Does any subset of a finitely generated group with positive upper density contain three points in arithmetic progression? Let $G$ be a countable finitely generated group, with word metric $d_S$ induced by some generating set $S$. Let $B_r$ be the ball of radius $r$ around the identity under $d_S$.
We say that $A \subset G$ has positive upper density if $\limsup_{n\to\infty} \frac{|A\cap B_n|}{|B_n|}>0$ for some, hence any word metric $d_S$.

Question: For any $A\subset G$ with positive upper density, does it hold that there exists some $a \in A$, and nonzero $g \in G$ such that $a, a + g, a + 2g$ are in $A$?

Edit: The original question asked about abelian groups, which is answered in the positive by Roth’s theorem as stated by Sean Eberhard in the comments.
 A: I hope this might be of interest, I'll consider a simpler problem and present a couple $(G,A)$ where one can't find a progression of length four: $a, ag, ag^2, ag^3$.
Example. Let $G$ be the free group $F_2$ on $2$ generators with generating set $x_1, x_1^{-1}, x_2, x_2^{-1}$. Take $A$ as the union of all words of length $\varphi(n)=4^n$ for $n\ge 2$. It is easy to see that $A$ has positive upper density. However one can not find elements $a, ag, ag^2,ag^3\in A$ with $g\ne 1$. The reason here is the following.
Observation. Fix $a$ and $g$ and  consider the sequence of words $\ldots, ag^{-1},a,ag,\ldots, ag^n,\ldots$. Let $l(n)$ be equal to the length of $ag^n$. Then $l(n)$ is given by formula $d|n-x|+b$, where $d>0$, $x$ a fixed (half integer) number $b\ge 0$.
I don't want to write a complete proof of this observation. But the idea is that the multiplication by $g$ is acting by isometry of the Cayley graph of $F_2$ and it sends a unique geodesic (in the graph) to itself, and points $ag^{-n}$ are on the same distance to this geodesic (maybe there is a more obvious reasoning).
From this observation it is very easy to deduce the claim. Consider for example the case when $l(a)=l(ag)<l(ag^2)<l(ag^3)$. Suppose by contradiction these elements are all in $A$. In this case distance in the Cayley graph between $a$ and $ag$ is at most $2\cdot 4^n$ for some $n$. But the distance between $ag$ and $ag^2$ is at least $3\cdot 4^{n}$, a contradiction (the distance should be the same). Other cases are similar.
