Let $G$ be finitely presented and let $H$ be finitely generated such that $G$ is quasi-isometric to $H$. Show that $H$ is finitely presented.
I am not 100% certain about the definition of quasi-isometry for groups, I believe we can take it to mean that their Cayley graphs are quasi-isometric (which I'm fine with, since I know how to show that if $S_1$, $S_2$ two finite generating sets of $G$ then $\Gamma(S_1 , G)$ is quasi-isometric to $\Gamma(S_2, G)$).
I started by writing $G$ as $\langle S \mid R\rangle$, and letting $T$ be a finite generating set of $H$. Let $f\colon G\to H$ be a $(K,A)$-quasi-isometry. I note that any $h\in H$ is at most a distance $L$ from $f(v)$ for some $v\in G$ ($L$ a positive constant independent of $h$).
I'm not sure how to proceed.
Best Answer
Read section "Connectivity and coarse connectivity" in Chapter 9 of
Druţu, Cornelia; Kapovich, Michael, Geometric group theory. With an appendix by Bogdan Nica, Colloquium Publications. American Mathematical Society 63. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-1104-6/hbk; 978-1-4704-4164-7/ebook). xx, 819 p. (2018). ZBL1447.20001.
The proof of the result you are asked to prove is there.
Of course, some of this material might be covered in your lecture notes, I have no way to tell.
Alternatively, read Theorem 18.2.12 (for groups of type $F_2$) in
Geoghegan, Ross, Topological methods in group theory, Graduate Texts in Mathematics 243. New York, NY: Springer (ISBN 978-0-387-74611-1/hbk). xiv, 473 p. (2008). ZBL1141.57001.