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?


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.