Timeline for Group cannot be the union of conjugates
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2010 at 11:54 | answer | added | David E Speyer | timeline score: 10 | |
| Oct 6, 2010 at 8:57 | history | edited | Bjørn Kjos-Hanssen |
edited tags
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| Sep 6, 2010 at 14:52 | vote | accept | C.S. | ||
| Aug 5, 2010 at 9:39 | answer | added | Primoz | timeline score: 2 | |
| Aug 5, 2010 at 8:30 | answer | added | coudy | timeline score: 2 | |
| Jul 31, 2010 at 19:58 | history | edited | Charles Matthews | CC BY-SA 2.5 |
downcase
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| Jul 31, 2010 at 19:58 | answer | added | Tony Scholl | timeline score: 8 | |
| Jul 31, 2010 at 19:08 | history | edited | S. Carnahan♦ | CC BY-SA 2.5 |
Minor corrections.
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| Jul 31, 2010 at 18:24 | comment | added | Robin Chapman | If $G$ is a finite group then all its subgroups have finite index. What the statement should say is that if $H$ is a proper finite index subset of $G$ then $G\ne\bigcup_{g\in G}gHg^{-1}$ (the case of infinite $G$ readily reduces to the case of finite $G$). As Keith shows, this is not always true for subgroups of infinite index. | |
| Jul 31, 2010 at 18:20 | comment | added | Jim Humphreys |
Yes, the statement is out of focus: $gHg^{−1}$ is intended (and "infinite index case"). The natural starting point is to ask whether the proof for finite index breaks down. –
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| Jul 31, 2010 at 18:16 | answer | added | KConrad | timeline score: 23 | |
| Jul 31, 2010 at 18:15 | comment | added | José Figueroa-O'Farrill | There's something I don't understand here: do you perhaps mean $gHg^{-1}$ instead of $ghg^{-1}$? | |
| Jul 31, 2010 at 18:05 | history | asked | C.S. | CC BY-SA 2.5 |