I have been trying to understand the famous theorem for Cayley graphs, which states that, essentially, most infinite groups have a $1$-ended Cayley graphs (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 $\langle x, y : xy=yx \rangle$?

  • $\begingroup$ Does the subgraph have to be induced? Or are you considering just ignoring edges in a way that loses connection with any subgroups of your group? $\endgroup$ – Charlie Cunningham Jul 21 at 16:13
  • $\begingroup$ As I am just trying to "visualize" the graph really just ignoring edges, but if there is any additional property I welcome it! $\endgroup$ – AnalysisStudent0414 Jul 21 at 16:17
  • 1
    $\begingroup$ The finitely generated groups with $0$, $2$, and $\infty$ ends all fit into verrrrry special classes: each $0$ ended group is finite; each $2$ ended group contains an infinite cyclic $\mathbb Z$ subgroup with finite index; and each $\infty$ ended group can be written as an amalgamated free product or HNN amalgamation over a finite group. Literally every finitely generated group not on one of those very special lists is one-ended. $\endgroup$ – Lee Mosher Jul 22 at 3:50

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 $[a,b] = a b a^{-1} b^{-1}$.

  • $\begingroup$ Here's a picture of what the Cayley graph of that last group looks like. (There are squares in this one, but no copy of $\mathbb{Z}^2$.) en.m.wikipedia.org/wiki/File:H2_tiling_245-4.png $\endgroup$ – Charlie Cunningham Jul 21 at 19:33
  • $\begingroup$ Thank you so much, it was incredibly helpful $\endgroup$ – AnalysisStudent0414 Jul 22 at 10:26
  • $\begingroup$ Alternatively, the Cayley graph of the surface group of genus $2$ under standard generators forms a tiling of the hyperbolic plane by regular octogons of angle $2\pi/8$. $\endgroup$ – YCor Jul 24 at 22:47

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