Timeline for Is there a non-Hopfian lacunary hyperbolic group?
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| Sep 22, 2011 at 14:06 | comment | added | user6976 | @Denis: This is what written in Update 2. It is not obvious how to make $\delta$ grow slower than the lengths of relations. In general there are two indirect ways to prove Hopf property. The first is to use residual finiteness which fails for many l.h. groups. The second is to use actions on R-trees as in Sela's paper and in our paper with Drutu. For this one needs at least that all cones are tree-graded by our result with Drutu. That is not true for some l.h. groups, although one would expect that groups given by small cancelation presentations (whose cones are circle-trees) are Hopfian. | |
| Sep 21, 2011 at 6:23 | comment | added | Denis Osin | @Mark: I think we discussed this idea 4 years ago and came to a conclusion that it does not work. The problem is the following. Let $n$ be the maximum of lengths of $P(U,V)$ and $Q(U,V)$ as words in $\{a^{\pm 1},b^{\pm 1}\}$. To make the map $a\mapsto U$, $b\mapsto V$ (call it $\phi $) a homomorphism, we have to add relations $\phi (S), \phi^2(S), \ldots $ for every relation $S(a,b)=1$. If the later has length $k$, we will get relations of length $k, nk, n^2k, \ldots $, which kills lacunar hyperbolicity, at least in the small cancellation case. Am I missing something? | |
| Sep 20, 2011 at 21:46 | comment | added | user6976 | Both constructions produce torsion-free groups. | |
| Sep 20, 2011 at 21:45 | history | edited | user6976 | CC BY-SA 3.0 |
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| Sep 20, 2011 at 9:31 | comment | added | HJRW | Thanks for this answer, Mark. It would be even more interesting if the lacunary hyperbolic group were a limit of torsion-free hyperbolic groups. If I understand your update correctly, it should be possible to perform your construction and keep things torsion-free. Does that sound right? | |
| Sep 19, 2011 at 15:44 | history | edited | user6976 | CC BY-SA 3.0 |
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| Sep 19, 2011 at 3:25 | history | answered | user6976 | CC BY-SA 3.0 |