Factorizations of groups into subsets

DefinitionFactorization of a group into subsets of given sizes#✓fully proved

Let \(G\) be a group, \(k\) a natural number and \(a_0, \dots, a_{k-1}\) natural numbers. The group \(G\) has a subset factorization of type \((a_0, \dots, a_{k-1})\) when there are finite subsets \(A_0, \dots, A_{k-1}\) of \(G\) with \(|A_i| = a_i\) for every \(i\) such that \(G = A_0 A_1 \cdots A_{k-1}\), that is, every \(g \in G\) can be written as \(g = x_0 x_1 \cdots x_{k-1}\) with \(x_i \in A_i\) for every \(i\).

def HasSubsetFactorization (G : Type) [Group G] (k : ℕ) (a : Fin k → ℕ) : Prop :=
  ∃ A : Fin k → Finset G, (∀ i, (A i).card = a i) ∧
    ∀ g : G, ∃ x : Fin k → G, (∀ i, x i ∈ A i) ∧ (List.ofFn x).prod = g

Used by 1

Used byKourovka Problem 19.35: counterexamples to subset factorization of every length

DefinitionKourovka Problem 19.35: counterexamples to subset factorization of every length#✓fully proved

The statement that for every natural number \(k \ge 3\) there are a finite group \(G\) and natural numbers \(a_0, \dots, a_{k-1}\), each at least \(2\), with \(a_0 a_1 \cdots a_{k-1} = |G|\), such that \(G\) has no subset factorization of type \((a_0, \dots, a_{k-1})\): there are no subsets \(A_0, \dots, A_{k-1}\) of \(G\) with \(|A_i| = a_i\) for every \(i\) and \(G = A_0 A_1 \cdots A_{k-1}\).

def subsetFactorizationCounterexampleEveryLength : Prop :=
  ∀ k : ℕ, 3 ≤ k → ∃ (G : Type) (_ : Group G) (_ : Finite G) (a : Fin k → ℕ),
    (∀ i, 2 ≤ a i) ∧ ∏ i, a i = Nat.card G ∧ ¬ HasSubsetFactorization G k a

Uses 1 · Used by 1

Statement usesFactorization of a group into subsets of given sizes

Used byFinite groups without subset factorizations exist for every length

TheoremFinite groups without subset factorizations exist for every length#●ready to prove

For every natural number \(k \ge 3\) there are a finite group \(G\) and natural numbers \(a_0, \dots, a_{k-1}\), each at least \(2\), with \(a_0 a_1 \cdots a_{k-1} = |G|\), such that there are no subsets \(A_0, \dots, A_{k-1}\) of \(G\) with \(|A_i| = a_i\) for every \(i\) and \(G = A_0 A_1 \cdots A_{k-1}\), that is, with every element of \(G\) a product \(x_0 x_1 \cdots x_{k-1}\) where \(x_i \in A_i\) for every \(i\).

In McCulloch's Theorem 2.3, \(G\) is a Frobenius group of order \(2^{k-1}(2^{k-1} - 1)\) and the factors are \((2, \dots, 2, 2^{k-1} - 1, 2)\); the case \(k = 3\) is Bergman's example in the alternating group \(A_4\). This answers Kourovka Problem 19.35 (M. H. Hooshmand) in the negative.

public theorem_wanted subsetFactorizationCounterexampleEveryLength_holds :
    subsetFactorizationCounterexampleEveryLength

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 19.35 (M. H. Hooshmand); resolution: R. McCulloch, Some answers regarding factorizations of finite groups, arXiv:2607.20569 (2026); counterexamples first given by M. Kabenyuk, Factorizations of finite groups, arXiv:2102.08605, Kourovka Archive, Problem 19.35: No (Bergman's A4 example for k = 3); counterexamples for all k ≥ 3 are constructed in (McCulloch); paper, Theorem 2.3: the Frobenius group of order 2^(k-1)(2^(k-1) - 1) has no k-fold factorization for (2, …, 2, 2^(k-1) - 1, 2), https://arxiv.org/abs/2607.20569.

Uses 1

Statement usesKourovka Problem 19.35: counterexamples to subset factorization of every length

2 fully proved1 ready to prove

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