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A conjugacy class of size coprime to $χ (1)$ forces either $χ (g) = 0$ or scalar action

Statement

Let $G$ be a finite group, let $V$ be an irreducible complex representation of $G$ with character $χ$ , and let $g \in G$ . If the size of the conjugacy class of $g$ is coprime to $χ (1)$ , then either $χ (g) = 0$ or $g$ acts as a scalar on $V$ .

Facts & Assumptions

Given: A finite group $G$ , an irreducible complex representation $V$ of $G$ with character $χ$ , and an element $g \in G$ .

[F1]

Central-character values are algebraic integers (The values of a central character are algebraic integers).

[F2]

An algebraic-integer average of roots of unity is either $0$ or constant (An algebraic-integer average of roots of unity is either $0$ or a common root of unity).

[F3]

The character value $χ (g)$ is the sum of the eigenvalues of $g$ , which are roots of unity, and equality in the modulus bound means scalar action (For a complex character, $χ (1) = \dim V$ , $χ$ is a class function, and $∣ χ (g) ∣ \leq χ (1)$ with equality exactly at scalars).

[F4]

For the conjugacy class $C$ of $g$ , the central character satisfies $ω_{χ} (\hat{C}) = ∣ C ∣ χ (g) / χ (1)$ (The central character of an irreducible complex character).

Proof

technique · direct

1.1

Let $m = ∣ {Cl}_{G} (g) ∣$ and $n = χ (1)$ . By [F4], $m χ (g) / n = ω_{χ} (\hat{C})$ for the class sum of ${Cl}_{G} (g)$ , so [F1] makes $m χ (g) / n$ an algebraic integer. Since $\gcd (m, n) = 1$ , choose integers $a, b$ with $a m + b n = 1$ . Then $χ (g) / n = a (m χ (g) / n) + b χ (g)$ is an algebraic integer because [F3] shows that $χ (g)$ is an algebraic integer.

F1F3F4givenchoosealgebra

2.1

Let $ζ_{1}, \dots, ζ_{n}$ be the eigenvalues of $ρ_{V} (g)$ . By [F3], they are roots of unity and $χ (g) / n = (ζ_{1} + \dots + ζ_{n}) / n$ . Applying [F2] to this average and step 1.1 gives either $χ (g) = 0$ or $ζ_{1} = \dots = ζ_{n}$ .

F2F3step 1.1algebra

3.1

If $ζ_{1} = \dots = ζ_{n}$ , then the diagonalizable operator $ρ_{V} (g)$ from [F3] is that common root of unity times the identity, so $g$ acts as a scalar on $V$ .

F3step 2.1algebra

4.1

Steps 2.1 and 3.1 prove the stated dichotomy.

step 2.1step 3.1∎

Depends on

Used by

A conjugacy class of prime-power size forces a proper nontrivial normal subgroup Theorem

Dependency tree · two levels

18 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

Pavel Etingof et al., Introduction to Representation Theory, Theorem 4.21 (standard reference, not scraped)
Anupam Singh, Representation Theory of Finite Groups, Lemma 16.2 (standard reference, not scraped)

A conjugacy class of size coprime to χ(1)χ(1) forces either χ(g)=0χ(g)=0 or scalar action