Let G be a finite abelian group, K an algebraically closed field, K[G] the group algebra of G over K and M a K[G] module. I would like to show that if M is simple, it has dimension 1 over K. I think that I can interpret this as showing that M is isomorphic to K as a K[G] module. You could interpret K itself as a module over K[G]. Take as addition the usual field addition and as scalar multiplication by $x \in K[G], x = \sum_{g \in G} a_g$ the multiplication by $a_e$, where e is the group identity. Now, if I can show that there is a module homomorphism between K and M which is not zero than $K \simeq M$ by Schur's Lemma.
Is this a helpful interpretation? How would a homomorphism look like?
If you could prove that $M$ was isomorphic to $K$ as a $K[G]$-module, you would have the result you're looking for, but I think even this is too strong. Instead, try this:
For $h\in G$, let $\phi_h:M\to M$ be the linear transformation given by $\phi_h(m)=hm$ for all $m\in M$. Since $K$ is algebraically closed, $\phi_h$ must have an eigenvalue $\lambda\in K$. Consider the linear transformation $\phi_h-\lambda I$, where $I$ is the identity on $M$. Since $G$ is Abelian, \begin{align*} (\phi_h-\lambda I)gm&=\phi_h (gm)-\lambda gm=hgm-\lambda g m = ghm-\lambda g m \\&= g\phi_h(m)-\lambda g m =g(\phi_h(m)-\lambda m)=g(\phi_h-\lambda I)m, \end{align*} so $\phi_h-\lambda I$ is a $K[G]$-module homomorphism, but, since $\lambda$ is an eigenvalue of $\phi_h$, it is not invertible, so by Schur's lemma it must be zero. It follows that $\phi_h=\lambda I$, or, in other words, $hm=\lambda m$ for every $m\in M$. Since this is true for every $h\in G$, it follows that every one-dimensional subspace of $M$ is a $K[G]$-submodule. This would contradict the assumption that $M$ is simple module unless $M$ were itself one-dimensional.