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?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Sep 21, 2019 at 19:19

1 Answer 1


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 $kerΦ \cap <a> = 1$ due to the construction of $φ$, which means that $kerΦ$ acts freely on the edges of the HNN tree (which are the conjugates of $<a>$ 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.

  • $\begingroup$ +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.) $\endgroup$
    – user1729
    Sep 21, 2019 at 12:35

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