Periodic Frobenius groups

DefinitionFrobenius group with given kernel and complement#✓fully proved

For subgroups \(F\) and \(H\) of a group \(G\), \(G\) is a Frobenius group with kernel \(F\) and complement \(H\) (Kourovka 6.53) if \(G = F \rtimes H\) (that is, \(F\) is normal in \(G\), \(F \cap H = 1\) and \(FH = G\)), \(F \ne 1\), \(H \ne 1\), \(H \cap gHg^{-1} = 1\) for every \(g \in G \setminus H\), and the non-trivial elements of \(F\) are exactly the elements of \(G\) that lie in no conjugate \(gHg^{-1}\) of \(H\).

def IsFrobeniusGroup {G : Type*} [Group G] (F H : Subgroup G) : Prop :=
  F.Normal ∧ F ⊓ H = ⊥ ∧ F ⊔ H = ⊤ ∧ F ≠ ⊥ ∧ H ≠ ⊥ ∧
    (∀ g : G, g ∉ H → H ⊓ H.map (MulAut.conj g).toMonoidHom = ⊥) ∧
    ∀ x : G, (x ∈ F ∧ x ≠ 1) ↔ ∀ g : G, x ∉ H.map (MulAut.conj g).toMonoidHom

Used by 1

Used byKourovka Problem 20.96 (periodic Frobenius groups)

DefinitionKourovka Problem 20.96 (periodic Frobenius groups)#✓fully proved

The statement, posed by A. I. Sozutov, that for every group \(G\) in which every element has finite order and every normal abelian subgroup \(A\) of \(G\) that is neither trivial nor all of \(G\) and such that for every \(a \in A\) with \(a \ne 1\) the centralizer \(C_G(a)\) of \(a\) in \(G\) is contained in \(A\), there are subgroups \(F\) and \(H\) of \(G\) such that \(G\) is a Frobenius group with kernel \(F\) and complement \(H\).

def periodicSelfCentralizingAbelianNormalFrobenius : Prop :=
  ∀ (G : Type) [Group G], IsMulTorsion G → ∀ A : Subgroup G, A.Normal → A ≠ ⊥ → A ≠ ⊤ →
    (∀ x ∈ A, ∀ y ∈ A, x * y = y * x) → (∀ a ∈ A, a ≠ 1 → Subgroup.centralizer {a} ≤ A) →
    ∃ F H : Subgroup G, IsFrobeniusGroup F H

Uses 1 · Used by 1

Statement usesFrobenius group with given kernel and complement

Used byLytkina–Mazurov theorem on periodic Frobenius groups

TheoremLytkina–Mazurov theorem on periodic Frobenius groups#●ready to prove

Let \(G\) be a group in which every element has finite order, and let \(A\) be a normal abelian subgroup of \(G\) that is neither trivial nor all of \(G\) and such that for every \(a \in A\) with \(a \ne 1\) the centralizer \(C_G(a)\) of \(a\) in \(G\) is contained in \(A\). Then there are subgroups \(F\) and \(H\) of \(G\) such that \(G\) is a Frobenius group with kernel \(F\) and complement \(H\). This answers Kourovka Problem 20.96 (A. I. Sozutov) in the affirmative.

public theorem_wanted periodicSelfCentralizingAbelianNormalFrobenius_holds :
    periodicSelfCentralizingAbelianNormalFrobenius

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 20.96 (A. I. Sozutov); resolution: D. V. Lytkina and V. D. Mazurov, Periodic Frobenius groups, Siberian Math. J. 64 (2023), no. 6, 1351-1355 (Sibirsk. Mat. Zh. 64 (2023), no. 6, 1224-1228), abstract: conditions for a strongly isolated normal subgroup of a periodic group to have a complement; Kourovka Notebook (21st issue), Problem 20.96: Yes, it is, https://www.mathnet.ru/eng/smj7826.

Uses 1

Statement usesKourovka Problem 20.96 (periodic Frobenius groups)

2 fully proved1 ready to prove

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