Ashot Minasyan
• Member for 11 years, 6 months
• Last seen more than a month ago
• Southampton, UK

Denis Osin [Osin, Denis, Small cancellations over relatively hyperbolic groups and embedding theorems, Ann. Math. (2) 172, No. 1, 1-39 (2010). ZBL1203.20031.] proved that every torsion-free countable ...

It is easy to construct a counter-example when $H$ and $K$ are not of finite index in $G$. Let $A$ be a finitely presented torsion-free group which is not residually finite (e.g., the Baumslag-...

The free Burnside group $B(2,n)$, of rank $2$ and exponent $n$, where $n$ is a sufficiently large power of an odd prime, satisfies both conditions: all of its abelian subgroups are cyclic (S. I. ...

The answer is negative even if $A$ is finitely generated. Here is a simple construction. Let $F$ be the free group on $\{x,y,z\}$ and let $A=\langle x,y \rangle$. Then there is a natural retraction $\... View answer Accepted answer 4 votes The answer is negative in general, even if you restrict to finitely generated subgroups. Indeed, in a hyperbolic group a f.g. subgroup can be conjugate to a proper subgroup of itself. For example, let ... View answer Accepted answer 4 votes The answer is yes. Let's prove this by induction on$n$. If$n=1$, then by M. Hall's theorem there is a finite index subgroup$K \leqslant F$such that$H_1 \subseteq K$and$|F:K|>1/\epsilon$. ... View answer 2 votes The following nice argument, due to Alexander Olshanskii, shows that the answer to both questions is negative. It uses the theorem of Golod-Shafarevich (see http://arxiv.org/abs/1206.0490 for a ... View answer 4 votes Complementing Misha's answer, the following is true for any non-(virtually cyclic) hyperbolic group$G$: Claim: for any sufficiently large$k \in \mathbb{N}$there is a free quasiconvex subgroup$...

Like with many problems solvable by small cancellation methods, the answer can be found by looking at A.Yu. Ol'shanskii's papers. Indeed, take any countable group $H$. Without loss of generality we ...

Assuming that you refer to Ivanov's construction from Ol'shanskii's book of a $2$-generated infinite group $G$ of exponent $p$, for a large prime $p$, having exactly $p$ conjugacy classes, then the ...

Let $F$ be a finitely generated non-abelian free group with a non-trivial normal subgroup $R \lhd F$. Suppose that $G:=F/R$ is finitely presented. Then the group $F/R'$ is finitely presented if and ...

Browsing through the archive of solved problems of Kourovka Notebook, I accidentally saw that the same question was asked by Yu.V. Kuz'min in 1999 (see question 14.52). Apparently the required ...

EDIT: The isomorphism problem for finitely presented solvable groups in the variety of all solvable groups of derived length $\le 7$ is undecidable. This was proved by Kirkinskiĭ and Remeslennikov (...

I think that Higman's group H has plenty of such normal subgroups. Indeed, let G be the extension of H with the automorphism h. Then H has index 4 in G. By Schupp's theorem, H is SQ-universal, hence ...

Apparently such sequences do exist. For example, one can take $G_i:=Alt(5^i)$ for $i=0,1,2,\dots$, where the embedding $\gamma_i: G_i \to G_{i+1}$ is the diagonal embedding of $Alt(5^i)$ into $Alt(5^{... View answer Accepted answer 17 votes The group$G$maps onto the free product$C_p*C_p*C_p$of three cyclic groups of order$p$(just send each$b_i^p$to$1$). This free product is virtually free, as a free product of finite groups (by ... View answer Accepted answer 14 votes A. Yu. Olshanskii in the paper "Periodic quotient groups of hyperbolic groups." ((Russian) Mat. Sb. 182 (1991), no. 4, 543--567; translation in Math. USSR-Sb. 72 (1992), no. 2, 519–541) proved that ... View answer 10 votes Yes, such groups exist. Consider the disjoint union of cyclic groups of odd order$\mathcal{C}:=\{ \mathbb{Z}/(2n+1)\mathbb{Z} \mid n \in \mathbb{N} \}$. By a theorem of A. Ol'shanskii (see Thm 35.... View answer 2 votes A modification of Guntram's example could produce a countable group with the required property, which is not an FC-group. Let$G$be the direct product of non-abelian symmetric groups$G=\times_{n\ge ...