 Solution

Let $\pi :G\to G/[G,G]$ be the abelianization map. Its image $\pi (A)$ is again a $K$-approximate group. Apply the large-progression form of the Freiman-Green-Ruzsa theorem to $\pi (A)$. It gives a finite subgroup $H$, elements $x_1,\dots ,x_r$, and lengths $L_1,\dots ,L_r$, withand

The large lifted product from a coset progression applied to this progression givesBriefly, choose a section of $\pi$ on $\pi (A^6)$. Multiplication by that section is multiplicative up to $A^{12}\cap [G,G]$; lifting successively the subgroup part and each progression direction therefore places every element of $A^6\cap {\pi }^{-1}(HP)$ in the displayed product. The fiber-counting lemma for a quotient map gives $|A^6\cap {\pi }^{-1}(HP)|\ge \exp (-K^{O(1)})|A|$, which proves the estimate.

SetThe intersection of an approximate group power with a subgroup shows that each $A_i$ is a $K^{O(1)}$-approximate group contained in $A^{O(1)}$. The preimage of a cyclic subgroup of $G/[G,G]$ has step less than $s$. The same is true of the preimage of the finite subgroup $H$ because $G$ is a torsion-free group. Consequently each $⟨A_i⟩$ has step less than $s$, and the displayed estimate is the required conclusion.