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Cyclic restrictions force a bounded denominator in the rational representation ring

Statement

Let G be a finite group and let x∈RQ(G). Suppose that Res⁡CGx∈R(C) for every cyclic subgroup C≤G. Then

∣G∣x∈R(G).

Facts & Assumptions

Given: A finite group G and an element x∈RQ(G) whose restriction to every cyclic subgroup lies in the integral character ring of that subgroup.

[F1]

There is an Artin relation ∣G∣1G=∑iaiInd⁡CiG1Ci with Ci≤G cyclic and ai∈Z (A positive integer multiple of the trivial character is an integral combination of cyclic permutation characters).

[F2]

Induction and restriction satisfy the projection formula: (Ind⁡HGχ)ψ=Ind⁡HG(χ Res⁡HGψ) (Induction and restriction satisfy the projection formula on character rings).

[F3]

The integral character ring R(G) is closed under integral linear combinations (Virtual characters and the character ring R(G) of a finite group).

[A1]

If θ is an honest complex character of a subgroup C≤G, then Ind⁡CGθ is an honest complex character of G; hence induction sends R(C) into R(G) by Z-linearity.

Proof

technique · direct
1.1F1given

Choose cyclic subgroups Ci≤G and integers ai with ∣G∣1G=∑iaiInd⁡CiG1Ci as in [F1]. Multiplying by x in R(G)⊗ZQ gives ∣G∣x=∑iai(Ind⁡CiG1Ci)x.

2.1F2F3step 1.1algebra

Applying [F2] termwise to the identity of step 1.1 yields ∣G∣x=∑iaiInd⁡CiG(Res⁡CiGx). By hypothesis each Res⁡CiGx lies in R(Ci), so [A1] places every induced summand in R(G). Since R(G) is closed under integral linear combinations, the whole right-hand side lies in R(G).

3.1step 2.1∎

Therefore ∣G∣x∈R(G), as claimed.

Depends on

Used by

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Dependency tree · two levels

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Sources