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.

  • 1
    $\begingroup$ 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 '13 at 19:59
  • 1
    $\begingroup$ 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 '13 at 20:20
  • 2
    $\begingroup$ 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 '13 at 20:23
  • 1
    $\begingroup$ PS: my first comment implies however that every infinite residually finite solvable group has an infinite abelian normal subgroup. $\endgroup$ – YCor Nov 1 '13 at 20:29

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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Alireza Abdollahi Nov 2 '13 at 10:52
  • 1
    $\begingroup$ 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 '13 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.