Let $x$ and $y$ be elements that are not conjugate in $G$. Then there is some irreducible character $\chi$ such that $\chi(x) \not = \chi(y)$.

Clearly the "irreducible" part isn't important, since any character can be written as the sum of irreducible characters, but I'm having trouble going beyond that. I'd appreciate a good hint over a full answer, and I'd be most interested in a way to construct a group representation $\varphi:G \to GL(V)$ of $G$ such that the character of the representation takes different values on $x$ and $y$.

  • 4
    $\begingroup$ Are you familiar with the orthogonality relations for columns of the character table? $\endgroup$ Sep 2, 2012 at 5:27
  • 5
    $\begingroup$ What you probably want to do is show the irreps form a basis for the class functions out of $G$. $\endgroup$
    – anon
    Sep 2, 2012 at 5:28

2 Answers 2


$\def\ZZ\mathbb{Z}$A direct construction: Let the order of $x$ be $n$. For $a \in \ZZ/n$, let $$f(a) = \# \{ h : hxh^{-1} = x^a \} \quad \mbox{and} \quad g(a) = \# \{ h : h y h^{-1} = x^a \}.$$ By hypothesis, $f(1) \geq 1$ and $g(1)=0$, so $f$ and $g$ are not equal. Consider the degree $n-1$ polynomial $\sum_{a=0}^{n-1} (f(a)-g(a)) x^a$. Since it has degree $n-1$, and is not the zero polynomial, there is some $n$-th root of unity $\zeta$ such that $\sum_{a=0}^{n-1} (f(a)-g(a)) \zeta^a \neq 0$.

Let $W$ be the one dimensional representation of $\langle x \rangle$ where $x$ acts by $\zeta$. Let $V = \mathrm{Ind}_{\langle x \rangle}^G W$. Then $$\chi_V(x) = \frac{1}{n} \sum_{a =0}^{n-1} f(a) \zeta^a \quad \mbox{and} \quad \chi_V(y) = \frac{1}{n} \sum_{a=0}^{n-1} g(a) \zeta^a.$$ So $\chi_V(x) \neq \chi_V(y)$.

  • $\begingroup$ Nice! For expository purposes, might it be good to start with the formulas for your induced char. on $x$ and $y$, and then introduce the rest as a means of forcing them to be distinct? $\endgroup$
    – Stephen
    Apr 19, 2013 at 23:45
  • $\begingroup$ Awesome, thanks. $\endgroup$
    – Carl
    Apr 24, 2013 at 1:59
  • $\begingroup$ What is $Ind_{<x>}^G W$? $\endgroup$
    – Ri-Li
    Feb 9, 2017 at 9:17
  • $\begingroup$ @user152715 $\langle x \rangle$ is the subgroup group generated by $x$, and $\mathrm{Ind}_{\langle x \rangle}^G$ is induction from that subgroup to $G$. en.wikipedia.org/wiki/Induced_representation $\endgroup$ Feb 9, 2017 at 11:50
  • $\begingroup$ Thank you. Now it is clear. $\endgroup$
    – Ri-Li
    Feb 9, 2017 at 13:20

I don't think you can "construct" one canonically, but think about this ; the indicator function $f : G \to \mathbb C$ defined by $f(g) = 1$ if $g \in \mathcal K$ and $0$ if not, where $\mathcal K$ is some conjugacy class of $G$, is a class function. You have a theorem which tells you that the irreducible characters form a basis for the vector space of all class functions over $\mathbb C$. Therefore, if every irreducible character would take equal values for $x \in \mathcal K$ and for $y \notin \mathcal K$, the function $f$, written as a linear combination of those characters, would necessarily have $f(x) = f(y)$ since this relation would hold for every irreducible character.

I know the theorem I quoted holds over $\mathbb C$ but I am not sure for other fields, so I can tell you my argument works over arbitrary fields if the theorem also holds there, but otherwise I don't know.

Hope that helps,

  • $\begingroup$ Thanks (to both you and the commenters). I'll hold off on accepting for a bit in case someone knows a canonical way to constructsuch a representation, but that at least answers the book's question! $\endgroup$
    – Carl
    Sep 2, 2012 at 5:44
  • 1
    $\begingroup$ Everything is still the same over a field $k$ of characteristic not dividing the order of $|G|$ such that $k$ contains all of the $|G|^{th}$ roots of unity (the keyword here is splitting field: groupprops.subwiki.org/wiki/Splitting_field). Away from this case it can happen that the irreducible characters do not form a basis for the space of class functions (the simplest case is something like $C_3$ over $\mathbb{R}$, where there are only two irreducible characters; here the group algebra is $\mathbb{R} \times \mathbb{C}$). $\endgroup$ Sep 2, 2012 at 6:03
  • 1
    $\begingroup$ math.stackexchange.com/a/188478 describes the elements that can be distinguished over a particular field whose characteristic does not divide the order of the group. $\endgroup$ Sep 2, 2012 at 6:22
  • $\begingroup$ I don't know what exactly you're looking for a construction, but constructing the character table is perhaps a good way to start, since it gives you such a character ; are you planning to use this in an argument and you're looking for a more precise type of construction? $\endgroup$ Sep 2, 2012 at 6:46
  • 1
    $\begingroup$ @Carl: If $G$ is abelian this won't work well. The set will only contain 2 elements, and so if the orders are odd then both must act trivially. $\endgroup$ Sep 2, 2012 at 7:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.