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If $S $ is infinite, locally finite graph which is not tree, $\tilde{S} $ is its universal cover, $p:\tilde{S}\rightarrow S $ is covering map,and $G $ is acting on $\tilde{S} $ with finite quotient $Y $, $q:\tilde{S}\rightarrow Y $ is the corresponding covering, can we find a group $H $ which acts on $S $ with quotient $Y$ such that the diagram commutes (i.e. $r\circ p=q $ where $r:S\rightarrow Y=S/H $ is natural map)?

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I don't understand where the "natural map" $r: S \to Y$ should come from without further assumptions. You may want to look at Bass's Covering theory for graphs of groups. After all, $\tilde{S}$ is a locally finite tree. – t.b. Aug 29 '11 at 10:29
I'm mentioning this since this question reminds me of this question on MO. For a good and elementary introduction to these ideas look at the appendix to Chapter 1 of Hatcher's book. Also, the book by Dicks and Dunwoody might be a place to look at. – t.b. Aug 29 '11 at 10:39
@Theo: Here natural map $r:S\rightarrow Y=S/H$ is the quotient map by action of $H$. The question is whether there exist $H$ acting on $S$ with quotient map $r:S\rightarrow Y=S/H$ with diagram commutative. I am looking the ref. in comment 1; but couldn't handle the question. – user8186 Aug 29 '11 at 11:17
I understand that that's what you want. But why should $S$ even be a cover of $Y$ (you're asking for much more)? – t.b. Aug 29 '11 at 11:22
@Theo: Here I have assumed $S$ is infinite graph which is not tree and $Y$ is a finite quotient of $\tilde{S}$. therefore, I was expecting whether there is such map from $S$ to $Y$. – user8186 Aug 29 '11 at 11:26

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