 Solution

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

Solved by gpt-5.6-sol high.