Is an extension of two abelian residually finite groups a residually finite group?

Question

We have a short exact sequence $1 \to H \to G \to K \to 1 $, where $H$ and $K$ are abelian residually finite groups. My question is: it is true that $G$ is then a residually finite group?.

Answer 1

In the article On the finiteness of certain soluble groups (1959), P. Hall proves that:

Theorem. Every finitely generated abelian-by-nilpotent group is residually finite. In particular, finitely generated metabelian groups are always residually finite.

Of course, the statement is false for groups that are not finitely generated. Even abelian groups may not be finitely generated (such as divisible groups, e.g. $\mathbb{Q}$ or $\mathbb{Z}(p^\infty)$).

Answer 2

Just slightly changing the notation: the question asks, given a group with normal subgroup $N$ and quotient $Q=G/N$, such that $N$ and $Q$ are abelian and residually finite, whether $G$ has to be residually finite.

(1) One counterexample is $K\rtimes K^*$ when $K$ is an infinite field of characteristic $p>0$. This is not residually finite (because $K$ is mapped injectively or trivially in every quotient), while $K$ is residually finite.

The group $K^*$ may fail to be residually finite (e.g., when $K$ is algebraically closed), but for $K=\mathbf{F}_p(t)$ it is residually finite (direct product of $\mathbf{F}_p^*$ with a free abelian group of infinite rank).

(2) Here's a counterexample, also split extension, with $Q$ finitely generated (infinite cyclic). Consider $M=\mathbf{F}_p[t^{\pm 1}]/\mathbf{F}_p[t]$, with $T$ the surjective locally nilpotent endomorphism induced by multiplication by $t$. Write $U=\mathrm{Id}+T$. It is thus invertible and we can consider the corresponding semidirect product $G=M\rtimes_U\mathbf{Z}$. Since for the action of $T$, $M$ is a locally nilpotent module, every finite quotient is a nilpotent module, but since $T$ is surjective, this means that the only finite quotient module of $M$ is reduced to $\{0\}$. So $M$ is not a residually finite module (for $T$, and hence for $U$) and thus $G$ is not residually finite.

(3) Here's a counterexample, with $N$ finite. Fix a prime $p$ and fix a non-abelian group $F$ of order $p^3$, with two non-commuting elements $x,y$. Let $F_n$ be a copy of $F$, with the corresponding elements $x_n,y_n$, and $z_n=[x_n,y_n]$. Let $G$ be quotient of the direct sum $\bigoplus_n F_n$ by identifying all central elements $z_n$ (thus defining a single central element $1\neq z=[x_n,y_n]$ of order $p$). Then $G/\langle z\rangle$ is abelian and residually finite. But $G$ is not residually finite: in a finite quotient, there exist $n\neq m$ such that $x_n$ and $x_m$ have the same image. Hence $z=[x_n,y_n]$ and $[x_m,y_n]=1$ have the same image. Hence $z$ is killed in every finite quotient.

(4) For a split extension, if $N$ is finitely generated, it is easy to see that $G$ is residually finite (without the abelian assumption).

(5) If $G$ itself is finitely generated, it is an observation of Ph. Hall that indeed $G$ is residually finite (as mentioned in Anthony's answer). Indeed, if $Q$ is finitely generated abelian, then $G$ is residually finite if and only if $N$ is a residually finite $Q$-module, and this indeed holds when $N$ is a finitely generated $Q$-module, but not in general as (2) above shows (even when the underlying group $N$ is residually finite).