New Lower Bounds for Permutation Arrays Using Contraction
Sergey Bereg, Zevi Miller, Luis Gerardo Mojica, Linda Morales, I.H. Sudborough
NNew Lower Bounds for Permutation Arrays UsingContraction
Sergey Bereg ∗ Zevi Miller † Luis Gerardo Mojica ∗ Linda Morales ∗ I.H. Sudborough ∗ August 28, 2018
Abstract A permutation array A is a set of permutations on a finite set Ω, say of size n . Givendistinct permutations π, σ ∈ Ω, we let hd ( π, σ ) = |{ x ∈ Ω : π ( x ) (cid:54) = σ ( x ) }| , called the Hamming distance between π and σ . Now let hd ( A ) = min { hd ( π, σ ) : π, σ ∈ A } . Forpositive integers n and d with d ≤ n , we let M ( n, d ) be the maximum number of permuta-tions in any array A satisfying hd ( A ) ≥ d . There is an extensive literature on the function M ( n, d ), motivated in part by suggested applications to error correcting codes for messagetransmission over power lines.A basic fact is that if a permutation group G is sharply k -transitive on a set of size n ≥ k , then M ( n, n − k + 1) = | G | . Motivated by this we consider the permutation groups AGL (1 , q ) and P GL (2 , q ) acting sharply 2-transitively on GF ( q ) and sharply 3-transitivelyon GF ( q ) ∪ {∞} respectively. Applying a contraction operation to these groups, we obtainthe following new lower bounds for prime powers q satisfying q ≡ M ( q − , q − ≥ ( q − / q odd, q ≥ M ( q − , q − ≥ ( q − q + 2) / q even, q ≥ M ( q, q − ≥ Kq log q for some constant K if q is odd, q ≥ M ( q − , q − ≥ q − q and M ( q, q − ≥ q − q for all prime powers q such that q (cid:54)≡ M ( n, d ) for a finite number of exceptional pairs n, d ,by applying this contraction operation to the sharply 4 and 5-transitive Mathieu groups. We consider permutations on a set Ω of size n . Given two such permutations π and σ , we let hd ( π, σ ) = |{ x ∈ Ω : π ( x ) (cid:54) = σ ( x ) }| , so hd ( π, σ ) is the number of elements of Ω at which π and σ disagree. When hd ( π, σ ) = d , we say that π and σ and are at Hamming distance d , or thatthe Hamming distance between π and σ is d . A permutation array A is a set of permutationson Ω. We say that hd ( A ) = d if d = min { hd ( π, σ ) : π, σ ∈ A } . For positive integers n and d ∗ Computer Science Dept., University of Texas at Dallas, Richardson, TX 75083, USA † Dept. of Mathematics, Miami University, Oxford, OH 45056, USA a r X i v : . [ m a t h . C O ] S e p ith d ≤ n we let M ( n, d ) be the maximum number of permutations in any permutation array A satisfying hd ( A ) ≥ d .Consider a fixed ordering x , x , · · · , x n of the elements of Ω. The image string of the per-mutation σ ∈ A is the string σ ( x ) σ ( x ) · · · σ ( x n ). Thus the permutation array A can also beregarded as an | A | × n matrix whose rows are the image strings of the permutations in A . When hd ( A ) = d , any two rows of A disagree in at least d positions and some pair of rows disagree inexactly d positions. In particular, if G is a permutation group acting on Ω, then we obtain a | G | × n permutation array whose rows consist of the mage strings of all the elements of G . Werefer to this array by G , and we use hd ( G ) to refer to the hamming distance of this array.The study of permutation arrays began (to our knowledge) with the papers [9] and [13], wheregood bounds on M ( n, d ) (together with other results) were developed based on combinatorialmethods, motivated by the Gilbert-Varshamov bounds for binary codes. In recent years therehas been renewed interest in permutation arrays, motivated by suggested applications in powerline transmission [12], [20], [28], [15], block ciphers [27], and in multilevel flash memories [17]and [18].We review here some of the known results and methods for estimating M ( n, d ).Some elementary exact values and bounds on M ( n, d ) are the following (summarized withshort proofs in [8]) ; M ( n,
2) = n !, M ( n,
3) = n !2 , M ( n, n ) = n , M ( n, d ) ≥ M ( n − , d ), M ( n, d ) ≥ M ( n, d + 1), M ( n, d ) ≤ nM ( n − , d ), and M ( n, d ) ≤ n !( d − . More sophisticatedbounds were developed in the above cited papers [9] and [13], with a recent improvement in [29].The smallest interesting case for d is d = 4. Here some interesting and non-elementary boundsfor M ( n,
4) were developed in [11], using linear programming, characters on the symmetric group S n , and Young diagrams. In [19] it is shown that if K > n > e /K and s < n − K , then M ( n, n − s ) ≥ θ ( s ! √ log n ). The lower bound is achieved by a polynomial timerandomized construction, using the Lovasz Local Lemma in the analysis.There are various construction methods for permutation arrays. First there is a connectionwith mutually orthogonal latin squares (MOLS). It was shown in [7] that if there are m MOLSof order n , then M ( n, n − ≥ mn . From this it follows that if q is a prime power, then M ( q, q −
1) = q ( q − M ( n, d ) for small n and d ,including clique search, and the use of automorphisms are described in [8], [16], and [23]. Thereare also constructions of permutation arrays that arise from the use of permutation polynomials,also surveyed in [8], which we mention briefly below.Additional construction methods are coset search [2] and partition and extension [3]. In thefirst of these, one starts with with a permutation group G on n letters with hd ( G ) = d , andwhich is a subgroup of some group H (for example H = S n ). Now for disjoint permutationarrays A, B on the same set of letters, let hd ( A, B ) = min { hd ( σ, τ ) : σ ∈ A, τ ∈ B } . For x / ∈ G we observe that the coset xG of G in H is a permutation array with hd ( xG ) = hd ( G ). Forcosets x G, x G, · · · , x k G of G , the Hamming distance of the permutation array ∪ ≤ i ≤ k x i G isthe minimum of d and m , where m = min { hd ( x i G, x j G ) : 1 ≤ i < j ≤ k } . The method of cosetsearch is to iteratively find coset representatives x i so that m , while in general less than d , isstill reasonably large. The partition and extension method is a way of obtaining constructivelower bounds M ( n + 1 , d + 1) from such bounds for M ( n, d ).Moving closer to the subject of this paper, we consider a class of optimal constructions whicharise through sharply transitive groups. We say that a permutation array A on a set Ω ofsize n is sharply k -transitive on Ω if given any two k -tuples x , x , · · · , x k and y , y , · · · , y k ofdistinct elements of Ω there exists a unique g ∈ A such that g ( x i ) = y i for all 1 ≤ i ≤ k . Inour applications A will be the set of image strings of a permutation group acting on Ω. Fromthe bound M ( n, d ) ≤ n !( d − we have for any positive integer k that M ( n, n − k + 1) ≤ n !( n − k )! .2ow if G is a sharply k -transitive group acting on Ω, then | G | = n !( n − k )! . Also, in such a G anytwo distinct elements g, h of G can agree in at most k − gh − is anonidentity element of G fixing at least k elements of Ω, contrary to sharp k -transitivity. Thus hd ( G ) ≥ n − k + 1. So a sharply k -transitive group G implies the existence of an optimal array(the set of image strings of elements of G ) realizing M ( n, n − k + 1) = n !( n − k )! . The followingtheorem gives a strong converse to the above, including the generalization to arbitrary arraysthat may not be groups. Theorem 1 [4] Let A be a permutation array on a set of n letters satisfying hd ( A ) ≥ n − k + 1 .Then | A | = n !( n − k )! = M ( n, n − k + 1) ⇐⇒ A is sharply k -transitive on this set. The sharply k -transitive groups (for k ≥
2) are known, and these are as follows [6], [10], [22]; k = 2 : the Affine General Linear Group AGL (1 , q ) acting on the finite field GF ( q ), consistingof the transformations { x → ax + b : x, a (cid:54) = 0 , b ∈ GF ( q ) } , k = 3 : the Projective Linear Group P GL (2 , q ) acting on GF ( q ) ∪ {∞} , consisting of thetransformations { x → ax + bcx + d : x, a, b, c, d ∈ GF ( q ) , ad − bc (cid:54) = 0 } , k = 4 : the Mathieu group M acting on a set of size 11, k = 5 : the Mathieu group M acting on a set of size 12,arbitrary k : the symmetric group S k acting on a set of size k is sharply k and ( k − A k acting on a set of size k is sharply ( k − M ( n, d ) for n and d near a prime power.Previous results of this kind are given in [8] where it is shown that for n = 2 k with n (cid:54)≡ M ( n, n − ≥ ( n + 2) n ( n −
1) and M ( n, n − ≥ n ( n − n + 3 n + 8). It isalso shown that for any prime power n with n (cid:54)≡ M ( n, n − ≥ n . Theseresults are based on permutation polynomials. Similar such results appearing in [2], are basedon a contraction operation applied to permutation arrays defined in the next section. The latterresults yield M ( n − , n − ≥ n − n (cid:54)≡ M ( n − , n − ≥ n ( n − n (cid:54)≡ n (cid:54)≡ , q is prime power satisfying q ≡ q ,we accomplish this by applying the contraction operation to the permutation arrays AGL (1 , q )and P GL (2 , q ) (acting on GF ( q ) and on GF ( q ) ∪ {∞} respectively), obtaining the followingconstructive lower bounds. for q ≥ M ( q − , q − ≥ ( q − / q odd and M ( q − , q − ≥ ( q − q + 2) / q even, and for q ≥ M ( q, q − ≥ Kq log q for some constant K if q is odd, and bounds for M ( n, d ) for a finite number of exceptional pairs n, d , obtained from the Mathieugroups.We will use standard graph theoretic notation. In particular for a graph G and S ⊆ V ( G )we let [ S ] G be the graph with vertex set S and edge set E ([ S ] G ) = { xy : x, y ∈ S, xy ∈ E ( G ) } ,and we call it the graph induced by S . When G is understood by context, we abbreviate [ S ] G by[ S ]. Consider a permutation array A acting on a set Ω = { x , x , · · · , x n } of size n , where the elementsof Ω are ordered by their subscripts. We distinguish some element, say x n , by renaming it F .3hus the image string of any element σ ∈ A will be σ ( x ) σ ( x ) · · · σ ( F ), and we say that σ ( x i )occurs in position or coordinate x i of the string. Now for any π ∈ A , define the permutation π (cid:52) on Ω by π (cid:52) ( x ) = π ( F ) if x = π − ( F ) ,F if x = F,π ( x ) otherwise.Thus the image string of π (cid:52) is obtained from the image string of π by interchanging thesymbols F and π ( F ) if π ( F ) (cid:54) = F , while π (cid:52) = π if and only if π ( F ) = F . In either case, π (cid:52) has F as its final symbol. We let π (cid:52)− be the permutation on n − π (cid:52) by dropping the last symbol F from π (cid:52) . As an example, if π = aF bcd , then π (cid:52) = adbcF , and π (cid:52)− = adbc . We call the operation π → π (cid:52)− contraction , and we call π (cid:52)− the contraction of thepermutation π . Further, for any subset R ⊂ A , let R (cid:52) = { π (cid:52) : π ∈ R } , and R (cid:52)− = { π (cid:52)− : π ∈ R } .So R (cid:52)− is a permutation array on the symbol set Ω −{ F } of size n −
1, and is called the contractionof R .We note some basic properties related to the contraction operation. Lemma 2
Let G be a permutation group acting on the set Ω of size n , and let π, σ ∈ G . a ) The only coordinates in either π or σ whose values are affected by the (cid:52) operation are π − ( F ) , σ − ( F ) , and F . Thus hd ( π (cid:52) , σ (cid:52) ) ≥ hd ( π, σ ) − . b ) Assume hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) − . Then πσ − contains a -cycle in its disjoint cyclefactorization, and | G | is divisible by . c ) Let S ⊆ G . Then | S (cid:52) | = | S (cid:52)− | and hd ( S (cid:52) ) = hd ( S (cid:52)− ) . If also hd ( S ) > , then | S | = | S (cid:52) | .Proof. Part a) follows immediately from the definition of the (cid:52) operation.For b), the assumption implies that there are positions x i , x j , F at which the image stringsof π and σ disagree and π (cid:52) and σ (cid:52) agree. So for some indices s, t we must have π ( x i ) = x s , π ( x j ) = F, π ( F ) = x t , while σ ( x i ) = F, σ ( x j ) = x t , σ ( F ) = x s . Then πσ − (composing leftto right) contains the 3-cycle ( x i , F, x j ) in its disjoint cycle factorization. Thus the subgroupof G generated by πσ − has order divisible by 3, and hence | G | is divisible by 3 by Lagrange’stheorem.Consider c). The first two equalities follow from the fact that all image strings in S (cid:52) have F as their last coordinate. To see | S | = | S (cid:52) | when hd ( S ) >
3, suppose to the contrary that π (cid:52) = σ (cid:52) for distinct π, σ ∈ S . As noted in the proof of part a), π (cid:52) and σ (cid:52) can agree in atmost three positions where π and σ disagreed. Thus π and σ already agreed in at least n − hd ( π, σ ) ≤
3, a contradiction.Consider a permutation array H on n symbols with hd ( H ) = d . The array H (cid:52)− is on n − hd ( H (cid:52)− ) ≥ d − H = AGL (1 , q ) , P GL (2 , q )and certain Mathieu groups, our goal is to find a subset I ⊂ H with hd ( I (cid:52)− ) ≥ d −
2; that is, asubset I whose contraction I (cid:52)− has Hamming distance larger by 1 than the lower bound d − hd ( H (cid:52)− ) given by Lemma 2a. The lower bound M ( n − , d − ≥ | I (cid:52)− | follows. Now ourunderlying arrays H will satisfy hd ( H ) >
3, so hd ( I ) ≥ hd ( H ) >
3. Thus by Lemma 2c we have hd ( I (cid:52)− ) = hd ( I (cid:52) ) and | I (cid:52)− | = | I (cid:52) | = | I | . So we get M ( n − , d − ≥ | I | , yielding the mainresults of this paper.To find such a subset I of H , we employ a graph C H defined as follows.4 efinition 3 Let H be a permutation array with hd ( H ) = d . Define the contraction graph for H ,denoted C H , by V ( C H ) = H and E ( C H ) = { πσ : π, σ ∈ H, hd ( π (cid:52) , σ (cid:52) ) = d − } . For π ∈ C H , notice that if π ( F ) = F , then π is an isolated point in C H . This is because then π (cid:52) = π , so that for any other σ ∈ C H we have hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ (cid:52) ) ≥ hd ( π, σ ) −
2, implyingno edge joining π and σ in C H . We thus have the following characterization of edges in C H : πσ ∈ E ( C H ) ⇐⇒ { π ( F ) (cid:54) = F, σ ( F ) (cid:54) = F, σ ( π − ( F )) = π ( F ) , π ( σ − ( F )) = σ ( F ) } . (1)This condition on edges is illustrated in Figure 1. Fxπ ( x ) σ ( x ) FbF a baπ − ( F ) σ − ( F ) Figure 1: Neighbors π, σ in C H ; σ ( π − ( F )) = π ( F ), and π ( σ − ( F )) = σ ( F )Since hd ( H (cid:52) ) ≥ d − I of vertices in C H must satisfy hd ( I (cid:52) ) ≥ d −
2. Now by using Lemma 2a,c (together with hd ( H ) > H ) get M ( n − , d − ≥ | I | as explained in the preceding paragraph. We are thus reducedto finding a large independent set in C H for each of the arrays H = AGL (1 , q ) , P GL (2 , q ), andMathieu groups considered in this paper. AGL (1 , q ) Let q be a prime power. Recall the Affine General Linear Group AGL (1 , q ) acting as a permu-tation group on the finite field GF ( q ) of size q , as the set of transformations { x → ax + b : a (cid:54) =0 , x, b ∈ GF ( q ) } under the binary operation of composition. For any π ∈ AGL (1 , q ) the permu-tation π (cid:52) on GF ( q ) is defined as in the previous section, based on some ordering x , x , · · · , x q ofthe elements of GF ( q ), where F = x q is a distinguished element. Clearly | AGL (1 , q ) | = q ( q − AGL (1 , q ) is sharply 2-transitive in this action, and it is straightforward tosee that hd ( AGL (1 , q )) = q − M ( q − , q −
3) for prime powers q ≥ q ≡ C AGL (1 ,q ) for AGL (1 , q ),which we henceforth abbreviate by C A ( q ).By definition we then have V ( C A ( q )) = AGL (1 , q ), and E ( C A ( q )) = { πσ : hd ( π (cid:52) , σ (cid:52) ) = q − } . Following the plan described in the previous section, we find an independent set I in C A ( q ). Once we have such an I , then I (cid:52)− is a permutation array on q − hd ( I (cid:52)− ) = hd ( I (cid:52) ) ≥ q −
3. This implies the lower bound M ( q − , q − ≥| I (cid:52)− | = | I (cid:52) | = | I | , the last equality by Lemma 2c, since q ≥ hd ( I ) ≥ q − >
3. Theactual size of I will then yield our precise lower bound.We are thus reduced to finding a large independent set I in C A ( q ), and from this we getthe bound M ( q − , q − ≥ | I | . We begin on that in the following Lemma, which establishesrelations in the the finite field GF ( q ) that correspond to edges in the graph C A ( q ).5 emma 4 Let π and σ be vertices of the graph C A ( q ) , q ≡ mod , say with σ ( x ) = ax + r and π ( x ) = bx + s . a) If a (cid:54) = b , then hd ( π, σ ) = q − . b) If π ( F ) = F , then π is an isolated point in C A ( q ) . There are q − points π satisfying π ( F ) = F . c) Suppose π and σ are neighbors in C A ( q ) . Then c1) hd ( π, σ ) = q − , and hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) − , and c2) ab and ba are the distinct roots of the quadratic t + t + 1 = 0 over GF ( q ) .Proof. For a), just observe that π ( x ) = σ ( x ) has the unique solution x = s − ra − b .For the first claim in b) suppose not, and let σ be a neighbor of π in C A ( q ). Then we have hd ( π (cid:52) , σ (cid:52) ) = q −
4, implying also that hd ( π, σ ) = q − i be the coordinateof agreement between π and σ . Since π ( F ) = F , we have π (cid:52) = π . Thus hd ( π, σ (cid:52) ) = q − σ (cid:52) can have at most two coordinates, apart from i , in which it agrees with π , these being F and j = σ − ( F ). So altogether π and σ (cid:52) agree in at most the 3 coordinates i, j, F . So q − hd ( π, σ (cid:52) ) ≥ q −
3, a contradiction.Now consider the second claim in b). For any fixed i ∈ GF ( q ), i (cid:54) = F Since π ( F ) = F , we have q − π ( i ) for any fixed i ∈ GF ( q ), i (cid:54) = F . Hencethere are q − π ( F )(= F ) , π ( i )), each such choice determining π uniquely by the sharp 2-transitivity of AGL (1 , q ) acting on GF ( q ). The claim follows.For c1), by the definition of edges in C A ( q ) we have q − hd ( π (cid:52) , σ (cid:52) ) ≥ hd ( π, σ ) − hd ( π, σ ) = q or q −
1, it follows that hd ( π, σ ) = q − hd ( π, σ ) = q − hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) −
3. So thereare distinct α, β ∈ GF ( q ), with neither α nor β being F , such that σ ( F ) = i , σ ( α ) = F , and σ ( β ) = j , and π ( F ) = j , π ( α ) = i , and π ( β ) = F for distinct i, j ∈ GF ( q ). This gives thefollowing set of equations in GF ( q ). σ ( α ) − σ ( β ) = F − j = a ( α − β ) σ ( α ) − σ ( F ) = F − i = a ( α − F ) π ( α ) − π ( β ) = i − F = b ( α − β ) π ( α ) − π ( F ) = i − j = b ( α − F ) . (2)The second and third equations of (2) imply a ( α − F ) = − b ( α − β ) . (3)Now starting with the first equation of (2) we obtain a ( α − β ) = F − j = ( F − i ) + ( i − j )= ( a + b )( α − F ) (by the second and fourth equations of (2)) . Multiplying equation (3) by a and the last equation by b , we obtain the equations (cid:26) a ( α − F ) = − ab ( α − β ) ab ( α − β ) = b ( a + b )( α − F ) . (4)Thus a ( α − F ) = − b ( a + b )( α − F ), and on dividing by α − F (since α (cid:54) = F ) we obtain a + b ( a + b ) = 0 . (5)6ividing equation (5) by a or by b , we obtain that a/b and b/a are both roots of the equation t + t + 1 = 0.We show that a/b and b/a are distinct. Assuming otherwise, then a/b = 1 or −
1. If q iseven then from 1 + t + t = 0 we get the contradiction 1 = 0 since the characteristic is 2. Nowassume q is odd. If a/b = 1, then we get 1 + 1 + 1 = 0, forcing q ≡ a/b = −
1, then we get 1 = 0, again a contradiction.Let t and t be the distinct roots of t + t + 1 = 0 in GF ( q ) for q ≡ π ∈ C A ( q ) with π ( x ) = ax + r , and let σ be a neighbor of π in C A ( q ). Then by Lemma4c2 we have σ ( x ) = at + s or σ ( x ) = a t + s , so far with s and s undetermined. The nextlemma shows that s and s are uniquely determined by π and t . Lemma 5
Let q be a prime power with q ≡ . Suppose π is not an isolated pointof C A ( q ) , say with π ( x ) = ax + r . Let t be a root of t + t + 1 = 0 in GF ( q ) . Then theneighbors of π in C A ( q ) are σ and σ , given by σ ( x ) = at x + ( a − t ) F + r (1 + t ) and σ ( x ) = a t x + ( a − t ) F + r (1 + t ) . In particular, each non-isolated point of C A ( q ) has degree in C A ( q ) .Proof. : Let N ( π ) be the set of neighbors of π in C A ( q ). First we verify that σ , σ ∈ N ( π ),giving details only for σ ∈ N ( π ) as the containment σ ∈ N ( π ) is proved similarly. To do this,we show that all conditions of (1) are satisfied with σ playing the role of σ . Clearly π ( F ) (cid:54) = F since π is not isolated. To show σ ( F ) (cid:54) = F , assume not. Suppose first that a (cid:54) = 1. Then σ ( F ) = F yields F = r − a . But now we get π ( F ) = ar − a + r = r − a = F , a contradiction. Nextsuppose a = 1 so π ( x ) = x + r . Then σ ( F ) = F together with a = 1 yields r (1 + t ) = 0.Combining this with t (cid:54) = − r = 0. But then π ( F ) = F , a contradiction.So it remains to show that σ ( π − ( F )) = π ( F ) and π ( σ − ( F )) = σ ( F ). For the firstequality, solving ax + r = F we obtain π − ( F ) = F − ra . Thus σ ( π − ( F )) = at (cid:0) F − ra (cid:1) + ( a − t ) F + r (1 + t ) = aF + r = π ( F ), as required. For the second equality, from the formula for σ we obtain σ − ( F ) = at (cid:0) F (1 − a + t ) − r (1 + t ) (cid:1) . Plugging this into π and simplifying, weobtain π ( σ − ( F )) = t (cid:0) F (1 − a + t ) − r (1 + t ) (cid:1) + r . Working backwards from the equality π ( σ − ( F )) = σ ( F ) we must show that t (cid:0) F (1 − a + t ) − r (1 + t ) (cid:1) = F (cid:0) a − t + at (cid:1) + rt . This is equivalent to F (cid:0) − a + t (cid:1) = r ( t + t + 1) + F ( at − t + at ) = F ( at − t + at ). Weare now reduced to showing 1 − a + t = at − t + at . This follows from t + t + 1 = 0.Now let σ ∈ N ( π ), and we show that σ = σ or σ . By Lemma 4, we know that σ ( x ) = at + s or σ ( x ) = a t + s for suitable s , s ∈ GF ( q ). Suppose first that σ ( x ) = at + s . Applying theequality σ ( π − ( F )) = π ( F ) together with π − ( F ) = F − ra , we get at (cid:0) F − ra (cid:1) + s = aF + r . so s = ( a − t ) F + r (1 + t ). Thus σ = σ . A very similar argument shows that if σ ( x ) = a t + s ,then s = ( a − t ) F + r (1 + t ), and thus σ = σ . So we have N ( π ) = { σ , σ } , completing theproof.Consider the subgroup Q = { x + b : b ∈ GF ( q ) } of AGL (1 , q ). Clearly | Q | = q , and for each h ∈ GF ( q ) , h (cid:54) = 0, Q has the coset hxQ = { hx + b : b ∈ GF ( q ) } , which we abbreviate by Q h . Theorem 6
Let q be a prime power with q ≡ . Then the connected components of C A ( q ) are as follows. a) There are q − isolated points, these being the points π satisfying π ( F ) = F . b) If q is odd, then each non-isolated point component is a cycle of length 6. c) If q is even, then each non-isolated point component is a cycle of length 3. roof. For part a), we show that π ∈ C A ( q ) is an isolated point if and only if π ( F ) = F . Thena) follows by Lemma 4b.If π ( F ) = F , then immediately π is isolated in C A ( q ) by Lemma 4b. For the converse,suppose to the contrary that π is isolated and that π ( F ) (cid:54) = F . Let σ be given by σ ( x ) = at x + ( a − t ) F + r (1 + t ) as in Lemma 5. Then the proof of Lemma 5, starting from theestablished claim π ( F ) (cid:54) = F (this claim being an assumption here), shows that σ is a neighborof π in C A ( q ). This contradicts π being isolated.Consider part b). By Lemma 5 each nontrivial component of C A ( q ) is a cycle. Let π be apoint on such a cycle C , say with π ∈ Q a . Let t be a fixed root of t + t + 1 = 0. Consider asequence of 4 vertices π π π π on C with π j π j +1 ∈ E ( C A ( q )) for 0 ≤ j ≤
2. We may supposethat π j ∈ Q at j by Lemma 5 and straightforward induction (otherwise replace t by t ). Thus π , π , π are distinct since they belong distinct cosets of Q . Since t = 1, we see also that π and π belong to the same coset Q a of Q . We now show that π (cid:54) = π . Writing π ( x ) = bx + c (so b = at ), we apply the first and third equations of (2) with π and π playing the roles of σ and π respectively, to get t = ba = − (cid:0) i − Fj − F (cid:1) = − (cid:0) π ( F ) − Fπ ( F ) − F (cid:1) . Applying this equation two moretimes we get 1 = t = − (cid:0) π ( F ) − Fπ ( F ) − F (cid:1) , so that (cid:0) π ( F ) − Fπ ( F ) − F (cid:1) = −
1. Thus π ( F ) (cid:54) = π ( F ) , so π (cid:54) = π .Thus each cycle component has length at least 4.Consider now a sequence of 7 vertices π π · · · π on C with π j π j +1 ∈ E ( C A ( q )) for 0 ≤ j ≤ π , π , · · · , π must be distinct as follows. Clearly any two vertices π j , π j +3 are distinct, 0 ≤ j ≤
2, by the same argument that showed π (cid:54) = π . But any twovertices π i , π j with i (cid:54)≡ j (mod 3) are distinct, since t (cid:54) = 1 and t (cid:54) = 1 imply that they belongto different cosets of Q , proving the claim. Finally note that 1 = t = (cid:0) π ( F ) − Fπ ( F ) − F (cid:1) , so that π ( F ) = π ( F ). Since also π and π also belong to the same coset Q a of Q , it follows that π = π . Thus the component C containing π is a cycle of length 6, as required.Now consider part c). Consider as above the sequence of 4 vertices π π π π in a nontrivialcomponent, with π j π j +1 ∈ E ( C A ( q )) for 0 ≤ j ≤
2. We get (cid:0) π ( F ) − Fπ ( F ) − F (cid:1) = − q is even.So since also π and π belong to the same coset Q a of Q , it follows that π = π . Thus thecycle containing π has length 3. Corollary 7
Let q be a prime power with q ≡ and q ≥ . Then a) if q is odd, then M ( q − , q − ≥ ( q − / , and b) if q is even, then M ( q − , q − ≥ ( q − q + 2) / .Proof. For part a) we form an independent set I in C A ( q ) by taking 3 independent pointsin each cycle component of of length 6, together with the set Y of isolated points. Then M ( q − , q − ≥ | Y | + ( | C A ( q ) − Y | ) = q − (cid:0) q ( q − − ( q − (cid:1) = ( q − / I in C A ( q ) by taking one point from each length 3cycle component, together with the set Y of isolated points. We then have M ( q − , q − ≥| Y | + ( | C A ( q ) − Y | ) = ( q − q + 2) / M ( q, q − ≥ q for prime powers q (cid:54)≡ P GL (2 , q ) Let q be a power of a prime. The permutation group P GL (2 , q ) is defined as the set of one toone functions σ : GF ( q ) ∪ {∞} → GF ( q ) ∪ {∞} , under the binary operation of composition,8iven by { σ ( x ) = ax + bcx + d : a, b, c, d ∈ GF ( q ) , ad (cid:54) = bc, x ∈ GF ( q ) ∪ {∞} } . (6)Here σ ( x ) is computed by the rules:1. if x ∈ GF ( q ) and x (cid:54) = − ( d/c ), then σ ( x ) = ax + bcx + d ,2. if x ∈ GF ( q ) and x = − ( d/c ), then σ ( x ) = ∞ ,3. if x = ∞ , and c (cid:54) = 0, then σ ( x ) = a/c , and4. if x = ∞ , and c = 0, then σ ( x ) = ∞ .We regard P GL (2 , q ) as a permutation group acting on the set GF ( q ) ∪ {∞} of size q + 1via the one to one map x (cid:55)→ σ ( x ). One can show that | P GL (2 , q ) | = ( q + 1) q ( q − P GL (2 , q ) is sharply 3-transitive in its action on GF ( q ) ∪ {∞} (see [24] for aproof). It is straightforward to verify that hd ( P GL (2 , q )) = q −
1, and by Theorem 1 we have M ( q + 1 , q −
1) = | P GL (2 , q ) | = ( q + 1) q ( q − GF ( q ) ∪ {∞} with ∞ as final symbol, say x , x , · · · , x q , ∞ wherethe x i are the distinct elements of GF ( q ). Then any element π ∈ P GL (2 , q ) is identified withthe length q + 1 string π ( x ) π ( x ) π ( x ) · · · π ( x q ) π ( ∞ ), which again we call the image string of π . For any such π ∈ P GL (2 , q ) the permutation π (cid:52) on GF ( q ) ∪ {∞} is defined as in thesection introducing contraction, where F = ∞ is the distinguished element of GF ( q ) ∪ {∞} in that definition. As an example, if π = a ∞ bcde , then π (cid:52) = aebcd ∞ , and π (cid:52)− = aebcd . Inthe same way, for any subset R ⊂ P GL (2 , q ), the sets R (cid:52) , and R (cid:52)− are defined as in thatsection, with F = ∞ . Since hd ( P GL (2 , q )) = q − q + 1 −
2, the image strings of any twoelements of
P GL (2 , q ) agree in at most two positions. It follows from Lemma 2a that for any π, σ ∈ P GL (2 , q ) we have hd ( π (cid:52) , σ (cid:52) ) ≥ hd ( π, σ ) − ≥ q −
4. That is, π (cid:52) and σ (cid:52) can agree inat most 5 positions; up to 2 occurring from the original π and σ , and up to 3 more occurringfrom the π (cid:52) and σ (cid:52) operation.As noted earlier, lower bounds for M ( q, q −
3) and M ( q, q −
4) when q (cid:54)≡ q ≡ q an odd prime power, where such bounds are not known. For technical reasonswe take q ≥ I ⊂ P GL (2 , q ) we will find a permutation array I (cid:52)− ⊂ P GL (2 , q ) (cid:52)− on q symbolswith hd ( I (cid:52)− ) ≥ q −
3, thus obtaining the lower bound on M ( q, q − ≥ | I (cid:52)− | . This set I will bean independent set in the contraction graph C P GL (2 ,q ) for P GL (2 , q ), which we abbreviate by C P ( q ).Since hd ( P GL (2 , q )) = q − C P ( q ) is given by V ( C P ( q )) = P GL (2 , q ), and E ( C P ( q )) = { πσ : hd ( π (cid:52) , σ (cid:52) ) = q − } . So edges πσ of C P ( q ) correspond to pairs π, σ ∈ P GL (2 , q ) forwhich hd ( π (cid:52) , σ (cid:52) ) achieves its least possible value of q −
4, occurring when π (cid:52) and σ (cid:52) agree in 5positions, so consequently hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) −
3. Thus a set I ⊆ V ( C P ( q )) is independentin C P ( q ) if and only if it satisfies hd ( I (cid:52) ) ≥ q −
3. By Lemma 2c, we get hd ( I (cid:52)− ) = hd ( I (cid:52) ) ≥ q − | I (cid:52)− | = | I (cid:52) | = | I | , with the last equality following from hd ( I (cid:52) ) = q − > q ≥ I in C P ( q ), from which M ( q, q − ≥ | I | follows.To do this, it will be useful to represent functions in P GL (2 , q ) in a form different than thestandard ax + bcx + d form. 9 efinition 8 Fix a prime power q . Let
K, r, i ∈ GF ( q ) with r (cid:54) = 0 . Define the function f : GF ( q ) ∪ {∞} → GF ( q ) ∪ {∞} by f ( x ) = K + rx − i for x / ∈ { i, ∞} , while f ( ∞ ) = K and f ( i ) = ∞ . Let P = { K + rx − i : K, r, i ∈ GF ( q ) , r (cid:54) = 0 } be the set of all functions defined in 1. Let N ⊂ P GL (2 , q ) be given by N = { π ∈ P GL (2 , q ) : π ( x ) = ax + bcx + d , c (cid:54) = 0 } . We will now see that P is the same set of functions as N . Lemma 9
Let the map α : N → P be defined as follows. For any π ∈ N with π ( x ) = ax + bcx + d , let α ( π ) ∈ P be given by α ( π )( x ) = ac + bc − adc x + dc . Then a) π and α ( π ) are the same function on GF ( q ) ∪ {∞} . b) | P | = | N | = q ( q − . c) The map α is one to one and onto.Proof. For a), straightforward manipulation shows that for x (cid:54) = − dc we have π ( x ) = ac + bc − adc x + dc = α ( π )( x ). Also by definition α ( π )( ∞ ) = ac = π ( ∞ ) and α ( π )( − dc ) = ∞ = π ( − dc ). so π and α ( π )are the same function.Consider b). Clearly | P | = q ( q −
1) since there are q − r , and q choices foreach of K and i , independent of each other. To show | N | = q ( q − π ∈ P GL (2 , q ) with π ( x ) = ax + bcx + d we have c = 0 ⇔ π ( ∞ ) = ∞ . The ⇒ direction isimmediate by definition. To see ⇐ , assume c (cid:54) = 0. Then π ( ∞ ) = ac (cid:54) = ∞ , completing theproof of the observation. Next we have π ( ∞ ) = ∞ ⇔ π ( x ) = Ax + B ∈ AGL (1 , q ) for all x for suitable A (cid:54) = 0 , B ∈ GF ( q ), by definition of computing in P GL (2 , q ). Thus we have | N | = | P GL (2 , q ) | − | AGL (2 , q ) | = ( q + 1) q ( q − − q ( q −
1) = q ( q − N are distinctas functions. As an alternative (constructive) proof, let f ( x ) = K + rx − i ∈ P be given. Then for π ( x ) = Kx + r − iKx − i ∈ N we have α ( π ) = f . Thus α is onto, and since | P | = | N | , α is also one toone.We now see how the above observations, together with results which come later, reduce thestudy of C P ( q ) to the set P of permutations.. It was shown above that for π ∈ P GL (2 , q ), wehave π ( ∞ ) = ∞ ⇔ c = 0. By condition (1) (with ∞ = F ) we see that π ( ∞ ) = ∞ implies that π is an isolated point in C P ( q ), and we will see later that for C P ( q ) the converse is also true.So to study the structure of C P ( q ) apart from its isolated points, we are reduced to studying itssubgraph induced by the permutations in N . By the bijection α : N ↔ P , under which π ∈ N and α ( π ) ∈ P are the same permutation on GF ( q ) ∪ {∞} , we are then reduced to studying P . Lemma 10
Let π, σ ∈ P with π ( x ) = a + rx − i , σ ( x ) = b + sx − j with r, s (cid:54) = 0 . Then hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) − ⇐⇒ ( b − a )( j − i ) = r and r = s .Proof. = ⇒ : By assumption we have π ( ∞ ) = a and π ( i ) = ∞ , together with σ ( j ) = ∞ and σ ( ∞ ) = b . By Lemma 2a the only coordinates of either π or σ whose values are affected bythe (cid:52) operation are the 3 coordinates π − ( ∞ ) = i, σ − ( ∞ ) = j , and ∞ . So the assumption hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) − σ ( i ) = a and π ( j ) = b . Thus we get π ( j ) = a + rj − i = b ,yielding ( b − a )( j − i ) = r as required. Now interchanging the roles of π and σ in this argument,specifically, using σ ( i ) = b + si − j = a , we get ( a − b )( i − j ) = s , so also r = s .10 . . . . . . . . (1 , a q ) (1 , a ) (1 , a ) (1 , a )... B g q − = B . . . . . . . . . ( g i , a q ) ( g i , a ) ( g i , a ) ( g i , a ) B g i . . . . . . . . . ( g, a q ) ( g, a ) ( g, a ) ( g, a ) B g . . . . . . . . . (0 , a q ) (0 , a ) (0 , a ) (0 , a ) B ...... Figure 2: The graph P , partitioned into levels B and B g i , ≤ i ≤ q − ⇐ = : Again by assumption we have π ( ∞ ) = a , σ ( ∞ ) = b , π ( i ) = ∞ , σ ( j ) = ∞ , and( b − a )( j − i ) = r . To prove hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) −
3, it remains only to show that π ( j ) = b and σ ( i ) = a . For simplicity we let r = s = 1, since the argument does not depend on r = s . Solving for b in ( b − a )( j − i ) = 1 we get b = j − i + a = π ( j ). Solving for a we get a = b − j − i = b + i − j = σ ( i ), as required. Lemma 11
Let q = p m , where p is an odd prime, with q ≡ mod , q ≥ . Let π, σ ∈ P ,say with π ( x ) = a + rx − i , σ ( x ) = b + sx − j , with r, s (cid:54) = 0 . Then hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) − ⇐⇒ πσ ∈ E ( C P ( q )) .Proof. ⇐ =: By definition of edges in C P ( q ) and Lemma 2a we have q − hd ( π (cid:52) , σ (cid:52) ) ≥ hd ( π, σ ) −
3. Now since q − ≤ hd ( π, σ ) ≤ q +1, equality is forced together with hd ( π, σ ) = q − hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) − ⇒ : By the assumption hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) − hd ( π, σ ) ≥ q − hd ( π, σ ) = q −
1; that is, that π and σ already agree in two coordinates.By assumption and Lemma 10 we have r = s , so write π ( x ) = a + rx − i and σ ( x ) = b + rx − j ,for a, b, i, j, k ∈ GF ( q ) with r (cid:54) = 0. Note also i (cid:54) = j , since otherwise by Lemma 10 we get r = 0,a contradiction.We now derive a quadratic equation over GF ( q ) whose distinct roots are the coordinatesof agreement between π and σ . Since hd ( π (cid:52) , σ (cid:52) ) = hd ( π, σ ) −
3, by Lemma 10 we have( b − a )( j − i ) = r . Thus b = rj − i + a . Now we set π ( x ) = σ ( x ) to find the possible coordinates x at which π and σ agree, understanding that x can be neither i nor j since π and σ canhave no agreements in any of the coordinates i = π − ( ∞ ) , j = σ − ( ∞ ), or ∞ by Lemma 2a.Substituting rj − i + a for b and simplifying we obtain x − i − x − j = j − i . Hence i − j ( x − i )( x − j ) = j − i ,and we get the quadratic x − ( i + j ) x + ij + ( i − j ) = 0 . By Corollary 23b there are twodistant roots to this equation, giving the two coordinates of agreement for π and σ as follows; x = [ i (1 + √−
3) + j (1 − √− x = [ i (1 − √−
3) + j (1 + √− πσ ∈ E ( C P ( q )), asrequired.The preceding two Lemmas yield the following.11 orollary 12 Let q = p m , where p is an odd prime, with q ≡ mod , q ≥ . a ) Let π, σ ∈ P , say with π ( x ) = a + rx − i , σ ( x ) = b + sx − j , r, s (cid:54) = 0 . Then πσ ∈ E ( C P ( q )) ⇐⇒ r = s and ( b − a )( j − i ) = r . b ) π ∈ P GL (2 , q ) is an isolated point in C P ( q ) ⇐⇒ π ( ∞ ) = ∞ .Proof. Part a) follows immediately from Lemmas 10 and 11.For part b), suppose first that π ( ∞ ) = ∞ . Then immediately π is isolated in C P ( q ) by theequivalence (1) (with ∞ = F ) applicable to any contraction graph.Conversely, suppose to the contrary that π is isolated in C P ( q ) and π ( ∞ ) = x (cid:54) = ∞ . Let i = π − ( ∞ ), and let j be any coordinate with j / ∈ { i, ∞} , and let π ( j ) = y . Then by sharp 3-transitivity of P GL (2 , q ) we can find an element σ ∈ P GL (2 , q ) satisfying σ ( j ) = ∞ , σ ( i ) = x ,and σ ( ∞ ) = y . Then we get hd ( σ (cid:52) , π (cid:52) ) = hd ( σ, π ) −
3. So by Lemma 11 we have πσ ∈ E ( C P ( q )), contradicting π being isolated.The next two theorems, which use the preceding Corollary, tell us more about C P ( q ). For S ⊂ C P ( q ), recall that [ S ] is the subgraph of C P ( q ) induced by S . When r is fixed by context,we denote a vertex π ∈ C P ( q ), π ∈ P , with π ( x ) = a + rx − i , by the abbreviation ( i, a ).Consider the partition of P given by P = ∪ r (cid:54) =0 P r , where for r ∈ GF ( q ) with r (cid:54) = 0, P r = { a + rx − i : a, i ∈ GF ( q ) } , so | P r | = q . Further consider the partition of P r given by P r = ∪ i ∈ GF ( q ) B i ( r ), where B i ( r ) = { a + rx − i : a ∈ GF ( q ) } . Theorem 13
Let q = p m , where p is an odd prime, with q ≡ mod , q ≥ . Then thefollowing hold in the graph C P ( q ) . a ) For any r (cid:54) = s , r, s (cid:54) = 0 , we have [ P r ] ∼ = [ P s ] . b ) For any r (cid:54) = 0 and i (cid:54) = j , [ B i ( r ) ∪ B j ( r )] is a perfect matching, which matches B i ( r ) to B j ( r ) . c ) For any r (cid:54) = 0 , the subgraph [ P r ] is regular of degree q − . d ) Let v ∈ C P ( q ) be a non isolated point, and N ( v ) the set of neighbors of v in C P ( q ) . Then [ N ( v )] is a disjoint union of cycles.Proof. For a), consider for any r ∈ GF ( q ), r (cid:54) = 0, the map ϕ : P → P r given by ϕ ( a + x − i ) = a + rx − ri . Let v, w ∈ P , say with v ( x ) = a + x − i and w ( x ) = b + x − j . Then vw ∈ E ([ P ]) ⇔ ( b − a )( j − i ) = 1 ⇔ ( b − a )( rj − ri ) = r ⇔ ϕ ( v ) ϕ ( w ) ∈ E ([ P r ]) . Thus ϕ is a graph isomorphism,and since r was arbitrary, it follows that for any s (cid:54) = 0 we have [ P r ] ∼ = [ P ] ∼ = [ P s ].Consider b). Fix r , and consider any two points ( i, a ) and ( j, b ) of P r . By Corollary 12we have ( i, a )( j, b ) ∈ E ( C P ( q )) if and only if i (cid:54) = j and ( b − a )( j − i ) = r in GF ( q ). Let H ij = [ B i ( r ) ∪ B j ( r )] for i (cid:54) = j . Note there can be no edge in H ij of the form ( i, a )( i, b ) since( b − a )( i − i ) = 0 (cid:54) = r , and similarly no edge of the form ( j, a )( j, b ). Now given ( i, a ) ∈ B i ( r ), apoint ( j, b ) ∈ B j ( r ) is a neighbor of ( i, a ) if and only if ( b − a )( j − i ) = r by Corollary 12.Thus for this fixed i and j we can uniquely determine b by the equation b = r ( j − i ) − + a ,showing that ( j, b ) is the only neighbor of ( i, a ) in B j ( r ). A symmetric argument shows thateach point in B j ( r ) has a unique neighbor in B i ( r ). Thus E ( H ij ) is a perfect matching, whichmatches B i ( r ) to B j ( r ).For c), let v ∈ C P ( q ), say with v ∈ B i ( r ) ⊂ P r for some r (cid:54) = 0. By Corollary 12, any neighborof v in C P ( q ) must also lie in P r . By part b), the neighbors of v are in one to one correspondencewith the sets B j ( r ), j (cid:54) = i , j ∈ GF ( q ). Thus v has exactly | GF ( q ) | − q − C P ( q ).For d), take v ∈ C P ( q ), and by the isomorphism of subgraphs [ P r ] from part a), we can take v = ( i, a ) ∈ P . By Corollary 12 we have N ( v ) ⊂ P . It suffices to show that [ N ( v )] is regular ofdegree 2. Let ( j, b ) ∈ N ( v ), so j (cid:54) = i by part b). Now any neighbor ( k, c ) of ( j, b ) in N ( v ) must12 B g B g i B g q − ......... Figure 3: Perfect matching between any two levels of P .lie in N (( i, a )) ∩ N (( j, b )). So to show that ( j, b ) has degree 2 in [ N ( v )], it suffices to show that( k, c ) ∈ P satisfies ( k, c ) ∈ N (( i, a )) ∩ N (( j, b )) if and only if k is a root in GF ( q ) of a quadraticequation over GF ( q ) having two distinct roots in GF ( q ).Suppose first that ( k, c ) ∈ N (( i, a )) ∩ N (( j, b )). By Corollary 12 we must have the equations( c − a )( k − i ) = 1 , ( b − c )( j − k ) = 1 , ( b − a )( j − i ) = 1 . Using the second and third equations we get c = ( j − i ) − − ( j − k ) − + a , and from the firstequation c = ( k − i ) − + a . Setting these two expressions for c equal we obtain ( k − i ) − + ( j − k ) − = ( j − i ) − . Some simplification leads to the quadratic k − k ( i + j ) + ij + ( j − i ) = 0with coefficients over GF ( q ) and unknown k . By Corollary 23b from the Appendix, we seethat that there are two distinct solutions for k ; namely k = [ i (1 + √−
3) + j (1 − √− k = [ i (1 − √−
3) + j (1 + √− . Conversely suppose that k is one of the two distinct solutions of k − k ( i + j )+ ij +( j − i ) = 0.Then ( k − i )( j − k ) = − k + k ( i + j ) − ij = ( j − i ) , and using k − i )( j − k ) = (cid:0) j − i (cid:1)(cid:0) k − i + j − k (cid:1) ,one can derive k − i + j − k = j − i . Now set c = k − i + a , so immediately we get ( c − a )( k − i ) = 1.Since ( i, a ) and ( j, b ) are neighbors we have ( b − a )( j − i ) = 1, so b = j − i + a . It follows that c = k − i + a = j − i − j − k + a = b − j − k . Hence we get ( b − c )( j − k ) = 1. Thus the threeequations ( c − a )( k − i ) = 1, ( b − c )( j − k ) = 1, and ( b − a )( j − i ) = 1 hold, showing that( k, c ) ∈ N (( i, a )) ∩ N (( j, b )) by Corollary 12.Note that once k is determined (as one of the two distinct roots), then the point ( k, c ) isuniquely determined by the perfect matching between B k (1) and B i (1) (or B j (1)). Thus weobtain that an arbitrary point ( j, b ) ∈ N ( v ) has exactly two neighbors in N ( v ), completing d).To round out the structure of C P ( q ) we consider the connected components of C P ( q ). Theorem 14
Let q = p m , where p is an odd prime, with q ≡ mod , q ≥ . Then theconnected components of C P ( q ) are as follows.1) the isolated points - these are of the form π ( x ) = ax + b , a (cid:54) = 0 , and there are q ( q − ofthem,2) the q − many connected components [ P r ] induced by the sets P r . roof. By Corollary 12b we have that π ∈ P GL (2 , q ) is an isolated point in C P ( q ) if and onlyif π ( ∞ ) = ∞ . This is equivalent to π ( x ) = ax + b , a (cid:54) = 0 and there are q ( q −
1) such points,completing part 1).The remaining permutations are all of the form π ( x ) = a + rx − i for suitable a, r, i ∈ GF ( q )with r (cid:54) = 0 as shown earlier. Hence it suffices to analyze the connected component structure of[ ∪ r (cid:54) =0 P r ]. By Corollary 12 and Theorem 13a, to prove part 2) it suffices to prove that any oneof the [ P r ], say [ P ], is connected.Recall the partition P = ∪ i ∈ GF ( q ) B i (1) defined above, and from now on we abbreviate B i (1)by B i . Let g by a generator of the multiplicative cyclic subgroup of nonzero elements in GF ( q ).Then we can write this partition as P = B ∪ ( ∪ ≤ k ≤ q − B g k ). We regard the sets in this partitionas “levels” of C P ( q ); where B is level 0 and B g k is level k , 1 ≤ k ≤ q −
1. See Figure 2 foran illustration of P from this viewpoint, where in that Figure we continue with the notation( i, a ) for a + x − i . In particular, ( g t , a ) refers to a + x − g t . By Theorem 13b the subgraph of [ P ]induced by any two levels has edge set which is a perfect matching, as illustrated in Figure 3.First we observe that to show that [ P ] is connected it suffices to show that any two verticesin B are joined by a path in [ P ]. For if that was true, then we can find a path in [ P ] from(0 ,
0) to any vertex w ∈ P (thus showing connectedness of [ P ]) as follows. If w ∈ B we aredone by assumption. So suppose w / ∈ B , say with w ∈ B ( g k ). Let v be the unique neighbor in B of w under the perfect matching E ([ B ∪ B ( g k )]). Let P be the path from (0 ,
0) to v in [ P ]which exists by assumption. Then P followed by the edge vw is a walk joining (0 ,
0) to w , so P contains a path from (0 ,
0) to w .By Theorem 13b there is a (unique) path in [ P ] starting at (0 ,
0) and passing through levels1 , , · · · , q − , − ( g, α ) − ( g , α ) − ... − ( g q − , α q − ) be this path,illustrated in bold lines in Figure 4, for suitable α k ∈ GF ( q ). For k ≥ , β k ) ∈ B bethe unique neighbor in level 0 of the vertex ( g k , α k ) in level k . The edges ( g k , α k )(0 , β k ) areillustrated by the dotted lines in in Figure 4.This path and the points (0 , β k ) are illustrated in Figure 4. Our first step is to obtain thevalues of α k and β k . Claim 1 : We havea) α = g , α = g − , and α k = g k − + g k − + g k − + ··· + g +1( g − g k − for k ≥ . b) β = 0, and β k = ( g − g +1)(1+ g + g + g + ··· + g k − ) g k ( g − for k ≥ r, a ) and ( s, b ) are adjacentvertices in the contraction graph C P ( q ), then ( s − r )( b − a ) = 1.For part a), since (0 , − ( g, α ) is an edge in C P ( q ) we have ( α − g −
0) = 1, so α = g . Since ( g, α ) − ( g , α ) is an edge we have ( α − g )( g − g ) = 1, yielding α = g − ,and similarly ( α − g − )( g − g ) = 1, yielding α = g +1( g − g . Now for k ≥ k = 3. Since ( g k , α k ) − ( g k − , α k − ) is an edge, we have( α k − α k − )( g k − g k − ) = 1. Solving for α k and applying the inductive hypothesis to α k − , weobtain α k = g k − g k − + g k − + g k − + g k − + ··· + g +1( g − g k − , which after simplification yields the claim.For part b), we have β = 0 since (0 , − ( g, α ) is an edge by definition. Since ( g , α ) − (0 , β )is an edge, we have ( g − − β )( g −
0) = 1, and solving for β and simplifying we get the claimfor k = 2. Consider now k ≥ . The existence of edge ( g k , α k ) − (0 , β k ) gives ( α k − β k ) g k = 1, so β k = α k − g k . Using the formula for α k from part a), we have β k = g k − + g k − + g k − + ··· + g +1( g − g k − − g k = g k + g k − + g k − + ··· + g +1( g − g k = ( g − g +1)(1+ g + g + g + ··· + g k − ) g k ( g − . QED Claim 2 : We have |{ β k : 1 ≤ k ≤ q − }| = q −
1; that is, the β k , 1 ≤ k ≤ q −
1, are pairwisedistinct. 14 g , α ) ... B g B g B g B g B ( g , α ) ( g , α )( g , α ) (0 ,
0) = (0 , β )(0 , β )(0 , β )(0 , β ) Figure 4: The path (0 , − ( g, α ) − ( g , α ) − · · · − ( g q − , α q − ) in P , where (0 , β i ) is the level0 neighbor of ( g i , α i ).Proof of Claim 2: In applying Claim 1, we note first that g could have been chosen so as not tobe a root of x − x + 1 = 0 as follows. The number of roots in GF ( q ) to this quadratic is atmost 2. Now the number of generators in the multiplicative cyclic group GF ( q ) − { } of order q − φ ( q − ≤ s ≤ q − q −
1. Since q is an odd prime power with q ≥
13, we know that φ ( q − >
2, so such a g exists.We show that for for any pair j, k with 1 ≤ j < k ≤ q − β k (cid:54) = β j .Consider first the case j = 1. Since β = 0, we need to show that β k (cid:54) = 0 for 2 ≤ k ≤ q −
1. Supposing the contrary and applying Claim 1b we get ( g − g +1)(1+ g + g + g + ··· + g k − ) g k ( g − = 0 . Canceling the nonzero factor g − g +1 g k ( g − (by the preceding paragraph) on the left side, we get0 = (1 + g + g + g + · · · + g k − ) = g k − − g − . This implies that g k − − g has order k − k − ≤ q − g , being a generator of the cyclic group GF ( q ) − { } ,must have order q − j ≥
2. Assuming the contrary that β k = β j and applying Claim 1b, weget after simplification that 1 + g + g + g + · · · + g k − = g k − j (1 + g + g + g + · · · + g j − ) = g k − j + g k − j +1 + · · · + g k − . Thus we have 0 = 1 + g + g + · · · + g k − j − = g k − j − g − . So g k − j = 1,which is impossible since k − j ≤ q −
3, while g has order q −
1. QEDWe introduce some notation in preparation for the rest of the argument. Let Z = { (0 , β k ) :1 ≤ k ≤ q − } ⊂ B . Since | B | = q , by Claim 2 we have | B − Z | = 1, and we let u be the unique vertex of B − Z . Further for any subset T of vertices in C P ( q ), we let N ( T ) = { v ∈ C P ( q ) : v / ∈ T, vt ∈ E ( C P ( q )) for some t ∈ T } be the neighbor set of T in C P ( q ).Recall also that [ T ] denotes the subgraph of C P ( q ) induced by T . Claim 3 : Let H = [ Z ∪ N ( Z )] C P ( q ) , and H (cid:48) = [ { u } ∪ N ( u )] C P ( q ) .a) H (cid:48) is connected.b) H is connected. 15) V ( H ) ∩ V ( H (cid:48) ) = ∅ d) We have the partition V ( P ) = V ( H ) ∪ V ( H (cid:48) ).Proof of Claim 3: For part a), we apply Theorem 13b to deduce that H (cid:48) has the spanning starsubgraph K ,q − , where the center is u and the leaves, one in each level B i , i (cid:54) = 0, form N ( u ).Thus H (cid:48) is connected.Consider part b). Since β = 0 we have (0 , ∈ Z ⊂ V ( H ). Thus it suffices to show that forany w ∈ V ( H ) there is a path in H joining (0 ,
0) to w .Suppose first that w ∈ Z , so w = (0 , β k ) for some k . Observe that ( g i , α i ) ∈ N ( Z ) forall i by definition. So the path (0 , − ( g, α ) − ( g , α ) − ... − ( g k , α k ) followed by the edge( g k , α k ) − (0 , β k ) is path in H joining (0 ,
0) to w .Next suppose w ∈ N ( Z ), say with w adjacent to (0 , β k ) ∈ Z . Then the path (0 , − ( g, α ) − ( g , α ) − ... − ( g k , α k ) followed by the length 2 path ( g k , α k ) − (0 , β k ) − w is a walk in H joining(0 ,
0) to w , and this walk contains the required path.Next consider c). Suppose not, and let z ∈ V ( H ) ∩ V ( H (cid:48) ), say with z ∈ B ( g k ), noting that k ≥ B , is an independent set in [ P ]. Then z has two distinctneighbors in B ; namely u and (0 , β j ), for some 1 ≤ j ≤ q −
1. This contradicts the fact thatthe edge set of [ B ( g k ) ∪ B ] is a perfect matching between the levels B ( g k ) and B by Theorem13b. Thus V ( H ) ∩ V ( H (cid:48) ) = ∅ .Consider now d). By part c), it suffices to show that | V ( P ) | = | V ( H ) | + | V ( H ) (cid:48) | . ByTheorem 13b, it follows that | V ( H ) | = | Z | q = ( q − q. For the same reason | V ( H (cid:48) ) | = q. Therefore | V ( P ) | = q = | V ( H ) | + | V ( H ) (cid:48) | as required. QEDWe can now complete the proof of the theorem by showing that P is connected. In view ofClaim 3, to do this we are reduced to showing that there is an edge vw ∈ E ([ P ]) with v ∈ H (cid:48) and w ∈ H . Suppose no such edge exists. Since [ P ] is ( q − H (cid:48) is a simple q − q vertices. Thus H (cid:48) = K q . Hence [ N ( u )] = K q − .But this is a contradiction for q ≥ C P ( q ) is regular of degree 2, while [ N ( u )] is regular of degree q − > q ≥ C P ( q ) as a consequence of our previous resultsand the following theorem of Alon [1]. Theorem 15 [1] Let G = (V,E) be a graph on N vertices with average degree t ≥ in whichfor every vertex v ∈ V the induced subgraph on the set of all neighbors of v is r -colorable. Thenthe maximum size α ( G ) of an independent set in G satisfies α ( G ) ≥ c log( r +1) Nt log t , for someabsolute constant c . Corollary 16
Let q be a power of an odd prime p , with q ≡ mod , q ≥ . a ) α ( C P ( q )) ≥ Kq log q for some constant K . b ) M ( q, q − ≥ Kq log q for some constant K .Proof. Consider a). By Corollary 12a there is no edge between any two subgraphs [ P r ] and [ P s ]for r (cid:54) = s . Since there are q such subgraphs, and by Theorem 13a) they are pairwise isomorphic,it suffices to show that α ( P ) ≥ Kq log q for some constant K .We now apply Alon’s theorem to the subgraph [ P ] of C P ( q ). Now [ P ] is ( q − q points. Since the neighborhood of every point is a disjoint union ofcycles by Theorem 13d, this neighborhood must be 3-colorable. It follows by Alon’s theoremthat [ P ] contains an independent set of size c log 4 q q − log( q − ∼ Kq log q , for some constant K .16or b), let I be an independent set in C P ( q ) of size Kq log q for suitable constant K , guar-anteed to exist by by part a). Then by the reduction made in the discussion preceding Lemma10 we have M ( q, q − ≥ | I | ≥ Kq log q . M ( n, d ) via the Mathieugroups In this section we consider the Mathieu groups M , M , M , M , M , discovered by E. Math-ieu in 1861 and 1873. These permutation groups are the earliest known example of sporadicsimple groups. See [10], [6], or [25] for a discussion of their construction. These groups act on11 , , , ,
24 letters respectively, with M being a 1 point stabilizer of M , while M and M are 1 and 2 point stabilizers of M respectively.In this section we apply the contraction operation to these permutation groups to obtain newpermutation arrays, with resulting lower bounds for M ( n, d ) for suitable n and d .Since M is sharply 5-transitive we have by Theorem 1 that hd ( M ) = 8 and M (12 ,
8) = | M | = 95040. Similarly since M is sharply 4-transitive we have M (11 ,
8) = | M | = 7920.For M we do not have sharp transitivity. But observe that for any permutation group G actingon some set, and three elements π, σ, τ ∈ G , we have hd ( π, σ ) = hd ( πτ, στ ) = hd ( τ π, τ σ ). Thus hd ( G ) =min { hd (1 , σ ) : σ ∈ G } . From the set of disjoint cycle structures of elements of M (available at [30]) we find that the largest number of 1-cycles in the disjoint cycle structure ofany nonidentity element of M is 8. Thus hd ( M ) = 24 − hd ( M ) = hd ( M ) = 16. We thus obtain M (24 , ≥ | M | = 24 , , M (23 , ≥ | M | = 10 , , M (22 , ≥ | M | = 443 , M on the 12-letter set Ω = { x , x , · · · , x } , we designate some element, say x , of Ω as thedistinguished element F in the definition of π (cid:52) . Then define for each π ∈ M the permutation π (cid:52) on the set Ω exactly as in the introduction. Thus, using the natural ordering of elements ofΩ by subscript, the image string of any σ ∈ M can be written σ ( x ) σ ( x ) · · · σ ( x ) σ ( F ).As before, we let π (cid:52)− be the permutation on 11 symbols obtained from π (cid:52) by dropping thefinal symbol F , and for any subset S ⊂ M , we let S (cid:52)− = { π (cid:52)− : π ∈ S } , sometimes writing thisas ( S ) (cid:52)− . Proposition 17 a ) hd (( M ) (cid:52)− ) ≥ . b ) M (11 , ≥ | M | = 95040 . c ) M (10 , ≥ .Proof. We start with a). Suppose not. Since hd ( M ) = 8, and for any α, β ∈ M we have hd ( α (cid:52) , β (cid:52) ) ≥ hd ( α, β ) − hd (( M (cid:52) ) − ) = 5.Thus there is a pair σ, τ ∈ M such that hd ( σ, τ ) = 8 and hd ( σ (cid:52) , τ (cid:52) ) = 5; so hd ( σ (cid:52) , τ (cid:52) ) = hd ( σ, τ ) −
3. Thus by Lemma 2b we know that πσ − has a 3-cycle in its disjoint cycle factorizationso the order of πσ − is divisible by 3.Since hd ( σ, τ ) = 8 and π and σ are permutations on 12 letters, it follows that there are fourpositions, call them x i , 1 ≤ i ≤
4, at which π and σ agree. Then πσ − belongs to the subgroup H of M fixing these four positions; that is H = { α ∈ M : α ( x i ) = x i , ≤ i ≤ } . This H ,denoted M , is known to be isomorphic to Q , the quaternion group of order 8 ([5], section 3.2).We can also verify this directly by making use of GAP (Groups, Algorithms, Programming), a17ystem for computational discrete algebra. The following output employing GAP shows that H ∼ = Q , the quaternion group of order 8 ([14])gap > G := M athieuGroup (12);;gap > H = Stabilizer ( G, [1 , , , , OnT uples );;gap > StructureDescription ( H );“ Q (cid:48)(cid:48) Now the order of πσ − is divisible by 3 as noted above. But 3 does not divide | Q | , acontradiction to Lagrange’s theorem.Consider next b). Using Lemma 2c and hd ( M ) = 8 >
3, we have | M | = | ( M ) (cid:52)− | . Thus( M ) (cid:52)− is a permutation array on 11 letters of size | M | with hd (( M ) (cid:52)− ) ≥
6. Part b) follows.For part c), we recall from the introduction the elementary bound M ( n − , d ) ≥ M ( n,d ) n .Using part a), we then obtain M (10 , ≥ M (11 , ≥ . We remark that using the same method as in part b) of the above proposition one can show M (10 , ≥ | M | = 7920. But this is obviously weaker than the bound we give in part c).We now consider the contraction of M and resulting special case bounds for M ( n, d ). Usingsimilar notation as for M above, we let M act on the set of 24 letters Θ = { x , x , · · · , x } ,and we designate x as the distinguished symbol F in the definition of π (cid:52) from the introduction.Now define π (cid:52) for any π ∈ M as in the introduction, along with accompanying definitions S (cid:52) and S (cid:52)− for S ⊆ M . Proposition 18 a ) hd (( M ) (cid:52)− ) ≥ . b ) M (23 , ≥ | M | = 244 , , . c ) M (22 , ≥ | M | = 10 , , . d ) M (21 , ≥ | M | . = 483 , .Proof. For a), suppose not. Since hd ( M ) = 16, and for any α, β ∈ M we have hd ( α (cid:52) , β (cid:52) ) ≥ hd ( α, β ) −
3, it follows that hd (( M ) (cid:52)− ) = 13. Thus there is pair σ, τ ∈ M such that hd ( σ, τ ) =16 and hd ( σ (cid:52) , τ (cid:52) ) = 13; so hd ( σ (cid:52) , τ (cid:52) ) = hd ( σ, τ ) −
3. Hence by Lemma 2b, τ σ − has a 3-cyclein its disjoint cycle structure factorization.Since hd ( σ, τ ) = 16, and σ and τ are permutations on 24 letters, it follows that σ and τ mustagree on 8 positions. Thus τ σ − belongs to the subgroup H of M fixing these 8 positions.From the structure theory of M , we know that if these 8 positions form an “octad” (amongthe 24 positions), then H = M ∼ = Z × Z × Z × Z , the elementary Abelian group of order 16([25] Theorem 3.21, and [26] pp. 197-208). Again, this can also be verified directly using GAPfrom the following output ([14]).gap > G := M athieuGroup (24);;gap > H := Stabilizer ( G, [1 , , , , , OnT uples );;gap > S = SylowSubgroup ( H, > octad := F iltered ([1 .. , x → not x in MovedPoints( S ));[1,2,3,4,5,8,11,13]gap > H := Stabilizer ( G, octad, OnT uples );;gap > StructureDescription ( H );“ C × C × C × C (cid:48)(cid:48) . 18f these 8 positions do not form an octad, then H is the identity ([25], Lemma 3.1). Now theorder of τ σ − is divisible by 3, so 3 must divide | H | . By Lagrange’s theorem, this contradictsthat | H | has order either 16 or 1.Consider next b). Using Lemma 2c and hd ( M ) = 16 >
3, we have | M | = | ( M ) (cid:52)− | . Thus( M ) (cid:52)− is a permutation array on 23 letters of size | M | , and by part a) we have hd (( M ) (cid:52)− ) ≥
14. Part b) follows.For part c), we again use the bound M ( n − , d ) ≥ M ( n,d ) n . Using part b), we then obtain M (22 , ≥ M (23 , ≥ | M | = 10 , , M ( n − , d ) ≥ M ( n,d ) n again we get M (21 , ≥ M (22 , ≥ | M | . = 483 , We mention some problems left open from our work. Recall that if I is an independent set in C P ( q ), then M ( q, q − ≥ | I | . To find a large such I , one can focus on any nontrivial connected component, say P , of C P ( q ). If P contains anindependent set of size k , then by the isomorphism of the connected components P i , 1 ≤ i ≤ q − k ( q −
1) + q ( q −
1) = ( q − k + q ) in C P ( q ), where q ( q − C P ( q ). Our lower bound M ( q, q − ≥ Kq log q implies,again by the isomorphism of components, that α ( P ) ≥ Cq log q (where α ( G ) is the maximumsize of an independent set in a graph G ), for some constant C . We therefore ask whether animprovement on this lower bound for α ( P ) can be found.Now V ( P ) can be viewed as a rectangular array { ( i, a ) : i, a ∈ GF ( q ) } as in Figure 2, wherewe let i be the row index, and a the column index. By Corollary 12a an independent set in P isjust a subset S of this array with the property that for any two points ( i, a ) , ( j, b ) ∈ S we have( b − a )( j − i ) (cid:54) = 1 in GF ( q ). Using the integer programming package GUROBI, we computedindependent sets in P of size k for various q . This k , together with the resulting lower bound( q − k + q ) for M ( q, q −
3) are presented in Table 1. The primes q = 41 , , , , , ,
89, forexample, are not included in this table since q (cid:54)≡ M ( q, q − ≥ ( q +1) q ( q − We also ask for good upper bounds on α ( P ). In this section we review some facts from number theory that were used in this paper.We start with some notation. For an odd prime p and integer r (cid:54)≡ p ), define theLegendre symbol ( rp ) to be 1 (resp. -1) if r is a quadratic residue (resp. nonresidue); that is asquare (resp. nonsquare) mod p . If r ≡ p ), then define ( rp ) = 0 . A couple of simple factsabout this symbol are these.
Lemma 19
For an odd prime p and integers r and s we have the following.a) ( − p ) = 1 if p ≡ (mod , and ( − p ) = − if p ≡ (mod .b) ( rsp ) = ( rp )( sp ) .Proof. For a), suppose p ≡ p = 4 k + 1, and consider the multiplicativegroup of nonzero elements mod p , which has order 4 k and is cyclic. Let x be a generator of this19roup. Then note that in this group we have 1 = x k = ( x k ) , while also ( − = 1 in thisgroup. Since the quadratic z − z = 1 or − GF ( q ), andsince x k (cid:54) = 1 since x is a generator, it follows that x k = −
1. Thus -1 is a square mod p .If p ≡ k + 2 for some integer k . This time wehave 1 = ( x k +1 ) , so that by the same reasoning as above we have x k +1 = −
1. This showsthat -1 is not a square mod p , since it is on odd power of the generator.Consider now b). Just observe that the product rs is a square mod p if and only if both r and s are squares mod p or if both r and s are non-squares mod p . Part b) then follows immediately.We now recall the quadratic reciprocity law. Theorem 20 (Gauss Quadratic Reciprocity Law) For odd primes p and q we have ( pq )( qp ) = ( − ( p − )( q − ) . There are lots of proof of quadratic reciprocity in the literature, so we omit the proof here.Now let’s apply these facts to determining ( − p ) for odd primes p. Theorem 21
Let p > be an odd prime. Thena) If p ≡ (mod ), then -3 is a quadratic residue mod p .b) If p ≡ (mod ), then -3 is a quadratic nonresidue mod p .Proof. By the lemma above we have ( − p ) = ( − p )( p ), while by quadratic reciprocity we have( p ) = ( p )( − p − . Thus ( − p ) = ( − p − ( − p )( p . The factors on the right depend on the residue classes of p mod 4 and p mod 3. Thus weconsider the four cases defined by the combinations of these two possibilities, obtaining resultsthat initially depend on the residue class of p mod 12. case 1 : p ≡ p ≡ p ≡ p ≡ p ) = 1. Also p ≡ − p − = 1 and by Lemma19 also implies ( − p ) = 1. So by the formula above we have ( − p ) = 1, showing that − p ≡ case 2 : p ≡ p ≡ p ≡ p ≡ p ) = −
1. Also p ≡ − p − = 1 and alsoLemma 19 implies ( − p ) = 1. So by the formula above we have ( − p ) = −
1, showing that − p ≡ case 3 : p ≡ p ≡ p ≡ p ≡ p ) = 1. Also p ≡ − p − = −
1, and alsoLemma 19 implies ( − p ) = −
1. So by the formula above we have ( − p ) = 1, showing that − p ≡ case 4 : p ≡ p ≡ p ≡ p ≡ p ) = −
1. Again p ≡ − p − = −
1, andalso that ( − p ) = −
1. So by the formula above we get ( − p ) = −
1, showing that − p ≡ − p when p ≡ − p when p ≡ Corollary 22
Consider the prime power q = p m , where p > is an odd prime. If q ≡ mod , then − is a square in the finite field GF ( q ) .Proof. Since p > p ≡ p ≡ p ≡ − GF ( p ) ⊆ GF ( q ) by Theorem 21, so − GF ( q ), as required.So suppose p ≡ GF ( p )( √−
3) of GF ( p ) obtainedby adjoining to GF ( p ) a root of the irreducible (by Theorem 21) polynomial x + 3 over GF ( p ).Then GF ( p )( √− ∼ = GF ( p ), and − GF ( p ).Since q ≡ p ≡ p ≡ m must be even. We recall the basic fact from finite fields that GF ( p r ) ⊆ GF ( p s ) if and only if r | s . It follows that GF ( p ) ⊆ GF ( q ). Thus since − GF ( p ), then − GF ( q ). Corollary 23
Let q = p m be a prime power, q ≡ mod . a) The equation x + x + 1 = 0 has two distinct solutions in GF ( q ) . If x is such a root, then x is the other distinct root. b) For q odd and distinct i, j ∈ GF ( q ) , the equation x − ( i + j ) x + ij + ( i − j ) = 0 has twodistinct roots in GF ( q ) .Proof. Consider a), and suppose first that p is odd. Since the characteristic of the fieldis odd, we may find the solutions by the standard quadratic formula. We obtain the solutions x = [ − √− , [ − −√− √− GF ( q ) by Corollary22. These solutions are distinct since p is odd.Now suppose p = 2. Recall the trace function T r GF ( q ) /GF (2) ( x ) = (cid:80) m − i =0 x i , defined forany x ∈ GF ( q ), which we abbreviate by T r ( x ). It can be shown (see [21]) that the quadraticequation ax + bx + c = 0, with a, b, c ∈ GF (2 m ), a (cid:54) = 0, has two distinct solutions in GF (2 m ) ifand only if b (cid:54) = 0 and T r ( acb ) = 0 . In our case we have a = b = c = 1, so acb = 1. Since p = 2 and q ≡ m must be even. Thus there are an even number of terms in the sum defining T r ( x ), each of them equal to 1. So since the characteristic is 2, we get T r ( acb ) = 0 in our case.It follows that x + x + 1 = 0 has two distinct solutions when p = 2, as required.Observe that if x is a root of of x + x + 1 = 0, then by direct substitution so is x . To showthat x and x are distinct, assume not. Then x = 1 or −
1. If q is even, then x + x + 1 = 0implies that 1 = 0 since the characteristic of the field is 2, a contradiction. Assume q is odd.Then if x = 1 we get 1 + 1 + 1 = 0, implying q ≡ x = −
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