 Solution

For a finite symmetric generating alphabet $\Sigma$, the growth function of a finitely generated group isIt counts distinct elements, not words. Relations affect which words coincide, while the resulting coarse growth type is independent of the finite generating set.

Let $f:G\to H$ be a quasi-isometry between the vertex word metrics, with constants $\lambda ,ϵ$. A quasi-isometry of the metric graphs can be rounded to vertices with only bounded changes of these constants. If $f(g)=f(g^{\prime })$, its lower distortion bound gives $d_G(g,g^{\prime })\le \lambda ϵ$. Thus every nonempty fibre has at most $M={\beta }_G(\lceil \lambda ϵ\rceil )$ elements, independently of its centre. Also $f(B_G(1,n))$ lies in the ball of radius $\lambda n+ϵ$ about $f(1)$. Left translation in $H$ preserves ball cardinalities, soApply the same argument to a quasi-inverse of a quasi-isometry to get the reverse comparison. Enlarging all constants to an integer $K$ gives the two inequalities required by the paper's $\lesssim$ relation, and therefore ${\beta }_G{\sim }_e{\beta }_H$. This proves quasi-isometry invariance of the stated growth class.

For the exponential upper bound, put $q=|\Sigma |$. Every element in the radius-$n$ ball has a representative of length at most $n$. Hencewhere the last inequality follows by comparison with the binomial expansion of $(q+1{)}^n$; it also holds for the empty alphabet of the trivial group. Thus every finitely generated group has at most exponential growth, with no finite-relator assumption needed.