Amenable groups with $\mathbb{N}*\mathbb{Z}$ as a subsemigroup It is known that there are examples of amenable groups with free semigroups $\mathbb{N}*\mathbb{N}$ sitting inside. On the other hand it is not possible for an amenable group to contain $\mathbb{Z}*\mathbb{Z}$. I wonder if the middle case is still possible: are there amenable groups with $\mathbb{N}*\mathbb{Z}$ as a subsemigroup?

Let $G$ be the wreath product of infinite cyclic groups $\langle x \rangle$ and
$\langle y \rangle$, with the base group being the normal closure of $\langle x \rangle$. So $G$ is metabelian and hence amenable. It looks to me as though the subsemigroup generated by $x$, $x^{-1}$ and $y$ is isomorphic to ${\mathbb Z} * {\mathbb N}$.
The base group of the wreath product is the free abelian group generated by the conjugates $x_i := z^ixz^{-i}$ (with $i \in {\mathbb Z}$) of $x$. Then the element $x^{i_1}y^{j_1} \cdots x^{i_n}y^{j_n}$ (with $i_k \in {\mathbb Z}$, $j_k \in {\mathbb N}$, and all terms except the first and last nontrivial) is equal to $x_0^{i_1}x_{m_1}^{i_2} \cdots x_{m_{k-1}}^{i_k}y^{m_k}$, where $m_1=j_1$, $m_2=j_1+j_2$, $\ldots$, $m_n=j_1+\cdots+j_n$, which is uniquely determined by the word.
