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Jan 8, 2015 at 1:59 comment added Igor Rivin @YCor of course, but I am just curious whether Greenberg's proof is actually different...
Jan 7, 2015 at 23:51 comment added YCor @Igor: often a particular case of a theorem is harder (say, more tricky) than the general case, because the general statement better isolates the possible ways to find the proof.
Jan 7, 2015 at 16:20 comment added HJRW Of course. But he did it before hyperbolic groups had been invented, and probably phrased his proof differently. I haven't looked at his paper.
Jan 7, 2015 at 16:13 comment added Igor Rivin But isn't the proof the same in that case?
Jan 7, 2015 at 16:02 comment added HJRW @Igor, he did Fuchsian groups.
Jan 7, 2015 at 14:54 comment added Igor Rivin What did Greenberg do?
Jan 7, 2015 at 14:14 comment added YCor The proof can be included: if $H$ is quasi-convex of infinite index, then $\partial H$ is a proper closed subset of $\partial G$, and $H$ normal implies that it's $G$-invariant. Since $G$ (which can be supposed non-elementary) is minimal on $\partial G$, this implies $\partial H=\emptyset$, hence $H$ is finite.
Jan 7, 2015 at 13:51 vote accept Pablo
Jan 7, 2015 at 13:31 history answered HJRW CC BY-SA 3.0