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

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.

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 $s$.

We induct on the nilpotency class $s$. For $s=1$, the group is Abelian, so take $C_1=A$ and no $X_i$. For $s>1$, part (i) writeswith $X$ small and every $⟨B_i⟩$ of class at most $s-1$. Apply the induction hypothesis to each $B_i$. Since $B_i\subseteq A^{O(1)}$ and its approximation parameter is $K^{O(1)}$, all resulting small sets lie in $A^{O_s(1)}$, all their sizes are at mostand all resulting approximate groups lie in $A^{O_s(1)}$ and have approximation parameter $K^{O_s(1)}$. Their generated subgroups are abelian groups.

There are $O({\log }^{O(1)}(2K))$ factors at each of at most $s$ induction levels. Absorbing the resulting products of the bounds into the $O_s$ notation givesKeeping the factors in the order supplied by the induction yields the required product of the $X_i$ and $C_j$ containing $A$.

Use the product from part (ii). Move each selected element of each small set $X_i$ to the left, conjugating every approximate-group factor that it crosses. For each tuple $\mathbf{x}\in X_1\times \cdots \times X_m$, the corresponding part of the product is therefore contained inwhere every $D_{j,\mathbf{x}}$ is a conjugate subset of some $C_j$. Conjugation preserves cardinality, the approximation parameter, and the property that the generated subgroup is abelian. The conjugating elements belong to $A^{O_s(1)}$, so $D_{j,\mathbf{x}}\subseteq A^{O_s(1)}$ after enlarging the implicit constant.

The number of tuples is at mostThese translated products cover $A$, so one has size at least the reciprocal fraction of $|A|$. For that tuple, put $D_j=D_{j,\mathbf{x}}$. Thenwhere $q=n\le O_s({\log }^{O_s(1)}(2K))$, and each $D_j$ is a $K^{O_s(1)}$-approximate group generating an abelian subgroup.