$G$ group, $H \triangleleft G$ such that $\frac{G}{H}$ is cyclic, $H$ is residually finite and $H$ is finitely generated Suppose $G$ is a group with a normal subgroup $H \triangleleft G$ such that $\frac{G}{H}$ is cyclic, $H$ is residually finite and $H$ is finitely generated. Show that $G$ is residually finite.
A group $H$ being residually finite means for all $h \in H$ there exists $N \triangleleft H$
such that $[H,N] < \infty$ and $h \notin N$.

So, i've been working on this forever, I think i'm on the right track but could definitely use some help finishing it up and smoothing out some details.
Let $g \in G$. Then, since $\frac{G}{H}$ is cyclic, $g = z^rh$ for some $z \in G$ and $h \in H$. Since $H$ is residually finite, $\exists$ $N \triangleleft H$ such that $h \notin N$ and $[H,N]=n < \infty$. Since $H$ is finitely generated, it has finitely many subgroups of index $n$. Since $H \triangleleft G$, we have that $N^{z^r} \triangleleft H^{z^r} = H$ for all $r \in \mathbb{N}$; thus $N$ has only finitely many conjugates in $G$. Let $I$ be the intersection of all these conjugates of $N$. Then $[H,I] < \infty$ and $I$ char $H$, and so $I \triangleleft G$
edit: STill trying to finish the proof!!
 A: The fact that $H$ is residually finite is not needed. If $G/H$ is generated by $zH$  then $G$ is generated by $z$ and generators of $H$.
The assumption that $H$ is residually finite implies that $G$ is also residually finite. Since that was the real question the proof is this. If the cyclic factor group is finite then $H$ has finite index in $G$ and so $G$ is residually finite (because every subgroup of finite index of $H$ contains a characteristic subgroup of finite index).
If the cyclic factor group is infinite then we have an extension of a finitely generated subgroup by a free group which is residually finite (proved by Baumslag): this extension necessarily splits and a semidirect product of two residually finite finitely generated groups is residually finite.
An alternative proof. Every finite index subgroup of $H$ contains a finite index characteristic subgroup. So $G$ is residually an extension of  a finite group by a cyclic group. It remains to show that an extension $G$ of a finite group $N$ by a cyclic group is residually finite. But in this group the derived subgroup is inside $N$ so it is finite. Then the center is of finite index. A finitely generated Abelian group is residually finite and we are done.
