Abstract

A permutation $\pi$ of an abelian group $G$ is said to destroy arithmetic progressions (APs) if, whenever $(a, \, b, \, c)$ is a non-trivial 3-term AP in $G$, that is $c-b=b-a$ and $a, \, b, \, c$ are not all equal, then $(\pi(a), \, \pi(b), \pi(c))$ is not an AP. In a paper from 2004, the first author conjectured that such a permutation exists of $\mathbb{Z}_n$, for all $n \not\in \{2, \, 3, \, 5, \, 7\}$. Here we prove, as a special case of a more general result, that such a permutation exists for all $n \geq n_0$, for some explicitly constructed number $n_0 \approx 1.4 \times 10^{14}$. We also construct such a permutation of $\mathbb{Z}_p$ for all primes $p > 3$ such that $p \equiv 3 \; ({\hbox{mod $8$}})$.

Highlights

  • Let G be an abelian group, S a subset of G

  • A bijection π : S → S is said to destroy1 arithmetic progressions (APs) if there is no triple (a, b, c) of elements of S satisfying (i) a, b, c are not all equal, (ii) c − b = b − a, (iii) π(c) − π(b) = π(b) − π(a)

  • Before stating our main result, we need to define an extension of the concept of AP-destroying permutation, in the special case of finite cyclic groups: Definition 1

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Summary

Introduction

Let G be an abelian group, S a subset of G. A bijection π : S → S is said to destroy arithmetic progressions (APs) if there is no triple (a, b, c) of elements of S satisfying (i) a, b, c are not all equal, (ii) c − b = b − a, (iii) π(c) − π(b) = π(b) − π(a) This notion was introduced by the first author in [2], though earlier Sidorenko [5] had given an example of such a permutation in the case G = Z, S = N. It remains interesting to try to prove the full conjecture and, in particular, to try to do so without resorting to any large-scale computer searches It follows from Lemma 3.5 of [2] that, if we let P denote the set of those n ∈ N for which Zn the electronic journal of combinatorics 22(4) (2015), #P4.39 admits an AP-destroying permutation, P is closed under multiplication.

Proof of Theorem 2
Proof of Theorem 3
Final remarks

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