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Let $G$ be a group wich acts on a tree $\Gamma$. Then $U$ acts on $\Gamma$ for every $U\leq G$.

Question: Why does the following hold?

If $|G:U|<\infty$. Then the minimal $U$-invariant subtree of $\Gamma$ coincides with the minimal $G$-invariant subtree of $\Gamma$. (minimal always with respect to inclusion)

Thanks for help.

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The question looks interesting but I think I'm missing some definitions, what do you mean by coincides and does the definition of a group action on a graph differ much from the definition for one on a set? Must adjacency be preserved by a fixed group element? Thanks – muzzlator Feb 4 '13 at 10:08
@user60855: There is a very well developed theory on groups acting on trees. The classic text is the book "Trees" by Serre (originally in French, if you speak it) but there is a gentle introduction by Meier called "Graphs, Groups and Trees" which is rather good. – user1729 Feb 4 '13 at 10:53
up vote 3 down vote accepted

Beware that there can be no minimal invariant tree. You have 3 cases:

(1) some $g\in G$ acts hyperbolically. Then the minimal invariant subtree is the (convex hull of the) union of axes of hyperbolic elements and this does not change when passing to a finite index subgroup.

(2) the action is bounded. Then [edit: I was a bit imprecise about this case in my first answer] there exists a minimal tree (an edge or a vertex), which can be non-unique. More precisely, if there's no fixed vertex, the unique minimal tree is an edge (which is minimal or not in restriction to a finite index subgroup); if there's a fixed vertex, any fixed vertex is a minimal tree (and of course remains minimal for each finite index subgroup). The case of an edge can be excluded if the action is with no inversion (this can be assumed by adding a vertex at the middle of each edge).

If $G$ is finitely generated, that's it. But otherwise there's:

(3) ("horocyclic" case) the action has no hyperbolic element, but is unbounded (so there's a unique fixed point at infinity). Then there is no minimal invariant tree for $G$ or for its finite index subgroups.

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Nice. Thank you very much. – Peter Feb 4 '13 at 17:30

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