= Solution
A <group> is a nonempty set $G$ with an operation $G\times G\to G$, $(g,h)\mapsto gh$, satisfying <associativity>, an <identity element> $e$ with $eg=ge=g$, and an <inverse element> $g^{-1}$ with $gg^{-1}=g^{-1}g=e$ for every $g$. These are the <group axioms>. An <Abelian group> additionally satisfies $gh=hg$ for every pair. A <subgroup> is a subset that is itself a <group> under the restricted operation; equivalently, it is nonempty and closed under $hk^{-1}$.
Let $R:G\to\operatorname{GL}(V)$ be a nonzero, finite-dimensional complex <irreducible representation> of an <Abelian group>. For each fixed $g$, commutativity gives
$$
R(g)R(h)=R(h)R(g)\qquad(h\in G).
$$
Thus $R(g)$ is an <intertwining operator> from $R$ to itself. By <Schur's lemma> over the <complex numbers>, $R(g)=\lambda_g I_V$. Every line in $V$ is therefore an <invariant subspace>. Since $R$ is <irreducible>, this forces
$$
\boxed{\dim_{\mathbb C}V=1.}
$$
This is the finite-dimensional convention used throughout the paper. Without it, the literal assertion for arbitrary <Abelian groups> needs qualification. A concrete counterexample is $G=\mathbb C(t)^\times$ acting by multiplication on the complex <vector space> $V=\mathbb C(t)$ of <rational functions>. Any nonzero invariant complex <vector subspace> containing $f$ contains $gf$ for every nonzero rational $g$, hence contains every rational function. This is an infinite-dimensional <irreducible representation>. The scalar conclusion of <Schur's lemma> used above depends on finite dimensionality.
Back to article page
