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

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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.


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.