Does a virtually abelian group contain a normal finite index abelian subgroup?

Question

I am trying to understand the proof of Lemma $3.1$ of the paper On Some of the Residual Properties of Finitely Generated Nilpotent Groups by Koberda. It states the following:

Let $N$ be a finitely generated torsion-free nilpotent group which is virtually abelian. Then $N$ is abelian.

The first step is to let $N' < N$ be a finite index normal subgroup which is abelian, but I'm not sure why such a subgroup exists. A group is virtually abelian if it has an abelian subgroup of finite index, but it doesn't necessarily have to be a normal subgroup does it?

If $G$ is a virtually abelian group, does it necessarily contain a finite index normal subgroup which is abelian?

If the answer is no, why does such a subgroup exist in this case?

Answer 1

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Yes, the point being that if $A$ is a subgroup of finite index of a group $G$, then it will have a finite number of conjugates, as this is the finite index of $N_{G}(A)$ in $G$.

Now the intersection of these conjugates is normal in $G$, and it has again finite index in $G$, because if $B, C$ are two subgroups of finite index, then $B \cap C$ also is, and actually $$ \Size{G : B \cap C} \le \Size{G : B} \cdot \Size{G : C}. $$ For the latter formula, one uses the fact that there is a bijection between the cosets of $B \cap C$ in $B$, and the cosets of $C$ in the (subset) $B C$. So that when the index of $C$ is finite, then there is a finite number of cosets of $C$ in $B C$, and thus $\Size{B : B \cap C}$ is finite, and at most $\Size{G : C}$.