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 ...

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 ...

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 ...

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://...

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 ...

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 ...

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).

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 }(...

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
$$ T_p(c,d) = \sum_{x=0}^{p-1} \left(\frac{x^5+cx^3+dx}{p}\right) \,. $$
...

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. ...

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
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 =
...

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 ...

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 ...

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 ...

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 ...

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/...

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 $G$ ...

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 ...

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}...

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 ...

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 ...

10
votes

Accepted

### Are the character degrees determined by the conjugacy class sizes?

SmallGroup(128,227) and SmallGroup(128,731)) are counterexamples.
...

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 $p_1,\dots p_n$ ...

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 ...

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 ...

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 ...

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 $...

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 $...

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....

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