25
votes
On permanents and determinants of finite groups
I won't discuss the permanent part of the question, but I think the other part can be done easily, even without Galois theory. Let $X = X(G)$ denote the character table of $G$ (rows indexed by complex ...
- 39.9k
21
votes
The character table of the symmetric group modulo m
This is true when $m$ is prime and false in general.
Counterexample. Take $S_8$ with $m=6$. Computer calculations show that the $\mathbb{Z}$-rank of the character table of $S_8$ with entries taken ...
- 10.4k
21
votes
The character table of the symmetric group modulo m
When $m$ is prime there is a simpler proof. The Smith normal form of the character table of $S_n$ is computed at Problem 14 here (solution here). From this it follows that the rank of the character ...
- 45k
18
votes
Accepted
Explicit version of the Burgess theorem
There are now at least two instances of such an explicit result.
Theorem 1.1 of Bordignon: https://arxiv.org/abs/2001.05114.
Theorem 1.1 (and Corollary 1.2) of Jain-Sharma, Khale, and Liu: https://...
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17
votes
Accepted
What did Frobenius prove about $M_{12}$?
It seems to me that Frobenius is using lots of specific facts about the permutation group $M_{12}$ (and $M_{24}$, respectively) here, and that he has no doubts about the existence of these groups. In ...
- 6,522
16
votes
Accepted
Tannakian Formalism for the Quaternions and Dihedral Group
Let $V_D$ and $V_Q$ be the two dimensional simple representations of $D_4$ and $Q_8$ respectively. Let $1_D$ and $1_Q$ denote their trivial representations.
Suppose that there is a tensor equivalence ...
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16
votes
What did Frobenius prove about $M_{12}$?
This summary of Frobenius' 1904 paper might be of use:
Thomas Hawkins, The Mathematics of Frobenius in Context (page 527-528).
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15
votes
Accepted
Conductor as volume of the integers ring
Apply the Fourier Inversion Formula to the characteristic function $\Phi(x) = \chi_\mathcal{O}(x)$ of the ring $\mathcal{O}$ of integers in $F$. The Fourier transform is the integral $\widehat{\Phi }(...
- 11k
14
votes
A series of conjectures on $\sum_{x=0}^{(p-1)/2}(\frac{x^5+cx^3+dx}p)$ (III)
Here is a proof of (i):
Since the relevant primes $p$ are $\equiv 1 \bmod 4$, we have
$S_p(c,d) = \frac{1}{2} T_p(c,d)$, where
$$ T_p(c,d) = \sum_{x=0}^{p-1} \left(\frac{x^5+cx^3+dx}{p}\right) \,. $$
...
- 10.7k
13
votes
Accepted
The finite groups with a zero entry in each column of its character table (except the first one)
Partial answer: the finite group $G$ is clearly in this class if it has a $p$-block of defect zero for every prime $p$ which divides $|G|$. This is a sufficient condition which may not be necessary. ...
- 39.9k
13
votes
Accepted
Finite groups with integral character table
There is no complete classification, but some structural results are known. To give you something to search for: such groups are called $\mathbb{Q}$-groups. There is a whole book devoted to their ...
- 12.4k
12
votes
Accepted
A sum over characters of the symmetric group
Since $\chi_\theta(\mu\cup\omega)=\langle s_\theta,p_\mu p_\omega\rangle$, your sum is given by
$$ \frac{1}{n!}\sum_{\mu\vdash n}
C_\mu\chi_\lambda(\mu)\langle s_\theta,p_\mu p_\omega\rangle =
...
- 45k
12
votes
Tannakian Formalism for the Quaternions and Dihedral Group
The categories ${\rm Rep}(Q_8)$ and ${\rm Rep}(D_8)$ are not equivalent as tensor categories. They have the same Grothendieck ring, but they have non equivalent associators. As far as I am aware, it ...
- 3,594
12
votes
Accepted
Is a $G$-invariant character $\theta$ of $H$ extendible to $G$?
The answer is yes (and some of the comments were moving in the right direction): Let $T$ be a transversal to $H$ in $G,$ and let $\sigma$ afford the representation of $H.$ For each $t \in T,$ there is ...
- 39.9k
11
votes
Accepted
Beyond Brauer's theorem
Jeremy Rickard's answer is perfectly correct, of course. It is also easy to see (by an argument close to Taketa's) that if $G$ is a finite simple group, and $\chi$ is a faithful (not assumed ...
- 39.9k
11
votes
Accepted
On the structure of a finite group of order $144$
Since $G$ has an irreducible character of degree $9 = |G|_{3},$ we have $O_{3}(G) = 1,$ so $G$ has more than one Sylow $3$-subgroup. If $G$ has only $4$ Sylow $3$-subgroups, then $G$ has a normal ...
- 39.9k
11
votes
Closed formulas for the character of the symmetric group
Giving explicit formulas for the characters is the content of the recent article "An explicit formula for the characters of the symmetric group" by
Michel Lassalle :
https://link.springer.com/...
- 22.4k
11
votes
The sum (with multiplicity) of the cubes of irreducible character degrees of a finite group
Here are a few remarks which may be helpful. I may be able to say more later. The condition that $\operatorname{cp}(G) > \frac{1}{4}$ already severely restricts the possibilities for $G$. If $G$ ...
- 39.9k
11
votes
Specific application of Cauchy-Schwarz and Large Sieve
The answer of Ofir Gorodetsky is perfectly fine, but one can also apply the Cauchy-Schwarz inequality for $L^2$ spaces directly.
Indeed, let us consider the $L^2$ space of functions on the set of ...
- 86.4k
10
votes
Beyond Brauer's theorem
Here is another way to think about this, where we consider nonnegative $real$ coefficients, and we do not restrict ourselves to integer coefficients as in Geoff Robinson's post.
If $\chi \in {\rm Irr}...
- 5,989
10
votes
Accepted
Extension of a formula for the quadratic Gauss sums
No, the relation is not as simple in this case. For example, if $k = 3$ and $p = 7$, the three different cubic Gauss sums are roots of $y^{3} - 21y - 7$, and the three roots of this polynomial do not ...
- 18k
10
votes
Accepted
About the existence of characters on $B(X)$
I guess that you mean that $B(H)$ has no character (=continuous unital algebra homomorphism into $\mathbf{C}$) if $H$ has dimension $\neq 1$ (idem for $M_n(\mathbf{C})$ for $n\neq 1$), and thus that ...
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10
votes
Accepted
Are the character degrees determined by the conjugacy class sizes?
SmallGroup(128,227) and SmallGroup(128,731)) are counterexamples.
...
- 30.2k
10
votes
Accepted
Pólya–Vinogradov inequality for Eisenstein integers
No.
Such a bound would imply a similar bound on
$$\displaystyle \left \lvert \sum_{N(z) = M} \left(\frac{z}{w} \right)_3 \right \rvert.$$
If $M$ is a product of distinct primes $p_1,\dots p_n$ ...
- 118k
9
votes
Accepted
Irreducible reps and characters of $G \rtimes A$
I might as well turn my comment into an answer. I will just write $GA$ for the semidirect product ( with the normal subgroup being $G$). Clifford's theorem outlines a procedure for computing the ...
- 39.9k
9
votes
Accepted
Relation of these two Dirichlet $L$-functions
Using the perspective of "pretentious" multiplicative number theory, one can see that these $L$ functions are correlated with each other through the Deuring-Heilbronn repulsion phenomenon, but there ...
- 91.7k
9
votes
About the existence of characters on $B(X)$
Examples were known before the Argyros-Haydon space mentioned in Yves Cornulier's answer. For instance, if $J$ denotes the James space, then the image of the canonical map $J\to J^{**}$ has ...
- 24k
9
votes
On permanents and determinants of finite groups
I'll address the remaining question 2.
Theorem: If $|G|=4r+2$ then the permanent of its character table vanishes.
Proof: We know $G$ has a normal subgroup $H$ of index $2$. Let's denote the cosets of $...
- 82.3k
8
votes
Accepted
A sum over characters of $S_{2n}$ and zonal spherical functions of $(S_{2n},H_n)$
The essential thing here is that the characters $\chi_{2\lambda}$ are exactly the irreducible constituents of the induced character $(1_{H_n})^{S_n}$. The result generalizes to an arbitrary subgroup $...
- 6,522
8
votes
Accepted
The lower bound of a group with characters of special degrees
Let me try to answer in some detail the case $p$. So $G$ is a finite group of minimal possible order subject to having a complex irreducible character $\chi$ of degree $p$, where $p$ is a chosen prime....
- 39.9k
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