Rational groups

DefinitionRational groups, conjugacy form#✓fully proved

A group \(G\) is rational if every element \(g \in G\) is conjugate to \(g^k\) for every natural number \(k\) coprime to the order of \(g\), where an element of infinite order has order \(0\), so that only \(k = 1\) is coprime to it. For finite groups this is the classical equivalent form of the condition that all ordinary characters of \(G\) are rational-valued.

def IsRationalGroup (G : Type*) [Group G] : Prop :=
  ∀ g : G, ∀ k : ℕ, Nat.Coprime k (orderOf g) → IsConj g (g ^ k)

Used by 1

Used byDarafsheh's question on Sylow 2-subgroups of rational groups (Kourovka 16.36)

DefinitionDarafsheh's question on Sylow 2-subgroups of rational groups (Kourovka 16.36)#✓fully proved

The conjecture that for every finite rational group \(G\), every Sylow \(2\)-subgroup of \(G\), viewed as a group in its own right, is rational. Here a group is rational if every element \(g\) is conjugate to \(g^k\) for every natural number \(k\) coprime to the order of \(g\).

def sylowTwoOfRationalIsRational : Prop :=
  ∀ (G : Type) [Group G] [Finite G], IsRationalGroup G →
    ∀ P : Sylow 2 G, IsRationalGroup (P : Subgroup G)

Uses 1 · Used by 1

Statement usesRational groups, conjugacy form

Used bySylow 2-subgroups of rational groups need not be rational (Kourovka 16.36)

TheoremSylow 2-subgroups of rational groups need not be rational (Kourovka 16.36)#●ready to prove

There is a finite rational group with a Sylow \(2\)-subgroup that is not rational as a group in its own right. Here a group is rational if every element \(g\) is conjugate to \(g^k\) for every natural number \(k\) coprime to the order of \(g\).

The Kourovka Archive answers 16.36 "No, not always", citing Isaacs and Navarro (2012).

public theorem_wanted sylowTwoOfRationalIsRational_refuted : ¬ sylowTwoOfRationalIsRational

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 16.36 (M. R. Darafsheh); resolution: I. M. Isaacs and G. Navarro, Math. Z. 272 (2012), 937-945, Kourovka Archive, Problem 16.36 (No, not always).

Uses 1

Statement usesDarafsheh's question on Sylow 2-subgroups of rational groups (Kourovka 16.36)

DefinitionRational group (conjugacy form, Kourovka 15.25)#✓fully proved

A group \(G\) is rational when every element \(g \in G\) is conjugate in \(G\) to \(g^k\) for every natural number \(k\) coprime to the order of \(g\) (an element of infinite order counts as having order \(0\), so for it only \(k = 1\) qualifies).

For finite groups this is equivalent to every irreducible complex character taking only rational values.

def IsRationalGroup (G : Type*) [Group G] : Prop :=
  ∀ (g : G) (k : ℕ), k.Coprime (orderOf g) → IsConj g (g ^ k)

Used by 1

Used byKourovka Problem 15.25 (statement)

DefinitionKourovka Problem 15.25 (statement)#✓fully proved

The statement that for every natural number \(n\), every Sylow \(2\)-subgroup \(P\) of the symmetric group \(S_{2^n}\) on \(2^n\) points is rational as a group in its own right: every \(g \in P\) is conjugate within \(P\) to \(g^k\) for every natural number \(k\) coprime to the order of \(g\).

def sylowTwoSubgroupSymmetricRational : Prop :=
  ∀ (n : ℕ) (P : Sylow 2 (Equiv.Perm (Fin (2 ^ n)))),
    IsRationalGroup (P : Subgroup (Equiv.Perm (Fin (2 ^ n))))

Uses 1 · Used by 1

Statement usesRational group (conjugacy form, Kourovka 15.25)

Used bySylow 2-subgroups of the symmetric groups of degree \(2^n\) are rational (Kourovka 15.25)

TheoremSylow 2-subgroups of the symmetric groups of degree \(2^n\) are rational (Kourovka 15.25)#●ready to prove

For every natural number \(n\), every Sylow \(2\)-subgroup \(P\) of the symmetric group \(S_{2^n}\) on \(2^n\) points is rational as a group in its own right: every \(g \in P\) is conjugate within \(P\) to \(g^k\) for every natural number \(k\) coprime to the order of \(g\).

A Sylow \(2\)-subgroup \(T_n\) of \(S_{2^n}\) is the wreath product of \(T_{n-1}\) with a group of order \(2\), which is rational when \(T_{n-1}\) is (Kletzing, Corollary 70), so induction on \(n\) applies.

public theorem_wanted sylowTwoSubgroupSymmetricRational_holds : sylowTwoSubgroupSymmetricRational

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 15.25 (Ya. G. Berkovich); resolution: A. Mann, letter of 1 October 2002, as recorded in the Kourovka Archive, Problem 15.25, using D. Kletzing, Structure and representations of Q-groups, Lecture Notes in Math. 1084, Springer, Berlin, 1984, Corollary 70, Kourovka Archive, Problem 15.25 (answer: Yes, they are; T_n = T_(n-1) wr C_2 and wreath products with C_2 preserve rationality), https://arxiv.org/abs/1401.0300v49.

Uses 1

Statement usesKourovka Problem 15.25 (statement)

4 fully proved2 ready to prove

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