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The sumset form of Plünnecke's inequality states: for nonempty finite sets $A,B$ in an abelian group, if $|A+B|\le K|A|$, there is a nonempty $X\subseteq A$ such thatHere $0B=\{0\}$. The same $X$ can be used for every $m$. This sumset statement follows from an elementary minimal-growth argument.

Choose $X$ minimizing $\kappa =|X+B|/|X|$ among nonempty subsets of $A$. Then $\kappa \le K$ and $|Y+B|\ge \kappa |Y|$ for every $Y\subseteq X$, including the empty set. We prove the Petridis minimal-growth lemmafor each finite set $C$. Enumerate $C=\{c_1,\dots ,c_t\}$. Put $C_i=\{c_1,\dots ,c_i\}$ and defineThe disjoint new contributions to $X+C$ have sizes $|X_i|$, so $|X+C|={\sum }_i|X_i|$. For $y\in Y_i$, the point $y+c_i$ was already present in $X+C_{i-1}$, hence $y+B+c_i\subseteq X+B+C_{i-1}$. Consequently the new contribution at stage $i$ to $X+B+C$ has size at mostSumming proves the lemma. Starting with $C=\{0\}$ and then taking $C=(m-1)B$ gives inductively $|X+mB|\le {\kappa }^m|X|\le K^m|X|$, proving the stated Plünnecke inequality.

One quantitative form of the Balog-Szemerédi-Gowers theorem is: if a nonempty finite set $A$ in an abelian group has $|A|=n$ and additive energy $E(A)\ge \eta n^3$, where $0<\eta \le 1$, then there is $B\subseteq A$ with $|B|\ge c_{\eta }n$ and both $|B-B|$ and $|B+B|$ bounded by $K_{\eta }|B|$. The constants depend only on $\eta$. We prove explicit, nonoptimal polynomial bounds, thereby proving the required Balog–Szemerédi statement.

Let $r(s)=|\{(a,a^{\prime })\in A^2:a+a^{\prime }=s\}|$. Then $E(A)={\sum }_{s}r(s{)}^2$. Declare $S=\{s:r(s)\ge \eta n/2\}$ to be the set of popular sums. The remaining sums contribute at most $(\eta n/2){\sum }_{s}r(s)=\eta n^3/2$. Since $r(s)\le n$, it follows that ${\sum }_{s\in S}r(s)\ge \eta n^2/2$. Also $|S|\le 2n/\eta$. Form a bipartite graph on two copies of $A$, joining $a$ to $a^{\prime }$ when $a+a^{\prime }\in S$. Its density is at least $\rho =\eta /2$.

We prove the needed four-step path lemma for a dense bipartite graph. Call an ordered pair of left vertices bad if its number of common right neighbors is less than $\tau n$, where $\tau ={\rho }^2/32$. For a uniformly chosen right vertex $y$, let $U=N(y)$ and let $b(U)$ count bad ordered pairs in $U^2$. ThenConsequently some $y$ satisfiesFor this $U$, $|U|\ge \rho n/2$ 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; the retained set $B$ has $|B|\ge |U|/2\ge \rho n/4$. For any $a,b\in B$, at least $|U|/2$ middle vertices $u\in U$ are good partners of both. Each such $u$ supplies at least $(\tau n{)}^2$ walks $a,v,u,w,b$ of four edges. Hence every pair in $B$ has at leastsuch walks. Repetition of vertices is allowed; it does not affect the counting argument.

Each walk has edge labelsall in $S$, and $a-b=s_1-s_2+s_3-s_4$. For a fixed pair $a,b$, these four labels determine $v,u,w$, so distinct walks have distinct label quadruples. For each element of $B-B$, fix one representing pair $a,b$. Label quadruples for distinct differences cannot overlap. ThereforeUsing $\rho =\eta /2$ and $|S|\le 2n/\eta$ yieldsThis proves the small-difference-set form of the Balog-Szemerédi-Gowers theorem. To obtain a small sumset too, write $K=2^{24}{\eta }^{-10}$ and apply the already proved Plünnecke inequality to starting set $-B$ and summand set $B$. Some nonempty $X\subseteq -B$ has $|X+2B|\le K^2|X|$. For any fixed $x\in X$, $x+2B\subseteq X+2B$, whence $|B+B|\le K^2|B|$. Thus the conventional small-sumset conclusion also follows with constants depending only on $\eta$.

The graph form of Plünnecke inequality statesEquivalently these roots are nonincreasing along the layers. If $D_n=0$ the conclusion is immediate. Otherwise put $C=D_n^{1/n}$ and use the weighted cut lemma for commutative graphs proved in part (i). For a minimum endpoint cut $S_0\cup S_n$,The bottom layer is a cut of this weight. Comparing with $(V_0\setminus Z)\cup {\mathrm{Im}}^{(k)}Z$ gives $|{\mathrm{Im}}^{(k)}Z|\ge C^k|Z|$, as required. In an addition graph, $D_1\le K$ when $|A+B|\le K|A|$, so $D_m\le K^m$: some nonempty bottom subset has growth at most $K^m$ at height $m$.

The useful simultaneous sumset form says that a nonempty $X\subseteq A$ can be chosen with $|X+mB|\le K^m|X|$ for every $m$. To verify the common-subset assertion directly, choose $X$ minimizing $K^{\prime }=|X+B|/|X|$ among subsets of $A$. For a finite set $T=\{t_1,\dots ,t_s\}$, let $X_i$ consist of those $x\in X$ for which $x+t_i$ is a new point of $X+T$. The translated $X_i$ partition $X+T$. Points contributed by $(X\setminus X_i)+B+t_i$ are already present in earlier translates of $X+B$, so the new contribution at stage $i$ has size at mostThe inequality also holds when the complementary set is empty. Summing gives $|X+B+T|\le K^{\prime }|X+T|$. Taking $T=(m-1)B$ and iterating proves $|X+mB|\le (K^{\prime }{)}^m|X|\le K^m|X|$. This is the Petridis minimal-growth lemma calculation and supplies a single subset for all needed heights.

We state the integer form of Freiman theorem for integer sets: for every $K\ge 1$ there are finite constants $r(K),C(K)$ such that every finite nonempty $A\subset \mathbb{Z}$ with doubling at most $K$ lies in a proper generalized arithmetic progression of rank at most $r(K)$ and cardinality at most $C(K)|A|$. An arbitrary abelian group requires the coset-progression version, allowing a subgroup factor. We prove the integer statement, with no claim of optimized constants.

We use the allowed Plünnecke inequality and basic Freiman isomorphism facts, and explicitly prove the modelling, Fourier and covering steps. Two geometry-of-numbers consequences are used as permitted: a cyclic rank-$d$ phase-radius-$\rho$ Bohr set contains a proper symmetric progression of rank at most $d+1$ and size at least $c(d,\rho )$ times the cyclic group order; and a rank-$r$ integer progression with parameter-box volume $V$ can be contained in a proper progression of rank at most $r$ and size at most $c_{r}V$. These follow from successive minima and lattice basis reduction. All these constants depend only on the displayed rank and radius, not on the ambient order or step sizes.

First construct a dense cyclic model of a large subset. Put $D=8A-8A$ and $q=8|D|$. Plünnecke gives $|D|\le K^{16}|A|$. Choose an auxiliary prime $p$ so large that every nonzero integer in $D$ stays nonzero modulo $p$, and $p>2q$. Arbitrarily large primes exist: a purported finite list would miss a prime divisor of one plus their product. For a uniform nonzero multiplier $\lambda (\mathrm{mod}p)$, each nonzero $d\in D$ has $\lambda d$ uniformly distributed among the nonzero residues. The forbidden residues are those represented by nonzero multiples of $q$ with absolute value less than $p$; there are at most $2(p-1)/q$ of them. The union bound gives failure probability at mostThus choose $\lambda$ avoiding them for every nonzero $d\in D$.

Represent the dilated elements of $A$ in $[0,p-1]$, divide this interval into nine consecutive subintervals, and retain a most-populated one. It defines $A^{\prime }\subseteq A$ with $|A^{\prime }|\ge |A|/9$. If $u(a)$ is the chosen representative of $\lambda a$, eight-term sums of representatives from this subinterval have a range shorter than $p$: eight times its width is less than $8p/9$. Consequently every original eight-term relation lifts from equality modulo $p$ to equality of the representative sums as integers.

Now reduce these representatives modulo $q$. If their two eight-term sums agree modulo $q$, their integer difference is a multiple of $q$ of absolute value below $p$. A nonzero such difference would give a forbidden residue $\lambda d$ for some nonzero $d\in D$. If $d=0$, its representative difference is already zero. Hence there are no new relations either. The map is therefore a Freiman s-isomorphism of order eight onto $B\subseteq \mathbb{Z}/q\mathbb{Z}$. It is injective, by padding shorter relations to eight terms with a fixed retained point. Its density satisfiesThis proves the cyclic Freiman model of a small-doubling integer set needed below.

Next we prove the cyclic Bogolyubov lemma. Use normalized finite-group Fourier analysis on $\mathbb{Z}/q\mathbb{Z}$. The convolutionhas Fourier expansion $r(x)={\sum }_{\chi }|\widehat{1_B}(\chi ){|}^4\chi (x)$. Let $\Gamma$ consist of those characters whose Fourier coefficient has modulus at least ${\beta }^{3/2}/\sqrt{2}$. Parseval gives $|\Gamma |\le 2{\beta }^{-2}$, and the fourth-power tail outside $\Gamma$ is at most ${\beta }^4/2$. The zero character belongs to $\Gamma$ and contributes ${\beta }^4$.

For the phase-distance Bohr set $\mathcal{B}=\{x:\Vert \arg \chi (x)/(2\pi )\Vert \le 1/10\text{ for all }\chi \in \Gamma \}$, every retained character has real part at least $\cos (\pi /5)>3/4$. ThereforeThe support of this convolution is $2B-2B$, so $\mathcal{B}\subseteq 2B-2B$. Apply the stated progression in a Bohr set from successive minima consequence. It supplies a symmetric proper progression $P_B\subseteq 2B-2B$ of rank bounded by a function of $K$ and size at least $c(K)q$, hence at least $c(K)|A|$ after adjusting the constant.

Lift this progression back to the integers. The inverse Freiman map inducesEquality of two representations uses a four-term relation on each side, so $\psi$ is well defined and injective. If $y,z,y+z$ all lie in this difference set, the relation between their representations involves at most six original terms on either side. The order-eight model therefore gives $\psi (y+z)=\psi (y)+\psi (z)$ and $\psi (0)=0$. Along every coordinate path within the symmetric progression box, all partial sums remain in $P_B$. It follows thatThus $P=\psi (P_B)$ is a symmetric proper integer progression of the same rank and cardinality, and $P\subseteq 2A-2A$. This proves the Freiman lifting of a progression step, including why a sufficiently high-order model was selected.

It remains to cover all of $A$, not merely its large modeled subset. Since $P\subseteq 2A-2A$, Plünnecke givesChoose $X\subseteq A$ maximal such that the translates $x+P$ are pairwise disjoint. Their union lies in $A+P$, so $|X|\le K^5|A|/|P|\le C_1(K)$. For every $a\in A$, maximality makes $a+P$ meet some $x+P$, giving $a\in x+P-P$. ConsequentlyThis proves the needed Ruzsa covering lemma directly.

Write $P=\{{\sum }_{i=1}^r{n}_i{w}_i:|n_i|\le L_i\}$ and $X=\{x_1,\dots ,x_s\}$. The setcontains $X+(P-P)$ and hence $A$. Its rank is at most $r+s$, bounded only by $K$, and its parameter-box volume is at most $2^{r+s}|P|$. Also $|P|\le |2A-2A|\le K^4|A|$, so this volume is at most $C_2(K)|A|$. The allowed integer properification consequence now contains $Q$ in a proper progression of bounded rank and size at most $C(K)|A|$. ThereforeThis proves Freiman's theorem, using geometry of numbers only in the two explicitly stated places.

For a finite nonempty set $A$, its additive energy iswhere $r_{A+A}(s)$ counts ordered representations. Since ${\sum }_s{r}_{A+A}(s)=|A{|}^2$, Cauchy-Schwarz givesThus $|A+A|\le K|A|$ implies $E(A,A)\ge |A{|}^3/K$: a small doubling constant forces many additive quadruples.

A polynomial form of the Balog-Szemerédi-Gowers theorem gives the converse after passing to a subset. There are absolute constants $c,C>0$ such that if $E(A,A)\ge \eta |A{|}^3$, $0<\eta \le 1$, then some $A^{\prime }\subseteq A$ satisfiesThe exponent and constants here need not be optimal.

Precisely, a graph form of the Balog-Szemerédi-Gowers theorem states the following. If $A,B$ are finite sets of size $n$ in an abelian group, $\Gamma \subseteq A\times B$ has at least $\delta n^2$ edges, and its restricted sumsethas size at most $Kn$, then there are $A^{\prime }\subseteq A$, $B^{\prime }\subseteq B$ with $|A^{\prime }|,|B^{\prime }|\ge c{\delta }^{C}n$ andThese are uniform absolute constants, valid for $0<\delta \le 1$ and $K\ge 1$.

To apply this theorem to additive energy, let $n=|A|$ and retain the sums with $r_{A+A}(s)\ge \eta n/2$. The discarded sums contribute at most $(\eta n/2){\sum }_{s}r(s)=\eta n^3/2$ to the additive energy. Since $r(s)\le n$, the retained pairs number at least $\eta n^2/2$. There are at most $2n/\eta$ retained sums. They therefore define a bipartite graph of density at least $\eta /2$ with restricted sumset size at most $2n/\eta$. The graph theorem supplies large subsets $A^{\prime },B^{\prime }$ with polynomially bounded $|A^{\prime }+B^{\prime }|$.

For completeness, turn this cross-sumset bound into a doubling constant bound. The Ruzsa triangle inequality gives $|A^{\prime }-A^{\prime }|\le |A^{\prime }+B^{\prime }{|}^2/|B^{\prime }|$. If $D=|A^{\prime }-A^{\prime }|/|A^{\prime }|$, the Plünnecke inequality, applied to the pair $A^{\prime },-A^{\prime }$, gives $|A^{\prime }+A^{\prime }|\le D^2|A^{\prime }|$. The minimal-growth proof below in Question 5 applies to arbitrary finite pairs and proves this particular expansion estimate as well. All losses are fixed powers of ${\eta }^{-1}$, proving the displayed converse after increasing $C$.

The requested dense bipartite common-neighbour lemma is a dependent random choice assertion. Call an ordered pair in $V^2$ bad when it has fewer than $ϵn$ common neighbours. Choose $w\in W$ uniformly, and put $S=N(w)\subseteq V$. ThenFor each bad pair, the probability that both endpoints lie in $S$ is its number of common neighbours divided by $n$, which is less than $ϵ$. Thus the expected number $b(S)$ of bad ordered pairs in $S^2$ is at most $ϵm^2$. ConsequentlyChoose a neighbourhood $V^{\prime }=S$ attaining at least this expectation. It has $|V^{\prime }|\ge \delta m/\sqrt{2}\ge \delta m/2$, andHence at least the proportion $1-2ϵ/{\delta }^2$ of its ordered pairs have at least $ϵn$ two-edge connections. As usual these connections are counted by common neighbours, including a return walk for a diagonal ordered pair; assume $\delta >0$, as the formula requires.

The relevance to the graph theorem is that many two-edge connections turn a sparse collection of permitted sums into additive control on a large vertex subset. On a connection $a\to b\to a^{\prime }$, $a-a^{\prime }=(a+b)-(a^{\prime }+b)$ expresses a difference using two elements of the restricted sumset. The abundance of connections, followed by standard cleaning and further path counting, bounds the number of differences without throwing away most vertices. This is the role of dependent random choice in obtaining polynomial losses in the graph form of the Balog-Szemerédi-Gowers theorem.

We first prove the needed Petridis minimal-growth lemma, including the expansion estimate rather than assuming it. For finite nonempty sets $A,B$, choose a nonempty $X\subseteq A$ minimizing $r=|X+B|/|X|$. For any finite set $C=\{c_1,\dots ,c_s\}$, defineThen $|X+C|={\sum }_i|X_i|$. An element from $(X\setminus X_i)+B+c_i$ has already appeared in an earlier $X+B+c_j$, so the new contribution of $X+B+c_i$ has size at mostThe inequality follows from minimality of $r$ on subsets of $A$, with an empty subset contributing zero. Summing proves $|X+B+C|\le r|X+C|$. Iteration gives $|X+jB|\le r^j|X|$, for every $j\ge 0$; in particular, $|jB|\le r^j|X|$ by translating $jB$ into $X+jB$.

Apply this with $B=A$, so $r\le K$. It follows thatThis is the required Plünnecke inequality.

We will also use the mixed-sumset bound $|mA-nA|\le K^{m+n}N$. To justify it, the Ruzsa triangle inequality saysChoose one representation $z=p_z-q_z$ for each $z\in P-Q$. The map $(z,t)\mapsto (p_z-t,t-q_z)$ is injective: adding the two coordinates recovers $z$, then the chosen representation recovers $t$. This proves the inequality. Apply it with $P=mA$, $Q=nA$, $T=-X$ and the minimal-growth set already chosen. The bounds on $X+mA$ and $X+nA$ give the mixed-sumset estimate, including $0A=\{0\}$.

The Ruzsa covering lemma states: if $S,T$ are finite, $T\ne \varnothing$, and $|S+T|\le L|T|$, then there is $D\subseteq S$, $|D|\le L$, such that $S\subseteq D+T-T$. Choose $D$ maximal with the translates $d+T$, $d\in D$, pairwise disjoint. They lie in $S+T$, so $|D||T|\le |S+T|$. Maximality says each $s+T$ meets some $d+T$, which writes $s=d+t_1-t_2$. This proves both assertions.

Let $T=A-A$. The mixed-sumset bounds give $|T|\le K^{2}N$ and $|2T+A|=|3A-2A|\le K^{5}N$. Apply the Ruzsa covering lemma to $S=2T$ and the set $A$. There is $D\subseteq 2T$ of size $q\le K^5$ withInduction then gives $\ell T\subseteq (\ell -1)D+T$. The number of possible sums of $\ell -1$ elements from the $q$-element set $D$ is at most the number of multiplicity vectors, namely $\left(\genfrac{}{}{0ex}{}{\ell +q-2}{q-1}\right)$. Since $A-a_0\subseteq T$ for any $a_0\in A$, we obtain the explicit polynomial growth of iterated sumsetsFor fixed $K>1$, this polynomial in $\ell$ is eventually smaller than $K^{ϵ\ell }$. For example, take $q_{\ast }=\lceil K^5\rceil$ and choose ${\ell }_0(K,ϵ)$ so that $2\log K+q_{\ast }\log (\ell +q_{\ast })\le ϵ\ell \log K$ for every $\ell \ge {\ell }_0$; such a threshold exists because $\log \ell /\ell \to 0$. ThereforeFor $K=1$, a finite integer set has $|A+A|\ge 2|A|-1$ (list the increasing sums $a_1+a_1,\dots ,a_1+a_N,a_2+a_N,\dots ,a_N+a_N$). Thus $N=1$, and the asserted bound holds for every $\ell$. Nonemptiness is implicit in the doubling constant hypothesis.

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.