# Degree of irreducible representation of nontrivial finite group is strictly less than order of group

## Statement

Suppose $G$ is a nontrivial finite group of order $n$ and $k$ is a field. Then, any irreducible representation of $G$ over $k$ has degree at most $n - 1$.

Equality is attained for infinitely many groups: namely, where $G$ is a group of prime order and $k$ is $\mathbb{Q}$, the field of rational numbers.

## Proof

Given: A group $G$ of order $n$. A nonzero vector space $V$, a homomorphism $\rho:G \to GL(V)$ such that $V$ has no proper nonzero $G$-invariant subspace.

To prove: The dimension of $V$ is at most $n - 1$.

Proof: Let $0 \ne v \in V$. Consider the set: $A = \{ g \cdot v \}_{g \in G}$

The vector space spanned by $A$ is $G$-invariant and nonzero, hence must equal $V$. Since it has a spanning set of size $n$, it has dimension at most $n$. Further, the dimension equals $n$ if and only if $A$ is linearly independent. However, if $A$ is linearly independent, the element: $\sum_{g \in G} g \cdot v$

is $G$-invariant and nonzero, hence it spans a nonzero $G$-invariant one-dimensional subspace of $V$. The subspace is proper since $|G| > 1$, contradicting the irreducibility assumption. Thus, $A$ cannot be linearly independent, and the dimension of $V$ is at most $n - 1$.