Assume that $X$ is a tree such that every vertex has infinite degree, and a discrete group $G$ acts on this tree properly (with finite stabilizers) and transitively.
Is it true that $G$ contains a non-abelian free subgroup?
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Sign up to join this communityAssume that $X$ is a tree such that every vertex has infinite degree, and a discrete group $G$ acts on this tree properly (with finite stabilizers) and transitively.
Is it true that $G$ contains a non-abelian free subgroup?
Yes.
You're assuming more than what's necessary.
For an isometric group action on a Gromov-hyperbolic space, you have 5 possibilities (Gromov's classification):
(a), (b), (c) are clearly ruled out in your setting (transitivity, and valency; valency $\ge 3$ would be enough). (d) is ruled out because discrete groups have no metrically proper focal action at all. So (e) holds.
The case of actions on trees is a useful motivating baby case illustrating the above "classification"; all cases can actually occur.
The geometry of simplicial trees can be extended in two directions: Gromov-hyperbolic spaces and CAT(0) cube complexes. Yves' answer is based on the first point of view. About cube complexes, we have the following statement (essentially due to Caprace and Sageev):
Theorem: Let $G$ be a group acting on a finite-dimensional CAT(0) cube complex $X$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index subgroup which decomposes as a short exact sequence $$1 \to L \to G \to \mathbb{Z}^n \to 1$$ where $n \leq \dim(X)$ and where $L$ is locally $X$-elliptic (ie., any finitely generated subgroup of $L$ has a bounded orbit in $X$).
As a consequence, if a Tits' alternative is known for cube-stabilisers, then it is possible to deduce a Tits' alternative in the entire group. In the particular case of trees, we have:
Corollary: Let $G$ be a group acting on a tree $T$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index* subgroup which decomposes as a short exact sequence $$1 \to L \to G \to Z \to 1$$ where $L$ is locally $T$-elliptic and where $Z$ is trivial or infinite cyclic.
Of course, a direct argument can be given here.
Sketch of proof of the corollary. We distinguish two cases.
First, assume that $G$ has a finite orbit in $\partial T$. So it contains a finite-index subgroup $H$ which fixes a point $\xi \in \partial T$. Fix a basepoint $x \in T$. Notice that, for every $g \in H$, the intersection $g [x,\xi) \cap [x,\xi)$ contains an infinite geodesic ray (as $g$ fixes $\xi$). Consequently, $g$ defines a translation of length $\tau(g)$ on a subray of $[x,\xi)$. (Here $\tau(g)$ is positive if the translation is directed to $\xi$ and negative otherwise.) The key observation is that $$\tau : H \to \mathbb{Z}$$ defines a homomorphism whose kernel contains only elliptic elements (as each such element fixes pointwise a subray of $[x,\xi)$). Thus, we get the desired short exact sequence.
Next, assume that $G$ has infinite orbits in $\partial T$. Play ping-pong to construct a non-abelian free subgroup. $\square$
*As noticed by Yves Cornulier in the comments, the subgroup can be always taken with index at most two.