An infinite polycyclic group has a free abelian normal subgroup The question is in the title: prove that an infinite polycyclic group has a free abelian normal subgroup. 
Can anybody give me a hint on how to go about proving this?
Things I know:


*

*A polycyclic group $G$ has a series $G = G_k \rhd G_{k-1} \rhd \dots \rhd G_1 \rhd G_0 = \{1\}$ with each $G_{i+1}/G_i$ cyclic for $0\le i \le k-1$.

*A free abelian group is an abelian group with a basis. 

*The factors in the series above are either infinite cyclic or finite. 

 A: Here is a sketch proof, and I leave you to fill in the details.
A polycyclic group is solvable, so the derived series of $G$ terminates in the trivial subgroup. Then, by considering the last infinite quotient in the successive factors of the derived series, we see that there are normal subgroups $N,M$ of $G$ with $N < M$, $N$ finite, and $M/N$ infinite abelian.
Let $K/N$ be the torsion subgroup of $M/N$. Then $K$ is finite and $M/K$ is infinite and free abelian. Since $K$ is characteristic in $M$, it is normal in $G$.
Now $C_M(K)$ is also normal in $G$ and, since $K$ is finite, has finite index in $G$. So, by replacing $M$ by $C_M(K)$ and $K$ by $Z(K)$, we may assume that $K \le Z(M)$.
Now, if $|K|= n$, then the subgroup $L:=\langle g^n : g \in M \rangle$ of $M$ is abelian,
because $[g,h] \in K\,\forall g,h, \in M$, and so $[g^n,h^n]=[g,h]^{n^2}=1$. So in fact $L = \{ g^n:g \in M \}$, which has trivial intersection with $K$. So $L$ it is free abelian. Also, it is characteristic in $M$ and hence normal in $G$.
