Timeline for Uniqueness of composition series for profinite groups
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 22, 2020 at 9:54 | vote | accept | Eusebio Gardella | ||
| Feb 19, 2014 at 2:19 | answer | added | Colin Reid | timeline score: 2 | |
| Jan 30, 2014 at 18:20 | comment | added | abz | Did you check whether the proof of the (Zassenhaus) Butterfly Lemma applies in your situation? It doesn't use much --- only the Noether isomorphism theorems and one of Dedekind's modular laws. That would imply that any two such sequences have a common refinement. | |
| Jan 30, 2014 at 1:36 | comment | added | Eusebio Gardella | @Ian: I would like a result that can handle infinite products of finite groups too; in particular, groups for which every finitely generated subgroup is finite. | |
| Jan 30, 2014 at 1:34 | history | edited | Eusebio Gardella | CC BY-SA 3.0 |
added 7 characters in body
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| Jan 30, 2014 at 1:33 | comment | added | Eusebio Gardella | @Yves: yes, the subgroups $G_n$ should be closed; I forgot to mention that. I'll edit the question. | |
| Jan 29, 2014 at 23:20 | comment | added | YCor | @Eusebio: do you assume that the $G_n$ are closed subgroups? | |
| Jan 29, 2014 at 21:41 | comment | added | Ian Agol | It's true if $G$ is (topologically) finitely generated. In this case, $G$ is determined uniquely by its collection of finite quotients, and the composition factors of the finite quotients determine those of $G$. | |
| Jan 29, 2014 at 20:37 | history | asked | Eusebio Gardella | CC BY-SA 3.0 |