# Tag Info

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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 ...

### 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 (...
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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 ...
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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 ...

### 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 ...
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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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I shouldn't have asked this question: turns out it is a straightforward computation if we decompose $\rho$ in terms of irreducible characters $\rho=\oplus_i n_i \rho_i$: $$\sum_i n_i^2=\left\langle \... 4 votes Accepted ### p-adic Norms and Character Values It is trivially correct when n = 1: the left side is 1 and the right side is |\chi(g)|_p^2, which is 1 since \chi(g) is a root of unity when n = 1. Is is never true for all g when n >... 4 votes Accepted ### Why do characters fail to characterize non-abelian LCH-groups? The short answer is that if G is a group and A an abelian group (like \mathbb T), then every group homomorphism \phi\colon G\to A factors through the abelianization G/[G,G]. Here [G,G] is ... 4 votes Accepted ### Formula for \chi(g)\chi(h), where \chi is an irreducible character of G Theorem Let \chi \in Irr(G), and x,y \in G, then$$\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 \... 3 votes Accepted ### 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 ... 3 votes Accepted ### 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 ... 3 votes ### 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 ... 3 votes ### 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\}. ... 2 votes Accepted ### 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.... 2 votes Accepted ### 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 ... 2 votes Accepted ### 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'... 2 votes Accepted ### 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.... 2 votes Accepted ### 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$$as Sean Eberhard wrote in the comment. (note: some ... 2 votes ### Is there some sort of irrep orthogonality for finite groups? What you wrote isn't right, and my guess is that basically any example will break it. It can be fixed though by adding a transpose:$$ \sum_{g \in G} \pi_1(g) \pi_2(g^{-1})^\top = 0$$Let's go back ... 2 votes Accepted ### Is there some sort of irrep orthogonality for finite groups? Let's generalize. Suppose (V_1,\pi_1) and (V_2,\pi_2) are two irreps of G. Define$$ 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 votes Accepted ### 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_{... 2 votes Accepted ### 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 \... 2 votes Accepted ### 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_{\...
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 ...
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 ...