# Is there a non-Hopfian lacunary hyperbolic group?

The question's in the title and is easily stated, but let me try to give some details and explain why I'm interested. First, a disclaimer: if the answer's not already somewhere in the literature then it could be rather hard; I'm asking this question here because MO is lucky enough to have some of the foremost experts on lacunary hyperbolic groups as active participants.

Definitions

1. A group $\Gamma$ is non-Hopfian if there is an epimorphism $\Gamma\to\Gamma$ with non-trivial kernel.

2. A group $\Gamma$ is lacunary hyperbolic if some asymptotic cone of $\Gamma$ is an $\mathbb{R}$-tree.

To motivate this second definition, note that a group is word-hyperbolic if and only if every asymptotic cone is an $\mathbb{R}$-tree.

Lacunary hyperbolic groups were defined and investigated in a paper of Ol'shanskii, Osin and Sapir (although examples of lacunary hyperbolic groups that are not hyperbolic already existed---I believe the first one was constructed by Simon Thomas). They construct examples that exhibit very non-hyperbolic behaviour, including torsion groups and Tarski monsters.

Question

Once again:-

Is there a non-Hopfian lacunary hyperbolic group?

Motivation

My motivation comes from logic, and the following fact.

Proposition: A lacunary hyperbolic group is a direct limit of hyperbolic groups (satisfying a certain injectivity-radius condition).

That is to say, lacunary hyperbolic groups are limit groups over the class of all hyperbolic groups. (Note: the injectivity-radius condition means that there are other limit groups over hyperbolic groups which are not lacunary hyperbolic. I'm also interested in them.) Sela has shown that limit groups over a fixed hyperbolic group $\Gamma$ (and its subgroups) tell you a lot about the solutions to equations over $\Gamma$. For instance, his result that a sequence of epimorphisms of $\Gamma$-limit groups eventually stabilises (which implies that all $\Gamma$-limit groups are Hopfian) has the following consequence.

Theorem (Sela): Hyperbolic groups are equationally Noetherian. That is, any infinite set of equations is equivalent to a finite subsystem.

In the wake of Sela's work we have a fairly detailed understanding of solutions to equations over a given word-hyperbolic group $\Gamma$. But it's still a matter of great interest to try to understand systems of equations over all hyperbolic groups.

Pathological behaviour in lacunary hyperbolic groups should translate into pathological results about systems of equations over hyperbolic groups. A positive answer to my question would imply that the class of hyperbolic groups is not equationally Noetherian. And that would be quite interesting.

Note: This paper of Denis Osin already makes a connection between equations over a single lacunary hyperbolic group and equations over the class of all hyperbolic groups.

This question attracted great answers from Yves Cornulier and Mark Sapir, as well as some excellent comments from Denis Osin. Let me quickly clarify my goals in answering the question, and try to summarise what the state of knowledge seems to be. I hope someone will correct me if I make any unwarranted conjectures!

My motivation came from the theory of equations over the class of all (word-)hyperbolic groups. For these purposes, it is not important to actually find a non-Hopfian lacunary hyperbolic group; merely a non-Hopfian limit of hyperbolic groups is enough. (That is, the injectivity radius condition in the above proposition can be ignored.) Yves Cornulier gave an example of a limit of virtually free (in particular, hyperbolic) groups which is non-Hopfian. From this one can conclude that the class of word-hyperbolic groups is not equationally Noetherian, as I had hoped.

[Note: I chose to accept Yves's answer. Mark's answer is equally worthy of acceptance.]

Clearly, the pathologies of Yves's groups derive from torsion---the class of free groups is equationally Noetherian---and there are some reasons to expect torsion to cause problems, so I asked in a comment for torsion-free examples. These were provided by Denis Osin, who referred to a paper of Ivanov and Storozhev. Thus, we also have that the class of all torsion-free hyperbolic groups is not equationally Noetherian.

Let us now turn to the question in the title---what if we require an actual lacunary hyperbolic group that is non-Hopfian. First, it seems very likely that such a thing exists. As Mark says, 'A short answer is "why not?"'. More formally, Denis claims in a comment that the subspace of the space of marked groups consisting of lacunary hyperbolic groups is comeagre in the closure of the subspace of hyperbolic groups. This formalises the idea that lacunary hyperbolic groups are not particularly special among limits of hyperbolic groups.

Mark also suggested two possible approaches to constructing a non-Hopfian lacunary hyperbolic group; however, in a comment, Denis questioned whether one of these approaches works. In summary, I feel fairly confident in concluding that a construction of a non-Hopfian lacunary hyperbolic group is not currently known, although one should expect to be able to find one with a bit of work.

Since you say that you are also interested in other limits of hyperbolic groups, here is an example. It is not lacunary hyperbolic.

The first observation is that every f.g. group $N\rtimes\mathbf{Z}$ is a limit of groups $G_n$ that are HNN-extensions over f.g. subgroups of $N$. This is due to Bieri-Strebel (1978) and can be checked directly (it is used in the paper of Olshanskii-Osin-Sapir to provide an elementary amenable lacunary hyperbolic group). Now assume in addition that $N$ is locally finite. Then the $G_n$ are virtually free, hence are hyperbolic. This shows that any (locally finite)-by-cyclic f.g. group is a limit of virtually free (hence hyperbolic) groups.

Now here's a non-Hopfian example. Recall that Hall's group is the group of invertible $3\times 3$ triangular matrices with $a_{11}=a_{33}=1$. Consider Hall's group $H$ over the ring $A=\mathbf{F}_p[t^{\pm 1}]$, where $\mathbf{F}_p=\mathbf{Z}/p\mathbf{Z}$ and $p$ is a fixed prime.

Its center $Z(A)$ corresponds to matrices differing to the identity at the entry $a_{13}$ only. Set $B=\mathbf{F}_p[t]$ and $G=H/Z(B)$. Then it can be shown that $G$ is non-Hopfian. Indeed, the conjugation by the diagonal matrix $(t,1,1)$ restricts to an automorphism of $H$ mapping $Z(B)$ strictly into itself and thus induces a non-injective surjective endomorphism of $G$. Now $G$ is a limit of virtually free groups, by the previous argument.

I don't know how to adapt the construction to yield a lacunary hyperbolic (LH) group, but limits of hyperbolic groups are in general much more ubiquitous than LH groups, which demand refined constructions. As far as I understood, the various constructions of LH groups were performed in the literature are specially manufactured to yield LH groups, and I'm not aware of any group that was explicitly constructed independently, and was then shown to be a LH group.

• @Yves: $B$ is a ring? What is $Z(B)$? As you said, many non-lacunary hyperbolic groups are limits of hyperbolic groups. Historically the first one was the infinite Burnside group $B_{m,n}$ from the paper of Novikov and Adian. The fact that the intermediate groups in their construction are hyperbolic is proved in their paper (many years before the concept of hyperbolic group was introduced). Also the lamplighter group is a limit of virtually free groups, etc. Groups with a non-trivial law or with a central element of infinite order cannot be lacunary hyperbolic by our result with Drutu. – Mark Sapir Sep 19 '11 at 3:38
• @Mark: $Z(B)$ is the set of $3\times 3$ matrices with 1 in the diagonal, $a_{12}=a_{23}=0$ and $a_{13}\in B$. Besides, if you pick "having a linear Dehn function" as a definition of being hyperbolic, C'(1/6) small cancelation groups are hyperbolic and this is due to Greendlinger in the 50s. This is before Novikov-Adian and is also less hard, and limits of C'(1/6)-groups are a rich source of infinitely presented limits of hyperbolic groups. – YCor Sep 19 '11 at 6:17
• Many thanks for your answer, Yves. My question focussed on lacunary hyperbolic groups because they seem well studies and of interest in their own right; but your answer is good enough for the application I have in mind, and has helped me a lot in appreciating the difference between lacunary hyperbolic groups and more general limits of hyperbolic groups. – HJRW Sep 20 '11 at 9:33
• @Henry: This is the answer to the follow up questions. 1. Such groups exist and are constructed by Ivanov and Storozhev in arXiv:math/0312491. These groups are limits of groups $G(i)$, which are hyperbolic by Lemma 1 from the paper. (The fact that condition R implies hyperbolicity can be easily extracted from Olshanskii's book cited in the paper). And the groups $G(i)$ are torsion free by Lemma 2. – Denis Osin Sep 21 '11 at 5:57
• @Henry, continuation: 2. In a sense, lacunary hyperbolic groups are generic. It is not hard to show the following. Fix an integer $n>1$ and let $\mathcal H_n$ (respectively, $\mathcal L_n$ be the subspace of presentations of non-elementary hyperbolic (respectively, lacunary hyperbolic) groups in the space of marked group presentations with $n$ generators. Then $\mathcal L_n$ is a dense $G_\delta$ subset of $\mathcal H_n$. In particular, it is comeagre. – Denis Osin Sep 21 '11 at 6:05

A short answer is "why not?". A longer answer would be to look at the known examples of non-Hopfian groups and try to make them lacunary hyperbolic. A quite general construction can be found in our paper with Dani Wise (Sapir, Mark; Wise, Daniel T. Ascending HNN extensions of residually finite groups can be non-Hopfian and can have very few finite quotients. J. Pure Appl. Algebra 166 (2002), no. 1-2, 191–202.). See Lemma 3.1 there, in particular. It is quite possible that this construction or its slight modification can be lacunary hyperbolic.

Update 1.Another way to construct an example is the following. Start with the free group $F_2=\langle a,b\rangle$. Pick two words $U(a,b), V(a,b)$ satisfying small cancelation. That will be the non-injective surjective endomorphism. To make it non-injective, pick a word $W(x,y)$ and impose the relation $W(U, V)=1$. To make it surjective, pick two words $P(a,b), Q(a,b)$, and impose the relations $P(U,V)=a, Q(U,V)=b$. Now to make the map $a\to U, b\to V$ an endomorphism, for every relation $S(a,b)=1$ introduced already, we need to add the relation $S(U,V)=1$, then apply the same operation to the resulting presentation, etc. This defines an infinite presentation naturally subdivided into finite subsets. It remains to choose the words $U,V,W, P, Q$ so that each finite piece of the presentation defines a hyperbolic groups and the whole presentation is lacunary hyperbolic. Some kind of small cancelation theory may help here.

Update 2. Both constructions give limits of hyperbolic groups. To prove lacunar hyperbolicity one needs to estimate the growth of hyperbolicity constants $\delta$ vs the growth of the length of defining relations. The problem could be that the hyperbolicity constants of the intermediate groups grow too fast comparing to the lengths of relations. It needs to be checked in both cases. The lengths of relations grow exponentially fast but so do the hyperbolicity constants $\delta$. One needs to compare the bases of exponents. Fortunately, in the second construction, it seems to me that the base of exponent of the rels growth is approximately the maximal length of $U,V$. And the base of growth of $\delta$ is a constant that is independent of $U,V$ (say, $4$). But it needs to be checked.

• Thanks for this answer, Mark. It would be even more interesting if the lacunary hyperbolic group were a limit of torsion-free hyperbolic groups. If I understand your update correctly, it should be possible to perform your construction and keep things torsion-free. Does that sound right? – HJRW Sep 20 '11 at 9:31
• Both constructions produce torsion-free groups. – Mark Sapir Sep 20 '11 at 21:46
• @Mark: I think we discussed this idea 4 years ago and came to a conclusion that it does not work. The problem is the following. Let $n$ be the maximum of lengths of $P(U,V)$ and $Q(U,V)$ as words in $\{a^{\pm 1},b^{\pm 1}\}$. To make the map $a\mapsto U$, $b\mapsto V$ (call it $\phi$) a homomorphism, we have to add relations $\phi (S), \phi^2(S), \ldots$ for every relation $S(a,b)=1$. If the later has length $k$, we will get relations of length $k, nk, n^2k, \ldots$, which kills lacunar hyperbolicity, at least in the small cancellation case. Am I missing something? – Denis Osin Sep 21 '11 at 6:23
• @Denis: This is what written in Update 2. It is not obvious how to make $\delta$ grow slower than the lengths of relations. In general there are two indirect ways to prove Hopf property. The first is to use residual finiteness which fails for many l.h. groups. The second is to use actions on R-trees as in Sela's paper and in our paper with Drutu. For this one needs at least that all cones are tree-graded by our result with Drutu. That is not true for some l.h. groups, although one would expect that groups given by small cancelation presentations (whose cones are circle-trees) are Hopfian. – Mark Sapir Sep 22 '11 at 14:06

Since Henry also discusses the property of being equationally Noetherian, I think the following observation is worth posting. And it is too long for a comment, so I post it as an answer.

The observation is that there exists a torsion free lacunary hyperbolic group that is not equationally Noetherian. In fact, such a group can be constructed directly. Alternatively, we can use the following theorem.

Theorem. Let $\mathcal H_n$ be the closure of the set of all $n$-generated torsion free non-cyclic hyperbolic groups in the space of marked group presentations. Then there exists a torsion free lacunary hyperbolic group $L$ such that the set of $n$-generator presentations of $L$ is dense in $\mathcal H_n$.

The proof uses the the same idea as in my paper mentioned by Henry. It is a bit too technical to be posted here (but it is just a page long and I did verify the details).

Now let us take the Ivanov-Storozhev non-hopfian group $G\in \mathcal H_2$. Since $G$ is not equationally Noetherian, there is a system of equations $S$ which is not equivalent to any finite subsystem on $G$. Let us assume that the coefficients in $S$ are written as words in generators, so we can consider the same system over $L$. Let $F$ be any finite subsystem of $S$.

Since $G$ is not equationally Noetherian, $S$ is not equivalent to $F$ over $G$. In particular, there is another finite subsystem $F_1$ of $S$ which contains $F$ and which is not equivalent to $F$ over $G$. Note that the property of a tuple of elements of $G$ to be a solution to a fixed finite system can be detected using some finite ball. Hence $F$ and $F_1$ are not equivalent over every group from some sufficiently small open neighborhood of $G$. In particular, $F$ is not equivalent to $F_1$ (and hence to $S$) over $L$. Thus we obtain

Corollary. The (torsion free lacunary hyperbolic) group $L$ is not equationally Noetherian.

• That's very interesting. Thanks, Denis! – HJRW Oct 8 '11 at 17:09

Contrary to I think the expectations of the earlier answers, in a recent paper, Coulon--Guirardel showed that the answer to the title question is 'no'!

Theorem (Coulon--Guirardel): Every lacunary hyperbolic group is Hopfian.

The proof is remarkably short, and although it of course uses the fact that hyperbolic groups are Hopfian, it doesn't need the details of the proof.

• That's very nice! (But in my answer there was no claim or suggestion that I expected a positive answer.) – YCor Jun 19 '16 at 9:13
• @YCor, I was really referring to Mark Sapir's answer, although one might also conjecture (incorrectly) from the conjunction of Denis Osin's proof that lacunary hyperbolic groups are generic in limits of hyperbolic groups and your example of a non-Hopfian limit that there should be a non-Hopfian lacunary hyperbolic group. But I didn't mean to imply that you had actually made that conjecture. – HJRW Jun 19 '16 at 14:12
• I had a look at the Coulon-Guirardel paper. Before using that hyperbolic groups are hopfian, it shows (Corollary 3.5) that endomorphisms can be lifted to the free group and factor to endomorphisms of the hyperbolic groups in the approximating family. The latter point fails for arbitrary forward limits of hyperbolic groups as in my example. – YCor Jun 19 '16 at 16:51
• @YCor, yes. Obviously the proof must use the definition. – HJRW Jun 20 '16 at 5:41