Derived length of p-groups with an automorphism

DefinitionGroups of sectional rank at most \(r\)#✓fully proved

Let \(G\) be a group and \(r\) a natural number. The group \(G\) has sectional rank at most \(r\) when every subgroup of \(G\) is generated by a set of at most \(r\) elements.

For a finite \(p\)-group \(G\), the least such \(r\) is the largest rank of an elementary abelian section of \(G\).

def SectionalRankLE (G : Type*) [Group G] (r : ℕ) : Prop :=
  ∀ H : Subgroup G, ∃ S : Finset G, S.card ≤ r ∧ Subgroup.closure (S : Set G) = H

Used by 1

Used byShalev and Khukhro's question on the derived length of \(p\)-groups with an automorphism

DefinitionShalev and Khukhro's question on the derived length of \(p\)-groups with an automorphism#✓fully proved

The statement that there is a function \(f : \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) such that for every prime \(p\), every finite \(p\)-group \(G\), all natural numbers \(r\) and \(m\) such that \(G\) has sectional rank at most \(r\) (every subgroup of \(G\) is generated by at most \(r\) elements), and every automorphism \(\varphi\) of \(G\) with exactly \(m\) fixed points, the derived length of \(G\) is at most \(f(r, m)\), that is, the term \(G^{(f(r, m))}\) of the derived series of \(G\) is trivial, where \(G^{(0)} = G\) and \(G^{(i+1)} = [G^{(i)}, G^{(i)}]\).

This is Kourovka Notebook Problem 13.56 (A. Shalev; E. I. Khukhro).

def derivedLengthBoundedByRankAndFixedPoints : Prop :=
  ∃ f : ℕ → ℕ → ℕ, ∀ (p : ℕ) (G : Type) [Group G] [Finite G], p.Prime → IsPGroup p G →
    ∀ r m : ℕ, SectionalRankLE G r → ∀ φ : G ≃* G, Nat.card {g : G // φ g = g} = m →
      derivedSeries G (f r m) = ⊥

Uses 1 · Used by 1

Statement usesGroups of sectional rank at most \(r\)

Used byThe derived length of a \(p\)-group is bounded by its sectional rank and the fixed points of an automorphism

TheoremThe derived length of a \(p\)-group is bounded by its sectional rank and the fixed points of an automorphism#●ready to prove

The derived length of a \(p\)-group is bounded by its sectional rank and the fixed points of an automorphism. There is a function \(f : \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) such that for every prime \(p\), every finite \(p\)-group \(G\), all natural numbers \(r\) and \(m\) such that \(G\) has sectional rank at most \(r\) (every subgroup of \(G\) is generated by at most \(r\) elements), and every automorphism \(\varphi\) of \(G\) with exactly \(m\) fixed points, the derived length of \(G\) is at most \(f(r, m)\), that is, the term \(G^{(f(r, m))}\) of the derived series of \(G\) is trivial, where \(G^{(0)} = G\) and \(G^{(i+1)} = [G^{(i)}, G^{(i)}]\).

Resolved true: the Kourovka Archive answers 13.56 "Yes, it is", citing Jaikin-Zapirain: the derived length of a finite \(p\)-group of sectional rank \(r\) with an automorphism having exactly \(m\) fixed points is bounded by a function of \(r\) and \(m\).

public theorem_wanted derivedLengthBoundedByRankAndFixedPoints_holds :
    derivedLengthBoundedByRankAndFixedPoints

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 13.56 (A. Shalev; E. I. Khukhro); resolution: A. Jaikin-Zapirain, Israel J. Math. 129 (2002), 209-220, Kourovka Archive, Problem 13.56 (answer: Yes, it is).

Uses 1

Statement usesShalev and Khukhro's question on the derived length of \(p\)-groups with an automorphism

2 fully proved1 ready to prove

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