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.)

undistortedorquasi-isometrically embeddedif 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. $\endgroup$ – AGenevois Apr 7 at 6:30