Why is every subgroup of a finitely generated nilpotent group closed in the profinite topology?

Question

This should be a well known claim, but what is the proof?

Why is every subgroup of a finitely generated nilpotent group closed in the profinite topology?

Answer 1

In a f.g. nilpotent group $G$, every subgroup is closed in the profinite topology.

This is proved by induction on the Hirsch length $h$ of $G$.

If $h=0$, $G$ is finite, OK. Otherwise, the center of $G$ is infinite and hence contains an infinite cyclic subgroup $Z=\langle z\rangle$. Write $Z^n=\langle z^n\rangle\subset Z$. Let $H$ be a subgroup of $G$. Then $HZ^n/Z^n$ is profinite-closed in $G/Z^n$, hence $HZ^n$ is profinite-closed in $G$. If some $Z^n$ for $n\ge 1$ is contained in $H$, then $HZ^n=H$ and we are done. Otherwise, it still holds that $\bigcap_n HZ^n$ is profinite-closed in $G$. Let's check that $H=\bigcap_n HZ^n$. Let $g$ be in the right-hand subgroup. Then for every $n$ there exists $n\neq m$ and $h,h'\in H$ such that $g=hz^n=h'z^m$. So $h^{-1}h'=z^{n-m}$, thus $z$ has a nontrivial power in $H$, which we excluded. So $H$ is profinitely closed.