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If $A$ is a finite $K$-approximate group in a torsion-free $s$-step nilpotent group, then there are $r\le K^{O(1)}$ approximate groups $A_0,\dots ,A_r\subseteq A^{O(1)}$, each with approximation parameter $K^{O(1)}$ and each generating a group of class less than $s$, such thatThe proof applies the large-progression form of the Freiman-Green-Ruzsa theorem in the abelianization, lifts its subgroup and cyclic directions, and uses the intersection of an approximate group power with a subgroup.

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.

Write $N=[G,G]$ and identify it with $(\mathbb{C},+)$ as in part (a). If every two members of $A$ commuted, then $⟨A⟩$ would be Abelian, contrary to hypothesis. Thus some commutator of two members of $A$ is a nonidentity translation in $A^4\cap N$. Consequently the translation-coordinate setcontains both zero and a nonzero element.

The identityshows that $T\pi (A)$ is contained in the translation coordinates of $A^6\cap N$, while $T+T$ is contained in those of $A^8\cap N$. The intersection of an approximate group power with a subgroup therefore givesApply the Solymosi sum-product theorem over the complex numbers with $U=V=T$ and $W=\pi (A)$. Since $1\in \pi (A)$, its hypotheses hold, andCancelling $|T{|}^2$ and absorbing the absolute constant proves

Choose one representative from $A$ above each point of $\pi (A)$ and collect them in $X$. Part (c) gives $|X|\le K^{O(1)}$. If $a\in A$ has the same image as $x\in X$, then $x^{-1}a\in A^2\cap N$, soThis is the required covering of $A$ by at most $K^{O(1)}$ left cosets of the abelian translation subgroup $N$.

The set $B=A^2\cap N$ is a $K^{O(1)}$-approximate group by the intersection of an approximate group power with a subgroup. Apply the Freiman-Green-Ruzsa theorem inside $N\cong (\mathbb{C},+)$. Because the additive group of the complex numbers is a torsion-free group, the finite subgroup part is trivial, so there is an abelian progression $P$ withSince $|B|\le |A^2|\le K|A|$, enlarging the implicit constant gives