$G$ is Abelian if it admits a one dimensional faithful representation over a two dimensional vector space $V$ I must check the following result:
Let G be a finite group. Assume $G$ admits a faithful reducible representation $\rho : G \to GL(V)$ where $V$ is a two dimensional vector space over a field $\mathbb{F}$ with zero characteristic. Prove that $G$ is commutative.
What I have managed to prove so far is that the commutator of $G$ is Abelian under such conditions. I haven't used the fact that $G$ is finite or nothing about the character of the field. However, seeing what happens with these additional constraints does not help to conclude that $G$ is commutative.
What would be some nice way to look at this task? Any hints are appreciated!
 A: Because we're considering representations over a field over characteristic zero of a finite group, by Maschke's theorem, the representation  $V$ decomposes into a direct sum of irreducible representations.
Because $V$ is two-dimensional, the only way for it to be reducible is if it is a sum of two one-dimensional representations $V=V_1 \oplus V_2$. This is just another way of saying that the family of operators $\rho(g)_{g \in G}$ is simultanously diagonalizable: Choose a basis $v_1,v_2$ for $V$ such that $v_1 \in V_1, v_2 \in V_2$, then for all $g \in G$, we have $\rho(g)v_1=\chi(g) v_1$ and $\rho(g)v_2=\psi(g)v_2$, where $\chi$ and $\psi$ are one-dimensional representations. But this means that with respect to this basis, $\rho$ has the form
$$\begin{pmatrix}\chi(g) & 0 \\ 0 & \psi(g) \end{pmatrix}$$
Finally, because diagonal matrices commute with each other, we see that the image $\rho(G)$ is an abelian group. As $\rho$ is faithful, this implies that $G$ is itself abelian.
The main part of this argument is something that holds for any group over any field and in any dimension, namely: a representation $\rho:G \to \mathrm{GL}(V)$ is a direct sum of one-dimensional subrepresentations if and only if the family of linear operators $\rho(g)_{g \in G}$ is simultanously diagonalizable. And if furthermore $\rho$ is faithful, this implies that $G$ is abelian, because as before, diagonal matrices commute with each other.
