Timeline for A free subgroup of GL(2,Z)?
Current License: CC BY-SA 2.5
10 events
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| Nov 22, 2010 at 0:32 | history | edited | user6976 | CC BY-SA 2.5 |
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| Nov 21, 2010 at 23:25 | comment | added | user6976 | @Nikita: It is in Kargapolov-Merzlyakov. | |
| Nov 21, 2010 at 22:24 | comment | added | Nikita Sidorov | Mark, thanks again! I didn't know about the Sanov subgroup. | |
| Nov 21, 2010 at 21:43 | comment | added | Kevin Buzzard | I see! The elements I was trying could also be used as part of a 2-element generating set for the group: the commutator is somehow "genuinely deeper in" so I can see the logic in your approach. | |
| Nov 21, 2010 at 21:36 | vote | accept | Nikita Sidorov | ||
| Nov 22, 2010 at 22:02 | |||||
| Nov 21, 2010 at 21:06 | comment | added | user6976 | @Kevin: I looked at the traces too. These are 1 and 2 - too small to generate a free group. I had done some computation with such groups in the past, and I usually look at the commutator and see if it is torsion first. I think that this is how Klimenko and Kopteva ruled out some cases in their paper as well (it might be even that I was the editor handling their paper in IJAC). | |
| Nov 21, 2010 at 21:03 | history | edited | user6976 | CC BY-SA 2.5 |
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| Nov 21, 2010 at 20:43 | comment | added | Nikita Sidorov | Mark, thanks! I have modified my question. | |
| Nov 21, 2010 at 20:43 | comment | added | Kevin Buzzard | Mark---how did you come to this answer? I am sure you know a lot more about this sort of thing than me. I just computed the char polys of $a^mb^n$ for some small values of $m$ and $n$ until I got lucky. I am sure there are far better ways to go about it! | |
| Nov 21, 2010 at 20:39 | history | answered | user6976 | CC BY-SA 2.5 |