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Cyclic restrictions force a bounded denominator in the rational representation ring

Statement

Let $G$ be a finite group and let $x \in R_{Q} (G)$ . Suppose that ${Res}_{C}^{G} x \in R (C)$ for every cyclic subgroup $C \leq G$ . Then

$∣ G ∣ x \in R (G) .$

Facts & Assumptions

Given: A finite group $G$ and an element $x \in R_{Q} (G)$ whose restriction to every cyclic subgroup lies in the integral character ring of that subgroup.

[F1]

There is an Artin relation $∣ G ∣ 1_{G} = \sum_{i} a_{i} {Ind}_{C_{i}}^{G} 1_{C_{i}}$ with $C_{i} \leq G$ cyclic and $a_{i} \in Z$ (A positive integer multiple of the trivial character is an integral combination of cyclic permutation characters).

[F2]

Induction and restriction satisfy the projection formula: $({Ind}_{H}^{G} χ) ψ = {Ind}_{H}^{G} (χ {Res}_{H}^{G} ψ)$ (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 $C \leq G$ , then ${Ind}_{C}^{G} θ$ is an honest complex character of $G$ ; hence induction sends $R (C)$ into $R (G)$ by $Z$ -linearity.

Proof

technique · direct

1.1

Choose cyclic subgroups $C_{i} \leq G$ and integers $a_{i}$ with $∣ G ∣ 1_{G} = \sum_{i} a_{i} {Ind}_{C_{i}}^{G} 1_{C_{i}}$ as in [F1]. Multiplying by $x$ in $R (G) \otimes_{Z} Q$ gives $∣ G ∣ x = \sum_{i} a_{i} ({Ind}_{C_{i}}^{G} 1_{C_{i}}) x$ .

F1given

2.1

Applying [F2] termwise to the identity of step 1.1 yields $∣ G ∣ x = \sum_{i} a_{i} {Ind}_{C_{i}}^{G} ({Res}_{C_{i}}^{G} x)$ . By hypothesis each ${Res}_{C_{i}}^{G} x$ lies in $R (C_{i})$ , 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 $R (G)$ .

F2F3step 1.1algebra

3.1

Therefore $∣ G ∣ x \in R (G)$ , as claimed.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Tammo tom Dieck, Representation Theory, Proposition (4.5.5) (standard reference, not scraped)