All Questions
30 questions
4
votes
1
answer
163
views
Examples of Noetherian integral group ring
I need to study the integral group ring of the fundamental group of a manifold. My knowledge of group and ring theory is very limited. I am looking for some examples of groups $G$ for which $\Bbb ZG$ ...
4
votes
2
answers
412
views
Units of the group algebra of a free group
Let $K$ be a field of characteristic zero and $F_n$ be a free group of rank $n$. What is known about the group of units $K[F_n]^\times$?
In the case of $n=1$, there are only trivial units: $K[F_1]^\...
1
vote
0
answers
74
views
The influence of the derived subgroup of the unit group of a group algebra
Let $FG$ be a group algebra in which $K$ is a field and $G$ is a group. Suppose that every element in the derived subgroup $\mathcal{U}(FG)'$ of the unit group $\mathcal{U}(FG)$ of $FG$ is a ...
2
votes
1
answer
329
views
Group rings of free abelian groups
Is it true that for free abelian finitely generated groups $G_1$ and $G_2$, if $\mathbb Z[G_1]\simeq \mathbb Z[G_2]$, then $G_1\simeq G_2$? If yes, is there any reference to such a fact?
12
votes
1
answer
312
views
Group ring with infinite stable rank
In searching for a counterexample in homological stability, I came across the following question:
Is there a known example of a finitely presented group $G$, so that the group ring $\mathbb{Z}[G]$ ...
9
votes
1
answer
507
views
When is the augmentation ideal projective as RG-module?
Let $G$ be a finite group and let $R$ be a commutative ring.
I'd like to ask, if there is a theorem of the following kind:
The augmentation ideal $I_G$ is projective as RG-module, if and only if ... ?...
9
votes
1
answer
580
views
Units of group algebra of dihedral group
Question:
Can we fully describe the group of units (=invertible elements) $(KG)^\times$ of the group algebra $KG$ for $K=\mathbf{F}_2$, $G=D_\infty=\langle s,t|s^2=t^2=1\rangle$, the infinite ...
4
votes
0
answers
158
views
Is the group ring of an amenable group, viewed as multiplicative monoid, amenable?
Motivated by this question, it seems natural to ask the following:
Question 1: Is there a [finitely generated discrete] torsion-free virtually Abelian (but not Abelian) group $G$ so that the ...
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, ...
3
votes
0
answers
180
views
Units in group rings
Let $F$ be any field with $p$ elements and $G$ be any finite $p$-group, combining together they form a group ring $FG$. And $V(FG)$ denotes group of units of coefficient-sum equal to 1 in $FG$. We ...
2
votes
0
answers
238
views
Flat augmentation ideal of a group-ring
If $G$ is a group and $I$ the augmentation ideal $I=Ker(\mathbb{Z}G\rightarrow \mathbb{Z})$ suppose that:
$I$ is a flat (right) $\mathbb{Z}G$-module.
$I$ is a finitely generated (right) $\mathbb{Z}G$...
3
votes
1
answer
489
views
augmentation ideal is always finitely generated?
$G$ is a finitely presented group (but not a finite group), and $\mathbb{Z}G$ is the corresponding group ring.
$I$ is the kernel of the augmentation morphism $\mathbb{Z}G\rightarrow \mathbb{Z}$.
Is $...
4
votes
1
answer
199
views
Groups for which all projections of $C^*_{\text{red}}G$ belong to $\mathbb{C}G$
Revision: According to comment of Wojowu we give a complete revise for this post.
A group $G$ is a pr-group if all projections of $C^*_{\text{red}} G$ are contained in its dense subalgebra $\mathbb{...
0
votes
1
answer
185
views
A subset (or subgroup) associated to a group
Edit: According to comment conversations we revise the question.
Let $G$ be a group. We consider the following subset of $G$:
$$\{g\in G \mid e^{\lambda_g} \in \mathbb{C}\lambda (G)\},$$
where $\...
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 $...
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 ...
3
votes
0
answers
109
views
Does this element belong to $\mathbb CG$?
Let $G$ be a torsion-free group. Let $\alpha$ be a symmetric element of $\mathbb CG$, i.e. $\alpha^*=\alpha$, with $\|\alpha\|_1=\sum|\alpha(g)|<1$, so $\beta:=\sum_{n\ge 0}(-1)^n\alpha^n$ is an ...
5
votes
0
answers
296
views
Are these element in a group algebra of a torsion-free group zero divisors?
Let $G$ be an arbitrary torsion-free group. For $x,y\in G$, which of these elements can be decided immediately not to be zero divisors in $\mathbb ZG$ (or in $\mathbb CG$)?
$$1+x+y,\quad 4+x+x^{-1}+y+...
38
votes
0
answers
1k
views
Groups whose complex irreducible representations are finite dimensional
By a complex irreducible representation of a group $G$, I mean a simple $\mathbb CG$-module. So my representations need not be unitary and we are working in the purely algebraic setting.
It is easy ...
14
votes
1
answer
581
views
Group rings such that every (countably generated) module has a maximal submodule
Every (non-zero) finitely generated module over a ring has a maximal proper submodule by a simple application of Zorn's lemma.
I am interested in the following question, with two variants.
...
7
votes
2
answers
485
views
Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring
Just recently I've stumbled across Warren Dicks' book Groups, trees and projective modules (1980) and I was pretty stunned. I know nothing of group cohomology, but I gather the "tree" ...
4
votes
1
answer
154
views
Right reversibility of submonoids of nilpotent groups
Let $G$ be a finitely generated group (optionally torsion-free). Let $N$ be a submonoid of $G$ (that is, a subsemigroup with $1$).
A (cancellative) monoid/semigroup $S$ is right reversible if for ...
6
votes
2
answers
483
views
In the group ring $\mathbb{Z}_p [G]$, what elements satisfy $(\sum a_g g)^p = \sum a_g g^p$?
Here $\mathbb{Z}_p$ is the ring of integers in $\mathbb{Q}_p$.
Preferably I would want to know this for a general group $G$, but I have been concentrating on the case $G = (\mathbb{Z} / p^n \mathbb{Z}...
6
votes
2
answers
210
views
groupring morphisms and bialgebra
Let $G_{1}$ and $G_{2}$ be two groups. Suppose that we have a morphism $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}] $ of bialgebras is it true that this morphism comes from a morphism of groups $G_{...
1
vote
2
answers
613
views
Jacobson radical of group algebra
For a finite group G and a finite field $\mathbb{F}_p$ of characteristic $p$,
J($\mathbb{F}_{p^k} \otimes_{\mathbb{F}_p }\mathbb{F}_p G ) = J(\mathbb{F}_{p^k}G)$? where $J(\mathbb{F}_{p^k}G)$ is the ...
11
votes
0
answers
400
views
Detecting a module for the free group algebra on a finite quotient
Let $F_2$ be the free group on two generators $x,y$ and let $R$ be the group algebra $\mathbf{Q}[F_2]$. Let $a,b,c$ be integers. Then define a right $R$-module
$M = R / (ax + by + c) R$.
I am ...
1
vote
0
answers
206
views
The normalizer problem for group rings
I recently studied about The Normalizer problem (NP) which states that given an integral group ring $\Bbb{Z}G$, $N_{\cal{U}}(G)=G\frak{z}$ where $\frak{z}$ denotes centre of $\cal{U}$ = $\cal{U}$$(\...
1
vote
1
answer
238
views
Why do we not lose any generality by proving it only for finitely generated groups [closed]
In the proof of following theorem, in a paper by Farkas-
Here $\Delta(G) = \{ g \in G : |G:C_G(g)| < \infty \}$ and $U_1(\mathbb{Z}G) $ is the set of normalized units of the integral group ring $\...
28
votes
1
answer
2k
views
Does GL_n(Z) have a noetherian group ring?
Has the (left, right, 2-sided) noetherian property of the integral group ring of arithmetic groups like $GL_n(Z)$ been considered in the literature?
Motivation: a recent trend has been to study "...
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 ...