# Are $G$-limits of a slender group $G$ in the space of marked groups small?

This question is a follow up to Are $G$-limits of a slender group $G$ also slender? The MO user @Ycor gave an excellent answer to this question, in which they show that $\mathbb{Z}\wr\mathbb{Z}$ embeds in a limit of $\mathbb{Z}^2 \rtimes \mathbb{Z}$, answering the question negatively.

Nonetheless $\mathbb{Z}\wr\mathbb{Z}$ is still small, i.e. it does not contain a non-abelian free group. Examples of f.g. small group with a non-small limit are groups containing a copy of every finite group. This includes the group $\mathrm{Sym}_0(\mathbb{Z})\rtimes\mathbb{Z}$, which is not clearly slender.

So my new question is:

1. Are $G$-limits of a slender group $G$ in the space of marked groups small?

2. What if $G$ is also amenable? (I know little about amenability.)

My motivation is to understand the actions of the limits of slender groups on trees.

Here's a scheme to produce slender (noetherian) 2-generated groups that contain all finite subgroups (and quasi-finite, in the sense that all proper subgroups are finite).

Given a group $G$, let $C(G)$ be the set of conjugacy classes of finite subgroups of $G$. A group homomorphism $f:G\to H$ induces a map $f_*:C(G)\to C(H)$. Let us say that $f$ is safe

1. $f_*$ is injective
2. for every $P\in C(G)\smallsetminus\{\{1\}\}$, $f(P)$ is contained in no $P'\in C(H)\smallsetminus f_*(C(G))$

We say that $Q$ is a safe quotient of $G$ if the projection $G\to Q$ is safe.

Remark: $f$ safe implies that $f$ is injective in restriction to any finite subgroup: indeed the first condition is that non-conjugate finite subgroups map to non-conjugate subgroups; applying this to $\{1\}$ and the cyclic subgroup generated by some element of finite order, we conclude.

The second condition means, in rough terms, that a finite subgroup of $G$ cannot be embedded in a larger finite subgroup in $H$, except in a way that comes from $G$.

"Conjecture": let $G$ be a non-elementary hyperbolic group.

1. If $H$ is a non-elementary subgroup, there exists a safe non-elementary hyperbolic quotient of $G$ such that the image of $H$ is all of $Q$.
2. If $H$ is a virtually infinite cyclic subgroup, there exists a safe non-elementary hyperbolic quotient of $G$ such that the image of $H$ is finite
3. If $F$ is any finite group, there exists a safe non-elementary hyperbolic quotient of $G$ containing a copy of $F$.

Statements close to (1) and (2) are stated in Gromov's original book and probably follow rigorously from later work, notably by Olshanski or Delzant. I'll update if I get more precise references in the comments. I guess (3) is of similar difficulty; the issue here is this safeness condition; however it's very plausible that it follows from the same methods.

Corollary of the "conjecture": every non-elementary hyperbolic group $G$ has a quotient in which all proper subgroups are finite, and containing isomorphic copy of all finite groups.

Proof of the corollary: enumerate the finitely generated subgroups of $G$ as ($L_n$). Define $G_0=G$. If $G_n$ is defined with a quotient map $G\to G_n$ and $G_n$ is hyperbolic, let $H_n$ be the image of $L_n$ in $G_n$. If $H_n$. Using (1) or (2), define a safe non-elementary hyperbolic quotient $G'_n$ of $G_n$ in which the image of $G_n$ is either finite or all of $G'_n$ (if $G_n$ is finite just take $G'_n=G_n$). Next, using (3), define a safe non-elementary hyperbolic quotient $G_{n+1}$ of $G'_n$ containing a copy of the symmetric group $S_n$.

Define $G_\infty$ as the inductive limit. Clearly it contains copies of all finite subgroups (as the successive maps are injective on finite subgroups). First let $H$ be a proper finitely generated subgroup. So $H$ is the image of $L_n$ for some $n$. Then the image of $L_n$ in $G_{n+1}$ is either finite or all of $G_{n+1}$, and hence $H$ is either finite or all of $G_\infty$. Finally, we have to prove that $G$ has no infinite locally finite subgroup. Indeed, suppose that $M$ is one. Let $M_1$ be some nontrivial finite subgroup in $M_1$; it can be lifted to an isomorphic copy $M'_1$ of itself in $G_n$ for large enough $n$. We fix such $n$. Let $W\subset G_n$ be the inverse image of $M$ in $G_n$ and let $M'_2$ be a maximal finite subgroup of $W$ containing $M'_1$, and $M_2$ its image in $G_\infty$. Let $M_3$ be a finite subgroup properly containing $M_2$ in $M$; it can be lifted in $G_m$ for some $m>n$, to a subgroup $M''_3$. Then the image of $M_2$ in $G_m$ is properly contained in $M''_3$. This contradicts the safeness assumption.

• I'll read over this more closely. This argument is not unlike Osin's construction of torsion groups using Dehn fillings. – NWMT Nov 4 '16 at 14:16
• For (3) of the conjecture one could first find a malnormal quasiconvex free subgroup $H$ of $\Gamma$. By Osin's Peripheral filling, we can take a Dehn filling where $H$ has a finite image and whose kernel avoids all finite subgroups of $\Gamma$, so that gives (1) of safeness. I think some work could be used to show (2) of safeness: you would have to show that none of the finite subgroups map to subgroups of "new" finite subgroups in the fnite quotient. Is this right? – NWMT Nov 4 '16 at 14:26
• Here's an example where (2) of safeness fails. Start from the free product with presentation $G=(x,y,z|z^4=1)$. Then take the quotient $Q$ by the relators $y^6=1$ and $y^3=z^2$. Note that $Q$ is amalgam of $(x,y|y^6)=1$ and $(z|z^4=1)$ over the cyclic group on 2 elements (so both $G$ are virtually free, hence non-elementary hyperbolic). $G\to Q$ is not safe because the finite subgroup $(z^2)$ is contained in a "new" finite subgroup, namely the subgroup of order 6 $(y)$ in $Q$. – YCor Nov 4 '16 at 15:32
• Okay. To get a slender group with a non-small limit, we can relax (3) of the conjecture a little. It is enough to produce a sequence $G_1 \to G_2 \to \ldots$ of safe quotients so that $G_i \geq F_i$ where $(F_i)$ is a sequence of finite subgroups converging to a free group. I'm pretty sure you can use Dehn fillings to get safe quotients: the new finite subgroups will be generated by "long" elements but I'll have to look at the geometry a bit more closely. – NWMT Nov 4 '16 at 15:35
• Proposition 5.29 in arxiv.org/pdf/1111.7048v5.pdf appears to give the answer, but there is some ambiguity in the language. I will contact an author. – NWMT Nov 4 '16 at 16:18

Thanks to Ycor's answer and an other answer due to Anton Klyatchko, the answer is no, limits of slender groups are not small.

In that Anton Klyatchkos's answer, Obraztsov's embedding theorem is used to produce 2-generated group $H$ in which the isomorphisms classes of its subgroups consists of infinite cyclic and all odd ordered finite subgroups. As explained in the excellent answer to Are $G$-limits of a slender group $G$ in the space of marked groups also slender? such a group has a limit containing a free limit. Indeed since free groups are residually $p$-groups for any prime $p$, there is a sequence of finite groups converging to the free group within the class of free groups with odd order.

All known slender (or noetherian) groups are either virtually polycyclic or "monstruous", i.e. weird and perfect. If $G$ is polycyclic then as explained again here it may have a non slender limit, however since polycyclic groups are Equationally Noetherian, every limit $L$ of $G$ must fully residually $G$ and therefore can be discriminated via maps $L \to G$. In particular if $G$ has derived length $n$ then so must $L$. It therefore follows that $L$ must also be small.