Subgroups with Infinite cyclic quotients of the Thompons's group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:57:59Z http://mathoverflow.net/feeds/question/87700 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87700/subgroups-with-infinite-cyclic-quotients-of-the-thomponss-group Subgroups with Infinite cyclic quotients of the Thompons's group Mustafa Gokhan Benli 2012-02-06T18:44:01Z 2012-02-15T06:59:41Z <p>A theorem in <a href="http://books.google.com/books/about/Topological_methods_in_group_theory.html?id=BwX6gblqV8MC" rel="nofollow">Geoghean's</a> book is the following (theorem 18.3.18):</p> <p>Let $G$ be a finitely presented group and let the rank of $G/G'$ (as a $\mathbb{Z}$-module) be at least 2. If $G$ has no non-abelian free subgroup, then there is a finitely generated normal subgroup $L$ of $G$ with $G/L$ infinite cyclic.</p> <p>What is (are generatos of) such a subgroup for the <a href="http://en.wikipedia.org/wiki/Thompson_groups" rel="nofollow">Thompson's group</a> $F$.</p> http://mathoverflow.net/questions/87700/subgroups-with-infinite-cyclic-quotients-of-the-thomponss-group/87755#87755 Answer by Steve D for Subgroups with Infinite cyclic quotients of the Thompons's group Steve D 2012-02-07T03:09:08Z 2012-02-07T03:09:08Z <p>The group $F$ has finite presentation $$F = \langle x_0,x_1\ |\ [x_0x_1^{-1},x_0^{-1}x_1x_0]=[x_0x_1^{-1},x_0^{-2}x_1x_0^2]=1\rangle.$$</p> <p>It is not hard to realize that one choice of such an $N$ could be the subgroup <em>normally generated</em> by $x_1$; this subgroup is good-old-fashioned generated by $x_1$ and $x_0^{-1}x_1x_0$.</p> <p>As is often the case, this is easier to see by looking at the <em>infinite</em> presentation $$F = \langle x_k,\ k\ge0\ |\ x_i^{-1}x_jx_i=x_{j+1},\ i&lt; j\rangle.$$</p> <p>For this presentation, $N=\langle x_k,\ k> 0\rangle$. It can be checked immediately that using $x_1$ and $x_2$ suffice.</p> http://mathoverflow.net/questions/87700/subgroups-with-infinite-cyclic-quotients-of-the-thomponss-group/88493#88493 Answer by Mustafa Gokhan Benli for Subgroups with Infinite cyclic quotients of the Thompons's group Mustafa Gokhan Benli 2012-02-15T06:59:41Z 2012-02-15T06:59:41Z <p>The first answer is incomplete and moreover I suspect that it is incorrect! </p> <p>Here is an answer submitted to me via email by Andrew Brunner, who asked me to post his answer for him since he is not signed up for MO.</p> <p>Let $F = \langle a,b| [ab^{-1},a^{-1}ba],[ab^{-1},a^{-2}ba^2] \rangle$ be the usual finite presentation of the Thompson group. Let $M$ be the normal closure of $a$. Then $M=\langle a_i\mid i \in \mathbb{Z} \rangle$ where $a_i=b^{-i}ab^i$.</p> <p>Take the relation $[ab^{-1},a^{-1}ba]=1$ and rewrite to get $(a_{-1})^{-2} (a_0)^2(a_1)^{-1}a_0=1$ which we call (*). Conjugation by $b$ gives $a_2=a_1(a_0)^{-2}(a_1)^2$, so we can deduce that $\langle a_i|i \geq 0 \rangle$ is contained in $\langle a_0,a_1 \rangle$.</p> <p>Take the relation $[ab^{-1},a^{-2}ba^2]=1$ and rewrite to get $(a_{-1})^{-3}(a_0)^3(a_1)^{-2}(a_0)^2=1$. Using this relation and (*) above we can now get $a_{-1}=(a_0)^3(a_1)^{-2}a_0a_1(a_0)^{-2}$, so $a_{-1}$ belongs to $\langle a_0,a_1 \rangle$. Deduce that $\langle a_i|i \leq 0 \rangle$ is contained in $\langle a_0,a_1 \rangle$.</p> <p>It follows that $M= \langle a,b^{-1}ab \rangle$ is a f.g. normal subgroup of the Thompson group with infinite cyclic quotient.</p>