# Sequence of epimorphisms of residually finite groups stabilizes

Let $$G_1 \to G_2 \to \cdots$$ be a sequence of epimorphisms of finitely generated residually finite groups. Does it eventually stabilize? That is, are all but finitely many epimorphisms actually isomorphisms?

Note that finitely generated residually finite groups are Hopfian, so this excludes the simple counterexample of each $$G_i$$ being a fixed group and each epimorphisms being a fixed one onto itself.

The analogous result holds when the groups are residually free: this is Proposition 6.8 in Charpentier Guirardel "Limit groups as limits of free groups". The proof only uses the fact that residually free groups are residually $$SL_2(\mathbb{C})$$, and it seems that it can be adapted to the case where each $$G_i$$ is residually $$GL_n(\mathbb{C})$$ for a fixed $$n$$. It seems unlikely that this holds for general residaully finite groups: the Jordan-Schur Theorem implies that for a general finite group the minimal degree $$n$$ such that it embeds into $$GL_n(\mathbb{C})$$ can be arbitrarily large.

Is there another way to adapt the proof? Is there a counterexample?

## 2 Answers

The answer is "no". The lamplighter group (which is infinitely presented) is a limit of a sequence of virtually free groups and surjective homomorphisms (see, for example, this question and answers there). All virtually free groups are residually finite.

• Ah, OK! The limit here is actually a colimit in category theoretic terminology? – Jeremy Rickard Jan 24 at 17:16
• It is the inductive limit in the standard terminology. – Mark Sapir Jan 24 at 17:45
• See the proof in the answer by YCor in the linked question. Of course it is not the original proof, – Mark Sapir Jan 24 at 20:04
• What I said at that question is that every (infinite locally finite)-by-finite finitely generated group is an inductive limit of a sequence of non-injective epimorphisms of virtually free finitely generated groups. It's a straightforward consequence of the 1978 Bieri–Strebel paper. However, I believe it was never said explicitly before a few years ago. – YCor Jan 24 at 22:13
• The proof uses only the fact that an HNN extension of a finite group is virtually free. That was known long ago (the 60s?). The argument about lamplighter groups appeared in print in arxiv.org/pdf/1206.2072.pdf, a year later than the question in MO.I am not sure there were papers containing that result before 2012. – Mark Sapir Jan 24 at 23:44

In the same vein as dodd's answer, a counterexample can also be deduced from the second Houghton group $$H_2$$, which is defined as the group of bijections $$L^{(0)} \to L^{(0)}$$ that preserves adjacency and non-adjacency for all but finitely pairs of vertices in the bi-infinite line $$L$$. A presentation of $$H_2$$ is $$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle$$ where $$t$$ corresponds to a unit translation and $$\sigma_i$$ to the permutation $$(i,i+1)$$. Now, truncate the presentation and define $$G_n$$ via $$\left\langle \sigma_i (i \in \mathbb{Z}), t \left| \array{ \sigma_i^2=1, \ i \in \mathbb{Z} \\ [\sigma_i,\sigma_j]=1, \ n \geq |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ i \in \mathbb{Z} \\ t\sigma_it^{-1}= \sigma_{i+1}, \ i \in \mathbb{Z}} \right. \right\rangle.$$ By using the relations $$t\sigma_it^{-1}=\sigma_{i+1}$$ in order to remove the generators $$\sigma_0,\sigma_{-1},\ldots$$ and $$\sigma_{n+2},\sigma_{n+3},\ldots$$, we find the following presentation of $$G_n$$: $$\left\langle \sigma_1, \ldots, \sigma_{n+1}, t \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \array{\sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \\ t\sigma_it^{-1}= \sigma_{i+1}, \ 1 \leq i \leq n} \right. \right\rangle.$$ Observe from this presentation that $$G_n$$ decomposes as an HNN extension of $$\left\langle \sigma_1,\ldots, \sigma_{n+1} \left| \array{ \sigma_i^2=1, \ 1 \leq i \leq n+1 \\ [\sigma_i,\sigma_j]=1, \ |i-j| \geq 2}, \ \sigma_i\sigma_{i+1}\sigma_i= \sigma_{i+1}\sigma_i \sigma_{i+1} = 1, \ 1 \leq i \leq n \right. \right\rangle,$$ which turns out to be isomorphic to the symmetric group $$\mathfrak{S}_{n+2}$$, where the stable letter conjugates $$\langle \sigma_1,\ldots, \sigma_n \rangle$$ to $$\langle \sigma_2, \ldots, \sigma_{n+1} \rangle$$. Thus, as the HNN extension of a finite group, $$G_n$$ must be virtually free.

The conclusion is that the canonical quotient maps $$G_1 \twoheadrightarrow G_2 \twoheadrightarrow \cdots$$ defines a sequence of epimorphisms between virtually free groups that does not stabilise.

Remark: By reproducing the above argument almost word for word with the lamplighter group $$\mathbb{Z}_2 \wr \mathbb{Z}$$ instead of the Houghton group $$H_2$$ provides the same conclusion. The reason is that these groups have a similar structure: they are of the form $$C \rtimes \mathbb{Z}$$ for some locally finite Coxeter group $$C$$ where $$\mathbb{Z}$$ acts on $$C$$ via an isometry of the graph defining $$C$$. (Loosely speaking, all the other groups of this form can be recovered from $$\mathbb{Z}_2 \wr \mathbb{Z}$$ and $$H_2$$, so there is no other interesting examples in this direction.)

• Just as a remark, the "second Houghton group" $H_2$ was discovered much before Houghton (as it's consider by B.H. Neumann in the 30's). The third Houghton group is a quite nontrivial analogue of $H_2$, and, quite surprisingly then, is finitely presented. Another remark is that this example is covered by my remark to dodd's post (it's an (infinite locally finite)-by-cyclic finitely generated group, so it's no surprise it's the same argument since it's the same in this generality, which is much broader than only locally finite Coxeter with action on the Coxeter graph). – YCor Jan 25 at 22:17
• As a further remark, $H_2$ is isomorphic to $\textrm{Sym}_{\textrm{fin}}(\mathbb{Z}) \rtimes \mathbb{Z}$ where $\textrm{Sym}_{\textrm{fin}}(X)$ are the finitely supported permutations on the set $X$ and $\mathbb{Z}$ acts on $X = \mathbb{Z}$ by translation (the "infinite cyclic permutation"). And another interesting property: the Houghton groups $H_n$ have the property $FP_{n-1}$ but not $FP_n$. – ARG Jan 27 at 8:23
• @Ycor Perhaps a separate (or stupid) question: Assume $G$ is finitely generated and $1 \to L \to G \to \mathbb{Z} \to 1$ where $L$ is infinite locally finite. Are there such $G$ which are not isomorphic to a subgroup of $H_2$? Perhaps need to require that $G$ is solvable and/or residually finite for a positive answer? – ARG Feb 17 at 20:55
• I would say that $\mathbb{Z}_2 \oplus \mathbb{Z}$ (and so $\mathbb{Z}_2 \oplus(\mathbb{Z}_2\wr\mathbb{Z})$ if you really want an (infinite locally finite)-by-$\mathbb{Z}$ group) does not embed in $H_2$. – AGenevois Feb 17 at 21:14
• @AGenevois the element of order 2 $(2n\leftrightarrow 2n+1)_{n\in\mathbf{Z}}$ and $n\mapsto n+2$ commute. – YCor Feb 18 at 7:39