Vector Spaces and Simple Modules

Question

Let $G$ be a finite group and let $R = \textbf{R}[G]$ be the group ring of $G$ with coefficients in the field $\textbf{R}$ of real numbers. Let $V$ be an $R$-module which is finite-dimensional as an $\textbf{R}$-vector space. How do I show that $V$ contains a simple $R$-module.

I am kind of lost here, however I know that that such a property is not true for an arbitrary ring $R$; for instance, $\textbf{Z}$ viewed as a module over itself contains no simple $\textbf{Z}$-submodules.

Thanks

Answer 1

Your notation has ambiguity. Use R for real numbers and $R$ for the group ring, R$[G]$. Now any R-module is also an R-vector space. As we are working with finite-dimensional ones how about that $R$-module that has the least dimension as an R-vector space ?