Timeline for A presentation for $GL(2,\mathbb{Z}/p^n \mathbb{Z})$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| 2 days ago | comment | added | SPDR | $a^2\neq1$ according to what you have associated $a$ to. Also the relations $(ab^2ab^r)^2=(abab^r)^3=1$ do not match. | |
| Feb 20, 2016 at 12:52 | history | edited | user48096 | CC BY-SA 3.0 |
added 76 characters in body
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| Feb 20, 2016 at 12:48 | comment | added | user48096 | Oh right, I see. Thanks, I'll change that now. | |
| Feb 19, 2016 at 18:06 | comment | added | alpoge | [The point is that diag(x,1) = (word in a,b)*const implies, on taking determinants, that x is a square if 2 is a square.] | |
| Feb 19, 2016 at 16:22 | comment | added | user48096 | Does it matter if 2 is a square? I thought that the important part is that it generates the group $\mathbb{Z}/p \mathbb{Z}$? | |
| Feb 19, 2016 at 16:17 | history | edited | user48096 | CC BY-SA 3.0 |
added 2 characters in body
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| Feb 19, 2016 at 0:33 | comment | added | Joël | There is something wrong with your presentation. Since $a,b$ have determinant $2$ and $1$, all their powers have determinant a power of $2$. If $2$ is a square modulo $p$, this implies that $a,b$ is not a system of generator. | |
| Feb 18, 2016 at 21:09 | answer | added | ahulpke | timeline score: 2 | |
| Feb 18, 2016 at 19:44 | history | asked | user48096 | CC BY-SA 3.0 |