All Questions
Tagged with gr.group-theory noncommutative-algebra
21 questions
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 ...
- 3,736
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 ...
- 63.9k
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 ...
- 923
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 $...
- 1,755
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 ...
- 2,388
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{\...
- 44.4k
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 ...
- 1,547
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 ...
- 479
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, ...
- 18.7k
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}...
- 2,160
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 ...
- 1,028
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 $...
- 3,948
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 ...
- 63.9k
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 $...
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 $A_1$, ...
- 2,837
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 ...
- 10.7k
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 ...
CommunityBot
- 1
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 \...
- 38.6k
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,...
- 161
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, ...
- 37.4k
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:))) $n,k,l,s\in\mathbb N$?
CommunityBot
- 1