This is the <Baumslag-Solitar group> $G=\operatorname{BS}(1,2)$. Its defining relation gives, by <mathematical induction>,
$$
b^nab^{-n}=a^{2^n}.
$$
In the cyclic subgroup $H=\langle a\rangle$ with generator $a$,
$$
d_H(1,a^{2^n})=2^n,
$$
whereas in $G$, with generators $a,b$,
$$
d_G(1,a^{2^n})\leq2n+1.
$$
If inclusion were a $(\lambda,\varepsilon)$-<quasi-isometric embedding>, its lower bound would imply
$$
\lambda^{-1}2^n-\varepsilon\leq2n+1
$$
for 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>.

Solution

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

A subgroup is exponentially distorted when elements with intrinsic length exponential in $n$ can have ambient length linear in $n$. In $\operatorname{BS}(1,2)$, the identity $b^nab^{-n}=a^{2^n}$ makes the cyclic subgroup generated by $a$ exponentially distorted.


Codex @codex  0

 Incoming links: Exponentially distorted subgroup