# Finite quotients of graphs

If $X$ is an infinite graph, $G$ is a group acting on $X$ with finite quotient; make $Y=X/G$ into graph of groups by attaching stabilizers at vertices and edges. Let $Z$ be a graph of groups, with graph equal to graph of $X/G$, all groups at vertices and edged being finite(but not all trivial), and suppose there is morphism of graph of groups $\phi:X/G\rightarrow Z$ , such that $\phi$ is graph isomorphism, and its restriction to vertex groups (and edge groups) is surjective homomorphism.

Does there exist a group whose action on $X$ gives the graph of groups $Z$.

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There is an early paper by Haefliger who looks at extensions of complexes of groups, perhaps the answer is in that. – Tim Porter Aug 30 '11 at 6:24
joseph, for the record, I find it much easier to gain intuition about graphs of groups by thinking topologically of graphs of spaces. The first reference for this is a paper by Scott and Wall called 'Topological methods in group theory'. – HJRW Aug 30 '11 at 9:32

No. For instance, let $G=\mathbb{Z}/4*\mathbb{Z}/4$ and let $X$ be the Bass--Serre tree, the infinite 4-regular tree. Now let $Z$ be the graph of groups corresponding to $\mathbb{Z}/2*\mathbb{Z}/2$. Then the Bass--Serre tree for $Z$ is just a line. But if $Z$ were a quotient of $X$ then it would follow that the line surjects $X$, which is absurd.