Is there a relation between the representative order of a conjugacy class and the corresponding irreducible character value? Thanks in advance.

  • 1
    $\begingroup$ Can you clarify the question you're asking? By "representative order" it sounds like you're asking whether there is a relationship between $\chi(x)$ and $|x|$ for $x\in G$. The other possibility is you're asking about a relationship between $\# C_g$ (where $C_g$ is a conjugacy class) and $\chi(C_g)$ (where this is the constant value taken by $\chi$ on $C_g$). Are either of these interpretations correct? If so, which? $\endgroup$ Feb 23, 2013 at 5:24
  • $\begingroup$ And to add to the question by Alex, in either interpretation my next question would be "which irreducible character?". $\endgroup$ Feb 23, 2013 at 5:44
  • $\begingroup$ I have no idea what you're asking. Voting to close $\endgroup$
    – Alexander Gruber
    Feb 23, 2013 at 5:58
  • $\begingroup$ Yeah I mean is there a relationship between χ(x) and |x| for x∈G $\endgroup$
    – Z. Fo
    Feb 23, 2013 at 6:09
  • $\begingroup$ And you did not specify over what fields are these representations taken. Per chance this matters not? $\endgroup$
    – awllower
    Feb 24, 2013 at 16:40

1 Answer 1


If I understand your question correctly, you are taking an element $x \in G$, and an irreducible character $\chi$, and asking what influence does the order $n = \lvert x \rvert$ of $x$ have on $\chi(x)$.

In this generality, I do not think you can say much more than the fact that $\chi(x)$ is the sum of some $n$-th roots of unity.

This is because, if $\rho$ is a representation corresponding to $\chi$, then $\rho(x)^n = \rho(x^n) = \rho(1) = I$, where $I$ is an appropriate unit matrix, so the minimal polynomial of $\rho(x)$ divides $x^{n} - 1$, and all eigenvalues of $\rho(x)$ are $n$-th roots of unity.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .