TITLE: Is Hopf property a quasi-isometry invariant?
QUESTION [17 upvotes]: 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. 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.

REPLY [23 votes]: 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.