# Infinite order elements with zero translation length on Cayley graph

Suppose $G$ is a group with a finite symmetric generating set $S$, and $g\in G$ an infinite order element of $G$. The asymptotic translation length of $g$ on the associated Cayley graph $Cay(G,S)$ is defined as $\tau(g)=\lim_{n\to \infty}||g^{n}||/n$ where $||.||$ denotes $S$-word length.

Is it possible for $g$ to be infinite order but for $\tau(g)$ to be zero?

• Yes: it's often called a "distorted element"; a search will point to many further examples.
– YCor
Nov 27 '17 at 0:02
Yes. Consider for example $BS(1,2)$, which is given by presentation $\langle a,b|bab^{-1}=a^2 \rangle$.
Then $||{a^{2^k}||/2^k} = ||b^{k} a b^{-k}||/2^k \leq (2k+1)/{2^k}$, which tends to $0$ as $k$ tends to infinity.
To see that $a$ has an infinite order consider action of the group on $\mathbb{R}$ given by $a(x)=x+1$ and $b(x)=2x$ for all $x \in \mathbb{R}$. By universal property this extends to an action provided $bab^{-1}(x)=a^2(x)$, which is easily seen to be satisfied.