Let $S$ be an infinite graph, $G$ is a group acting (effectively) on $S$ with finite quotient graph $S/G$. Make $S/G$ into graph of groups in obvious way by assigning stabilizers at vertices and edges.

Let $\tilde{S}$ be universal cover of $S$ and $H$ be a group acting (effectively) on $\tilde{S}$ with same quotient as graph $S/G$, but the graph of group may be different.

Question: Does there exist a subgroup $K$ of $G$ which acts on $S$, with quotient graph of groups $S/K$ and $\tilde{S}/H$ isomorphic?

[Here I am failing to use covering space theory directly, because here I am considering quotients spaces with some algebraic structures on them, namely graph of groups. This problem arise when I was studying Serre's "Trees".]

  • $\begingroup$ In your third paragraph, should $H$ be $G$? $\endgroup$ – HJRW Aug 29 '11 at 6:58

If I understand your question correctly then the answer is 'no'. Indeed, whenever countable $\Gamma=A*_C B$ with $|A:C|=|B:C|=\infty$ then the Bass--Serre tree $S$ is a countably branching regular tree and the quotient graph $S/\Gamma$ is just a single edge. So, taking $S=\widetilde{S}$ and $K=G$, if the answer to your question were 'yes' then this would imply that, for any two such amalgamated free products, one contains another as a subgroup. But this is absurd.

For instance, let $G$ be the fundamental group of the orientable surface of Euler characteristic -2, and let $H$ be the fundamental group of the non-orientable surface of Euler characteristic -2. Then neither contains the other as a subgroup. But cutting either along a non-separating simple closed curve realises is as an amalgamated free product as above.

The moral of the story is that the combinatorial type of the Bass--Serre tree, without the data of the group action, doesn't contain much information.

  • $\begingroup$ @HW: True! To avoid countably branching in $S$ or $\tilde{S}$, we can consider $S$ to be locally finite. Then can we hope for positive answer? $\endgroup$ – joseph Aug 29 '11 at 8:03
  • $\begingroup$ No. Take A and B to be non-isomorphic finite groups of the same order. Then $A*A$ and $B*B$ will exhibit the same problem. $\endgroup$ – HJRW Aug 29 '11 at 11:32

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