Vtyjkyuk

Let $G$ group and $left(mathbbZ/nmathbbZright)cong Hleq Z(G)$ 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 $gin Gbackslashe$ there exists $gnotin Ntriangleleft G$). It makes a lot of sense, but I couldn't prove it or find an counterexample. Any help is appreciated.

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.

There's also an example — which is most likely folklore, but explicitly written down by Deligne.

Claim. Any central extension $Bbb Z/k to widetildeSp(2n, Bbb Z) to Sp(2n, Bbb Z)$ 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)