The characteristic of the field does not divide the dimension of an irreducible representation Let $k$ be an algebraically closed field of characteristic $p$, and $G$ be a finite group whose order is not divisible by $p$. I would like to prove the following: if $V$ is an irreducible representation of $kG$, then $\dim V \neq 0$ in $k$, i.e. $p$ does not divide the dimension of $V$.
Here is one explanation of why it is true: for representations over $\mathbb{C}$, the dimension of a representation must divide the order of the group. Then, there is a claim that the dimensions of the irreducible representations of $G$ over $k$ are equal to the dimensions of the irreducible representations of $G$ over $\mathbb{C}$. (I do not know why this claim is true). This then implies that over $kG$, the dimension of any irreducible representation $V$ divides $|G|$, and hence is coprime to $p$.
Is there a shorter argument, which avoids the character theory over $\mathbb{C}$?
 A: The claim that the dimensions of irreducible representations over $k$ are the same as over $\mathbb{C}$ is not particularly hard. It follows from these three general facts about representations in characteristic p:

*

*If an irreducible representation over $k$ lifts to characteristic zero, the lift is irreducible too.

*Any representation of a finite group $G$ which is a projective object in the category of $G$ representations over $k$ can be lifted to characteristic zero.

*If $p \nmid |G|$ representations of $G$ are semisimple, and therefore every object is projective.

If you insist on not using that fact you could use the general theory of blocks and defect groups in characteristic p, although I'm not sure that's an easier theory. Here are some general facts from that theory that imply what you want:

*

*Each irreducible representation lies in a unique block.

*Each block has an associated defect group, a p-subgroup of $G$.

*A block contains a representation of dimension prime to $p$ iff the defect group is a Sylow subgroup.

*Since $p \nmid |G|$, each irreducible representation forms its own block.

Probably neither of these explanations is quite as simple as you'd like, but given that the characteristic zero result is fairly nontrivial I'm somewhat doubtful a slick proof of this will be found.
