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Sep 23, 2021 at 1:45 history edited David Roberts CC BY-SA 4.0
fixed arxiv front-end link
Oct 4, 2010 at 14:06 history edited user6976 CC BY-SA 2.5
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Oct 3, 2010 at 23:00 comment added user6976 @Yo: See Edit 4.
Oct 3, 2010 at 22:14 history edited user6976 CC BY-SA 2.5
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Oct 3, 2010 at 20:45 comment added seldom seen Mark, I can't figure out how to ensure that $G'$ has rank $2$. In Rips' original paper he adds two generators. Is there a another reference? I'm not having much luck combing through mathscinet. -yo
Oct 3, 2010 at 20:30 vote accept seldom seen
Oct 3, 2010 at 20:14 history edited user6976 CC BY-SA 2.5
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Oct 3, 2010 at 20:01 comment added seldom seen You are supposed to go the other way, and build $G_{n+1}$, $G_{n+2}$, $G_{n+3}$, etc.
Oct 3, 2010 at 19:55 comment added user6976 Example with hyperbolic groups give you the sequence in the opposite order: $G_n, G_{n-1},...$ where $G=G_n$. Here $G_n/free group=G_{n-1}, G_{n-1}/free group=G_{n-2}$, etc.
Oct 3, 2010 at 19:49 comment added seldom seen Regarding Edit 2, I am trying to extend $G_0$ to $G_1$, $G_1$ to $G_2$, etc., not take further and further quotients of $G_0$. I think examples where $\mathrm{rank}(G_i)$ is bounded independently of $i$, with the ranks of $K_i$ not necessarily finite, would be very interesting as well.
Oct 3, 2010 at 19:34 history edited user6976 CC BY-SA 2.5
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Oct 3, 2010 at 19:08 comment added seldom seen Again, I should learn to be more specific. The $G_i$ should be commutative transitive. Editing to reflect. Thanks.
Oct 3, 2010 at 18:57 history edited user6976 CC BY-SA 2.5
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Oct 3, 2010 at 18:36 comment added seldom seen I should have been more specific. $K_i$ should be nonabelian. I apologize.
Oct 3, 2010 at 18:34 history answered user6976 CC BY-SA 2.5