Suppose, $G = \mathbb{Z} \ast H$, where $H$ is an arbitrary group. Suppose, $g \in G$ and $g \notin \langle\langle H \rangle \rangle $.
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)


Yesterday, I have asked this question on math.stackexchange.com and was advised to re-ask it there:
[Is the intersection of two subgroups, defined below, always trivial? v2.0](https://math.stackexchange.com/q/2738440)


Any help will be appreciated.



Is the intersection of two subgroups, defined below, always trivial?
Your question is related to a famous conjecture:



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> **Kervaire Conjecture:** Given a non-trivial group $H$ and an element $g \in H \ast \mathbb{Z}$, the quotient $(H \ast \mathbb{Z} ) / \langle \!\langle g \rangle\!\rangle$ is non trivial.
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In fact, a positive answer to your question turns out to be equivalent to the strong Kervaire conjecture:



> 
> **Strong Kervaire Conjecture:** Given a non-trivial group $H$ and an element $g \in H \ast \mathbb{Z}$, the canonical map $$H \to (H \ast \mathbb{Z}) / \langle\!\langle g \rangle\!\rangle$$ is injective if $\exp(g) \neq 0$.
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In the statement, $\exp(g)$ refers to the sum of exponents $\sum\limits\_{i=1}^r \epsilon\_i$ if you write $g$ as a reduced word $h\_1 t^{\epsilon\_1} \cdots h\_r t^{\epsilon\_r} h\_{r+1}$ where $h\_1, \ldots, h\_r \in H$ and $t$ is a fixed generator of $\mathbb{Z}$. Notice that the condition $\exp(g) = 0$ is equivalent to $g \in \langle \langle H \rangle \rangle$, and that the map above is injective precisely when $\langle\! \langle g \rangle\!\rangle \cap H$ is trivial.


You can find relevant references in this document: [Presentation of the Kervaire conjecture (notes by A. Ould Houcine)](http://math.univ-lyon1.fr/~ould/Fiches-papiers/kervaire2.pdf). 


