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.

Opposite 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.

Applications

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

Facts used

 * 1) uses::Character orthogonality theorem
 * 2) uses::Maschke's averaging lemma
 * 3) uses::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).