Yes. Take $G=F_2$. The <free product> satisfies
$$
G*G\cong F_2*F_2\cong F_4.
$$
A connected three-sheeted cover of the two-petal rose has <fundamental group> of rank $1+3(2-1)=4$ by the <Nielsen–Schreier formula>. Thus $F_4$ is isomorphic to an index-three subgroup of $F_2$. A <finite-index subgroup quasi-isometry> then gives
$$
\boxed{F_2\simeq_{\mathrm{qi}}F_4\cong F_2*F_2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 143 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-143/2/d/solution