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.

Yaghoub,

This is Alex Youcis, just out of curiosity what book are you using for rep. theory?

Hi Alex

Nothing yet.

Well, if you happen to find any hidden gems, I would appreciate if you’d let me know 🙂

Good luck with your bloggings, I enjoy them and will enjoy seeings your take on rep. theory.

Well, I have just started writing these stuff up and, yes, there are very nice stuff in character theory of finite groups that I will definitely get to them in details. And, I’m glad you like my blog.