Automorphisms of order four with few fixed points

DefinitionKourovka Problem 11.126 (automorphisms of order four with few fixed points)#✓fully proved

The statement that there are a natural number \(h\) and a function \(f : \mathbb{N} \to \mathbb{N}\) such that, for every natural number \(m\), every finite soluble group \(G\) and every automorphism \(\varphi\) of \(G\) of order \(4\) with at most \(m\) fixed points, there are normal subgroups \(M \le N\) of \(G\) with \(|G : N| \le f(m)\), \([[N, N], N] \le M\) (so \(N/M\) is nilpotent of class at most \(2\)) and \(\gamma_{h+1}(M) = 1\) (so \(M\) is nilpotent of class at most \(h\)), where \(\gamma_1(M) = M\) and \(\gamma_{i+1}(M) = [\gamma_i(M), M]\).

def orderFourAutomorphismBoundedSeries : Prop :=
  ∃ (h : ℕ) (f : ℕ → ℕ), ∀ (m : ℕ) (G : Type) [Group G] [Finite G], Group.IsSolvable G →
    ∀ φ : MulAut G, orderOf φ = 4 → Nat.card {x : G // φ x = x} ≤ m →
      ∃ M N : Subgroup G, M.Normal ∧ N.Normal ∧ M ≤ N ∧ N.index ≤ f m ∧
        ⁅⁅N, N⁆, N⁆ ≤ M ∧ M.lowerCentralSeries h = ⊥
Used by 1

TheoremBounded normal series for soluble groups with an automorphism of order four#●ready to prove

There are a natural number \(h\) and a function \(f : \mathbb{N} \to \mathbb{N}\) such that, for every natural number \(m\), every finite soluble group \(G\) and every automorphism \(\varphi\) of \(G\) of order \(4\) with at most \(m\) fixed points, there are normal subgroups \(M \le N\) of \(G\) with \(|G : N| \le f(m)\), \([[N, N], N] \le M\) (so \(N/M\) is nilpotent of class at most \(2\)) and \(\gamma_{h+1}(M) = 1\) (so \(M\) is nilpotent of class at most \(h\)), where \(\gamma_1(M) = M\) and \(\gamma_{i+1}(M) = [\gamma_i(M), M]\). That is, the answer to Kourovka Problem 11.126 is positive.

The Kourovka Archive records the answer "Yes, there do", given by Makarenko and Khukhro.

public theorem_wanted orderFourAutomorphismBoundedSeries_holds : orderFourAutomorphismBoundedSeries
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 11.126 (P. V. Shumyatskiĭ); resolution: N. Yu. Makarenko and E. I. Khukhro, Algebra and Logic 45 (2006), 326-343, Kourovka Archive, Problem 11.126: "Yes, there do", https://arxiv.org/abs/1401.0300v49.
Uses 1
1 fully proved1 ready to prove

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