Jordan decomposition in integral group rings

DefinitionElements with integer coefficients#✓fully proved

An element \(x\) of the monoid algebra \(\mathbb{Q}[G]\) lies in \(\mathbb{Z}[G]\) when all its coefficients are integers: for every \(g \in G\), the coefficient of \(x\) at \(g\) is an integer.

def HasIntegralCoeffs {G : Type*} (x : MonoidAlgebra ℚ G) : Prop :=
  ∀ g : G, ∃ z : ℤ, x.coeff g = z

Used by 2

Used byAdditive Jordan decomposition property · Multiplicative Jordan decomposition property

DefinitionAdditive Jordan decomposition property#✓fully proved

For a group \(G\), let \(\mathbb{Z}[G]\) be the set of elements of the rational group algebra \(\mathbb{Q}[G]\) all of whose coefficients are integers. Then \(\mathbb{Z}[G]\) has the additive Jordan decomposition property (AJD) when, for all \(a, s, n \in \mathbb{Q}[G]\) such that \(a \in \mathbb{Z}[G]\), \(a = s + n\), \(sn = ns\), \(n\) is nilpotent and the minimal polynomial of \(s\) over \(\mathbb{Q}\) is squarefree, both \(s\) and \(n\) lie in \(\mathbb{Z}[G]\).

For finite \(G\) every \(a \in \mathbb{Q}[G]\) has exactly one such decomposition \(a = s + n\), its Jordan decomposition, so AJD says that the semisimple and nilpotent parts of every \(a \in \mathbb{Z}[G]\) lie in \(\mathbb{Z}[G]\).

def HasAJD (G : Type*) [Group G] : Prop :=
  ∀ a s n : MonoidAlgebra ℚ G, HasIntegralCoeffs a → a = s + n → Commute s n → IsNilpotent n →
    Squarefree (minpoly ℚ s) → HasIntegralCoeffs s ∧ HasIntegralCoeffs n

Uses 1 · Used by 1

Statement usesElements with integer coefficients

Used byKourovka Problem 16.104

DefinitionMultiplicative Jordan decomposition property#✓fully proved

For a group \(G\), let \(\mathbb{Z}[G]\) be the set of elements of the rational group algebra \(\mathbb{Q}[G]\) all of whose coefficients are integers. Then \(\mathbb{Z}[G]\) has the multiplicative Jordan decomposition property (MJD) when, for all \(a, b, s, u \in \mathbb{Q}[G]\) such that \(a, b \in \mathbb{Z}[G]\), \(ab = ba = 1\), \(a = su\), \(su = us\), \(u - 1\) is nilpotent and the minimal polynomial of \(s\) over \(\mathbb{Q}\) is squarefree, both \(s\) and \(u\) lie in \(\mathbb{Z}[G]\).

Equivalently, for every unit \(a\) of \(\mathbb{Z}[G]\) and every factorization \(a = su\) in \(\mathbb{Q}[G]\) into commuting factors with \(s\) semisimple over \(\mathbb{Q}\) and \(u\) unipotent, both factors lie in \(\mathbb{Z}[G]\); for finite \(G\) such a factorization exists and is unique.

def HasMJD (G : Type*) [Group G] : Prop :=
  ∀ a b s u : MonoidAlgebra ℚ G, HasIntegralCoeffs a → HasIntegralCoeffs b → a * b = 1 →
    b * a = 1 → a = s * u → Commute s u → IsNilpotent (u - 1) → Squarefree (minpoly ℚ s) →
      HasIntegralCoeffs s ∧ HasIntegralCoeffs u

Uses 1 · Used by 1

Statement usesElements with integer coefficients

Used byKourovka Problem 16.104

DefinitionKourovka Problem 16.104#✓fully proved

The proposition that there is a natural number \(B\) such that every finite \(2\)-group \(G\) whose integral group ring \(\mathbb{Z}[G]\) has the multiplicative Jordan decomposition property (MJD) but not the additive Jordan decomposition property (AJD) has order at most \(B\).

A. W. Hales and I. B. S. Passi asked whether only finitely many isomorphism classes of finite \(2\)-groups \(G\) have MJD but not AJD for \(\mathbb{Z}[G]\); a bound on the orders is equivalent.

def finitelyManyTwoGroupsMJDNotAJD : Prop :=
  ∃ B : ℕ, ∀ (G : Type) [Group G] [Finite G], IsPGroup 2 G → HasMJD G → ¬ HasAJD G →
    Nat.card G ≤ B

Uses 2 · Used by 1

Statement usesAdditive Jordan decomposition property · Multiplicative Jordan decomposition property

Used byFinitely many \(2\)-groups have MJD but not AJD

TheoremFinitely many \(2\)-groups have MJD but not AJD#●ready to prove

There is a natural number \(B\) such that every finite \(2\)-group \(G\) whose integral group ring \(\mathbb{Z}[G]\) has the multiplicative Jordan decomposition property (MJD) but not the additive Jordan decomposition property (AJD) has order at most \(B\). In particular, only finitely many isomorphism classes of finite \(2\)-groups have MJD but not AJD, which answers Kourovka Problem 16.104 in the affirmative.

public theorem_wanted finitelyManyTwoGroupsMJDNotAJD_holds : finitelyManyTwoGroupsMJDNotAJD

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 16.104 (A. W. Hales, I. B. S. Passi); resolution: A. W. Hales, I. B. S. Passi and L. E. Wilson, The multiplicative Jordan decomposition in group rings, II, J. Algebra 316 (2007), no. 1, 109-132; corrigendum, J. Algebra 371 (2012), 665-666, Kourovka Archive, Problem 16.104 (answer: Yes, it is true).

Uses 1

Statement usesKourovka Problem 16.104

4 fully proved1 ready to prove

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