If $G/\left(\mathbb{Z}/n\mathbb{Z}\right)$ is residually finite and $\left(\mathbb{Z}/n\mathbb{Z}\right)\leq Z(G)$, then $G$ is residually finite Let $G$ group and $\left(\mathbb{Z}/n\mathbb{Z}\right)\cong H\leq Z(G)$ a subgroup of the center of $G$ isomorphic to the integers modulo $n$. Is it true that if $G/H$ is residually finite, then $G$ is residually finite? (def: $G$ is residually finite if for every $g\in G\backslash\{e\}$ there exists $g\not\in N\triangleleft G$). It makes a lot of sense, but I couldn't prove it or find an counterexample. Any help is appreciated.
 A: I should make my comment into an answer.
Let $p$ be a prime and let $X = \{x_i,y_i : i \in {\mathbb Z}\} \cup \{z\}$, and $$R = \{x_i^p,y_i^p,[x_i,y_i]z^{-p}:i \in {\mathbb Z}\} \cup \{[x_i,x_j],[y_i,y_j]:i,j \in {\mathbb Z}\} \cup \{[x_i,y_j]:i,j \in {\mathbb Z}, i \ne j\},$$
and let $G$ be the group defined by the presentation $\langle X \mid R \rangle$.
So $G$ is the central product of countably infinitely many copies of a nonabelian group of order $p^3$.
Then it is not hard to see that any subgroup of $G$ of finite index must contain $z$, but $G/\langle z \rangle$ is elementary abelian and is residually finite.
I think you can get a finitely generated example $H = \langle t \rangle$ by adjoining a new generator $t$ of infinite order together with the relations $x_i^t = x_{i+1}$, $y_i^t = y_{i+1}$ for all $i$. Then we still have $z \in Z(H)$, $H/\langle z \rangle$ is residually finite, but $H$ is not.
A: There's also an example — which is most likely folklore, but explicitly written down by Deligne. 
Claim. Any central extension $\Bbb Z/k \to \widetilde{Sp(2n, \Bbb Z)} \to Sp(2n, \Bbb Z)$ corresponding to $k$-sheeted cover of $Sp(2n, \Bbb R)$ except for $k = 2$ is not rFin.
This example is finitely presented, because it is central extension of arithmetic group by cyclic — arithmetic groups are f. p. and being f. p. is preserved by extensions.
(P. Deligne, Extensions centrales non résiduellement finies de groupes arithmétiques, 1978)
