Timeline for Countability of conjugacy classes of closed subgroups
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Sep 30, 2020 at 14:37 | history | edited | Qayum Khan | CC BY-SA 4.0 |
simplified the basic step for the induction
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| Nov 27, 2019 at 21:01 | history | edited | Qayum Khan | CC BY-SA 4.0 |
added 4 characters in body
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| Nov 27, 2019 at 1:10 | history | edited | Qayum Khan | CC BY-SA 4.0 |
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| Nov 26, 2019 at 2:06 | history | edited | Qayum Khan | CC BY-SA 4.0 |
totally revised so as to incorporate the correction of the flaw pointed out by YCor and also to put the citation details at the end for easier reading
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| Nov 23, 2019 at 3:13 | history | edited | Qayum Khan | CC BY-SA 4.0 |
added 176 characters in body
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| Nov 23, 2019 at 2:49 | history | edited | Qayum Khan | CC BY-SA 4.0 |
deleted 115 characters in body
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| Nov 23, 2019 at 1:30 | history | edited | Qayum Khan | CC BY-SA 4.0 |
added 110 characters in body
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| Nov 23, 2019 at 0:26 | history | edited | Qayum Khan | CC BY-SA 4.0 |
added 13 characters in body
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| Nov 22, 2019 at 18:17 | history | edited | Qayum Khan | CC BY-SA 4.0 |
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| Nov 21, 2019 at 0:41 | history | edited | Qayum Khan | CC BY-SA 4.0 |
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| Nov 20, 2019 at 6:15 | history | edited | Qayum Khan | CC BY-SA 4.0 |
added 3 characters in body
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| Nov 20, 2019 at 6:05 | history | edited | Qayum Khan | CC BY-SA 4.0 |
I corrected that YCor is not the original poster.
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| Nov 20, 2019 at 5:58 | comment | added | Qayum Khan | @YCor : I've added an edit to address your two comments. I hope that my amended answer is satisfying from a purely classical point of view. | |
| Nov 20, 2019 at 5:53 | history | edited | Qayum Khan | CC BY-SA 4.0 |
"3 Proof" is corrected in response to the two comments of the original poster on this answer
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| Nov 19, 2019 at 10:21 | comment | added | YCor | Probably you can prove with such an argument that there exists a sequence of closed proper subgroups $(U_n)$ such that every proper subgroup is conjugate into one of the $U_n$, and then the argument works as soon as you know that there is no properly decreasing sequence of closed subgroups. | |
| Nov 19, 2019 at 10:19 | comment | added | YCor | The statement "There exists $n$ and proper closed subgroups $H_1,\dots,H_n$ of $G$ such that $CptSgp(K)\subset U_{H_1}\cup\dots\cup U_{H_n}$ is clearly false (since $K$ is missing), but even if you mean $CptSgp(K)\subset U_{H_1}\cup\dots\cup U_{H_n}\cup\{K\}$, it's false when $K$ is the circle group. (I'm also confused you define $U(H)$ and then denote $U_H$.) | |
| Nov 19, 2019 at 5:34 | history | edited | Qayum Khan | CC BY-SA 4.0 |
added Item 1 and renumbered accordingly
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| Nov 19, 2019 at 5:10 | history | edited | Qayum Khan | CC BY-SA 4.0 |
changed "G" to "K' in the first paragraph, for notation consistent with the rest of the answer
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| Nov 19, 2019 at 5:05 | history | edited | Qayum Khan | CC BY-SA 4.0 |
changed "G" to "K' in the first paragraph, for notation consistent with the rest of the answer
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| Nov 19, 2019 at 4:47 | history | answered | Qayum Khan | CC BY-SA 4.0 |