Source labels openChecked July 26, 2026

Conjecture 20.76: 76»

Let $G$ be a finite $p$ -group and assume that all abelian normal subgroups of $G$ have order at most $p^k$ . Is it true that every abelian subgroup of $G$ has order at most $p^{2k}$ ?

Group theory AMS 20

Mathematical statement

Let $G$ be a finite $p$ -group and assume that all abelian normal subgroups of $G$ have order at most $p^k$ . Is it true that every abelian subgroup of $G$ has order at most $p^{2k}$ ?

Statement source: Kourovka Notebook statement material

Statement terms: Source-specific

Source-specific terms. Therefore does not assert reuse rights beyond attributed display.

Statement artifacts, not proofs

These records expose exact Lean propositions and statement-only wrappers. Defining a proposition does not supply a proof of it. A placeholder-bearing target also contains no proof. Elaboration checks syntax and types; it does not certify that a formalization perfectly captures every nuance of the informal problem.

Pinned Lean formulation 1

kourovka.«20.76»

Canonical source

Complete statement target, proof intentionally absentLean 4

theorem kourovka.«20.76» : answer(sorry) ↔    ∀ᵉ (p : ℕ) (hp : p.Prime) (G : Type) (_ : Group G) (hg : IsPGroup p G) (_ : Finite G) (k : ℕ)    (h : ∀ H: Subgroup G, H.Normal ∧ IsMulCommutative H → Nat.card H ≤ p ^ k),    (∀ H : Subgroup G, IsMulCommutative H → Nat.card H ≤ p ^ (2 * k)) := by  sorry

Statement source: Formal Conjectures
Lean version: v4.27.0
Placeholder: Present; no proof artifact
Source evidence: Pinned source index
Fidelity review: Community formulation

References

Pinned Formal Conjectures sourceformalization The Kourovka Notebookbackground