Timeline for Is there a residually finite non-elementary hyperbolic group whose profinite completion is boundedly generated?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2017 at 21:12 | answer | added | HJRW | timeline score: 10 | |
| Apr 20, 2017 at 21:11 | comment | added | Pablo | @HJRW good! I will think about it. | |
| Apr 20, 2017 at 20:28 | comment | added | HJRW | It seems important to know the answer to this question! I'll take the liberty of writing an answer with the conclusions of this conversation. | |
| Apr 20, 2017 at 14:44 | comment | added | Pablo | @HJRW I am ashamed to say that I can't come up with an example, but I feel that infinite index subgroups of boundedly generated (profinite) groups may be arbitrary. | |
| Apr 20, 2017 at 14:40 | comment | added | HJRW | Apologies, you mentioned this above. What's an example of a boundedly generated profinite group containing a free profinite subgroup? | |
| Apr 20, 2017 at 14:39 | comment | added | Pablo | @HJRW I am afraid that this is may not be quite enough, because if there exists a closed free profinite subgroup in the profinite completion, then this property follows. | |
| Apr 20, 2017 at 14:35 | comment | added | HJRW | It means that, for every finite group, there exists a finite-index subgroup that surjects it. It's enough to contradict the congruence subgroup property. Is it enough to contradict bounded generation? | |
| Apr 20, 2017 at 14:33 | comment | added | Pablo | @HJRW does that mean that there exists a finite index subgroup that surjects onto every finite group? | |
| Apr 20, 2017 at 14:31 | comment | added | HJRW | Ah-ha! Another relevant fact (also following from Agol--Groves--Manning) is that if every hyperbolic group is residually finite then every non-elementary hyperbolic group virtually surjects every finite group. I suspect that's enough to deduce that the profinite completion is not boundedly generated. | |
| Apr 20, 2017 at 14:19 | comment | added | Pablo | @HJRW I think that the pro-congruence completion is always boundedly generated. There is a theorem saying that an arithmetic group has the congruence subgroup property if and only if the profinite completion is boundedly generated. | |
| Apr 20, 2017 at 14:16 | comment | added | HJRW | The next natural question is whether the congruence completion of a lattice in $Sp(n,1)$ is boundedly generated. As you know, it's open whether or not this is the full profinite completion. | |
| Apr 20, 2017 at 14:09 | comment | added | Pablo | @HJRW It seems that you are right. If a group $G$ surjects onto a nonabelian free group, then its profinite completion is not boundedly generated. | |
| Apr 20, 2017 at 14:06 | comment | added | HJRW | In that case, I guess (although perhaps my intuition about bounded generation is incorrect again) that the profinite completion of a non-elementary, hyperbolic, virtually special group $G$ is never boundedly generated, since $\widehat{G}$ virtually retracts to a non-abelian profinite free group. | |
| Apr 20, 2017 at 9:12 | comment | added | Pablo | @HJRW The fact that the profinite completion of a free group isn't boundedly generated is also explained in that link: Just note that $S_n$ is the image of your free group and use Landau's bound on the order of a cyclic subgroup. | |
| Apr 20, 2017 at 9:10 | comment | added | HJRW | Thank you! This is a very nice question. I assume it's known that the profinite completion of a free group isn't boundedly generated? | |
| Apr 20, 2017 at 9:01 | comment | added | Pablo | @HJRW The proof uses boundary dynamics, and a sketch appears in the link (in the question). | |
| Apr 20, 2017 at 9:00 | comment | added | HJRW | I guess you're right... I'm not used to thinking about bounded generation. Can you remind me of the proof that non-elementary hyperbolic groups aren't boundedly generated? | |
| Apr 20, 2017 at 8:59 | comment | added | Pablo | @HJRW I am afraid that I am not convinced. There are boundedly generated (discrete and profinite) groups which contain free subgroups. | |
| Apr 20, 2017 at 8:57 | comment | added | HJRW | Every non-elementary hyperbolic group contains a quasiconvex non-abelian free subgroup. If every hyperbolic group were residually finite then this subgroup, and all its finite-index subgroups, would be separable (by a theorem of Agol--Groves--Manning). I think it follows that the profinite completion would not be boundedly generated. So I'm fairly certain no such example is known to exist. | |
| Apr 20, 2017 at 8:52 | history | asked | Pablo | CC BY-SA 3.0 |