Suppose $\Gamma$ is a finitely generated countable discrete torsion free group with a generating set $S$. Let $l$ be the word length function given by $S$. 
Let $F\_n=\{s\in\Gamma| l(s)\leq n\}$.


Assume that $\Lambda$ is a subset of $\Gamma$ such that $$\limsup\_{n\to\infty} \frac{|\Lambda\cap F\_n|}{|F\_n|}>0.$$


Question: For every positive integer $k$, do there exist $b$ and $a$ in $\Gamma$ such that $\{b^{j}a\}\_{j=0}^{k-1}\subseteq\Lambda$?


When $\{b^{j}a\}\_{j=0}^{k-1}$has $k$ distinct elements, we call it a **left arithmetic progression of length $k$** in $\Gamma$.


Remark: One would avoid the trivial case that $b=e\_\Gamma$. In fact, let $\Gamma=\mathbb{Z}$ with $S=\{1\}$, the answer to the above question is affirmative by Szemeredi's theorem, which says that a subset of $\mathbb{Z}$ with positive upper density contains arbitrarily long arithmetic progressions.



Arithmetic progressions in finitely generated groups
Let $\Gamma$ be a free group over the alphabet $X$, $|X|\geq 2$, and put $S=X\cup X^{-1}$. Pick an increasing sequence of integers $n\_i$, and put $\Lambda=\{g\in\Gamma|\exists i:\ell(g)=n\_i\}$. Under some mild restrictions on the sequence $n\_i$ we have that any arithmetic progression of length $\geq 3$ consists of elements of equal length. But then $a$, $ba$ and $b^2a$ have equal length. Hence $\ell(b)$ is even, and we decompose the reduced words for $a$ and $b$ as $a=vw$, $b=uv^{-1}$, where $\ell(u)=\ell(v)$, that is, $ba=uw$. In the same way $\ell(b^2a)=\ell(ba)$ implies $v=u$, thus $b=uu^{-1}=e$. But arithmetic progressions are assumed to consist of different elements, so this choice of $b$ is excluded. On the other hand the set of elements of length $=n\_i$ is of positive density among all elements of length $\leq n\_i$, so we obtain a set of positive upper density without arithmetic progressions of length 3.


I would assume that a similar argument works for any group of exponential word growth.


