Timeline for embeddings of finite group into permutation groups
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2022 at 16:42 | comment | added | YCor | Derek answers up to global conjugation of the image. But it's easier to find a counterexample if you ask about up to isomorphism of $G$-set. Namely, if $G=\mathrm{Alt}_6$, then there are two non-isomorphic faithful $G$-sets of cardinal $6$ (since there are two conjugacy classes of subgroups of index 6, and no non-trivial non-transitive action on 6 elements). | |
| S Mar 14, 2022 at 13:33 | history | suggested | CommunityBot | CC BY-SA 4.0 |
Typesetting
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| Mar 14, 2022 at 12:13 | review | Suggested edits | |||
| S Mar 14, 2022 at 13:33 | |||||
| Jul 18, 2013 at 19:57 | answer | added | Phil | timeline score: -4 | |
| Jul 18, 2013 at 18:56 | answer | added | Igor Rivin | timeline score: 7 | |
| Jul 18, 2013 at 17:22 | answer | added | Derek Holt | timeline score: 18 | |
| Jul 18, 2013 at 16:39 | comment | added | Arturo Magidin | @Shaki: That's a rather big condition... and you should have edited the question to include it; I've added it for you now. | |
| Jul 18, 2013 at 16:39 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
added condition mentioned in the comments
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| Jul 18, 2013 at 15:48 | review | First posts | |||
| Jul 18, 2013 at 15:54 | |||||
| Jul 18, 2013 at 15:46 | comment | added | Shaki | yes, it is truth, sorry, I forgot to add the condition that the group G is simple ... | |
| Jul 18, 2013 at 15:39 | comment | added | nsrt | $Z/2\times Z/2\to S_4$ seems to be a counterexample? | |
| Jul 18, 2013 at 15:33 | history | asked | Shaki | CC BY-SA 3.0 |