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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 $y$.

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Are you familiar with the orthogonality relations for columns of the character table? –  Qiaochu Yuan Sep 2 '12 at 5:27
What you probably want to do is show the irreps form a basis for the class functions out of $G$. –  anon Sep 2 '12 at 5:28

2 Answers 2

up vote 8 down vote accepted

$\def\ZZ\mathbb{Z}$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 $$\chi_V(x) = \frac{|G|}{n} \sum_{a =0}^{n-1} f(a) \zeta^a \quad \mbox{and} \quad \chi_V(y) = \frac{|G|}{n} \sum_{a=0}^{n-1} g(a) \zeta^a.$$ So $\chi_V(x) \neq \chi_V(y)$.

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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? –  S123 Apr 19 '13 at 23:45
Awesome, thanks. –  Carl Apr 24 '13 at 1:59

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 be equal 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,

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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! –  Carl Sep 2 '12 at 5:44
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 $\mathbb{R} \times \mathbb{C}$). –  Qiaochu Yuan Sep 2 '12 at 6:03
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. –  Jack Schmidt Sep 2 '12 at 6:22
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? –  Patrick Da Silva Sep 2 '12 at 6:46
@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. –  Jack Schmidt Sep 2 '12 at 7:45

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