Ordered groups with three conjugacy classes

DefinitionKourovka Problem 11.68 (ordered groups with three conjugacy classes)#✓fully proved

The statement that for every fully ordered group \(G\) there are a fully ordered group \(H\) with exactly three conjugacy classes and a strictly increasing homomorphism \(G \to H\) (an order-preserving embedding). Here a fully ordered group is a group with a linear order \(\le\) such that \(a \le b\) implies \(ca \le cb\) and \(ac \le bc\) for all \(a\), \(b\), \(c\).

def orderedEmbedsInThreeClasses : Prop :=
  ∀ (G : Type) [Group G] [LinearOrder G] [MulLeftMono G] [MulRightMono G],
    ∃ (H : Type) (_ : Group H) (_ : LinearOrder H), MulLeftMono H ∧ MulRightMono H ∧
      Nat.card (ConjClasses H) = 3 ∧ ∃ f : G →* H, StrictMono f

Used by 1

Used byNot every fully ordered group embeds in one with three conjugacy classes

TheoremNot every fully ordered group embeds in one with three conjugacy classes#●ready to prove

There is a fully ordered group \(G\) such that for every fully ordered group \(H\) with exactly three conjugacy classes, no homomorphism \(G \to H\) is strictly increasing. Here a fully ordered group is a group with a linear order \(\le\) such that \(a \le b\) implies \(ca \le cb\) and \(ac \le bc\) for all \(a\), \(b\), \(c\). That is, the answer to Kourovka Problem 11.68 is negative.

The Kourovka Archive records the answer "No, not every", given by Bludov.

public theorem_wanted orderedEmbedsInThreeClasses_refuted : ¬ orderedEmbedsInThreeClasses

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 11.68 (B. Neumann); resolution: V. V. Bludov, Algebra and Logic 44, no. 6 (2005), 370-380, Kourovka Archive, Problem 11.68: "No, not every", https://arxiv.org/abs/1401.0300v49.

Uses 1

Statement usesKourovka Problem 11.68 (ordered groups with three conjugacy classes)

1 fully proved1 ready to prove

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