HNN Embedding Theorem for Amenable Groups? Does there exist an analog of the HNN Embedding Theorem for the class of countable amenable groups? In other words, is it true that every countable amenable group embeds into a 2-generator amenable group? Perhaps easier, is it true that every countable amenable group embeds into a finitely generated amenable group?
 A: Theorem 2 of Hall, P.
On the finiteness of certain soluble groups.
Proc. London Math. Soc. (3) 9 1959 595--622.
shows that there is a finitely generated solvable group of derived
length 3 with a subgroup isomorphic to ℚ

The group is 3-generated.  I guess it shouldn't be too difficult to
provide a 2-generated example.  A related question: is every
countable elementary amenable group embeddable in a 2-generated
elementary amenable group?
A: By a result of B.H. Neumann and H. Neunmann (Neumann, B. H.; Neumann, Hanna Embedding theorems for groups.  J. London Math. Soc.  34  1959 465--479.), every countable solvable group of class $c$ embeds into a 2-generated solvable group of class $c+2$. For  finite groups one can use the following construction. Consider the group $S_\infty$ of finitary permutations of $\mathbb N$ (all permutations with finite support). It is generated by transpositions $(1,2), (2,3),...,(n,n+1),...$. The shift $n\mapsto n+1$ induces an injective endomorphism of $S_\infty$ into itself. Consider the (ascending) HNN extension of $S_\infty$ corresponding to this endomorphism. The resulting group is elementary amenable, 2-generated, and contains all finite groups as subgroups. I do not know of any results about embeddings of countable non-elementary amenable groups into finitely generated ones.   
A: Yes, The Grigorchuk group embeds into a 2 generated amenable group which is also finitely presented. For a reference you can look at the paper of Grigorchuk titled "Solved and unsolved problems around one group".
A: If I am not mistaken, the answer is "yes".
Theorem. Every countable amenable (respectively, elementary amenable) group embeds into a $2$-generated amenable (respectively, elementary amenable) group.
The proof is based on the following lemma, which admits a quite elementary proof using wreath products (see [P. Hall, The Frattini subgroups of finitely generated groups, Proc. London Math. Soc. 11 (1961), 327-352]). Given a group $X$, we denote by $X^\omega$ the restricted direct product of countably many copies of $X$. 
Lemma (P. Hall).  Let $H$ be a countable group. Then there exists a short exact sequence
$$
1\longrightarrow M \longrightarrow G \longrightarrow \mathbb Z \longrightarrow  1,
$$
where $G$ is $2$-generated and $[M,M]=[H,H]^\omega$. 
In particular, the lemma implies the theorem when the countable group is perfect. To prove the theorem in the general case we use the following trick which goes back, I believe, to the paper by B. H. Neumann and H. Neumann cited by Mark.
Starting with a countable group $K$, consider the subgroup $H$ of the Cartesian (unrestricted) wreath product $K \, {\rm Wr}\, \mathbb Z$ generated by $\mathbb Z$ and the set of all elements of the base group $ f_k\colon \mathbb Z\to K$, $k\in K$, such that $f_k(n)=1$ for $n\le 0$ and $f_k(n)=k$ for $n> 0$. Let $t$ be a generator of $\mathbb Z$. For definiteness let $t=1$. Then $t^{-1}f_ktf_k^{-1}$, considered as a function $\mathbb Z\to K$, takes only one nontrivial value $k$ (at $0$). This obviously gives an embedding $K\le [H,H].$
Moreover, it is easy to see that the intersection of $H$ with the base $B$ of the wreath product consists of functions $f\colon \mathbb Z\to K$ with the following property: There exists $N_f\in \mathbb N $ such that $f(n)=1$ whenever $n\le -N_f$ and $f(n)=f(N_f)$ whenever $n\ge N_f$. Obviously the map $\varepsilon\colon H\cap B\to K$, which maps every function $f\colon \mathbb Z\to K$ as above to $f(N_f)$, is a homomorphism and $Ker\, \varepsilon$ is isomorphic to a subgroup of $K^\omega $. In particular, if $K$ is amenable (or elementary amenable), then so is $H\cap B$ and, consequently, so is $H$. Now applying the lemma to $H$ yields the theorem as $K\le [H,H]$.
