Algebra in superextensions of groups, I: zeros and commutativity

Given a group X 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={C⊂X:{x∈X:x−1C∈B}∈A} that extends the group operation of X. We characterize right zeros of...

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Дата:	2008
Автори:	Banakh, T.T., Gavrylkiv, V., Nykyforchyn, O.
Формат:	Стаття
Мова:	English
Опубліковано:	Інститут прикладної математики і механіки НАН України 2008
Назва видання:	Algebra and Discrete Mathematics
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Назва журналу:	Digital Library of Periodicals of National Academy of Sciences of Ukraine
Цитувати:	Algebra in superextensions of groups, I: zeros and commutativity / T.T. Banakh, V. Gavrylkiv, O. Nykyforchyn // Algebra and Discrete Mathematics. — 2008. — Vol. 7, № 3. — С. 1–29. — Бібліогр.: 13 назв. — англ.

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spelling	irk-123456789-1533732019-06-15T01:25:44Z Algebra in superextensions of groups, I: zeros and commutativity Banakh, T.T. Gavrylkiv, V. Nykyforchyn, O. Given a group X 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={C⊂X:{x∈X:x−1C∈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 right zero if and only if each element of X has odd order. On the other hand, the semigroup λ(X) contains a left zero if and only if it contains a zero if and only if X has odd order \|X\|≤5. The semigroup λ(X) is commutative if and only if \|X\|≤4. We finish the paper with a complete description of the algebraic structure of the semigroups λ(X) for all groups X of cardinality \|X\|≤5. 2008 Article Algebra in superextensions of groups, I: zeros and commutativity / T.T. Banakh, V. Gavrylkiv, O. Nykyforchyn // Algebra and Discrete Mathematics. — 2008. — Vol. 7, № 3. — С. 1–29. — Бібліогр.: 13 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 20M99, 54B20. http://dspace.nbuv.gov.ua/handle/123456789/153373 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	Given a group X 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={C⊂X:{x∈X:x−1C∈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 right zero if and only if each element of X has odd order. On the other hand, the semigroup λ(X) contains a left zero if and only if it contains a zero if and only if X has odd order \|X\|≤5. The semigroup λ(X) is commutative if and only if \|X\|≤4. We finish the paper with a complete description of the algebraic structure of the semigroups λ(X) for all groups X of cardinality \|X\|≤5.
format	Article
author	Banakh, T.T. Gavrylkiv, V. Nykyforchyn, O.
spellingShingle	Banakh, T.T. Gavrylkiv, V. Nykyforchyn, O. Algebra in superextensions of groups, I: zeros and commutativity Algebra and Discrete Mathematics
author_facet	Banakh, T.T. Gavrylkiv, V. Nykyforchyn, O.
author_sort	Banakh, T.T.
title	Algebra in superextensions of groups, I: zeros and commutativity
title_short	Algebra in superextensions of groups, I: zeros and commutativity
title_full	Algebra in superextensions of groups, I: zeros and commutativity
title_fullStr	Algebra in superextensions of groups, I: zeros and commutativity
title_full_unstemmed	Algebra in superextensions of groups, I: zeros and commutativity
title_sort	algebra in superextensions of groups, i: zeros and commutativity
publisher	Інститут прикладної математики і механіки НАН України
publishDate	2008
url	http://dspace.nbuv.gov.ua/handle/123456789/153373
citation_txt	Algebra in superextensions of groups, I: zeros and commutativity / T.T. Banakh, V. Gavrylkiv, O. Nykyforchyn // Algebra and Discrete Mathematics. — 2008. — Vol. 7, № 3. — С. 1–29. — Бібліогр.: 13 назв. — англ.
series	Algebra and Discrete Mathematics
work_keys_str_mv	AT banakhtt algebrainsuperextensionsofgroupsizerosandcommutativity AT gavrylkivv algebrainsuperextensionsofgroupsizerosandcommutativity AT nykyforchyno algebrainsuperextensionsofgroupsizerosandcommutativity
first_indexed	2025-07-14T04:36:00Z
last_indexed	2025-07-14T04:36:00Z
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fulltext	Jo u rn al A lg eb ra D is cr et e M at h . Algebra and Discrete Mathematics RESEARCH ARTICLE Number 3. (2008). pp. 1 – 29 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 group X 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 = {C ⊂ X : {x ∈ X : x−1C ∈ 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 right zero if and only if each element of X has odd order. On the other hand, the semigroup λ(X) contains a left zero if and only if it contains a zero if and only if X has odd order \|X\| ≤ 5. The semigroup λ(X) is commutative if and only if \|X\| ≤ 4. We finish the paper with a complete description of the algebraic structure of the semigroups λ(X) for all groups X of cardinality \|X\| ≤ 5. Introduction 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 space X can be extended to a right-topological semigroup operation on β(X), the Stone-Čech com- pactification of X. The extension of the operation from X to β(X) can 2000 Mathematics Subject Classification: 20M99, 54B20. Key words and phrases: Superextension, right-topological semigroup. 1Unpublished, see [HS, p.102], [H2] Jo u rn al A lg eb ra D is cr et e M at h .2 Algebra in superextensions of groups, I be defined by the simple formula: U ◦ V = {A ⊂ X : {x ∈ X : x−1A ∈ V} ∈ U}, (1) where U ,V are ultrafilters on X and x−1A = {y ∈ X : xy ∈ A}. Endowed with the so-extended operation, the Stone-Čech 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-Čech 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 over X. By definition, a family F of non-empty subsets of a discrete space X is called an inclusion 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 set G(X) there is an important transversality operation assigning to each inclusion hyperspace F ∈ 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} and U− = {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 ∗ from X to G(X) can be defined in the same way as for ultrafilters, i.e., by the formula (1) applied to any two inclusion hyperspaces U ,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-Čech 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⊥} Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 3 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 a system of subsets L of X is linked if A ∩ B 6= ∅ for all A, B ∈ L. An inclusion hyperspace A ∈ G(X) is linked if and only if A ⊂ 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 of G(X). The superextension λ(X) consists of all maximal elements of N2(X), see [G1], [G2]. In this paper we start a systematic investigation of the algebraic struc- ture of the semigroup λ(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 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 apply maximal linked systems to Combinatorics and Ramsey Theory. In this paper we concentrate on describing zeros and commutativity of the semigroup 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 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 semigroup λ(X) rarely is commutative: this holds if and only if the group X 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 of X. We use those bounds to characterize groups X with sl(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 group X. In Theorem 2.2 we show that for an infinite group X the cardinality of ↔ λ(X) equals 22\|X\| . In Proposition 2.3 and Theorem 2.6 we Jo u rn al A lg eb ra D is cr et e M at h .4 Algebra in superextensions of groups, I calculate the cardinality of ↔ λ(X) for all finite groups X 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 groups X of order \|X\| ≤ 5. Now a couple of words about notations. Following the algebraic tradi- tion, by Cn we denote the cyclic group of order n and by D2n the dihedral group of cardinality 2n, that is, the isometry group of the regular n-gon. For a group X by e we denote the neutral element of X. For a real number x 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 subset A of a group G is self-linked if A∩xA 6= ∅ for each x ∈ G. In fact, this notion can be defined 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 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 subset A ⊂ X of a G-space X defined to be self-linked if A∩gA 6= ∅ for all g ∈ G. Let us observe that a subset A ⊂ G of a group G is self- linked if and only if AA−1 = G. For a G-space X by sl(X) we denote the smallest cardinality \|A\| of a self-linked subset A ⊂ 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. Then 1) sl(G) ≥ (1 + √ 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 Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 5 \|G\| = \|G \ {e}\| + 1 ≤ \|A2 \ ∆A\| + 1 = sl(G)2 − sl(G) + 1, which just implies that sl(G) ≥ (1 + √ 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 set C = BA is self-linked. Given arbitrary x ∈ G we should prove that the intersection C ∩ xC is not empty. Since B is self-linked, the intersection B ∩ xB contains the coset bH = xb′H for some b, b′ ∈ B. It follows that b−1xb′ ∈ H = AA−1. The latter equality follows from the fact that the set A ⊂ H is self-linked in H. Consequently, b−1xb′ = a′a−1 for some a, a′ ∈ A. Then xC ∋ xb′a = ba′ ∈ C and thus C ∩ xC 6= ∅. The self-linkedness of C implies the desired upper bound sl(G) ≤ \|C\| ≤ \|A\| · \|B\| = sl(H) · sl(G/H). 2b. Next, we show that sl(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 ∈ 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)/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 (because B ∩ H is a singleton). Theorem 1.2. For a finite group G (i) sl(G) = ⌈(\|G\|+1)/2⌉ > \|G\|/2 if and only if G is 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 inequality sl(G) < \|G\|/2 for all groups G not isomorphic to the groups appearing in the items (i), (ii). Given such a group G we should find a self-linked subset A ⊂ G with \|A\| < \|G\|/2. We consider 8 cases. 1) 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) ≤ 2 · 2 < 9/2 = \|G\|/2. 2) \|G\| /∈ {9, 12, 15} and G contains a subgroup H of order n = \|H\| ≥ 3 and index m = \|G/H\| ≥ 3. It this case n + m − 1 < nm/2 and sl(G) ≤ \|H\| + \|G/H\| − 1 = n + m − 1 < nm/2 by Proposition 1.1(3). Jo u rn al A lg eb ra D is cr et e M at h .6 Algebra in superextensions of groups, I 3) G is cyclic of order n = \|G\| ≥ 9. Given a generator a of G, construct a sequence (xi)2≤i≤n/2 letting x2 = a0, x3 = a, x4 = a3, x5 = a5, and xi = xi−1a i 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 that A = {e, a, a3} is a 3-element self-linked subset and thus sl(G) ≤ 3 < \|G\|/2. 5) G contains a cyclic subgroup H ⊂ G of prime order \|H\| ≥ 7. By the preceding two cases, sl(H) < \|H\|/2 and then sl(G) ≤ sl(H) · sl(G/H) < \|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 subgroup H ⊂ G of order \|H\| = 4 and 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, then G is 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, then G is isomorphic to C10 or D10. If G is 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 hence sl(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 〈a, b\|a4 = b3 = 1, aba−1 = b−1〉. If G is isomorphic to C12, C6 ⊕ C2 or A4, then G contains a normal 4-element subgroup H. By Sylow’s Theorem, G contains 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. If G is isomorphic to C3 ⋊ C4, then G contains a normal subgroup H 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. Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 7 Finally, consider the case of the dihedral group D12. It contains an element a generating a cyclic subgroup of order 6 and an element b of order 2 such that bab−1 = a−1. Consider the 5-element set A = {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 inequality sl(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 that sl(G) = ⌈(\|G\|+ 1)/2⌉ > \|G\|/2 for all groups G with \|G\| ≤ 5. It remains to check that sl(G) > \|G\|/2 if G is isomorphic to D6 or C3 2 . First we consider the case G = D6. In this case G contains a normal 3- element subgroup T . Assuming that sl(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 to A (otherwise replace A by a suitable shift xA). Taking into account that AA−1 = G, we conclude that A 6⊂ T and thus we can find an element a ∈ A \ T . This element has order 2. Then 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 C3 2 . 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 element e 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 of A. 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 yields sl(G) = 3 = \|G\|/2. If \|G\| = 8, then sl(G) ≥ 4 by Proposition 1.1(1). If G is cyclic of order 8 and a is a generator of G, then the set A = {e, a, a3, a4} is self-linked and thus sl(C8) = 4. If G is isomorphic to C4 ⊕C2, then G has two commuting generators a, b such that a4 = b2 = 1. One can check that the set A = {e, a, a2, b} is self-linked and thus sl(C4 ⊕ C2) = 4. Jo u rn al A lg eb ra D is cr et e M at h .8 Algebra in superextensions of groups, I If G is isomorphic to the dihedral group D8, then G has 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 G is isomorphic to the group Q8 = {±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 cardinal sl(G) for all groups G of cardinality \|G\| ≤ 13. Proposition 1.3. The number sl(G) for a group G of size \|G\| ≤ 13 can be found from the table: G C2 C3 C5 C4 C2 ⊕ C2 C6 D6 C8 C2 ⊕ C4 D8 Q8 C3 2 sl(G) 2 2 3 3 3 3 4 4 4 4 4 5 G C7 C11 C13 C9 C3 ⊕ C3 C10 D10 C12 C2 ⊕ C6 D12 A4 C3 ⋊ C4 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 de- termined by the lower bound sl(G) ≥ 1+ √ 4\|G\|−3 2 from Proposition 1.1(1) and the upper bound from Theorem 1.2. It remains to consider the groups G of order 11 ≤ \|G\| ≤ 13. 1. If G is cyclic of order 11 or 13, then take a generator a of G and check that the 4-element set A = {e, a4, a5, a7} is self-linked, witnessing that sl(C12) = 4. 2. If G is cyclic of order 12, then take a generator a for G and check that the 4-element subset A = {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 bound sl(G) ≤ 5. So, we need to show that sl(G) > 4 for all non-cyclic groups G with \|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 subset A ⊂ G. Since AA−1 = 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 6⊂ H, there is a point b ∈ A \ H. Since the quotient group G/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. If c ∈ b−1H, then H = H ∩ AA−1 = {e, a}, which is absurd. So, c ∈ bH and thus c = bh for some h ∈ H. Since h = h−1, we get cb−1 = bhb−1 = Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 9 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 and A4. 4. Assume that G is isomorphic to the dihedral group D12. Then G contains a normal cyclic subgroup H of order 6, and for each b ∈ G \ H and a ∈ H we get b2 = e and 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 group H. Since a ∈ AA−1 = G, we can find two element x, y ∈ A such that a = xy−1. Then the shift Ay−1 contains e and a. Replacing A by Ay−1, if necessary, we can assume that e, a ∈ A. Since A 6⊂ 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−1 H b−1, which implies bA−1 H = AHb−1 = bA−1 H ∪ AHb−1 = AA−1 ∩ bH = bH. This is a contradiction, because \|H\| = 4 > 3 = \|bA−1 H \|. Then c ∈ bH and hence H = 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 presentation 〈a, b\|a4 = b3 = 1, aba−1 = b−1〉. Then the cyclic subgroup H generated by b is normal in G and the quotient G/H is cyclic of order 4. Assuming that 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 6⊂ H, there is an element c ∈ A \ H. We claim that the fourth element d ∈ A \ {e, b, c} does not belong to H ∪ cH ∪ c−1H. Otherwise, AA−1 ⊂ H ∪cH ∪c−1H 6= G. This implies that one of the elements, say c belongs to the coset a2H and the other to aH or a−1H. We lose no generality assuming that d ∈ aH. Then c = a2bi, d = abj for some i, 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 = −1 and 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 + √ 4\|G\| − 3)/2⌉ for all finite cyclic groups G? Jo u rn al A lg eb ra D is cr et e M at h .10 Algebra in superextensions of groups, I 2. Maximal invariant linked systems In this section we study (maximal) invariant linked systems on groups. An inclusion hyperspace A on a group X is called invariant if xA = 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 minimal ideal of G(X). The rectangularity of ↔ G(X) means that A ◦ B = B for all A,B ∈ ↔ G(X). Let ↔ N2(X) = N2(X) ∩ ↔ G(X) denote the set of all invariant linked systems on X and ↔ λ(X) = max ↔ N2(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 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 observation implies the set ↔ λ(X) is not empty. Next, we show that the subsemigroup ↔ λ(X) is closed in G(X). Since the set ↔ N2(X) = N2(X)∩ ↔ G(X) is closed in G(X), it suffices to show that ↔ λ(X) is closed in ↔ N2(X). Take any invariant linked system L ∈ ↔ 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 that B /∈ L and B ∈ M ⊃ L implies X \B ∈ L⊥ and B ∈ L⊥. We claim that O(L) = B− ∩ (X \ B)− ∩ ↔ N2(X) is a neighborhood of L in ↔ N2(X) that misses the set ↔ λ(X). Indeed, for any A ∈ O(L), we get that A is an invariant linked system such that B ∈ A⊥. Observe that for every x ∈ X and A ∈ A we get x−1A ∈ A by the invariantness of A and hence the set B ∩ x−1A and its shift xB ∩ 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 A is not maximal invariant linked. Next, we shall evaluate the cardinality of ↔ λ(X). Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 11 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 < κ such that x0 is the neutral element of X. For every α < κ let Bα = {xβ , x−1 β : β < α}. By transfinite induction, choose a transfinite sequence (aα)α<κ such that a0 = x0 and aα /∈ B−1 α BαA<α where A<α = {aβ : β < α}. Consider the set A = {aα : α < κ}. By [HS, 3.58], the set Uκ(A) of κ-uniform ultrafilters on A has cardinality \|Uκ(A)\| = 22κ . 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 ∈ Uκ(A) assign the invariant filter FU = ⋂ x∈X xU . 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, \|U \| = \|V \| = κ. 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 = ⋃ α<κ xαUα ∈ FU and FV = ⋃ α<κ xαVα ∈ FV . Let us show that FU ∩ FV = ∅. Otherwise there would exist two ordinals α, β and points u ∈ Uα, v ∈ Vβ such that xαu = xβv. It follows from u 6= v that α 6= β. Write the points u, v as u = aγ and v = aδ for some γ > α and δ > β. Then we have the equality xαaγ = xβaδ. The inequality u 6= vimplies that γ 6= δ. We lose no generality assuming that δ > γ. Then aδ = x−1 β xαaγ ∈ B−1 δ BδA<δ which contradicts the choice of aδ. Therefore, FU ∩FV = ∅. Taking into account that the linked systems LU ⊃ FU ∋ FU and LV ⊃ FV ∋ FV contain disjoint sets FU , FV , we Jo u rn al A lg eb ra D is cr et e M at h .12 Algebra in superextensions of groups, I conclude that LU 6= LV . Consequently, \| ↔ λ(X)\| ≥ \|{LU : U ∈ Uκ(A)}\| = \|Uκ(A)\| = 22κ . The preceding theorem implies that \| ↔ λ(G)\| = 2c 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 L0 = {A ⊂ X : 2\|A\| > \|G\|} and the subset ↑L0 = {A ∈ ↔ λ(G) : A ⊃ L0} of ↔ λ(G). Proposition 2.3. Let G be a finite group. If sl(G) ≥ \|G\|/2, then ↔ λ(G) = ↑L0. 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 set A ∈ A is self-linked, we conclude that \|A\| ≥ \|G\|/2 and hence A intersects each shift xL of L (because \|A\|+\|xL\| > \|G\|). Since the set L is self-linked, we get that the invariant linked system A ∪ {xL : x ∈ G} is equal to A by the maximality of A. Consequently, L ∈ A and 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 ⊂ G of cardinality \|A\| = \|G\|/2. On the family S consider the 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 relation and s = \|S/∼\| stand for the cardinality of S/∼. Proposition 2.4. \| ↔ λ(G)\| ≥ \|↑L0\| = 2s. Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 13 Proof. First we show that ∼ indeed is an equivalence relation on S. So, assume that S 6= ∅. Let us show that G \ 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 equality A∩x−1A = ∅ implies x−1 /∈ AA−1 = G, which is a contradiction. Taking into account that A = eA for every A ∈ S, we conclude that ∼ is a reflexive relation on S. 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 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 → 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 groups G of 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 C2 ⊕ C2 C5 D6 C6 C7 C3 2 D8 C4 ⊕ C2 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 5 cases. 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\|/2 if and only if \|G\| ≤ 5 or G is isomorphic to D6 or C3 2 . 2. If sl(G) = \|G\|/2, then \| ↔ λ(G)\| = 2s where s = \|S/∼\|. So it remains to calculate the number s for the groups C6, D8, C4 ⊕ C2, C8, and Q8. 2a. If G is cyclic of order 6, then we can take any generator a on G and by routine calculations, check that S = {xT, x(G \ T ) : x ∈ G} Jo u rn al A lg eb ra D is cr et e M at h .14 Algebra in superextensions of groups, I where T = {e, a, a3}. It follows that s = \|S/∼\| = 1 and thus \| ↔ λ(G)\| = \|↑L0\| = 2s = 2. 2b. If G is cyclic of order 8, then we can take any generator a 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 to C4 ⊕ C2 and let G2 = {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 ∩ G2\| is odd. To prove the “if” part of this claim, assume that \|A∩G2\| = 3. We claim that A is self-linked. Let A2 = A∩G2 and note that G2 = A2A −1 2 ⊂ AA−1 because \|A2\| = 3 > 2 = \|G2\|/2. Now take any element a ∈ A \ G2 and note that AA−1 ⊃ aA−1 2 ∪ A2a −1. Observe that both aA−1 2 = aA2 and A2a −1 = a−1A2 are 3-element subsets in the 4-element coset aG2. Those 3-element sets are different. Indeed, assuming that aA−1 2 = A2a −1 we would obtain that a2A2 = A2 which implies that \|A2\| = 3 is even. Consequently, aG2 = aA−1 2 ∪ 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 that Aa−1 is self-linked and so is A. To prove the “only if” part of the claim assume that \|A∩G2\| is even. If \|A∩G2\| = 4, then A = G2 and AA−1 = G2G −1 2 = G2 6= G. If \|A∩G2\| = 0, then A = G2a for any a ∈ A and hence AA−1 = G2aa−1G−1 2 = G2 6= G. If \|A ∩ G2\| = 2, then \|G2 ∩ AA−1\| ≤ 3 and again AA−1 6= G. Thus 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 ∼. Therefore s = 2 and \| ↔ λ(G)\| = \|↑L0\| = 2s = 4. 2d. Assume that G is isomorphic to the dihedral group D8 of isome- tries of the square. Then G contains an element a of order 4 generating Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 15 a normal cyclic subgroup H. The element a2 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−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, replace A by its shift Ax−1 where x, y ∈ A are such that yx−1 = a2). Now take any element b ∈ A \ H. Since G is 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 form c = ba or c = 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 that G is 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 element b ∈ A \ {1,−1, a} of A is not equal to −a (otherwise, A is a subgroup of G). Conversely, one can easily check that each set A = {1,−1, a, b} with a, b ∈ G \ H and a 6= −b is self-linked. This means that S± = {{−1, 1, a, b} : a 6= −b and a, b ∈ G \ H} and thus \|S±\| = C2 6 − 3 = 12. Observe that for each A ∈ S2 the set −A ∈ S2 and there axactly two shifts of X \ A that belong to S2. This means that the equivalence class [A]∼ of any set A ∈ S intersects S2 in four sets. Consequently, s = \|S/∼\| = \|S±\|/4 = 12/4 = 3 and \| ↔ λ(G)\| = \|↑L0\| = 2s = 8. 3. If \|G\| = 7, then L0 is one of three elements of ↔ λ(G). The other two elements can be found as follows. Consider the invariant linked system L1 = {A ⊂ G : \|A\| ≥ 5} Jo u rn al A lg eb ra D is cr et e M at h .16 Algebra in superextensions of groups, I and observe that L1 ⊂ A for each A ∈ ↔ 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\| ≤ 2 we can find x ∈ G \BB−1. It follows that B, xB are two disjoint sets in A which 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 TT−1 = G and T−1 ∩ T = ∅. By a routine calculation, one can check that L3 = {xT, xT−1 : x ∈ G}. Since T and T−1 are disjoint, the invariant linked system A cannot con- tain both the sets T and T−1. If A contains none of the sets T, T−1, then A = L0. If A contains T , then A = (L0 ∪ {xT : x ∈ G}) \ {y(G \ T ) : y ∈ G}. If T−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 group G the following conditions are equiva- lent: 1) \| ↔ λ(G)\| = 1; 2) sl(G) > \|G\|/2; 3) \|G\| ≤ 5 or else G is isomorphic to D6 or C3 2 . Proof. (2) ⇒ (1). If sl(G) > \|G\|/2, then L0 = {A ⊂ G : \|A\| > \|G\|/2} is a unique maximal invariant linked system on G (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 Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 17 then any maximal invariant linked system A containing the self-linked set A 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 that B∩xB = ∅, which implies that A = G\B = xB and 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}∪L0 and B = {xB : x ∈ G}∪L0 are invariant linked systems that can be enlarged to maximal linked systems Ã and B̃, respectively. Since the sets A ∈ A ⊂ Ã and B ∈ B ⊂ B̃ are disjoint, Ã 6= B̃ are two distinct maximal invariant systems on G 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 groups X whose superextensions λ(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) zero in S if xz = z (resp. zx = z) for every x ∈ S. This is equivalent to saying that the singleton {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 and only 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 compact rectangular topological semigroup. We recall that a semigroup S 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 L is a right zero of the semi- group 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). 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 Jo u rn al A lg eb ra D is cr et e M at h .18 Algebra in superextensions of groups, I 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 ≥ 1 such that xn coincides with the neutral element 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 can be written as ↔ λ(X) ∩ λ(X) 6= ∅); 3) each maximal invariant linked system is maximal linked (which can be written as ↔ λ(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 −1 i = X. Indeed, for every x ∈ X the invariantness of A implies that xAi ∈ A and hence Ai ∩ xAi 6= ∅, which implies x ∈ AiA −1 i . (4) ⇒ (3) Assume that for every partition X = A∪B either AA−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 xA of A does not intersect A and 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 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 linked system, 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. Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 19 ¬(5) ⇒ ¬(4) Assume that X \ {e} contains a point a whose 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 −1 i for i ∈ {1, 2}. Since A1 ∪ A2 = 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−1 and b /∈ BB−1 for some a, b ∈ X. Then aA ⊂ X \A = B and bB ⊂ X \B = A. Observe that baA ⊂ bB ⊂ A and by induction, (ba)iA ⊂ A for all i > 0. Since all elements of X have finite order, (ba)n = e for 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 that bA = B. Thus x ∈ A if 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 because b(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 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) the following 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. Jo u rn al A lg eb ra D is cr et e M at h .20 Algebra in superextensions of groups, I 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 on X, which implies that the group X is odd according 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 unique invariant 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 xZ◦A = Z◦A, which means that Z◦A is an invariant maximal linked system. By Proposition 3.1, Z ◦ A is a right zero and hence Z ◦A = Z because Z is a unique right zero. This means that Z is a left zero, and being a right zero, a zero in λ(X). (2) ⇒ (1) is trivial. Theorem 4.2. The superextension λ(X) of a group X has a zero if and only if X is isomorphic to C1, C3 or C5. Proof. If X is a group of odd order \|X\| ≤ 5, then ↔ λ(X) ⊂ λ(X) because X is odd and \| ↔ λ(X)\| = 1 by Theorem 2.6. This means that λ(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 to D3 or C3 2 . Since X 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 superextension λ(X) of a group X is commutative if and only if \|X\| ≤ 4. Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 21 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 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 Ã ⊃ A and B̃ ⊃ B. Because of the commutativity of λ(X), we get A = B ◦ A ⊂ B̃ ◦ Ã = Ã ◦ B̃ ⊃ 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 con- clude that \|X\| ≤ 5 or X is isomorphic to C3 2 . It remains to show that the semigroups λ(C5) and λ(C3 2 ) are not com- mutative. The non-commutativity of λ(C5) will be shown in Section 6. To see that λ(C3 2 ) is not commutative, take any 3 generators a, b, c of C3 2 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 C3 2 . For every i ∈ {1, 2} consider the linked system Ai = 〈{H1, H2} ∪ {xA : x ∈ Hi}〉 and extend it to a maximal linked system Ãi on C3 2 . We claim that the maximal linked systems Ã1 and Ã2 do not com- mute. Indeed, Ã2 ◦ Ã1 ∋ ⋃ x∈H1 x ∗ (x−1bA) = bA = {e, b, ba, ac}, Ã1 ◦ Ã2 ∋ ⋃ x∈H2 x ∗ (x−1bcA) = bcA = {a, c, bc, abc}. It follows from bA ∩ bcA = ∅ that Ã1 ◦ Ã2 6= Ã2 ◦ Ã1. 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 Jo u rn al A lg eb ra D is cr et e M at h .22 Algebra in superextensions of groups, I 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 group G. 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 semigroup S (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 subsemigroup T ⊂ S will be called a transversal semigroup if the restric- tion π : T → S/G is an isomorphism of the semigroups. If S admits a transversal semigroup T , 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 of S from the structure of a transversal semigroup. For a system B of subsets of a set X by 〈B〉 = {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 semigroup λ(Cn)/Cn is trivial. Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 23 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 ⊲ = 〈{1, z}, {1,−z}, {z,−z}〉 which is the zero in λ(C3). The superextension λ(C3) is isomorphic to the multiplicative semigroup C0 3 = {z ∈ C : z4 = z} of the complex plane. The latter semigroup has zero 0 and unit 1 which are the unique idempotents. The transversal semigroup λ(C3)/C3 is isomorphic to the semilattice 2 = {0, 1} endowed with the min-operation. 6.3. The semigroups λ(C4) and λ(C2 ⊕ C2) The semigroup λ(C4) contains 12 elements while the orbit semigroup λ(C4)/C4 contains 3 elements. The semigroup λ(C4) contains a transver- sal semigroup λT (G) = {1,△, �} where 1 is the neutral element of C4 = {1,−1, i,−i}, △ = 〈{1, i}, {1,−i}, {i,−i}〉 and � = 〈{1, i}, {1,−i}, {1,−1}, {i,−i,−1}〉. The transversal semigroup is isomorphic to the extension C1 2 = C2 ∪ {e} of the cyclic group C2 by an external unit e /∈ C2 (such that ex = x = xe for all x ∈ C1 2 ). The action of the group C4 on λ(C4) is free so, λ(C4) is isomorphic to λT (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) where e is 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: △ = 〈{(1, 1), (1,−1)}, {(1, 1), (−1, 1)}, {(1,−1), (−1, 1)}〉 and � = 〈{(1, 1), (1, -1)}, {(1, 1), (-1, 1)}, {(1, 1), (-1, -1)}, {(1, -1), (-1, 1), (-1, -1)}〉. The transversal semigroup λT (C2 ⊕C2) is isomorphic to C1 2 and λ(C2 ⊕ C2) is isomorphic to C1 2 ⊕ C2 ⊕ C2. We summarize the obtained results on the algebraic structure of the semigroups λ(C4) and λ(C2 ⊕ C2) in the following proposition. Proposition 6.1. Let G be a group of cardinality \|G\| = 4. Jo u rn al A lg eb ra D is cr et e M at h .24 Algebra in superextensions of groups, I 1. The semigroup λ(G) is isomorphic to C1 2 ⊕ G and thus is commu- tative; 2. λ(G) contains two idempotents; 3. λ(G) has a unique proper ideal λ(G) \ G isomorphic to the group C2 ⊕ G. 6.4. The semigroup λ(C5). Unlike to λ(C4), the semigroup λ(C5) has complicated algebraic struc- ture. It contains 81 elements. One of them is zero Z = {A ⊂ C5 : \|A\| ≥ 3}, which is invariant under any bijection of C5. All the other 80 elements have 5-element orbits under the action of C5, which implies that the orbit semigroup λ(C5)/C5 consists of 17 elements. Let π : λ(C5) → λ(C5)/C5 denote the orbit projection. It will be convenient to think of C5 as the field {0, 1, 2, 3, 4} with the multiplicative subgroup C∗ 5 = {1,−1, 2,−2} of invertible elements (here −1 and −2 are identified with 4 and 3, respectively). Also for elements x, y, z ∈ C5 we shall write xyz instead of {x, y, z}. The semigroup λ(C5) contains 5 idempotents: U =〈0〉, Z, Λ4 =〈01, 02, 03, 04, 1234〉, Λ =〈02, 03, 123, 014, 234〉, 2Λ =〈04, 01, 124, 023, 143〉, which commute and thus form an abelian subsemigroup E(λ(C5)). Being a semilattice, E(λ(C5)) carries a natural partial order: e ≤ f iff e◦f = e. The partial order Z ≤ Λ, 2Λ ≤ Λ4 ≤ U on the set E(λ(C5)) is designed at the picture: Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 25 Z r � � @ @ rΛ r 2Λ Λ4 r @ @ @ @ � � r U The other distinguished subset of λ(C5) is √ E(λ(C5)) = {L ∈ λ(C5) : L ◦ L ∈ E(λ(C5))} = = {L ∈ λ(C5) : L ◦ L ◦ L ◦ L = L ◦ L}. We shall show that this set contains a point from each C5-orbit in λ(C5). First we show that this set has at most one-point intersection with each orbit. Indeed, if L ∈ √ E(λ(C5)) and L ◦ L 6= Z, then for every a ∈ C5 \ {0}, we get (L + a) ◦ (L + a) ◦ (L + a) ◦ (L + a) = L ◦ L ◦ L ◦ L + 4a = =L ◦ L + 4a 6= L ◦ L + 2a = (L + a) ◦ (L + a). witnessing that L + a /∈ √ λT (C5). By a direct calculation one can check that the set λT (C5) contains the following four maximal linked systems: ∆ =〈02, 03, 23〉, Λ3 =〈02, 03, 04, 234〉, Θ =〈14, 012, 013, 123, 024, 034, 234〉, Γ =〈02, 04, 013, 124, 234〉. For those systems we get ∆ ◦ ∆ = ∆ ◦ ∆ ◦ ∆ = Λ, Λ3 ◦ Λ3 = Λ3 ◦ Λ3 ◦ Λ3 = Λ, F ◦ Θ = F ◦ Γ = Z for every F ∈ λ(C5) \ C5. All the other elements of λ(C5) can be found as images of ∆, Θ, Γ, Λ3 under the affine transformations of the field C5. Those are maps of the form fa,b : x 7→ ax + b mod 5, Jo u rn al A lg eb ra D is cr et e M at h .26 Algebra in superextensions of groups, I where a ∈ {1,−1, 2,−2} = C∗ 5 and b ∈ C5. The image of a maximal linked system L ∈ λ(C5) under such a transformation will be denoted by aL + b. One can check that aΛ4 = Λ4 for each a ∈ C∗ 5 while Λ = −Λ, and Θ = −Θ. Since the linear transformations of the form fa,0 : C5 → C5, a ∈ C∗ 5 , are authomorphisms of the group C5 the induced transformations λfa,0 : λ(C5) → λ(C5) are authomorphisms of the semigroup λ(C5). This implies that those transformations do not move the subsets E(λ(C5)) and √ E(λ(C5)). Consequently, the set √ E(λ(C5) contains the maximal linked systems: a∆, aΘ, aΛ3, aΓ, a ∈ Z ∗ 5, which together with the idempotents form a 17-element subset T17 = E(λ(C5)) ∪ { a∆, aΘ : a ∈ {1, 2} } ∪ {aΛ3, aΓ : a ∈ Z ∗ 5} that projects bijectively onto the orbit semigroup λ(C5)/C5. The set T17 looks as follows (we connect an element x ∈ T17 with an idempotent e ∈ T17 by an arrow if x ◦ x = e): Z � � @ @ Λ 2Λ Λ4 @ @ @ @ � � U −Γ Θ - 2Θ� � �� Γ � �� 2Γ B BBM −2Γ @ @I −Λ3 � �3 ∆ - Λ3Q Qs −2Λ3 Q Qk 2∆� 2Λ3� �+ The set √ E(λ(C5)) includes 24 elements more and coincides with the union T17 ∪ √ Z where √ Z = {aΘ + b, aΓ + b : a ∈ Z ∗ 5, b ∈ C5}. Since each element of λ(C5) can be uniquely written as the sum L+ b for some L ∈ T17 and b ∈ C5, the multiplication table for the semigroup λ(C5) can be recovered from the Cayley table for multiplication of the elements from T17: Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 27 ◦ Λ4 Λ ∆ Λ3 −Λ3 2Λ 2∆ 2Λ3 −2Λ3 aΘ, aΓ Λ4 Λ4 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 Λ3 Λ3 Λ Λ Λ Λ 2Θ+2 2Θ+2 2Θ+2 2Θ+2 Z −Λ3 −Λ3 Λ Λ Λ Λ 2Θ−2 2Θ−2 2Θ−2 2Θ−2 Z 2Λ 2Λ Z Z Z Z 2Λ 2Λ 2Λ 2Λ Z 2∆ 2∆ Θ Θ Θ Θ 2Λ 2Λ 2Λ 2Λ Z 2Λ3 2Λ3 Θ−1 Θ−1 Θ−1 Θ−1 2Λ 2Λ 2Λ 2Λ Z −2Λ3 −2Λ3 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−1 Θ−1 Θ−1 Θ−1 2Θ−2 2Θ−2 2Θ−2 2Θ−2 Z 2Γ 2Γ Θ−1 Θ−1 Θ−1 Θ−1 2Θ+2 2Θ+2 2Θ+2 2Θ+2 Z −2Γ −2Γ Θ+1 Θ+1 Θ+1 Θ+1 2Θ−2 2Θ−2 2Θ−2 2Θ−2 Z Looking at this table we can see that T17 is not a subsemigroup of λ(C5) and hence is not a transversal semigroup for λ(C5). This is not occasional. Proposition 6.2. The semigroup λ(C5) contains no transversal semi- group. Proof. Assume conversely that λ(C5) contains a subsemigroup T that projects bijectively onto the orbit semigroup λ(C5)/C5. Then T must include the set E(λ(C5)) of idempotents and also the subset √ E(λ(C5))\√ Z. Consequently, T ⊃ {U ,Z, Λ,−Λ, ∆, 2∆, Λ3,−Λ3, 2Λ3,−2Λ3}. Since 2Λ3 ◦ Λ = Θ − 1 6= Θ = 2∆ ◦ Λ, then there are two different points in the intersection T ∩ (Θ+C5) which should be a singleton. This contradiction completes the proof. Analysing the Cayley table for the set T17 we can establish the fol- lowing properties of the semigroup λ(C5). Proposition 6.3. 1. The maximal linked system Z is the zero of λ(Z). 2. λ(C5) contains 5 idempotents: U , Z, Λ4, Λ, 2Λ, which commute. 3. The set of central elements of λ(C5) coincides with C5 ∪ {Z}. 4. All non-trivial subgroups of λ(C5) are isomorphic to C5. Jo u rn al A lg eb ra D is cr et e M at h .28 Algebra in superextensions of groups, I 6.5. Summary table The obtained results on the superextensions of groups G with \|G\| ≤ 5 are summed up in the following table in which K(λ(G)) stands for the minimal ideal of λ(G). \|G\| \|λ(G)\| λ(G) \|E(λ(G))\| K(λ(G)) maximal group 2 2 C2 1 C2 C2 3 4 C3 ∪ {⊲} 2 {⊲} C3 4 12 C1 2 × G 2 C2 × G C2 × G 5 81 T17 · C5 5 {Z} C5 References [BG2] T. Banakh, V. Gavrylkiv. Algebra in superextension of groups, II: cancelativity and centers, Algebra Discrete Math. [BG3] T. Banakh, V. Gavrylkiv. Algebra in superextension of groups, III: the minimal ideal of λ(G), preprint (arXiv:0805.1583). [BG4] T. Banakh, V. Gavrylkiv. Algebra in superextension of groups, IV: representa- tion theory, preprint. [De] R. Dedekind, Über Zerlegungen von Zahlen durch ihre grüssten gemeinsammen Teiler // In Gesammelte Werke, Bd. 1 (1897), 103–148. [G1] V. Gavrylkiv. The spaces of inclusion hyperspaces over noncompact spaces, Matem. Studii. 28:1 (2007), 92–110. [G2] V. Gavrylkiv, Right-topological semigroup operations on inclusion hyperspaces, Matem. Studii. 29:1 (2008), 18–34. [H1] N. Hindman, Finite sums from sequences within cells of partition of N // J. Com- bin. Theory Ser. A 17 (1974), 1–11. [H2] N. Hindman, Ultrafilters and combinatorial number theory // Lecture Notes in Math. 751 (1979), 49–184. [HS] N. Hindman, D. Strauss, Algebra in the Stone-Čech compactification, de Gruyter, Berlin, New York, 1998. [vM] J. van Mill, Supercompactness and Wallman spaces, Math. Centre Tracts. 85. 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. Jo u rn al A lg eb ra D is cr et e M at h .T. Banakh, V. Gavrylkiv, O. Nykyforchyn 29 Contact information Taras Banakh Ivan Franko National University of Lviv, Universytetska 1, 79000, Ukraine E-Mail: tbanakh@yahoo.com URL: www.franko.lviv.ua/faculty/mechmat/ Departments/Topology/bancv.html V. Gavrylkiv, O. Nykyforchyn Vasyl Stefanyk Precarpathian National Uni- versity, Ivano-Frankivsk, Ukraine E-Mail: vgavrylkiv@yahoo.com Received by the editors: 14.02.2008 and in final form 14.10.2008.

Algebra in superextensions of groups, I: zeros and commutativity

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