# Irreducible representations of an abelian group $G$ are $1$-dimensional.

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.