Orders of 2-groups of bounded abelian rank
DefinitionSambale's question on the order of \(2\)-groups of bounded abelian rank (Kourovka 19.75)#✓fully proved
The statement that for all natural numbers \(e\) and \(r\), every finite \(2\)-group \(P\) of exponent \(2^e\) in which every abelian subgroup is generated by at most \(r\) elements has order \(|P| \le 2^{r(e+1)}\).
def twoGroupOrderBoundByAbelianRank : Prop :=
∀ (P : Type) [Group P] [Finite P], IsPGroup 2 P → ∀ e r : ℕ, Monoid.exponent P = 2 ^ e →
(∀ A : Subgroup P, (∀ x ∈ A, ∀ y ∈ A, x * y = y * x) →
∃ s : Finset P, s.card ≤ r ∧ Subgroup.closure (s : Set P) = A) →
Nat.card P ≤ 2 ^ (r * (e + 1))
TheoremSambale's order bound for \(2\)-groups of bounded abelian rank fails (Kourovka 19.75)#●ready to prove
There exist natural numbers \(e\) and \(r\) and a finite \(2\)-group \(P\) of exponent \(2^e\) in which every abelian subgroup is generated by at most \(r\) elements, such that \(|P| > 2^{r(e+1)}\).
The Kourovka Archive answers 19.75 "No, it is not true", as follows from Olshanskii's examples of finite \(p\)-groups whose abelian subgroups are small compared to the group (pointed out by Mann).
public theorem_wanted twoGroupOrderBoundByAbelianRank_refuted : ¬ twoGroupOrderBoundByAbelianRank
def twoGroupOrderBoundByAbelianRank : Prop :=
∀ (P : Type) [Group P] [Finite P], IsPGroup 2 P → ∀ e r : ℕ, Monoid.exponent P = 2 ^ e →
(∀ A : Subgroup P, (∀ x ∈ A, ∀ y ∈ A, x * y = y * x) →
∃ s : Finset P, s.card ≤ r ∧ Subgroup.closure (s : Set P) = A) →
Nat.card P ≤ 2 ^ (r * (e + 1))
WantedExt/GroupTheory/TwoGroupAbelianRankOrderBoundKourovka19Problem75Wanted.lean:37
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 19.75 (B. Sambale); resolution: A. Yu. Olshanskii, The number of generators and orders of abelian subgroups of finite p-groups, Math. Notes 23 (1978), no. 3, 183-185; A. Mann, letter of 23 April 2018, abstract: the least f(n) such that every finite p-group whose abelian subgroups are generated by at most n elements has at most f(n) generators grows quadratically in n; Kourovka Archive, Problem 19.75: No, it is not true, as follows from the examples constructed by Olshanskii (Mann), https://doi.org/10.1007/BF01651428.
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