Timeline for When is the torsion subgroup of an abelian group a direct summand?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22, 2022 at 8:13 | history | edited | CommunityBot |
replaced http://math.uga.edu/~pete with http://alpha.math.uga.edu/~pete
|
|
| Oct 28, 2021 at 20:15 | comment | added | Todd Trimble | @KConrad I guess it got moved around; here it is again: people.brandeis.edu/~igusa/Math131b/tsubgp.pdf. Although I might copy it and store it someplace. | |
| Oct 28, 2021 at 19:27 | comment | added | KConrad | @ToddTrimble your link to Igusa's pdf file is now broken. The story of the internet... | |
| Jan 12, 2019 at 12:12 | history | edited | Jeremy Rickard | CC BY-SA 4.0 |
fixed link
|
| S Dec 24, 2016 at 14:31 | history | suggested | Martin Sleziak | CC BY-SA 3.0 |
changed dead link http://www.math.uga.edu/* to working link http://math.uga.edu/* (the question has been bumped anyway)
|
| Dec 24, 2016 at 13:24 | review | Suggested edits | |||
| S Dec 24, 2016 at 14:31 | |||||
| Dec 3, 2013 at 2:25 | comment | added | Todd Trimble | In case someone sees Pete's example of a non-split torsion sequence but doesn't see the proof easily -- perhaps addressing a future version of myself? -- see Igusa's notes people.brandeis.edu/~igusa/Math101b/tsubgp.pdf example 17.6. | |
| Apr 4, 2011 at 20:21 | comment | added | Kevin Buzzard | It just gives a new way of looking at things, that's all. I guess the dual of your exact sequence is: $0\to D\to C\to S\to 0$ with $C$ a general compact abelian top group, $D$ the maximal divisible sub, and $S$ topologically finitely-generated and profinite, and you're asking whether it always splits... | |
| Apr 4, 2011 at 19:24 | comment | added | Pete L. Clark | @Kevin: yes, I know this, hence the (ad hoc) term "cofinite type". But does it actually help to solve the problem? | |
| Apr 4, 2011 at 18:37 | comment | added | Kevin Buzzard | [the bit I'm missing is "what does the Pont. dual of a torsion-free discrete group look like?"] | |
| Apr 4, 2011 at 18:36 | comment | added | Kevin Buzzard | Pete: you can translate your question into an equivalent one by taking Pontrjagin duals. Your discrete group $G$ becomes a compact group $C$, the torsion subgroup of cofinite type becomes a topologically finitely generated profinite group which is a quotient of $C$, and the torsion-free part is...umm...some sub of $C$ which I don't understand very well but perhaps google could help...maybe at least it gives you another way of thinking about the problem. | |
| Apr 4, 2011 at 11:54 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
added 4 characters in body
|
| Apr 4, 2011 at 11:52 | vote | accept | Pete L. Clark | ||
| Apr 4, 2011 at 11:41 | comment | added | Pete L. Clark | @S, @Alex: thanks; I changed this to what I really meant. | |
| Apr 4, 2011 at 11:41 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
deleted 118 characters in body
|
| Apr 4, 2011 at 10:55 | comment | added | Alex B. | In fact, by the same argument, the torsion subgroup seems to be uncountable, so very far from $\bigoplus\mathbb{Z}/p^n\mathbb{Z}$. | |
| Apr 4, 2011 at 9:31 | answer | added | Gjergji Zaimi | timeline score: 23 | |
| Apr 4, 2011 at 8:40 | comment | added | S. Carnahan♦ | I am having difficulty seeing why the subgroup in your nonsplit extension is in fact the torsion subgroup. Can't you have an element of order $p$ of the form $(1,p,p^2,\ldots)$? | |
| Apr 4, 2011 at 7:10 | history | asked | Pete L. Clark | CC BY-SA 2.5 |