Irreducible Representations of Abelian Group
Theorem
Let $\struct {G, \cdot}$ be a finite abelian group.
Let $V$ be a non-null vector space over an algebraically closed field $K$.
Let $\rho: G \to \GL V$ be a linear representation.
Then $\rho$ is irreducible if and only if $\map {\dim_K} V = 1$, where, $\dim_K$ denotes dimension.
Proof
Sufficient Condition
Suppose that $\map {\dim_K} V = 1$.
That $\rho$ is irreducible is shown on Representation of Degree One is Irreducible.
$\Box$
Necessary Condition
Suppose that $\rho$ is an irreducible linear representation.
Let $g \in G$ be arbitrary. Now, for all $h \in G$, have:
\(\ds \map \rho g \map \rho h\) | \(=\) | \(\ds \map \rho {g h}\) | $\rho$ is a group homomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \rho {h g}\) | $G$ is an abelian group | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \rho h \map \rho g\) | $\rho$ is a group homomorphism |
Now, combining Commutative Linear Transformation is G-Module Homomorphism and Corollary to Schur's Lemma (Representation Theory) yields that:
- $\exists \lambda_g \in K: \map \rho g = \lambda_g \operatorname{Id}_V$
That is, there is a $\lambda_g \in K$ such that $\map \rho g$ is the linear mapping of multiplying by $\lambda_g$.
Hence, $\forall v \in V: \map {\map \rho g} v = \lambda_g v$.
It follows that any vector subspace of $V$ of dimension $1$ is invariant.
So, had $V$ any proper vector subspace of dimension $1$, $\rho$ would not be irreducible.
Since $V$ is non-null, it follows from Trivial Vector Space iff Zero Dimension that $\map {\dim_K} V > 0$.
Hence necessarily $\map {\dim_K} V = 1$.
$\blacksquare$