# Equivalent linear representations of finite group over field are equivalent over subfield in characteristic zero

## Contents

## Statement

### In terms of linear representations

Suppose is a finite group. Suppose and are fields of characteristic zero with a subfield of . Suppose are linear representations of over . We can view and as linear representations over .

The claim is that if are Equivalent linear representations (?) over , they are also equivalent linear representations over .

### In terms of finite subgroups and conjugacy-closed

Suppose and are fields of characteristic zero with a subfield of . Then, the general linear group is a finite subgroup-conjugacy closed subgroup inside the general linear group . In other words, any two finite subgroups of that are conjugate inside are also conjugate inside .

## Related facts

### Conjugacy-closed angle

- General linear group over subfield is conjugacy-closed: This is for elements, possibly of infinite order.

## Facts used

- Character determines representation in characteristic zero, which in turn relies on the orthogonal projection formula which in turn relies on the character orthogonality theorem, which in turn uses Schur's lemma.

## Proof

### Short proof

The proof follows directly from Fact (1). If are equivalent over , they in particular have the same character. By Fact (1), they must be equivalent over .

### What's going on

This superficial-looking proof can obscure what's going on. The real meat of the proof lies in the following aspect of the proof of the character orthogonality theorem: two irreducible representations that are inequivalent over have characters that are orthogonal to each other. However, if they were equivalent over , then applying the character orthogonality theorem inside , we'd get a nonzero inner product value of the characters. This explains why the statement is true for irreducible characters, and the rest follows by piecing things together.