Number of simple $\mathbb{C}[G]$-modules is finite via Jordan-Hölder? Is it possible to show that the number of simple $\mathbb{C}[G]$-modules is finite with Jordan-Hölder? $G$ is assumed to be finite.
 A: Yes, it is possible. The idea is similar to Mark's but Jordan-Hölder can be used to avoid using Maschke.
Let $S = (v)$ be a simple $\mathbb{C}[G]$-module. We consider the $\mathbb{C}[G]$-algebra homomorphism $$ \varphi : \mathbb{C}[G] \to S, \ x \mapsto xv.$$ Since $S$ is simple, this map is surjective and therefore $S \cong \mathbb{C}[G]/\ker{\phi}$.
Therefore, $S$ appears in a composition series of $\mathbb{C}[G]$ by prolonging $\ker{\phi} \subseteq \mathbb{C}[G]$. So every simple module appears in a composition series and so by Jordan-Hölder, all simple modules appear in the finite (up to equivalence) composition series of $\mathbb{C}[G]$. Hence, there are only finitely many simple $\mathbb{C}[G]$-modules.
A: Let $A=\mathbb{C}G$. In particular $A$ is a module over itself with the operation of left multiplication. From Maschke's theorem we know that there are simple modules $S_1,...,S_k$ such that $A=S_1\oplus...\oplus S_k$ as $A$-modules. (I think of it as an inner direct sum) Now it is sufficient to show that any simple $A$-module is isomorphic to one of the modules $S_1,...,S_k$.
So let $S$ be a simple module. Choose any $0\ne s\in S$. We can define a homomorphism of modules $\varphi:A\to S$ by $\varphi(a)=as$. This homomorphism is nontrivial, because $\varphi(1)=s\ne 0$. Since $A=S_1\oplus...\oplus S_k$ there must be some $1\leq i\leq k$ such that $\varphi|_{S_i}:S_i\to S$ is nontrivial. Since this is a nontrivial homomorphism of simple modules it follows from Schur's lemma that it is an isomorphism.
Note that we only used that $A$ is an algebra with identity where $A$ as a module has a decomposition into a finite direct sum of simple submodules.
