Subgroups permuting with their conjugates

DefinitionPermuting conjugates and simplicity (Kourovka 10.66)#✓fully proved

The conjecture that a group \(G\) containing nontrivial subgroups \(A\) and \(B\) such that the product set \(AB\) is not all of \(G\) and \(A \cdot gBg^{-1} = gBg^{-1} \cdot A\) for every \(g \in G\) is not simple. Products of subgroups are products of subsets of \(G\).

This is Kourovka Problem 10.66 (A. N. Fomin, V. P. Shunkov).

def permutingConjugatesNotSimple : Prop :=
  ∀ (G : Type) [Group G] (A B : Subgroup G), A ≠ ⊥ → B ≠ ⊥ → (A : Set G) * (B : Set G) ≠ Set.univ →
    (∀ g : G, (A : Set G) * (B.map (MulAut.conj g).toMonoidHom : Set G) =
      (B.map (MulAut.conj g).toMonoidHom : Set G) * (A : Set G)) →
    ¬ IsSimpleGroup G

Used by 1

Used byA simple group with permuting conjugate subgroups (Kourovka 10.66 refuted)

TheoremA simple group with permuting conjugate subgroups (Kourovka 10.66 refuted)#●ready to prove

There exist a simple group \(G\) and nontrivial subgroups \(A\) and \(B\) of \(G\) such that the product set \(AB\) is not all of \(G\) and \(A \cdot gBg^{-1} = gBg^{-1} \cdot A\) for every \(g \in G\), where products of subgroups are products of subsets of \(G\). So the conjecture of Kourovka Problem 10.66 is false.

Bludov's example takes a simple linearly ordered group \(G\) and nontrivial proper convex subgroups \(A\) and \(B\). Conjugates of convex subgroups are convex and the convex subgroups form a chain, so \(A \cdot gBg^{-1}\) and \(gBg^{-1} \cdot A\) both equal the larger of \(A\) and \(gBg^{-1}\), and \(AB \ne G\). Such subgroups exist: a nontrivial simple ordered group is not archimedean (archimedean ordered groups are abelian, by Hölder), so it has nontrivial proper convex subgroups.

public theorem_wanted permutingConjugatesNotSimple_refuted : ¬ permutingConjugatesNotSimple

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 10.66 (A. N. Fomin, V. P. Shunkov); resolution: V. V. Bludov, letter of 12 February 1997, recorded in the Kourovka Archive, Problem 10.66, Archive answer: "No; for example, for G being any simple linearly ordered group and A and B arbitrary proper convex subgroups of G".

Uses 1

Statement usesPermuting conjugates and simplicity (Kourovka 10.66)

1 fully proved1 ready to prove

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