Timeline for Zero divisors with support of size 3 in group algebras of finite groups
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Sep 1, 2015 at 10:19 | answer | added | Geoff Robinson | timeline score: 6 | |
| Aug 31, 2015 at 15:33 | comment | added | Ilya Bogdanov | If such elements exist for some group, they also exist for every overgroup. So it is more reasonable to make the least prime divisor of $|G|$ large. | |
| Aug 31, 2015 at 14:23 | comment | added | Alireza Abdollahi | So maybe a question as follows is interesting: Is there a function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that there is no finite group $G$ of order coprime to 3 and size greater than $f(n)$ and no field $\mathbb{F}$ such that the group algebra $\mathbb{F}[G]$ contains elements $\alpha$ and $\beta$ with the following properties: $\alpha \beta=0$, $|supp(\alpha)|=3$, $|supp(\beta)|=n$? | |
| Aug 31, 2015 at 13:58 | comment | added | Andreas Thom | Yes exactly, so why not assume something on the other factor as well, like that its support is much smaller than $|G|$ or not related to the torsion in $G$. | |
| Aug 31, 2015 at 13:51 | comment | added | Alireza Abdollahi | @AndreasThom: Another motivation to propose the question is the question of the existence of zero divisors with support of size 3 in the group algebra of torsion-free groups which are residually finite. The number ``3" is the first unsettled. | |
| Aug 31, 2015 at 9:02 | comment | added | Andreas Thom | This is a very legitimate question, however, you should assume that the other factor (in the factorization of zero) also has support-size coprime to the order of the group. | |
| Aug 31, 2015 at 5:42 | vote | accept | Alireza Abdollahi | ||
| Aug 30, 2015 at 10:28 | answer | added | Ilya Bogdanov | timeline score: 11 | |
| Aug 30, 2015 at 10:12 | comment | added | Ilya Bogdanov | One of the definitions of a group algebra regards its elements as functions from $G$ to $\mathbb F$. | |
| Aug 30, 2015 at 8:56 | comment | added | Stefan Kohl♦ | @GabrielC.Drummond-Cole: I'd think the support of an element of $\mathbb{F}[G]$ is meant to be the set of group elements whose coefficient is $\neq 0$ -- but the notation $\alpha(x) \neq 0$ looks also confusing to me. | |
| Aug 30, 2015 at 8:32 | comment | added | Gabriel C. Drummond-Cole | Why isn't the support of every element all of $G$? Aren't elements of $G$ units in $\mathbb{F}[G]$? Do I misunderstand the notation $\alpha(x)$? | |
| Aug 30, 2015 at 8:04 | history | asked | Alireza Abdollahi | CC BY-SA 3.0 |