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

A quasi-isometry $f:X\to Y$ has constants $\lambda \ge 1$, $ϵ\ge 0$ such thatfor all $x,x^{\prime }$, and every $y\in Y$ is within $ϵ$ of some point of $f(X)$. One may use a separate coarse-surjectivity constant and enlarge $ϵ$ to include it. No continuity is required.

Choose $f^{\prime }(y)\in X$ with $d_Y(f(f^{\prime }(y)),y)\le ϵ$. The triangle inequality givesCombining this with the distortion inequalities for $f$ yieldsAlso $d_Y(f(f^{\prime }(f(x))),f(x))\le ϵ$, so the lower distortion inequality gives $d_X(f^{\prime }(f(x)),x)\le 2\lambda ϵ$. Hence every point of $X$ is within $2\lambda ϵ$ of the image of $f^{\prime }$. Taking the larger common error constant shows that $f^{\prime }$ is a $(\lambda ,3\lambda ϵ)$ quasi-isometry. It is a quasi-inverse of a quasi-isometry; both composites are uniformly close to their respective identities.

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.

Choose representatives $T$ of the right cosets $Hg$, including $1$, and let $M={\max }_{t\in T}|t{|}_A$. This is finite because the index is finite. For $t\in T$ and $a\in A^{\pm 1}$, let $\overline{ta}\in T$ represent $Hta$. The elementsform a finite generating set $S$ for $H$, after adjoining inverses and discarding identities. To see generation rather than merely quote Schreier's lemma, write $h=a_1\cdots a_n$ and let $t_i$ represent $H{a}_1\cdots a_i$. Then $t_0=t_n=1$ andEach factor is a Schreier generator. For a shortest $A$-word this gives $|h{|}_S\le |h{|}_A$. Conversely every Schreier generator has $A$-length at most $2M+1$, so $|h{|}_A\le (2M+1)|h{|}_S$. Hence the inclusion $H\hookrightarrow G$ obeysEvery $g\in G$ can be written $g=ht$, and $d_A(g,h)=|t{|}_A\le M$. Thus the inclusion is coarsely onto and is a finite-index subgroup quasi-isometry.

Only finite generation of $G$ was needed for the metric argument. Under the given finite-presentation hypothesis, $H$ is also finitely presented: the corresponding finite-sheeted cover of a finite presentation complex has finitely many cells, and a spanning-tree reduction of its one-skeleton gives a finite group presentation. Therefore it belongs to the class of groups used in part (ii).