Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 5 / 2 / ii / Solution

For a finite generating set $A$, use the symmetric alphabet $\Sigma =A\cup A^{-1}$. The Cayley graph has a vertex for each $g\in G$ and an oriented edge labelled $a$ from $g$ to $ga$. Give every edge length one. Its vertex word metric isTwo finitely presented groups are quasi-isometric when their Cayley graphs, with these path metrics, are quasi-isometric. The choice of finite generating set is immaterial, as the following proof shows; the relators identify vertices but do not otherwise enter the metric definition.

Given finite generating sets $A,B$ of the same group, put $L=\max (1,{\max }_{a\in A^{\pm 1}}|a{|}_B)$ and $M=\max (1,{\max }_{b\in B^{\pm 1}}|b{|}_A)$. Replace each letter in a shortest $A$-word by a fixed $B$-word to obtain $|g{|}_B\le L|g{|}_A$. Reversing the roles gives $|g{|}_A\le M|g{|}_B$. Therefore the vertex identity map satisfiesIt is onto, and hence a quasi-isometry. Map each edge to the fixed word path for its label, or pass through the uniformly dense vertex sets, to obtain the corresponding graph quasi-isometry with a bounded additive error. This proves equivalence of finite word metrics and independence of the two finite group presentations.