Hopfian and Co-Hopfian groups (examples)

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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.).

Answer 4

$\mathbb{Q}$ is a classical example.

Answer 5

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.)

Answer 6

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.

Answer 7

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).