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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4$\begingroup$ 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 StarrCommented Jun 7, 2017 at 13:27
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$\begingroup$ 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 PolizziCommented Jun 7, 2017 at 13:31
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1$\begingroup$ @FrancescoPolizzi, there's no requirement that $H$ shouldn't be trivial in the question. $\endgroup$– HJRWCommented Jun 7, 2017 at 13:33
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$\begingroup$ @Jason Starr: Oh, right! Not what I had in mind, but it works fine, thanks. $\endgroup$– abxCommented Jun 7, 2017 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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4$\begingroup$ The non-residual finiteness of that extension is also the reason that my answer to the following question is wrong: mathoverflow.net/questions/108211/… $\endgroup$ Commented Jun 7, 2017 at 13:38
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3$\begingroup$ Deligne's examples regularly reappear in MathOF; in addition to Jason's answer: mathoverflow.net/questions/22814/…, mathoverflow.net/questions/78341/…, mathoverflow.net/questions/136431/… $\endgroup$– YCorCommented Jun 7, 2017 at 21:04