I want to prove that any irreducible complex representation of an abelian group is 1-dimensional. I will use the following:
Definition 1: A nonzero $G$ module $V$ is reducible if it has a basis in which every $g\in G$ is assigned a block matrix of the form $$ X(g)=\begin{bmatrix} A(g)&B(g)\\ 0 & C(g)\end{bmatrix}, $$ where $A(g)$ is square and of the same size for each $g$ and $0$ is a nonempty zero matrix.
Proposition 2: Let $X$ be an irreducible matrix representation of $G$ over $\mathbb{C}$. Then the only matrices $T$ that commute with $X(g)$ for all $g\in G$ are those of the form $T=cI$, with $c\in \mathbb{C}$ and $I$ the identity.
Proof: Let $X$ be an irreducible complex representation of an abelian group $G$. Then for arbitrary $g,h\in G$, we have: $$\tag{1} X(g)X(h)=X(gh)=X(hg)=X(h)X(g), $$ which shows that $X(h)$ commutes with $X(g)$. Since this is true for any $g$, Proposition 2 tells us that $X(h)$ is diagonal. Since $h$ was arbitrary, Definition 1 tells us that $X$ is reducible unless $\dim{X}=1$. But we assumed that $X$ was irreducible, so we must have $\dim{X}=1$. Is this conclusion correct?
Edit: I was asked to specify what I am unsure about in my proof. Basically, it is just the last part. I know equation $(1)$ is correct, because $G$ is abelian. So I would like to confirm that I am using definition 1 and proposition 2 correctly in my conclusion that $\dim{X}=1$.