Homology of finitely generated metabelian groups

DefinitionKourovka Problem 10.37 (statement)#✓fully proved

The statement that every finitely generated metabelian group \(G\) (that is, one whose second derived subgroup \(G''\) is trivial) all of whose integral homology groups \(H_n(G, \mathbb{Z})\) with trivial coefficients, \(n \ge 0\), are finitely generated has finite Prüfer rank: there is a natural number \(r\) such that every finitely generated subgroup of \(G\) is generated by at most \(r\) elements.

def metabelianHomologyFiniteRank : Prop :=
  ∀ (G : Type) [Group G], Group.FG G → derivedSeries G 2 = ⊥ →
    (∀ n : ℕ, Module.Finite ℤ (groupHomology (Rep.trivial ℤ G ℤ) n)) →
    ∃ r : ℕ, ∀ H : Subgroup G, H.FG →
      ∃ S : Finset G, S.card ≤ r ∧ Subgroup.closure (S : Set G) = H
Used by 1

TheoremFinitely generated metabelian groups with finitely generated integral homology have finite rank#●ready to prove

Finitely generated metabelian groups with finitely generated integral homology have finite rank. Let \(G\) be a finitely generated metabelian group, that is, its second derived subgroup \(G''\) is trivial, such that for every \(n \ge 0\) the integral homology group \(H_n(G, \mathbb{Z})\) with trivial coefficients is finitely generated. Then \(G\) has finite Prüfer rank: there is a natural number \(r\) such that every finitely generated subgroup of \(G\) is generated by at most \(r\) elements. This answers Kourovka Problem 10.37 positively.

The Kourovka Archive records this answer, citing Kochloukova (2003); Groves had proved it when \(G\) splits over its derived subgroup.

public theorem_wanted metabelianHomologyFiniteRank_holds : metabelianHomologyFiniteRank
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 10.37 (G. A. Noskov); resolution: D. H. Kochloukova, Groups with finitely generated integral homologies, in: Groups St. Andrews 2001 in Oxford, Vol. II, Cambridge Univ. Press, 2003, 332-343, Kourovka Archive, Problem 10.37 ("Yes, it is").
Uses 1
1 fully proved1 ready to prove

The dependency explorer needs JavaScript and its graph data.

Download the complete graph data.