Subgroups of free groups with normal derived subgroup

DefinitionKourovka Problem 11.124 (free-group subgroups with normal derived subgroup)#✓fully proved

The statement that for every set \(X\) such that the free group \(F\) on \(X\) is not cyclic, every subgroup \(R\) of \(F\) that is not cyclic and whose derived subgroup \([R, R]\) is normal in \(F\) is itself normal in \(F\).

def derivedNormalImpliesNormal : Prop :=
  ∀ (α : Type), ¬ IsCyclic (FreeGroup α) → ∀ R : Subgroup (FreeGroup α), ¬ IsCyclic R →
    (⁅R, R⁆).Normal → R.Normal
Used by 1

TheoremA non-cyclic subgroup of a free group with normal derived subgroup is normal#●ready to prove

Let \(F\) be a free group on an arbitrary set of generators, and suppose that \(F\) is not cyclic. Then every subgroup \(R\) of \(F\) that is not cyclic and whose derived subgroup \([R, R]\) is normal in \(F\) is itself normal in \(F\).

Resolved true: Tursunbaev proves that subgroups \(H\), \(K\) of a free group of arbitrary rank whose derived subgroups \(H' = [H, H]\) and \(K' = [K, K]\) are equal and nontrivial coincide; for a non-cyclic \(R\) with \(R' = [R, R]\) normal in \(F\) and any \(f \in F\), the conjugate \(R^f = f^{-1} R f\) satisfies \((R^f)' = (R')^f = R' \ne 1\), so \(R^f = R\). The Kourovka Notebook (21st issue) marks 11.124 solved, "Yes, it is true".

public theorem_wanted derivedNormalImpliesNormal_holds : derivedNormalImpliesNormal
Sources
  • E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 11.124 (V. E. Shpilrain); resolution: A. Tursunbaev, Subgroups of free groups with the same derived subgroup, arXiv:2609.06755 (2026), main theorem (subgroups H, K of a free group of arbitrary rank with H' = K' ≠ 1 are equal) and its corollary answering 11.124; the Kourovka Notebook (21st issue) marks 11.124 solved, https://arxiv.org/abs/2609.06755.
Uses 1
1 fully proved1 ready to prove

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