I believe I read somewhere that residually finite-by-$\mathbb{Z}$ groups are residually finite. That is, if $N$ is residually finite with $G/N\cong \mathbb{Z}$ then $G$ is residually finite.
However, I cannot remember where I read this, and nor can I find another place which says it. I was therefore wondering if someone could confirm whether this is true or not, and if it is give either a proof or a reference for this result? (If not, a counter-example would not go amiss!)
Note that I definitely know it is true if $N$ is f.g. free (this can be found in a paper of G. Baumslag, "Finitely generated cyclic extensions of free groups are residually finite" (Bull. Amer. Math. Soc., 5, 87-94, 1971)).
Best Answer
The modified question has a positive answer if $N$ is finitely generated.
Consider an extension $1 \to N \to G \to \mathbb Z \to 1$ and take a lift $u \in G$ of the generator of $\mathbb Z$. If $N$ is finitely generated and $H' \subset N$ is a subgroup of finite index, then the intersection of all subgroups of index $[N:H']$ (call it $H$) is invariant under conjugation by $u$. Hence, for all $m \in \mathbb Z$, the subgroup $Hu^{m \mathbb Z}$ is a finite index normal subgroup of $G$.
Hence, if $N$ is finitely generated and residually finite, then $G$ is residually finite as well.