Conjugacy separability

DefinitionKapovich's conjugacy separability question (Kourovka 19.41)#✓fully proved

The assertion that for every natural number \(n\) and every automorphism \(\varphi\) of the free group \(F_n\) of rank \(n\), the semidirect product \(F_n \rtimes_\varphi \mathbb{Z}\), in which \(k \in \mathbb{Z}\) acts on \(F_n\) by \(\varphi^k\), is conjugacy separable.

def freeByCyclicConjugacySeparable : Prop :=
  ∀ (n : ℕ) (φ : MulAut (FreeGroup (Fin n))),
    IsConjugacySeparable
      (FreeGroup (Fin n) ⋊[zpowersHom (MulAut (FreeGroup (Fin n))) φ] Multiplicative ℤ)

Uses 1 · Used by 1

Statement usesConjugacy separable groups

Used byFinite-rank free-by-cyclic groups are conjugacy separable

TheoremFinite-rank free-by-cyclic groups are conjugacy separable#●ready to prove

For every natural number \(n\) and every automorphism \(\varphi\) of the free group \(F_n\) of rank \(n\), the semidirect product \(F_n \rtimes_\varphi \mathbb{Z}\), in which \(k \in \mathbb{Z}\) acts on \(F_n\) by \(\varphi^k\), is conjugacy separable: any two of its elements whose images are conjugate in every quotient by a normal subgroup of finite index are conjugate.

public theorem_wanted freeByCyclicConjugacySeparable_holds : freeByCyclicConjugacySeparable

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 19.41 (I. Kapovich); resolution: F. Dahmani, S. Hughes, M. Kudlinska and N. Touikan, Free-by-cyclic groups are conjugacy separable, preprint, arXiv:2509.22346v2 (2026), abstract: all finitely generated free-by-cyclic groups are conjugacy separable, resolving Question 19.41; the Kourovka Notebook (21st issue) marks 19.41 solved: Yes, it is, https://arxiv.org/abs/2509.22346.

Uses 1

Statement usesKapovich's conjugacy separability question (Kourovka 19.41)

1 fully proved1 ready to prove

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