Timeline for Is a finitely generated subgroup of a free profinite group virtually a retract?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2015 at 19:35 | comment | added | HJRW | One obstruction might come from finitely generated (abstract) subgroups of $F$ which are not themselves free. Parafree groups give examples of these. | |
| Jan 28, 2015 at 17:15 | comment | added | Pablo | @HJRW For the pro-$p$ case this is a Frattini argument given in Ribes-Zalsskii (2-nd edition) Theorem 9.1.19 which shows that $H$ is a free factor of some open $H \leq U \leq F$ so it is a retract since we can map $H$ to itself and its free complement in $U$ to $1$ producing a retraction. | |
| S Jan 28, 2015 at 17:08 | history | suggested | Seirios | CC BY-SA 3.0 |
corrected spelling
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| Jan 28, 2015 at 16:56 | review | Suggested edits | |||
| S Jan 28, 2015 at 17:08 | |||||
| Jan 28, 2015 at 16:38 | comment | added | HJRW | Do you have a reference for the free pro-p case? | |
| Jan 28, 2015 at 16:37 | comment | added | HJRW | In answer to your '(which?)', as usual the right hypothesis on the subgroup is not 'finitely generated' but 'quasiconvex'. Haglund and Wise showed that any quasiconvex subgroup of any virtually special group is a virtual retract. | |
| Jan 28, 2015 at 15:14 | history | asked | Pablo | CC BY-SA 3.0 |