# Subgroups with Infinite cyclic quotients of the Thompons's group

A theorem in Geoghean's book is the following (theorem 18.3.18):

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.

What is (are generatos of) such a subgroup for the Thompson's group $F$.

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The group $F$ has finite presentation $$F = \langle x_0,x_1\ |\ [x_0x_1^{-1},x_0^{-1}x_1x_0]=[x_0x_1^{-1},x_0^{-2}x_1x_0^2]=1\rangle.$$

It is not hard to realize that one choice of such an $N$ could be the subgroup normally generated by $x_1$; this subgroup is good-old-fashioned generated by $x_1$ and $x_0^{-1}x_1x_0$.

As is often the case, this is easier to see by looking at the infinite presentation $$F = \langle x_k,\ k\ge0\ |\ x_i^{-1}x_jx_i=x_{j+1},\ i< j\rangle.$$

For this presentation, $N=\langle x_k,\ k\> 0\rangle$. It can be checked immediately that using $x_1$ and $x_2$ suffice.

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How to see that $x_0x_1x_0^{-1} \in N$ –  Mustafa Gokhan Benli Feb 8 '12 at 1:36

The first answer is incomplete and moreover I suspect that it is incorrect!

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.

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 $a_i=b^{-i}ab^i$.

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 $\langle a_0,a_1 \rangle$.

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 $\langle a_0,a_1 \rangle$.

It follows that $M= \langle a,b^{-1}ab \rangle$ is a f.g. normal subgroup of the Thompson group with infinite cyclic quotient.

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Easier: F' is clearly in M and M is clearly f.g. Now F/F' is abelian so M is normal in F. Then F/M is (F/F') mod (a) since the two gens of M are equal mod F'. But F -> F/F' is given by (slope near 0, slope near 1). So (F/F') mod (a) is seen to be iso to Z. –  Matt Brin Aug 23 '13 at 4:36