Order-divisibility bijections onto cyclic groups

DefinitionKourovka Problem 18.1 (statement)#✓fully proved

The statement that for every finite group \(G\), of order \(n\) say, there is a bijection \(f\) from \(G\) onto the cyclic group \(\mathbb{Z}/n\mathbb{Z}\) such that for every \(x \in G\) the order of \(x\) divides the additive order of \(f(x)\).

def orderDivisibilityBijectionToCyclic : Prop :=
  ∀ (G : Type) [Group G] [Finite G], ∃ f : G ≃ ZMod (Nat.card G),
    ∀ x : G, orderOf x ∣ addOrderOf (f x)

Used by 1

Used byAmiri's theorem (Kourovka Problem 18.1)

TheoremAmiri's theorem (Kourovka Problem 18.1)#●ready to prove

For every finite group \(G\), of order \(n\) say, there is a bijection \(f\) from \(G\) onto the cyclic group \(\mathbb{Z}/n\mathbb{Z}\) such that for every \(x \in G\) the order of \(x\) divides the additive order of \(f(x)\).

public theorem_wanted orderDivisibilityBijectionToCyclic_holds : orderDivisibilityBijectionToCyclic

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 18.1 (I. M. Isaacs); resolution: M. Amiri, J. Pure Appl. Algebra 228 (2024), no. 7, 107632 (preprint arXiv:2002.11455), abstract (for any finite group G of order n there is a bijection f from G onto C_n with o(x) | o(f(x)) for all x); the Kourovka Notebook (21st issue) marks 18.1 solved (Yes, it does), https://arxiv.org/abs/2002.11455.

Uses 1

Statement usesKourovka Problem 18.1 (statement)

1 fully proved1 ready to prove

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