Prime-order automorphisms with centralizer of finite rank
DefinitionGroups of bounded rank#✓fully proved
A group \(G\) has rank at most \(r\), for a natural number \(r\), when every finitely generated subgroup of \(G\) is generated by some set of at most \(r\) elements.
For periodic nilpotent groups this rank is the sectional rank.
def RankLE (G : Type*) [Group G] (r : ℕ) : Prop :=
∀ H : Subgroup G, H.FG → ∃ S : Finset G, S.card ≤ r ∧ Subgroup.closure (S : Set G) = H
DefinitionFixed-point subgroup of an automorphism#✓fully proved
For an automorphism \(\varphi\) of a group \(G\), the centralizer \(C_G(\varphi)\) is the subgroup \(\{x \in G : \varphi(x) = x\}\) of fixed points of \(\varphi\).
def fixedSubgroup {G : Type*} [Group G] (φ : G ≃* G) : Subgroup G :=
(φ : G →* G).eqLocus (MonoidHom.id G)
DefinitionNilpotent-by-bounded-rank problem for automorphisms of prime order (Kourovka 13.58)#✓fully proved
The statement that there are functions \(f \colon \mathbb{N} \to \mathbb{N}\) and \(g \colon \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) with the following property. Let \(G\) be a nilpotent group in which every element has finite order, let \(\varphi\) be an automorphism of \(G\) of prime order \(p\), and let \(r\) be a natural number such that the fixed-point subgroup \(C_G(\varphi)\) has rank at most \(r\), that is, every finitely generated subgroup of \(C_G(\varphi)\) is generated by at most \(r\) elements. Then \(G\) has a normal subgroup \(N\), nilpotent of class at most \(f(p)\), such that the quotient \(G/N\) has rank at most \(g(r, p)\).
For periodic nilpotent groups this rank is the sectional rank.
def nilpotentByBoundedRankOfCentralizerRank : Prop :=
∃ (f : ℕ → ℕ) (g : ℕ → ℕ → ℕ), ∀ (G : Type) [Group G], Group.IsNilpotent G →
(∀ x : G, IsOfFinOrder x) → ∀ (φ : G ≃* G) (p r : ℕ), p.Prime → orderOf φ = p →
RankLE (fixedSubgroup φ) r →
∃ (N : Subgroup G) (_ : N.Normal), Subgroup.upperCentralSeries N (f p) = ⊤ ∧
RankLE (G ⧸ N) (g r p)
def RankLE (G : Type*) [Group G] (r : ℕ) : Prop :=
∀ H : Subgroup G, H.FG → ∃ S : Finset G, S.card ≤ r ∧ Subgroup.closure (S : Set G) = H
WantedExt/GroupTheory/PrimeOrderAutomorphismRankNilpotentKourovka13Problem58Wanted.lean:42
def fixedSubgroup {G : Type*} [Group G] (φ : G ≃* G) : Subgroup G :=
(φ : G →* G).eqLocus (MonoidHom.id G)
WantedExt/GroupTheory/PrimeOrderAutomorphismRankNilpotentKourovka13Problem58Wanted.lean:46
Uses 2 · Used by 1
TheoremKhukhro's theorem on automorphisms of prime order with centralizer of finite rank#●ready to prove
There are functions \(f \colon \mathbb{N} \to \mathbb{N}\) and \(g \colon \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) with the following property. Let \(G\) be a nilpotent group in which every element has finite order, let \(\varphi\) be an automorphism of \(G\) of prime order \(p\), and let \(r\) be a natural number such that the fixed-point subgroup \(C_G(\varphi)\) has rank at most \(r\), that is, every finitely generated subgroup of \(C_G(\varphi)\) is generated by at most \(r\) elements. Then \(G\) has a normal subgroup \(N\), nilpotent of class at most \(f(p)\), such that the quotient \(G/N\) has rank at most \(g(r, p)\). This is the positive answer to Kourovka Problem 13.58; the case \(p = 2\) was proved earlier by Shumyatsky.
public theorem_wanted nilpotentByBoundedRankOfCentralizerRank_holds :
nilpotentByBoundedRankOfCentralizerRank
def nilpotentByBoundedRankOfCentralizerRank : Prop :=
∃ (f : ℕ → ℕ) (g : ℕ → ℕ → ℕ), ∀ (G : Type) [Group G], Group.IsNilpotent G →
(∀ x : G, IsOfFinOrder x) → ∀ (φ : G ≃* G) (p r : ℕ), p.Prime → orderOf φ = p →
RankLE (fixedSubgroup φ) r →
∃ (N : Subgroup G) (_ : N.Normal), Subgroup.upperCentralSeries N (f p) = ⊤ ∧
RankLE (G ⧸ N) (g r p)
WantedExt/GroupTheory/PrimeOrderAutomorphismRankNilpotentKourovka13Problem58Wanted.lean:53
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 13.58 (E. I. Khukhro); resolution: E. I. Khukhro, J. London Math. Soc. 77 (2008), 130-148 (the case p = 2: P. Shumyatsky, Arch. Math. 71 (1998), 425-432), Kourovka Archive, Problem 13.58 (answer: Yes, it does).
Uses 1
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