It is well-known theorem that for an elementary amenable group $G$, $G$ has exponential growth rate iff $G$ contains non cyclic free semigroup.
Now I am interested in the following questions:
Let $G=\mathbb{Z}^2\rtimes_{\sigma}\mathbb{Z}=\langle y,z,x|yz=zy, xyx^{-1}=y^2z, xzx^{-1}=yz\rangle$, which is a polycyclic group and has exponential growth rate, so by the above theorem, we can find a pair $(a,b)\in G$ such that $a, b$ generates a free subsemigroup.
1, Are there any concrete examples of these $(a,b)$ in the above group?
2, Especially, can we find such one pair $(a,b)$ with the following additional property:
For any $n\geq 1$ and any $g_1,\cdots, g_n\in \langle a,b\rangle,$ we can find some $g\in \langle a,b\rangle$, such that $gg_i\in\langle a,b\rangle^{+},\forall 1\leq i\leq n$.
Here, $\langle a,b\rangle$ denotes the subgroup generated by $a,b$, and $\langle a,b\rangle^{+}$ denotes the free semigroup generated by $a,b$.
3, If the above question has a negative answer, could anyone give me a polycyclic group $G$ where the above answer is positive?
Thanks in advance!
For 1: yes there are concrete free subsemigroups: if you write the matrix $\sigma=\begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}$, then your group embeds into $\mathbf{R}^2\rtimes_\sigma\mathbf{Z}$, and $\sigma$ is $\mathbf{R}$-diagonalizable. Hence by projecting modulo a real eigenline, you get a dense embedding into $\mathbf{R}\rtimes_t\mathbf{Z}$, where the action is by multiplication by $t$ (one of the eigenvalues of $\sigma$). Of course the latter embeds into the affine group $\mathbf{R}\rtimes\mathbf{R}$.
Now there are easy-to-check concrete criteria for two elements in the affine group to generate a free subsemigroup (see for instance, although it's certainly not the first reference, Lemma 3.1 in http://www.normalesup.org/~cornulier/tq.pdf).
For 2, I don't know. the ping-pong lemma prove freeness of a pair of elements, but does not characterize the subsemigroup they generate. In the context of isometries of the hyperbolic plane, there is maybe an interpretation of your condition in 2., but I haven't thought about it.