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Can you give me an example of a finitely generated infinitely presented amenable group which is a quotient of a finitely presented amenable group?

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up vote 12 down vote accepted

Take the finitely presented solvable group $G$ with undecidable word problem, constructed by Kharlampovich. That group has infinite center that is a direct product of infinite number of cyclic group. The center has uncountably many subgroups $N_\alpha$, each normal in $G$, most groups $G/N_\alpha$ are not finitely presented but amenable. The description of an easier construction of Kharlampovich's group is in our survey Kharlampovich, O. G. Sapir, M. V. Algorithmic problems in varieties. Internat. J. Algebra Comput. 5 (1995), no. 4-5, 379–602. Another, easier, example, is Abels' group and its quotients by central subgroups. Abels' group can be found here: H. Abels, An example of a finitely presented solvable group, Homological group theory (Proc. Sympos., Durham, 1977), London Math. Soc. Lecture Note Ser., 1979, p. 205–211. It has been mentioned on MO before (see Cornulier's answers).

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Indeed, Abels's group was the first example of a f.p. solvable group without max-n (i.e. with an ascending sequence of normal subgroups, or equivalently with an infinitely presented quotient), solving a question of P. Hall from the fifties. I guess it's also the first example of a f.p. amenable group without max-n. – YCor Dec 6 '11 at 22:07
No, you are right: she constructed the group in 1980. – Mark Sapir Dec 6 '11 at 23:07
Take a finite number of relations $R$ of the lamplighter group $G$. To deduce these relations one needs, say, first $n$ of the standard relations $U$ of $G$. Therefore the group $H$ given by the relations $R \cup U$ is also given by $U$. Hence $H$ is a homomorphic image of the group $T$ given by $R$. But it is well known that $H$ is virtually free and non-amenable. Hence $T$ is not amenable either. Thus $G$ is not a homomorphic image of a finitely presented amenable group. The same works for the Grigorchuk group and for our lacunary hyperbolic group. – Mark Sapir Dec 7 '11 at 1:12
@Benjamin There is a theorem of Bieri and Strebel which says: If G is an infinitely presented metabelian group then any finitely presented covering group contains a nonabelian free group. – Mustafa Gokhan Benli Dec 7 '11 at 2:02
By the way, it would be interesting to know what kind of finitely presented groups can cover infinitely presented metabelian groups. It may be that these finitely presented groups satisfy much stronger hyperbolicity properties than just "contain a free non-Abelian subgroup". Of course they are not necessarily hyperbolic because they can be direct products, for example. But still ... . Somebody may want to analyse the old result of Bieri and Strebel again. – Mark Sapir Dec 7 '11 at 3:11

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