Recall that a group $G$ is called *Hopfian* if every surjective endomorphism $G\to G$ is injective. Malcev observed that all finitely-generated (f.g.) residually finite groups are Hopfian. It is well-known that residual finiteness is not a *coarse invariant*, i.e. a residually finite f.g. group can be quasi-isometric to a non-residually finite one. For instance, Burger and Mozes proved that 
$F\_2\times F\_2$ is quasi-isometric to a simple group. Earlier examples, due to Deligne, were of non-residually finite central extensions of residually finite groups, with kernel of order 2. Deligne's examples imply that residual finiteness is not even a *virtual isomorphism* invariant. 


*Question 1:* Is Hopfian property of preserved by quasi-isometries of f.g. groups? 


Natural candidates would be examples of non-Hopfian CAT(0) groups constructed by Dani Wise in his [thesis](http://www.gidon.com/dani/tl.cgi?athe=pspapers/nonhopfCAT(0).ps). However, I do not know if such groups are quasi-isometric to, say, residually finite groups. A subquestion of Question 1 is:


*Question 2*. Suppose that $G$ is a group acting geometrically on a product of simplicial trees of finite valence. Can $G$ be non-Hopfian? 


Note that such $G$ is necessarily quasi-isometric to a product of free groups and such products are residually finite. 



Is Hopf property a quasi-isometry invariant?
The answer to Question 1 is "no". The group $\langle x, y \mid x^{12}y = yx^{18}\rangle$
is Hopﬁan but contains a non-Hopﬁan subgroup of ﬁnite index (see Baumslag, Gilbert; Solitar, Donald Some two-generator one-relator non-Hopfian groups. Bull. Amer. Math. Soc. 68 1962 199–201.).


 **Edit.**  Since the paper of Baumslag and Solitar contains almost no proofs, and some results of it turned out to be wrong, it is probably better to refer to Proposition A of Collins, Donald J.; Levin, Frank
Automorphisms and Hopficity of certain Baumslag-Solitar groups. 
Arch. Math. (Basel) 40 (1983), no. 5, 385–400 in combination with S.Meskin, On residually finite one-relator groups. Trans. Amer. Math. Soc. 164, 105--114 
(1972). The simplest Hopfian group with non-Hopfian subgroup of finite index would then be 
$\langle a,b \mid ab^2=b^4a\rangle$. Collins and Levin proved that this group is Hopfian and Meskin proved that it contains a non-Hopfian subgroup of finite index. 


