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 ${\phi }_1,{\phi }_2,\dots ,{\phi }_s$ 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:

${\rho }_1\cong a_{11}{\phi }_1\oplus a_{12}{\phi }_2\oplus \dots \oplus a_{1s}{\phi }_s$

and

${\rho }_2\cong a_{21}{\phi }_1\oplus a_{22}{\phi }_2\oplus \dots \oplus a_{2s}{\phi }_s$

Let ${\chi }_i$ denote the character of ${\phi }_i$ and denote by $m_i$ the value $⟨{\chi }_i,{\chi }_i{⟩}_G$ (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	$m_i{a}_{1i}=⟨\chi ,{\chi }_i{⟩}_G$ and $m_1{a}_{2i}=⟨\chi ,{\chi }_i{⟩}_G$ for each $1\le i\le s$	Fact (3)			Direct application of fact
2	$a_{1i}=\frac{1}{m}⟨\chi ,{\chi }_i{⟩}_G$ and $a_{2i}=\frac{1}{m}⟨\chi ,{\chi }_i{⟩}_G$ for each $1\le i\le s$		$K$ has characteristic zero, so the manipulation makes sense	Step (1)
3	$a_{1i}=a_{2i}$ for each $1\le i\le s$		$K$ 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)