Timeline for Does there exists a group structure on $\circ$ on $(\mathbb{R},\circ)$ such that $(\mathbb{R},\circ)$ is non-isomorphic to $(\mathbb{R},+)$?
Current License: CC BY-SA 4.0
17 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Oct 7, 2018 at 21:52 | comment | added | YCor | Related: mathoverflow.net/questions/312177/… | |
| Aug 26, 2018 at 5:09 | history | edited | Martin Sleziak |
Removed the deprecated (abstract-algebra) tag - see the tag info: https://mathoverflow.net/tags/abstract-algebra/info (if there are some other suitable tags, choose them instead.)
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| Aug 25, 2018 at 16:26 | vote | accept | CommunityBot | ||
| Aug 25, 2018 at 14:27 | answer | added | YCor | timeline score: 32 | |
| Aug 25, 2018 at 13:26 | comment | added | Uri Bader | @YCor yes, my first "index" should have been "intersection". Further, I underestimated $\mathbb{Z}$: the solution of the analogue question there is non-trivial as you can permute the primes (and signs), I think. | |
| Aug 25, 2018 at 13:23 | comment | added | YCor | @UriBader I think there's a typo or missing word ("intersection"?) in your last comment? | |
| Aug 25, 2018 at 13:03 | comment | added | Uri Bader | The index of a maximal subgroup in $\mathbb{Z}^2$ with each cyclic group $C$ is either $C$ or an index $p$ subgroup for some prime. So you can track the index, as you do know $C$ as a group. | |
| Aug 25, 2018 at 12:59 | comment | added | YCor | @UriBader yes it should be doable; the index should indeed be preserved (in this setting) although this requires some argument. | |
| Aug 25, 2018 at 12:54 | comment | added | Uri Bader | For the record: it is enough to show that $\circ$ and $+$ coinside on 2-generated groups, these are isomorphic either to $\mathbb{Z}$ or $\mathbb{Z}^2$ in $(\mathbb{R},+)$. Thus it is enough to solve the analogue question for $\mathbb{Z}$ and $\mathbb{Z}^2$. The first one is very easy. I expect the second one to be easy as well. | |
| Aug 25, 2018 at 12:46 | comment | added | YCor | Still the group will be torsion-free, the notion of cyclic subgroup is preserved. One indeed needs to characterize $Z^2$ by its lattice of subgroups. | |
| Aug 25, 2018 at 12:44 | comment | added | YCor | @UriBader but the index is not a priori preserved | |
| Aug 25, 2018 at 12:41 | comment | added | Uri Bader | the $\circ$ group generated by each element is isomorphic to the + group generated by it, as Z is the unique group having a unique subgroup of each index. A similar argument for Z^2 solves the problem. | |
| Aug 25, 2018 at 12:40 | comment | added | Derek Holt | I deleted the comment because I was assuming that the other group was abelian. | |
| Aug 25, 2018 at 12:39 | comment | added | YCor | @DerekHolt this is my expectation and I can check it when $\circ$ is assumed to be abelian. | |
| Aug 25, 2018 at 12:36 | comment | added | YCor | The second condition is equivalent to the requirement that subgroups are the same for both group laws. | |
| Aug 25, 2018 at 12:23 | history | asked | user57432 | CC BY-SA 4.0 |