# Finitely generated solvable groups all of whose abelian normal subgroups are finite

Is there a classification for infinite finitely generated solvable groups all of whose abelian normal subgroups are finite?

I mean by classification something like presentation.

Edited: Is there an infinite finitely generated solvable group $G$ all of whose abelian normal subgroups are finite and $G$ is not residually finite?

Thanks in advance for any help.

• I think a classification and a presentation are two different things -- maybe you can clarify what kind of classification you are looking for? -- Up to isomorphism (seems asking a lot here), or what else? – Stefan Kohl Nov 1 '13 at 19:59
• The naive way to try to prove that these are only finite solvable groups results in: every infinite solvable group has an infinite step-2 nilpotent normal subgroup whose derived subgroup is finite. – YCor Nov 1 '13 at 20:20
• However, the classical example of a group with finite derived subgroup but center of infinite index (see Derek's post in MathSE: math.stackexchange.com/questions/272152/…) is probably a source of infinite f.g. solvable groups with no infinite abelian normal subgroups, by taking the semidirect product of Derek's example with $\mathbf{Z}$ shifting the $x_i,y_i$. – YCor Nov 1 '13 at 20:23
• PS: my first comment implies however that every infinite residually finite solvable group has an infinite abelian normal subgroup. – YCor Nov 1 '13 at 20:29

There is an infinite finitely generated solvable group with no infinite normal abelian subgroup.

The example is an extension of $$\mathbf{Z}/p\mathbf{Z}$$ by the lamplighter $$(\mathbf{Z}/p\mathbf{Z})^2\wr\mathbf{Z}$$.

Fix a prime $$p$$ congruent to -1 modulo 4 (so that -1 has no square root mod $$p$$). Consider a group $$H_p$$ made up of formal sums $$\sum a_ne_n\oplus\sum b_nf_n\oplus c$$, where $$a_n,b_n,c\in\mathbf{F}_p$$ (all but finitely being 0), and $$e_n,f_n$$ is a fixed "basis". Define the "Heisenberg-like product" $$(\sum a_ne_n\oplus\sum b_nf_n\oplus c)(\sum a'_ne_n\oplus\sum b'_nf_n\oplus c')=\sum a''_ne_n\oplus\sum b''_nf_n\oplus c''$$, where $$a''_n=a_n+a'_n$$, $$b''_n=b_n+b'_n$$, $$c''=c+c'+\sum a_nb'_n-a'_nb_n$$. This is a group (it is the quotient of an infinite direct sum of copies of the Heisenberg by a central subgroup, at least if $$p\neq 2$$).

We denote by $$z$$ the element $$0\oplus 0\oplus 1$$ of $$H_p$$. There is an obvious automorphism $$\alpha$$ of $$H_p$$ shifting the indices, namely mapping $$e_n\mapsto e_{n+1}$$, $$f_n\mapsto f_{n+1}$$, $$z\mapsto z$$. Let us also consider the involution $$s: e_n\mapsto f_n\mapsto e_n$$, $$z\mapsto -z$$, which commutes with $$\alpha$$.

Consider the semidirect product $$G_p=H_p\rtimes(\mathbf{Z}\times\mathbf{Z}/2)$$, where $$\mathbf{Z}\times\mathbf{Z}/2$$ is identified to $$\langle\alpha,s\rangle$$. Then $$G$$ is 3-step solvable, and finitely generated (by 3 generators, say $$\alpha$$, $$s$$, and $$e_0$$).

I claim that $$G_p$$ has no infinite abelian normal subgroup, and more precisely that every abelian normal subgroup $$N$$ of $$G_p$$ is contained in the normal cyclic subgroup $$\langle z\rangle$$ of order $$p$$. Otherwise, the projection $$N'$$ of $$N$$ on $$G_p$$ is a nontrivial abelian normal subgroup. Since $$H'_p=H_p/Z$$ is its own centralizer in $$G'_p=G_p/Z$$, we have $$N'\cap H'_p$$ nontrivial. So we can suppose that $$N\subset H_p$$. Thus $$N'$$ is an $$\langle\alpha,s\rangle$$-invariant subgroup of $$H'_p$$, i.e. a $$\mathbf{F}_{p^2}[\alpha^{\pm 1}]$$-submodule of $$H'_p$$. Here we defined this module structure by letting a given square root of -1 (which is in $$\mathbf{F}_{p^2}\smallsetminus\mathbf{F}_{p}$$) act as $$s$$. Since $$N'$$ is nonzero, it contains some element of the form $$w=\lambda_0e_0+\lambda_1e_1+\dots \lambda_ke_k$$, where $$k\ge 0$$, $$\lambda_i\in\mathbf{F}_{p^2}$$, and $$\lambda_0\lambda_k\neq 0$$. So it also contains the element $$w'=\lambda_0e_{-k}+\dots \lambda_ke_0$$. But any two lifts in $$N$$ of $$w$$ and $$w'$$ do not commute and we get a contradiction.

Every infinite solvable group has an infinite 2-step nilpotent characteristic subgroup with finite derived subgroup. In particular, every infinite solvable group $$G$$ that is residually finite has an infinite abelian normal subgroup (which can be chosen characteristic if $$G$$ is finitely generated).
(A group $$G$$ is by definition 2-step nilpotent if its derived subgroup $$G'$$ is central; this includes abelian groups.)
The second assertion follows from the first: choose $$N$$ as in the first assertion; by residual finiteness, there is a finite index normal subgroup $$M$$ such that $$N'\cap M=\{1\}$$; if $$G$$ is residually finite then $$M$$ can be chosen to be characteristic. Hence $$N\cap M$$ is an infinite abelian normal subgroup, and is characteristic if $$M$$ is characteristic.
For the first assertion, let $$N$$ be the last infinite term in the derived series. So $$N'$$ is finite. Hence, the centralizer of $$N'$$ in $$N$$ has finite index in $$N$$, is an infinite characteristic subgroup with central finite derived subgroup, that is, 2-step nilpotent.
• Many thanks. Is it possible to find such a group $G$ of any derived length $d>2$ such that $G$ is indecomposable? By an indecomposable group, I mean a group which cannot be written as a direct product of two nontrivial subgroups. – Alireza Abdollahi Nov 2 '13 at 10:52
• Probably yes: just consider a finite solvable group $F$ and consider the wreath product $G_p\wr F=G_p^F\rtimes F$. If $F$ has derived length $d$ then this has derived length between $d$ and $d+3$ (I'm lazy to check the exact number). – YCor Nov 2 '13 at 11:24