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Sep 24, 2012 at 23:53 vote accept i. m. soloveichik
Sep 14, 2012 at 0:04 comment added Misha Mark, you are right, I was sloppy here.
Sep 13, 2012 at 22:08 comment added user6976 cont: generated by $S_n$ is f.p. but there is no algorithm that, given $n$ computes a finite presentation of $A_n$. They use a Dani Wise version of the Rips' construction to construct $\Gamma_n$. So each $\Gamma_n$ is linear but the degree of the linear presentation of $\Gamma_n$ depends on $n$.
Sep 13, 2012 at 22:06 comment added user6976 @Misha: Your second comment cannot be correct. For any given finitely generated subgroup the question of finding a finite presentation is not a "mass" problem. So the statement "we cannot..." does not make sense. Moreover, if you consider the mass problem where the input is a finitely presented subgroup of $SL_n(\mathbb{Z})$ and the output its finite presentation, then it is not clear that this problem is undecidable for any given $n$. Bridson and HW proved that here are recursive sequences of hyperbolic groups $\Gamma_n$ and of finite sets $S_n⊂\Gamma_n×\Gamma_n$ for which the group $A_n$ ...
Sep 13, 2012 at 21:47 vote accept i. m. soloveichik
Sep 16, 2012 at 2:11
Sep 13, 2012 at 20:48 answer added Misha timeline score: 5
Sep 13, 2012 at 13:25 comment added Misha Even more interestingly, there are examples due to Bridson and HW where you have a fp subgroup for which one cannot compute a finite presentation!
Sep 13, 2012 at 13:10 comment added Misha Yes, it's impossible for $n>3$ by Mikhailova's construction. Open problem, essentially due to Serre, for $n=3$.
Sep 13, 2012 at 12:29 history asked i. m. soloveichik CC BY-SA 3.0