Abelian progression 2026-10-03
An abelian progression is a generalized arithmetic progression inside an abelian group. It is the image of an integer box under a group homomorphism.
Coset progression 2026-10-03
A coset progression is a sumset , where is a finite subgroup of an abelian group and is a generalized arithmetic progression.
Dense Bogolyubov-Ruzsa lemma 2026-09-28
For every , if has density at least in a cyclic group of prime order, then contains a proper generalized arithmetic progression of rank and size .
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 25 5 Solution Created 2026-10-03 Updated 2026-10-14
We state the integer form of Freiman theorem for integer sets: for every there are finite constants such that every finite nonempty with doubling at most lies in a proper generalized arithmetic progression of rank at most and cardinality at most . 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- phase-radius- Bohr set contains a proper symmetric progression of rank at most and size at least times the cyclic group order; and a rank- integer progression with parameter-box volume can be contained in a proper progression of rank at most and size at most . 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 and . Plünnecke gives . Choose an auxiliary prime so large that every nonzero integer in stays nonzero modulo , and . Arbitrarily large primes exist: a purported finite list would miss a prime divisor of one plus their product. For a uniform nonzero multiplier , each nonzero has uniformly distributed among the nonzero residues. The forbidden residues are those represented by nonzero multiples of with absolute value less than ; there are at most of them. The union bound gives failure probability at mostThus choose avoiding them for every nonzero .
Represent the dilated elements of in , divide this interval into nine consecutive subintervals, and retain a most-populated one. It defines with . If is the chosen representative of , eight-term sums of representatives from this subinterval have a range shorter than : eight times its width is less than . Consequently every original eight-term relation lifts from equality modulo to equality of the representative sums as integers.
Now reduce these representatives modulo . If their two eight-term sums agree modulo , their integer difference is a multiple of of absolute value below . A nonzero such difference would give a forbidden residue for some nonzero . If , 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 . 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 . The convolutionhas Fourier expansion . Let consist of those characters whose Fourier coefficient has modulus at least . Parseval gives , and the fourth-power tail outside is at most . The zero character belongs to and contributes .
For the phase-distance Bohr set , every retained character has real part at least . ThereforeThe support of this convolution is , so . Apply the stated progression in a Bohr set from successive minima consequence. It supplies a symmetric proper progression of rank bounded by a function of and size at least , hence at least 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 is well defined and injective. If 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 and . Along every coordinate path within the symmetric progression box, all partial sums remain in . It follows thatThus is a symmetric proper integer progression of the same rank and cardinality, and . This proves the Freiman lifting of a progression step, including why a sufficiently high-order model was selected.
It remains to cover all of , not merely its large modeled subset. Since , Plünnecke givesChoose maximal such that the translates are pairwise disjoint. Their union lies in , so . For every , maximality makes meet some , giving . ConsequentlyThis proves the needed Ruzsa covering lemma directly.
Write and . The setcontains and hence . Its rank is at most , bounded only by , and its parameter-box volume is at most . Also , so this volume is at most . The allowed integer properification consequence now contains in a proper progression of bounded rank and size at most . ThereforeThis proves Freiman's theorem, using geometry of numbers only in the two explicitly stated places.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 129 4 c Solution 2026-09-28
The Dense Bogolyubov-Ruzsa lemma says that, for every , if has density at least in a cyclic group of prime order, then contains a proper generalized arithmetic progression of rank and size .
Now suppose and . The Ruzsa modelling lemma, taken at a sufficiently high fixed Freiman order, supplies with and a Freiman isomorphism from to a subset , where is prime, , and . Apply the dense Bogolyubov-Ruzsa lemma to and transfer the resulting progression back through the Freiman model. We obtain a proper progressionof rank and size .
The Plünnecke-Ruzsa inequality givesThere are pairs with and , distributed among the sums in . Some therefore has at least representations . Equivalently,Set . Translation and negation preserve properness and rank. Moreover and another use of Plünnecke gives . Thusas required.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 129 4 e Solution 2026-09-28
Consider the graphThe hypothesis says exactly that has at least additive quadruples. By the Balog-Szemerédi-Gowers theorem, it contains with and .
Part b gives a Freiman s-isomorphism from to a set , where it is enough to take any fixed . The set has bounded doubling, with a bound depending only on . Part c therefore provides a proper generalized arithmetic progression of rank such that
Write as a proper parameter box. Since , one of its side lengths tends to infinity with . Averaging over all lines parallel to that side gives a line on which has density bounded below in terms of . The Szemerédi theorem quoted in the question then gives, once is sufficiently large in terms of and , a nonconstant -term arithmetic progression in .
The inverse Freiman isomorphism sends it to a -term arithmetic progression in , because each relation between three consecutive terms is an additive-quadruple relation. Write this progression asIts first-coordinate difference cannot be zero: the graph of a function has only one point above each . Hence . On the nonconstant progressionwe haveTaking and , both in , proves the claim.