A recursive presentation for Mihailova's subgroup

We give an explicit recursive presentation for Mihailova's subgroup $M(H)$ of $F_n \times F_n$ corresponding to a finite, concise and Peiffer aspherical presentation $H=$. This partially answers a question of R.I. Grigorchuk, [8, Problem 4.14]. As a corollary, we construct a finitely generated recursively presented orbit undecidable subgroup of $Aut(F_3)$.Comment: 9 page

Orbit decidability, applications and variations

We present the notion of orbit decidability into a more general framework, exploring interesting generalizations and variations of this algorithmic problem. A recent theorem by Bogopolski-Martino-Ventura gave a renovated protagonism to this notion and motivated several interesting algebraic applications

Orbit decidability and the conjugacy problem for some extensions of groups

Given a short exact sequence of groups with certain conditions, $1\to F\to G\to H\to 1$, we prove that $G$ has solvable conjugacy problem if and only if the corresponding action subgroup $A\leqslant Aut(F)$ is orbit decidable. From this, we deduce that the conjugacy problem is solvable, among others, for all groups of the form $\mathbb{Z}^2\rtimes F_m$, $F_2\rtimes F_m$, $F_n \rtimes \mathbb{Z}$, and $\mathbb{Z}^n \rtimes_A F_m$ with virtually solvable action group $A\leqslant GL_n(\mathbb{Z})$. Also, we give an easy way of constructing groups of the form $\mathbb{Z}^4\rtimes F_n$ and $F_3\rtimes F_n$ with unsolvable conjugacy problem. On the way, we solve the twisted conjugacy problem for virtually surface and virtually polycyclic groups, and give an example of a group with solvable conjugacy problem but unsolvable twisted conjugacy problem. As an application, an alternative solution to the conjugacy problem in $Aut(F_2)$ is given

On abstract commensurators of groups

We prove that the abstract commensurator of a nonabelian free group, an infinite surface group, or more generally of a group that splits appropriately over a cyclic subgroup, is not finitely generated. This applies in particular to all torsion-free word-hyperbolic groups with infinite outer automorphism group and abelianization of rank at least 2. We also construct a finitely generated, torsion-free group which can be mapped onto Z and which has a finitely generated commensurator.Comment: 13 pages, no figur

On abstract commensurators of groups

Abstract We prove that the abstract commensurator of a nontrivial free group, an infinite surface group, or more generally a group that splits appropriately over a cyclic subgroup is not finitely generated. This applies in particular to all torsion-free word-hyperbolic groups with infinite outer automorphism group and abelianization of rank at least 2. We also construct a finitely generated group which can be mapped onto Z and which has a finitely generated commensurator

The mean dehn functions of Abelian groups

"Vegeu el resum a l'inici del document del fitxer adjunt"

Subgroups of small index in Aut(Fn) and Kazhdan's property (T)

We give a series of interesting subgroups of finite index in Aut(Fn). One of them has index 42 in Aut(F3) and infinite abelianization. This implies that Aut(F3) does not have Kazhdan's property (T) (see [3] and [6] for another proofs). We proved also that every subgroup of finite index in Aut(Fn), n >= 3, which contains the subgroup of IA-automorphisms, has a finite abelianization