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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.2012-03-21

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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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    Sorry about the misunderstanding of $G/N$, it all makes sense now. Using Fundamental Theorem of Finitely Generated Abelian Group, I can see why finitely generated abelian groups are polycyclic. Thanks!2012-03-24