Timeline for Profinite closure of characteristic subgroup
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 23, 2018 at 9:00 | comment | added | ChanaG | Here is something in answe to myself. I'm not 100% sure it's correct, and I don't know yet if it can be used to find concrete examples. Let $K_{n}$ be the intersection of all normal subgroups of $F$ of index $n$. Then $\hat{F}$ is the inverse limit of $F/K_{n}$, and the automorphism group of $\hat{F}$ is the inverse limit of automorphism groups of these quotients. Hence thw closure of $H\subset F$ is characteristic in $\hat{F}$ iff the image of $H$ is characteristic in $F/K_{n}$ for every $n$, | |
| Mar 15, 2018 at 12:24 | comment | added | YCor | The other point is that finding an example can be quite energy-consuming (unless there's a pointer to the literature, or a simple example I would fail to see), so it would be more motivating if it's part of a single question. | |
| Mar 15, 2018 at 12:10 | comment | added | ChanaG | @YCor The linked question considered general $F$, I was hoping that restricting to free $F$ might help to gather new answers. | |
| Mar 15, 2018 at 11:10 | comment | added | YCor | The first question is precisely a duplicate of the linked question (mathoverflow.net/questions/250809) by @rtz. It's a pity that rtz accepted his/her own answer, because it would be better gather new answers there, and because it would be worth isolating the second question (about existence of an example). | |
| Mar 15, 2018 at 10:45 | comment | added | YCor | "(i.e., a free profinite group of the same rank)" The profinite completion is not just a profinite group, it's a profinite group along with a homomorphism from the original group. | |
| Mar 15, 2018 at 10:41 | history | asked | ChanaG | CC BY-SA 3.0 |