User yo - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:58:58Z http://mathoverflow.net/feeds/user/2225 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94689/finite-subgroups-of-relatively-hyperbolic-groups Finite subgroups of relatively hyperbolic groups yo 2012-04-20T20:59:54Z 2012-04-21T22:00:50Z <p>It is well known that in a given $\delta$-hyperbolic group there are only finitely many conjugacy classes of finite subgroups. This is clearly false for relatively hyperbolic groups since we have no control over the parabolic subgroups. Fix a relatively hyperbolic group $\Gamma$ with parabolic subgroups $P_i$. Are there only finitely many conjugacy classes of finite subgroups $F&lt;\Gamma$ with the property that $F$ is contained in a two-ended subgroup $H&lt;\Gamma$ which is not conjugate into any $P_i$? Perhaps weaker: are there only finitely many isomorphism types of two-ended subgroups which are not conjugate into any parabolic? Weaker still: Is there an upper bound to the orders of elements in such subgroups $F$?</p> <p>(Motivation: I am trying to find a law obeyed by non-parabolic two-ended subgroups of relatively hyperbolic groups, or at least a partition of them into finitely many families, each of which obeys a law.)</p> http://mathoverflow.net/questions/50261/determinants-of-almost-identity-matrices Determinants of "almost identity" matrices. yo 2010-12-23T23:32:28Z 2010-12-25T20:58:34Z <p>Suppose that $A$ is a real square matrix with all diagonal entries $1$, all off-diagonal entries non-positive, and all column sums positive and non-zero. Does it follow that $\det(A)\neq0$? Is this just an exercise? Are these matrices well-known?</p> http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups Ranks of iterated extensions of a group by free groups. yo 2010-10-03T18:10:58Z 2010-10-04T14:06:37Z <p>Let $G_0$ be a finitely generated group, and suppose there are groups $G_i$ and $K_i$ as in the following short exact sequences</p> <p>$1\to K_i\to G_{i+1}\to G_i\to 1$</p> <p>with $K_i$ free and nonabelian (you may assume finitely generated), and $G_i$ commutative transitive. (If $a$ is nontrivial and $b$ and $c$ both commute with $a$, then $b$ and $c$ commute.) Does it follow that $\mathrm{rank}(G_i)\to\infty$ as $i\to\infty$? Are there examples of extensions of this sort where the rank doesn't increase?</p> http://mathoverflow.net/questions/115937/non-generating-sets-in-a-free-group Comment by yo yo 2012-12-10T14:24:17Z 2012-12-10T14:24:17Z Sorry, I was referring to the second paragraph of the question, not the first (It seems like Dror is perhaps aware of Nielsen's criterion.). There is a paper by Kharlampivic and Myasnikov claiming to give a proof of BF, but I don't know if it works in this case. Either way, you can probably use Sela's envelopes to get something similar. http://mathoverflow.net/questions/115937/non-generating-sets-in-a-free-group Comment by yo yo 2012-12-10T13:42:01Z 2012-12-10T13:42:01Z This question is about definable subsets of $F_n$, $n&gt;2$ not folding. (The case $n=2$ is due to Nielsen, and there the answer is yes.) The set of $a_1$ such that there are $a_2,\dotsc,a_n$ so that $x_i^{a_i}$ generates is neither negligible nor co-negligible (see <a href="http://andromeda.rutgers.edu/~feighn/luminy.pdf" rel="nofollow">andromeda.rutgers.edu/~feighn/luminy.pdf</a>) and is therefore not definable without coefficients. I don't know exactly, but my gut-feeling is that this set isn't definable with coeffcients either. Bestvina-Feighn probably works, and there is really no simple way to check, other than to go ahead and do it with folding. http://mathoverflow.net/questions/94689/finite-subgroups-of-relatively-hyperbolic-groups Comment by yo yo 2012-04-21T19:01:17Z 2012-04-21T19:01:17Z Good point. In any case, now there is a law and I'm very happy. http://mathoverflow.net/questions/94689/finite-subgroups-of-relatively-hyperbolic-groups/94731#94731 Comment by yo yo 2012-04-21T18:58:27Z 2012-04-21T18:58:27Z Thanks, Lee. I was trying to do something along these lines, but with the action on a fine hyperbolic graph, which is of course totally wrongheaded since I was never going to get $N(p)$ finite. http://mathoverflow.net/questions/80998/groups-with-no-bounds-on-the-size-of-abelian-subgroups-without-infinite-ones Comment by yo yo 2011-11-15T19:15:38Z 2011-11-15T19:15:38Z This contains lots of $\mathbb{Z}$'s. http://mathoverflow.net/questions/75816/the-generalization-of-semidirect-product-and-hnn-extension Comment by yo yo 2011-09-19T11:56:08Z 2011-09-19T11:56:08Z (Sorry, I can't make the symbols work.) Isn't this just $G*_A A\rtimes B*_B H$? http://mathoverflow.net/questions/72734/when-is-an-hnn-extension-a-free-group/72740#72740 Comment by yo yo 2011-08-16T03:46:42Z 2011-08-16T03:46:42Z spring: there is a newer, easier to read version of 'recent paper...'. I'll send you a copy if you're interested. http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups/40946#40946 Comment by yo yo 2010-10-03T20:45:27Z 2010-10-03T20:45:27Z 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 http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups/40946#40946 Comment by yo yo 2010-10-03T20:30:43Z 2010-10-03T20:30:43Z Clever! Thanks! http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups/40946#40946 Comment by yo yo 2010-10-03T20:01:08Z 2010-10-03T20:01:08Z You are supposed to go the other way, and build $G_{n+1}$, $G_{n+2}$, $G_{n+3}$, etc. http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups/40946#40946 Comment by yo yo 2010-10-03T19:49:16Z 2010-10-03T19:49:16Z 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. http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups/40946#40946 Comment by yo yo 2010-10-03T19:08:19Z 2010-10-03T19:08:19Z Again, I should learn to be more specific. The $G_i$ should be commutative transitive. Editing to reflect. Thanks. http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups/40946#40946 Comment by yo yo 2010-10-03T18:36:18Z 2010-10-03T18:36:18Z I should have been more specific. $K_i$ should be nonabelian. I apologize. http://mathoverflow.net/questions/40944/ranks-of-iterated-extensions-of-a-group-by-free-groups Comment by yo yo 2010-10-03T18:30:43Z 2010-10-03T18:30:43Z Minimal number of elements in a generating set.