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A rational-valued irreducible character need not come from a $Q$ -representation

Statement refuted

Every rational-valued irreducible character of a finite group is afforded by a representation over $Q$ .

Let $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ be the quaternion group (The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions). Consider the complex representation $ρ : Q_{8} \to {GL}_{2} (C)$ given by

$ρ (i) = (\begin{array}{cc} i & 0 \\ 0 & - i \end{array}), ρ (j) = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}), ρ (k) = (\begin{array}{cc} 0 & i \\ i & 0 \end{array}) .$

Then $ρ (- 1) = - I_{2}$ , and its character $χ$ has values

$χ (1) = 2, χ (- 1) = - 2, χ (\pm i) = χ (\pm j) = χ (\pm k) = 0.$

So $χ$ is rational-valued. However, the cited Schur-index computation shows that $χ$ has Schur index $2$ over $Q$ , so no $Q$ -representation affords it.

Facts & Assumptions

Given: The quaternion group $Q_{8}$ and the matrices displayed above.

[F1]

The group $Q_{8}$ is the eight-element group ${\pm 1, \pm i, \pm j, \pm k}$ (The quaternion group $Q_{8} = {\pm 1, \pm i, \pm j, \pm k}$ inside the nonzero quaternions).

[F2]

In $Q_{8}$ , the element $- 1$ is the unique element of order $2$ , while $\pm i$ , $\pm j$ , and $\pm k$ all have order $4$ ( $Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$ ).

Counterexample

technique · direct

1.1

The displayed matrices satisfy $ρ (i)^{2} = ρ (j)^{2} = ρ (k)^{2} = - I_{2}$ and $ρ (i) ρ (j) = ρ (k)$ , $ρ (j) ρ (i) = - ρ (k)$ . Hence they obey the same relations as the generators of $Q_{8}$ from [F1], so they define a two-dimensional complex representation of $Q_{8}$ . Their traces are $2, - 2, 0, 0, 0$ on the conjugacy classes $1$ , $- 1$ , ${\pm i}$ , ${\pm j}$ , and ${\pm k}$ , so the character is rational-valued.

F1F2givenalgebra

2.1

The Magma Schur-index example in the cited source computes that this irreducible character has Schur index $2$ over $Q$ . A character with Schur index greater than $1$ is not afforded by any $Q$ -representation. Therefore this rational-valued irreducible character is not defined over $Q$ , refuting the statement.

step 1.1algebra∎

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Sources

The Magma Handbook, The Schur Index, Example: Schur Index (standard reference, not scraped)
Kay Yang, Rational Valued Characters, Introduction (standard reference, not scraped)

A rational-valued irreducible character need not come from a QQ-representation