A group $G$ is *slender* if every subgroup $H \leq G$ is finitely generated. This includes polycyclic-by-finite groups. Such groups are also called *noetherian*.


Suppose that $L$ is a $G$-limit group in the sense that $L$ has a finite generating set $X$ such that for every $n$, the ball $B\_n$ of radius $n$ in the Cayley graph is isomorphic to some ball of radius $n$ in a Cayley graph of some subgroup of $G$. Then is it true that $L$ is itself slender? Does it change anything if $G$ is equationally noetherian?


What about smallness? If $G$ is small, then must a $G$-limit group also be small?


I suspect this may be a tough question, but maybe I'm missing something obvious. Any response would be much appreciated.



Are $G$-limits of a slender group $G$ in the space of marked groups also slender?
1) Consider the semidirect product $G=\mathbf{Z}^2\rtimes\mathbf{Z}$ with action by the matrix $A=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$. It has the presentation
$$\langle t,x,y\mid txt^{-1}=x^2y,\; tyt^{-1}=xy,\; [x,y]=1\rangle.$$
(It is actually generated by $(t,x)$).



> 
> Claim: some limit of markings on $G$ fails to be noetherian (although $G$ is)
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[Added: this result was proved by Luc Guyot (*Limits of metabelian groups. Internat. J. Algebra Comput. 22 (2012), no. 4, 1250031*; [ArXiv link](https://arxiv.org/abs/1007.1356), [MR link](http://www.ams.org/mathscinet-getitem?mr=2946296)), showing (2nd item of Theorem A therein) that suitable markings accumulate to the free metabelian group on 2 generators.]


*Proof of the claim:* [Edit: It's enough for purposes here to prove something weaker than Guyot's theorem, namely showing that some limit admits $\mathbf{Z}\wr\mathbf{Z}$ as a subgroup.] Consider the generating subset $(t^{n^2+1},t^n,x)$. There is a unique homomorphism $j\_n$ from the wreath product $\langle s,x\mid [s^mxs^{-m}, x]=1,\forall m \in \mathbb{Z}\rangle$ to $G$ mapping $s$ to $s:=t^n$ and $x$ to $x$. We claim that it is injective in a ball of radius $p\_n$ going to infinity with $n$. Once this is proved, this shows that some limit of copies of $G$ (any accumulation point for this choice of marking) contains a copy of $\mathbf{Z}\wr\mathbf{Z}$ and hence is not noetherian.


To prove the claim: assume the contrary. This means that some nontrivial element $w$ of the wreath product such that $j\_n(w)=1$ for infinitely many $n$ (say $n\in I$, infinite set of integers). Clearly $w$ should be in the kernel of the canonical map to $\mathbf{Z}$, and therefore can be written $w=P(x)$ with $P\in\mathbf{Z}[u^{\pm 1}]$ (Laurent polynomial ring). Here for $P=\sum a\_nu^n$, $P(x)$ means $\prod s^nx^{a\_n}s^{-n}$. (This is a standard identification, since $\mathbf{Z}\wr\mathbf{Z}\simeq\mathbf{Z}[u^{\pm 1}]\rtimes\mathbf{Z}$). 


That $j\_n(w)=1$ means that, in $\mathbf{Z}^2$, $P(A^n)x=0$. Since $A$ is irreducible, that $P(A^n)$ has nontrivial kernel implies that it's zero. So for all $n\in I$, we have $P(A^n)=0$. But the $A^n$ have distinct eigenvalues (except for $n$ and $-n$), so infinitely of them cannot be killed by a single polynomial, contradiction.


Note that $G$ is linear over $\mathbf{Z}$, hence equationally Noetherian.


2) Small groups. Consider, for instance, a f.g. group $H$ containing all finite groups as subgroups (e.g., Neumann's group $\mathrm{Sym}\_0(\mathbf{Z})\rtimes\mathbf{Z}$, or its finitely presented analogue due to Houghton), with no free subgroup.



> 
> Claim: some limit of markings on $H$ contains a free subgroup (although $H$ doesn't)
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Fix a sequence of finite marked groups $(V\_n,x\_n,y\_n)$ tending to the free groups on 2 generators. Then consider a sequence of generating subset of $H$, as a concatenation of a fixed generating subset of $H$, and a pair $(x'\_n,y'\_n)$ in $H$ generating a subgroup isomorphic to $(V\_n,x\_n,y\_n)$ as marked group. Then any accumulation point of this sequence of marked groups has the property that the last two generators freely generated a free group.


