Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 133 / 3

For each generator $t\in T$, let $|t{|}_S$ be its length in the generating set $S$, and putA shortest $T$-word for $h_1^{-1}{h}_2$ has $d_T(h_1,h_2)$ letters. Replacing each letter by an $S$-word of length at most $C$ givesThus inclusion of any finitely generated subgroup is Lipschitz continuous for the corresponding word metrics.

Let $r:G\to H$ be a retraction. Part (a) supplies a constant $C$ such thatThe finite set $r(S)$ generates $H$. PutIf a shortest $S$-word represents $h_1^{-1}{h}_2\in H$, applying the group homomorphism $r$ gives a $T$-word for the same element of length at most $D{d}_S(h_1,h_2)$. HenceThe inclusion is therefore bilipschitz and in particular a quasi-isometric embedding. Thus every finitely generated retract subgroup is quasi-isometrically embedded.

This is the Baumslag-Solitar group $G=\mathrm{BS}(1,2)$. Its defining relation gives, by mathematical induction,In the cyclic subgroup $H=⟨a⟩$ with generator $a$,whereas in $G$, with generators $a,b$,If inclusion were a $(\lambda ,\epsilon )$-quasi-isometric embedding, its lower bound would implyfor every $n$, which is impossible because an exponential function eventually dominates every linear function. Hence $H$ is not quasi-isometrically embedded; it is an exponentially distorted subgroup.