Suppose we have a group $G$ which is finitely generated , and let $|\cdot |$ denote some word metric on it. Must there be an element $a\in G$ such that $|a^n|\ge c\cdot n$ for some $c>0$?

My intuition says that the answer is "yes", but I wasn't able to find a proof in the general case. There are many cases that I know that the answer is yes - abelian and nilpotent groups, free products, Baumslag-solitar subgroups and many more.

**What I managed to do** - If there is a finite index subgroup $H$ of $G$ such that the abelianization $\frac{H}{[H,H]}$ is infinite, then one can find such an element.

The reason that this is true is not so hard - by quasi-isometry, one might as well ask the question for H. Now, we know that $\frac{H}{[H,H]}$ is a finitely generated abelian group (as $H$ is, beacuse $G$ is). Therefore it must have a further quotient which is isomorphic to $\mathbb{Z}^d$. We let $a\in G$ be some element of $G$ which does not vanish after projecting it to $\mathbb{Z}^d$, and let $v$ be its image there

Suppose $|\cdot|$ is a word metric which corresponds to $S \subset G$. Let $T$ be the image of $S$ in $\mathbb{Z}^d$, and denote the associated word metric by $|\cdot |'$. Then it's clear that:

$$ |a^n| \ge |n\cdot v|' \ge \frac{||n\cdot v||}{\max_{t\in T}{||t||}}=c_{v,T}\cdot n $$

for any norm $||\cdot ||$ on $\mathbb{R}^d$. This gives the linear lower bound we wanted.

** Edit**
I forgot to mention that $G$ is not the trivial group or a torsion group. Thank you Derek Holt for mentioning it!