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.)

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    $\begingroup$ What if all elements have finite order? $\endgroup$ – Fedor Petrov Apr 6 at 19:48
  • $\begingroup$ 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. $\endgroup$ – AGenevois Apr 7 at 6:30
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    $\begingroup$ No, I am really interested in infinite-order elements. $\endgroup$ – AGenevois Apr 7 at 9:51
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    $\begingroup$ The product a distorted element with its inverse is $1$. How is $1$ distorted? $\endgroup$ – AGenevois Apr 7 at 18:48
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    $\begingroup$ The question is among those many open questions about finitely presented groups. $\endgroup$ – YCor Apr 7 at 19:11

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