Involutions in quotient groups

DefinitionSozutov's question on involutions in quotients (Kourovka 18.94)#✓fully proved

This is the statement that whenever \(G\) is a group without involutions (if \(g^2 = 1\) then \(g = 1\)), \(a \in G\) is not the square of any element of \(G\), and \(n\) is an odd natural number, the quotient of \(G\) by the normal closure of \(a^n\) has no involutions: every element \(y\) of the quotient with \(y^2 = 1\) is \(1\).

def involutionFreeQuotientByNonSquarePower : Prop :=
  ∀ (G : Type) [Group G], (∀ g : G, g ^ 2 = 1 → g = 1) → ∀ a : G, (¬ ∃ x : G, x ^ 2 = a) →
    ∀ n : ℕ, Odd n → ∀ y : G ⧸ Subgroup.normalClosure {a ^ n}, y ^ 2 = 1 → y = 1

Used by 1

Used byQuotients by the normal closure of an odd power of a non-square can have involutions

TheoremQuotients by the normal closure of an odd power of a non-square can have involutions#●ready to prove

There exist a group \(G\) without involutions (if \(g^2 = 1\) then \(g = 1\)), an element \(a \in G\) that is not the square of any element of \(G\), and an odd natural number \(n\) such that the quotient of \(G\) by the normal closure of \(a^n\) contains an involution, that is, an element \(y \ne 1\) with \(y^2 = 1\). This answers Kourovka Problem 18.94 in the negative.

Besides an example of V. I. Trofimov, the group \(G = \langle a, b \mid a^2 = b^2 \rangle\) with \(n = 1\) is an example: its quotient by the normal closure of \(a\) has order \(2\).

public theorem_wanted involutionFreeQuotientByNonSquarePower_refuted :
    ¬ involutionFreeQuotientByNonSquarePower

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 18.94 (A. I. Sozutov); resolution: V. I. Trofimov (example), as recorded in the Kourovka Archive, Problem 18.94, Kourovka Archive, Problem 18.94: No, not always, by an example of V. I. Trofimov; another example: G = ⟨a, b | a² = b²⟩ modulo ⟨a^G⟩ has order 2, https://arxiv.org/abs/1401.0300v49.

Uses 1

Statement usesSozutov's question on involutions in quotients (Kourovka 18.94)

1 fully proved1 ready to prove

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