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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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How can a request for examples have an answer? That's a good criterion to motivate turning a question into Community Wiki, I think. –  Mariano Suárez-Alvarez Feb 12 '10 at 15:57
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@Portland, would you explain why SL(n,R) is co-Hopfian? The proof I came up with is too heavy-handed. –  Igor Belegradek Feb 12 '10 at 19:01
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Igor, isn't it an immediate consequence of super-rigidity? –  HJRW Feb 12 '10 at 19:04
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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 –  Portland Feb 13 '10 at 1:16
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On reflection, I think Mariano has a point. I have no idea why one of the answers below is more "correct" than another. –  HJRW Feb 13 '10 at 6:38
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6 Answers 6

up vote 14 down vote accepted

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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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. –  Yiftach Barnea Mar 28 '11 at 9:59
    
@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. –  Igor Belegradek Mar 28 '11 at 11:59
    
Igor, thanks for the clarification. –  Yiftach Barnea Mar 28 '11 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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Thank you Richard, I'll look into these. In the meantime, do you have examples of "one-ended torsion free hyperbolic groups"? –  Portland Feb 12 '10 at 15:42
    
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. –  Richard Kent Feb 12 '10 at 15:46
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You could look at Bridson and Haefliger's book "Metric spaces of nonpositive curvature" for more about hyperbolic groups. –  Richard Kent Feb 12 '10 at 15:51
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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. –  HJRW Feb 12 '10 at 15:53
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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. –  HJRW Feb 12 '10 at 21:32
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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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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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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. –  Colin Reid Mar 28 '11 at 17:02
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$\mathbb{Q}$ is a classical example.

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