Why are complex finite-dimensional irreducible representations of abelian groups one-dimensional?

Question

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!

Answer 1

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.

Answer 2

As an alternative approach:
Since all homomorphisms from a finite 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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Edit:

I am not sure now what my argument was for the equation $\lvert\operatorname{Hom}(G,\mathbb C^*)\rvert=\lvert G\rvert$, but I think for a finite abelian group, we can prove this by writing $G$ as a direct product of cyclic groups, by means of the fundamental theorem of finitely generated abelian groups, and then showing that this equation holds for cyclic groups, and both sides of the equation are multiplicative with respect to the direct product.

Answer 3

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, $D_i=k$].