Equivalence of finite word metrics

Changing a finite generating set of a group changes individual distances, but it cannot introduce arbitrarily large relative distortion. For two finite symmetric generating sets of a group $S,T$ of the same group, setEach $S$-letter can be replaced by at most $K$ $T$-letters, and conversely each $T$-letter by at most $L$ $S$-letters. Applying these substitutions to shortest representatives of $g^{-1}h$ givesThus the identity map is a bilipschitz equivalence between the two word metrics, and hence a quasi-isometry. The constants depend on the chosen generating sets of a group.^[ref]

In the Cayley graphs of the integers example, $S=\{-1,1\}$ and $T=\{-2,-1,1,2\}$ give $K=1$ and $L=2$: a unit move costs one in either alphabet, and a move of size two costs two unit moves. This recovers $d_T\le d_S\le 2d_T$. The diamond and square in word metric balls in the integer lattice obey the same comparison, because $\max \{|m|,|n|\}\le |m|+|n|\le 2\max \{|m|,|n|\}$.

Finiteness cannot simply be dropped from this universal statement. For the additive integers, take the infinite symmetric generating set of a group $U=\mathbb{Z}\setminus \{0\}$. Every nonzero displacement is a single allowed move, so $d_U(a,b)=1$ whenever $a\ne b$, whereas $d_S(0,n)=|n|$ for $S=\{-1,1\}$. There is no finite $L$ with $d_S\le L{d}_U$: choosing an integer $n>L$ violates it. This is a counterexample to unrestricted equivalence, not a claim that every infinite generating set of a group fails; two such sets with uniformly bounded generator lengths can still satisfy the same substitution bounds.^[ref]