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?



Representations in characteristic p
Yes, you are correct. The point is that a $p$-group acting in char. $p$ always has
a fixed point (and so acts trivially on an irrep.). So *every* irrep. of $G$
in char. $p$ factors through $G'$, as you anticipated.


The proof of the claim about $p$-groups is not hard. One approach (in general,
even when $P$ is not nec. abelian) is to prove it
by induction on the order of $P$, and so (using the fact that the descending central
series is non-trivial) reduce to the case when $P$ is cyclic of order $p$. In this
case, one can for example look at the group ring $k[P] = k[t]/(t^p - 1) = k[t-1]/(t-1)^p,$
and observe that it has a unique maximal ideal, and hence that $P$ has a unique
irrep., namely the trivial one. 


