Must the powers of some element always grow linearly with respect to a word metric?

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!

• I see this question has been reopened, but it still has the trivial group as a counterexample, so you really should edit it to make it more sensible. I would suggest assuming that $G$ is nontrivial and torsion-free. – Derek Holt Apr 14 '15 at 10:15

(Probably the question would be more suitable for MathSE)

The answer is no. Well, the trivial group is a counterexample. Also finite groups are counterexamples. So first and for all you should have specified that you assume the group to be infinite. But then the answer is still no.

In a finitely generated group, an element $g$ is called distorted if $\lim |g^n|_S/n=0$ (the limit exists, and that the limit vanishes does not depend on the choice of finite generating subset $S$). So you're asking about the existence of an undistorted elements.

Torsion elements are distorted. Hence any finitely generated torsion group have all its elements distorted, and many infinite examples are known (Golod-Shafarevich, Alëshin-Grigorchuk, Burnside groups by work of Adian, etc).

There also exist infinite torsion-free finitely generated groups with all elements distorted: it includes groups with finitely many conjugacy classes constructed by Osin (and maybe Ivanov before but I don't remember if his examples are torsion-free).

There are positive results for nice (restricted) classes of groups: for instance any infinite finitely generated group that is either elementary amenable (e.g., solvable), Gromov-hyperbolic, or has a faithful finite-dimensional representation over some field, has an undistorted element.

No, as there are infinite finitely generated periodic groups.

I followed YCor's suggestion and googled the relevant paper of Osin: "Small cancellations over relatively hyperbolic groups and embedding theorems" available on the arXiv. You want the following.

Corollary 1.2. Any countable torsion–free group can be embedded into a (torsion–free) 2–generated group with exactly 2 conjugacy classes.

In particular, there is a 2-generated, torsion-free, infinite group $G$ with exactly two conjugacy classes. So, as YCor observes, every element in $G$ is conjugate to its square. Thus every cyclic subgroup of $G$ is distorted.

• Actually in Distortion functions for subgroups, in Geometric Group Theory Down Under, Proc. of a Special Year in Geometric Group Theory, Canberra, Australia, 1996, Ed. J.Cossey, Walter de Gruyter, Berlin - New York, 1999, 281-291, Olshanskii already constructed a torsion-free 2-generated group in which every element $g$ satisfies $|g^n|=O(\log(n))$. (Reference 56 in math.vanderbilt.edu/~olsh/publ.html) – YCor Apr 14 '15 at 21:12
• PS: $|g^n|=O(\log n)$ follows if $g$ is conjugate to its square. – YCor Apr 15 '15 at 5:04