Timeline for How a profinite group can be obtained from its normal open subgroups?
Current License: CC BY-SA 4.0
8 events
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| Dec 31, 2019 at 12:07 | comment | added | YCor | Do you have some motivation? It's hard to imagine an answer that is not a tautological restatement of the condition. Also as regards the connection with the title, necessarily $G$ should be isomorphic to some element of $\Delta$. So $G$ is not really obtained from the collection of normal subgroups, it just appears among them. | |
| Dec 31, 2019 at 12:06 | history | edited | YCor | CC BY-SA 4.0 |
clarified question, added tag
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| Dec 31, 2019 at 11:12 | comment | added | MSMalekan | @YCor Yes, that's exactly what I mean. | |
| Dec 31, 2019 at 9:58 | comment | added | YCor | So, if I understand correctly $\Delta$ is a set of profinite groups, and you're asking whether there is an isomorphism-preserving bijection from $\Delta$ to the set of normal open subgroups of some profinite group. | |
| Dec 31, 2019 at 5:24 | comment | added | MSMalekan | @YCor Thanks for your useful comment. But really I want $\Delta$ to be all of my open normal subgroups, not only the isomorphism classes. | |
| Dec 30, 2019 at 17:39 | comment | added | YCor | I think OP means that the data is the a set $\Delta$ of isomorphism classes of profinite groups, and the question is whether there exists a profinite groups whose set of isomorphism classes of open normal subgroups is exactly $\Delta$. (In this case, the answer is plainly: if and only if there is some $G\in\Delta$ such that every $H\in\Delta$ is isomorphic to some open normal subgroup of $G$, and such that every open normal subgroup of $G$ is isomorphic to some element of $\Delta$.) | |
| Dec 30, 2019 at 16:12 | comment | added | LSpice | Certainly they need to form a directed system, in the sense that, for each $N, N' \in \Delta$, there is $N'' \in \Delta$ that injects into both with normal, cofinite index. Do you mean $\Delta$ to consist of the proper normal open subgroups? Otherwise it seems possible that directed + has a (unique) maximal element is enough. | |
| Dec 30, 2019 at 15:25 | history | asked | MSMalekan | CC BY-SA 4.0 |