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.



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



Amenable inverse limits of torsionfree amenable groups
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](https://en.wikipedia.org/wiki/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.](http://arxiv.org/abs/1511.07798) 


