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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?

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$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$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 $\Size{G : C}$.

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  • $\begingroup$ I like this proof. The argument I was familiar with was to consider the intersection $K$ of all finite index subgroups of the same index as $A$. It is clear that $K$ is normal and abelian. If $G$ is finitely generated, then $K$ is again finite index. But your argument does not need $G$ to be finitely generated. On the other hand, $K$ is fully invariant, i.e. $\phi(K) = K$ for all automorphisms $\phi$ of $G$. Is it possible to find a fully invariant finite index abelian subgroup when $G$ is infinitely generated? $\endgroup$ – Robert Bell Apr 16 '18 at 16:36

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