Groups acting on trees

Question

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?

Answer 1

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.

Answer 2

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.