Suppose that $G$ is a commutative group with $|G|<\infty$. Show that any irreducible finite $G$-module over $\mathbb{C}$ must be one dimensional.
A question first: Is it possible that this is not one dimensional? Simple example is really appreciated. Thanks.
Attempt:
Let $V$ be the irreducible $G$-module. Hence, $V$ must be nonzero by definition, and $G$ acts on $V$ with linearity property. For every $g\in G$, consider the map $f_g:V\rightarrow V$ given by $f_g(v)=g.v$ (the action of $g$ on $v$). This map cannot be zero, otherwise for some nonzero $v\in V$, $g.v=0$ implies $v=(g^{-1}g).v=0$, a contradiction.
We show that $f_g$ is a $G$-module map (for using Schur's Lemma). Let $x\in G$. We have $f_g(x.v)=g.(x.v)=(gx).v=(xg).v$ by commutativity of $G$. Hence, $f_g(x.v)=x.(g.v)=x.f_g(v)$.
Therefore, by Schur's Lemma, $f_g=k_gI$ (Wait, do we show $f_g$ as a linear map too, or Schur's Lemma actually works for any auto $G$-module map?) for some $k_g\in\mathbb{C}$ and $I$ is the identity map on $V$. Recall that our $g$ is arbitrary.
Now, consider the subspace $H:=span(v)$ for some nonzero $v\in V$. We show that $V=H$ by showing that $H$ is a $G$-submodule of $V$ and use the irreducibility of $V$ to conclude.
To do that, let $g\in G$ and $mv\in H$, where $m\in\mathbb{C}$. We see that $g.mv=m(g.v)=mf_g(v)=m(k_g v)=(mk_g)v\in H$. Thus, $H=span(v)$ is a $G$-submodule of $V$.
Hopefully there is any correction or improvement (too much/too little). Thank you very much.
EDIT: I should have said that I have searched and found answers in MSE before posting this, but I would like to try my own solution and would like to know whether mistakes exist as I haven't really understood other concepts well. Thank you.