The degree of an irreducible representation of a finite solvable group

Question

I know that the degree of an irreducible representation of a finite group $G$ over a field of characteristic $0$ divides the order of $G$. I also know that such a result is false when the characteristic $p$ of the field is positive (even if $p$ does not divide the order of $G$). However, such examples seem all to be non-solvable and actually searching online I've found clues of the fact that for a solvable group it is indeed possible that the degree of an irreducible representation (over a field of positive characteristic) divides the order of the group itself. Since I cannot find anything better than clues, I'm asking: is this true? Where is it possible to find a proof of this fact?

Answer 1

I'll convert my comments to an answer though whether it sufficiently answers the question might be a bit subjective. If $k$ is an algebraically closed field of characteristic not dividing $|G|$, then the dimensions of the irreducible representations of $G$ over $k$ should always divide $|G|$. In fact, I don't think it should matter which $k$ you choose: over algebraically closed fields with nonmodular characteristic, the general structure of representation theory is always pretty much the same; the representation categories are semisimple, for instance.

If the field is not algebraically closed, funny things can happen. I mentioned that the cyclic group $C_3$ has an irreducible $2$-dimensional representation over $\mathbb{R}$. In positive characteristic, equally funny things can happen. For instance, if we take the cyclic group $C_n$ of order $n$, and we let $p$ be a prime not dividing $n$, then the irreducible representations of $C_n$ over the finite field of $\mathbb{F}_p$ are pretty fun to study, and their degrees aren't so well-behaved.

I do not know what structural statements one can make about degrees of irreducible representations over non-algebraically closed fields (though I'd be interested to learn about it). In addition, I'm not aware of any connections with solvability of the group considered. (For what it's worth all examples I gave are abelian and hence trivially solvable. This raises the question for me: What does the representation theory of $A_5$ look like over say finite fields of characteristic $p > 5$?)

Answer 2

It is true (with the caveats given in the other answer): Let $G$ be a finite solvable group, and $p$ a prime dividing the order of $G$. Then the degree of every absolutely irreducible representation (that is, the field sufficiently large) of $G$ in characteristic $p$ divides the order of $G$.

The only proof for this that I am aware of, is a consequence of the Fong-Swan theorem, which states that for solvable groups actually the representation itself must come from reducing a representation in characteristic zero (and then using the result for characteristic zero). This theorem is (not the original reference, but probably the easiest accessible place) Theorem 38 on p. 135 of Serre, Linear Representations of Finite Groups.