# Degree of irreducible representation divides order of group

This article gives the statement, and possibly proof, of a constraint on numerical invariants that can be associated with a finite group

This article states a result of the form that one natural number divides another. Specifically, the (degree of a linear representation) of a/an/the (irreducible linear representation) divides the (order) of a/an/the (group).
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## Statement

Let $G$ be a finite group and $\varphi$ an irreducible representation of $G$ over an algebraically closed field of characteristic zero (or, more generally, over any splitting field of characteristic zero for $G$). Then, the degree of $\varphi$ divides the order of $G$.

## Related facts

### Other facts about degrees of irreducible representations

Further information: Degrees of irreducible representations

### Breakdown for a field that is not algebraically closed

Let $G$ be the cyclic group of order three and $\R$ be the field. Then, there are two irreducible representations of $G$ over $\R$: the trivial representation, and a two-dimensional representation given by the action by rotation by multiples of $2\pi/3$. The two-dimensional representation has degree $2$, and this does not divide the order of the group, which is $3$.

We still have the following results:

## Facts used

1. Character orthogonality theorem: The relevant part for us is that
2. Size-degree-weighted characters are algebraic integers: This states that for an irreducible linear representation $\varphi$ of a finite group $G$ over an algebraically closed field of characteristic zero (or more generally, over any splitting field), a conjugacy class $c$ in $G$ and an element $g \in c$, the number $|c|\chi(g)/\chi(1)$ (with $1$ denoting the identity element of the group) is an algebraic integer.
3. Characters are algebraic integers

## Proof

Given: A finite group $G$, an irreducible linear representation $\varphi$ of $G$ over a splitting field of characteristic zero for $G$, with character $\chi$ and degree $d$. Note that $d$ equals $\chi(1)$, i.e., the value of $\chi$ at the identity element of $G$.

To prove: $d$ divides the order of $G$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The following holds: $\sum_c |c| \chi(c) \overline{\chi(c)} = |G|$ where the sum is over all conjugacy classes $c$ of $G$, and where $\chi(c)$ denotes the value of $\chi$ at any element of $c$. Fact (1) $\varphi$ is irreducible over a splitting field of characteristic zero, with character $\chi$. Follows from fact (1). The $|c|$ comes because for each conjugacy class $c$, $|c|$ elements of the class appear in the full statement of the column orthogonality theorem.
2 $\sum_c |c|\chi(c) \overline{\chi(c)}/\chi(1) = |G|/d$ Step (1) Divide both sides of step (1) by $d = \chi(1)$.
3 Each $|c|\chi(c)/\chi(1)$ is an algebraic integer for each conjugacy class $c$. Fact (2) $\varphi$ is irreducible over a splitting field of characteristic zero, with character $\chi$.
4 Each $\overline{\chi(c)}$ is an algebraic integer for each conjugacy class $c$. Fact (3) $\chi$ is a character. The complex conjugate of an algebraic integer is also an algebraic integer.
5 $\sum_c |c|\chi(c) \overline{\chi(c)}/\chi(1)$ is an algebraic integer. Steps (3), (4) The set of algebraic integers forms a ring, so a finite sum of products of algebraic integers is an algebraic integer.
6 $|G|/d$ is an algebraic integer. Steps (2), (5) By Step (5), the left side of Step (2) is an algebraic integer, hence so is the right side.
7 $|G|/d$ is a positive integer, so $d$ divides $|G|$. Step (6) Both $|G|$ and $d$ are positive integers, hence their quotient is a positive rational number. The only way a rational number can be an algebraic integer is if it is an integer, hence the conclusion.