Consider the infinite group $G=(\mathbb Z,+)$ and let $V=\mathbb C^2$. Consider the operation defined for $n\in \mathbb Z$ and $v \in V$ by: $$ nv:= \left(\begin{array}{cc} 1&n \\ 0 & 1 \end{array} \right)v$$ I have already shown that this makes $V$ into an $FG$-module ($F=\mathbb C, G=\mathbb Z$). I also have shown that the following space is a submodule of $V$:
$W_1=$span $\binom{1}{0}$
Now we have introduced the preliminaries of the problem, I am doubting myself about:
Show that there is no submodule $W_2$ of $V$ such that $V=W_1 \oplus W_2$.
As usual is with such problems, I assume that there exists such a $W_2$ and try to arrive at a contradiction. for any $v\in V$ we may write: $$ v= w_1 + w_2 = \binom{x_1}{0} + w_2. $$ where $x_1 \in \mathbb C$ and $w_1 \in W_1$, likewise $w_2 \in W_2$. I suppose the problem arises with the zero vector am I correct? So we let $v=0$, then: $$\binom{0}{0}=\binom{x_1}{0} + w_2 \implies w_2 =\binom{-x_1}{0} \in W_1 $$ Contradiction. These two spaces cannot overlap.
Edit: This argument breaks down whenever $x_1=0$, we then have $\binom{0}{0}=\binom{0}{0} + \binom{0}{0} \dots$ So my argument is incorrect, can someone help me?
The zero vector is not the problem, all subspaces intersect at zero since every subspace is assumed to contain zero. In your case, these reason it is not a contradiction is that $w_2$ is the zero vector and $x_1 = 0$.
A correct proof goes as follow. Assume, for the sake of contradiction, that there exists such a $W_2$, then by dimension counting $\operatorname{dim}_{\mathbb{C}}(W_2) = 1$. But that means every non-zero element of $W_2$ is simultaneously an eigenvector for every $g \in G$. But for $n \neq 0$, the matrix $$ A_n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}, $$
has a unique eigenvector which is $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, which is contained in $W_1$, and so we're done.