Timeline for On the symmetric group of 2^n elements
Current License: CC BY-SA 3.0
17 events
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| Apr 12, 2016 at 6:54 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 12, 2016 at 4:45 | answer | added | Amritanshu Prasad | timeline score: 3 | |
| Apr 11, 2016 at 10:16 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 11, 2016 at 9:43 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 11, 2016 at 7:42 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 11, 2016 at 5:29 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 11, 2016 at 5:18 | comment | added | DLN | @S. Carnahan, for any $ x\in X_k^n $, $ \phi_k(x) $ has $ 2^{k-1} $ elements and $ \phi_k(x) $ is a subset of $ X_1^n $. So $ X_k^n=\phi $ for $ k>n+1 $. | |
| Apr 11, 2016 at 5:13 | comment | added | DLN | @Amritanshu Prasad, No. because $ \{1,3,2,5\}\cap \{1,5,2,3\}\ne \phi $ | |
| Apr 11, 2016 at 4:32 | comment | added | Amritanshu Prasad | Just checking if I understand correctly: is $ \{\{\{1,3\},\{2,5\}\},\{\{1,5\},\{2,3\}\}\} $ also an element of $X^4_3$? | |
| Apr 11, 2016 at 3:19 | comment | added | S. Carnahan♦ | It looks like $X_k^n$ is empty whenever $2^{k-1} > n$. Am I mistaken? | |
| Apr 10, 2016 at 22:26 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 10, 2016 at 21:51 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 10, 2016 at 21:46 | comment | added | DLN | Edited the first paragraph to make it clear. | |
| Apr 10, 2016 at 21:45 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 10, 2016 at 21:33 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 10, 2016 at 21:21 | history | edited | DLN | CC BY-SA 3.0 |
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| Apr 10, 2016 at 21:12 | history | asked | DLN | CC BY-SA 3.0 |