Large automorphism groups of p-groups

DefinitionKourovka Problem 15.32: large automorphisms and cyclic subgroups#✓fully proved

For every prime \(p\) and all natural numbers \(k\) and \(m\) with \(m \ge 64k + 2\), every finite group \(G\) of order \(p^m\) that has an automorphism of order \(p^{m-k}\) has a cyclic subgroup of index \(p^k\).

Kourovka 15.32 (Ya. G. Berkovich) asks for some threshold \(f(k)\) beyond which this holds; the threshold \(f(k) = 64k + 2\) is the one of its resolution.

def cyclicSubgroupOfIndexFromLargeAutomorphism : Prop :=
  ∀ p k m : ℕ, p.Prime → 64 * k + 2 ≤ m → ∀ (G : Type) [Group G] [Finite G], Nat.card G = p ^ m →
    (∃ φ : MulAut G, orderOf φ = p ^ (m - k)) → ∃ C : Subgroup G, IsCyclic C ∧ C.index = p ^ k
Used by 1

TheoremLarge automorphisms of p-groups force cyclic subgroups of given index#●ready to prove

For every prime \(p\) and all natural numbers \(k\) and \(m\) with \(m \ge 64k + 2\), if a finite group \(G\) of order \(p^m\) has an automorphism of order \(p^{m-k}\), then \(G\) has a cyclic subgroup of index \(p^k\).

Resolved true: Jayadevan proved this with the threshold \(f(k) = 64k + 2\); the Kourovka Notebook (21st issue) marks 15.32 solved, "Yes, one can take \(f(k) = 64k + 2\)" (AI solution confirmed by A. Caranti).

public theorem_wanted cyclicSubgroupOfIndexFromLargeAutomorphism_holds :
    cyclicSubgroupOfIndexFromLargeAutomorphism
Sources
  • E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 15.32 (Ya. G. Berkovich); resolution: A. Jayadevan, Large-order automorphisms of finite p-groups force cyclic subgroups of prescribed index, preprint, Zenodo record 22910914 (2026), AI-generated solution confirmed by A. Caranti, main theorem (for every k ≥ 0, a group of order p^m with m ≥ 64k + 2 and an automorphism of order p^(m-k) has a cyclic subgroup of index exactly p^k); the Kourovka Notebook (21st issue) marks 15.32 solved, https://zenodo.org/records/22910914.
Uses 1
1 fully proved1 ready to prove

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