Timeline for Schreier's formula and supersolvable groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2015 at 21:43 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
added 11 characters in body
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| Jan 8, 2015 at 21:42 | comment | added | Benjamin Steinberg | Oh that is a typo. I'll fix. | |
| Jan 8, 2015 at 20:18 | comment | added | Pablo | So in this case I suspect that every occurrence of prosolvable in your answer should be prosupersolvable. | |
| Jan 8, 2015 at 20:03 | comment | added | Benjamin Steinberg | It means it is the pro-C completion of a free group with respect to a variety of finite groups in the sense of the book of Ribes and Zalesski. Or it is the free object in the class of profinite groups satisfying some profinite identity. | |
| Jan 8, 2015 at 19:24 | comment | added | Pablo | What is a relatively free profinite group? | |
| Jan 8, 2015 at 19:21 | comment | added | Benjamin Steinberg | Supersolvable is crucial here. Every finite supersolvable group is a subdirect product of groups with a normal p-subgroups corresponding quotient an abelian group of order dividing p-1. This is what we used for the relatively free case. | |
| Jan 8, 2015 at 19:06 | comment | added | Pablo | Thank you very much for the detailed explaination, I was aware of the reduction to the case of infinitely many primes dividing the order. There do exist freely indexed prosolvable groups with infinitely many prime divisors, constructed by Lubotzky and v.d. Dries. But their example is not prosupersolvable... | |
| Jan 8, 2015 at 18:56 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |