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)?



Distance to a geodesic ray
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 $|x|\_S=3$.


