Number of irreducible representations of a finite group over a field of characteristic 0

Question

Let $G$ be a finite group and $K$ a field with $\mathbb{Q} \subseteq K \subseteq \mathbb{C}$.

For $K=\mathbb{C}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.

For $K=\mathbb{R}$ the number of irreducible representations of $KG$ is equal to $\frac{r+s}{2}$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.

For $K=\mathbb{Q}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.

(You can find quick proofs of those results in the very recent book "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Caroline Gruson and Vera Serganova, where a nice quick overview of representation theory of finite groups in characteristic 0 is given in chapters 1 and 2.)

Question:

Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $\mathbb{Q}$.

Answer 1

There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $\zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(\zeta)/K)$. We can identify $H$ with a subgroup of $(\mathbb Z/n\mathbb Z)^*$. Call $a,b\in G$ $K$-conjugate if $b^j=gag^{-1}$ for some $g\in G$ and $j\in H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.

In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.