TITLE: Is the intersection of two subgroups, defined below, always trivial?
QUESTION [25 upvotes]: 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
Any help will be appreciated.

REPLY [2 votes]: As pointed out, the Kervaire conjecture requires an assumption about the exponent sum on $g$. So, in fact, the answer to the original question is"no".
Take $H$ be the cyclic group of order 6 generated by $a$. Take $\mathbb{Z}$ to be generated by $t$. Lastly, take $g=a^2ta^3t^{-1}$. With these assumptions, we get $H\subset \langle\langle g\rangle\rangle$.
This is most easily seen by looking in the quotient group $G=(H*\mathbb{Z})/\langle\langle g\rangle\rangle$. Since $g=1$, we have $a^2=ta^3t^{-1}$. 
Now, since $a^6=1$ we have $a^4=(a^2)^2=(ta^3t^{-1})^2=1$.
Also, $1=(a^2)^3=(ta^3t^{-1})^3=ta^3t^{-1}$ whence $a^3=1$. So $a=1$. It follows that $H\subseteq \langle\langle g\rangle\rangle$.