Throughout is a finite group. Recall that we defined the degree of a character to be the degree of the representation which affords
Remark 1. Let be a representation of and let be its character. Then
i)
ii) for all
iii) for all
Proof. i) Let By definition, But the matrix of is the identity matrix and so its trace is Thus
ii) Recall from linear algebra that for any matrices and we have Thus
iii) By ii),
Remark 2. Let be two representations of If and are equivalent, then
Proof. By in here, there exists a -module isomorphism such that for all Choose some -basis for and and look at and as matrices. Then, since for any matrices and we have
and so
The converse of the statement in Remark 2 is also true, i.e. if then and are equivalent. We will prove this later.