Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 149 / 2 / b / i / Solution

WriteUnder the improved bounds allowed in the question, $r=O({\log }^{O(1)}(2K))$ andMoreover $AC\subseteq A^{O(1)}$, whose size is at most $K^{O(1)}|A|$ by the higher product bound for an approximate group. ThusThe Ruzsa covering lemma supplies $X\subseteq A$ of size at most $\exp (O({\log }^{O(1)}(2K)))$ such thatTaking the $B_i$ to be the factors in this last product giveswhere $k\le O({\log }^{O(1)}(2K))$ and every $B_i$ is a $K^{O(1)}$-approximate group in $A^{O(1)}$ generating a subgroup of step less than $s$.