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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.

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    $\begingroup$ Further evidence for Question 1: ams.org/mathscinet-getitem?mr=1878322 and the fact that hyperbolic groups are Hopfian and QI invariant $\endgroup$
    – Ian Agol
    Commented Jul 11, 2013 at 18:43
  • $\begingroup$ Tag open-problem? $\endgroup$
    – Alexander Chervov
    Commented Jul 11, 2013 at 18:54
  • $\begingroup$ Incidentally, the class of relatively hyperbolic groups is known to be QI rigid and Hopfian property in this class is currently unclear (one has to impose Hopfian condition on parabolic subgroups to make this reasonable). This is an open problem due to Osin, with partial results due to Belegradek and Szczepanski. (My knowledge here might be outdated though.) $\endgroup$
    – Misha
    Commented Jul 11, 2013 at 18:54
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    $\begingroup$ @AlexanderChervov: Maybe, I am not sure yet: I might be missing some known examples, which is why I am asking. There was a major activity in the last 5 years on groups acting on cubical CAT(0) complexes and I know only a fraction of these papers. $\endgroup$
    – Misha
    Commented Jul 11, 2013 at 18:55
  • $\begingroup$ Don't you mean every surjective endomorphism is injective? $\endgroup$
    – Benjamin Steinberg
    Commented Jul 11, 2013 at 19:12

1 Answer 1

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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.

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    $\begingroup$ The example here (mathoverflow.net/questions/59209/hopfian-property/60392#60392) (as corrected by BS in the comments there) also shows that hopfian is not stable under taking quotients by finite normal subgroups. $\endgroup$
    – YCor
    Commented Jul 11, 2013 at 23:07
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    $\begingroup$ Note that the condition for BS(m,n) given by Baumslag-Solitar to be hopfian is not correct, it was corrected by Collins and Levin 16 years later. The corrected version is that, for $|m|,|n|>1$, BS(m,n) is hopfian iff $n$ and $m$ have the same set of prime divisors. So this indeed implies that $BS(12,18)$ is hopfian, but maybe there is a double-check to know whether it has a non-hopfian subgroup of finite index. $\endgroup$
    – YCor
    Commented Jul 11, 2013 at 23:14
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    $\begingroup$ here's the link to the Collins-Levin paper ams.org/journals/tran/1972-164-00/S0002-9947-1972-0285589-5/… (the title is "nonresidually finite one-relator groups). They indeed observe that in $BS(2,4)=(t,x|tx^2t^{-1}=x^4)$, the normal subgroup generated by $t$ and $x^2$, which has index 2, has the presentation $(t,u,v|tvt^{-1}=v^2,uvu^{-1}=v^2)$ (where $u=xtx^{-1}$, $v=x^2$); this is Higman's 1951 example of a non-Hopfian group (using the onto endomorphism mapping $(t,u,v)$ to $(t,u,v^2)$, whose kernel contains $[tu^{-1},v]$). $\endgroup$
    – YCor
    Commented Jul 12, 2013 at 11:34
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    $\begingroup$ (Sorry, the previous link is for Meskin's paper. Collins-Levin is here, but in restricted access: link.springer.com/article/10.1007%2FBF01192801 ) $\endgroup$
    – YCor
    Commented Jun 7, 2017 at 16:22
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    $\begingroup$ In contrast, for f.g. groups (or more generally groups with finitely many subgroups of each given index), being Hopfian is inherited from finite index subgroups. Reference: R. Hirshon. Some theorems on hopficity. Trans. Amer. Math. Soc. 141 (1969) 229-244. $\endgroup$
    – YCor
    Commented Jun 7, 2017 at 16:23

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