4
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On the proof of: If $0<\frac{|χ(g)|}{χ(1)}<1$ then $\frac{χ(g)}{χ(1)}\notin\overline{\mathbb{Z}}$
This has something to do with Galois theory. Since the minimal polynomial is irreducible, its roots are permuted transitively by the Galois group of the splitting field. But the Galois group sends ...
- 2,698
4
votes
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Definition of a character of a linear algebraic group
Yes, please don't feel bad about this confusion. It is for semi-historical reasons as KReiser rightly pointed out above, and can be quite confusing.
Namely, let us fix a field $K$ and an (affine) ...
- 51.9k
3
votes
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Same character values iff related by outer automorphism, for perfect groups
There are plenty of counterexamples. The ATLAS of finite groups is a good place to look. I think the smallest is for the group ${\rm PSL}(2,11)$, for which there are two characters of degree $12$ with ...
- 84.8k
3
votes
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If $\chi$ is a complex-valued character of a representation of a finite group, is it always true that $\overline{\chi(g)}=\chi(g^{-1})$?
Another way to see that is as follows:
Let $G$ be a finite group, $ρ:G\rightarrow \text{GL}_n(\mathbb{C})$ a representation and $χ$ the corresponding character.
We know that for any $g\in G$, all the ...
- 405
2
votes
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If a hom. $\phi:G\to H$ of diagonalisable linear algebraic groups is injective, then the induced hom. $\phi^*:X^*(H)\to X^*(G)$ is surjective
Unless I'm missing something here, the exercise is wrong as stated.
Let $k$ of characteristic $p>0$ and let $G=H=\Bbb G_m$. Let $\phi:G \to G$ be the map that raises everything to the $p$-th power....
- 17k
2
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Legendre symbol as a group character
The character group $\hat{G}$ of a group $G$ is defined as
$$\hat{G}=\lbrace \chi:G\to \mathbb{C}^\times \mid \chi\text{ homomorphism}\rbrace$$
as Sean Eberhard wrote in the comment. (note: some ...
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