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Consider a central extension of groups $\mathcal{E}$ : $$1\rightarrow A\rightarrow \widetilde{G}\rightarrow G \rightarrow 1$$with $A$ finite cyclic. Does there always exist a finite index subgroup $H$ of $G$ such that the restriction of $\mathcal{E}$ to $H$ becomes trivial?

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    $\begingroup$ I think the following short exact sequence is a counterexample. $$0\to (1/2)\mathbb{Z}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z} \xrightarrow{2} \mathbb{Q}/\mathbb{Z} \to 0$$ $\endgroup$ – Jason Starr Jun 7 '17 at 13:27
  • $\begingroup$ What if you consider the central extension $$1 \to \mathbb{Z}_2 \to Q_8 \to \mathbb{Z_2} \times \mathbb{Z}_2 \to 1,$$ where $Q_8= \{\pm1, \, \pm i, \, \pm j, \, \pm k \}$ is the group of unities of quaternions? It seems to me that all restrictions to the subgroups of $G$ are in this case of the form $$1 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 1.$$ $\endgroup$ – Francesco Polizzi Jun 7 '17 at 13:31
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    $\begingroup$ @FrancescoPolizzi, there's no requirement that $H$ shouldn't be trivial in the question. $\endgroup$ – HJRW Jun 7 '17 at 13:33
  • $\begingroup$ @Jason Starr: Oh, right! Not what I had in mind, but it works fine, thanks. $\endgroup$ – abx Jun 7 '17 at 14:26
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No. Deligne gave a famous example of a central extension $$ 1\to\mathbb{Z}/2\to \widetilde{G}\to G\to 1 $$ such that $G$ is a finite-index subgroup of $PSp(2n,\mathbb{Z})$ but $\widetilde{G}$ is not residually finite. (See, for instance, the references in this MO question.) But it's easy to see that if the extension were to virtually split then $\widetilde{G}$ would be residually finite.

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