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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

The Plünnecke-Ruzsa inequality says that if $A,B$ are finite nonempty subsets of an abelian group andthen for all nonnegative integers $m,n$,

Choose a nonempty $X\subseteq A$ minimizingthen $K^{\prime }\le K$. We first prove the Petridis minimal-growth lemmafor every finite $C$, by induction on $|C|$. Remove $c\in C$, write $C^{\prime }=C\setminus \{c\}$, and letThe new points contributed to $X+C$ by $X+c$ are exactly $|X\setminus X^{\prime }|$. Moreover, $X^{\prime }+B+c\subseteq (X+B+c)\cap (X+B+C^{\prime })$, soThe induction hypothesis, the identity $|X+B|=K^{\prime }|X|$, and minimality, which gives $|X^{\prime }+B|\ge K^{\prime }|X^{\prime }|$, yieldIteration with $C=(r-1)B$ gives

Finally, the Ruzsa triangle inequality givesas required.

The Freiman-Ruzsa theorem over a finite field states that if $A\subseteq {\mathbb{F}}_p^n$ and $|A+A|\le K|A|$, then $A$ is contained in a vector subspace $H$ withAfter translating $A$, assume $0\in A$. Put $S=A-A$, and choose $X\subseteq A+S$ maximal subject to the translates $x+A$, $x\in X$, being pairwise disjoint. Since $X+A\subseteq 2A+S=3A-A$, the Plünnecke-Ruzsa inequality givesso $|X|\le K^4$.

Maximality gives $A+S\subseteq X+S$: if $y\in A+S$ is not already in $X$, then $(y+A)\cap (x+A)\ne \varnothing$ for some $x\in X$, whence $y\in x+A-A=x+S$. Inductively, $mA+S\subseteq ⟨X⟩+S$ for every positive integer $m$. Because $0\in A$, every element of $⟨A⟩$ belongs to some $mA$ in the finite vector space, and thereforeFinally, $|⟨X⟩|\le p^{|X|}$ and $|S|\le K^2|A|$ by the Plünnecke-Ruzsa inequality, so

If a nonempty finite set $X$ minimizes $|X+B|/|X|$ among the nonempty subsets of $A$, with minimum $K^{\prime }$, thenfor every finite set $C$. Iteration is a short proof of the Plünnecke-Ruzsa inequality.