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Hi, I'm looking for examples of groups that are both Hopfian and Co-Hopfian. I have a non trivial (and beautiful, at least to me) example: $\mathrm{SL}(n,\mathbb{Z})$ (with $n>2$).

Do you know others (non trivial)?

Thank you.

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    $\begingroup$ @Portland, would you explain why SL(n,R) is co-Hopfian? The proof I came up with is too heavy-handed. $\endgroup$ Feb 12, 2010 at 19:01
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    $\begingroup$ Igor, isn't it an immediate consequence of super-rigidity? $\endgroup$
    – HJRW
    Feb 12, 2010 at 19:04
  • $\begingroup$ Henry, I think co-Hopf property for SL(n,Z) goes back to Borel, and learned about this from Prasad's 1976 paper. In fact, they worked in much more general setting of irreducible lattices in semisimple Lie groups. This is what I referred to as heavy-handed, and hence I was curious whether there is a more elementary solution. $\endgroup$ Feb 12, 2010 at 20:27
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    $\begingroup$ Igor, yes there is a more elementary solution, please see "Some consequences of the elementary relations of $SL_n$" from Steinberg ams.org/bookstore?fn=20&arg1=alggeom&ikey=CWORKS-7 $\endgroup$
    – Portland
    Feb 13, 2010 at 1:16

7 Answers 7

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Mapping class groups of closed surfaces are both Hopfian and co-Hopfian (the former follows from residual finiteness, and the latter is due to Ivanov-McCarthy).

Out(F_n) also has both properties (residual finiteness and a theorem of Farb-Handel).

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Before going to examples, here are some general comments:

a) Proving that a finitely generated group is Hopfian is usually pretty hard unless the group is residually finite e.g. finitely generated subgroups of $GL(n,\mathbb R)$ are residually finite, hence Hopfian by an old result of Mal'cev.

b) A common method in proving that a group $G$ is co-Hopfian is to use an invariant of $G$ that is multiplicative under passing to finite index subgroups. If $G$ has a nonzero such invariant, then $G$ has no finite index subgroups isomorphic to itself. For example, if $G$ is the fundamental group of a finite aspherical CW-complex of nonzero Euler characteristic, then $G$ has no finite index subgroups isomorphic to itself.

c) If $G$ is the fundamental group of a closed aspherical manifold, then $G$ has no infinite index subgroups isomorphic to $G$ (look at top-dimensional homology).

d) Euler characteristic, signature, $L^2$-Betti numbers, simplicial volume are are multiplicative under finite covers of closed aspherical manifolds, so if $G$ is the fundamental group of a closed aspherical manifold with say nonzero signature, then $G$ is co-Hopfian.

Here are some specific examples to add to Richard's example of one-ended torsion free hyperbolic groups. All of the following groups are linear, hence residually finite, hence Hopfian.

  1. The fundamental groups of closed locally symmetric spaces of nonpositive curvature without local flat factors are co-Hopfian because they have nonzero simplicial volume thanks to a result of Lafont-Schmidt.

  2. If memory serves me, it is possible to figure out which geometric 3-manifold groups are co-Hopfian. For example, the $SL_2(\mathbb R)$-Seifert fibered spaces have a certain invariant detecting volume of the base $2$-orbifold which is multiplicative under finite covers. Check papers of Pierre Derbez in arXiv.

  3. Fundamental groups of some nilmanifolds are co-Hopfian, see my paper here.

  4. Delzant-Potyagalo classified co-Hopfian Kleinian groups (in real hyperbolic space of any dimension). See here.

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  • $\begingroup$ Igor, your answer confused me. What is the definition of co-Hopfian? Part b of your answer implies that it means not to be isomorphic to a subgroup of finite index. However, that is not the definition in Planenmath planetmath.org/encyclopedia/HopfianGroup.html. Here journals.cambridge.org/… it is called finite co-Hopfian. $\endgroup$ Mar 28, 2011 at 9:59
  • $\begingroup$ @Yiftach, I did not want to use the term "finitely co-Hopfian", but checking this property is part of the task of checking whether the group is co-Hopf, and in b) I explain a most common way to do so. The two properties are equivalent for the fundamental groups of closed aspherical $n$-manifolds (as the cohomological dimension of the group is $n$, while any infinite index subgroup has cohomological dimension $<n$ being the fundamental group of a noncompact aspherical $n$-manifold). On the other hand a (finitely generated nonabelian) free group is finitely co-Hopf but not co-Hopf. $\endgroup$ Mar 28, 2011 at 11:59
  • $\begingroup$ Igor, thanks for the clarification. $\endgroup$ Mar 28, 2011 at 12:59
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Torsion-free $\delta$-hyperbolic groups are hopfian, and it's a theorem of Sela that one-ended torsion free hyperbolic groups are co-hopfian (Z. Sela. Structure and rigidity in (Gromov) hyperbolic groups and discrete groups in rank 1 Lie groups. II. Geom. Funct. Anal., 7(3):561–593, 1997.).

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  • $\begingroup$ Thank you Richard, I'll look into these. In the meantime, do you have examples of "one-ended torsion free hyperbolic groups"? $\endgroup$
    – Portland
    Feb 12, 2010 at 15:42
  • $\begingroup$ The simplest examples are fundamental groups of closed hyperbolic manifolds, but for those, it is much easier to prove hopfian and cohopfian (hopfian since they're residually finite, cohopfian by considering cover spaces and volume). More interesting examples can be found by adding a "random relation" to such a thing. $\endgroup$ Feb 12, 2010 at 15:46
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    $\begingroup$ You could look at Bridson and Haefliger's book "Metric spaces of nonpositive curvature" for more about hyperbolic groups. $\endgroup$ Feb 12, 2010 at 15:51
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    $\begingroup$ Also, a 'randomly chosen' finite presentation is 'almost surely' a one-ended, torsion-free hyperbolic group. And you can build lots of explicit examples using small-cancellation theory. $\endgroup$
    – HJRW
    Feb 12, 2010 at 15:53
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    $\begingroup$ One more comment. The fact that hyperbolic groups are Hopfian is also a theorem of Sela. This is a really deep result - remember that hyperbolic groups are not known to be residually finite. The reference is: Endomorphisms of hyperbolic groups. I. The Hopf property. Topology 38 (1999), no. 2, 301--321. $\endgroup$
    – HJRW
    Feb 12, 2010 at 21:32
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$\mathbb{Q}$ is a classical example.

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Torsion-free finitely generated nilpotent groups are Hopfian. Although the easiest (nontrivial) ones (such as abelian ones, Heisenberg, those occurring in unipotent radicals of parabolics of reductive groups...) are not cohopfian, many are cohopfian too.

An iff condition is that the Malcev completion (say, the complex one, but it works equally with the rational or real one) has a nontrivial non-negative grading. See my paper. The initial observation showing their existence, with a sufficient (but not necessary) condition, namely that the Malcev completion is characteristically nilpotent, appears in I. Belegradek arXiv link and earlier in D. Segal's MR review of Smith, Geoff C. Compressibility in nilpotent groups. Bull. London Math. Soc. 17 (1985), no. 5, 453–457. Quoth:

It is perhaps not so well known that finitely generated torsion-free nilpotent groups need not be compressible: this follows from J. L. Dyer's construction of a nilpotent Lie algebra all of whose automorphisms are unipotent [Bull. Amer. Math. Soc. 76 (1970), 52–56.

(In turn, Dyer 1970 is not the original reference for the latter fact, but Dixmier, J.; Lister, W., Derivations of nilpotent Lie algebras. Proc. Amer. Math. Soc. 8 (1957), 155–158.)

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I think compact $p$-adic analytic groups that have no abelian normal subgroups are Hopfian and co-Hopfian as topological groups, but I haven't seen this explicitly stated anywhere.

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  • $\begingroup$ I think arbitrary compact $p$-adic Lie groups are Hopfian (as top. groups). A sufficient condition for a compact $p$-adic Lie group to be cohopfian is that the automorphism group of its Li algebra is volume preserving. This is indeed the case when the Lie algebra is semisimple (which is equivalent to your assumption). $\endgroup$
    – YCor
    Sep 16, 2021 at 17:48
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Every open subgroup of the Nottingham group (p>3) is both Hopfian and co-Hopfian. On the one hand, the Nottingham group is hereditarily just infinite. So all it open subgroup are just infinite, that is their proper quotients are finite. On the other hand, Mikhail Ershov showed that the commensurator of the Nottingham group (p>3) is its automorphism group. So if two open subgroups are isomorphic the isomorphism extends to an automorphism of the Nottingham group. In particular, the indices in the Nottingham group of such subgroups are the same. Thus, if one is a subgroup of the other, they are equal.

EDIT: I should be a bit more careful. The fact that the commensurator of the Nottingham group is its automorphism group, does not say that the isomorphism extends to an automorphism, but that the isomorphism restricted to an open subgroup extends to an automorphism. This is good enough for the claim above. It is also may be that Mikhail actually proved the stronger claim that I made (I am not 100% sure).

EDIT2: I have got confused about the definition of co-Hopfian. This argument shows that the Nottingham group is finite co-Hopfian. It is not true that it is co-Hopfian from results of Rachel Camina (and also a paper of Fesenko).

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    $\begingroup$ In this paper (Theorem E) I show that all just infinite profinite groups that are not virtually abelian are open co-Hopfian: arxiv.org/abs/0906.1771 On the other hand, isomorphisms between a group and its closed subgroups are much more difficult to control, for instance self-reproducing branch groups are clearly not co-Hopfian. $\endgroup$
    – Colin Reid
    Mar 28, 2011 at 17:02

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