Ranks in elementary abelian extensions

DefinitionElementary abelian subgroups#✓fully proved

A subgroup \(H\) of a group \(G\) is elementary abelian if any two of its elements commute and there is a prime \(p\) with \(x^p = 1\) for all \(x \in H\).

def IsElementaryAbelian {G : Type} [Group G] (H : Subgroup G) : Prop :=
  (∀ x ∈ H, ∀ y ∈ H, x * y = y * x) ∧ ∃ p : ℕ, p.Prime ∧ ∀ x ∈ H, x ^ p = 1

Used by 1

Used byZenkov's rank bound in elementary abelian extensions (Kourovka 21.150)

DefinitionZenkov's rank bound in elementary abelian extensions (Kourovka 21.150)#✓fully proved

Call a subgroup elementary abelian if it is abelian and there is a prime \(p\) with \(x^p = 1\) for all its elements \(x\). This is the proposition that for every finite group \(G\), all elementary abelian subgroups \(A\) and \(B\) of \(G\) (possibly for different primes) with \(A\) normal, \(A \cap B = 1\) and \(\langle A, B \rangle = G\) (so \(B\) is a complement to \(A\) and \(B \cong G/A\)), and every \(a \in A\) whose centralizer \(C_B(a)\) in \(B\) is trivial, the following holds for \(H = \langle a, B \rangle\), with centre \(Z(H)\) and derived subgroup \(H' = [H, H]\): whenever \(B\) is generated by a finite set of \(k\) elements, \(Z(H) \cap H'\) (with \(Z(H)\) viewed as a subgroup of \(G\)) is generated by a finite set of at most \(k\) elements. That is, the rank of \(Z(H) \cap H'\) is at most the rank of \(B\).

The question is due to V. I. Zenkov.

def centreDerivedRankLeComplementRank : Prop :=
  ∀ (G : Type) [Group G] [Finite G] (A B : Subgroup G) (a : G), A.Normal →
    IsElementaryAbelian A → IsElementaryAbelian B → A ⊓ B = ⊥ → A ⊔ B = ⊤ → a ∈ A →
    (∀ b ∈ B, b * a = a * b → b = 1) →
    ∀ s : Finset G, Subgroup.closure (s : Set G) = B →
      ∃ t : Finset G, t.card ≤ s.card ∧ Subgroup.closure (t : Set G) =
        (Subgroup.center (Subgroup.closure {a} ⊔ B : Subgroup G)).map
            (Subgroup.closure {a} ⊔ B : Subgroup G).subtype ⊓
          ⁅Subgroup.closure {a} ⊔ B, Subgroup.closure {a} ⊔ B⁆

Uses 1 · Used by 1

Statement usesElementary abelian subgroups

Used byZenkov's rank bound fails (Kourovka 21.150)

TheoremZenkov's rank bound fails (Kourovka 21.150)#●ready to prove

Call a subgroup elementary abelian if it is abelian and there is a prime \(p\) with \(x^p = 1\) for all its elements \(x\). There are a finite group \(G\), elementary abelian subgroups \(A\) and \(B\) of \(G\) with \(A\) normal, \(A \cap B = 1\) and \(\langle A, B \rangle = G\), and an element \(a \in A\) whose centralizer \(C_B(a)\) in \(B\) is trivial, such that, for \(H = \langle a, B \rangle\), with centre \(Z(H)\) and derived subgroup \(H' = [H, H]\), some finite generating set of \(B\) has fewer elements than every finite generating set of \(Z(H) \cap H'\) (here \(Z(H)\) is viewed as a subgroup of \(G\)).

Counterexample (Monticone; van Doorn, Judin, Monticone and Morrison, Theorem 9.1): \(A = \mathbb{F}_3[x, y]/(x, y)^3\), \(B = \langle 1 + x, 1 + y \rangle\) acting by multiplication, \(G = A \rtimes B\) and \(a = 1\); then \(C_B(a) = 1\), while \(Z(H) \cap H' = \langle x^2, xy, y^2 \rangle\) has rank \(3 > 2 = \operatorname{rank} B\). The 21st issue of the Kourovka Notebook marks 21.150 solved ("No, not always").

public theorem_wanted centreDerivedRankLeComplementRank_refuted :
    ¬ centreDerivedRankLeComplementRank

Sources

• E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Problem 21.150 (V. I. Zenkov); resolution: P. Monticone, letter of 21 March 2026 and solution note, Kourovka Notebook site; W. van Doorn, E. Judin, P. Monticone and D. Morrison, On some problems from the Kourovka Notebook, preprint, arXiv:2607.17477 (2026), Kourovka Notebook (21st issue), Problem 21.150: No, not always: A = F₃[x, y]/I³ (I = (x, y)), B = ⟨1 + x, 1 + y⟩ acting by multiplication, G = A ⋊ B, a = 1; then C_B(a) = 1 and Z(H) ∩ H' = ⟨x², xy, y²⟩ has rank 3 while B has rank 2; van Doorn-Judin-Monticone-Morrison, Theorem 9.1, https://arxiv.org/abs/2607.17477.

Uses 1

Statement usesZenkov's rank bound in elementary abelian extensions (Kourovka 21.150)

2 fully proved1 ready to prove

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