Extension of a virtually polycyclic group by a virtually polycyclic group I want to prove that an extension of a virtually polycyclic group by a virtually polycyclic group is virtually polycyclic, and I see no reference. I do see a proof of the analogous statement with "virtually" removed.
My attempt:
Given $1\to N\to G\to G/N\to 1$, we may assume $G/N$ is polycyclic, since any subgroup $T$ of $G/N$ is of the form $G'/N$ with $G'$ a subgroup of $G$ of the same index as $T$ (hence finite if $T$ was the finite index polycyclic subgroup). Now, given $S<N$ polycylic, if we further assume that $S$ is normal then we may use $1\to S\to G\to G/S\to 1$ to show that $G$ is virtually polycyclic (we show that $G/S$ is virtually polycyclic using the extension $1\to N/S \to G/S\to G/N\to 1$ and then repeat the trick above lifting a finite index subgroup of $G/S$ to a finite index subgroup of $G$).
Of course, $S$ need not be normal, and hence the natural candidate would be to use its normalizer in $G$. The problem is I can't prove that the normalizer has finite index in $G$ (using that $S$ has finite index with respect to a normal subgroup). Any help?
 A: Here is a proof of the question in the title.
Suppose we have the extension
$$ 1\rightarrow N\rightarrow G\rightarrow G/N\rightarrow1$$
where both $N$ and $G/N$ are virtually polycyclic.  As already claimed in the OP, we can safely assume $G/N$ is polycyclic.  By the discussion in the comments, there is a subgroup $S\le N$ such that $S$ is normal in $G$ and $N/S$ is finite. We thus have the extension
$$ 1\rightarrow N/S \rightarrow G/S\rightarrow G/N\rightarrow1$$
and we want to prove $G/S$ is virtually polycyclic (since then, with $S$ polycylic, $G$ would be virtually polycyclic).
So to rephrase, we have
$$ 1\rightarrow F\rightarrow H\rightarrow P\rightarrow1$$
with


*

*$F$ finite

*$P$ polycyclic


and we want to show $H$ is virtually polycyclic.  We do so by induction on the the cyclic series length of $P$; that is, we can write
$$ 1 < P_1 < P_2 < \cdots < P_{n-1} < P_n=P$$
where the successive quotients $P_{i+1}/P_i$ are cyclic. We are then inducting on $n$.
When $n=0$, $P$ is the trivial group, and then $H=F$ is a finite group, and thus clearly virtually polycyclic.
Now take the pre-image of $P_{n-1}$ in $H$ (call it $K$): this is a normal subgroup of $H$, which by the induction hypothesis, is virtually polycyclic.
If $H/K$ is finite, then $H$ is also virtually polycyclic, and we are done.
If $H/K$ is infinite, we have the extension
$$ 1\rightarrow K\rightarrow H\rightarrow \mathbb{Z}\rightarrow1$$
and since $\mathbb{Z}$ is free, this extension splits: there is an infinite cyclic subgroup $L\le H$ such that $KL=H$. But if $Q\le K$ is the normal polycyclic subgroup that makes $K$ virtually p.c., then $QL$ is a finite index polycyclic subgroup of $H$.
