Half-size prime cyclic Freiman model ($|A^{\prime }|\ge |A|/2,p>4K^4|A|$)

For a nonempty integer set with doubling constant at most $K$, Plünnecke-Ruzsa inequality bounds $|2A-2A|$ by $K^4|A|$. Embed the set modulo a sufficiently large auxiliary prime $q$, randomly dilate it, and select the more populated of two half-intervals of representatives. Differences of two pair sums on that half have absolute value less than $q$, so the lift preserves original relations. Every nonzero original difference is uniformly dilated among nonzero residues; avoiding all residues represented by nonzero multiples of $p$ of absolute value less than $q$ has positive probability when $p>4|2A-2A|$. Reduction modulo $p$ then yields a Freiman 2-isomorphism on the selected half.

For a finite nonempty integer set $A$, put $D=2A-2A$. If an integer $N>|D|$, some $A^{\prime }\subseteq A$ of size at least $|A|/2$ has a Freiman 2-isomorphism into $\mathbb{Z}/N\mathbb{Z}$. Choose an auxiliary prime $P>N$ so large that reduction is injective on $A$ and every nonzero element of $D$ remains nonzero. Dilate modulo $P$ by a uniformly chosen nonzero multiplier. The forbidden residues are those represented by nonzero multiples of $N$ of absolute value less than $P$, at most $2\lfloor (P-1)/N\rfloor$ residues. The nonzero elements of $D$ occur in opposite pairs; the two elements in each pair have the same bad event. The union bound therefore gives failure probability at most $(|D|-1)/N<1$. Choose a multiplier avoiding all these events. Select the more populated of two half-intervals of residues modulo $P$. Differences of two pair sums of representatives then have absolute value less than $P$. An original additive relation must therefore lift to equality in the integers; conversely an equality modulo $N$ cannot have a nonzero lifted difference, by the forbidden-residue choice. Reduction modulo $N$ is the required isomorphism.