Let G be a finite group and let F be an algebraically closed field. If the characteristic of F is 0, then the number of irreducible F-representations of G is given by the number of conjugacy classes of elements of G. A paper I'm reading says that if the characteristic of F is p>0, then the number of F-irreps of G is the same as the number of conjugacy classes of elements whose order is not divisible by p.

If G is abelian, it seems to me that this should say that the p-sylow subgroup of G acts trivially on every characteristic p irrep. This is because I can split G into G'x P (non-p and p-sylow subgroups), and then any irrep of G' extends to one of G by letting P act trivially. Since the formula mentioned above would say that they have the same number of irreps these must be all of them.

My question is: Is this true? If not, then where is my reasoning going off track?