I am currently looking for a reference to a proof (or counterexample) to the following statement:

**Statement:** Assume $G$ is a finitely generated solvable group of exponential growth, then there is a finitely generated subgroup $H$ of exponential growth which is either polycyclic or (virtually-)abelian-by-cyclic (i.e. it admits a short exact sequence $1 \to V \to H \to \mathbb{Z} \to 1$ where $V$ is virtually abelian).

It seems this result should be known (or known to false, or known to be open), since I could already sketch a strategy to produce the subgroup. Please understand that I'm interested in a reference (and I'm not asking anyone to check if the following strategy is sound). I wrote the strategy below just in case it helps ring a bell.

*Strategy:* First look at the largest integer $k$ so that $G^{(k)}$ (the $k^\text{th}$ derived group) contains a finitely generated subgroup which has exponential growth. Let $K$ be such a subgroup, then $[K,K]$ is locally virtually nilpotent.
  Case 1: if Si $[K,K]$ is finitely generated, then this is a polycyclic group.
  Case 2: Else, $Q = K/[K,K]$ is finitely generated abelian. Let $F_n$ be an increasing sequence of subgroups of $[K,K]$ such that $\cup F_n = [K,K]$. Since $K$ is finitely generated, there is an element $z$ of $K$ whose image in $Q$ is non-trivial. 

The element $z$ has an action on $[K,K]$ by conjugation. Look at the orbits of elements of $[K,K]$ under this action. This action must have an unbounded orbit (not contained in some $F_n$). Since the FC-centre of $[K,K]$ is a caracteristic subgroup and is not contained in some $F_n$, there is even an orbit $\mathcal{O}$ (of the conjugation action of $z$) which is unbounded in $F_C([K,K])$. 

The desired subgroup $H$ is given by $1 \to U \to H \to Z \to 1$ where $Z$ is the subgroup generated by $z$ and $U$ is the subgroup generated by the orbit $\mathcal{O}$.