Timeline for $G=\langle a\rangle H$ for subgroup $H$ [closed]
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16, 2013 at 23:49 | comment | added | Venkataramana | It is unfortunate that this was closed. I remember in my graduate student days that the same question for a higher rank lattice, say, a lattice in $SL_3({\mathbb R})$ was an unsolved problem (and was attributed to Robert Zimmer). | |
| Feb 16, 2013 at 20:42 | history | closed |
user10534 Martin Brandenburg user6976 HJRW Ian Agol |
too localized | |
| Feb 16, 2013 at 12:53 | comment | added | Derek Holt | Since this site is intended to be for research level problems, you should really ask questions like this on math.stackexchange | |
| Feb 16, 2013 at 12:05 | answer | added | Geoff Robinson | timeline score: 5 | |
| Feb 16, 2013 at 6:40 | comment | added | Victor | Well, about Todd-Coxeter -- you know there is famous Todd-Coxeter enumeration procedure to find out the coset reps for f.p. groups wrt to their f.g. subgroups. Can we draw from there when we can find the coset reps forming a subgroup? | |
| Feb 16, 2013 at 6:05 | comment | added | Mariano Suárez-Álvarez | It is not clear how to make sense of the question in your comment, victor. | |
| Feb 16, 2013 at 5:59 | comment | added | Mariano Suárez-Álvarez | (Finite groups with a unique maximal proper subgroup are all cyclic, so Jack's examples are infinite) | |
| Feb 16, 2013 at 5:58 | comment | added | Victor | Ok, can I ask then -- when Todd-Coxeter gives us a subgroup? | |
| Feb 16, 2013 at 5:56 | comment | added | Jack Huizenga | More generally, this is obviously false for any non-cyclic group with a unique maximal proper subgroup. | |
| Feb 16, 2013 at 5:55 | comment | added | Mariano Suárez-Álvarez | You should reserve your death for something of more moment. :-) | |
| Feb 16, 2013 at 5:54 | comment | added | Victor | You are right! Please kill me somebody! | |
| Feb 16, 2013 at 5:51 | comment | added | Mariano Suárez-Álvarez | Assuming you do want $H$ to be proper (otherwise this is silly), no: take $G=\mathbb Z/4\mathbb Z$ and $a$ the class of $2$. | |
| Feb 16, 2013 at 5:48 | history | asked | Victor | CC BY-SA 3.0 |