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Suppose $A \cong \mathbb{Z^n}$ and $B$ is a finitely generated abelian group, let $G=A \rtimes B$. Is $G$ a polycyclic group?

I know that $G$ is a solvable group because of the sequence $G \rhd A \rhd 1$, and I know that a group is polycyclic iff it's solvable and satisfy the maximal condition (equivalent to every subgroup is finitely generated), but I can't seem to get further.

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Alex B. has answered this in the affirmative. In general, if $G \rhd A \rhd 1$ and $G/A$ and $A$ are both polycyclic, then so is $G$. A supersolvable group is one with a finite descending series of normal subgroups with cyclic factor groups. There are examples of the groups that you describe that are not supersolvable. – Derek Holt Mar 21 '12 at 8:54
up vote 2 down vote accepted

One of the characterisations of polycyclic groups is that there is a subnormal series $$\{1\}=G_n\leq G_{n-1}\leq \ldots \leq G_1\leq G_0=G,$$ where each $G_i$ is normal in $G_{i-1}$ with cyclic quotient. This tells you immediately that if $G\rhd N$ and both $N$ and $G/N$ are polycyclic, then $G$ is polycyclic. Finitely generated abelian groups are polycyclic, in particular $A$ and $B$ in your question are, so that gives you what you want.

Edit: The most conceptual explanation of why finitely generated abelian groups are polycyclic is that $\mathbb{Z}$ is a Noetherian ring. An abelian group is the same as a $\mathbb{Z}$-module, and finitely generated modules over Noetherian rings are Noetherian. In other words, any submodule of such a module is finitely generated. In particular, any subgroup of a finitely generated abelian group is itself finitely generated.

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Finitely generated abelian groups are polycyclic... But I don't think $\mathbb{Q}$ is; doesn't matter for the purposes of this question, of course. – Arturo Magidin Mar 21 '12 at 2:30
Good point! Editing... – Alex B. Mar 21 '12 at 2:34
Can you elaborate on why finitely generated abelian groups are polycyclic? I also have problem understanding the structure of $G/N$ here. – nullgraph Mar 22 '12 at 4:17
@nullgraph I have added some remarks concerning finitely generated abelian groups. As for $G/N$, in your example, $N=A$, and so $G/A\cong B$. I am not sure what you mean by "trouble understanding the structure". – Alex B. Mar 23 '12 at 2:54
@nullgraph: For a "nonconceptual reason", consider the Fundamental Theorem of finitely generated abelian groups: every finitely generated abelian group is a direct sum of cyclic groups. If $G = C_1\oplus\cdots\oplus C_n$, then you can take $G_i = C_{i+1}\oplus\cdots \oplus C_n$ in the definition of polycyclic. – Arturo Magidin Mar 23 '12 at 2:59

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