On the size of autotopism groups of Latin squares.

Transcription

1 On the size of autotopism groups of Latin squares. Douglas Stones and Ian M. Wanless School of Mathematical Sciences Monash University VIC 3800 Australia {douglas.stones, 1 Introduction A Latin square of order n is a n n array L = (l i j ) of n symbols such that each symbol occurs exactly once in each row and column. For example, the Cayley table of any group is a Latin square. We shall usually take the row and column indices and the symbol set to be Z n. A Latin square on the symbols Z n is called reduced if the first row is (0, 1,..., n 1) and the first column is (0, 1,..., n 1) T. Let L n denote the number of Latin squares of order n and R n denote the number of reduced Latin squares of order n. Then L n = n!(n 1)!R n. (1) In [6] it is shown that R n is divisible by n/! and if n is odd and n/ + 1 is composite then ( n/ + 1)! divides R n. In [1] we show R n 1 (mod n) if n is prime and R n 0 (mod n) if n is composite. In [11] we find a congruence for R n (mod p) involving the number of partial orthomorphisms, for prime p satisfying p < n < p. In [10] we show that if n is an odd prime number then R n+1 (mod n) and if n is a composite number then R n+1 0 (mod n). For any Latin square, L, an ordered triplet of permutations ζ = (α, β, γ) shall denote a mapping of L such that the rows of L are permuted according to α, the columns of L are permuted according to β and the symbols of L are permuted according to γ. The mapping ζ is called an isotopism. If α = β = γ, then ζ is said to be an isomorphism. The identity permutation shall be denoted ε. Any isotopism other than (ε, ε, ε) is non-trivial. Let L and L be Latin squares. If there exists an isotopism, ζ, such that ζ(l) = L then L and L are said to be isotopic. The set of all Latin squares isotopic to L is called the isotopy class of L, denoted Red(L). If I is an isotopy class, define Red(I) = Red(L) for any L I. If ζ(l) = L, then ζ is said to be an autotopism of L. The group of all autotopisms of L is called the autotopism group of L, denoted Is(L). Supported by ARC grant DP

2 If ζ is an isomorphism and an autotopism of L then ζ is said to be an automorphism of L. The group of all automorphisms of L is called the automorphism group of L, denoted Aut(L). The notions of group isomorphism and isomorphism of Latin squares that are group tables are equivalent. In fact, we shall use Is(G) to denote the autotopism group of the Cayley table of the group G. Let L be a Latin square of order n. It follows that Define Red(L) = n!n Is(L). () τ(n) = gcd( Red(L) ), (3) L where the gcd is over all Latin squares L of order n. Therefore τ(n) divides R n. The values of τ(n) for 1 n 10 are given in Table 1 and were obtained from the data by McKay, Meynert and Myrvold [4]. By using the template in [1] it is straight-forward to show that ( n/ 1)! divides τ(n) for n 3. We also showed in [1] that Red(Z p ) = (p )! for prime p. n τ(n) Table 1: The values of τ(n) in (3) for 1 n 10 along with its prime factorisation. Factorisation of the θ function For all primes p and a 0, let θ(p a ) = (p a 1)(p a p)... (p a p a 1 ) and for relatively prime d, t N, let θ(dt) = θ(d)θ(t) (Sloane s [8] A061350). In particular, θ(p a ) = Aut(Z a p). Let G be a group of cardinality n. Neumann [7] completed the proof of a conjecture of Birkhoff and Hall [] that states Aut(G) divides nθ(n). Therefore, Is(G) divides n 3 θ(n) and so (n 1)! divides nθ(n) Red(G), by (). For any prime q and natural number t, let ORD q (t) be the largest integer such that q ORD q(t) divides t. Hence ORD q ( Red(G) ) ORD q ((n 1)!) ORD q (θ(n)) ORD q (n). (4) Equation (4) motivates us to seek the value of ORD q (θ(n)). Since θ is multiplicative, it is sufficient to find ORD q (θ(p a )) for arbitrary prime p. For prime p and d indivisible by p let Ord(d, p) be the multiplicative order of p modulo d.

3 Theorem 1. Let p and q be primes and a 0. Then 1a(a 1) if p = q ORD q (θ(p a )) = l 1 a + l a + a k if q = p k l a + a k 1 otherwise, Ord(p,q) Ord(p,q) q k where l 1 and l are the largest integers satisfying p 1 (mod l 1 ) and p 1 (mod l 1+l ) and l is the largest integer such that p Ord(p,q) 1 (mod q l ). Proof. Since θ satisfies θ(p a ) = (p a 1)p a 1 θ(p a 1 ), we get and therefore θ(p a ) = p a(a 1) ORD p (θ(p a )) = 1 a(a 1). So now assume that p and q are distinct primes. It follows from (5) that a (p i 1) (5) i=1 ORD q (θ(p a )) = m 1 a Ord(p, q m ). (6) Case I: q = p. For m 1 observe that Ord(p, m+1 ) {Ord(p, m ), Ord(p, m )} and that Ord(p, ) = 1. Claim: Ord(p, m+1 ) = Ord(p, m ) whenever Ord(p, m ) 4. Assume, for the sake of contradiction, that Ord(p, m ) 4 and p Ord(p,m ) 1 (mod m+1 ). Then p Ord(p,m) 1 = (p Ord(p,m )/ + 1)(p Ord(p,m )/ 1) is divisible by m+1. Since 4 divides Ord(p, m ) we have that m divides p Ord(p,m )/ 1 contradicting that Ord(p, m ) is the order of p modulo m. Thus proving the claim. So (Ord(p, m )) m 1 = (1, } 1, {{..., } 1, },, {{..., },, 3,... ). l 1 l The result now follows from (6). Case II: q. Observe that Ord(p, q m+1 ) {Ord(p, q m ), qord(p, q m )} for all m 1. Claim: Ord(p, q m+1 ) = qord(p, q m ) whenever Ord(p, q m ) = qord(p, q m 1 ). Assume, for the sake of contradiction, that Ord(p, q m+1 ) = Ord(p, q m ) = qord(p, q m 1 ). Then p qord(p,qm 1) 1 = (p Ord(p,qm 1) 1) q 1 i=0 piord(p,qm 1) is divisible by q m+1. Since q 1 i=0 (pord(p,qm 1) ) i q 1 i=0 1 = q (mod q ), we know q does not divide q 1 )i i=0 pord(p,qm 1. Therefore q m divides p Ord(p,qm 1) 1 giving a contradiction. Thus proving the claim. Therefore the sequence (Ord(p, q m )) m 1 is given by The result now follows from (6). Ord(p, q)(1, } 1, {{..., } 1, q, q, q 3, q 4,... ). l 3

4 Theorem. Let q be a prime. If G is a group of cardinality n then ORD q ( Red(G) ) = O(log (n)). Proof. Observe that ORD q ((n 1)!) = n 1 k 1 q = n O(log(n)). By (4) it is sufficient to k q 1 n q 1 show that ORD q (θ(n)) O(log (n)). Let P n denote the set of prime divisors of n and for p P n let a = a(p, n) be the largest integer such that p a divides n. Since θ is a multiplicative function, ORD q (θ(n)) = p P n ORD q (θ(p a )). So ORD q (θ(n)) 1 log q (n)(log q (n) 1) + P n max p P n \{q} (ORD q(θ(p a ))) by Theorem 1. Since P n O(log(n)), it is now sufficient to show that ORD q (θ(p a )) O(log(p a )) whenever q and p are distinct primes. However this is implied by the simple upper bounds l 1 < log (p), l < log (p) and l < Ord(p, q) log q (p) and Theorem 1. In Figure 1 we show the value of ORD q (R n ) for q {, 3} and n 11. It seems reasonable to suspect that group tables have asymptotically the largest autotopism groups of all Latin squares. In that case, they would have asymptotically the fewest reduced Latin squares within each isotopy class and the right-hand side of (4) would provide a lower bound on ORD q (R n ) as n. Unfortunately, there are far too few known values of R n for Figure 1 to provide significant guidance. The peculiarly large value of ORD (R n ), as noticed by Alter [1] over forty years ago, remains one of the largest unsolved mysteries of the study of Latin squares. q = q = Figure 1: The value of ORD q (R n ) for q {, 3} and n 11. For 0 n 0, the dotted line is the lower bound for ORD q (R n ) in [6] and the solid line is the right-hand side of (4). 3 The maximum number of subsquares of a Latin square If a submatrix M of a Latin square L is also a Latin square then M is called a subsquare of L. Let I k (n) denote the maximum number of subsquares of order k in a Latin square of order n. An entry of L is an ordered triplet (i, j, l i j ) corresponding to row i, column j and symbol l i j. A subsquare of order is called an intercalate. Heinrich and Wallis [3, 5] showed that 1 45 n3 I (n) 1 4 n3 1 4 n for n 4. The asymptotic expectation of the number of intercalates 4

5 are given by [5]; in particular, for ɛ > 0, most Latin squares of order n have n 3/ ɛ intercalates. The values of I (n) for 1 n 9 are given in Table (Sloane s [8] A0937). n I (n) Table : The values of I (n) for 1 n 9 and a lower bound on I (n) for 10 n 1. 1 Lemma 1. Let ζ = (α, β, γ) be an autotopism of L such that α, β and γ all have at least one fixed point. Then the rows and columns fixed by ζ define a subsquare of L. Lemma 1 follows from a result of McKay, Meynert and Myrvold [4]. We shall also need the following simple observation. Lemma. If L is a Latin square of order n and M is a subsquare of L of order m. Then 1 m n/ or m = n. Consider now subsquares of order k. In a Latin square of order n, we can choose a pair of entries containing the same symbol a in n ( n ) ways. These two entries are contained within a unique smallest subsquare M 1. If M 1 is of order k then stop, otherwise choose another entry containing the symbol a outside of M 1. These three entries are contained within a unique smallest subsquare M and the order of M is at least twice the order of M 1. We continue this process until we find a subsquare M t of order k. The order of M t is at least t, so we require no more than log (k) steps. Therefore I k (n) 1 n (n 1)(n )(n )... (n log (k) ) n log (k)+ (7) In the case when k = 3, any two entries containing the same symbol a can exist only within a unique subsquare of order 3. Furthermore, every subsquare would be counted 9 times in this process. Therefore we can improve (7) to I 3 (n) 1 18 n n. It follows from (7) that n/ k=1 I k (n) n/ n log ( n/ )+ n log (n)+. (8) A formula for the number of subgroups of prime power order in an Abelian group is given by Stehling [9], for example. In particular, the number of subgroups of cardinality p r in (Z p ) a is asymptotic to p ra r. It follows that I k (n) O(n log(k) ) when k is a prime power and therefore n/ k=1 I k(n) O(n log(n) ). Theorem 3. Let q be a prime. Then ORD q (R n ) n q 1 O(log3 (n)). Proof. Let L be an arbitrary Latin square of order n and let S = Is(L). Let x be an entry of L and let S x be the stabiliser of x in S, that is S x = {ζ S : ζ(x) = x}. The Orbit-Stabiliser Theorem implies that S n S x since there are n entries in L. 5

6 Let M be the set of subsquares of L containing x. For any ζ S x there exists a subsquare M M of L formed by the fixed rows and columns of ζ, by Lemma 1. For any M M let S M denote the stabiliser of M in S, that is S M = {ζ S : ζ(m) = m for all entries m M}. Therefore n/ S x S M S M I k (n) n log (n)+ S M, (9) M M k=1 by (8), for any M M such that S M = max M M S M. If M = L then S M = 1 and we are done. Otherwise, let y be an entry of L in the same row as x outside of M. Let S M,y be the stabiliser of y in S M, that is S M,y = {ζ S M : ζ(y) = y}. For all ζ S M we have S M,y = {ζ ζ 1 : ζ S M and ζ(y) = ζ (y)}. Therefore S M (n 1) S M,y. Similarly to (9), it follows that S M,y S M n log (n)+ S M, M M M M y M by (8), for any M M such that M M, y M and S M = max S M. M M M M y M Importantly, Lemma implies that the order of M is at least twice the size of the order of M. We can repeat this process of choosing an entry z outside of the current subsquare in the same row as x and finding a subsquare containing z with a stabiliser of maximal size. After at most log (n) steps we will have found a subsquare at least half the size of the Latin square L itself. By Lemma 1 we can then stop. Combining all these upper bounds yields S n ( (n 1)n log (n)+) log (n) n log (n)+3 log (n)+. Therefore ORD q ( S ) log q (n)(log (n) + 3 log (n) + ). Note that n ORD q (n!) = n q k q 1 log q (n). By () and (3), ORD q (R n ) k 1 n log q 1 q (n)( 1 log (n) + 3 log (n) + 3). References [1] R. ALTER, How many Latin squares are there?, Amer. Math. Monthly, 8 (1975), pp [] G. BIRKHOFF AND P. HALL, On the order of groups of automorphisms, Trans. Amer. Math. Soc., 39 (1936), pp

7 [3] K. HEINRICH AND W. D. WALLIS, The maximum number of intercalates in a Latin square, in Combinatorial Mathematics VIII, Geelong, 1980, vol. 884, Springer, 1981, pp [4] B. D. MCKAY, A. MEYNERT, AND W. MYRVOLD, Small Latin squares, quasigroups, and loops, J. Combin. Des., 15 (007), pp [5] B. D. MCKAY AND I. M. WANLESS, Most Latin squares have many subsquares, J. Combin. Theory Ser. A, 86 (1999), pp [6], On the number of Latin squares, Ann. Comb., 9 (005), pp [7] P. M. NEUMANN, Proof of a conjecture by Garret Birkhoff and Philip Hall on the automorphisms of a finite group, Bull. London Math. Soc., 7 (1995), pp. 4. [8] N. J. A. SLOANE, On-line encyclopedia of integer sequences. njas/sequences/. [9] T. STEHLING, On computing the number of subgroups of a finite Abelian group, Combinatorica, 1 (199), pp [10] D. STONES AND I. M. WANLESS, Compound orthomorphisms of the cyclic group. In preparation. [11], A congruence connecting Latin rectangles and partial orthomorphisms. Submitted, 10 pp. [1], Divisors of the number of Latin rectangles. Submitted, 16 pp. 7