On A Conjecture Regarding Permutations Which Destroy Arithmetic Progressions

Hegarty conjectured for $n\neq 2, 3, 5, 7$ that $\mathbb{Z}/n\mathbb{Z}$ has a permutation which destroys all arithmetic progressions mod $n$. For $n\ge n_0$, Hegarty and Martinsson demonstrated that $\mathbb{Z}/n\mathbb{Z}$ has an arithmetic-progression destroying permutation. However $n_0\approx 1.4\times 10^{14}$ and thus resolving the conjecture in full remained out of reach of any computational techniques. However, this paper using constructions modeled after those used by Elkies and Swaminathan for the case of $\mathbb{Z}/p\mathbb{Z}$ with $p$ being prime, establish the conjecture in full. Furthermore our results do not rely on the fact that it suffices to study when $n<n_0$ and thus our results completely independent of the proof given by Hegarty and Martinsson.


Background
In 2004 Hegarty [2] introduced the notion of permutations that destroy arithmetic progressions in finite cyclic groups. Definition 1. Given a permutation π : Z{nZ Ñ Z{nZ, a three term arithmetic progression pa, a`r, a`2rq, with not all terms equal, is called preserved in π if π pa`2rq´2π pa`rq`π paq " 0. A permutation is said to destroy all arithmetic progressions if it has no preserved arithmetic progressions.
For the sake of simplicity, a three-term arithmetic progression will be denoted an AP and a permutation that destroys all APs will be called AP-Destroying. This notion can be extended to permutations which destroy k-term arithmetic progressions and Hegarty [2] demonstrated that for n ‰ 3, 4 there exists a permutation of Z{nZ that destroys all k-term arithmetic progressions for all k ě 4. However, classifying which cyclic groups have an AP-Destroying permutation has been resistant to proof. In particular Hegarty [2] gave the following conjecture regarding AP-Destroying permutations based on computational evidence. Conjecture 2. For n R t2, 3, 5, 7u, there exists an AP-Destroying permutation π : Z{nZ Ñ Z{nZ.
This conjecture was proved for sufficiently large n by Hegarty and Martinsson [3] in 2015.
Theorem 3. For n ě p9ˆ11ˆ16ˆ17ˆ19ˆ23q 2 , there exists a AP-Destroying permutation π : Z{nZ Ñ Z{nZ. n 2 a 1 3 a 2 5 a 3 7 a 4 p 1 in order to represent n an find a permutation for each of the integers independently.
To prove the result for the cases t2p, 3p, 5p, 7p | p prime and p ě 11u we model our construction based on the one used by Elkies and Swaminathan [1] to demonstrate Theorem 4. The key similarity is the following lemma of Elkies and Swaminathan from [1], which we will rely heavily on as well. Note that in the statement below, and elsewhere in this paper we will not distinguish between an arithmetic progression and its reverse.
Lemma 8. Suppose that π : Z{pZ Ñ Z{pZ is the following permutation with t ‰ 0: Then the only APs preserved by π are`0, 3 2 , 3˘,`1 3 , 2 3 , 1˘. Elkies and Swaminathan [1] then performed two transpositions in order to eliminate these preserved APs and this demonstrated the case when n is prime. In the case n " 2p we will "glue" together two such permutations in a careful manner so that there is exactly one preserved AP, and then using a single transposition we eliminate preserved AP. In the remaining cases however we are able to significantly simplify this approach by directly giving an AP that has no arithmetic progression, avoiding the need for a transposition. In each of these cases however we will not simply be able to show 2p, 3p, 5p, 7p for all primes p directly; instead, certain character estimates will show it for p sufficiently large. Thus we show the conjecture to be true for all n ď 2500 using computational techniques, and this will be a starting point for the analysis in the remaining cases. Note that the 5p case, where p ą 500 is assumed, is the limiting case here. All mentioned computational files can be found on the corresponding arXiv submission.
One piece of machinery that is used multiple times in this paper is the Hasse-Weil bound. (Elkies and Swaminathan [1] similarly require such character estimates, but they can make do with the elementary Hasse bound.) Note that the version we are using is equivalent to counting the number of points on the hyperelliptic curve y 2 " g pxq over a finite field and the bound we are using was proven for curves by Weil in [4].
Theorem 9. Let F p be the field with p elements, p being prime, and let´¨p¯be the Legendre symbol. If f P F p rxs is a polynomial of degree 2g`1 or 2g`2 such that g is not a constant times a perfect square in F p rxs, thenˇˇˇÿ yPZ{pZˆf pyq p˙ˇˇˇď 2g ? p`1.
Proof. We proceed via contradiction. Suppose that t satisfies the above properties, and that some other three term arithmetic progression T is preserved. Let U be the image of T . Furthermore denote by T 2 and T p the mod 2 and mod p components of T , respectively, and define U 2 and U p similarly. We separate cases based on the numbers of elements of T p which are in t0, 1u. Case 1. T p is of the form pa´r, a, a`rq with ta´r, a, a`ru X t0, 1u " H. We take cases which exhaust the possible values of T 2 . Case 1.a. T 2 " p0, 0, 0q or p1, 1, 1q. Then U p "`1 a´r , 1 a , 1 a`r˘o r U p "`t a´r , t a , t a`r˘d epending on T 2 . In either case, since t ı 0, U p being an AP is equivalent to 2 a " 1 a´r`1 a`r mod p, which is equivalent to r 2 " 0. However, this is impossible as T would then be a degenerate AP. Case 1.b. T 2 " p0, 1, 0q . Then U p "`1 a´r , t a , 1 a`r˘a nd U p being an AP is equivalent to 2t a " 1 a´r`1 a`r mod p. This is equivalent to`r a˘2 " 1´1 t , which is impossible as´1´1 t p¯"´1 . Case 1.c. T 2 " p1, 0, 1q . Then U p "`t a´r , 1 a , t a`r˘. Hence 2 a " t a´r`t a`r mod p, which is equivalent with`r a˘2 " 1´t. However, this is impossible as´1´t p¯"´1 . Case 2. We now consider the case where |T p X t0, 1u| " 1. It therefore follows, reversing the AP if necessary, that either T p " p1, 1`r, 1`2rq, T p " p1´r, 1, 1`rq, T p " p0, r, 2rq, or T p " p´r, 0, rq. Case 2.a. T p " p1, 1`r, 1`2rq . Note that if T 2 " p0, w 1 , w 2 q then U 2 " p1, w 1 , w 2 q or vice versa and both of these can not be APs. Case 2.b. T p " p1´r, 1, 1`rq . There are now four possible cases of T 2 . If T 2 " p0, 0, 0q or p0, 1, 0q then U p "`1 1´r , 0, 1 1`r˘. This being an AP is equivalent to 1 1´r`1 1`r " 0. Simplifying, this is equivalent to 2 1´r 2 " 0 which is impossible. If T 2 " p1, 0, 1q or p1, 1, 1q then U p "`t 1´r , 0, t 1`r˘. This being an AP is equivalent to t 1´r`t 1`r " 0. Simplifying, this is equivalent to 2t 1´r 2 " 0 which is impossible as t ı 0. Case 2.c. T p " p0, r, 2rq . There are now four possible cases of T 2 . If T 2 " p0, 0, 0q then U 2 " p1, 0, 0q which is not an AP modulo 2. If T 2 " p1, 1, 1q then U 2 " p0, 1, 1q which is not an AP modulo 2. If T 2 " p0, 1, 0q then U 2 " p1, 1, 0q which is not an AP modulo 2. Finally if T 2 " p1, 0, 1q then U 2 " p0, 0, 1q which is not an AP modulo 2. Case 2.d. T p " p´r, 0, rq . There are now four possible cases of T 2 . If T 2 " p0, 0, 0q then U p "`´1 r , t, 1 r˘w hich is not an AP as t ı 0. If T 2 " p0, 1, 0q then U p "`´1 r , 1, 1 rw hich is not an AP as 1 ı 0. If T 2 " p1, 1, 1q then U p "`´t r , 1, t r˘w hich is not an AP as 1 ı 0. If T 2 " p1, 0, 1q then U p "`´t r , t, t r˘w hich is not an AP as t ı 0. Case 3. In the final case we have that at least two elements of T p are in t0, 1u. Reversing the AP if necessary, this gives the cases T p " p0, 0, 0q, p1, 1, 1q,`0, 1 2 , 1˘, p0, 1, 2q, or p´1, 0, 1q. The second case gives exactly the APs mentioned in the statement of the lemma and therefore it suffices to study the other four cases. Case 3.a. T p " p0, 0, 0q. In order for T to not be a trivial progression, T 2 " p0, 1, 0q or p1, 0, 1q . In the first case, U p " pt, 1, tq which is not an AP as t ı 1. In the second case, U p " p1, t, 1q which is not an AP as t ı 1. Case 3.b. T p "`0, 1 2 , 1˘. If T 2 " p0, 0, 0q then U p " pt, 2, 0q but t ı 4. If T 2 " p1, 1, 1q then U p " p1, 2t, 0q but t ı 1 4 . If T 2 " p0, 1, 0q then U p " pt, 2t, 0q but t ı 0. Finally if T 2 " p1, 0, 1q then U p " p1, 2, 0q which is never an AP. Case 3.c. T p " p0, 1, 2q . If T 2 " p0, 0, 0q then U 2 " p1, 1, 0q. If T 2 " p1, 1, 1q then U 2 " p0, 0, 1q. If T 2 " p0, 1, 0q then U 2 " p1, 0, 0q. Finally if T 2 " p1, 0, 1q then U 2 " p0, 1, 1q. In none of these cases is U 2 an AP.
In none of these cases is U 2 an AP.
Now we claim that a t with the conditions of the previous lemma exists for every prime p ě 31.
Lemma 11. For p ě 31, there exists a t such that Proof. First note that´1´1 t p¯"´t pt´1q p¯f or t ı 0. Then note that where we have used that´p t´1q 2 p¯" 1 for t ‰ 1 and the Hasse-Weil bound. Therefore the number of t P Z{pZ which satisfy´t pt´1q p¯"´1´t p¯"´1 is at least p´5 4 as t " 0, 1 together contribute exactly 1 in total to the sum. For p ě 31, we have p´5 4 ą 6 so for such p there exists a t outside of those in the set t0, 1, 1 4 , 4, 1 9 , 9u as required. Now choose any such fixed t satisfying the above conditions. Consider the following adjustment of π 2 : x˘, x R t0, 1, yu . We claim that there exists a y for which π y 2 , which is π 2 with the values of p0, yq and p0, 1q exchanged, is AP-Destroying permutation for some choice of y. In particular, we claim the following.
Then π y 2 is AP. Proof. Note that the only difference between π 2 and π y 2 is the exchange of p0, 1q and p0, yq. Note that this transposition destroys the APs tp1, 0q , p0, 0q , p1, 0qu and tp0, 0q , p1, 0q , p0, 0qu, and it suffices to demonstrate that we created no new APs. Due to Lemma 10 these APs must contain p0, 1q or p0, yq. We have four cases. Case 1. T contains p0, yq and T p " py, y`r, y`2rq. We take two cases based on the possibilities for T 2 . Case 1.a. T 2 " p0, 0, 0q. First note that r ı 0, as otherwise T is a trivial AP. Then if y`2r ı 0 it follows that U 2 " p1,¨, 0q which is never an AP. If y`2r " 0 then T p "`y, y 2 , 0˘. Since y ‰ 2, U p "´0, 2 y , t¯but y ı 4 t so this is not an AP. Case 1.b. T 2 " p0, 1, 0q. If r " 0, then U p "´0, t y , 0¯, which is never an AP since t ı 0. Otherwise, if y`2r ı 0 then U 2 " p1,¨, 0q which is not an AP. If y`2r " 0 then T p "`y, y 2 , 0˘. Since y ı 2 then U p "´0, 2t y , t¯but y ı 4 so this is not an AP. 5 Case 2. T contains p0, yq and T p " py´r, y, y`rq. We take two cases based on the possibilities for T 2 . Case 2.a. T 2 " p0, 0, 0q. First note that r ı 0, as otherwise T is a trivial AP. If ty`r, y´ru X t0, 1u " H, then U p "´1 y´r , 0, 1 y`r¯, which is not an AP as y ı 0. By symmetry, it suffices to check the cases y´r " 0, 1. If y´r " 0 then y`r " 2y ı 1 as y ı 1 2 . Therefore we have U 2 " p1, 1, 0q, which is not an AP. In the case y´r " 1, we have y`r " 2y´1 R t1, yu. Now 2y´1 ı 0, as y ı 1 2 . Therefore, U p "´1 y , 0, 1 2y´1¯, which is not an AP as y ı 1 3 . Case 2.b. T 2 " p1, 0, 1q. If r " 0, U p "´t y , 0, t y¯w hich is never an AP as t ‰ 0 and thus r ı 0 suffices. If ty`r, y´ru X t0, 1u " H, then U p "´t y´r , 0, t y`r¯, which is not an AP as yt ı 0. If y´r " 0, then y`r ı 1 as y ı 1 2 . Furthermore since y`r ı 0 it follows that U 2 " p0, 1, 1q which is not an AP. If y´r " 1, then y`r ı 0 as y ı 1 2 . Since y`r ı 1 it follows that U 2 " p0, 1, 1q which is never an AP. Note that in the following two cases, we may assume that T does not contain p0, yq as these have been handled. Case 3. T contains p0, 1q and T p " p1, 1`r, 1`2rq. We take two cases based on the possibilities for T 2 . Case 3.a. T 2 " p0, 0, 0q. First note that r ı 0, as otherwise T is a trivial AP. If 1`2r P t0, yu, then U 2 " p1,¨, 0q is not an AP. If 1`r " 0, then U p "´1 y , t,´1¯, which is impossible as y ı 1 2t`1 . Similarly, 1`r " y yields U p "´1 y , 0, 1 2y´1¯, which is not an AP since y ı 1 3 . Therefore it suffices to study the general case where U p "´1 y , 1 1`r , 1 1`2r¯. The condition for this being an AP is a quadratic in r and has discriminant p9y´1q py´1q. This is not a perfect square as´1´9 y p¯" 1 and´1´y p¯"´1 by assumption.
Case 3.b. T 2 " p0, 1, 0q. If r " 0, then U p "´1 y , 0, 1 y¯, which is not an AP. If 1`2r P t0, yu, then U 2 " p0,¨, 1q is not an AP. If 1`r " 0, then since y ı´1, we have U p "´1 y , 1,´1¯, which is not an AP as y ı 1 3 . Finally, in the general case we have U p "´1 y , t 1`r , 1 1`2r¯. The condition for this sequence being an AP is a quadratic in r, and its discriminant is p1´4tq 2 y 2´2 p4t`1q y`1. However this is not a perfect square by assumption. Case 4. T contains p0, 1q and T p " p1´r, 1, 1`rq. We take two cases based on the possibilities for T 2 . Case 4.a. T 2 " p0, 0, 0q. First note that r ı 0, as otherwise T is a trivial AP. If 1´r " 0, then 2 " 1`r ı y. It follows that U 2 " p1, 0, 0q, which is not an AP. Since we can assume 1´r ı y due to previous cases and we can reverse the AP as necessary, it suffices to consider t1´r, 1`ru X t0, 1, yu " H. In the remaining cases it follows that U p "´1 1´r , 1 y , 1 1`r¯, which implies y " 1´r 2 . But this is impossible since´1´y p¯"´1 . 6 Case 4.b. T 2 " p1, 0, 1q. If r " 0, then U p "´0, 1 y , 0¯, which is never an AP. If 1´r " 0, then 1`r " 2, so U 2 " p0, 0, 1q which is not an AP. Finally, in the general case U p "´t 1´r , 1 y , t 1`r¯. The condition for this being an AP is equivalent to r 2 " 1´ty, which is impossible as´1´t y p¯"´1 . This exhausts all possible cases, so the proof is complete.
Having shown this, we finally proceed to showing the existence of y which satisfies the hypotheses of Lemma 12.
Lemma 13. For p ą 500 and a fixed t which satisfies the hypotheses of Lemma 10, there exists a y which satisfies the hypotheses of Lemma 12.
Expanding this product yields the p plus 15 terms of the form ř p´1 y"0˘´˘g pyq p¯w here g pyq is the product of some terms in the set t1´y, 9y´1, 1´ty, f 1 u.
We claim that none of the g pyq which arise are perfect squares. To see this we instead prove the stronger claim that no two terms share a root and thus it suffices to show that the discriminant of the product is nonzero. In particular ∆ ppy´1q p9y´1q pty´1q f 1 q " 2 28 t 3 pt´9q 2 pt´4q 2 pt´1q 8 p9t´1q 2 and all roots of the discriminant are in the set of excluded t. Hence each of the 15 sums is at most 4 ?
p`1 in absolute value using the Hasse-Weil bound, so the entire sum is at least p´60 ? p´15. When p ą 10000, we have p´60 ? p´15 16 ě 40 ? p´15 16 ą 9. So, more than 9 values of y contribute a nonzero term to the above sum, which means that some y outside of the required exceptional set satisfiesˆ1´t y p˙"ˆ1´y p˙"˜p 1´4tq 2 y 2´2 p4t`1q y`1 p¸"´1 andˆ1´9 y p˙" 1 as required. Hence there exists an AP-Destroying permutation for n " 2p, p ą 10000. In the cases 500 ă p ă 10000, the existence of y is verified in LegrendeSymbol2p.java.
Proof. Suppose for sake of contradiction that some arithmetic progression T is preserved, and let U be its image. Denote by T 3 , T p , U 3 , U p the projections of T and U modulo 3 and p respectively. We take three cases: Case 1. Three elements of T p are in t0, 1u. Then since T p is an AP and p ą 2, this implies T p " p0, 0, 0q or T p " p1, 1, 1q. In the former case, U p is a permutation of p0, 1, tq, which is not an AP as t R t´1, 1 2 , 2u. In the latter case, U p is a permutation of p1, 0, 0q, which is not an AP. Hence case 1 is impossible. Case 2. One or two elements of T p are in t0, 1u. Consider the triple T 1 3 obtained by incrementing each of the three elements in T 3 . Note that π 3 increments the mod 3 value of its input if that input is 0 or 1 mod p, and otherwise the mod 3 value stays the same. It follows that if one element of T p is in t0, 1u, then U 3 differs from T 3 in exactly one element, and if two elements of T p are in t0, 1u, then U 3 differs from T 1 3 in exactly one element. In both cases, U 3 cannot be an AP. Case 3. None of the elements of T p are in t0, 1u. Let T p " pa´r, a, a`rq. Then we take four cases based on the possible values of T 3 . Case 3.a. T 3 " p0, 0, 0q or p1, 1, 1q. Then U p "`1 a´r , 1 a , 1 a`r˘. It follows that 1 a´r`1 a`r " 2 a , so r " 0, which is impossible. Case 3.b. T 3 " p2, 2, 2q. Then U p "`t a´r , t a , t a`r˘. Since t ı 0, this reduces to the previous case. Case 3.c. T 3 " p0, 1, 2q , p1, 0, 2q , p2, 0, 1q , or p2, 1, 0q. By symmetry, we may suppose T 3 is of one of the first two triplets. Then U p "`1 a´r , 1 a , t a`r˘. Solving the AP condition as a quadratic in r, we obtain a discriminant pt´1q pt´9q. This, however, is not a perfect square mod p by assumption. Case 3.d. T 3 " p0, 2, 1q or p1, 2, 0q. Then U p "`1 a´r , t a , 1 a`r˘. Solving the AP condition as a quadratic in r, we obtain a discriminant 16t pt´1q. This, however, is not a perfect square mod p by assumption. Now we prove that for p ě 31, some prime satisfying the conditions of Lemma 8 exists.
Proof. We may calculate where we have used the Hasse-Weil Bound and that´p t´1q 2 p¯" 1 for t ı 1. It follows that the number of solutions to´t pt´1q p¯"´p t´1qpt´9q p¯"´1 over t P Z{pZ is at least p´5 4 ą 6, so that there is in particular some t outside of the exceptional set satisfying these conditions. For this value of t, π 3 is an AP-Destroying permutation, as desired.

Lemma 16. Suppose that t P Z{pZ such that
Expanding this product yields the p plus 63 terms of the form ř tPZ{pZ˘´˘f ptq p¯w here f ptq is the product of some terms in the set t9t´16, 9´t, 1`t, pt´1q pt´9q , pt´1q p9t´1q , tu.
We claim that none of the f pyq which arise are perfect squares. To see this it suffices note that the roots 16 9 , 9 16 , 0, 1,´1, 9, 1 9 ( are all distinct for p ą 500 and no terms involving both pt´1q pt´9q and p9t´1q pt´1q give perfect squares. Upon expanding it can be verified that we get 4 terms of degree 1, 9 terms of degree 2, 16 terms of degree 3, 19 terms of degree 4, 12 terms of degree 5, and 3 terms of degree 6. Using the Hasse Weil bound it follows that ? p`1q´15 p4 ? p`1q while the sum over the excluded t is at most 22 p64q " 1408 and the sum over the roots not in the excluded set is at most 4 p64q " 256. It follows that for p ą 21000 that the sum in question is greater than 1408`256 so such a t exists and for 500 ă p ď 21000 the existence of such t is verified in LegrendeSymbol5p.java.
However note that where the Hasse Weil-Bound and that´p t´1q 2 p¯" 1 for t ı 1 is used. Since p´4 ą 61 the result follows.

Computational Techniques
In the previous sections, computational techniques are often required to ensure the existence of AP-Destroying permutations. For n " 2p, 3p, 5p, 7p corresponding to n ď 2500, we verified the existence of an AP-Destroying permutation via a descent algorithm; see DescentPermutation.java. In particular, we choose a random starting permutation, and only administer random transpositions if they decrease the total number of APs preserved. This condition can be checked in time linear in n for each iteration. Empirically, the running time of this algorithm appeared to be roughly quadratic in n, which suggests that a random permutation descends to an AP-Destroying permutation with positive probability.
Whenever larger permutations were required, we calculated the necessary value of t or y, t directly, and this appears in many of the lemmas scattered throughout the proof. This was done instead of directly generating permutations due to the run time of this algorithm being empirically linear in n versus quadratic for the above.

Application to Finite Abelian Groups
In the previous sections, we've classified which finite cyclic groups have AP-Destroying permutations. One particularly useful result is the following result of Hegarty which allows one to quotient out by subgroups with an AP-Destroying permutation.
Theorem 20. If there exists an AP-Destroying permutation for H and G{H, there exists an AP-Destroying permutation for G.
Previously Elkies and Swaminathan [1] demonstrated that all finite p-groups with odd order have an AP-Destroying permutation. We extend their result by classifying all finite abelian groups with odd order that have an AP destroying permutation.
Theorem 21. Let G be a finite abelian group with odd order greater than 7. Then G has an AP-Destroying permutation.
Proof. We first claim that the result holds if Ω p|G|q ď 2, where Ω pnq denotes the number of prime factors of n with multiplicity. Indeed, if G is of the form Z{pZˆZ{qZ for primes p ‰ q or Z{pZ or Z{p 2 Z for a prime p, then the result follows from the main theorem. Finally the case G " pZ{pZq 2 follows from the result of Elkies and Swaminathan [1]. Now we consider the case Ω p|G|q " 3. If G is in the set below, then the direct verification of the existence of an AP-Destroying permutation is in FiniteAbelian.java.
Other than the above set and cyclic groups, all other groups G of odd order with Ω p|G|q " 3 have Z p as a subgroup for some prime p ą 11 or are G " pZ{pZq 3 for p " 3, 5, 7. In the latter case the result follows from the result of Elkies and Swaminathan [1] while in the former G has an AP-Destroying permutation due to Theorem 20 along with the case Ω p|G|q ď 2.
Finally, we prove the full result using strong induction on Ω p|G|q, with base cases Ω p|G|q P t2, 3u established. Suppose the result holds for 2 ď Ω p|G|q ď k, an that Ω p|G|q " k`1. Then there exists some product pq of two possibly equal primes p, q such that there is an order pq subgroup H of G. Then H and G{H both have an AP-Destroying permutation by the inductive hypothesis, so G does as well by Theorem 20. This completes the induction.
We remark that there are infinite families of even-order abelian groups which do not have an AP-Destroying permutation. For example, the following is true, which is mentioned in Remark 4.2 in [2] Proposition 22. Suppose that H is an abelian group with |H| ă 2 k . Then G " pZ{2Zq kˆH has no AP-Destroying permutation.
Proof. Suppose otherwise, and let σ : pZ{2Zq kˆH Ñ pZ{2Zq kˆH be such a permutation. Let π H : G Ñ H be the projection of G onto the second coordinate. Then since 2 k ą |H|, there exist some a ‰ b P pZ{2Zq k such that π H˝σ pa, 0q " π H˝σ pb, 0q. But then tpa, 0q , pb, 0q , pa, 0qu is an AP preserved by σ, a contradiction. So no such AP-Destroying permutation exists as required.
In particular, if the largest odd number dividing a positive integer n is less than ? n, then there exists a finite abelian group of order n which does not have an AP-Destroying permutation.

Conclusion
In this paper, we have resolved a conjecture of Hegarty. In particular, we proved that there exists an AP-Destroying permutation for all cyclic groups of order not in the set t2, 3, 5, 7u. However, as the last section demonstrates, this result does not immediately resolve the case for all finite abelian groups, and in fact for every positive integer k there is a finite abelian group whose order is a multiple of k which does not have any AP-Destroying permutation. In light of this, the following question is still open.
Question 23. For which even order finite abelian groups do there exist AP-Destroying permutations?