Eulerian polynomials of groups

DefinitionOrder polynomial of a subgroup#✓fully proved

For a subgroup \(H\) of a group, the order polynomial of \(H\) is \(f(H) = \prod_p x_p^{v_p(|H|)}\) in the integer polynomial ring \(\mathbb{Z}[x_0, x_1, \dots]\), where \(p\) runs over the primes, \(x_p\) is the variable indexed by \(p\), and \(v_p(|H|)\) is the exponent of \(p\) in the order \(|H|\) of \(H\). If \(H\) is infinite, \(f(H) = 1\).

noncomputable def orderPolynomial {G : Type*} [Group G] (H : Subgroup G) : MvPolynomial ℕ ℤ :=
  (Nat.card H).factorization.prod fun p a => MvPolynomial.X p ^ a
Used by 1

DefinitionEulerian function of a group (G. E. Wall)#✓fully proved

Let \(G\) be a group and let \(f\) be the order polynomial. A function \(\varphi\) from the subgroups of \(G\) to the integer polynomial ring \(\mathbb{Z}[x_0, x_1, \dots]\) is an Eulerian function of \(G\) if \(\sum_{K \le H} \varphi(K) = f(H)\) for every subgroup \(H\) of \(G\), the sum running over the subgroups \(K\) of \(H\). The value \(\varphi(G)\) is the Eulerian polynomial of \(G\).

For finite \(G\) this relation determines \(\varphi\) uniquely, by induction on \(H\). If infinitely many subgroups \(K \le H\) have \(\varphi(K) \ne 0\), the sum is read as \(0\).

def IsEulerianFunction {G : Type*} [Group G] (φ : Subgroup G → MvPolynomial ℕ ℤ) : Prop :=
  ∀ H : Subgroup G, ∑ᶠ (K : Subgroup G) (_ : K ≤ H), φ K = orderPolynomial H
Uses 1 · Used by 2

Definition\(p\)-solvable group#✓fully proved

For a natural number \(p\), a group \(G\) is \(p\)-solvable if it has a chain of subgroups \(1 = N_0 \le N_1 \le \dots \le N_n = G\) in which each \(N_i\) is normal in \(N_{i+1}\) and each index \([N_{i+1} : N_i]\), taken to be \(0\) when it is infinite, is a power \(p^k\) with \(k \ge 0\) or is coprime to \(p\). For finite \(G\) and prime \(p\) this says that every factor \(N_{i+1}/N_i\) is a \(p\)-group or has order prime to \(p\).

def IsPSolvable (p : ℕ) (G : Type*) [Group G] : Prop :=
  ∃ (n : ℕ) (N : ℕ → Subgroup G), N 0 = ⊥ ∧ N n = ⊤ ∧ ∀ i < n, N i ≤ N (i + 1) ∧
    ((N i).subgroupOf (N (i + 1))).Normal ∧
    ((∃ k : ℕ, (N i).relIndex (N (i + 1)) = p ^ k) ∨ Nat.Coprime ((N i).relIndex (N (i + 1))) p)
Used by 1

DefinitionKourovka Problem 11.104(a) (Eulerian polynomials and solvability, G. E. Wall)#✓fully proved

This is the statement that for every finite group \(G\) and every Eulerian function \(\varphi\) of \(G\), with values in the integer polynomial ring \(\mathbb{Z}[x_0, x_1, \dots]\): if there are a finite set \(S\) of primes and, for each \(q \in S\), an integer polynomial \(P_q\) involving no variable other than \(x_q\), such that the Eulerian polynomial \(\varphi(G)\) equals \(\prod_{q \in S} P_q\), then \(G\) is solvable.

def eulerianPolynomialSplitsSolvable : Prop :=
  ∀ (G : Type) [Group G] [Finite G] (φ : Subgroup G → MvPolynomial ℕ ℤ), IsEulerianFunction φ →
    (∃ (s : Finset ℕ) (P : ℕ → MvPolynomial ℕ ℤ), (∀ q ∈ s, q.Prime ∧ (P q).vars ⊆ {q}) ∧
      φ ⊤ = ∏ q ∈ s, P q) →
    Group.IsSolvable G
Uses 1 · Used by 1

DefinitionKourovka Problem 11.104(b) (Eulerian polynomials and \(p\)-solvability)#✓fully proved

This is the statement that for every finite group \(G\), every prime \(p\) and every Eulerian function \(\varphi\) of \(G\), with values in the integer polynomial ring \(\mathbb{Z}[x_0, x_1, \dots]\): if the Eulerian polynomial \(\varphi(G)\) is a product \(PQ\) of integer polynomials where \(P\) involves no variable other than \(x_p\) and \(x_p\) does not occur in \(Q\), then \(G\) is \(p\)-solvable.

def eulerianPolynomialSeparatesPSolvable : Prop :=
  ∀ (G : Type) [Group G] [Finite G] (p : ℕ), p.Prime → ∀ φ : Subgroup G → MvPolynomial ℕ ℤ,
    IsEulerianFunction φ →
    (∃ P Q : MvPolynomial ℕ ℤ, P.vars ⊆ {p} ∧ p ∉ Q.vars ∧ φ ⊤ = P * Q) → IsPSolvable p G
Uses 2 · Used by 1

TheoremSolvability from one-variable factorizations of the Eulerian polynomial#●ready to prove

Let \(G\) be a finite group and let \(\varphi\) be an Eulerian function of \(G\), with values in the integer polynomial ring \(\mathbb{Z}[x_0, x_1, \dots]\). If there are a finite set \(S\) of primes and, for each \(q \in S\), an integer polynomial \(P_q\) involving no variable other than \(x_q\), such that the Eulerian polynomial \(\varphi(G)\) equals \(\prod_{q \in S} P_q\), then \(G\) is solvable.

Resolved true: the Kourovka Archive answers 11.104(a) Yes, citing Detomi and Lucchini (2004).

public theorem_wanted eulerianPolynomialSplitsSolvable_holds : eulerianPolynomialSplitsSolvable
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 11.104 (G. E. Wall); resolution: E. Detomi and A. Lucchini, J. London Math. Soc. (2) 70 (2004), 165-181 (a); E. Damian and A. Lucchini, Commun. Algebra 35 (2007), 3451-3472 (b), Kourovka Archive, Problem 11.104 ("Yes, the converses are true").
Uses 1

TheoremEulerian polynomial factorization implies \(p\)-solvability#●ready to prove

Let \(G\) be a finite group, let \(p\) be a prime, and let \(\varphi\) be an Eulerian function of \(G\), with values in the integer polynomial ring \(\mathbb{Z}[x_0, x_1, \dots]\). If the Eulerian polynomial \(\varphi(G)\) is a product \(PQ\) of integer polynomials where \(P\) involves no variable other than \(x_p\) and \(x_p\) does not occur in \(Q\), then \(G\) is \(p\)-solvable.

Resolved true: the Kourovka Archive answers 11.104(b) Yes, citing Damian and Lucchini (2007).

public theorem_wanted eulerianPolynomialSeparatesPSolvable_holds :
    eulerianPolynomialSeparatesPSolvable
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 11.104 (G. E. Wall); resolution: E. Detomi and A. Lucchini, J. London Math. Soc. (2) 70 (2004), 165-181 (a); E. Damian and A. Lucchini, Commun. Algebra 35 (2007), 3451-3472 (b), Kourovka Archive, Problem 11.104 ("Yes, the converses are true").
Uses 1
5 fully proved2 ready to prove

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