Approximate group

A $K$-approximate group is a finite symmetric subset of a group $A$ for which some set $X$ with $|X|\le K$ satisfies $A^2\subseteq XA$. In additive notation the covering condition is $A+A\subseteq X+A$.

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 $A^{2m}\cap H$.

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 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.

Let $\pi :G\to G/N$ be a quotient group map, let $A$ be finite and symmetric, and suppose $P\subseteq \pi (A^m)$ has $|P|\ge \delta |\pi (A)|$. ThenChoose one lift in $A^m$ of each member of $P$ and multiply those lifts by $A^2\cap N$. Distinct fibers are disjoint, while $|\pi (A)||A^2\cap N|\ge |A|$.

A subset $A$ of a multiplicatively written group is symmetric when it contains the identity element and $A^{-1}=A$. Consequently $⟨A⟩={\bigcup }_{m\ge 1}{A}^m$.

If finite sets satisfy $|S+T|\le K|T|$, then a maximal family of disjoint translates $x+T$, with $x\in S$, has at most $K$ members and gives