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 $\...

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 ...

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 $|F:K|>1/\epsilon$. ...

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 ...

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^{...

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 ...

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 ...

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....

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 ...

Well, any finitely generated group G of exponent 3 is finite by a classical theorem of Burnside. And since the order of every element is 3, the order of G must be a power of 3 by Cauchy's theorem. It ...