Timeline for Product-one sets in non-commutative groups
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2021 at 10:04 | vote | accept | Taras Banakh | ||
| Jul 6, 2021 at 9:42 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added Problem 2.
|
| Jul 5, 2021 at 0:11 | answer | added | Taras Banakh | timeline score: 3 | |
| Jul 4, 2021 at 9:26 | answer | added | YCor | timeline score: 5 | |
| Jul 4, 2021 at 7:43 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Removed statements to a partial answer.
|
| Jul 4, 2021 at 7:41 | answer | added | Taras Banakh | timeline score: 0 | |
| Jul 4, 2021 at 7:36 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 1983 characters in body
|
| Jul 3, 2021 at 15:00 | comment | added | Taras Banakh | @YCor Thank you for the comment but this has been already done (with the same graph angument) in this answer mathoverflow.net/a/38997/61536 to the question of Gjergji Zaimi. | |
| Jul 3, 2021 at 13:44 | comment | added | Taras Banakh | @verret I added a Proposition explaining why we can assume that D contains no elements of order 2. | |
| Jul 3, 2021 at 13:43 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added Proposition
|
| Jul 3, 2021 at 12:24 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 228 characters in body
|
| Jul 3, 2021 at 7:27 | comment | added | YCor | Possibly you view it as trivial, but anyway a simple remark is that there exist $0<m\le |D|$ and $x_1,\dots,x_m\in S$ (possibly not distinct) such that $\prod x_i=1$. Indeed, make $D$ an oriented graph with $x\to y$ if $x\in yD$. By assumption for every $x$ there exists $y$ with $x\to y$. So there is an oriented simple loop of size $0<m\le |S|$: $y_0,\dots,y_{m-1}$, with $y_i=y_{i+1}x_i$ for some $x_i\in D$, $i$ modulo $m$. Hence $y_0=y_1x_0=y_2x_1x_0=\dots=y_{m-1}x_{m-2}\dots x_0=y_0x_{m-1}\dots x_0$, so $x_{m-1}\dots x_0=e$. | |
| Jul 3, 2021 at 7:04 | history | edited | YCor |
edited tags
|
|
| Jul 3, 2021 at 7:02 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added a theorem
|
| Jun 28, 2021 at 5:18 | comment | added | spin | @verret: I think I have a mistake in my argument. (Also in my previous comment, where I claim minimal $D$ has property $D \subseteq Dx \cup xD \cup \{x\}$) | |
| Jun 27, 2021 at 22:43 | comment | added | verret | @TarasBanakh Can you expand on that last comment? | |
| Jun 27, 2021 at 15:24 | comment | added | Taras Banakh | @spin We can also assume that $D$ contains no elements of order 2. | |
| Jun 27, 2021 at 9:23 | comment | added | spin | Few obvious comments, for finite $G$. For a minimal counterexample (or an inductive proof), we can assume that $1 \not\in D$ and that $D$ is not contained in any maximal subgroup of $G$. With minimality we can also assume that no nonempty proper subset of $D$ is decomposable, so for all $x \in D$, the set $D \setminus \{x\}$ is not decomposable. Then for all $x \in D$, we must have $D \subseteq Dx \cup xD \cup \{x\}$. | |
| Jun 27, 2021 at 7:54 | history | edited | Aaron Meyerowitz | CC BY-SA 4.0 |
deleted 7 characters in body
|
| Jun 26, 2021 at 23:08 | history | edited | Aaron Meyerowitz | CC BY-SA 4.0 |
added 7 characters in body
|
| Jun 26, 2021 at 21:11 | comment | added | Joseph Van Name | Suppose $G=\text{Sym}(X)$ for some finite $X$ and $d$ is the metric on $G$ defined by $d(f,g)=|\{x\in X\mid f(x)\neq g(x)\}|$, and $L$ is the loss function where $L(D)=\sum_{g\in D}d(g,DD)$. Then $L$ measures how close $D$ is to being decomposable and $L(D)=0$ iff $D$ is decomposable. I have therefore tried to find decomposable subsets of $\text{Sym}(X)$ simply by minimizing $L(D)$ using evolutionary algorithms and artificial intelligence, but all my examples were trivial in the sense that they always had subsets of the form $\{x,x^{-1}\}$ where $x^{2}\neq e$ or $x=e$. | |
| Jun 26, 2021 at 17:00 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 7 characters in body
|
| Jun 26, 2021 at 15:02 | history | asked | Taras Banakh | CC BY-SA 4.0 |