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.
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.