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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 ...
- 843
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Validating a Character Table for a Given Finite Group
Adding to the excellent answer of @BrauerSuzuki I wanted to comment on the point:
For $H≤G$, one gets a permutation character $π$ from the (transitive) action of $G$ on the set of (left) cosets by (...
- 51
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Does any valid character table correspond to a group?
In his famous report for the ICM, Richard Brauer proposed a number of questions on representations of finite groups. His Problem 6 asks to give necessary and sufficient criteria on a complex matrix to ...
- 3,597
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Confused about the domain of characters
Usually, when people ask about potential errors in books, either the book is wrong or they’re wrong. In this case, you’re both a bit wrong.
The book is wrong in that it introduces the circle group ...
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Character table guarantees certain property of groups
This happens if and only if
$$\sum_{\chi\in\mathrm{Irr}(G)}\frac{\chi(x)\chi(y)\chi(z)\chi(g)}{\chi(1)^2}$$
is constant (i.e. does not depend on $g$). See Corollary 4.14 in [Navarro, Characer theory ...
- 3,597
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If $g$ is commutator then so is $g^m$ for $(m,o(g))=1$
A character-free proof of this result of K. Honda (1953) has recently been found by Hendrik Lenstra. It has been published in Operations Research Letters, Powers of Commutators, Volume 51, Issue 1, ...
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Why does character of identity element always positive integer?
The character of the identity element is the trace of the identity matrix, i.e. the dimension (the "degree") of the representation.
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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) ...
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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 ...
- 3,597
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Why is the determinant of a character well-defined?
For any two choices $\rho$ and $\rho'$ giving the same character $\chi$, there exists $P\in GL_n(\mathbb{C})$ such that $\rho'(g)=P\rho(g)P^{-1}$ for all $g\in G$: in other words, $\rho$ and $\rho'$ ...
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Is there a group $G$ with $|G|=8$ and $\chi$ assuming values $-2,-1,0,0,0,0,1,2?$
I shouldn't have asked this question: turns out it is a straightforward computation if we decompose $\rho$ in terms of irreducible characters $\rho=\oplus_i n_i \rho_i $:
$$\sum_i n_i^2=\left\langle \...
- 1,306
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p-adic Norms and Character Values
It is trivially correct when $n = 1$: the left side is $1$ and the right side is $|\chi(g)|_p^2$, which is $1$ since $\chi(g)$ is a root of unity when $n = 1$.
Is is never true for all $g$ when $n >...
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Why do characters fail to characterize non-abelian LCH-groups?
The short answer is that if $G$ is a group and $A$ an abelian group (like $\mathbb T$), then every group homomorphism $\phi\colon G\to A$ factors through the abelianization $G/[G,G]$. Here $[G,G]$ is ...
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Formula for $\chi(g)\chi(h)$, where $\chi$ is an irreducible character of $G$
Theorem Let $\chi \in Irr(G)$, and $x,y \in G$, then
$$\chi(x)\chi(y)=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi(xy^z)$$
Before proving this theorem we need to set notation and an observation. Write $\...
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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 ...
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Help with proof of $\langle\chi,\psi\rangle=\delta_{\chi,\psi}$ with $\chi$ and $\psi$ characters of a group.
From my understanding, it comes from the fact that the matrix $X=(x_{i,j})$ is a matrix of indeterminants, so these $x_{i,j}$ are just variables, and the quantity $Y$ will be a polynomial in the ...
- 2,128
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The set of characters
Instead of regarding your algebra as a subalgebra of $C_b(\mathbb{R})$, I'll further enlarge the ambient algebra to $L^\infty(\mathbb{R})$. We note that any character on $C_\infty(\mathbb{R})$ can be ...
- 1,489
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Determine conjugacy classes from group representation for character table
A conjugacy class of $x$ in $G$ is a set $\{gxg^{-1} : g \in G\}$. In the case of an abelian group $G$, we have $gxg^{-1} = x$ for all $g \in G$, so every conjugacy class is a singleton $\{x\}$. ...
- 361
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Permutation analog of a character table
Every group action is the “union” of transitive actions. Every transitive action of a group $G$ is equivalent to the action on the set of (left) cosets $G/H$ for some $H\le G$ by (left) multiplication....
- 3,597
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Same character table implies same composition factors
Theorem 4 of
Kimmerle, Wolfgang; Sandling, Robert, Group theoretic and group ring theoretic determination of certain Sylow and Hall subgroups and the resolution of a question of R. Brauer, J. Algebra ...
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Let $\chi(g)$ is a nonnegative real number. Show that if $\chi$ is irreducible, then $\chi$ is the trivial character
Edit: I'll rewrite this in a much simpler way - basically what MarkH wrote in the comment. What I wrote last night was extremely convoluted.
Since $\chi$ is irreducible, $\chi$ is one of your $\chi_r$'...
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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....
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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$$
as Sean Eberhard wrote in the comment. (note: some ...
- 2,561
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Is there some sort of irrep orthogonality for finite groups?
What you wrote isn't right, and my guess is that basically any example will break it. It can be fixed though by adding a transpose:
$$ \sum_{g \in G} \pi_1(g) \pi_2(g^{-1})^\top = 0$$
Let's go back ...
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Is there some sort of irrep orthogonality for finite groups?
Let's generalize. Suppose $(V_1,\pi_1)$ and $(V_2,\pi_2)$ are two irreps of $G$. Define
$$ T(M)=\frac{1}{|G|}\sum_{g\in G} \pi_1(g)M\pi_2(g^{-1}). \tag{$\color{green}{\triangle1}$}$$
Observe $T$ is a ...
- 2,467
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Character of a representation and dimensione of $G$-invariant space
It seems like the step that's giving you trouble is proving that $(\overline{\chi_V},\chi_V)=(\chi_0,\chi_V^2)$.
The calculation is:
\begin{align*}
(\overline{\chi_V},\chi_V) &= \frac{1}{|G|}\sum_{...
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Character of induced representation
Assuming that $\{t_{\lambda}\}$ is a set of coset representatives for $G/H$.
First of all, we need to verify that $\dot{\alpha}$ satisfies something like equation $(1)$, namely we want to check that $\...
- 862
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Character of restricted representation.
Your interpretation seems correct to me.
It also easily follows using the inner product.
If $\rho$ is a non-linear character then $[$$\rho_{\{e\}}$,$\rho_{\{e\}}$$]$$=\rho(e)^2>1$ and hence $\rho_{\...
- 638
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A question about hypothesis of Clifford's Theorem in Isaacs Character Theory book
The irreducibility of $\chi$ is necessary to assert that $\chi$ is a constituent of $\vartheta^G$. By Frobenius reciprocity we have that $0\not=[\chi_H,\vartheta]=[\chi,\vartheta^G]$. Then $\chi$ is a ...
- 638
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The set of characters
Consider $C$ be an ${}^*$-algebra (not $C^*$ yet) generated by $C_\infty$ and $\{e^{it}\}.$ $C = C_\infty \oplus P(e^{it}, e^{-it})$ where $P$ is polynomial algebra. Direct sum is viewed in category ...
- 1,085
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