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

Hence

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

Therefore for all 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.