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

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.

Your question is related to a famous conjecture:

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.

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 $\exp(g) \neq 0$.

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).

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 $H\subseteq \langle\langle g\rangle\rangle$.

• This doesn't seem to satisfy $g\notin \langle\langle H\rangle\rangle$, though? That's the condition on the original question that corresponds to the exponent condition. – Harry Altman Apr 18 '18 at 0:24
• Ah yes. Read it too fast. Thanks for the correction. – Andrew Clifford Apr 22 '18 at 1:58