Subgroups with all prime divisors of a simple group

DefinitionKourovka Problem 11.64 (Monakhov)#✓fully proved

The statement that there is a natural number \(N\) such that every finite simple group \(G\) that is not isomorphic to the alternating group \(A_n\) for any natural number \(n\) and has a proper subgroup \(H\) with \(\pi(H) = \pi(G)\) satisfies \(|G| \le N\), where \(\pi(X)\) denotes the set of prime divisors of the order \(|X|\).

Since there are finitely many groups of each order, this says that only finitely many such groups \(G\) exist up to isomorphism.

def finitelyManySimpleWithFullSpectrumSubgroup : Prop :=
  ∃ N : ℕ, ∀ (G : Type) [Group G] [Finite G], IsSimpleGroup G →
    (∀ n : ℕ, IsEmpty (G ≃* alternatingGroup (Fin n))) →
    (∃ H : Subgroup G, H ≠ ⊤ ∧ (Nat.card H).primeFactors = (Nat.card G).primeFactors) →
    Nat.card G ≤ N

Used by 1

Used byInfinitely many simple groups with proper subgroups of equal prime set

TheoremInfinitely many simple groups with proper subgroups of equal prime set#●ready to prove

Resolved false: the Kourovka Archive records (V. I. Zenkov) that the symplectic groups G = S_{4k}(2^s) have proper subgroups H ≅ Ω⁻_{4k}(2^s) with π(H) = π(G), an infinite family of non-alternating finite simple groups. Source: 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 11.64 (V. S. Monakhov); resolution: V. I. Zenkov, letter of 10 March 1994, Kourovka Archive, Problem 11.64: "No, there are infinitely many such groups. If G = S_{4k}(2^s) and H ≅ Ω⁻_{4k}(2^s), then π(G) = π(H)", https://arxiv.org/abs/1401.0300v49.

public theorem_wanted finitelyManySimpleWithFullSpectrumSubgroup_refuted :
    ¬ finitelyManySimpleWithFullSpectrumSubgroup

Uses 1

Statement usesKourovka Problem 11.64 (Monakhov)

1 fully proved1 ready to prove

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