4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Apr 14 '15 at 10:15
21
$\begingroup$

(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.

$\endgroup$
5
$\begingroup$

No, as there are infinite finitely generated periodic groups.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – YCor Apr 14 '15 at 21:12
  • $\begingroup$ PS: $|g^n|=O(\log n)$ follows if $g$ is conjugate to its square. $\endgroup$ – YCor Apr 15 '15 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.