I was reading some notes on Geometric Group Theory from a seminar, and I came across a statement that reminded me of a pumping lemma. First of all I'll give some definitions and then I'll ask a couple of question on that statement and one of its implications.

Notation: Let $G$ an infinite finitely generated group. For $g\in G$ and a finite generating set $X\subseteq G$, I denote $|g|_X$ the distance from $1_G$ to $g$ on the Cayley graph of $G$ with respect to $X$. By $w_g$ I denote the word of minimum lenght representing $g$ in $G$ and by $l(w)$ I denote the length of the word $w$.

Definition: A word $w=x_1\cdots x_n$ is said to be $p$-periodic if $x_i=x_{i+p}$ for $1\leq i\leq n-p$.

Definition: The growth function of $G$ with respect to $X$ is defined by $\beta_{(G,X)}(n)=\#\{g\in G\mid |g|_X\leq n\}$.

The statement is the following

Given $m\in\mathbb{N}$ and $g\in G$ with $|g|_X\geq 2m$, there exist words $u_g,v_g,s_g$ such that $l(u_g),l(v_g)\leq m$, $s_g$ is $p$-periodic with $p\leq m$ and $w_g=u_gs_gv_g$.

This statement appears in a proof of: the growth function of $G$ is of the order of a degree 1 polynomial implies that $G$ contains an infinite cyclic subgroup of finite index.

First of all, this statement reminds me of the pumping lemma for regular languages, so I would like to know whether this is a known result with a name and if it is indeed related with some pumping lemma.

Sencondly, there is the following explanation:

We observe that the above statement implies the existence of elements of strictly high order. We have $l(s_g)\geq |g|_X-2m$ and $s_g=t_g^ks_g'$ with $t_g\in G$ such that $|t_g|_X=p\leq m$ and $t_g$ has order $>k$. Since there are finitely many choices of $t_g$, there must be one of infinite order.

I think it should be $l(w_{t_g})=p$ and not necessarily $|t_g|=p$, since $t_g$ is represented by the word that is repeated in $s_g$. Furtheremore, I don't understand what is meant by "there are finitely many choices of $t_g$" since there is only one $t_g$ for each word $s_g$, and I don't see why this implies that there is some element of infinite order.

  • $\begingroup$ Yes, this appears inside a proof, so I'll add those hypothesis @LeeMosher $\endgroup$ – Javi Jul 18 '18 at 15:51
  • $\begingroup$ It feels like there's an inequality or quantifier reversed somewhere in the first statement - as written we can just take $m=1$ and say that all words $w$ in $G$ other than the generators are of the form $x_ax_b^nx_c$ and that surely isn't correct. $\endgroup$ – Steven Stadnicki Jul 18 '18 at 16:07
  • $\begingroup$ I copied it just like it was stated so I don't know what is wrong. Perhaps some quantifier on $m$ is needed. @StevenStadnicki $\endgroup$ – Javi Jul 18 '18 at 16:11
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    $\begingroup$ To answer your question about whether this is a known result with a name, I know of no such result or name. Also, knowing that this lemma is appearing only in the middle of the proof you mention, I'm sure there is no need for a general name, because the class of groups that possess a cyclic subgroup of finite index is an extremely special class of groups. $\endgroup$ – Lee Mosher Jul 20 '18 at 17:34

When it says there are finitely many choices for $t_g$ it just means there are finitely many words of length $p$. If all elements represented by words in the finite set $\{t_g|\,g\in G\}$ have finite order, then for any $g\in G$ \begin{align} |g|_X&=l(w_g)\\ &=l(u_g)+l(v_g)+l(s'_g)+l(t_g^k)\\ &\leq 2m+p+p\max_{g\in G} o(t_g),\\ &<\infty \end{align} where $o(w_g)$ denotes the order of $g$, but this contradicts $G$ being infinite.


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