Timeline for Generalized conjugacy classes in (topological) groups
Current License: CC BY-SA 4.0
17 events
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| Sep 25, 2023 at 19:28 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 25, 2023 at 8:45 | comment | added | YCor | In a discrete group $G$, $g,h$ are equivalent iff $g$, $h$ have the same order and the (possibly infinite) index of $\langle g\rangle$ and $\langle h\rangle$ is the same. (If $G$ is not virtually infinite cyclic, the latter condition can be dropped.) | |
| Sep 25, 2023 at 8:42 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 25, 2023 at 7:59 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 25, 2023 at 7:26 | comment | added | Ali Taghavi | @MoisheKohan please read my previous comment | |
| Sep 25, 2023 at 7:25 | comment | added | Ali Taghavi | @BenjaminSteinberg I wonder if the measurability of each algebraic conjugqcy class or topological conjugacy class is an obvious question | |
| Sep 24, 2023 at 15:30 | comment | added | Moishe Kohan | Most likely the answer is positive but I do not have a proof. | |
| Sep 24, 2023 at 14:53 | comment | added | Ali Taghavi | @BenjaminSteinberg What would be an apprpriate analogues of Caley theorem? On the other jand assume that $H$ is a closed subgroup of $G$. Assume that $a,b\in H$ are equivalent as elements of H are they equivalent as elements of $G$? this question is obvious for the algebraic conjugacy but what about topological congugacy? | |
| Sep 24, 2023 at 14:47 | comment | added | Ali Taghavi | @MoisheKohan I would appreciate if you give comment on the following question: | |
| Sep 24, 2023 at 11:33 | comment | added | Ali Taghavi | @BenjaminSteinberg In fact the order can a non natural number as in the case of circle $\theta$ can be count as an order of $e^{i\theta}$ in $S^1$ | |
| Sep 24, 2023 at 11:31 | comment | added | Ali Taghavi | @MoisheKohan Thank you for your correction | |
| Sep 22, 2023 at 22:49 | comment | added | Moishe Kohan | It is an equivalence relation and an equivalence class. | |
| Sep 22, 2023 at 12:57 | comment | added | Ali Taghavi | @BenjaminSteinberg Thank you for your interesting comment. Since "order" plays a crucial role so your comment is a motivation to consider a concept of order for elements of a topological group. The order is not a number but is an equivalent class. | |
| Sep 22, 2023 at 12:30 | comment | added | Benjamin Steinberg | The number of elements of order dividing n divides the order of the group but not the number of elements of order exactly n. | |
| Sep 22, 2023 at 12:26 | comment | added | Benjamin Steinberg | For a finite group two elements are equivalent if and only if they have the same order. This is because if g has order n then it acts as a disjoint union of |G|/n n-cycles corresponding to the cosets | |
| Sep 22, 2023 at 11:34 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 22, 2023 at 11:23 | history | asked | Ali Taghavi | CC BY-SA 4.0 |