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.

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    $\begingroup$ 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? $\endgroup$
    – Stefan Kohl
    Nov 1, 2013 at 19:59
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Nov 1, 2013 at 20:20
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    $\begingroup$ 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 $x_i,y_i$. $\endgroup$
    – YCor
    Nov 1, 2013 at 20:23
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    $\begingroup$ PS: my first comment implies however that every infinite residually finite solvable group has an infinite abelian normal subgroup. $\endgroup$
    – YCor
    Nov 1, 2013 at 20:29

1 Answer 1


To your edited question:

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.

Addendum to give details of my comments to the question:

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.

  • $\begingroup$ 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. $\endgroup$ Nov 2, 2013 at 10:52
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    $\begingroup$ 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). $\endgroup$
    – YCor
    Nov 2, 2013 at 11:24

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