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For every finite $K$-approximate group $A$ in a group $G$, there are subgroups $H⊴C<G$ such that$C/H$ is a nilpotent group of class $O_K(1)$, and $A$ is contained in the union of $O_K(1)$ left cosets of $C$.

If $A$ is a finite $K$-approximate group, thenfor every positive integer $m$, where $A^2\subseteq XA$ and $|X|\le K$. This follows immediately by mathematical induction.

If $A$ is a $K$-approximate group in $G$ and $H\le G$, then $A^m\cap H$ is a $K^{O(m)}$-approximate group. Indeed, the at most $K^{m-1}$ translates of $A$ covering $A^m$ induce at most that many translates of $A^2\cap H$ covering $A^m\cap H$, and the same argument covers its square inside $A^{2m}\cap H$.

Let $\pi :G\to G/[G,G]$ be the abelianization map and let $A$ be a finite $K$-approximate group. Suppose a coset progression $HP(x_1,\dots ,x_r;L_1,\dots ,L_r)$ lies in $\pi (A^4)$ and has size at least $\delta |\pi (A)|$. ThenA section of $\pi$ is multiplicative up to a bounded power of $A$ inside the commutator subgroup; successively separating the subgroup and progression coordinates proves the inclusion needed for this estimate.

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.

Choose a covering set $X$ of size at most $K$ such that $A^2\subseteq XA$, as allowed by the definition of an approximate group. Discard every $x\in X$ for which $xA$ does not meet $A^2$. Each remaining $x$ belongs to $A^2{A}^{-1}=A^3$, so $X\subseteq ⟨A⟩$. Induction givesBecause $A$ is symmetric and contains the identity, $⟨A⟩={\bigcup }_{m\ge 1}{A}^m$. Hence

We use the bounded-exponent finitely generated nilpotent group order bound. In an $s$-step nilpotent group, a subgroup generated by $k$ elements is generated in collected form by the simple group commutators in those generators of weights at most $s$. There are at mostsuch commutators. Every one has order at most $r$, soTaking $H=⟨A⟩$ now gives

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.

Assume $|A+A|\le K|A|$ and putThis set is symmetric, contains zero, and the Plünnecke-Ruzsa inequality givesIt also contains $A-A$, so for every $x\in A$,

To control $H+H=4A-4A$, observe again by Plünnecke-Ruzsa thatApply the Ruzsa covering lemma with $S=H+H$ and $T=A$. There is a set $Y$ with $|Y|\le K^9$ such thatConsequently $H$ is a $K^9$-approximate group. Taking a sufficiently large absolute constant $C$ gives