Throughout is a finite group.** **

**Definition**. Let be a representation of The map defined by is called the **character** of and is denoted by We say that is an **irreducible** character if is irreducible. We also define the **degree** of to be

**Note 1**. If is understood, we will just write instead of

**Note 2**. The term (or ) **affords** is also used.

**Remark 1**. In the above definition, menas the trace of the linear transformation with respect to a basis of To be precise, is the trace of the matrix of with respect to a basis of Note that, since the trace of similar matrices are equal, does not depend on the basis we choose for and so our definition makes sense.

**Remark 2**. If then clearly and so for all

**Remark 3**. Recall from linear algebra that the trace of a linear transformation is the sum of its eigenvalues. So is the sum of eigenvalues of

**Theorem**. If and then where are (not necessarily distinct) -th roots of unity.

*Proof*. By Remark 3, we only need to show that every eigenvalue of is an -th root of unity. Since the degree of the characteristic polynomial of is and thus this polynomial has roots, which are not necessarily distinct. Now if is an eigenvalue of then for some Since we have and hence Thus and hence because So each eigenvalue of is an -th root of unity.

**Corollary 1**. is an algebraic integer for all

*Proof*. By the theorem, is a sum of some roots of unity. Obviously any root of unity is an algebraic integer and we know that the set of algebraic integers is a ring and so it is closed under addition.

**Corollary 2**. where is the complex conjugate of

*Proof*. Let be a representation which affords Suppose that By the theorem, where are the eigenvalues of and they are all roots of unity. Since the eigenvalues of are Also, because are roots of unity. Thus