Character determines representation in characteristic zero

Statement

Suppose $G$ is a finite group and $K$ is a field of characteristic zero. Then, the character of any finite-dimensional representation of $G$ over $K$ completely determines the representation, i.e., no two inequivalent finite-dimensional representations can have the same character.

Note that $K$ does not need to be a splitting field.

Related facts

• Character does not determine representation in any prime characteristic: The problem is that we can construct representations whose character is identically zero simply by adding $p$ copies of an irreducible representation to itself.

Facts used

1. Character orthogonality theorem
2. Maschke's averaging lemma
3. Orthogonal projection formula

Proof

Given: A group $G$, two linear representations $\rho _{1},\rho _{2}$ of $G$ with the same character $\chi$ over a field $K$ of characteristic zero.

To prove: $\rho _{1}$ and $\rho _{2}$ are equivalent as linear representations.

Proof: By Fact (2), both $\rho _{1}$ and $\rho _{2}$ are completely reducible, and are expressible as sums of irreducible representations. Suppose $\varphi _{1},\varphi _{2},\dots ,\varphi _{s}$ is a collection of distinct irreducible representations obtained as the union of all the representations occurring in a decomposition of $\rho _{1}$ into irreducible representations and a decomposition of $\rho _{2}$ into irreducible representations. In other words, there are nonnegative integers $a_{11},a_{12},\dots ,a_{1s},a_{21},a_{22},\dots ,a_{2s}$ such that:

$\rho _{1}\cong a_{11}\varphi _{1}\oplus a_{12}\varphi _{2}\oplus \dots \oplus a_{1s}\varphi _{s}$

and

$\rho _{2}\cong a_{21}\varphi _{1}\oplus a_{22}\varphi _{2}\oplus \dots \oplus a_{2s}\varphi _{s}$

Let $\chi _{i}$ denote the character of $\varphi _{i}$ and denote by $m_{i}$ the value $\langle \chi _{i},\chi _{i}\rangle _{G}$ (note: this would be 1 if $K$ were a splitting field, and in general it is the sum of squares of multiplicities of irreducible constituents over a splitting field).

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$m_{i}a_{1i}=\langle \chi ,\chi _{i}\rangle _{G}$ and $m_{1}a_{2i}=\langle \chi ,\chi _{i}\rangle _{G}$ for each $1\leq i\leq s$	Fact (3)			Direct application of fact
2	$a_{1i}={\frac {1}{m}}\langle \chi ,\chi _{i}\rangle _{G}$ and $a_{2i}={\frac {1}{m}}\langle \chi ,\chi _{i}\rangle _{G}$ for each $1\leq i\leq s$		$K$ has characteristic zero, so the manipulation makes sense	Step (1)
3	$a_{1i}=a_{2i}$ for each $1\leq i\leq s$		$K$ has characteristic zero		[SHOW MORE] Note that if $K$ had characteristic $p$, we could only conclude equality modulo $p$, and not equality as nonnegative integers.
4	$\rho _{1}$ and $\rho _{2}$ are equivalent			Step (3)