Partitions of groups of singular cardinality

DefinitionPartitions of groups of singular cardinality (Kourovka 13.45)#✓fully proved

The statement that every infinite group \(G\) whose cardinality \(\mathfrak{m} = |G|\) is singular (not regular) can be split into two disjoint sets \(A_1\) and \(A_2\) with \(A_1 \cup A_2 = G\) such that \(A_1 F \ne G\) and \(A_2 F \ne G\) for every subset \(F \subseteq G\) with \(|F| < \mathfrak{m}\), where \(A F = \{a f : a \in A, f \in F\}\).

def singularCardinalityPartition : Prop :=
  ∀ (G : Type) [Group G], Infinite G → ¬ (#G).IsRegular →
    ∃ A₁ A₂ : Set G, A₁ ∪ A₂ = Set.univ ∧ Disjoint A₁ A₂ ∧
      ∀ F : Set G, #F < #G → A₁ * F ≠ Set.univ ∧ A₂ * F ≠ Set.univ
Used by 1

TheoremSome group of singular cardinality has no such partition#●ready to prove

There is an infinite group \(G\) whose cardinality \(\mathfrak{m} = |G|\) is singular (not regular) such that for all disjoint sets \(A_1\) and \(A_2\) with \(A_1 \cup A_2 = G\) there is a subset \(F \subseteq G\) with \(|F| < \mathfrak{m}\) and \(A_1 F = G\) or \(A_2 F = G\), where \(A F = \{a f : a \in A, f \in F\}\). That is, the statement of Kourovka Problem 13.45 fails.

Protasov and Slobodianiuk show that the answer depends on the algebraic structure of \(G\): such a partition exists for every free group, but not for abelian groups of singular cardinality. The Kourovka Archive answers 13.45 "No, not always".

public theorem_wanted singularCardinalityPartition_refuted : ¬ singularCardinalityPartition
Sources
  • E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Archive of Solved Problems, Problem 13.45 (I. V. Protasov); resolution: I. Protasov and S. Slobodianiuk, A note on partitions of groups, Quest. Answers Gen. Topology 33 (2015), no. 2, 61-70, arXiv:1408.5607, abstract and Theorem 2(iii): the statement fails for abelian groups of singular cardinality (it holds for free groups); Kourovka Archive, Problem 13.45, https://arxiv.org/abs/1408.5607.
Uses 1
1 fully proved1 ready to prove

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