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

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 $$\mathbb{Q}$$.

• To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $\mathbb{Q}$ and $\mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.) – Jim Humphreys Nov 4 '18 at 14:30
• @JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven. – Mare Nov 4 '18 at 15:49

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.
• In particular, this says that every irrep can be defined over $\mathbb{Q}(\zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem. – Kevin Casto Nov 4 '18 at 17:58