Timeline for free group of finite rank can contain free groups of infinite rank as a subgroup
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2011 at 14:02 | comment | added | HJRW | Benjamin - I never actually look at the proofs in Lyndon and Schupp. | |
| Sep 12, 2011 at 11:43 | vote | accept | Rekha Biswal | ||
| Sep 12, 2011 at 11:44 | |||||
| Sep 10, 2011 at 19:12 | comment | added | Benjamin Steinberg | HW, thanks for reminding me of the name. There is a "horrible" combinatorial proof of this at the beginning of Lyndon and Schupp using Marshall Hall's theorem that finitely generated subgroups of a free group are free factors of finite index subgroups. When I was a grad student I realized Greenberg's theorem had a one-line proof with Stalling's graphs (or more generally the fact that finitely generated subgroups of infinite index do not contain non-trivial normal subgroups). I think everybody else did, too. | |
| Sep 9, 2011 at 20:35 | comment | added | HJRW | Benjamin - FYI, this is sometimes called 'Greenberg's theorem'. | |
| Sep 9, 2011 at 0:32 | comment | added | Autumn Kent | I think that's why Igor said "Even better". | |
| Sep 8, 2011 at 21:58 | comment | added | Benjamin Steinberg | In any event, it is trivial to prove that a finitely generated subgroup of a free group that is normal must have finite index. The covering space of a wedge of circles associated to a finitely generated subgroup of infinite index has a finite core (the Stallings graph) with a bunch of infinite trees attached to it. The covering space associated to a normal subgroup is a Cayley graph of the quotient group and looks the same at each vertex. Clearly, these two situations cannot hold at the same time. | |
| Sep 8, 2011 at 20:37 | comment | added | Igor Rivin | @HW and @Richard: indeed. | |
| Sep 8, 2011 at 20:36 | history | edited | Igor Rivin | CC BY-SA 3.0 |
corrected the statement
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| Sep 8, 2011 at 20:00 | comment | added | Autumn Kent | $\mathbb{Z} \times \mathbb{Z}$ is even better. ;) | |
| Sep 8, 2011 at 19:58 | comment | added | HJRW | This isn't quite right: $F\times\mathbb{Z}$ is a counterexample! I think Bieri proved that no non-free normal subgroup of infinite index of a group of cohomological dimension two is finitely presented. | |
| Sep 8, 2011 at 18:01 | history | answered | Igor Rivin | CC BY-SA 3.0 |