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.