# Enquiry about the number of irreducible representations of a finite group

In the representation theory of groups, it is a common assertion that the number of irreducible representations of a finite group G over the complex numbers is equal to the number of conjugacy classes of G. I have two quires:

1- Do we still have such a result if the field we are working on is of characteristic zero?

2- Till now the books I read, the authors stated such a fact with either the field of complex numbers $\mathbb{C}$ or the algebraically closed field F; what is significant of requiring algebraically closed field?

• How many conjugacy classes does a cyclic group have? Must all irreducible representations of a cyclic group be degree 1? Aug 25, 2018 at 2:03

There are some important represantation theoric properties which are lost when $F$ is not algebraically closed.

Lemma $1$: Let $A$ be a finite dimensional $F$-algebra where $F$ is algebraicly closed and $V$ be an irreducable $A$-module. Then $End_A(V)\cong F$.

Lemma $2$: Let $A$ be a finite dimensional commutative $F$algebra where where $F$ is algebraically closed and $V$ be an irreducible $A$-module. Then $dim_{ F}(V)=1$.

Above lemma says that if $G$ is abelian and $V$ is an $FG$ module where $F$ is algebraicly closed then $dim_{ F}(V)=1$. On the other hand, if $F=\mathbb{R}$ then we can only say that $dim_{ \mathbb R}(V)\leq 2.$

Thus, number of the linear characters of $G$ need not to be equal $|G|$ in that case. (Notice that when $G$ is abelian number of the conjugacy classes is $|G|$).

Notice that Mackey's theorem works for fields whose characteristic is zero. Thus the group algebra $FG$ is completely reducible when $F=\mathbb R$ or $F=\mathbb C$. From now on assume characteristic of $F$ is zero.

Now let $V$ be an irreducible $A=FG$ (right) module and $0\neq v\in V$. Since $vA$ is $A$-invariant space, we get $V=vA$ due to the simplicity of $V$. Then $V\cong A/ann(v)$. Since $A$ is completly reducable $A$-modue, $A=Ann(v)\oplus U$ where $U$ is $A$-submodue of $A$. Then we see that $U$ is isomorphic to $V$, that is, every irreducible $A$-module is isomorphic to a submodule of $A$ (this also show that there are finitely many distinct irreducible $A$-module).

Now suppose that $F=\mathbb C$. There are many ways of showing that the number of the irreducable character of $G$ (equivalently, the number of the irreducable $A$-module up to isomorphism) is equal to the number of conjugacy classes of $G$.

Each author follows a distinct way. Some of them defines the concept of "characters" and "class function" then shows that the set of irreducable characters constiyes a ortanormal base for the spaces of all class functions which has a dimension equal to the number of the conjugacy classes of $G$. (You can read the chapter 15 of "Algebra: A Graduate Course - I. Martin Isaacs". The the theorem you are looking for is Theorem 15.5) Another approach is purely by "represantation theory".

Now assume $F=\mathbb R$ and $G$ is abelian so that $FG$ is a commutative algebra. Since Lemma $2$ is no longer true, the number of the irreducable character of $A$ could be less than the dimension of $A$, which is eqaul to $|G|$, as some of the irreducable $A$ moduel can be of dimension $2$.

• Does this fully answer the question? Yes, the number of linear characters may have changed, but it doesn't follow (at least not immediately) that the number of characters has changed. Aug 25, 2018 at 6:24
• @SteveD: Notice that $FG$ is semisimple for both $F=\mathbb C$ and $F=\mathbb R$. Now assume that $G$ is abelian then total number of character is $|G|$ when $F=\mathbb C$. On the other hand, if $F=\mathbb R$, above answer shows that number of the chracter can be less than $|G|$. (Since dimension of $FG$ is $|G|$, $V$ is an irreducucable FG module of dimension $2$ then $FG$ may not include $|G|$ different character.) Aug 25, 2018 at 10:45
• yes, I know. But you haven't mentioned semisimple anywhere in the answer. You also haven't given an actual example, but I think that's ok if you don't want to spell it out. Aug 25, 2018 at 13:55
• You may criticize the answer. However, i do not like the way you did. Yes, i did not mention that FG is completly reducable as it is Mackey theorem and it is one of the first theorem in representatin theory. By considering the level of question, i did no think that i should mention this. But if someone ask such an explanation, i would provide. When i would have time, i may add more explanation and a concrete example. Aug 25, 2018 at 16:19
• Thank you so much Mesel for your answer and also Steve for your comment. Could you please Mesel add that explanation and example to your answer so it can be a full and nice answer for all mathematicians who encounter such a concern! Of course if time permits Aug 25, 2018 at 16:24

over $\mathbb{Q}$, there are two irreducible representations of $C_3=\langle c\mid c^3=1\rangle$, the cyclic group of order 3. One representation is the trivial one, the other is given by $$c\mapsto\begin{pmatrix}0&-1\\1&-1\end{pmatrix}.$$ You can understand this by considering the isomorphisms $$\mathbb{Q}C_3\cong\mathbb{Q}[c]/(c^3-1)\cong\mathbb{Q}[c]/(c-1)\oplus\mathbb{Q}[c]/(c^2+c+1).$$