Timeline for Group which "resembles" the free product of a cyclic group of order two and a cyclic group of order three, but isn't.
Current License: CC BY-SA 2.5
9 events
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| Oct 13, 2010 at 7:58 | comment | added | user6976 | @Vagabond: In fact they probably always generate $PSL(2,\mathbb Z)$ if $x,y,\iota$ are sufficiently independent. But see updates 2 and 3 in my answer: there are two matrices of sizes 13 with integer coeff. satisfying $a^2=b^3=1$ that generate the whole $SL(13,\mathbb Z)$. | |
| Oct 12, 2010 at 20:16 | history | edited | Vagabond | CC BY-SA 2.5 |
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| Oct 12, 2010 at 20:00 | comment | added | Vagabond | True. Now, How does one find a relation ? I think I have ended up asking the same question as you originally asked !! | |
| Oct 12, 2010 at 19:37 | comment | added | Todd Trimble | Well, I surely believe that you get the desired relations if $x$ and $y$ are algebraically independent, but Mark's objection would still need to be addressed. | |
| Oct 12, 2010 at 19:26 | comment | added | Vagabond | I am still wondering how to prove that the order of A.B is infinite in a neat way. There must be a way to describe the action in a simple geometric way and conclude . | |
| Oct 12, 2010 at 19:24 | history | edited | Vagabond | CC BY-SA 2.5 |
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| Oct 12, 2010 at 19:16 | comment | added | user6976 | These two matrices may generate a group isomorphic to $PSL(2,\mathbb Z)$ (at least for some $x,y,\iota$? | |
| Oct 12, 2010 at 19:03 | history | edited | Vagabond | CC BY-SA 2.5 |
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| Oct 12, 2010 at 18:27 | history | answered | Vagabond | CC BY-SA 2.5 |