Torsion-free groups with three conjugacy classes

DefinitionKourovka Problem 11.42 (statement)#✓fully proved

The statement that there is a torsion-free group \(G\) (no element of \(G\) other than the identity has finite order) that has exactly three conjugacy classes and has a subgroup of index \(2\).

The problem was posed by A. V. Izosov.

def torsionFreeThreeClassesIndexTwo : Prop :=
  ∃ (G : Type) (_ : Group G), (∀ g : G, g ≠ 1 → ¬ IsOfFinOrder g) ∧
    Nat.card (ConjClasses G) = 3 ∧ ∃ H : Subgroup G, H.index = 2

Used by 1

Used byMinasyan's theorem (Kourovka Problem 11.42)

TheoremMinasyan's theorem (Kourovka Problem 11.42)#●ready to prove

There is a torsion-free group \(G\) (no element of \(G\) other than the identity has finite order) that has exactly three conjugacy classes and has a subgroup of index \(2\).

public theorem_wanted torsionFreeThreeClassesIndexTwo_holds : torsionFreeThreeClassesIndexTwo

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 11.42 (A. V. Izosov); resolution: A. Minasyan, Comment. Math. Helv. 84 (2009), 259-296, Kourovka Archive, Problem 11.42: "Yes, there does", https://arxiv.org/abs/1401.0300v49.

Uses 1

Statement usesKourovka Problem 11.42 (statement)

DefinitionKourovka Problem 11.67 (statement)#✓fully proved

The statement that there is a torsion-free group \(G\) (no element of \(G\) other than the identity has finite order) with exactly three conjugacy classes in which no element \(g \neq 1\) is conjugate to its inverse \(g^{-1}\), that is, no nontrivial conjugacy class contains a pair of mutually inverse elements.

The problem was posed by B. Neumann.

def torsionFreeThreeClassesNoInverseConj : Prop :=
  ∃ (G : Type) (_ : Group G), (∀ g : G, g ≠ 1 → ¬ IsOfFinOrder g) ∧
    Nat.card (ConjClasses G) = 3 ∧ ∀ g : G, g ≠ 1 → ¬ IsConj g g⁻¹

Used by 1

Used byBagayoko's theorem (Kourovka Problem 11.67)

TheoremBagayoko's theorem (Kourovka Problem 11.67)#●ready to prove

There is a torsion-free group \(G\) (no element of \(G\) other than the identity has finite order) with exactly three conjugacy classes in which no element \(g \neq 1\) is conjugate to its inverse \(g^{-1}\).

Bagayoko's example is a linearly ordered group with exactly three conjugacy classes, the positive transseries under composition; an ordered group is torsion-free, and conjugation preserves positivity, so no \(g \neq 1\) is conjugate to \(g^{-1}\).

public theorem_wanted torsionFreeThreeClassesNoInverseConj_holds :
    torsionFreeThreeClassesNoInverseConj

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 11.67 (B. Neumann); resolution: V. Bagayoko, Ordered groups of formal series, and a conjugacy problem, arXiv:2509.09186 (2025), main result: a linearly ordered group (𝕋^{>ℝ}, ∘, <) of transseries under composition with exactly three conjugacy classes; the Kourovka Notebook (21st issue) marks 11.67 solved, https://arxiv.org/abs/2509.09186.

Uses 1

Statement usesKourovka Problem 11.67 (statement)

2 fully proved2 ready to prove

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