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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?

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I must be missing something. What f.g. group does ${\mathbb Q}$ embed into? –  Yemon Choi Jul 5 '10 at 19:42
    
The HNN Embedding Theorem says that every countable group embeds into a 2-generator group. Of course, $\mathbb{Q}$ doesn't embed into a f.g. abelian group. However, I don't know whether $\mathbb{Q}$ embeds into an amenable f.g. group. –  Simon Thomas Jul 5 '10 at 19:45
    
Suppose we restrict our attention to finitely generated groups. Does Grigorchuk's group embed into a 2-generator amenable group? –  Jon Bannon Jul 5 '10 at 19:48
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@ Jon: As you would expect, I don't know the answer to your question ... –  Simon Thomas Jul 5 '10 at 19:54
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4 Answers 4

up vote 14 down vote accepted

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 $K\le [H,H]$.

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Very nice! It would probably be a little clearer if you used the phrase "restricted direct product" instead of "direct product". –  Simon Thomas Sep 24 '11 at 12:34
    
I added "restricted". –  Denis Osin Sep 24 '11 at 17:44
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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.

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Could you say a little more about why every countable locally finite group embeds in your group? (The construction of Hall's universal locally finite group proceeds by iterating the left regular permutation representation ... how does the shift manage this?) –  Simon Thomas Sep 22 '10 at 11:37
    
In fact, isn't your group the finitary permutations of $\mathbb{Z}$ extended by the shift? If so, it certainly doesn't contain all countable locally finite groups .... –  Simon Thomas Sep 22 '10 at 12:43
    
@Simon: OK, I was too quick. I meant all finite groups (because they are all in $S_\infty$. I have modified the answer. –  Mark Sapir Sep 22 '10 at 13:06
    
@Mark: Fair enough! –  Simon Thomas Sep 22 '10 at 13:23
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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?

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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".

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