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The printed statement needs $A\ne \varnothing$: for the empty set every linear subspace has cardinality at least one, so the requested bound is impossible. We prove the intended nonempty version of the Freiman-Ruzsa theorem over a finite field. Necessarily $C\ge 1$, because any one translate of $A$ inside $A+A$ has size $|A|$.

We first prove all sumset estimates needed below. For finite nonempty $X$ and finite $U,V$ in an abelian group, choose one representation $z=u_z-v_z$ for each $z\in U-V$. The mapis injective: subtracting the outputs recovers $z$, hence its chosen representation and then $x$. This proves the Ruzsa triangle inequality

Choose nonempty $X\subseteq B$ minimizing $c^{\prime }=|X+B|/|X|\le C$, where $B$ is any set with $|B+B|\le C|B|$. We prove the Petridis minimal-growth lemma $|X+B+D|\le c^{\prime }|X+D|$ for every finite $D$. List $D=\{d_1,\dots ,d_s\}$, and let $X_i$ consist of those $x\in X$ for which $x+d_i$ was not already in $X+\{d_1,\dots ,d_{i-1}\}$. The disjoint pieces $X_i+d_i$ partition $X+D$.

Every $(X\setminus X_i)+B+d_i$ was already covered in $X+B+\{d_1,\dots ,d_{i-1}\}$. The new part at step $i$ therefore has size at mostHere minimality of $X$ applies to every nonempty subset $X\setminus X_i$, and the empty case has size zero. Summing proves the lemma. Iteration gives $|X+kB|\le C^k|X|$. Apply the Ruzsa triangle inequality with $U=kB$, $V=\ell B$ to obtainThis derives the required Plünnecke-Ruzsa inequality. In characteristic two, subtraction is addition, so $|rB|\le C^r|B|$ for every positive integer $r$.

Fix $a\in A$ and translate to $B=A+a$, which contains zero, has size $|A|$, and has the same doubling constant. Choose $m=\lceil {\log }_2(2|12B|)\rceil$. A uniformly random linear map $L:{\mathbb{F}}_2^N\to G^{\prime }={\mathbb{F}}_2^m$ kills each specified nonzero vector with probability $2^{-m}$. The union bound shows that some $L$ satisfiesBecause $0\in B$, all lower iterated sumsets lie in $12B$. In particular $L$ is injective on $B$, since the difference of two elements belongs to $2B$. For $B^{\prime }=L(B)$, the density $\alpha =|B^{\prime }|/|G^{\prime }|$ is at least $1/(4C^{12})$. This proves injective linear modelling of a small sumset.

We next prove the Finite-field Bogolyubov lemma in this model. Write $f=1_{B^{\prime }}$ and use normalized real Fourier coefficients on a finite abelian group $\hat{f}(\xi )={\mathbb{E}}_{x}f(x)(-1{)}^{\xi \cdot x}$. LetBy Parseval identity, $|S|\le 2{\alpha }^{-2}$. Thus $W$ is a vector subspace of codimension at most $D=\lceil 32C^{24}\rceil$.

For $w\in W$, Fourier inversion of the fourfold normalized convolution givesAll terms with $\xi \in S$ are nonnegative, and the zero-frequency term is ${\alpha }^4$. The absolute contribution from the remaining frequencies is at mostHence the convolution is positive and $W\subseteq 4B^{\prime }$. Also $|W|\ge 2^{-D}|G^{\prime }|\ge 2^{-D}|B|$.

For each $w\in W$, choose its lift $\sigma (w)\in 4B$. It is unique because two lifts differ by an element of $8B\cap \ker L$, which is zero. For $w_1,w_2\in W$, the discrepancybelongs to $12B\cap \ker L$, so it vanishes. Also $\sigma (0)=0$. Therefore $\sigma$ is a linear map and its image $\tilde{W}\subseteq 4B$ is a vector subspace with $|\tilde{W}|=|W|\ge 2^{-D}|B|$. This establishes lifting a subspace through a Freiman model without assuming that arbitrary choices of preimages preserve addition.

Finally, $B+\tilde{W}\subseteq 5B$ is a union of $\tilde{W}$-cosets. The number of these cosets is at mostChoose representatives $b_1,\dots ,b_t$ from $B$. Then $B\subseteq \tilde{W}+\{b_1,\dots ,b_t\}$, so adjoining these representatives and $a$ to $\tilde{W}$ gives a vector subspace $V$ containing $A$. Over ${\mathbb{F}}_2$, adjoining each vector multiplies the size by at most two. Since $|\tilde{W}|\le |4B|\le C^4|B|$, the Freiman-Ruzsa bound from Bogolyubov and linear modelling yieldsThis bound depends only on $C$, as required, and every sumset estimate used in its derivation has been proved.

We will prove a polynomial progression in a high-energy fourfold difference set. The route is to extract a large subset with a small difference set, construct a dense Freiman s-isomorphism model, and use the Bogolyubov lemma and the pigeonhole principle. To respect the proof requirement, every ingredient beyond the elementary rules for Freiman homomorphisms is established below.

Assume $n\ge 1$. Since three entries of an additive quadruple determine the fourth, $E(A)\le n^3$, so a nonvacuous hypothesis has $0<\theta \le 1$. Let $r(t)=|\{(a,b)\in A^2:a+b=t\}|$. Then $E(A)={\sum }_{t}r(t{)}^2$ and ${\sum }_{t}r(t)=n^2$. Define the popular sum set $S=\{t:r(t)\ge \theta n/2\}$. The contribution to additive energy from its complement is at most $\theta n^3/2$. Since $r(t)\le n$, the bipartite graph on two copies of $A$ whose edges have sums in $S$ has at least $\theta n^2/2$ edges. Moreover $|S|\le 2n/\theta$.

Put $\delta =\theta /2$ and $\tau ={\delta }^2/16$. Here is a dependent random choice argument producing the small difference set. Independently choose two right vertices, allowing repetition, and let $X$ be their common left neighbourhood. The Cauchy-Schwarz inequality gives $\mathbb{E}|X|\ge {\delta }^{2}n$. Call an ordered pair of left vertices bad if its common right neighbourhood has fewer than $\tau n$ elements. The expected number $b(X)$ of bad ordered pairs in $X$ is at most ${\tau }^2{n}^2$, because a fixed bad pair survives both random choices with probability at most ${\tau }^2$. ThusChoose a realization attaining at least this expectation. Then $|X|\ge 3{\delta }^{2}n/4$, and $b(X)\le {\delta }^{2}n|X|/64\le |X{|}^2/48$. Remove from $X$ every vertex having more than $|X|/4$ bad partners in $X$. At most $|X|/12$ vertices are removed. The remaining set $B$ has $|B|\ge {\delta }^{2}n/2$, and any $a,b\in B$ have at least $|X|/2$ vertices $z\in X$ for which both $(a,z)$ and $(z,b)$ are good.

For such a $z$, there are at least $\tau n$ right vertices $y$ adjacent to $a,z$, and at least $\tau n$ right vertices $y^{\prime }$ adjacent to $z,b$. These paths givea representation with four members of $S$. For fixed $a,b$, the tuple of four sums determines $y,z,y^{\prime }$ uniquely, so there are at least $|X|{\tau }^2{n}^2/2$ distinct tuples. Select one ordered pair $a,b$ for each member of $B-B$. Tuples belonging to different differences are disjoint. Counting all possible tuples in $S^4$ provesIn particular, with $m=|B|$, we have $m\ge {\theta }^{2}n/8$ and $|B-B|\le Km$ for $K=2^{25}{\theta }^{-12}$. This is the required small-difference set form of the Balog-Szemerédi-Gowers theorem, proved by counting paths rather than cited.

We next control higher iterated sumsets. Choose a nonempty $X\subseteq B$ minimizing $\rho =|X-B|/|X|$, so $\rho \le K$. We prove the Petridis minimal-growth lemma in the form $|X-B+C|\le \rho |X+C|$ for finite $C$. Order $C=\{c_1,\dots ,c_t\}$, and let $X_i$ consist of those $x\in X$ for which $x+c_i$ is new in the union of the translates $X+c_j$. Then ${\sum }_i|X_i|=|X+C|$. Every member of $(X\setminus X_i)-B+c_i$ already occurs in an earlier translate of $X-B$. By minimality, $|(X\setminus X_i)-B|\ge \rho |X\setminus X_i|$, also when the complement is empty. Thus the new contribution from $X-B+c_i$ is at most $\rho |X_i|$. Sum over $i$ to prove the lemma. Iterating with $C=-(j-1)B$ gives $|X-jB|\le K^j|X|$.

For completeness, the Ruzsa triangle inequality in the needed form isFor each $d\in U-W$, fix $d=u_d-w_d$. The map $(d,v)\mapsto (u_d-v,v-w_d)$ is injective: adding the coordinates recovers $d$, and then $v$. Take $U=kB$, $W=\ell B$, $V=X$ and apply the preceding bounds. We obtain the Plünnecke-Ruzsa inequalityIn particular $T=8B-8B$ has at most $K^{16}m$ elements.

We now prove a slightly wasteful version of the Ruzsa modeling lemma, avoiding any assumption about the diameter of $B$. Set $q=8|T|$. Choose a prime $p$ larger than all $|t|$, $t\in T$, and much larger than $q$. Arbitrarily large primes exist by the elementary argument that a prime divisor of one plus the product of all primes up to a given bound exceeds that bound. For a uniformly chosen nonzero multiplier $r$ modulo $p$, each nonzero $t\in T$ has $rt$ uniform among the nonzero residues. Call a multiplier forbidden if some such $rt$ agrees modulo $p$ with $jq$, where $|jq|<p$. There are at most $2p/q+1\le 3p/q$ such residues. For $p\ge 4$, a union bound therefore makes the probability of being forbidden less thanChoose a nonforbidden $r$. Represent $rB$ by integers in $[0,p-1]$ and partition that interval into sixteen consecutive pieces of diameter less than $p/16$. One piece contains the images of a subset $B^{\prime }\subseteq B$ with $|B^{\prime }|\ge m/16$. Let $\varphi (b)$ be the integer representative of $rb$ in that piece, reduced modulo $q$.

A difference of eight sums of these representatives has absolute value less than $p/2$. If its original difference in $B$ is zero, it is a multiple of $p$, hence zero. Conversely, if its reduction modulo $q$ is zero, it equals $jq$ with $|jq|<p$; our choice of $r$ forces its original difference to be zero. This proves that $\varphi$ is a Freiman s-isomorphism of order eight. Injectivity follows by padding a one-term equality to eight terms with a fixed element. Consequently $D=\varphi (B^{\prime })\subseteq G=\mathbb{Z}/q\mathbb{Z}$ has subset densityNo primality of $q$ is needed.

Here is the cyclic Bogolyubov lemma with all its Fourier analysis on a finite abelian group details. Use $\hat{f}(r)={\mathbb{E}}_{x}f(x)e(-rx/q)$, $f=1_D$, and normalized convolution on a finite group. A finite geometric series proves orthogonality, and expanding the finite sums proves the Parseval identity on a finite group, Fourier inversion on a finite group, and $\widehat{f\ast g}=\hat{f}\hat{g}$. For $\tilde{f}(x)=f(-x)$, the nonnegative functionhas support exactly $2D-2D$ and Fourier coefficients on a finite abelian group $|\hat{f}(r){|}^4$. Put $\Gamma =\{r\ne 0:|\hat{f}(r)|\ge {\alpha }^{3/2}/\sqrt{8}\}$. The Parseval identity on a finite group gives $|\Gamma |\le 8{\alpha }^{-2}$. The total fourth moment of the frequencies outside $\Gamma \cup \{0\}$ is at most ${\alpha }^4/8$.

If $\Vert rx/q\Vert \le 1/16$ for every $r\in \Gamma$, then $\mathrm{Re}e(rx/q)\ge \cos (\pi /8)>3/4$. The zero frequency contributes ${\alpha }^4$ with no loss. Separating the large and small frequencies in Fourier inversion on a finite group yieldsThus $2D-2D$ contains the Bohr set expressed in distance coordinates by $\Vert rx/q\Vert \le 1/16$. This is consistent with the existing character-distance convention for a Bohr set: here it is enough that each character lies in the corresponding short arc around one.

We finally prove the needed nonwrapping progression in a cyclic Bohr set, including for composite $q$. Write $d=|\Gamma |$ and $Q=\lfloor q^{1/(d+1)}\rfloor$. Partition the $d$-dimensional unit cube into $Q^d$ boxes. The $Q^d+1$ points with coordinates $jr/q$ modulo one, $0\le j\le Q^d$, give by the pigeonhole principle an integer $1\le t\le Q^d$ with $\Vert rt/q\Vert \le 1/Q$ for all $r\in \Gamma$. ThenThese points are distinct modulo $q$, since their integer representatives lie between zero and $Q^{d+1}/16\le q/16$. Their number is at least $q^{1/(d+1)}/32$. For $d=0$ the same argument simply chooses $t=1$.

The Freiman s-isomorphism induces a bijection $\Phi :2B^{\prime }-2B^{\prime }\to 2D-2D$: define it on a representation by taking the same two positive and two negative images. Equality of representations uses an order-four relation, and injectivity uses the converse relation. If three consecutive members $y_{j-1},y_j,y_{j+1}$ of $P$ satisfy $y_{j-1}+y_{j+1}=2y_j$, their preimages satisfy the same equality, since its expansion is an equality of eight-term sums in $B^{\prime }$. Hence the inverse image of $P$ is an ordinary arithmetic progression in $2B^{\prime }-2B^{\prime }\subseteq 2A-2A$, with distinct terms.

To remove the multiplicative constant from the length bound, let $R=\lceil 8{\alpha }_0^{-2}\rceil$ and $\beta =1/(R+1)$. The length just proved is at least $c_{\theta }{n}^{\beta }$, where $c_{\theta }={\theta }^{2\beta }/32>0$. Choose $N_0\ge 3$ so that $c_{\theta }{n}^{\beta }\ge n^{\beta /2}$ for $n\ge N_0$, and set $\gamma =\min \{\beta /2,\log 3/\log N_0\}$. For $2\le n<N_0$, two distinct members of $A$ give a nonzero $u\in A-A$, and $\{-u,0,u\}\subseteq 2A-2A$ has length three, at least $n^{\gamma }$. For $n=1$, the singleton $\{0\}$ suffices. In every nonempty case,with $\gamma >0$ depending only on $\theta$. If empty sets are permitted, the required length is zero and there is no substantive assertion.

An additive form of the Plünnecke inequality is the following. If $A,B$ are nonempty finite subsets of an abelian group and $|A+B|\le K|A|$, then there is one nonempty $X\subseteq A$ such that, simultaneously for every integer $m\ge 0$,Here $mB$ is an iterated sumset and $0B=\{0\}$. In particular, its Plünnecke-Ruzsa inequality consequence isWe prove both statements, so that the sumset and difference set formulations are covered.

Choose a nonempty $X\subseteq A$ minimizing the ratio $|X+B|/|X|$, and denote this minimum by $\kappa$. Such a minimizer exists because $A$ is finite, and $\kappa \le K$. Every $Z\subseteq X$ satisfies $|Z+B|\ge \kappa |Z|$, with the empty case also valid. We first prove the Petridis minimal-growth lemmafor every finite nonempty $C$.

List $C=\{c_1,\dots ,c_t\}$. Let $U_i=X+\{c_1,\dots ,c_i\}$ and $U_0=\varnothing$. DefineThe new points of $U_i$ are exactly $X_i+c_i$, so $|X+C|={\sum }_i|X_i|$. Since $Z_i+c_i\subseteq U_{i-1}$, the sumset $Z_i+B+c_i$ is already contained in $U_{i-1}+B$. It follows that the new points introduced into $U_i+B$ are contained inAs $Z_i+B\subseteq X+B$, their number is at mostSumming these increments proves the Petridis minimal-growth lemma. Taking $C=(m-1)B$ for $m\ge 1$ and iterating now givesThis proves the Plünnecke inequality with the same minimizing set $X$ for every $m$.

To obtain the difference set bound, we also prove the required Ruzsa triangle inequality. For finite $U,V,Z$ with $Z\ne \varnothing$, choose one representation $d=u_d-v_d$ of each $d\in U-V$. The mapis an injection: the sum of the output coordinates recovers $d$, and the first coordinate then recovers $z$ because $u_d$ was fixed. ThusUse $U=kB$, $V=\ell B$, and $Z=-X$. The already proved Plünnecke inequality yieldsThis finishes the proof of the stated Plünnecke-Ruzsa inequality. The use of an abelian group is essential in commuting the translates and sumsets in the minimal-growth argument.

The Plünnecke-Ruzsa inequality states that, for all nonnegative integers $m,n$,Here $mA$ is the iterated sumset of $m$ copies of $A$, with $0A=\{0\}$. In particular, $|mA|\le K^m|A|$.

If $E(A)\ge \theta |A{|}^3$, a subset $B\subseteq A$ satisfies $|B|\ge c_{\theta }|A|$ and $|B-B|\le K_{\theta }|B|$. A dependent random choice argument in the popular sum graph finds many four-edge paths representing each member of the difference set. The Petridis minimal-growth lemma then bounds all higher iterated sumsets and difference sets.