# Finitely generated group where $|g^n|=o(n)$ for all $g\in G$.

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?
• @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. May 3, 2023 at 15:58
• That's better. Thank you. May 3, 2023 at 16:42

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 $$\tau(g)=0$$.
I do not know if this is the case for $$G$$ finitely presentable.