Permutations destroying arithmetic structure

Given a linear form C1X1 + · · · + CnXn, with coefficients in the integers, we characterize exactly the countably infinite abelian groups G for which there exists a permutation f that maps all solutions (α1, . . . , αn) ∈ Gn (with the αi not all equal) to the equation C1X1+ · · ·+CnXn = 0 to non-solutions. This generalises a result of Hegarty about permutations of an abelian group avoiding arithmetic progressions. We also study the finite version of the problem suggested by Hegarty. We show that the number of permutations of Z/pZ that map all 4-term arithmetic progressions to non-progressions, is asymptotically e−1p!.


Introduction
Hegarty [8] characterized the countably infinite abelian groups for which there exists a bijection mapping arithmetic progressions to non-arithmetic progressions.He also considered other problems regarding permutations f of an abelian group that ruin arithmetic structures.In particular, for large enough N , he gave a construction of a permutation of Z/N Z that mapped all 4-term progressions to non-progressions.
In the sequel we address the following problem.If G is a countably infinite abelian group and C 1 , C 2 , . . ., C n ∈ Z, under what conditions does there exist a bijection f : G → G such that, for any (α 1 , α 2 , . . ., α n ) ∈ G n , with the α i not all equal, if n i=1 C i α i = 0 then n i=1 C i f (α i ) = 0? We obtain Hegarty's result as a special case of our result for C 1 = C 2 = 1 and C 3 = −2.

Breaking linear forms in countably infinite abelian groups
For this section we take G to be a countably infinite abelian group and L(X 1 , . . ., X n ) = C 1 X 1 + • • • C n X n to be a linear form with integer coefficients.Before stating our main result, Theorem 1, we introduce the following terms.We say that (α 1 , . . .α n ) ∈ G is a proper solution to the linear equation 0 and the elements α 1 , α 2 , . . ., α n are not all equal.A solution is called improper otherwise.We say that a permutation f of G breaks the linear form L(X 1 , . . ., X n ) over G if all proper solutions to L(X 1 , . . ., X n ) = 0 over G get mapped to non-solutions of L(X 1 , . . ., X n ) = 0.That is, f breaks L over G if f is a permutation of G such that if (α 1 , . . ., α n ) ∈ G n is a proper solution to L(X 1 , . . ., X n ) = 0 then (f (α 1 ), . . ., f (α n )) is not a solution.We call a linear form C i = 0 and i∈B C i = 0 for any nonempty proper subset B of [n].For example, the linear form X 1 +X 2 −2X 3 is minimal, while the linear form X 1 +X 2 −X 3 −X 4 is not.
We remark that, for the purposes of breaking the linear form It follows that for any bijection f : G → G and any t, t ∈ G, and f does not break L. Hence the question of what groups have a permutation breaking L is only interesting for minimal linear forms.
be a minimal linear form, let B be the set of all non-empty proper subsets of [n], and let B = { i∈B C i : B ∈ B}.For m ∈ N, let Ω m denote the m-torsion subgroup of G, i.e., the subgroup of G consisting of all elements that have order that divides m.We can now state the main result of this section.
Theorem 1.The minimal linear form L is breakable over a countably infinite abelian group G if and only if for any β ∈ B the quotient group G/Ω |β| is infinite.
We need the following definition in the course of the proof of Theorem 1.We say that a group G has the (q 1 , . . ., q n )-property, where the q i are integers, if for any finite set A ⊂ G there exists some g ∈ G so that q i g ∈ A for all i = 1, . . ., n.The following four lemmas give some sufficient conditions under which a group G has the (q 1 , . . ., q n )-property.
Proof.Suppose, for a contradiction, that there exists a finite set A so that for every g ∈ G we have q i g ∈ A for some i ∈ [n].Now write lcm(q 1 , . . ., q n ) = kq i so we have lcm(q 1 , . . ., q n )g = kq i g = ka for some a ∈ A. Hence |lcm(q 1 , . . ., q n )G| < ∞, a contradiction.
Lemma 4. If G has an element of infinite order then G has the (q 1 , . . ., q n )-property for any non-zero integers q 1 , . . .q n .Proof.Suppose that G has an element of infinite order.Then G has a subgroup and we appeal to Lemma 3.
In the proof of the next lemma we will use the following notation.For a positive integer n and a prime p, let ν p (n) denote the largest integer k for which p k |n.
Lemma 5. Let G be an abelian group and let q 1 , . . ., q n be integers.If |G/Ω q i | = ∞ for all i ∈ [n] then G has the (q 1 , . . ., q n )-property.
Proof.We argue by induction on n.If n = 1, assume for a contradiction that for every g ∈ G we have q 1 g ∈ A, for some finite set A. Thus, it follows that qG ⊆ A. But, since A is finite and qG ∼ = G/Ω |q 1 | , a contradiction follows.Now assume that n > 1.By Lemma 4, we may assume that every element of G has finite order.Thus G admits the decomposition where T p denotes the subgroup of elements of G that have order a power of p.
Note that If there are infinitely many primes p such that T p is non-trivial, then we may write G ∼ = H 1 × H 2 , with H 1 = p T p , where p ranges over all primes p that divide at least one the electronic journal of combinatorics 22(2) (2015), #P2.5 of the q i 's and H 2 = p T p , where p ranges over all primes that divide none of the q i 's.Since |H 2 | = ∞ and since lcm(q 1 , . . .q n )H 2 ∼ = H 2 it follows that |lcm(q 1 , . . .q n )H 2 | = ∞.By Lemma 3, H 2 has the (q 1 , . . ., q n )-property and consequently G has the (q 1 , . . ., q n )property.
Suppose that there are only finitely many primes p such that T p = {e}.Then G ∼ = k j=1 T p j , for some primes p 1 , . . ., p k , and . This together with |q i G| = ∞ implies that for each q i there is some prime p j for which We distinguish two cases.If f is not a constant function we partition {p 1 , . . ., p k } = R∪S so that f ({q 1 , . . ., q n })∩ R = ∅ and f ({q 1 , . . ., q n }) ∩ S = ∅.Next we define H 1 = p j ∈R T p j and H 2 = p j ∈S T p j and note that By induction, H 1 has the (q i ) f (q i )∈R -property and H 2 has the (q i ) f (q i )∈S -property.By Lemma 2 it follows that G has the (q 1 , . . ., q n )-property.
If f is a constant function then there is a prime p for which . We write G ∼ = T p ×H and assume that q 1 is such that ν p (q 1 ) is maximum over {q 1 , . . ., q n }.From lcm(q 1 , . . ., q n )T p ∼ = q 1 T p it follows that |lcm(q 1 , . . ., q n )T p | = ∞.By Lemma 3, T p has the (q 1 , . . ., q n ) property and therefore G has the (q 1 , . . ., q n ) property.
The proof of Theorem 1 follows.
Proof.We define a permutation f : G → G in a "back-and-forth" manner.Let G = {x 1 , x 2 , . ..} be an enumeration of G.We start the recursion by setting f (x 1 ) = x 1 .We now define f in stages.At stage t we have the following: 1.The function f has been defined on a finite set D t that contains x 1 , . . ., x t .
2. The function f −1 has been defined on a finite set D t that contains x 1 , . . ., x t .
3. There are no proper pairs of solutions (α 1 , . . ., α n ), (f (α 1 ), . . ., f (α n )), with Assuming inductively, that the above hold for the first t stages we proceed to stage t + 1 and show that we can preserve the above properties.At stage t + 1 we choose the smallest value in the enumeration of G \ D t , say x k , for which the function f has not yet been defined.We want to choose a value for f (x k ) among G \ f (D t ) such that no proper solutions are mapped to solutions.To this end, for each I ∈ B we define It is clear that A is a finite set.Thus A ∪ f (D t ) is a finite set.By Lemma 5, G has the (q : q ∈ B )-property and, consequently, has the (q : q ∈ B ∪{1})-property.It follows that we can find a g ∈ G such that qg ∈ A ∪ f (D t ) for every q ∈ B ∪{1}.We define f (x k ) = g.We now check the desired properties.First observe that f (x k ) ∈ f (D t ).Now suppose that there is some proper solution (α 1 , α 2 , . . ., α n ) to the linear equation L(X 1 , . . ., X n ) = 0 for which f (α i ) is defined for i = 1, . . ., n.We may assume that x appears among α 1 , . . ., α n as previous elements are left unaltered by the choice of f (x k ).Hence we may define S ⊂ [n] to be the non-empty, proper subset of indices i for which α i = x.Now suppose that (f (α 1 ), . . ., f (α n )) is a solution to the equation L(X 1 , . . ., X n ) = 0.In this case we have and hence We now take the smallest element of G \ (D t ∪ {g}) for which the value of f −1 is not yet defined and then argue exactly as above with the role of f −1 taking the place of f .Notice that since f breaks L if and only f −1 breaks L the argument is indeed identical.
The above defines a bijection on G that satisfies the desired properties.This proves one direction of the argument.
Conversely suppose that for some B ∈ B and β = i∈B C i the factor group G/Ω |β| is finite.Let f : G → G be a bijection.Since G is infinite, the cosets of Ω |β| are infinite.This, together with the fact that G/Ω |β| is finite, implies that there are t, t ∈ G, t = t , such that t and t belong to the same coset of Ω |β| and that f (t) and f (t ) belong to the same coset of Ω |β| .For each i ∈ [n] we define We obtain the result of Hegarty [8] as a special case of Theorem 1.
Corollary 6. (Hegarty) Let G be a countably infinite abelian group, then there exists a 3-term arithmetic progression avoiding permutation of G if and only if G/Ω 2 is infinite.

Permutations avoiding 4-term arithmetic progressions
In what follows, we restrict ourselves to the study of the groups Z/pZ where p is a prime.We also assume that p > 3 in all that follows.We are interested in the problem of estimating the number of permutations of Z/pZ that map 4-term progressions to nonprogressions.Let α(p) denote the number of such permutations.We show that To show this we note that the quantity α(p) p! is the same as the probability that a permutation of f ∈ Z/pZ, sampled uniformly at random, maps all 4-term progressions to non-progressions.We then estimate this probability by showing that the random variable that counts the number of 4-term progression preserved by f is asymptotically Poisson.This is done by the "method of moments".
In what follows, we consider all 4-term progressions as ordered quadruples (x, x+d, x+ 2d, x + 3d) and define For a permutation f of Z/pZ, we abuse notation slightly and define f ((a, b, c, d It is easy to check that this agrees with the equational definition of a permutation that avoids 4-term progressions given by Hegarty [8].
We now sample a permutation f ∈ Z/pZ uniformly at random and consider the random variable X, that counts the number of progressions that are mapped to a progression by f .More precisely, Following standard notation, for k ∈ N we write (X) k = X(X − 1) • • • (X − k + 1) and call E((X) k ) the kth factorial moment of X.In what follows, we show that the random variable X is asymptotically Poisson by way of the following well-known lemma.See, for example, Theorem 1.22 in [3].Lemma 7. Suppose that X n is a sequence of non-negative, integer valued, random variables and that λ is a real number.If for each k we have E((X n ) k ) → λ k as n → ∞ then X n weakly converges to a random variable X that is Poisson distributed with parameter λ.In other words, for every non-negative integer l To estimate the moments we require following simple lemma.Let Q be a nonempty set of 4-tuples (a, b, c, d) where a, b, c, d are distinct elements of a finite ground set V .For some fixed ordering of the ground set V = {x 1 , . . ., x n } of Q, we say that an element x i ∈ V is fixed with respect to Q (or if it clear that we are speaking of Q we shall simply say fixed ) if there is a quadruple (a, b, c, d) ∈ Q with one of a, b, c, d equal to x i and two more of a, b, c, d among {x 1 , . . ., x i−1 }.We call an element x i ∈ V free with respect to Q (or simply free) if it is not fixed.Lemma 8.If Q is a non-empty set of 4-tuples of distinct elements of a finite ground set V such that every element of V appears in some quadruple of Q then one of the following holds.
1.There exists an ordering of the ground set V = {x 1 , . . ., x n } so that the number of fixed elements of V is strictly greater than the number of free elements 2. Every element of V is contained in at most one quadruple of Q.
Proof.Choose a linear ordering π of the ground set uniformly at random over all possible orderings.Define the random variable Y as the number of x i that are fixed by Q, with respect to the ordering.We write 1 (x i is fixed in the ordering π) .
Now if q i ∈ Q is an arbitrarily chosen quadruple that x i appears in, we have P (x i is fixed in the ordering π) P (at least 2 coordinates of q i appear before where the last equality holds by symmetry and the fact that P ( 2 coordinates of q i appear before x i )+P ( 2 coordinates of q i appear after x i ) = 1.
Now assume that there exists some x i that appears in more than one quadruple (i.e.we have excluded alternative 1 of the Lemma).In this case, it is easy to check that Thus we have the strict inequality Since Y is integer-valued there must be some choice of π for which the number of fixed x i exceeds the number of free x i by at least 1.This completes the proof of the lemma.
the electronic journal of combinatorics 22(2) (2015), #P2.5 define the notion of isomorphism between two sets of quadruples in a natural way.We say that Q, Q are isomorphic if there exists a bijection φ between the grounds sets of Q, Q so that φ((a, b, c, d)) ∈ Q if and only if (a, b, c, d) ∈ Q.We will also use the notation We are now in a position to estimate the kth factorial moment of X.
Lemma 9. Let k ∈ N and suppose that X is the random variable, as defined above, that counts the number of 4-term progressions that are mapped to 4-term progressions by f , a permutation sampled uniformly at random over all permutations of Z/pZ, then Proof.Let f be a permutation of Z/pZ, sampled uniformly at random over all permutations of Z/pZ.For a progression P ∈ A p we define E(P ) to be the event "f (P ) ∈ A p ". Hence we may express the kth factorial moment as To estimate the value of E((X) k ) we split the sum up according to isomorphism classes of the {P i 1 , . . ., P i k } and then estimate the probability that a given configuration of progressions {P i 1 , . . ., P i k } gets mapped to a collection of progressions by f .We have where the outer sum runs over a collection of representatives from each isomorphism class.Crucially, we note that there are only finitely many such isomorphism classes.Our main contribution will come from configurations where every point is contained in at most one quadruple.We call this configuration Q 0 .Hence the above is equal to To estimate the first sum, we fix some Q, and consider it on some fixed (abstract) ground set V = {x 1 , . . ., x l }.Let Free(Q) denote the number of free x i s and Fixed(Q) denote the number of fixed x i s.We start with the question: how many terms are in the following sum?
This is the same as asking for the number of subsets of A p of size k that are isomorphic to Q.We let S i be the number of maps φ : {x 1 . . ., x i } → Z/pZ such that there exists some isomorphism φ between Q and a subset of A p that is identical to φ when restricted to {x 1 , . . ., x i }.We will call a map such as φ a partial isomorphism.Notice that S l is an the electronic journal of combinatorics 22(2) (2015), #P2.5 upper bound for the number of of A p that are isomorphic to Q -the quantity that we want to estimate.
We now proceed to bound S l .We do this by way of two simple observations.First we have the trivial bound as there are at most p choices for x i once x 1 , . . .x i−1 have been chosen.More carefully, we observe that if x i is a fixed element with respect to Q then we have the sharper bound.
To see this, we count the number of ways of extending a fixed partial isomorphism φ on x 1 , . . ., x i−1 of Q to a partial isomorphism φ on x 1 , . . ., x i .Since x i is fixed there are two other elements x j , x k among x 1 , . . ., x i−1 that appear in a 4-tuple with x i .Since this 4-tuple must eventually get mapped to a member of A p we must ensure that φ (x i ) is contained in a progression with x j , x k .Finally, note that φ(x j ), φ(x k ) are are mutually contained in 4 2 = 6 tuples of A p .Hence there are at most 6 • 2 = 12 possible choices for φ(x j ).The claim follows.
Inductively applying these estimates, we arrive at the following estimate on S l .
S l 12 l p Free(Q) 12 4k p Free(Q) Where the last inequality follows as l 4k.Hence the number of terms in the sum under consideration is at most 12 4k p Free(Q) .We now turn to estimate the quantity P(E(P i 1 ) ∩ . . .∩ E(P i k )) for {P i 1 , . . ., P i k } ∼ = Q.In particular, we claim that P(E(P i 1 ) ∩ . . .∩ E(P i k )) We prove this in a way very similar to the above.We fix {P i 1 , . . ., P i k } ∼ = Q on the ground set {y 1 , . . ., y l } ⊆ Z/pZ.We may assume that we chose the ordering of {y 1 , . . ., y l } so that the map sending x i to y i , i = 1, . . ., l, is an isomorphism of Q and {P i 1 , . . ., P i k }.
Now we define the event E i for i = 1, . . ., l to be the event: "there exists some isomorphism θ : Z/pZ → Z/pZ of {P i 1 , . . ., P i k } that agrees with f on x 1 , . . ., x i ".Thus We bound each term P(E i | E i−1 ) trivially by 1 if y i is free and by 6 p−i if y i is fixed.After noting that p → ∞ and l 4k, the claim follows.
Putting the above together, it follows that the electronic journal of combinatorics 22(2) (2015), #P2.5 by Lemma 8, we may work with some ordering on the ground set A of Q that has more fixed elements than free elements, the above sum is bounded by C/p, for some constant C (which depends on k).And since there are only finitely many isomorphism classes, we have Where the C is a constant that depends on k, but not p.That is, the contribution from the Q ∼ = Q 0 terms in the sum that represents E((X) k ) is negligible.
We now need to determine the contribution from the main term Q ∼ = Q 0 .Let us first count the number of ways of choosing a set of 4-term progressions of Z/pZ that have no common elements.We claim that there are such sets.To see this, observe that there are |A p | = p(p − 1) ways of choosing the first set.Then after we have chosen i sets to be in our collection, there are O(ip) sets of A p that share an element with the previously chosen sets.Hence there are p(p − 1) − O(ip) choices for the (i + 1)th set.We then must divide by k!, to remove the order from our choice.
To determine P {E(P i 1 ) ∩ • • • ∩ E(P i k )} we use a trick.We realize the uniform probability measure on the space of permutations (henceforth P π ) as the probability measure uniform on all functions (henceforth P f ) f : Z/pZ → Z/pZ conditioned on the event "f (0), . . ., f (p − 1) are distinct" .Before we begin the calculation, note that without loss we may assume that the elements appearing in P i 1 , . . .P i k are exactly x 1 , . . ., x 4k .We have where the second equality holds by independence and the basic property of conditional probability P(E|A ∩ B) = P(E|A) if B is independent of E. The last equality holds by the fact that P f {x 1 , . . ., x 3k are distinct} = (1 − o(1)) as p tends to infinity while k is fixed.Putting this estimate together with our earlier estimate on the number of distinct sums gives .
Our main result of this section follows.
Theorem 10.Let p be a prime and let α(k, p) be the number of permutations of Z/pZ that avoid k-term arithmetic progressions.Then Proof.We sample a permutation f of Z/pZ uniformly at random and let X k,p denote the random variable that counts the number of k-term progressions that f maps to k-term progressions.It is also useful to define for k 2 the set A k,p of all (ordered) arithmetic progressions (x, x+d, x+2d, . . ., (k−1)d), where x, d ∈ Z/pZ and d = 0. Before proceeding we need an easy calculation.For 2 k < p, let P = (x, x + d, . . ., x + (k − 1)d) ∈ A k,p and let E(P ) be the event "f (P ) ∈ A k,p ".We observe that .
To see this, first note that if P ∈ A 2,p then P {E(P )} = 1.Now for k 3 we may expand the quantity P {E(P )} as

Open questions
Perhaps the most interesting open question is whether or not there exists a permutation f : Z/N Z → Z/N Z that avoids 3-term progressions.Hegarty [8] has conjectured that such permutations exist for N = 2, 3, 5, 7.
Let us also note that the type of question explored in this paper is related to a question about the existence of a map (not necessarily a permutation) f : N → N that maps arithmetic progressions to non-arithmetic progressions with |f (n) − f (n + 1)| < C, where C has no dependence on n.This question originated in a different but equivalent formulation in the setting of infinite words [2,4,5,6,7,9].The question of the existence of such a map is open and appears to be quite difficult.