Fully invariant subgroups

DefinitionMacHale's question on fully invariant subgroups (Kourovka 16.61)#✓fully proved

The statement that every finite group \(G\) having, for every divisor \(d\) of \(|G|\), a fully invariant subgroup of order \(d\) (a subgroup \(H\) with \(\varphi(H) \le H\) for every endomorphism \(\varphi\) of \(G\)) is cyclic.

def fullyInvariantSubgroupsOfEachOrderCyclic : Prop :=
  ∀ (G : Type) [Group G] [Finite G],
    (∀ d : ℕ, d ∣ Nat.card G → ∃ H : Subgroup G, Nat.card H = d ∧ ∀ φ : G →* G, H.map φ ≤ H) →
    IsCyclic G

Used by 1

Used byFully invariant subgroups of every order do not force cyclicity

TheoremFully invariant subgroups of every order do not force cyclicity#●ready to prove

Not every finite group \(G\) having, for every divisor \(d\) of \(|G|\), a fully invariant subgroup of order \(d\) (a subgroup \(H\) with \(\varphi(H) \le H\) for every endomorphism \(\varphi\) of \(G\)) is cyclic.

The Kourovka Archive answers 16.61 "No" (Abdollahi): take \(G = C_p \times C_{p^2}\), where \(C_p\) and \(C_{p^2}\) are cyclic \(p\)-groups of orders \(p\) and \(p^2\).

public theorem_wanted fullyInvariantSubgroupsOfEachOrderCyclic_refuted :
    ¬ fullyInvariantSubgroupsOfEachOrderCyclic

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 16.61 (D. MacHale); resolution: A. Abdollahi, letter of 4 March 2009, as recorded in the Kourovka Archive, Kourovka Archive, Problem 16.61 (No: G = C_p × C_{p²}).

Uses 1

Statement usesMacHale's question on fully invariant subgroups (Kourovka 16.61)

1 fully proved1 ready to prove

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