Counts of elements of each order

DefinitionShi's question on element-order counts (Kourovka 18.112)#✓fully proved

The proposition that if a finite group \(G\) has the property that, for every divisor \(d > 1\) of \(|G|\) and every subgroup \(H\) of \(G\) whose order is coprime to \(d\), \(|H|\) divides the number of elements of \(G\) of order \(d\), then every non-identity element of \(G\) has prime-power order.

def elementOrderCountPrimePowerOrders : Prop :=
  ∀ (G : Type) [Group G] [Finite G],
    (∀ d : ℕ, 1 < d → d ∣ Nat.card G → ∀ H : Subgroup G, Nat.Coprime (Nat.card H) d →
      Nat.card H ∣ Nat.card {g : G // orderOf g = d}) →
    ∀ g : G, g ≠ 1 → IsPrimePow (orderOf g)

Used by 1

Used byCounting elements of each order forces prime-power orders

TheoremCounting elements of each order forces prime-power orders#●ready to prove

Let \(G\) be a finite group such that for every divisor \(d > 1\) of \(|G|\) and every subgroup \(H\) of \(G\) whose order is coprime to \(d\), the order \(|H|\) divides the number of elements of \(G\) of order \(d\). Then every non-identity element of \(G\) has prime-power order.

Buturlakin, Shen and Shi proved this in 2017, answering Kourovka Problem 18.112 (W. J. Shi); the Kourovka Archive answers 18.112 "Yes, it is true".

public theorem_wanted elementOrderCountPrimePowerOrders_holds : elementOrderCountPrimePowerOrders

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 18.112 (W. J. Shi); resolution: A. A. Buturlakin, R. Shen and W. Shi, Siberian Math. J. 58 (2017), 405-407, Kourovka Archive, Problem 18.112 (Yes, it is true).

Uses 1

Statement usesShi's question on element-order counts (Kourovka 18.112)

1 fully proved1 ready to prove

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