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This is a question from J.P.Serre's book 'Linear representation of finite groups',section 2.4

The question: Let $G$ be a finite group. Show that each character of $G$ which is zero for all $g \ne 1$ is an integral multiple of the character $r_G$ of the regular representation.

What I have done so far: $r_G$ satisfies $r_G(g) = 0$ for all $g \ne 1$, and $r_G(1) = |G|$, the order of $G$. If $\chi$ denotes the character, then $\chi(g) = r_G(g) = 0$ for all $g \ne 1$, so it is enough to show that $|G|$ divides $\chi(1)$. If $\chi_1,...,\chi_k$ denotes all the irreducible characters of $G$, with dimension of the representations $n_1,...,n_k$ respectively, then we can write $\chi = \sum_{i=1}^k \langle \chi,\chi_i\rangle \chi_i$, where $\langle \chi,\chi_i\rangle$ is the inner product. And it is easy to calculate $\langle \chi,\chi_i\rangle = (\chi(1)/|G|)\,n_i$. So each of these values must be integers for all $i$. But how does one conclude that in fact $\chi(1)/|G|$ is an integer?

thanks in advance.

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Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. –  Zev Chonoles Jun 8 '13 at 15:55
    
Thank you! I don't know LaTeX, so sorry for my cumbersome notations. –  mathmansujo Jun 8 '13 at 15:57
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Note, that the trivial representation is irreducible and one-dimensional. –  martini Jun 8 '13 at 16:08
    
@martini : Can you please give some more details? I can't see why it is important here. –  mathmansujo Jun 8 '13 at 17:32
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@usersujo Let $\chi_k$ be the character of a one-dimensional representation then $n_k = 1$. By your Argument above $\langle \chi, \chi_k\rangle = \chi(1)/|G|\cdot n_k = \chi(1)/|G|$ is an integer. –  martini Jun 8 '13 at 17:35
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up vote 2 down vote accepted

One of the irreducible representations is the trivial, one-dimensional one, say $\chi_i$ is its character. Then $n_i = 1$, and your argument above gives $\def\<#1>{\left\langle#1\right\rangle}$that $$\<\chi, \chi_i> = \chi(1)/|G| \cdot n_i = \chi(1)/|G| $$ is an integer.

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