Embeddings into simple periodic groups
DefinitionKourovka Problem 11.106 (periodic groups in simple periodic groups)#✓fully proved
MathlibExt.GroupTheory.PeriodicGroupSimpleEmbeddingKourovka11Problem106Wanted.periodicGroupEmbedsInSimplePeriodic
The statement, posed by R. Phillips, that every periodic group embeds in a simple periodic group: for every group \(G\) in which every element has finite order, there are a simple group \(S\) in which every element has finite order and an injective homomorphism \(G \to S\).
def periodicGroupEmbedsInSimplePeriodic : Prop :=
∀ (G : Type) [Group G], (∀ g : G, IsOfFinOrder g) →
∃ (S : Type) (_ : Group S), IsSimpleGroup S ∧ (∀ s : S, IsOfFinOrder s) ∧
∃ f : G →* S, Function.Injective f
Used by 1
TheoremEvery periodic group embeds in a simple periodic group#●ready to prove
MathlibExt.GroupTheory.PeriodicGroupSimpleEmbeddingKourovka11Problem106Wanted.periodicGroupEmbedsInSimplePeriodic_holds
For every group \(G\) in which every element has finite order, there are a simple group \(S\) in which every element has finite order and an injective homomorphism \(G \to S\). This answers Kourovka Problem 11.106 (R. Phillips) in the affirmative; the result is due to A. Yu. Olshanskii (1992).
public theorem_wanted periodicGroupEmbedsInSimplePeriodic_holds :
periodicGroupEmbedsInSimplePeriodic
def periodicGroupEmbedsInSimplePeriodic : Prop :=
∀ (G : Type) [Group G], (∀ g : G, IsOfFinOrder g) →
∃ (S : Type) (_ : Group S), IsSimpleGroup S ∧ (∀ s : S, IsOfFinOrder s) ∧
∃ f : G →* S, Function.Injective f
WantedExt/GroupTheory/PeriodicGroupSimpleEmbeddingKourovka11Problem106Wanted.lean:36
Sources
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 11.106 (R. Phillips); resolution: A. Yu. Olshanskii, Embedding periodic groups in simple periodic groups, Ukr. Mat. Zh. 44 (1992), 845-847 (Russian); English translation: Ukrain. Math. J. 44 (1992), 761-763, Kourovka Archive, Problem 11.106 ("Yes").
Uses 1
1 fully proved1 ready to prove
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