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The additive energy of a finite set $S\subseteq G$ in an abelian group $G$ counts additive quadruples, with order and repeated members allowed:At a fixed sum $s$, independently choosing two of its $r_{S+S}(s)$ ordered pairs gives $r_{S+S}(s{)}^2$ additive quadruples. Thus additive energy weights heavily represented sums more strongly than the cardinality of the sumset does.

For $A=\{0,1,2,3\}$ and $B=\{0,1,4,10\}$, the sumset representation functions giveThe arithmetic progression has additional coincidences such as $0+3=1+2$. The set $B$ still has coincidences forced by commutativity, including $0+1=1+0$, so its additive energy is not merely the sixteen identical-pair comparisons.

The same additive energy can be counted using the difference set. Define $r_{S-S}(t)=|\{(a,b)\in S\times S:a-b=t\}|$. The identity $a+b=c+d$ is equivalent to $a-c=d-b$, and the coordinate rearrangement $(a,b,c,d)\mapsto (a,c,d,b)$ is a bijection. ConsequentlyFor $A$, the difference counts from $-3$ through $3$ are again $1,2,3,4,3,2,1$. For $B$, the zero difference has four representations and each of the twelve nonzero differences has one, giving $4^2+12\cdot 1^2=28$. The sumset representation functions for sums and differences can look different while their squared totals agree.

For every nonempty finite set $S\subseteq G$ in an abelian group $G$,The lower bound follows from the Cauchy-Schwarz inequality and the total sumset representation function count:For the upper bound, any three members $a,b,c$ determine at most one fourth member $d=a+b-c$ of an additive quadruple. See Yufei Zhao's MIT lecture 25, pages 1-3, for representation counts and the energy bounds.

In the two integer examples, the lower bounds are $256/7$ and $256/10$, and the upper bound is $64$. For the subgroup $H=\{0,2,4\}$ of $\mathbb{Z}/6\mathbb{Z}$, each element of $H$ has three sum representations, so $E(H)=3\cdot 3^2=27$ and both bounds are equalities.

For nonempty $S$, dividing additive energy by $|S{|}^3$ gives a relative measure between $1/K(S)$ and one. Small doubling constant therefore forces relatively large additive energy. A converse requires care: the Balog-Szemerédi-Gowers theorem extracts a large subset with small doubling constant, rather than asserting that all of $S$ has small doubling constant.

For finite sets $A,B\subseteq G$ in an abelian group $G$, their difference set isHere $-b$ is the additive inverse in $G$. The difference set records possible displacements from a member of $B$ to a member of $A$; it is distinct from the set difference $A\setminus B$.

For the integer sets $A=\{0,1,2,3\}$ and $B=\{0,1,4,10\}$,The positive differences for $B$ are $1,3,4,6,9,10$, each obtained once; reversing each ordered pair supplies the corresponding negative difference. Every nonempty finite set $S$ has $0\in S-S$, with $|S|$ representations $(s,s)$. Swapping the members of an ordered pair also shows $S-S=-(S-S)$. In the cyclic group $\mathbb{Z}/6\mathbb{Z}$, the subgroup $H=\{0,2,4\}$ satisfies $H-H=H$ as well as $H+H=H$.

Assume $A$ is finite and nonempty; the empty case is immediate. Here $2A=A+A$. Choose a nonempty subset $X\subseteq A$ minimizingTaking $X=A$ as a competitor gives $K\le C$, and every subset $Y\subseteq X$ satisfies $|Y+A|\ge K|Y|$, including $Y=\varnothing$.

We first prove the Petridis minimal-growth lemma in the particular form needed: for every finite set $S$,List $S=\{s_1,\dots ,s_t\}$ and let $X_j$ consist of those $x\in X$ for which $x+s_j$ does not belong to an earlier translate $X+s_h$, $h<j$. The sets $X_j+s_j$ partition $X+S$, so $|X+S|={\sum }_j|X_j|$.

If $x\in X\setminus X_j$, then $x+s_j=x^{\prime }+s_h$ for some earlier $h$ and $x^{\prime }\in X$. Hence every point of $x+A+s_j$ already belongs to $X+A+s_h$. Therefore the new points contributed by $(X+A)+s_j$ are contained inTheir number is at mostSumming over $j$ proves the lemma. With $S=A$, it gives

We also prove the Ruzsa triangle inequality directly. For each $d\in U-V$, choose one representation $d=u_d-v_d$. The mapfrom $(U-V)\times X$ into $(X+U)\times (X+V)$ is injective: subtracting the output coordinates recovers $d$, after which either coordinate recovers $x$. Thus $|U-V||X|\le |X+U||X+V|$. Taking $U=V=2A$ now yieldsThis proves the required growth and difference set estimates directly.

Assume $n>0$; the empty case is immediate. Write $r(s)=\#\{(a,b)\in A^2:a-b=s\}$. The additive energy in the hypothesis is ${\sum }_{s}r(s{)}^2$: rearranging $x+y=z+w$ gives $x-z=w-y$. Since $r(s)\le n$, the energy is at most $n^3$, so the substantive range is $0<\alpha \le 1$.

Let $S=\{s:r(s)\ge \alpha n/2\}$. The contribution of the other differences to the energy is at most $(\alpha n/2){\sum }_{s}r(s)=\alpha n^3/2$. ConsequentlyMake a bipartite graph with two copies of $A$, joining a left vertex $a$ to a right vertex $b$ when $a-b\in S$. Put $\rho =\alpha /2$. This graph has at least $\rho n^2$ edges, and $|S|\le n/\rho$.

We prove the needed dense-graph selection rather than quote the Balog-Szemerédi-Gowers theorem. A pair of left vertices is bad if it has fewer than $\tau n$ common right neighbours, where $\tau ={\rho }^2/32$. Choose a right vertex $y$ uniformly, let $U=N(y)$, and let $b(U)$ be the number of ordered bad pairs in $U$. Averaging the degrees gives $\mathbb{E}|U|\ge \rho n$. Every bad pair lies in fewer than $\tau n$ of these neighbourhoods, so $\mathbb{E}b(U)\le \tau n^2$. Therefore some $y$ satisfiesFor this neighbourhood, $|U|\ge \rho n/2$ and $b(U)\le \rho n|U|/16\le |U{|}^2/8$. Delete vertices having more than $|U|/4$ bad partners in $U$. At most half are deleted, so the remaining set $B$ has $|B|\ge |U|/2\ge \rho n/4$.

For any $b,b^{\prime }\in B$, at least $|U|/2$ vertices $v\in U$ are good partners of both. Each such $v$ has at least $\tau n$ common neighbours with $b$ and at least $\tau n$ with $b^{\prime }$. Independently choosing those neighbours gives at leastfour-edge walks $b,u,v,w,b^{\prime }$. Repeated vertices cause no difficulty. This proves the four-step path lemma for a dense bipartite graph in the required form.

The labels of such a walk are $s_1=b-u$, $s_2=v-u$, $s_3=v-w$, $s_4=b^{\prime }-w$, all in $S$, andFor a fixed pair $(b,b^{\prime })$, its label tuple uniquely recovers $u=b-s_1$, $v=u+s_2$, and $w=v-s_3$, so distinct walks have distinct label tuples. Choose one pair representing each element of the difference set $B-B$. Tuples belonging to different differences are disjoint because their alternating sum determines that difference. HenceWe have obtained explicit constants for the small-difference-set form of the Balog-Szemerédi-Gowers theorem:

We take $\alpha ,\beta >0$, as is necessary for the assertion. The nonvacuous cases have $0<\alpha \le 1/2$ and $0<\beta \le 1$. Put $n=|A|$ andWe prove a lower bound for $K$ by transferring small sumset growth through algebraic operations on the scalars. This gives a self-contained polynomial skew-sumset expansion over a prime field proof using the permitted Ruzsa triangle inequality, Ruzsa covering lemma and Plünnecke-Ruzsa inequality.

Since $|B|>1$, choose $b_0\in B\setminus \{0\}$. The bound $|b_{0}A+A|\le K|b_{0}A|$ and the mixed-set Plünnecke-Ruzsa inequality giveIn particular $|A+A|\le K^{2}n$ and $|3A-2A|\le K^{5}n$. Here repeated-set sums such as $3A$ allow independent choices from each copy.

For a scalar $t$, write $D(t)=|A+tA|/n$, and put $Q=A-A$ for the difference set. The Ruzsa covering lemma gives $tA\subseteq T_t+Q$ with $|T_t|\le D(t)$. Indeed choose a maximal family of disjoint translates $u+A$, with $u\in tA$; counting their union bounds the family size, and maximality forces every remaining translate to meet one of them.

For any $r,s$, the containments $A+(r+s)A\subseteq A+rA+sA$ and $A+rsA\subseteq r{T}_s+A+rQ$ give, after covering each copy of $rA,sA$,For the second bound, specifically $rQ=rA-rA\subseteq T_r-T_r+2Q$, and $A+2Q=3A-2A$. The Ruzsa triangle inequality also gives $|A-rA|\le |A+A||A+rA|/n$, so $D(-r)\le K^{2}D(r)$. These are the closure bounds for small skew-sumset scalars.

We next prove a growth lemma for the scalar sets. For $S\subseteq {\mathbb{F}}_p$ with $m=|S|\ge 2$, letIf $R\ne {\mathbb{F}}_p$, then $R+1$ is not contained in $R$: otherwise equal cardinality would make $R$ invariant under addition of one, which generates the entire prime field. Choose $r=1+(a-b)/(c-d)\notin R$. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Indeed a collision with different second coordinates would express $r$ as a ratio of two differences from $S$, while equal second coordinates force equal first coordinates. Clearing the nonzero denominator puts its image insideso the latter set has at least $m^2$ elements. Here $SS$ is the product set, and $3SS-3SS$ denotes three independent product terms minus three others, not multiplication by the scalar three.

If $R={\mathbb{F}}_p$, let $E_t$ count the collisions $x_1+t{y}_1=x_2+t{y}_2$ with all four variables in $S$. Summing over $t$ givesThe first term comes from equal second coordinates, which force equal first coordinates. For unequal second coordinates, each quadruple determines exactly one scalar. Some $t$ consequently satisfies $E_t\le m^2+m^4/p$. The Cauchy-Schwarz inequality on the representation counts givesRepresenting $t$ as a ratio from $R$ and clearing its denominator embeds a dilate of this sumset in $2SS-2SS\subseteq 3SS-3SS$, padding with the same product in the positive and negative sums. We have proved the quadratic growth of a sixfold product difference set estimate

Starting from $S_0=B$, define $S_{j+1}=3S_j{S}_j-3S_j{S}_j$. ChooseFor $p\ge \max (4,2^{2/\beta })$, the growth lemma implies $|S_k|\ge p/2$. To see the constants explicitly, put $v_j=|S_j|/2$. Then $v_0\ge p^{\beta /2}$ and $v_{j+1}\ge \min (p/4,v_j^2)$. Induction gives $v_j\ge \min (p/4,p^{\beta 2^{j-1}})$, and the chosen $k$ makes the exponent at least one.

Every $t\in S_0$ has $D(t)\le K$. If $D(t)\le K^{L_j}$ on $S_j$, the closure bounds show that each product has exponent at most $3L_j+5$, its negative at most $3L_j+7$, and a sum of six such terms has exponent at most $18L_j+67$. Thus, withwe have $|A+tA|\le K^{L_k}n$ for every $t\in S_k$. The integer $L_k$ depends only on $\beta$.

Now apply the collision count to $A$ instead of $S$. Summing over all scalars gives at most $p{n}^2+n^4$. Since $|S_k|\ge p/2$, some $t\in S_k$ has collision count at most $2n^2+2n^4/p$. The Cauchy-Schwarz inequality therefore givesIt follows thatFor $p\ge 4^{2/\alpha }$ this gives $K\ge p^{\alpha /(2L_k)}\ge n^{\alpha /(2L_k)}$, proving the desired positive exponent for all sufficiently large primes.

For completeness, the same conclusion with a possibly smaller exponent holds for the remaining primes. Choose an integerIf $p<P_0$, choose any nonzero $b\in B$. We have $1<n<p$. If $|A+bA|=n$, all its translates $A+u$ for $u\in bA$ would coincide. Two distinct such $u$ would make $A$ invariant under a nonzero translation, forcing $A={\mathbb{F}}_p$, a contradiction. Thus $|A+bA|\ge n+1\ge n^{1+c_{\alpha ,\beta }}$ by the definition of the second exponent. Together with the large-prime case this provesPositivity of the parameters cannot be dropped: at $\alpha =0$, taking $A={\mathbb{F}}_p$ prevents any expansion; at $\beta =0$, the allowed set $B=\{0\}$ prevents expansion of any proper $A$ with more than one element. No external sum-product theorem was invoked: the scalar growth and transfer arguments were proved above.

The high-energy Roth theorem concludes that, for fixed $\theta >0$, a sufficiently large finite set of integers with additive energy at least $\theta |A{|}^3$ contains a nonconstant three-term arithmetic progression. The size threshold depends on $\theta$, and there is no assumption about the diameter of $A$.

Here is why additive energy replaces interval subset density. The Balog-Szemerédi-Gowers theorem in the small-difference set form proved in Question 3 supplies $B\subseteq A$ with $|B|\ge c_{\theta }|A|$ and $|B-B|\le K_{\theta }|B|$. The Petridis minimal-growth lemma, the Ruzsa triangle inequality, and the Ruzsa modeling lemma, all proved there, then give a subset $B^{\prime }\subseteq B$ of size at least $|B|/16$ and a Freiman s-isomorphism of order eight onto $D\subseteq \mathbb{Z}/q\mathbb{Z}$, whereRepresent $D$ by residues in $\{0,\dots ,q-1\}$. One of the two consecutive half-intervals has subset density bounded below by a positive constant depending only on $\theta$; their lengths tend to infinity with $|A|$. Apply the Roth theorem on three-term arithmetic progressions proved in part (i) to this interval. The resulting three distinct residues satisfy $d_0+d_2=2d_1$ modulo $q$. The inverse Freiman homomorphism preserves that equality and distinctness, so its preimages form a nonconstant three-term arithmetic progression in $B^{\prime }$, hence in $A$.

Large additive energy therefore gives the same three-term progression conclusion after passing to a dense finite model. A forward reference to the detailed lemmas in Question 3 avoids reproving them here.

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.

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.