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64 votes
4 answers
8k views

What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Let $K$ be a field and $G$ a group. The so called zero-divisor conjecture for group rings asserts that the group ring $K[G]$ is a domain if and only if $G$ is a torsion-free group. A couple of good ...
Johan Öinert's user avatar
15 votes
2 answers
1k views

A Non-Commutative Nullstellensatz

In studying presentations of pro-$p$-groups via generators and relations, one is led (via the so-called Magnus embedding) to questions involving power series in non-commuting variables. Results from ...
Cam McLeman's user avatar
  • 8,467
12 votes
3 answers
849 views

Subalgebra of a group algebra

Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra. Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$. Question: Is there any ...
Student's user avatar
  • 5,230
7 votes
1 answer
281 views

Question concerning the coefficients of block idempotents

Let $G$ be a finite group. Let $p$ be a prime number such that $p \mid |G|$. Let Irr$(G)$ denote the set of ordinary irreducible characters of $G$. For $\chi\in$ Irr$(G)$ define $e_{\chi} := \frac{\...
Bernhard Boehmler's user avatar
6 votes
1 answer
437 views

G = [G,G] with two generators

Is it true that groups $\langle a,b \mid a^n b^k=b^ka^{n+1}, b^la^s=a^sb^{l+1}\rangle$ are non-trivial for almost all (in any sense:))) $n,k,l,s\in\mathbb N$?
Nikita Kalinin's user avatar
5 votes
1 answer
132 views

Integral monoid rings and Ore conditions

Consider a cancellative monoid $S$ satisfying the left Ore condition, so it embeds in a group $G=S^{-1}S$. Consider also the integral monoid rings $\mathbb Z[S]$ and $\mathbb Z[G]$. I have two, ...
Simone Virili's user avatar
5 votes
1 answer
196 views

Expressing a element of a Matrix subgroup in terms of subgroup generators

I'm no (computational) algebraist, and my searches have been pretty unyielding (probably due to the vast amounts written on the key words), but perhaps someone may know if this is possible, and if so, ...
philiph's user avatar
  • 153
5 votes
1 answer
273 views

'Lie correspondence' for formal power series in non-commuting indeterminates

This is related to an earlier question of mine. I would like an argument or a reference (or a missing hypothesis if needed) for the following. Let $\mathbb{F}\langle\langle \alpha\rangle\rangle$ and ...
shane.orourke's user avatar
5 votes
0 answers
187 views

Any f.p. faithful simple module over a primitive group ring?

Recall that a ring $R$ is primitive if it has a faithful simple left module. Let $G$ be a countable discrete group and $R=\mathbb{k}G$, where $\mathbb{k}$ denotes some field or $\mathbb{Z}$. There ...
Jiang's user avatar
  • 1,528
4 votes
0 answers
164 views

non-abelian tensor products of several groups

R. Brown and J-L. Loday had defined the tensor product of two arbitrary groups acting on each other. Let $G,H$ be groups with actions on each other on the right. each group act on itself by ...
M masa's user avatar
  • 479
3 votes
1 answer
220 views

Intersection of Maximal Left Ideals with Finite Dimensional Quotient

Let $\Gamma$ be a finitely generated group and let $A=\mathbb{C}[\Gamma]$ be the corresponding group algebra over $\mathbb{C}$. Let $X$ be the set of all maximal left ideals of $A$ and let $X_0=\{I \...
Hans's user avatar
  • 3,031
3 votes
0 answers
317 views

How to go from primitive idempotents in $\text{End}_A(M)$ to primitive idempotents in $A$?

Let $K$ be a large enough finite field, let $A$ be a finite-dimensional $K$-algebra. Moreover, let $M$ be a finitely generated $A$-module and let $M = M_1\oplus ... \oplus M_n$ be a decomposition of $...
Bernhard Boehmler's user avatar
3 votes
0 answers
65 views

Intersections of generating sets of subalgebras

Let $A$ be a finitely generated, finitely presented, Noetherian, unital algebra over the complex numbers, which has no zero divisors. We do not assume that $A$ is commutative however. Moreover, let $...
Hans gluckmann's user avatar
3 votes
0 answers
115 views

The group of automorphisms and anti-automorphisms of the first Weyl algebra

Let $k$ be a field of characteristic zero, and let $A_1=A_1(k)$ be the first Weyl algebra. It is well known (first proved by Dixmier, if I am not wrong) that the group of automorphisms of $A_1$, ...
user237522's user avatar
  • 2,837
3 votes
0 answers
269 views

Computing the Abelianization of an Automorphism Group

Setup: We are working in a Henselian local ring $(R, \mathfrak m, k)$ that way may assume is Cohen-Macaulay, admits a canonical module and is of finite type (so is an isolated singularity). Let $M_1,...
Floresza's user avatar
  • 161
2 votes
1 answer
310 views

Flatness of submodules of free modules

Are submodules of free $\mathbb{Z}[G]$-modules flat? if not what conditions on $G$ makes it true? $G$ is an infinite group. If $\mathbb{Z}[G]$ is a Prüfer domain then this is true. Can a group ring $...
user127776's user avatar
  • 5,901
2 votes
0 answers
144 views

Zero divisors in the extra-special group algebra $\mathbb{R}[2^{1+6}_+]$

Can you characterize the unit-group of the real group-algebra of the extraspecial plus-type 2-group of order 128? (That is $\mathbb{R}[2_+^{1+6}]$ using Conway's notation.) (Please choose any irrep ...
Eric Downes's user avatar
2 votes
0 answers
103 views

Lattices with trivial coinvariants for finite groups

Let $G$ be a finite group. A $\mathbb{Z}G$-lattice is a $\mathbb{Z}G$-module that is (as abelian group) a free abelian group of finite rank. Question: Is there a finite group $G$ and a $\mathbb{Z}...
tj_'s user avatar
  • 2,160
1 vote
0 answers
88 views

Sequences generated from commuted quaternions and general commuted linear transformations

Given a pair of non-commuting linear transformations, $A$ and $B$, define the "next pair" in a sequence as $A*B$ and $B*A$. I am interested in finite cycles (i.e., the sequence eventually ...
bobuhito's user avatar
  • 1,547
0 votes
0 answers
58 views

An action on multiplicatively antisymmetric matrix

A matrix $ Q=(q_{ij})$ is called multiplicatively antisymmetric over a field $ F $ if $ q_{ii}=1 $ and $ q_{ij}={q_{ji}}^{-1} $.Let $ \mathcal{Q} $ be the set of all $ n \times n $ multiplicatively ...
Sky's user avatar
  • 923
0 votes
0 answers
308 views

Basic question about power series and complete group algebras

This is a pretty basic question, but I suspect it might be too exotic for math.stackexchange. Let $\mathbb{Z}_p$ be the $p$-adic integers. For free pro-$p$ group $F_r$ of rank $r$, we can consider ...
Will Chen's user avatar
  • 10.7k