A question on representation of commutative groups I am recently reading Naimark and Stern's book "Theory of Group Representations". On page 37, an exercise is to show that a finite dimensional representation of a finite commutative group is a direct sum of one-dimensional representations. I know that such representation has at least one 1-dimentional representation, but don't have a clue how to prove this statement. Can anyone give me a hint? Thanks.
 A: You can use induction on the dimension (and Maschke's theorem).
If $\dim V = 1$, it's over. If not, then $V$ is not irreducible (for an abelian group, only one-dimensional representations are), so there's a sub-representation, say $U \subset V$. Then because of Maschke's theorem, $U$ admits an invariant direct complement $W$. So you've written
$$V = W \oplus U$$
where both $W$ and $U$ are sub-representation of lower dimension, which enables you to use induction and conclude.
To prove that irreducible representations of an abelian group are one-dimensional, use Schur's Lemma : let $G$ be an abelian group, and $(\pi, V)$ be an irreducible representation of $G$. Let $g \in G$, you can check (using commutativity) that $\pi(g) : V \to V$ is an intertwining from $(\pi, V)$ to itself. So from Schur's lemma, there exists $\lambda_g$ such that $g.v = \lambda_g v$ for all $v \in V$. If $v \in V$ is non-zero, the linear span of $v$ is one-dimensional and stable under the action of $G$, so it's $V$ (from irreducibility of $V$), and $\dim V = 1$.
