Amenable inverse limits of torsionfree amenable groups

Question

Let $$ \cdots \to \Gamma_n \to \Gamma_{n-1} \to \cdots \to \Gamma_0$$ be an inverse system countable groups and let's assume (for this post) that all homomorphisms in such an inverse system are surjective. We say that $\Gamma$ is an inverse limit if there exists compatible surjective homomorphisms $\varphi_n \colon \Gamma \to \Gamma_n$, such that $\varphi_n(g)\neq e$ for every $g \neq e$ eventually.

Question: Is there an inverse system of torsionfree amenable, such that all inverse limits are non-amenable?

What about the inverse system of free nilpotent or free solvable groups? I conjecture that the answer to the question is yes, but it seems hard to find an example.

Answer 1

Suppose that $\Gamma_n$ is the free nilpotent group on $k$ generators of nilpotency class $n$. Any finite $k$-generated $p$-group ($p$ a prime) is nilpotent, hence the free $k$-generated pro-$p$ group is surjected by the (universal) inverse limit of $\Gamma_n$. Hence any inverse limit of $\Gamma_n$ maps to a dense subgroup of the free pro-$p$ group on $k$ generators.

But the free pro-$p$ $k$-generator group surjects a non-solvable linear group for large enough $k$ (say $ker\{ GL_m(\mathbb{Z}_p)\to GL_m(\mathbb{Z}/p)\}$, $m\geq 2$). This group is linear and not virtually solvable, hence any of its dense subgroups has a free subgroup by the Tits alternative. So any inverse limit of the free $k$-generated nilpotent groups will have a free subgroup, hence is not amenable.

Addendum: In fact, we may take $k=2$ above, since any finite-index subgroup of $SL_n(\mathbb{Z})$, $n\geq 3$, contains a finite-index rank 2 subgroup.