Timeline for Can we construct an isomorphism between $\mathrm{BS}(1,n)$ and $\mathbb{Z}[1/n]\rtimes\mathbb{Z}$ such that it preserve the order?
Current License: CC BY-SA 4.0
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| 4 hours ago | history | edited | navashree chanania | CC BY-SA 4.0 |
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| 4 hours ago | comment | added | navashree chanania | yeah .sorry typing mistake. It's n. Thank you. | |
| 4 hours ago | comment | added | Corentin B | I guess $q=n$ (?) | |
| 4 hours ago | comment | added | navashree chanania | The elements of the group $\mathbb{Z}[1/n]\rtimes\mathbb{Z}$ are pairs $(r,m),$ where $r\in \mathbb{Z}[1/n]$ and $m\in\mathbb{Z}.$ The multiplication is defined as $(r,m)(r',m') = (r + q^m\cdot r',m + m').$ | |
| 4 hours ago | history | edited | navashree chanania | CC BY-SA 4.0 |
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| 5 hours ago | answer | added | Corentin B | timeline score: 1 | |
| 5 hours ago | comment | added | Corentin B | In order for your question to be well-posed, one still need to define how $\mathbb Z$ acts on $\mathbb Z[\frac1n]$: is $(0,1)(1,0)(0,1)^{-1}=(n,0)$ or $(\frac1n,0)$? | |
| 6 hours ago | comment | added | navashree chanania | In $\mathbb{Z}[1/n]\rtimes\mathbb{Z}$ , $b^2$ is greater than $b$, but i am not sure in BS(1,n). | |
| 7 hours ago | comment | added | HenrikRüping | Is $b^2$ bigger than $b$ in the order on $BS(1,n)$ ? | |
| 8 hours ago | history | edited | navashree chanania | CC BY-SA 4.0 |
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| 9 hours ago | comment | added | YCor | Your last sentence seems to be truncated. You need to specify orders on both groups for the question to make sense. | |
| 9 hours ago | history | edited | YCor | CC BY-SA 4.0 |
fixed multiple typos
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| 9 hours ago | history | edited | navashree chanania | CC BY-SA 4.0 |
added 4 characters in body; edited title
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| 9 hours ago | history | asked | navashree chanania | CC BY-SA 4.0 |