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No. Your group has Serre's Property FA, meaning that any action on a tree has a global fixed point. (This can be deduced from the fact that it has a generating set such that every element is torsion …

If I understand your question correctly then the answer is 'no'. Indeed, whenever countable $\Gamma=A*_C B$ with $|A:C|=|B:C|=\infty$ then the Bass--Serre tree $S$ is a countably branching regular tr …

Here's a silly counter-example (though I doubt this is what you have in mind). Take $G_1=G_2=\mathbb{Z}$ with $C_1=2\mathbb{Z}$ and $C_2=3\mathbb{Z}$. Certainly $G_1,G_2$ are limit groups. Then $G= …

You're right that hyperbolic graphs of free groups with cyclic edge groups are locally quasiconvex. This can be proved by combining subgroup separability with results about combination of quasiconvex …

For finitely presented groups, you're close to two well known open questions/conjectures. Question 1: If a finitely presented group is Noetherian, is it virtually polycyclic? (This question is FP11 …

This is one of those 'well known' things that every one does in different ways. I believe the notion of a covering space of a graph of groups was worked out by Bass. The details are rather technical …

A random group at density less than 1/6 is known to be the fundamental group of a compact, non-positively curved cube complex, by Ollivier--Wise. By Agol's theorem, all such hyperbolic groups are vir …

As Derek Holt indicates in comments, this is standard. You should do the following easy exercise. If $G = A*_CB$ or $G=A*_C$ is hyperbolic and $C$ is quasiconvex, then so is $A$ (and $B$). Since …

In paragraph 1.1.7 of of a recent preprint of Kharlampovich--Myasnikov--Sapir, the authors write: We expect the approach used in this paper to be useful in solving other problems that are still op …

The simplest candidate I know of is Higman's group $\langle a,b,c,d\mid a^b=a^2, b^c=b^2, c^d=c^2, d^a=d^2\rangle$ (where, as usual, $a^b$ means $b^{-1}ab$). Terry Tao wrote a nice blog post about …

The following is an elaboration of the last paragraph of Misha's answer. For me, the thing that makes non-linear (discrete) groups interesting is that we are not very good at constructing them! Line …

No: T is infinite, finitely presented and simple. Fg linear groups are residually finite, by Mal'cev's theorem. QED.

Let $\alpha: \Gamma\to\Gamma$ be an injection sending $\Gamma$ to $\Gamma'$. Then the $\Gamma''$ you're looking for is the infinite amalgamated product $\cdots *_{\Gamma}\Gamma *_{\Gamma}*\Gamma*_{\ …

Regarding the question of `whether the CAT(0) boundary works, if the space doesn't contain $\mathbb{R}^2$ as a subspace' (see comments above), the Flat Plane Theorem asserts that any CAT(0) group that …

No. For instance, let $G=\mathbb{Z}/4*\mathbb{Z}/4$ and let $X$ be the Bass--Serre tree, the infinite 4-regular tree. Now let $Z$ be the graph of groups corresponding to $\mathbb{Z}/2*\mathbb{Z}/2$. …