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Given the Cayley graph of a group $G$ (for some fixed generating set $S$), consider the set $J_S$ of all the elements which lie on some infinite geodesic ray starting at the identity element of $G$, $e_G$. Let $d$ denotes the distance in the Cayley graph

A simple example where $J_S \neq G$ is $G = \mathbb{Z} \times \mathbb{Z}_2$ with the generating set $S = \lbrace (\pm 1,0), (\pm 1 ,1) \rbrace$. In that case, $(0,1)$ does not lie on an infinite geodesic ray starting at $(0,0)$.

Question: Is there a choice of $G$, $S$ and $x \in G \setminus S$ so that $d(x,J_S) > \tfrac{1}{2} d(x,e_G)$.

[Edit: As David pointed out below, the interesting question would be: does $\limsup \frac{d(x_n,J_S)}{ d(x_n,e_G)} \leq \tfrac{1}{2}$ (where $x_n$ is some enumeration of the vertices)?

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    $\begingroup$ Since $x \in S$, and assuming $x \ne e_G$, it follows that $d(x,e_G)=1$. Your inequality therefore simplifies to $d(x,J_s) > \frac{1}{2}$. Is that correct? $\endgroup$
    – Lee Mosher
    Commented Nov 10, 2017 at 1:25
  • $\begingroup$ As Lee points out, your example seems to be a solution to the question as stated. Perhaps you want such $x$ to exist arbitrarily far from the identity? For instance, in a lamplighter, I would expect that for any sequence $x_n$, we should have $\limsup(d(x_n,J_S)/|x_n|)\leq 1/4$. $\endgroup$
    – David Cohen
    Commented Nov 10, 2017 at 4:11
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    $\begingroup$ Perhaps $x \in G$ rather than $x \in S$ was intended? $\endgroup$
    – Derek Holt
    Commented Nov 10, 2017 at 7:55
  • $\begingroup$ For $\mathbb{Z}\times\mathbb{Z}_2$, $S=\{(\pm1,0),(0,\pm1)\}$ and $x= (0,1)$ we have $\mathrm{dist}_{J_S}x=\mathrm{dist}_{e}x$... $\endgroup$
    – Anton Petrunin
    Commented Nov 11, 2017 at 0:54
  • $\begingroup$ @Lee: In my example, $d(x,e_G) = 2$ you have to move one up on the $\mathbb{Z}$ part [$(1,0)$ move] then one diagonally [$(-1,1)$ move]. In particular, $d(x,e_G) = 2$ and $d(x,J_S) = 1$ so it is not a counterexample. $\endgroup$
    – ARG
    Commented Nov 11, 2017 at 9:12

1 Answer 1

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Let $G=\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$. Let $S$ consist of $(1,1),(1,0),(0,1)$ and their inverses. Let $x=(0,3)$.

It seems clear that the neighbors of $x$ are contained in $\{-1,0,1\}\times\{2,3,4\}$, since each coordinate of a neighbor of $x$ will differ by $0,1$, or $-1$ from the corresponding coordinate of $x$.

If $(n,k)\in G$ and $n>7$, then it is clear that the word norm $|(n,k)|_S$ is equal to $n$ (representing $k$ by some element of $\{0,1,\ldots, 6\}$, we have that $a^kb^{n-k}$ represents $(n,k)$, where $a=(1,1)$ and $b=(1,0)$, and it is obvious that no shorter word can represent $(n,k)$). Hence, any geodesic word representing $(n,k)$ uses only the generators $(1,1)$ and $(1,0)$. Similarly, if $n<-7$, any geodesic word representing $(n,k)$ uses only the generators $(-1,-1)$ and $(-1,0)$. It follows that $J_S$ contains no neighbors of $x$, as

$J_S\cap (\{-1,0,1\}\times\mathbb{Z}/7\mathbb{Z})=\{(-1,-1),(-1,0),(0,0),(1,0),(1,1)\}$.

Thus, $d(x,J_S)\geq 2$. But $|x|_S=3$.

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