Irreducible complex representation of abelian group is one dimensional

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Statement

Every irreducible complex representation of an abelian group $G$ is one-dimensional.

Proof

Let $(\rho ,V)$ be an irreducible complex representation of an abelian group $G$.

Schur's lemma states that for an irreducible representation $\rho :G\to GL(V)$ over an algebraically closed field, the only elements of $GL(V)$ that commute with everything in $\rho (G)$ are the scalar multiples of the identity.

$G$ is abelian and $\mathbb {C}$ is algebraically closed, so, each $\rho (g)$ is a scalar multiple of the identity map on $V$. Then, for $v\in V$ non-zero, $\langle v\rangle$ is a subrepresentation of $V$. But $V$ is irreducible. So $V=\langle v\rangle$. So $V$ is one dimensional.

Warning

This is certainly not true over other fields, since this is a corollary of Schur's lemma, in particular the part that requires the field be algebraically closed, which $\mathbb {C}$ is. For example, there exist irreducible two-dimensional representations of abelian groups over $\mathbb {R}$. A similar fact, that an irreducible representation of an abelian group over the real numbers is one or two dimensional, can however be shown.