A theorem in [Geoghean's](http://books.google.com/books/about/Topological_methods_in_group_theory.html?id=BwX6gblqV8MC) 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](http://en.wikipedia.org/wiki/Thompson_groups) $F$.



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


