Timeline for Schreier's index formula
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 22, 2014 at 22:44 | vote | accept | Pablo | ||
| Apr 17, 2014 at 13:05 | comment | added | Pablo | Suppose $H$ is a given subgroup containing the k-th derived subgroup of $F$. Must there be a finite index subgroup above $H$ violating Schreier's formula? I'm interested in this in the profinite case too. | |
| Apr 16, 2014 at 10:46 | vote | accept | Pablo | ||
| Apr 16, 2014 at 10:46 | |||||
| Apr 16, 2014 at 10:46 | vote | accept | Pablo | ||
| Apr 16, 2014 at 10:46 | |||||
| Apr 15, 2014 at 13:02 | comment | added | Benjamin Steinberg | Free pro-solvable groups and free pro-p groups satisfy Schreier's formula. | |
| Apr 15, 2014 at 7:27 | comment | added | Pablo | Ok, I think I've got you. What if instead F is the free profinite group on a finite number of generators? In this case the aforementioned intersection is not trivial. | |
| Apr 14, 2014 at 20:17 | comment | added | Benjamin Steinberg | The intersection of all the terms of the derived series is trivial because the free group is residually a finite p-group and hence residually solvable. So you do get a free group and there is no contradiction. | |
| Apr 14, 2014 at 19:59 | comment | added | Pablo | If you take the quotient of F by the intersection of all the terms in its derived series you get a finitely generated residually finite group, which is not free but satisfies Schreier's index formula. Does this contradict your claim? | |
| Apr 14, 2014 at 16:43 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |