# Complex finite dimensional irreducible representation of abelian group

I'm supposed to show that each Complex finite dimensional irreducible representation of an abelian group is one dimensional.

For any map $\phi: V \rightarrow V$ it holds that $\phi(\rho(g)v) = \rho(g) \phi(v)$. Also since the group $\rho(h) \rho(g) v = \rho(g) \rho(h) v$. From a previous exercise I know that $\phi = \lambda \cdot id_V$ for some $\lambda \in \mathbb{C}$. This transforms the previous equation into $\lambda \cdot id_V \cdot (\rho(g)v) = \rho(g) \lambda \cdot id_V \cdot v$ which implies that $\lambda \cdot id_V \cdot (\rho(g)v) =\lambda \cdot \rho(g) \cdot v$. Now I'm not quite sure how to bring into play that $G$ is abelian. Could someone give me a hint?

Cheers!

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Let us sort out things a bit.

Let $\rho: G \to \operatorname{GL}(V)$ be an irreducible representation of any group $G$.

You have seen that if $\varphi : V \to V$ commutes with all $\rho(g)$, for $g \in G$, that is $$\varphi (\rho(g) v) = \rho(g) \varphi (v)\tag{comm}$$ for all $g \in G$ and $v \in V$, then $\varphi = \lambda \operatorname{id}_{V}$ for some $\lambda \in \mathbf{C}$.

Now if $G$ is abelian we have $\rho(x) \rho(g) = \rho(g) \rho(x)$ for all $g, x \in G$, so that $\varphi = \rho(x)$ satisfies (comm).

It follows that for any $x \in G$ there is $\lambda \in \mathbf{C}$ such that $$\rho(x) = \lambda \operatorname{id}_{V},$$ so all $\rho(x)$ are scalars, and then leave every subspace invariant.

Since the representation is irreducible, $V$ must have dimension $1$ then.

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As an alternative approach:
Since all homomorphisms from an abelian group $G$ to $C^*$ are irreducible characters of $G$, and there are $|G|$ many of these, by dint of orthogonality relations, we conclude that they are all irreducible characters of $G$, and hence all irreducible representations of $G$ are of dimension $1$.
Inform me of any error. Thanks.

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+1 This would be my approach when $G$ is finite. –  Alexander Gruber Apr 9 '13 at 1:11

For some variety, here is a ring-theoretic argument.

For a finite abelian group and an algebraically closed field $k$, $R:=kG\cong k^{|G|}$ as rings by Artin-Wedderburn.

[$R\cong\prod_iM_{n_i}(D_i)$ is a finite product of matrix rings over division rings, $D_i\cong\operatorname{End}_R(V_i)$ a division ring (finite dimensional over $k$), $V_i$ runs over the isomorphism classes of the simple left $R$-modules, $n_i$ their multiplicity in $R$ as a left module over itself.]

Hence the dimensions of the simple left $R$-modules must all be one [since $R$ is commutative, $n_i=1$ and $D_i$ is a field, because $k$ is algebraically closed, $D_i=k$].

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