# Free normal subgroup of an HNN-extension

Suppose $$F$$ is a finitely generated free group and $$a,b$$ are not in $$F'$$ but $$b^{-1}a \in F'$$. By taking the HNN extension $$G=\langle F,t | t^{-1}atb^{-1}\rangle$$, is there a way to find a normal free subgroup of $$G$$ so that their quotient is cyclic?

I'm trying to define a homomorphism from $$G$$ to $$\Bbb Z$$ so that the kernel acts freely on the vertices of the HNN tree, which have the conjugates of $$F$$ as stabilizers but with no success. Moreover by defining $$f_1:F \to \Bbb Z$$ in general and let $$f_2:\langle t\rangle \to \Bbb Z$$ be trivial I have a map from $$G\to \Bbb Z$$ but this map can't be injective on the conjugates of $$F$$ because $$f_1$$ can never be injective due to $$f_1(a)=f_1(b)$$.

Is there something I'm missing?

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 21, 2019 at 19:19

Like in the comments, we factor a map through the abelianization (of $$F$$), which is finitely generated free abelian, and then project on $$\Bbb Z$$ making sure we map $$a$$ to a non trivial element. We now have a map $$φ:F \to \Bbb Z$$ with $$φ(a)=φ(b) \neq 0$$. We expand this map to a map $$Φ:G\to \Bbb Z$$ by killing $$t$$. The $$kerΦ \cap = 1$$ due to the construction of $$φ$$, which means that $$kerΦ$$ acts freely on the edges of the HNN tree (which are the conjugates of $$$$ in $$G$$). This means that G is a free product with free factors since the stabilizers are subgroups of conjugates of $$F$$ which are free as subgroups of free groups. Hence $$kerΦ$$ is free.
• +1 although you need to prove that $F$ maps onto the integers in such a way that $a$ is not killed. (Although this is not hard to see, by considering exponent sums.) Sep 21, 2019 at 12:35