Products of periodic subgroups
DefinitionKourovka Problem 16.38 (subgroups inside a product of periodic subgroups)#✓fully proved
The statement, posed by F. de Giovanni, that for every soluble group \(G\) and all subgroups \(A\) and \(B\) of \(G\) in which every element has finite order, every subgroup \(H\) of \(G\) contained in the set \(AB\) of all products \(ab\) with \(a \in A\) and \(b \in B\) has every element of finite order.
def subgroupInPeriodicProductIsPeriodic : Prop :=
∀ (G : Type) [Group G], Group.IsSolvable G → ∀ A B : Subgroup G,
(∀ a ∈ A, IsOfFinOrder a) → (∀ b ∈ B, IsOfFinOrder b) →
∀ H : Subgroup G, (H : Set G) ⊆ (A : Set G) * (B : Set G) → ∀ h ∈ H, IsOfFinOrder h
TheoremSubgroups inside a product of two periodic subgroups of a soluble group are periodic#●ready to prove
Let \(G\) be a soluble group and let \(A\) and \(B\) be subgroups of \(G\) in which every element has finite order. Then every subgroup \(H\) of \(G\) contained in the set \(AB\) of all products \(ab\) with \(a \in A\) and \(b \in B\) is periodic: every element of \(H\) has finite order. This answers Kourovka Problem 16.38 (F. de Giovanni) in the affirmative; the proof is due to Kinyon and Urban (Proposition 3.8).
public theorem_wanted subgroupInPeriodicProductIsPeriodic_holds :
subgroupInPeriodicProductIsPeriodic
def subgroupInPeriodicProductIsPeriodic : Prop :=
∀ (G : Type) [Group G], Group.IsSolvable G → ∀ A B : Subgroup G,
(∀ a ∈ A, IsOfFinOrder a) → (∀ b ∈ B, IsOfFinOrder b) →
∀ H : Subgroup G, (H : Set G) ⊆ (A : Set G) * (B : Set G) → ∀ h ∈ H, IsOfFinOrder h
WantedExt/GroupTheory/PeriodicSubgroupsInProductSetKourovka16Problem38Wanted.lean:40
Sources
- E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 16.38 (F. de Giovanni); resolution: M. Kinyon and J. Urban (with Codex gpt-6-astra), Results on problems from the Kourovka Notebook, preprint, version 4 (19 September 2026), Proposition 3.8 (if G is soluble and A, B ≤ G are periodic, every subgroup contained in the set AB is periodic); the Kourovka Notebook (21st issue) marks 16.38 solved (Yes, AI solution confirmed by B. Amberg), https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/09/kourovka-mathematics.pdf.
Uses 1
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