aa r X i v : . [ m a t h . G R ] A ug A CONJECTURE ON PARTITIONS OF GROUPS
IGOR PROTASOV, SERGII SLOBODIANIUK
Abstract.
We conjecture that every infinite group G can be partitioned into countably manycells G = S n ∈ ω A n such that cov ( A n A − n ) = | G | for each n ∈ ω . Here cov ( A ) = min {| X | : X ⊆ G, G = XA } . We confirm this conjecture for each group of regular cardinality and for somegroups (in particular, Abelian) of an arbitrary cardinality. Introduction
For any finite partition of an infinite group, at last one cell of the partition has a richcombinatorial structure. The Ramsey Theory of groups gives a plenty of concrete examples(see [4], [7])On the other hand, the subsets of a group could be classified by their size. For correspondingpartition problems see the survey [8].Given a group G and a subset A of G , we denote cov ( A ) = min {| X | : X ⊆ G, G = XA } . The covering number cov ( A ) evaluates a size of A inside G and, if A is a subgroup, coincideswith the index | G : A | .It is easy to partition each infinite group G = A ∪ A so that cov ( A ) and cov ( A ) are infinite.Moreover, if | G | is regular, there is a partition G = S α< | G | H α such that cov ( G \ H α ) = | G | for each α < | G | . In particular, there is a partition G = A ∪ A such that cov ( A ) = cov ( A ) = | G | .See [5], [9], [10] for these statements, their generalizations and applications.However, for every n ∈ N , there is a (minimal) natural number Φ( n ) such that, for everygroup G and every partition G = A ∪ ... ∪ A n , cov ( A i A − i ) ≤ Φ( n ) for some cell A i of thepartition. It is still an open problem posed in [6, Problem 13.44] whether Φ( n ) = n . For thehistory and results behind this problem see the survey [1].In [2, Question F], J. Erde asked whether, given a partition P of an infinite group G suchthat |P| < | G | , there is A ∈ P such that cov ( AA − ) is finite. After some simple examplesanswering this question extremely negatively, we run into the following conjecture. Conjecture.
Every infinite group G of cardinality κ can be partitioned G = S n<ω A n so that cov ( A n A − n ) = κ for each n ∈ ω . In this note we confirm Conjecture for every group of regular cardinality and for some groups(in particular, Abelian) of an arbitrary cardinality.
Mathematics Subject Classification.
Key words and phrases. partition of groups, covering number.
Results
For a cardinal κ , we denote by cf ( κ ) the cofinality of κ . Theorem 1.
Let G be an infinite group of cardinality κ . Then there exists a partition G = S n ∈ ω A n such that cov ( A n A − n ) ≥ cf ( κ ) for each n ∈ ω .Proof. If G is countable, the statement is trivial: take any partition of G into finite subsets.For κ > ℵ , we choose a family { G α : α < κ } of subgroups of G such that(1) G = { e } and G = S α< κ G α , e is the identity of G ;(2) G α ⊂ G β for all α < β < κ ;(3) S α<β G α = G β for each limit ordinal β < κ ;(4) | G α | < κ for each α < κ .Following [11], for each α < κ , we decompose G α +1 \ G α into right cosets by G α and choosesome system X α of representatives so G α +1 = G α X α . Take an arbitrary element g ∈ G \ { e } and choose the smallest subgroup G α with g ∈ G α . By (3), α = α + 1 for some ordinal α < κ .Hence g ∈ G α +1 \ G α and there exists g ∈ G α , x α ∈ X α such that g = g x α . If g = e ,we choose the ordinal α and elements g ∈ G α +1 \ G α and x α ∈ X α such that g = g x α .Since the set of ordinals { α : α < κ } is well-ordered, after finite number s ( g ) of steps, we getthe representation g = x α s ( g ) x α s ( g ) − ...x α x α , x α i ∈ X α i . We note that this representation is unique and put γ ( g ) = α , γ ( g ) = α , ..., γ s ( g ) ( g ) = α s ( g ) , max( g ) = γ ( g ) . Each ordinal α < κ can be written uniquely as α = β + n where β is a limit ordinal and n ∈ ω . We put f ( α ) = n and denote by Seq ( ω ) the set of all finite sequences of elements of ω .Then we define a mapping χ : G \ { e } → Seq ( ω ) by χ ( g ) = f ( γ s ( g ) ( g )) f ( γ s ( g ) − ( g )) ...f ( γ ( g )) f ( γ ( g )) , and, for each s ∈ Seq ( ω ), put H s = χ − ( s ). Since the set Seq ( ω ) is countable, it suffices toprove that cov ( H s H − s ) ≥ cf κ for each s ∈ Seq ( ω ).We take an arbitrary s ∈ Seq ( ω ) and an arbitrary K ⊆ G such that | K | < cf κ . Then wechoose γ < κ such that γ > max g for each g ∈ K and f ( γ ) / ∈ { s , ..., s n } . We pick h ∈ X γ andshow that KH s ∩ hH s = ∅ .If g ∈ KH s and γ i ( g ) ≥ γ then f ( γ i ( g )) ∈ { s , ..., s n } . To prove this we take an arbitrary g ∈ KH s and fix k ∈ K , x ∈ H s such that g = kx . Let i ∈ ω be the length of the representationof x after G γ , which means that there exist x ∈ X γ ( x ) , ..., x i ∈ X γ i ( x ) such that x = yx i ...x ,for some y ∈ G γ . As x ∈ H s it follows that f ( x j ) ∈ { s , ..., s n } , for 1 ≤ j ≤ i and since k ∈ K ⊆ G γ we have that g = kx = zx α i ...x α , where z ∈ G γ . So for any i ≤ s ( g ) for which γ i ( g ) ≥ γ we get that f ( γ i ( g )) ∈ { s , ..., s n } .Now if g ′ ∈ hH s then the representations of g ′ and h − g after G γ +1 are equal with thesame length i due to the previous argument as h ∈ G γ +1 . But they are different after G γ as f ( γ i +1 ( h − g ′ )) ∈ { s , ..., s n } , and then since, h / ∈ G γ , f ( γ i +1 ( g ′ )) = f ( γ ) / ∈ { s , ..., s n } .Hence KH s ∩ hH s = ∅ , so h / ∈ KH s H − s and cov ( H s H − s ) can not be less then cf κ . (cid:3) CONJECTURE ON PARTITIONS OF GROUPS 3
Theorem 2.
Let λ, κ be infinite cardinals, λ < κ and let { H α : α < κ } be a family of groupssuch that | H α | ≤ λ for each α < κ . Let G be a subgroup of the direct product H = ⊗ α< κ H α such that | G | = κ . Then there exists a partition G = S n<ω A n such that cov ( A n A − n ) = κ foreach n ∈ ω .Proof. For each g ∈ G , supt ( g ) denotes the number of non-identity coordinates of g , χ ( g ) = | supt ( g ) | . For each h ∈ ω we put A n = G ∩ χ − ( n )and show that cov ( A n A − n ) = κ .We take an arbitrary K ⊂ G such that | K | < κ and denote S = { α < κ : pr α g = e α for some g ∈ K } , T = κ \ S,G T = G ∩ ⊗ α ∈ T H α . Since λ < κ and | G | = κ , we have | G T | = κ . If g ∈ KA n A − n ∩ G T then | supt ( g ) | ≤ n .On the other hand, for every m ∈ ω , there is h ∈ G T such that | supt ( h ) | > m . Hence, G T \ KA n A − n = ∅ . (cid:3) Remark 1.
It is well-known [3, Theorems 23.1 and 24.1] that each Abelian group can be embeddedinto the direct product of countable groups. Applying Theorem 2, we confirm conjecture forAbelian groups. Moreover, if a group G of cardinality κ has an Abelian homomorphic imageof cardinality κ (in particular, if G is a free group of rank κ ) then Conjecture is valid for G . Remark 2.
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