TITLE: Representations in characteristic p
QUESTION [13 upvotes]: 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?

REPLY [5 votes]: Brauer's proof that the number of similarity classes of irreducible representations of $G$ over an algebraically closed field of characteristic $p$ is equal to the number of $p$-regular conjugacy classes of $G$ is ring-theoretic in flavor, and rather tricky. There is also an easy character theoretic proof based on the following ideas. First, the set IBr$(G)$ of irreducible Brauer characters is in bijective correspondence with the irreducible representations, and this set of functions lives in the space $V$ of all complex-valued class functions defined on the set of p-regular elements. Since $\dim(V)$ equals the number of $p$-regular classes, it suffices to show that IBr($G$) is a basis for $V$. The linear independence of IBr$(G)$ is a standard result. To see that IBr$(G)$ spans, use the facts that Irr$(G)$ spans the space of all class functions and that on each $p$-regular class, the value of an ordinary character is a linear combination (and in fact, a sum) of values of Brauer characters.