Timeline for Quotients of finitely generated nilpotent groups
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2019 at 23:27 | comment | added | YCor | Unfortunately the question hasn't yet been solved (and doesn't seem out of reach) but an answer was accepted. | |
| Nov 26, 2019 at 19:48 | history | edited | darij grinberg | CC BY-SA 4.0 |
edited body
|
| May 30, 2015 at 13:10 | vote | accept | Miel Sharf | ||
| May 29, 2015 at 20:48 | history | edited | YCor | CC BY-SA 3.0 |
for clarification, edited to remove the statement that Woess claimed the stronger statement
|
| May 29, 2015 at 19:16 | answer | added | Arturo Magidin | timeline score: 4 | |
| May 29, 2015 at 10:00 | history | edited | Miel Sharf | CC BY-SA 3.0 |
added 12 characters in body
|
| May 29, 2015 at 9:28 | history | edited | Miel Sharf | CC BY-SA 3.0 |
Making this a question instead of a reference request
|
| May 29, 2015 at 8:59 | vote | accept | Miel Sharf | ||
| May 29, 2015 at 9:23 | |||||
| May 29, 2015 at 8:44 | review | Close votes | |||
| May 29, 2015 at 12:27 | |||||
| May 29, 2015 at 8:25 | comment | added | Derek Holt | Ah well, I shall stop thinking about the original question now, but I would still be interested to know whether it is correct! I played with a few examples, and I could not find any easily described procedure for finding the torsion-free subgroup with the calimed properties. | |
| May 29, 2015 at 8:22 | answer | added | Alireza Abdollahi | timeline score: 5 | |
| May 29, 2015 at 8:10 | history | edited | Miel Sharf | CC BY-SA 3.0 |
Clarification of my own stupidty
|
| May 29, 2015 at 7:26 | comment | added | Alireza Abdollahi | @MielSharf: Could you please let us know on which page of Woess' article your claim is mentioned (or is used)? For upper central series is true. As every f.g. nilpotent group has a torsion-free nilpotent subgroup of finite index. | |
| May 29, 2015 at 6:38 | comment | added | Arturo Magidin | @DaveWitteMorris: I'm not sure I follow what you are writing... are you reversing the roles of $N$ (original group) and $H$ (group we are looking for)? If we take the original group to be the integral Heisenberg group, then since $N/N_2\cong \mathbb{Z}^2$, $N_2/N_3\cong \mathbb{Z}$, and $N_3=\{e\}$, we can just take $H=N^1=N$. How did I get $H^2$ and $H^4$? | |
| May 29, 2015 at 6:03 | comment | added | Dave Witte Morris | @ArturoMagidin: The solution requires more care than that. The OP's condition is satisfied for the integral Heisenberg group $H$. However, for $N = H^k$, the quotient $N/[N,N]$ contains the torsion subgroup $[H,H]^k/[H^k,H^k]$ of order $k$. Therefore, an example where your suggestion does not work is $H^2$. Your suggestion would replace this with $(H^2)^2 = H^4$, but that makes the situation worse, not better. | |
| May 29, 2015 at 3:23 | comment | added | Arturo Magidin | Is there some reason that you cannot just take $N^k$, where $k$ is a common multiple of the exponents of the torsion subgroups of all $N_i/N_{i+1}$ ? | |
| May 28, 2015 at 20:24 | comment | added | YCor | here's the MathSE link: math.stackexchange.com/questions/1301400/… | |
| May 28, 2015 at 20:00 | history | edited | Miel Sharf | CC BY-SA 3.0 |
More accurate description
|
| May 28, 2015 at 19:59 | comment | added | Arturo Magidin | Note that what you call "its central series" is usually called the lower central series. | |
| May 28, 2015 at 19:24 | comment | added | Miel Sharf | I have asked this question in math.se, but failed to get an answer, which I do need urgently. Thanks for the help! | |
| May 28, 2015 at 19:24 | history | asked | Miel Sharf | CC BY-SA 3.0 |