[proofplan] Let $V$ be an irreducible complex representation of the finite abelian group $G$. For each $g \in G$, multiplication by $g$ is a $G$-endomorphism of $V$ — abelianness of $G$ is exactly the condition that $g$ commutes with every other group element, hence its action commutes with the actions of every other element. By part 2 of Schur's lemma, this $G$-endomorphism is a scalar multiple of the identity, so every group element acts on $V$ by a scalar. Then any one-dimensional subspace spanned by a non-zero vector is $G$-invariant, and irreducibility forces $V$ to coincide with this line. [/proofplan] [step:Show every group element acts on $V$ as a $G$-endomorphism] Let $V$ be an irreducible complex $G$-representation, where $G$ is a finite abelian group. By definition $V$ is a non-zero finite-dimensional $\mathbb{C}$-vector space with a $G$-action. Fix $g \in G$. Define \begin{align*} \theta_g : V &\to V \\ v &\mapsto gv. \end{align*} This is $\mathbb{C}$-linear (the action $\rho(g)$ is a linear map). We verify $\theta_g$ is a $G$-homomorphism. For any $h \in G$ and $v \in V$, using $gh = hg$ (since $G$ is abelian): \begin{align*} \theta_g(hv) = g(hv) = (gh)v = (hg)v = h(gv) = h\theta_g(v). \end{align*} Hence $\theta_g \in \operatorname{End}_G(V)$. [guided] The hypothesis that $G$ is abelian is doing precisely one thing: it makes "multiplication by $g$" commute with the action of every other group element on $V$. Concretely, define $\theta_g(v) := gv$. This is automatically $\mathbb{C}$-linear. We need $\theta_g(hv) = h\theta_g(v)$ for all $h$, which is equivalent to $(gh)v = (hg)v$ for all $v$, which is equivalent to $gh = hg$ in $G$. Abelianness gives this for all pairs $(g, h)$. Without abelianness, $\theta_g$ would generally not be a $G$-endomorphism — the proof breaks at this step. (For instance, in $S_3$ acting on its standard 2-dimensional irreducible, no individual transposition acts by a scalar.) [/guided] [/step] [step:Apply Schur's lemma to write the action of each $g$ as a scalar] Since $V$ is an irreducible complex $G$-representation and $\mathbb{C}$ is algebraically closed, part 2 of [Schur's Lemma](/theorems/2414) applies to $\theta_g \in \operatorname{End}_G(V)$, yielding a unique scalar $\lambda_g \in \mathbb{C}$ such that \begin{align*} \theta_g = \lambda_g \iota_V. \end{align*} Concretely, $gv = \lambda_g v$ for every $v \in V$. [guided] Schur 2 requires $V$ irreducible and the base field algebraically closed. Both are direct hypotheses ($V$ is an irreducible complex representation, $\mathbb{C}$ is algebraically closed). Applied to the $G$-endomorphism $\theta_g$, it produces a unique scalar $\lambda_g \in \mathbb{C}$ with $\theta_g = \lambda_g \iota_V$. So every group element acts on $V$ as a scalar. This is a strong constraint — once $\rho(g) = \lambda_g \iota_V$, the action of $g$ preserves every subspace of $V$. Combined with irreducibility, this will force $V$ to be one-dimensional. [/guided] [/step] [step:Pick a non-zero vector and show its span is $G$-invariant] Since $V \neq 0$ (irreducibles are non-zero by definition), choose $v \in V$ with $v \neq 0$. Let \begin{align*} W := \langle v \rangle = \{\mu v : \mu \in \mathbb{C}\} \subseteq V \end{align*} be the $\mathbb{C}$-linear span of $v$, a one-dimensional subspace. We check $W$ is $G$-invariant. For any $g \in G$ and $\mu \in \mathbb{C}$, \begin{align*} g(\mu v) = \mu (gv) = \mu \lambda_g v = (\mu \lambda_g) v \in W, \end{align*} using the previous step in the second equality. Hence $W$ is a $G$-invariant non-zero subspace of $V$. [guided] We need a $G$-invariant subspace of $V$ to leverage irreducibility. Pick any non-zero $v \in V$ — such a $v$ exists because $V$ is non-zero by definition. Its span $W = \mathbb{C}v$ is one-dimensional. Why is $W$ $G$-invariant? Because every $g$ acts as a scalar: $gv = \lambda_g v \in W$, and any $\muv \in W$ gets sent to $g(\muv) = \mu \lambda_g v \in W$. So the action of $G$ preserves $W$. This is where the previous step pays off. If $g$ did not act as a scalar, then $gv$ might leave the line through $v$, and $W$ might not be $G$-invariant. [/guided] [/step] [step:Conclude $V = W$ is one-dimensional by irreducibility] We have $W \subseteq V$ a non-zero $G$-invariant subspace. Since $V$ is irreducible, the only $G$-invariant subspaces of $V$ are $0$ and $V$ itself. As $W \neq 0$, we conclude $W = V$. Hence \begin{align*} V = W = \langle v \rangle \end{align*} is one-dimensional over $\mathbb{C}$. This completes the proof. [guided] Irreducibility of $V$ says: the only $G$-invariant subspaces are $\{0\}$ and $V$. We have constructed a non-zero $G$-invariant $W$, so $W$ cannot be $\{0\}$, leaving $W = V$. Since $W$ is one-dimensional, so is $V$. The dichotomy of Schur 2 (every endomorphism is a scalar) plus the dichotomy of irreducibility (only $0$ and $V$ are invariant) collapses the dimension of $V$ to $1$. The result is the abelian-group analogue of the fact that for a commuting family of operators on a complex vector space, the simultaneous eigenspaces are one-dimensional in the irreducible case — here all of $G$ is the commuting family, and $V$ is itself the eigenspace. [/guided] [/step]