Related papers: Verbally closed subgroups of free groups

We prove that every verbally closed two-generated subgroup of a free solvable group G of a finite rank is a retract of G.

A theorem of Myasnikov and Roman'kov says that any verbally closed subgroup of a finitely generated free group is a retract. We prove that all free (and many virtually free) verbally closed subgroups are retracts in any finitely generated…

We study verbally closed subgroups of free solvable groups. A number of results is proved that give sufficient conditions under whose a verbally closed subgroup is turned to be a retract and so algebraically closed of the full group.

Any virtually free group $H$ containing no non-trivial finite normal subgroup (e.g., the infinite dihedral group) is a retract of any finitely generated group containing $H$ as a verbally closed subgroup.

Answering a question of A. V. Vasil'ev, we show that each finite symmetric (or alternating) group $H$ is a retract of any group containing $H$ as a verbally closed subgroup.

We show, in particular, that, if a finite group $H$ is a retract of any finite group containing $H$ as a verbally closed subgroup, then the centre of $H$ is a direct factor of $H$.

We prove that the finitely presentable subgroups of residually free groups are separable and that the subgroups of type $\mathrm{FP}_\infty$ are virtual retracts. We describe a uniform solution to the membership problem for finitely…

Let $F_n$ be a free group of finite rank $n \geq 2$. We prove that if $H$ is a subgroup of $F_n$ with $\textrm{rk}(H)=2$ and $R$ is a retract of $F_n$, then $H \cap R$ is a retract of $H$. However, for every $m \geq 3$ and every $1 \leq k…

It was recently proven that all free and many virtually free verbally closed subgroups are algebraically closed in any group. We establish sufficient conditions for a group that is an extension of a free non-abelian group by a group…

For a subgroup of a free product of finite groups, we obtain necessary conditions (on its Kurosh decomposition) to be verbally closed.

In a number of recent works, it has been established that many virtually free groups, almost all fundamental groups of surfaces and all groups which are nontrivial free products of groups satisfying a non-trivial law are algebraically…

We prove that in every finitely generated profinite group, every subgroup of finite index is open; this implies that the topology on such groups is determined by the algebraic structure. This is deduced from the main result about finite…

The assertion in the title implies that many interesting groups (e.g., all non-abelian braid groups or ${\bf SL}_{100}(\mathbb Z)$) are not strongly verbally closed, i.e., they embed into some finitely generated groups as verbally closed…

We give new and improved results on the freeness of subgroups of free profinite groups: A subgroup containing the normal closure of a finite word in the elements of a basis is free; Every infinite index subgroup of a finitely generated…

If $G$ is a group, a virtual retract of $G$ is a subgroup which is a retract of a finite index subgroup. Most of the paper focuses on two group properties: property (LR), that all finitely generated subgroups are virtual retracts, and…

In this note we prove that if $G$ is a finitely generated profinite group then the verbal subgroup $G^{q}$ is open. Equivalently in a $d$-generator finite group every product of $q$th powers is a product of $f(d,q)$ $q$th powers.

Let $F$ be a free non-abelian group. We show that for any group word $w$ the set $w[F]$ of all values of $w$ in $F$ is rational in $F$ if and only if $w[F] = 1$ or $w[F] = F.$ We generalize this to a wide class of free products of groups.

We prove that a subgroup of a finitely generated free nilpotent group F is undistorted if and only if it is a retract of a subgroup of finite index in F.

We prove that for any automorphism $\alpha$ of a free group F of finite rank, one can efficiently compute a basis of the fixed point subgroup Fix(\alpha).

We prove that the outer automorphism group of a free group of countably infinite rank is complete.