Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y$.

share|improve this question
4  
Are you familiar with the orthogonality relations for columns of the character table? –  Qiaochu Yuan Sep 2 '12 at 5:27
4  
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
add comment

2 Answers

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 $\chi_V(x) \neq \chi_V(y)$.

share|improve this answer
    
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
add comment

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,

share|improve this answer
    
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
1  
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
1  
@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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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