This question arose after I gave a general talk on the sub-additive ergodic theorem, where I gave the (simple) proof that in a finitely generated group, if one has a stationary sequence of random elements of the group, $g_1$, $g_2$, $\ldots$ where the expected word length of $g_i$ is finite, then defining $h_n$ to be the product $g_1g_2\cdots g_n$, one has $|h_n|/n$ converges almost surely to a constant. A question arose concerning the relationship between volume growth in the group and the speed of growth. I believe there is no relationship between the two, and wanted to claim that in any group with elements of infinite order, one can find a $g$ such that $|g^n|$ grows linearly.

Let $G$ be a finitely generated group containing elements with infinite order. Let $S$ be a generating set. For $g\in G$, write $|g|=\min\{n\colon \text{$g$ can be expressed as the product of $n$ elements in $S$}\}$ (the word metric).

Must there exist $g\in G$ such that $\liminf_{n\to\infty}|g^n|/n$ (which is equal to $\inf_n|g^n|/n$ by Fekete's lemma) is positive?
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    $\begingroup$ @shaun:I have now provided some context. It's basically a question that arose through general interest. I am not an expert in geometric group theory, but I expect that someone in that field could give me an immediate answer. $\endgroup$ May 3, 2023 at 15:58
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    $\begingroup$ That's better. Thank you. $\endgroup$
    – Shaun
    May 3, 2023 at 16:42

1 Answer 1


The number $\tau(g)=\inf_n|g^n|/n$ is called the translation length of $g$.

No, there are finitely generated groups $G$ all of whose elements have $0$ translation length. This is a consequence of two facts:

  1. If an infinite-order element $g$ is conjugate to a proper power of itself, then $\tau(g)=0$.
  2. There are finitely generated, torsion-free groups with precisely two conjugacy classes. These were first constructed in the paper D. V. Osin, Small Cancellations over Relatively Hyperbolic Groups and Embedding Theorems, Ann Math. 172, no. 1 (2010), 1-39, (journal), (arXiv).

I do not know if this is the case for $G$ finitely presentable.

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    $\begingroup$ Thanks for this! $\endgroup$ May 4, 2023 at 3:39

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