ∈∆A = {(x, y) ∈A2 : x =y}, we get


Journal Algebra Discrete Math.

|G| = |G\ {e}|+ 1 ≤ |A2\∆A|+ 1 = sl(G)2 −sl(G) + 1, which just implies thatsl(G)≥(1 +p

4|G| −3)/2.

2a. Let 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 setC=BAis self-linked. Given arbitraryx∈Gwe should prove that the intersection C∩xC is not empty. Since B is self-linked, the intersectionB ∩xBcontains the cosetbH =xbH for someb, b ∈B.

It follows thatb−1xb ∈H=AA−1. The latter equality follows from the fact that the setA⊂H is self-linked inH. Consequently,b−1xb =aa−1 for some a, a ∈ A. Then xC ∋ xba = ba ∈ C and thus C∩xC 6= ∅. The self-linkedness ofC implies the desired upper bound

sl(G)≤ |C| ≤ |A| · |B|=sl(H)·sl(G/H).

2b. Next, we show thatsl(G/H)≤ ⌈(|G/H|+ 1)/2⌉. Take any subset A ⊂G/H of size |A|=⌈(|G/H|+ 1)/2⌉ and note that |A|>|G/H|/2.

Then for each x∈Gthe shiftxA 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 andsl(G/H)≤ |A|=⌈(|G/H|+ 1)/2⌉.

3. Pick a subset B ⊂ G of size |B| = |G/H| such that BH = G and observe that the set A = H ∪B is self-linked and has size |A| ≤

|H|+|B| −1 (becauseB∩H is a singleton).

Theorem 1.2. For a finite group G

(i) sl(G) =⌈(|G|+ 1)/2⌉>|G|/2if and only ifGis isomorphic to one of the groups: C1, C2, C3,C4, C2×C2,C5, D6,(C2)3;

(ii) sl(G) =|G|/2 if and only if G is isomorphic to one of the groups:

C6,C8, C4×C2, D8, Q8.

Proof. I. First we establish the inequalitysl(G)<|G|/2for all groupsG not isomorphic to the groups appearing in the items (i), (ii). Given such a groupG we should find a self-linked subsetA⊂Gwith|A|<|G|/2.

We consider 8 cases.

1) Gcontains a subgroup H of order |H|= 3 and index |G/H|= 3.

Thensl(H) = 2 and we can apply Proposition 1.1(2) to conclude that sl(G)≤sl(H)·sl(G/H)≤2·2<9/2 =|G|/2.

2)|G|∈ {/ 9,12,15}andGcontains a subgroupHof ordern=|H| ≥3 and index m=|G/H| ≥3. It this case n+m−1< nm/2and sl(G) ≤

|H|+|G/H| −1 =n+m−1< nm/2 by Proposition 1.1(3).


Journal Algebra Discrete Math.

3)Gis cyclic of ordern=|G| ≥9. Given a generatoraofG, construct a sequence (xi)2≤i≤n/2 letting x2 =a0, x3 = a, x4 = a3, x5 = a5, and xi =xi−1ai for 5 < i ≤n/2. Then the set A ={xi : 2 ≤ i≤n/2} has size |A|< n/2 and is self-linked.

4) G is cyclic of order |G| = 7. Given a generator a of G observe thatA={e, a, a3}is a 3-element self-linked subset and thussl(G)≤3<


5)Gcontains a cyclic subgroupH⊂Gof prime order|H| ≥7. By the preceding two cases,sl(H)<|H|/2and thensl(G)≤sl(H)·sl(G/H)<


2 ·|G||H|=|G|/2.

6) |G| > 6 and |G| ∈ {/ 8,10,12}. If |G| is prime or |G| = 15, then G is cyclic of order |G| ≥ 7 and thus has sl(G) < |G|/2 by the items (3), (4). If |G|= 2p for some prime number p, then G contains a cyclic subgroup of order p≥7 and thus has sl(G)<|G|/2 by the item (5). If

|G| = 4n for some n≥4, then by Sylow’s Theorem (see [OA, p.74]), G contains a subgroupH ⊂Gof order|H|= 4and index|G/H| ≥4. Then sl(G)<|G|/2 by the item (2). If the above conditions do not hold, then

|G| = nm 6= 15 for some odd numbers n, m ≥ 3 and we can apply the items (1) and (2) to conclude that sl(G)<|G|/2.

7) If|G|= 8, thenGis isomorphic to one of the groups: C8,C2×C4, (C2)3,D8,Q8. All those groups appear in the items (i), (ii) and thus are excluded from our consideration.

8) If|G|= 10, thenGis isomorphic toC10 orD10. IfGis isomorphic to C10, then sl(G) <|G|/2 by the item (3). If G is isomorphic to D10, then G contains an element a of order 5 and an element b of order 2 such that bab−1 = a−1. Now it is easy to check that the 4-element set A={e, a, b, ba2} is self-linked and hencesl(G)≤4<|G|/2.

9) In this item we consider groups G with|G|= 12. It is well-known that there are five non-isomorphic groups of order 12: the cyclic group C12, the direct sum of two cyclic groups C6 ⊕C2, the dihedral group D12, the alternating group A4, and the semidirect product C3⋊C4 with presentation ha, b|a4 =b3 = 1, aba−1=b−1i.

If G is isomorphic to C12,C6⊕C2 or A4, then G contains a normal 4-element subgroupH. By Sylow’s Theorem,Gcontains also an element a of order 3. Taking into account that a2 ∈/ H and Ha−1 =a−1H, we conclude that the 5-element set A = {a} ∪H is self-linked and hence sl(G)≤5<|G|/2.

IfGis isomorphic toC3⋊C4, then Gcontains a normal subgroupH of order 3 and an element a ∈ G such that a2 ∈/ H. Observe that the 5-element set A=H∪ {a, a2} is self-linked. Indeed, AA−1 ⊃H∪aH ∪ a2H∪Ha−1 =G. Consequently, sl(G)≤5<|G|/2.


Journal Algebra Discrete Math.

Finally, consider the case of the dihedral group D12. It contains an elementagenerating a cyclic subgroup of order 6 and an elementbof order 2 such thatbab−1 =a−1. Consider the 5-element setA={e, a, a3, b, ba} and note that AA−1 ={e, a, a3, b, ba} · {e, a5, a3, b, ba}=G. This yields the desired inequality sl(G)≤5<6 =|G|/2.

Therefore we have completed the proof of the inequalitysl(G)<|G|/2 for all groups not appearing in the items (i),(ii) of the theorem.

II. Now we shall prove the item (i).

The lower bound from Proposition 1.1(1) implies thatsl(G) =⌈(|G|+ 1)/2⌉>|G|/2 for all groupsGwith|G| ≤5.

It remains to check thatsl(G)>|G|/2ifGis isomorphic toD6orC23. First we consider the caseG=D6. In this caseG contains a normal 3- element subgroupT. Assuming thatsl(G)≤ |G|/2 = 3, find a self-linked 3-element subset A. Without loss of generality we can assume that the neutral element e of G belongs toA (otherwise replace A by a suitable shiftxA). Taking into account thatAA−1=G, we conclude thatA6⊂T and thus we can find an element a∈ A\T. This element has order 2.


AA−1 ={e, a, b} · {e, a, b−1}={e, a, b, a, e, ba, b−1, ba, e} 6=G, which is a contradiction.

Now assume that G is isomorphic to C23. In this case G is the 3- dimensional linear space over the field C2. Assuming that sl(A) ≤ 4 =

|G|/2, find a 4-element self-linked subset A ⊂ G. Replacing A by a suitable shift, we can assume that A contains a neutral elemente of G.

Since AA−1 = G, the set A contains three linearly independent points a, b, c. Then

AA−1 ={e, a, b, c} · {e, a, b, c}={e, a, b, c, ab, ac, bc} 6=G, which contradicts the choice ofA.

III. Finally, we prove the equality sl(G) = |G|/2 for the groups ap- pearing in the item (ii).

If G = C6, then sl(G) ≥ 3 by Proposition 1.1(1). On the other hand, we can check that for any generator a of G the 3-element subset A={e, a, a3}is self-linked in G, which yieldssl(G) = 3 =|G|/2.

If|G|= 8, thensl(G)≥4by Proposition 1.1(1).

If G is cyclic of order 8 and a is a generator ofG, then the set A = {e, a, a3, a4}is self-linked and thussl(C8) = 4.

IfGis isomorphic toC4⊕C2, then Ghas two commuting generators a, bsuch thata4=b2= 1. One can check that the setA={e, a, a2, b}is self-linked and thus sl(C4⊕C2) = 4.


Journal Algebra Discrete Math.

IfGis isomorphic to the dihedral groupD8, thenGhas two generators a, b connected by the relations a4 =b2 = 1 and bab−1 =a−1. One can check that the 4-element subset A={e, a, b, ba2} is self-linked.

If Gis isomorphic to the groupQ8 ={±1,±i,±j,±k} of quaternion units, then we can check that the 4-element subset A = {−1,1, i, j} is self-linked and thus sl(Q8) = 4.

In the following proposition we complete Theorem 1.2 calculating the values of the cardinalsl(G)for all groupsG of cardinality |G| ≤13.

Proposition 1.3. The number sl(G) for a group Gof size |G| ≤13 can be found from the table:

G C2 C3 C5 C4 C2C2 C6 D6 C8 C2C4 D8 Q8 C23

sl(G) 2 2 3 3 3 3 4 4 4 4 4 5

G C7 C11 C13 C9 C3C3 C10 D10 C12 C2C6 D12 A4 C3C4

sl(G) 3 4 4 4 4 4 4 4 5 5 5 5

Proof. For groups Gof order|G| ≤10 the value ofsl(G)is uniquely de- termined by the lower boundsl(G)≥ 1+


2 from Proposition 1.1(1) and the upper bound from Theorem 1.2. It remains to consider the groups Gof order11≤ |G| ≤13.

1. If Gis cyclic of order 11 or 13, then take a generator a of Gand check that the 4-element set A ={e, a4, a5, a7} is self-linked, witnessing thatsl(C12) = 4.

2. If Gis cyclic of order 12, then take a generator afor Gand check that the 4-element subsetA={e, a, a3, a7} is self-linked witnessing that sl(G) = 4.

It remains to consider all other groups of order 12. Theorem 1.2 gives us an upper boundsl(G)≤5. So, we need to show thatsl(G)>4for all non-cyclic groups Gwith|G|= 12.

3. If G is isomorphic to C6 ⊕C2 or A4, then G contains a normal subgroup H isomorphic to C2⊕C2. Assuming that sl(G) = 4, we can find a 4-element self-linked subsetA⊂G. SinceAA−1 =G, we can find a suitable shiftxAsuch thatxA∩H contains the neutral elementeofG and some other elementaofH. Replacing AbyxA, we can assume that e, a∈ A. Since A 6⊂ H, there is a point b ∈ A\H. Since the quotient groupG/H has order 3,bH∩Hb−1 =∅.

Concerning the forth element c ∈ A\ {e, a, b} there are three pos- sibilities: c ∈ H, c ∈ b−1H, and c ∈ bH. If c ∈ H, then bH = bH∩AA−1 =b(A∩H)−1 consists of 3 elements which is a contradiction.

Ifc∈b−1H, thenH =H∩AA−1 ={e, a}, which is absurd. So, c∈bH and thus c=bh for some h∈H. Since h=h−1, we getcb−1 =bhb−1 =


Journal Algebra Discrete Math.

bh−1b−1 =bc−1. Then H =H∩AA−1 ={e, a, cb−1, bc−1} has cardinal- ity|H|=|{e, a, cb−1=bc−1}| ≤3, which is not true. This contradiction completes the proof of the inequality sl(G) >4 for the groups C6⊕C2 andA4.

4. Assume that G is isomorphic to the dihedral group D12. ThenG contains a normal cyclic subgroup H of order 6, and for each b∈G\H and a∈H we getb2 =eand bab−1 =a−1. Assuming that sl(D12) = 4, we can find a 4-element self-linked subset A ⊂G. Let a be a generator of the groupH. Since a∈AA−1 =G, we can find two element x, y∈A such thata=xy−1. Then the shiftAy−1 containseand a. ReplacingA by Ay−1, if necessary, we can assume that e, a∈A. Since A6⊂H, there is an element b∈A\H. Concerning the forth element c∈ A\ {e, a, b} there are two possibilities: c ∈ H and c /∈ H. If c ∈ H, then the set AH =A∩H ={e, a, c}contains three elements and is equal to bA−1H b−1, which implies bA−1H = AHb−1 = bA−1H ∪AHb−1 = AA−1 ∩bH = bH.

This is a contradiction, because|H|= 4>3 =|bA−1H |. Thenc∈bH and henceH =H∩AA−1 ={e, a, a−1, bc−1, cb−1}which is not true because

|H|= 6>5.

5. Assume that G is isomorphic to the semidirect product C3 ⋊C4 and hence has a presentationha, b|a4 =b3 = 1, aba−1 =b−1i. Then the cyclic subgroupHgenerated bybis normal inGand the quotientG/His cyclic of order 4. Assuming thatsl(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 6⊂

H, there is an element c ∈ A\H. We claim that the fourth element d∈A\ {e, b, c} does not belong toH∪cH∪c−1H. Otherwise, AA−1 ⊂ H∪cH∪c−1H 6=G. This implies that one of the elements, saycbelongs to the coset a2H and the other to aH or a−1H. We lose no generality assuming thatd∈aH. Thenc=a2bi,d=abj for somei, j∈ {−1,0,1}. It follows that

aH =aH∩AA−1 ={d, db−1, cd−1}=

={abj, abj−1, a2bi−ja−1}={abj, abj−1, abj−i}

which implies that i=−1and thus c=a2b−1. In this case we arrive to a contradiction looking at

a2H∩AA−1 ={c, cb−1, c−1, bc−1}={a2b−1, a2b−2, ba2, b2a2} 6∋a2.

Problem 1.4. What is the value of sl(G) for other groups G of small cardinality? Is sl(G) = ⌈(1 +p

4|G| −3)/2⌉ for all finite cyclic groups G?


Journal Algebra Discrete Math.

2. Maximal invariant linked systems

In this section we study (maximal) invariant linked systems on groups.

An inclusion hyperspaceAon a groupX is calledinvariantifxA=Afor allx∈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 minimal ideal of G(X). The rectangularityof G(X) means thatA ◦ B=Bfor all A,B ∈G(X).

Let N2(X) = N2(X)∩G(X) denote the set of all invariant linked systems on X andλ(X) = maxN2(X) be the family of all maximal ele- ments of N2(X). Elements of λ(X) are called maximal invariant linked systems. The reader should be conscious of the fact that maximal invari- ant linked systems need not be maximal linked!

Theorem 2.1. For every group X the set λ(X) is a non-empty closed rectangular subsemigroup of G(X).

Proof. The rectangularity of λ(X) implies from the rectangularity of

G(X) established in [G2, §5] and the inclusionλ(X)⊂G(X).

The Zorn Lemma implies that each invariant linked system onX (in particular,{X}) can be enlarged to a maximal invariant linked system on X. This observation implies the set λ(X) is not empty. Next, we show that the subsemigroup λ(X) is closed inG(X). Since the set N2(X) = N2(X)∩G(X)is closed inG(X), it suffices to show thatλ(X)is closed in N2(X). Take any invariant linked systemL ∈N2(X)\λ(X). Being not maximal invariant, the linked system L can be enlarged 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 thatB /∈ LandB ∈ M ⊃ LimpliesX\B∈ L andB ∈ L. We claim thatO(L) =B∩(X\B)∩N2(X) is a neighborhood ofL inN2(X) that misses the setλ(X). Indeed, for anyA ∈O(L), we get thatAis an invariant linked system such thatB ∈ A. Observe that for everyx∈X andA∈ Awe getx−1A∈ Aby the invariantness ofAand hence the set B∩x−1A and its shiftxB∩A both are not empty. This witnesses that xB ∈ A for every x ∈ X. Then the maximal invariant linked system generated by A ∪ {xB :x∈X} is an invariant linked enlargement of A, which shows that Ais not maximal invariant linked.

Next, we shall evaluate the cardinality of λ(X).


Journal Algebra Discrete Math.

Theorem 2.2. For any infinite group X the semigroup λ(X)has cardi- nality|λ(X)|= 22|X|.

Proof. The upper bound |λ(X)| ≤ 22|X| follows from the chain of inclu- sions:

λ(X)⊂G(X)⊂ P(P(X)).

Now we prove that |λ(X)| ≥ 22|X|. Let |X|=κ and X ={xα :α <

κ} be an injective enumeration of X by ordinals< κsuch thatx0 is the neutral element of X. For every α < κ let Bα = {xβ, x−1β : β < α}. By transfinite induction, choose a transfinite sequence(aα)α<κ such that a0 =x0 and

aα∈/ Bα−1BαA

whereA={aβ :β < α}.

Consider the set A ={aα :α < κ}. By [HS, 3.58], the set Uκ(A) of κ-uniform ultrafilters onAhas cardinality|Uκ(A)|= 22κ. We recall that an ultrafilterU isκ-uniform if for every setU ∈ U and any subsetK ⊂U of size|K|< κthe setU \K still belongs toU.

To each κ-uniform ultrafilter U ∈ Uκ(A) assign the invariant filter FU = T

x∈XxU. This filter can be extended to a maximal invariant linked system LU. We claim that LU 6= LV 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,


For every α < κ consider the sets Uα ={aβ ∈ U :β > α} ∈ U and Vα ={aβ ∈V :β > α} ∈ V.

It is clear that FU = [


xαUα∈ FU and FV = [


xαVα ∈ FV.

Let us show that FU ∩FV = ∅. Otherwise there would exist two ordinalsα, β and points u∈Uα,v∈Vβ such thatxαu=xβv. It follows from u6= v that α 6=β. Write the points u, v asu =aγ and v=aδ for some γ > α and δ > β. Then we have the equality xαaγ = xβaδ. The inequalityu6=vimplies that γ6=δ. We lose no generality assuming that δ > γ. Then

aδ =x−1β xαaγ ∈B−1δ BδA which contradicts the choice ofaδ.

Therefore,FU∩FV =∅. Taking into account that the linked systems LU ⊃ FU ∋ FU and LV ⊃ FV ∋ FV contain disjoint sets FU, FV, we


Journal Algebra Discrete Math.

conclude thatLU 6=LV. Consequently,

|λ(X)| ≥ |{LU :U ∈ Uκ(A)}|=|Uκ(A)|= 22κ.

The preceding theorem implies that |λ(G)| = 2c for any countable groupG. Next, we evaluate the cardinality of λ(G)for finite groupsG.

Given a finite group Gconsider the invariant linked system L0 ={A⊂X: 2|A|>|G|}

and the subset

↑L0 ={A ∈λ(G) :A ⊃ L0} of λ(G).

Proposition 2.3. LetGbe a finite group. Ifsl(G)≥ |G|/2, thenλ(G) =


Proof. We should prove that each maximal invariant linked system A ∈

λ(G) contains L0. Take any set L ∈ L0. Taking into account that sl(G)≥ |G|/2 and each setA∈ A is self-linked, we conclude that |A| ≥

|G|/2and henceAintersects each shiftxLofL(because|A|+|xL|>|G|).

Since the set L is self-linked, we get that the invariant linked system A ∪ {xL:x∈ G} is equal toA by the maximality of A. Consequently, L∈ Aand hence L0⊂ A.

In light of Proposition 2.3 it is important to evaluate the cardinality of the set ↑L0. In |G| is odd, then the invariant linked system L0 is maximal linked and thus ↑L0 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−1=G, |A|=|G|/2}

of self-linked subsets A⊂Gof cardinality|A|=|G|/2. On the familyS consider the equivalence relation ∼ letting A ∼B for A, B ∈ S if there is x∈G such thatA=xB or X\A=xB. LetS/ the quotient set of S by this equivalence relation ands=|S/|stand for the cardinality of S/.

Proposition 2.4. |λ(G)| ≥ |↑L0|= 2s.


Journal Algebra Discrete Math.

Proof. First we show that ∼indeed is an equivalence relation on S. So, assume that S 6=∅. Let us show thatG\A ∈ S for every A ∈ S. Let B = G\A. Assuming that B /∈ S, we conclude that B∩xB = ∅ for some x ∈ G. Since |B| = |A| = |G|/2, we conclude that xB = A and G\A=B =x−1A. The equalityA∩x−1A=∅impliesx−1 ∈/ AA−1 =G, which is a contradiction.

Taking into account that A=eA for everyA ∈ S, we conclude that

∼ is a reflexive relation onS. If A∼ B, then there is x ∈X such that A=xB or G\A=xB. This implies that B=x−1A or X\B =x−1A, that isB ∼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 thatA=xB or G\A=xB and B =yC or G\B =yC. It is easy to check that in these casesA=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 →2 ={0,1}consider the maximal invariant linked system

Lf =L0∪ {xT :x∈G, T ∈f−1(0)} ∪ {x(G\T) :x∈G, T ∈f−1(1)}. It can be shown that

|↑L0|=|{Lf :f ∈2T}|= 2|T |= 2s.

This proposition will help us to calculate the cardinality of the set

λ(G) for all finite groupsGof order|G| ≤8:

Theorem 2.5. The cardinality of λ(G) for a group G of size |G| ≤ 8 can be found from the table:

G C2 C3 C4 C2C2 C5 D6 C6 C7 C32 D8 C4C2 C8 Q8

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 5cases.

1. If sl(G) > |G|/2, then L0 is a unique maximal invariant linked system and thus|λ(X)|= 1. By Theorem 1.2, sl(G)>|G|/2if and only if |G| ≤5or Gis isomorphic to D6 or C23.

2. Ifsl(G) =|G|/2, then |λ(G)|= 2swheres=|S/|. So it remains to calculate the number sfor the groupsC6,D8,C4⊕C2,C8, and Q8.

2a. If G is cyclic of order 6, then we can take any generator aon G and by routine calculations, check that

S={xT, x(G\T) :x∈G}


Journal Algebra Discrete Math.

whereT ={e, a, a3}. It follows thats=|S/|= 1 and thus

|λ(G)|=|↑L0|= 2s= 2.

2b. If G is cyclic of order 8, then we can take any generatora on G and by routine verification check that

S={xA, G\xA, xB, G\xB, C, G\xC:x∈G}

where A = {e, a, a2, a4}, B = {e, a, a2, a5}, and C = {e, a, a3, a5}. It follows that s=|S/|= 3 and thus

|λ(G)|=|↑L0|= 2s= 8.

2c. Assume that the group G is isomorphic toC4⊕C2 and letG2 = {x ∈ G : xx = e} be the Boolean subgroup of G. We claim that a 4-element subset A⊂Gis self-linked if and only if|A∩G2|is odd.

To prove the “if” part of this claim, assume that|A∩G2|= 3. We claim thatAis self-linked. LetA2 =A∩G2and note thatG2=A2A−12 ⊂AA−1 because |A2|= 3> 2 =|G2|/2. Now take any element a∈ A\G2 and note that AA−1 ⊃ aA−12 ∪ A2a−1. Observe that both aA−12 = aA2

and A2a−1 = a−1A2 are 3-element subsets in the 4-element coset aG2. Those 3-element sets are different. Indeed, assuming thataA−12 =A2a−1 we would obtain that a2A2 = A2 which implies that |A2| = 3 is even.

Consequently,aG2=aA−12 ∪A2a−1⊂AA−1 and finally G=AA−1. If |A∩G2|= 1, then we can take any a ∈ A\G2 and consider the shift Aa−1 which has |Aa−1∩G2|= 3. Then the preceding case implies thatAa−1 is self-linked and so isA.

To prove the “only if” part of the claim assume that|A∩G2|is even. If

|A∩G2|= 4, thenA=G2 andAA−1=G2G−12 =G2 6=G. If|A∩G2|= 0, then A=G2afor any a∈A and henceAA−1 =G2aa−1G−12 =G2 6= G. If|A∩G2|= 2, then|G2∩AA−1| ≤3 and againAA−1 6=G.


S ={A⊂G:|A|= 4 and |A∩G2|is odd}.

Each set A ∈ S has a unique shift aA with aA∩G2 ={e}. There are exactly four subsets A ∈ S with A∩G2 ={e} forming two equivalence classes with respect to the relation∼. Therefores= 2and

|λ(G)|=|↑L0|= 2s= 4.

2d. Assume that Gis isomorphic to the dihedral group D8 of isome- tries of the square. Then Gcontains an element aof order 4 generating


Journal Algebra Discrete Math.

a normal cyclic subgroup H. The element a2 commutes with all the elements of the groupG.

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 thenAA−1 =Abb−1A−1 =H 6=G. If |A∩H| equals 1 or 3, then replacing A by a suitable shift, we can assume that A∩H = {e} and hence A = {e} ∪B for some 3-element subset B ⊂ G\H. It follows that G\H = AA−1 \H = (B ∪B−1) = B 6= G\H. This contradiction shows that|A∩H|= 2. Without loss of generality, we can assume that A∩H={e, a2} (if it is not the case, replaceA by its shift Ax−1 wherex, y ∈ A are such that yx−1 = a2). Now take any element b∈A\H. Since Gis not commutative, we get ab=ba3. Observe that ba2 ∈/ A (otherwise A = {e, b, a2, ba2} would be a subgroup of G with AA−1 =A 6=G). Consequently, the 4-th element c∈A\ {e, a2, b} of A should be of the formc=baorc=ba3=ab. Observe that both the sets A1 = {e, a2, b, ba} and A2 = {e, a2, b, ab} are self-linked. Observe also that

a3(G\A1) =a3· {a, a3, ba2, ba3}={e, a2, ab, b}=A2. Consequently,s=|S/|= 1 and|λ(G)|= 2s= 2.

2e. Finally assume thatGis isomorphic to the group Q8={±1,±i,±j,±k}

of quaternion units. The two-element subset H = {−1,1} 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\ {1,−1}. Observe that the 4-th elementb∈A\ {1,−1, a}of Ais not equal to −a(otherwise, Ais a subgroup ofG).

Conversely, one can easily check that each set A={1,−1, a, b} with a, b∈G\H and a6=−b is self-linked. This means that

S±={{−1,1, a, b}:a6=−band a, b∈G\H}

and thus |S±| = C62 −3 = 12. Observe that for each A ∈ S2 the set

−A∈ S2 and there axactly two shifts of X\A that belong toS2. This means that the equivalence class [A] of any set A ∈ S intersects S2 in four sets. Consequently,s=|S/|=|S±|/4 = 12/4 = 3and

|λ(G)|=|↑L0|= 2s= 8.

3. If|G|= 7, thenL0 is one of three elements ofλ(G). The other two elements can be found as follows. Consider the invariant linked system

L1 ={A⊂G:|A| ≥5}


Journal Algebra Discrete Math.

and observe thatL1 ⊂ Afor eachA ∈λ(G). Indeed, assuming that some A ∈ L1 does not belong to A, we would conclude that B =G\A ∈ A by the maximality of A. Since|G\B| ≤2we can find x∈G\BB−1. It follows thatB, xB are two disjoint sets inAwhich is not possible. Thus L1 ⊂ A.

Observe that L1 ⊂ A ⊂ L0∪ L3, where

L3 ={A⊂G:|A|= 3, AA−1 =G}.

Given a generator a of the cyclic group G, consider the 3-element set T ={a, a2, a4}and note that T T−1 =Gand T−1∩T =∅. By a routine calculation, one can check that

L3 ={xT, xT−1 :x∈G}.

SinceT andT−1 are disjoint, the invariant linked system Acannot con- tain both the setsT andT−1. IfAcontains none of the sets T, T−1, then A=L0. IfA contains T, then

A= (L0∪ {xT :x∈G})\ {y(G\T) :y ∈G}. IfT−1 ∈ A, then

A= (L0∪ {xT−1 :x∈G})\ {y(G\T−1) :y∈G}.

And those are the unique 3 maximal invariant systems in λ(G).

In the following theorem we characterize groups possessing a unique maximal invariant linked system.

Theorem 2.6. For a finite groupG the following conditions are equiva- lent:

1) |λ(G)|= 1;

2) sl(G)>|G|/2;

3) |G| ≤5 or elseG is isomorphic to D6 or C23.

Proof. (2)⇒(1). If sl(G) >|G|/2, thenL0 ={A⊂G:|A|>|G|/2} is a unique maximal invariant linked system onG(because invariant linked systems compose of self-linked sets).

(1) ⇒ (2) Assume that sl(G) ≤ |G|/2 and take a self-linked subset A ⊂ G with |A| ≤ |G|/2. If |G| is odd, then L0 is maximal linked and


Journal Algebra Discrete Math.

then any maximal invariant linked system A containing the self-linked setA is distinct from L0, witnessing that|λ(G)|>1.

If G is even, then we can enlarge A, if necessary, and assume that

|A| = |G|/2. We claim that the complement B = G\A of A is self- linked too. Assuming the converse, we would find some x /∈ BB−1 and conclude thatB∩xB =∅, which implies thatA=G\B =xBand hence x−1A = B. Then the sets A and x−1A are disjoint which contradicts x−1 ∈ AA−1 = G. Thus BB−1 = G which implies that {xB : x ∈ G} is an invariant linked system. Since |G|= 2|A|is even, the unions A = {xA:x∈G}∪L0andB={xB :x∈G}∪L0are invariant linked systems that can be enlarged to maximal linked systems A˜ and B˜, respectively.

Since the sets A ∈ A ⊂ A˜and B ∈ B ⊂ B˜ are disjoint, A 6˜= ˜B are two distinct maximal invariant systems onG and thus|λ(G)| ≥2.

The equivalence (2)⇔(3)follows from Theorem 1.2(i).

3. Right zeros in λ(X)

In this section we return to studying the superextensions of groups and shall detect groupsX whose superextensionsλ(X)have right zeros. We shall show that for every group X the right zeros of λ(X) coincide with invariant maximal linked systems.

We recall that an element z of a semigroup S is called a right(resp.

left)zeroinS ifxz=z(resp. zx=z) for everyx∈S. This is equivalent to saying that the singleton {x} is a left (resp. right) ideal ofS.

By [G2, 5.1] an inclusion hyperspace A ∈ G(X) is a right zero in G(X)if and only ifAis 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 compact rectangular topological semigroup. We recall that a semigroupS is called rectangular 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 Lis a right zero of the semi- groupλ(X) if and only if L is invariant.

Proof. IfLis invariant, then by proposition 5.1 of [G2],L is a right zero inG(X) and consequently, a right zero inλ(X).

Assume conversely that L is a right zero in λ(X). Then for every x∈X we get xL=L, which means that L is invariant.

Unlike the semigroup G(X) which always contains right zeros, the semigroup λ(X) contains right zeros only for so-called odd groups. We


Journal Algebra Discrete Math.

define a group X to be odd if each element x ∈ X has odd order. We recall that theorderof an elementxis the smallest integer numbern≥1 such thatxn coincides with the neutral elementeof 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 onXis maximal linked (which can be written asλ(X)∩λ(X)6=∅);

3) each maximal invariant linked system is maximal linked (which can be written asλ(X)⊂λ(X));

4) for any partition X=A∪B either AA−1 =X or BB−1 =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 =A1∪A2, use the maximality of A to find i ∈ {1,2} with Ai ∈ A. We claim that AiA−1i = X. Indeed, for every x∈X the invariantness of A implies that xAi ∈ A and hence Ai∩xAi 6=∅, which implies x∈AiA−1i .

(4)⇒(3)Assume that for every partitionX=A∪B eitherAA−1 = X or BB−1 =X. We need to check that each maximal invariant linked system L is maximal linked. In the other case, there would exist a set A∈ L\ L. Since L 6∋A is maximal invariant linked system, some shift xAof Adoes not intersect Aand thus x /∈AA−1. Then our assumption implies that B = X \A has property BB−1 = 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 = ∅ and conclude that A∈ L becauseL ⊂X\B = A. But this contradicts the choice of A∈ L\ L. ThereforeB ∈ L and

L ∪ {L⊂X :∃x∈X (xB⊂L)}

is an invariant linked system that enlarges L. Since L is a maximal invariant linked system, we conclude that B ∈ L, which is not possible becauseB 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.


Journal Algebra Discrete Math.

¬(5)⇒ ¬(4) Assume that X\ {e} contains a point awhose order is even or infinity. Then the cyclic subgroup H = {an :n∈Z} generated by a decomposes into two disjoint sets H1 = {an : n ∈ 2Z+ 1} and H2 = {an : n ∈ 2Z} such that aH1 = H2. Take a subset S ⊂ X meeting each coset Hx, x ∈ X, in a single point. Consider the disjoint sets A1 = H1S and A2 = H2S and note that aA1 = A2 = X \A1 and aA2 = X\A2, which implies that a /∈ AiA−1i for i∈ {1,2}. Since A1∪A2=X, we get a negation of (4).

(5)⇒(4)Assume that each element ofX has odd order and assume thatX admits a partitionX =A⊔B such thata /∈AA−1 andb /∈BB−1 for somea, b∈X. ThenaA⊂X\A=B andbB ⊂X\B =A. Observe that

baA⊂bB ⊂A

and by induction,(ba)iA⊂Afor all i >0. Since all elements ofX have finite order,(ba)n=efor some n∈N. Then (ba)n−1A⊂A implies

A= (ba)nA⊂baA⊂bB ⊂A and hence bB=A. It follows from

X =bA⊔bB =bA⊔A=B⊔A thatbA=B. Thusx∈Aif and only if bx∈B.

Let H = {bn : n ∈ Z} ⊂ X be the cyclic subgroup generated by b.

By our assumption it is of odd order. On the other hand, the equality bB = A = b−1B implies that the intersections H ∩A and H∩B have the same cardinality becauseb(B∩H) =A∩H. But this is not possible because of the odd cardinality of H.

4. (Left) zeros of the semigroup λ(X)

An elementzof a semigroupSis called azero inS ifxz =z=zxfor all x∈S. This is equivalent to saying thatz 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) the following conditions are equivalent:

1) L is a left zero inλ(X);

2) L is a zero in λ(X);

3) L is a unique invariant maximal linked system onX.


Journal Algebra Discrete Math.

Proof. (1) ⇒ (3) Assume that Z is a left zero in λ(X). Then Zx =Z for all x∈X and thus

Z−1 ={Z−1 :Z ∈ Z}

is an invariant maximal linked system onX, which implies that the group X is odd according to Theorem 3.2. Note that for every right zero Aof λ(X) we get

Z=Z ◦ A=A

which implies thatZis a unique right zero inλ(X)and by Proposition 3.1 a unique invariant maximal linked system on X.

(3)⇒(2)Assume thatZ is a unique invariant maximal linked system on X. We claim that Z is a left zero of λ(X). Indeed, for everyA ∈ A andx∈Xwe getxZ ◦A=Z ◦A, which means thatZ ◦Ais an invariant maximal linked system. By Proposition 3.1, Z ◦ A is a right zero and henceZ ◦ A=Z becauseZ is a unique right zero. This means thatZ is a left zero, and being a right zero, a zero inλ(X).

(2)⇒(1)is trivial.

Theorem 4.2. The superextension λ(X) of a groupX has a zero if and only if X is isomorphic to C1,C3 or C5.

Proof. IfX is a group of odd order|X| ≤5, thenλ(X)⊂λ(X) because Xis odd and|λ(X)|= 1by Theorem 2.6. This means thatλ(X)contains a unique invariant maximal 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. By Proposition 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)| ≤1. By Theorem 2.6, X has order |X| ≤5 or is isomorphic toD3 or C23. SinceX is odd,X must be isomorphic to C1,C3 or C5.

5. The commutativity of λ(X)

In this section we detect groups X with commutative superextension.

Theorem 5.1. The superextensionλ(X)of a groupX is commutative if and only if |X| ≤4.


Journal Algebra Discrete Math.

Proof. The commutativity of the superextensions λ(X) of groups X of order|X| ≤4 will be established in Section 6.

Now assume that a group X has commutative superextension λ(X).

Then X is commutative. 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 ofG(X) is a right zero inG(X)(see [G2, 5.1]) we conclude that

A ◦ B=B 6=A=B ◦ A.

Extend the linked systems systemsA,B to maximal linked systemsA ⊃˜ Aand 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 extendingA, which is not possible because of the maximality of A. This contradiction shows that |λ(X)| = 1. Applying Theorem 2.6, we con- clude that |X| ≤5 or X is isomorphic to C23.

It remains to show that the semigroupsλ(C5)andλ(C23)are not com- mutative. The non-commutativity ofλ(C5) will be shown in Section 6.

To see that λ(C23) is not commutative, take any 3 generators a, b, c of C23 and consider the sets A = {e, a, b, abc}, H1 = {e, a, b, ab}, H2 = {e, a, bc, abc}. Observe that H1, H2 are subgroups in C23. For every i∈ {1,2} consider the linked system Ai =h{H1, H2} ∪ {xA:x ∈Hi}iand extend it to a maximal linked system A˜i on C23.

We claim that the maximal linked systems A˜1 and A˜2 do not com- mute. Indeed,

2◦A˜1 ∋ [


x∗(x−1bA) =bA={e, b, ba, ac},

1◦A˜2∋ [


x∗(x−1bcA) =bcA={a, c, bc, abc}.

It follows frombA∩bcA=∅ thatA˜1◦A˜26= ˜A2◦A˜1. 6. The superextensions of finite groups

In this section we shall describe the structure of the superextensions λ(G) of finite groups G of small cardinality (more precisely, of cardi- nality |G| ≤ 5). It is known that the cardinality of λ(G) growth very


Journal Algebra Discrete Math.

quickly as |G| tends to infinity. The calculation of the cardinality of

|λ(G)|seems to be a difficult combinatorial problem related to the still unsolved Dedekind’s problem of calculation of the number M(n) of in- clusion hyperpspaces on an n-element subset, see [De]. We were able to calculate the cardinalities of λ(G) only for groups G of cardinality

|G| ≤ 6. The results of (computer) calculations are presented in the following table:

|G| 1 2 3 4 5 6

|λ(G)| 1 2 4 12 81 2646

|λ(G)/G| 1 1 2 3 17 447

Before describing the structure of superextensions of finite groups, let us make some remarks concerning the structure of a semigroup S containing a group G. In this case S can be thought as a G-space en- dowed with the left action of the groupG. So we can consider the orbit space S/G = {Gs : s ∈ S} and the projection π : S → S/G. If G lies in the center of the semigroupS (which means that the elements of G commute with all the elements of S), then the orbit space S/G ad- mits a unique semigroup operation turning S/G into a semigroup and the orbit projection π :S → S/G into a semigroup homomorphism. A subsemigroupT ⊂S will be called a transversal semigroupif the restric- tion π : T → S/G is an isomorphism of the semigroups. If S admits a transversal semigroupT, then it is a homomoprhic image of the product G×T under the semigroup homomorphism

h:G×T →S, h: (g, t)7→gt.

This helps to recover the algebraic structure ofS from the structure of a transversal semigroup.

For a system B of subsets of a setX 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| ≤ 5 is abelian and is isomorphic to one of the groups: C1, C2, C3, C4, C2 ⊕C2, C5. It will be convenient to think of the cyclic group Cn as the multiplicative subgroups {z ∈C:zn= 1} of the complex plane.

6.1. The semigroups λ(C1) and λ(C2)

For the groups Cn with n ∈ {1,2} the semigroup λ(Cn) coincides with Cn while the orbit semigroupλ(Cn)/Cn is trivial.


Journal Algebra Discrete Math.

6.2. The semigroup λ(C3)

For the group C3 the semigroup λ(C3) contains the three principal ul- trafilters 1, z,−z where z = e2πi/3 and the maximal linked inclusion hyperspace ⊲ = h{1, z},{1,−z},{z,−z}i which is the zero in λ(C3).

The superextensionλ(C3) is isomorphic to the multiplicative semigroup C30 ={z∈ C:z4 =z} of the complex plane. The latter semigroup has zero0 and unit1which are the unique idempotents.

The transversal semigroup λ(C3)/C3 is isomorphic to the semilattice 2 ={0,1}endowed with themin-operation.

6.3. The semigroups λ(C4) and λ(C2⊕C2)

The semigroup λ(C4) contains 12 elements while the orbit semigroup λ(C4)/C4contains 3 elements. The semigroupλ(C4)contains a transver- sal semigroup

λT(G) ={1,△,}

where1is the neutral element of C4 ={1,−1, i,−i},

△=h{1, i},{1,−i},{i,−i}iand

=h{1, i},{1,−i},{1,−1},{i,−i,−1}i.

The transversal semigroup is isomorphic to the extension C21=C2∪ {e} of the cyclic groupC2 by an external unite /∈C2 (such thatex=x=xe for allx ∈C21). The action of the group C4 on λ(C4)is free so, λ(C4) is isomorphic toλT(C4)⊕C4.

The semigroup λ(C2⊕C2) has a similar algebraic structure. It con- tains a transversal semigroup

λT(C2⊕C2) ={e,△,} ⊂λ(C2⊕C2)

whereeis the principal ultrafilter supported by the neutral element(1,1) of C2⊕C2 and the maximal linked inclusion hyperspaces △and are defined by analogy with the case of the group C4:

=h{(1,1),(1,1)},{(1,1),(1,1)},{(1,1),(1,1)}i and

=h{(1,1),(1,-1)},{(1,1),(-1,1)},{(1,1),(-1,-1)},{(1,-1),(-1,1),(-1,-1)}i. The transversal semigroupλT(C2⊕C2)is isomorphic to C21 andλ(C2⊕ C2)is isomorphic to C21⊕C2⊕C2.

We summarize the obtained results on the algebraic structure of the semigroupsλ(C4) andλ(C2⊕C2)in the following proposition.

Proposition 6.1. Let G be a group of cardinality |G|= 4.



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