Endomorphisms of free and free metabelian groups

AbbreviationThe free metabelian group \(M_n\)#✓fully proved

For a natural number \(n\), the free metabelian group of rank \(n\) is the quotient \(F_n / F_n''\) of the free group \(F_n\) on \(n\) generators by its second derived subgroup \(F_n'' = [F_n', F_n']\), where \(F_n' = [F_n, F_n]\) is the derived subgroup of \(F_n\).

abbrev FreeMetabelianGroup (n : ℕ) : Type :=
  FreeGroup (Fin n) ⧸ derivedSeries (FreeGroup (Fin n)) 2
Used by 1

DefinitionGroups whose endomorphisms with non-cyclic images are determined by two elements#✓fully proved

A group \(F\) has this property when there exist elements \(g, h \in F\) such that any two endomorphisms \(\varphi, \psi\) of \(F\) whose images are not cyclic subgroups and which satisfy \(\varphi(g) = \psi(g)\) and \(\varphi(h) = \psi(h)\) are equal.

def DeterminedByTwoElements (F : Type) [Group F] : Prop :=
  ∃ g h : F, ∀ φ ψ : F →* F, ¬ IsCyclic φ.range → ¬ IsCyclic ψ.range → φ g = ψ g →
    φ h = ψ h → φ = ψ
Used by 2

DefinitionShpil'rain's two-element problem for free groups (Kourovka 13.66(a))#✓fully proved

For every natural number \(n\), the free group \(F_n\) of rank \(n\) contains two elements \(g\) and \(h\) such that any two endomorphisms of \(F_n\) whose images are not cyclic subgroups and which agree at \(g\) and at \(h\) are equal; that is, \(F_n\) has its endomorphisms with non-cyclic images determined by two elements.

def freeGroupEndomorphismsDeterminedByTwoElements : Prop :=
  ∀ n : ℕ, DeterminedByTwoElements (FreeGroup (Fin n))
Uses 1 · Used by 1

DefinitionShpil'rain's two-element problem for free metabelian groups (Kourovka 13.66(b))#✓fully proved

For every natural number \(n\), the free metabelian group \(M_n = F_n / F_n''\) of rank \(n\) (the quotient of the free group \(F_n\) of rank \(n\) by its second derived subgroup) contains two elements \(g\) and \(h\) such that any two endomorphisms of \(M_n\) whose images are not cyclic subgroups and which agree at \(g\) and at \(h\) are equal.

def freeMetabelianEndomorphismsDeterminedByTwoElements : Prop :=
  ∀ n : ℕ, DeterminedByTwoElements (FreeMetabelianGroup n)
Uses 2 · Used by 1

TheoremEndomorphisms of free groups with non-cyclic images are determined by two elements#●ready to prove

For every natural number \(n\), the free group \(F_n\) of rank \(n\) contains two elements \(g\) and \(h\) such that any two endomorphisms \(\varphi, \psi\) of \(F_n\) whose images are not cyclic subgroups and which satisfy \(\varphi(g) = \psi(g)\) and \(\varphi(h) = \psi(h)\) are equal.

public theorem_wanted freeGroupEndomorphismsDeterminedByTwoElements_holds :
    freeGroupEndomorphismsDeterminedByTwoElements
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 13.66 (V. E. Shpil'rain); resolution: (a) D. Lee, J. Algebra 247 (2002), no. 2, 509-540; (b) E. I. Timoshenko, Math. Notes 62 (1998), no. 5-6, 767-770, Kourovka Archive, Problem 13.66 (answers: a) Yes, one can; b) No, not always).
Uses 1

TheoremEndomorphisms of free metabelian groups are not always determined by two elements#●ready to prove

For a natural number \(n\), let \(M_n = F_n / F_n''\) be the free metabelian group of rank \(n\), the quotient of the free group \(F_n\) of rank \(n\) by its second derived subgroup. There is an \(n\) such that for all elements \(g, h \in M_n\) there are endomorphisms \(\varphi \ne \psi\) of \(M_n\) whose images are not cyclic subgroups and which satisfy \(\varphi(g) = \psi(g)\) and \(\varphi(h) = \psi(h)\).

public theorem_wanted freeMetabelianEndomorphismsDeterminedByTwoElements_refuted :
    ¬ freeMetabelianEndomorphismsDeterminedByTwoElements
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 13.66 (V. E. Shpil'rain); resolution: (a) D. Lee, J. Algebra 247 (2002), no. 2, 509-540; (b) E. I. Timoshenko, Math. Notes 62 (1998), no. 5-6, 767-770, Kourovka Archive, Problem 13.66 (answers: a) Yes, one can; b) No, not always).
Uses 1
4 fully proved2 ready to prove

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