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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 …

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 co …

Let $G$ be a virtually compact special group, in the terminology of Haglund and Wise (i.e., $G$ has a finite index subgroup $H$ which is isomorphic to the fundamental group of some compact special non …

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 $ …

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-Solita …

Let $\Gamma$ be a connected graph, let $\lambda \ge 1$ and $c \ge 0$ be some constants. Recall that a combinatorial path $p$ in $\Gamma$ is said to be $(\lambda,c)$-quasigeodesic if for every combinat …

Is there an elementary and efficient algorithm for testing the membership to a double coset of f.g. subgroups in a free group? Has this membership problem been implemented in GAP/Magma? More precise …

Although this is an older question, I think that the answer may still be of interest. I realised that a positive answer follows from the work of Grunewald and Segal [Grunewald, Fritz; Segal, Daniel, …

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 $|F:K|>1/\epsilon$. Choo …