# A group of all whose elements are distorted

Does there exist a finitely presented group, not torsion, all of whose infinite-order elements are distorted?

An infinite-order element $$g$$ of a finitely generated group $$G$$ is undistorted if there exist constants $$A,B \geq 1$$ such that $$\|g^n\|_S \geq \frac{n}{A} -B$$ for every $$n \in \mathbb{Z}$$, where $$\| \cdot \|_S$$ denotes the word length associated to a fixed finite generating set $$S$$ of $$G$$. An element which is not undistorted is distorted.

I am mainly interested in the finitely presented case, but a finitely generated example would be interesting as well.

Edit: I realised that a finitely generated example exists as a consequence of Osin's paper Small cancellations over relatively hyperbolic groups and embedding theorems. (Since an infinite-order element which is conjugate to its square is necessarily distorted.)

• What if all elements have finite order? – Fedor Petrov Apr 6 at 19:48
• More generally, a finitely generated subgroup $H \leq G$ is undistorted or quasi-isometrically embedded if the inclusion $H \hookrightarrow G$ induces a quasi-isometric embedding when $H$ and $G$ are both endowed with a word metric associated to a finite generating set. So finite-order elements are always undistorted. – AGenevois Apr 7 at 6:30
• No, I am really interested in infinite-order elements. – AGenevois Apr 7 at 9:51
• The product a distorted element with its inverse is $1$. How is $1$ distorted? – AGenevois Apr 7 at 18:48
• The question is among those many open questions about finitely presented groups. – YCor Apr 7 at 19:11