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

## Statement

### In terms of linear representations

Suppose $G$ is a finite group. Suppose $K$ and $L$ are fields of characteristic zero with $K$ a subfield of $L$. Suppose $\varphi_1,\varphi_2:G \to GL(n,K)$ are linear representations of $G$ over $K$. We can view $\varphi_1$ and $\varphi_2$ as linear representations over $L$.

The claim is that if $\varphi_1,\varphi_2$ are Equivalent linear representations (?) over $L$, they are also equivalent linear representations over $K$.

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

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

## Facts used

1. 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 $\varphi_1,\varphi_2$ are equivalent over $L$, they in particular have the same character. By Fact (1), they must be equivalent over $K$.

### 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 $K$ have characters that are orthogonal to each other. However, if they were equivalent over $L$, then applying the character orthogonality theorem inside $L$, 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.