Cfrgtkky

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?

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) bounded orbits


• (b) horocyclic (fixes a unique point at infinity, no loxodromic element; preserves "horospheres")


• (c) axial (preserves an axis, on which some element acts loxodromically)


• (d) focal (fixes a unique point at infinity, existence of a loxodromic element)


• (e) general type (= other): implies the existence of a non-abelian free subgroup acting metrically properly.

(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.