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 $H⊴C\le G$ such that $H\subseteq A^4$, the quotient group $C/H$ is nilpotent of class $O_K(1)$, and $A$ is contained in the union of $O_K(1)$ left cosets of $C$.

The structural subgroup $C$ may equal $G$. For example, if $A=G$ is a cyclic group of large prime order, then $A$ is a $1$-approximate group. Its only proper subgroup is trivial, so a bounded number of cosets of a proper subgroup cannot cover $A$ uniformly in the prime. The full structure theorem provides finer nilpotent progression control; the statement here records its quotient-and-covering consequence.

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.

Let $A$ be a finite $K$-approximate group in $G$, let $H\le G$, and let $m\ge 2$ be an integer. Then $B=A^m\cap H$ is a $K^{2m-1}$-approximate group.

The higher product bound for an approximate group covers $A^{2m}$ by at most $K^{2m-1}$ left translates of $A$. In every translate meeting $B^2$, choose one representative $b\in B^2$. If $z$ is another point in that intersection, then $b^{-1}z\in A^{-1}A\cap H=A^2\cap H\subseteq B$. Thus the chosen representatives give $B^2\subseteq YB$ with $|Y|\le K^{2m-1}$. Identity containment and inversion symmetry are inherited from $A^m\cap H$.

The restriction $m\ge 2$ matters. In $\mathbb{Z}\times {\mathbb{Z}}_2$, put $A=([-N,N]\times \{1\})\cup (S\times \{0\})$, where $S\subseteq [-N,N]$ is any symmetric set containing zero. Four translates of $A$, with first-coordinate shifts $\pm N$ and both second-coordinate shifts, cover $[-2N,2N]\times {\mathbb{Z}}_2$, hence cover $A^2$. Thus $A$ is a $4$-approximate group, while $A\cap (\mathbb{Z}\times \{0\})=S\times \{0\}$ can have arbitrarily large doubling constant.

Suppose $s\ge 2$, and let $A$ be a finite $K$-approximate group in a torsion-free $s$-step nilpotent group. 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 nilpotency class less than $s$, such thatThe reduction 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.

This is a reduction for $s\ge 2$, followed by an abelian structural theorem as the base case. If $s=1$, class less than $s$ means the trivial group. Products of such factors have size one, which cannot give the displayed positive proportional lower bound for arbitrarily large intervals in $\mathbb{Z}$ with bounded doubling constant.

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 $⟨A⟩={\bigcup }_{m\ge 1}{A}^m$.

For finite subsets $S,T$ of an abelian group, with $T$ nonempty, suppose $|S+T|\le K|T|$. Then some $X\subseteq S$ satisfiesChoose a maximal family of pairwise disjoint translates $x+T$ with $x\in S$. Their union has size $|X||T|$ inside $S+T$, proving the cardinality bound. Maximality means that every $s+T$ meets some $x+T$. An equation $s+t=x+t^{\prime }$ gives $s\in x+T-T$, proving the covering. If $S$ is empty choose $X$ empty. An empty $T$ is excluded: it would make the assumed inequality vacuous while $X+T-T$ remained empty.