Let $G$ be a group with generators $a, b\in G$.


Define $\mathrm{len}:G\to\mathbb{Z}\_{\geq 0}$ by sending $g$ to the minimum length of a word in $a, b, a^{-1}, b^{-1}$ equal to $g$.


Assume that for all $g\neq e\in G$ there is infinitely many $n\in \mathbb{Z}\_{\geq 0}$ such that $\mathrm{len}(g^{n+1})<\mathrm{len}(g^n)$.


Must every element of $G$ be of finite order then?



Infinite oscillation of minimum word length in 2-generated group
This answer is partly inspired by HJRW's comment.


**Definition 1.** Let $G$ be a finitely generated group and $g\in G$, and word length $|\cdot|$ with respect to some finite generating subset. Say that $g$ is very distorted if $\liminf |g^n|/\log(n)=0$, or equivalently if $u\_g(n)=\sup\{m:|g^m|\le n\}$ satisfies $\limsup\log(u\_g(n))/n=\infty$.


It is not hard to check that this implies that $|g^n|>|g^{n+1}|$ for infinitely many $n$ (otherwise the growth would be more than exponential).


**Proposition.** Let $G$ be a finitely generated group with finitely many conjugacy classes. Then every element $g$ of $G$ is very distorted.


Proof: if $g$ has finite order this is trivial. Otherwise, $g$ has infinite order. Then there exists $n>1$ and $h\in G$ such that $hgh^{-1}=g^n$. Hence $h^kgh^{-k}=g^{n^k}$ for all $k$, and hence $h$ has infinite order. In turn there exists $f\in G$ and $m>1$ such that $fhf^{-1}=h^m$, so $|f^khf^{-k}|=|h^{m^k}|=O(k)$. In turn, $|h^{m^k}gh^{-m^k}|=|g^{n^{m^k}}|=O(k)$. Hence $\liminf |g^k|/\log\log k<\infty$, so $g$ is very distorted.


The existence of torsion-free finitely generated groups with finitely many conjugacy classes was proved by Ivanov in the late 80s, and later Osin found examples with two conjugacy classes (I think both can be chosen generated by a pair). So this answers the question.


