Irreducible Representations of Abelian Group

Theorem
Let $\left({G, \cdot}\right)$ be a finite abelian group and $\{0\}\ne V$ a vector space over an algebraically closed field $k$. Let $\rho: G \to \operatorname{GL} \left({V}\right)$ be a linear representation.

Then $\rho$ is irreducible iff $\dim \left({V}\right) = 1$, where, $\dim$ denotes dimension.

Sufficient Condition
Suppose that $\dim \left({V}\right) = 1$.

That $\rho$ is irreducible is shown on Representation of Degree One is Irreducible.

Necessary Condition
Suppose that $\rho$ is an irreducible linear representation.

Let $g \in G$ be arbitrary. Now, for all $h \in G$, have:

Now, combining Commutative Linear Transformation is G-Module Homomorphism and Corollary to Schur's Lemma (Representation Theory) yields that:


 * $\exists \lambda_g \in k: \rho \left({g}\right) = \lambda_g \operatorname{Id}_V$

That is, there is a $\lambda_g \in k$ such that $\rho \left({g}\right)$ is the linear mapping of multiplying by $\lambda_g$.

Hence, for all $v \in V$, $\rho \left({g}\right) \left({v}\right) = \lambda_g v$.

It follows that any vector subspace of $V$ of dimension $1$ is invariant.

So, had $V$ any proper vector subspace of dimension $1$, then $\rho$ would not be irreducible.

Hence necessarily $\dim \left({V}\right) = 1$, as $V$ is nontrivial.