I want to show that irreducible representations of an abelian group $G$ are $1$-dimensional where $G$ is either a finite or a Lie group. As I understand, it can be shown by using Schur's Lemma. But, I have a question if the following idea works for the finite abelian groups and if it can be extended to a Lie groups:
Let $G$ be a finite group, and $\rho:G\to GL(V)$ be a representation of $G$ where $V$ is a finite vector space over complex numbers. Since $G$ is abelian, then $\rho(a)\rho(b)=\rho(b)\rho(a)$. So, if we are going to fix a basis for $V$, then we are going to obtain a finite set of commuting matrices $\{\rho(a)|a\in G\}$ which are simultaneously diagonazible i.e. a representation $\rho$ is completely reducible and every subrepresentation is $1$-dimensional. So, the only irreducible representations are $1$-dimensional.
In a case of a Lie group $G$, the set $\{\rho(a)|a\in G\}$ can have an uncountable cardinality. I see no problem arguing that a countable set of a commuting matrices is simultaneously diagonazible, but I am not sure about an uncountable set.
Any possible feedback would be appreciated. Thank you!
Proof is fine for finite groups, assuming you know $\rho(g)$s are diagonalizable in the first place.
For infinite $G$, each $\rho(g)$ induces a decomposition of $V$ into subspaces, and any two of these decompositions have a mutual refinement. If $V$ is finite-dimensional, among these there must be a maximal refinement (you can't keep refining forever) with respect to which all the operators are diagonalizable. Hope this works.