Is the intersection of two subgroups, defined below, always trivial?

Question

Suppose, $G = \mathbb{Z} \ast H$, where $H$ is a torsion-free group. Suppose, $g \in G$ and $g \notin H$. Is $\langle\langle g \rangle \rangle \cap H$ always trivial? ($\ast$ stands for free product, and $\langle \langle \dots \rangle \rangle$ stands for normal closure)

I failed to construct a counterexample, but I have no idea how to prove this statement too.

Any help will be appreciated.

Answer 1

Let $a\in G$ be the image of $1\in\mathbb{Z}$. Pick some nontrivial $h\in H$, and take $g=aha^{-1}$. Then $g\not\in H$, but $h\in H\cap\langle\langle g\rangle\rangle$.