I have been trying to understand the famous theorem for Cayley graphs, which states that, essentially, most infinite groups have a $1$-ended Cayley graph (by characterizing precisely the cases in which the graph is $0$, $2$ or infinite-ended).
Now, one-generated groups (cyclic) have $2$-ended graphs, so these are out… and the "stupidest" example of a 1-ended graph I can think of is $\mathbb{Z}^2$, generated by two generators that commute. I was wondering if $\mathbb{Z}^2$ is a "toy example" or if, maybe, there are different $1$-ended graphs. Hence, my question:
Is there an example of a group $G$ and a set of generators such that the associated Cayley graph is $1$-ended, but does not contain a subgraph isomorphic to the Cayley graph of $\mathbb{Z}^2$ with the presentation $\langle x, y : xy=yx \rangle$?
Best Answer
Start by taking your favorite tiling of the hyperbolic plane by convex, compact polygons. Take the dual graph (a vertex in the center of every face of the tiling, and an edge between adjacent vertices), and this will be a Cayley graph for the symmetry group of the tiling. The group will be delta-hyperbolic, so it will contain no subgroups isomorphic to $\mathbb{Z}^2$. In fact, if your polygons have at least 5 sides and meet at least 5 together at a vertex, your Cayley graph will have no induced squares at all, so there's no copy of the standard Cayley graph of $\mathbb{Z}^2$ in it. It's also 1 ended because your original tiling was by compact (and hence non-ideal) polygons.
As a specific example, consider:
$\langle a, b, c, d, e \mid a^2 = b^2 = c^2 = d^2 = e^2 = [a,b] = [b,c] = [c,d] = [d,e] = [e,a] = 1\rangle$
where $[a,b] = a b a^{-1} b^{-1}$.