Dear Kelvin,

if one has an irreducible representation r: G->GL(V), V being a vector space over a field F, then V is a simple F(G)-module  over the group ring F(G). In fact, each G-invariant (with respect to the G-action  given by the representation r) subspace W of V is a F(G)-submodule  of V and vice versa. If one has only {0} and V as G-invariant subspaces of V then F(G)-module V has only trivial F(G)-submodules  {0} and V. Thus, V is a simple F(G)-module. I think, the answer is "yes".

Best regards, Yuri

The answer is YES: Every n-dimensional FG-module M affords a matrix representation T:G to GL(n,F) ( of degree n) provided by the G-action on the F-space M. It is true that M is simple FG-module if and only if the matrix representation T is irreducible.

yes