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?

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

2 Answers 2


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 $2$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Steve D
    Aug 25, 2018 at 6:24
  • $\begingroup$ @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.) $\endgroup$
    – mesel
    Aug 25, 2018 at 10:45
  • $\begingroup$ 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. $\endgroup$
    – Steve D
    Aug 25, 2018 at 13:55
  • $\begingroup$ 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. $\endgroup$
    – mesel
    Aug 25, 2018 at 16:19
  • $\begingroup$ 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 $\endgroup$
    – Mal JA
    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 $$\mathbb{Q}C_3\cong\mathbb{Q}[c]/(c^3-1)\cong\mathbb{Q}[c]/(c-1)\oplus\mathbb{Q}[c]/(c^2+c+1).$$


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