The case of equality in the Livingstone-Wagner Theorem

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1 J Algebr Comb 009 9: 15 7 DOI /s The case of equality in the Livingstone-Wagner Theorem David Bundy Sarah Hart Received: 1 January 007 / Accepted: 1 February 008 / Published online: 4 March 008 Springer Science+Business Media, LLC 008 Abstract Let G be a permutation group acting on a set of size n N and let 1 k<n 1/. Livingstone and Wagner proved that the number of orbits of G on k-subsets of is less than or equal to the number of orbits on k + 1-subsets. We investigate the cases when equality occurs. Keywords Livingstone-Wagner Theorem Permutation groups Orbits Partitions 1 Introduction Throughout this article we let G be a permutation group acting on a set of size n N and let 1 k<n 1/. In [7] Livingstone and Wagner proved the following theorem. Theorem 1.1 Livingstone, Wagner [7] The number of orbits of G on k-subsets of is less than or equal to the number of orbits on k + 1-subsets. Alternative proofs were subsequently given by Robinson [8] and Cameron [1]who extended the result to infinite. An investigation of the cases when equality occurs for infinite was then made by Cameron [], [] and Cameron and Thomas [5]. The case of equality also follows from a stronger interchange property examined by Cameron, Neumann and Saxl [4]. In this article, we will prove some similar results about the case of equality when is finite. D. Bundy Mathematisches Seminar, Universität zu Kiel, Ludewig-Meyn Straße 4, Kiel 4098, Germany bundy@math.uni-kiel.de S. Hart School of Economics, Mathematics and Statistics, Birkbeck, University of London, Malet Street, London WC1E 7HX, UK

2 16 J Algebr Comb 009 9: 15 7 In Section we consider the case when G is intransitive. We show see Lemma.1 that G must have one orbit of length at least n k and see Proposition. that the action of G on this orbit satisfies a strong condition which in almost all cases forces G to be k-homogeneous on this orbit. Transitive but imprimitive groups are then investigated in Section. In this case there are too many examples for a complete classification to be feasible, so we concentrate on finding a necessary condition for the sizes and number of blocks in a system of imprimitivity. This quickly reduces to a combinatorial problem of determining when the number of partitions of k into at most r parts of size at most s is the same as for k + 1. This problem is also of independent interest in invariant theory, where such partitions can be used to count the number of linearly independent semi-invariants of degree r and weight k of a binary form of degree s. We are able to determine all the cases of equality for r 4 see Theorem.1 and conjecture that for s r 5, there are only finitely many cases of equality see Conjecture. for details. Theorem.7 shows that for s r 5, equality can only occur when k rs 1 1, that is k is close to half n. We have strong experimental evidence for believing Conjecture. to be true. We observe that for large enough fixed r and s the number of partitions of k into at most r parts of size at most s approximates to a Gaussian distribution whose peak becomes sharper for larger r and s. In the final section we make some observations about the case when G is primitive. Aside from k + 1-homogeneous groups the only examples we know are the affine general linear groups over a field of size see Proposition 4. and a list of 19 further examples of degree at most 4, many of which are subgroups of M 4. The absence in [4] of any examples of degree greater than 4 suggests that such examples may also be rare or non-existent in our situation. Notation and preliminary results For each 0 l n, letσ l G be the number of orbits of G on the set of l-subsets of. A permutation group is said to be l-homogeneous if it is transitive in its action on l-subsets, that is σ l G = 1. Let be a G-invariant subset of. Then G will denote the permutation group induced by G in its action on. Let H be a subgroup of a group G, χ be a character of G and ψ a character of H. Then χ H will denote the restriction of χ to H and ψ G will denote the character induced by ψ on G. Furthermore 1 G will denote the trivial character on G. Lemma 1. Let G Symn, 0 l n and ψ l be the character of Symn induced by the trivial character on Syml Symn l. Then ψ l G, 1 G is the number of orbits of G on l-subsets of {1,...,n} and if 0 l<n 1/, then ψ l+1 ψ l is an irreducible character of Symn. Proof See [8]. Lemma 1. Let H G Symn and 1 k<n 1/. Then σ k+1 G σ k G σ k+1 H σ k H. In particular, if σ k+1 H = σ k H, then σ k+1 G = σ k G.

3 J Algebr Comb 009 9: Proof Let χ := ψ k+1 ψ k Lemma 1.. Then be the irreducible character in the conclusion of σ k+1 G σ k G = χ G, 1 G χ H,1 H =σ k+1 H σ k H. In particular, if σ k+1 H = σ k H, then the right-hand side is zero and by Theorem 1.1 the left-hand side is non-negative, so must also be zero. Intransitive groups with equality In this section we investigate intransitive permutation groups which achieve equality in the Livingstone-Wagner Theorem. Lemma.1 Let G Symn and suppose σ k G = σ k+1 G for some 1 k< n 1/. Then G has an orbit of length at least n k. Proof Suppose G has no orbit of length at least n k.ifg Symn l Syml =: M, forsomek<l<n k, then σ k+1 M = k + >k+ 1 = σ k M, which contradicts Lemma 1.. So the sum of the lengths of any set of orbits of G is either at most k or at least n k. In particular, each orbit of G has length at most k and G has at least three orbits. We claim that there exists a subgroup M of Symn containing G isomorphic to Syml 1 Syml Syml, where n = l 1 + l + l and l i k for each 1 i. Let g 1 and g be the lengths of two distinct orbits of G. Ifg 1 + g n k, then the total length of the remaining orbits is at most k. Then there exists M isomorphic to Symg 1 Symg Symn g 1 g, which fulfils the claim. Otherwise g 1 +g k and there exists a group G containing G which operates as Symg 1 + g on the union of these two orbits and operates in the same way as G on the other orbits. We then replace G by G and repeat the argument to find M. We may assume without loss that l 1 l l. It remains to show that σ k M < σ k+1 M, which yields a contradiction to Lemma 1.. Note that σ k M is the number of tuples a 1,a,a such that k = a 1 + a + a and a i l i, for each 1 i. For each such tuple, a 1 + 1,a,a corresponds to a suitable partition of k + 1, except in those cases when a 1 = l 1. The number of such exceptional tuples is k l 1 + 1, because l 1 + l n k>kimplies that k l 1 <l. The tuples of the form 0,a,a, where a + a = k + 1 do not correspond to any partition of k as above. The number of such tuples is l + l k, because a can range between k + 1 l and l. Since l + l k k l = n k + 1>0, this shows that σ k+1 M > σ k M as required. Proposition. Let G Symn and 1 k<n 1/ with σ k G = σ k+1 G. Let be an orbit of G of length at least n k. Then σ l G = σ l+1 G, for all k n l mink, k. Proof Note that an orbit of length at least n k exists by Lemma.1. LetM := G Sym \ G and let m :=. Fort N, twot-subsets of are in the

4 18 J Algebr Comb 009 9: 15 7 same M-orbit if and only if their intersections with are in the same G -orbit. In particular, these intersections must be of the same size. Hence σ t M = mint,m l=max0,t n m σ l G. Now m n k k + 1 k = k + 1. Also k n m k + n k n = 0. Therefore 0 = σ k+1 M σ k M = k+1 l=k+1 n m = σ k+1 G σ k n m G. σ l G k l=k n m σ l G That is, σ k+1 G = σ k n m G.Ifk<m 1 then the Livingstone-Wagner Theorem forces σ l G = σ l+1 G, for each k n m l k. On the other hand, suppose k m 1. Then σ k+1 G = σ m k+1 G and m k + 1 is within the range to which the Livingstone-Wagner Theorem applies. We also have that m k + 1 k n m = n 1 k>0. Hence, by the Livingstone-Wagner Theorem, σ l G = σ l+1 G, for each k n m l m k. Note that mink, m k is k precisely when k<m 1and m k otherwise, so the proof is complete. Proposition. provides the means to reduce the case of equality for an intransitive group to that of equality for a transitive group. Indeed if G is intransitive with an orbit satisfying the condition of Proposition., then we nearly always have equality σ l G = σ l+1 G for several consecutive values of l. If there is just one value of l then either G is already transitive or n = k +. This almost forces G to be k- homogeneous. The only known exceptions with k<n 1/ are where G = M 4 or M. Imprimitive groups with equality There is an abundance of imprimitive groups which achieve equality in the Livingstone-Wagner Theorem and a complete classification of them seems intractable. Nevertheless, we are able to give a condition on the block sizes which is necessary if equality in the Livingstone-Wagner Theorem holds. Observe that by Lemma 1., if σ k H = σ k+1 H holds for an imprimitive group H with r blocks of size s, then σ k G = σ k+1 G, where G = Syms Symr is the full stabiliser in Symrs of the blocks of H. Note also that the number of orbits of G on k-subsets is equal to the number of ways, Pr,s,k, to partition k into at most r parts of size at most s. We require Pr,s,k= Pr,s,k+ 1. The following result is established by Lemma.5, Proposition.6 and Proposition.9.

5 J Algebr Comb 009 9: Theorem.1 Let r {,, 4} with r s and 1 k<rs 1/. Then Pr,s,k= Pr,s,k+ 1 if and only if one of the following holds. a r = and k is even. b r = and k = s, if s is odd, c r = 4 and k = s or r = s = k = 4. We also make the following conjecture. s 4, if s 0 mod 4, s or s 6, if s mod 4. Conjecture. Let 1 <r s, 1 k<rs 1/ and suppose Pr,s,k = Pr,s,k+ 1. Then one of the following holds: a r {,, 4} and the possibilities for s and k are as in Theorem.1; or b r, s and k have the values given by a column of the following table r s k Remark. The quantity Pr,s,k Pr,s,k 1 is of interest in invariant theory. By a theorem of Cayley and Sylvester see Satz.1 of [9] it is equal to the number of linearly independent semi-invariants of degree r and weight k of a binary form of degree s. Conjecture., if proven, would then give the values of r, s and k for which no such semi-invariant exists. We now define some more notation which we will use in this section. Let Pr,s,k be the set of partitions of k into at most r parts of size at most s, sopr,s,k= Pr,s,k. We will use the convention that Pr,s,k= 0ifk<0ork>rs.By considering dual partitions we observe that Pr,s,k= Ps,r,k, so without loss we will assume that r s. Elements of Pr,s,k will be written a 1,a,...,a r where ri=1 a i = k and s a 1 a r 0. Let Ar,s,k be the subset of Pr,s,k consisting of all partitions of the form s, a,...,a r and let Br,s,k + 1 be the subset of Pr,s,k+ 1 consisting of all partitions of the form x,x,a,...,a r,for some x s. Furthermore, let Ar, s, k = Ar,s,k and Br,s,k = Br,s,k. Note that Ar, s, k = Pr 1,s,k s. We will define a bijection from a subset of Pr,s,kto a subset of Pr,s,k+ 1.Leta 1,a,...,a r Pr,s,kwith s>a 1 a... a r 0, and define fa 1,a,...,a r = a 1 + 1,a,...,a r. Then f is a bijection from Pr,s,k\ Ar,s,k to Pr,s,k+ 1 \ Br,s,k+ 1. In particular we have the following result. Lemma.4 Let r, s, k 1. Then Pr,s,k+ 1 Pr,s,k= Br,s,k + 1 Ar, s, k.

6 0 J Algebr Comb 009 9: 15 7 So the problem of determining when Pr,s,k= Pr,s,k+ 1 reduces to that of determining when Br,s,k+1 = Ar, s, k. We now consider in turn the cases when r =, and 4. Lemma.5 Let s 0. Then 0, if k>s, or k<0, P,s,k= s k + 1, if s k s, k + 1, if 0 k s. In particular, if 1 k<s, then P,s,k= P,s,k+ 1 if and only if k is even. Proof Elementary. Proposition.6 Let s and 1 k<s 1/. Then P,s,k= P,s,k+ 1 if and only if one of the following holds: a s is odd and k = s /, b s 0 mod 4 and k = s 4/, c s mod 4 and k = s / or s 6/. Proof Let d k = P,s,k+1 P,s,k= B,s,k+1 A,s,k. By Lemma.5, { k s A,s,k= P,s,k s = + 1 if s k<s 1/, 0 if k<s. Moreover, k + 1 B,s,k+ 1 = {a,a,b s a b,a + b = k + 1} = Hence B,s,k+ 1 k k = k 6 > 0. So if A,s,k= 0, then d k k/6 > 0. We may therefore assume that k s s k<s 1/ and A,s,k= + 1. Thus k + 1 k + 1 k s d k =. Suppose s is odd. Then k + 1 k s mod. Hence k + 1 k s k + 1 d k = = s + 1 k + 1. k

7 J Algebr Comb 009 9: Therefore { s + 1 d k = 0 k, s 1, s }. Since k<s 1/, this forces k = s. Suppose s is even. Then d k = d k k k + k s = 1 s k 6. 6 Assume d k = 0. Then k s 6. Thus s/ k s/ 1 and so Therefore { k s k s = 0, if k is even k+1 s k s 1 = 1, if k is odd, a contradiction. Thus k is even, s 6 k s and hence { s 4 k = if s 0 mod 4 s or s 6 if s mod 4. k+1 = s. Theorem.7 Let 4 r s and 1 k<rs 1/. If Pr,s,k= Pr,s,k + 1, then k rs 1 1/ or r = s = k = 4. Proof Suppose first that k<s. Then Ar, s, k = 0butBr,s,k > 0, since r 4. Therefore by Lemma.4 Pr,s,k <Pr,s,k + 1. Now suppose that k = s 5. Then Pr,s,k= Pr,k,k = + Pr,k,k and Pr,s,k+ 1 = Pr,k,k+ 1 = + Pr,k,k+ 1. Since rk 1/ 4k 1/ = k 9/ >k, applying Theorem 1.1 yields Pr,k,k Pr,k,k + 1 and so Pr,s,k<Pr,s,k + 1 in this case. It remains to show for s < k < rs 1 1/ that Pr,s,k<Pr,s,k + 1. So we assume for a contradiction that Pr,s,k= Pr,s,k+ 1 in this case. Observe that Pr,s,k= Pr,s 1,k+ Pr 1,s,k s. Since k < rs 1 1/ and k s < rs 1 1 s/ < r 1s 1/, by Theorem 1.1, Pr,s 1,k Pr,s 1,k+ 1 and Pr 1,s,k s Pr 1, s,k s + 1. So under our assumption we have Pr 1,s,k s = Pr 1,s, k s + 1. We now proceed by induction on r.

8 J Algebr Comb 009 9: 15 7 Suppose first that r = 4. Then by Proposition.6, P,s,k s = P,s,k s +1 implies s/ k s s/ 1. However k<4s 1 1/ = s 5/, so k s s < s/, a contradiction. Now suppose r>4 and the result holds for r 1 in place of r. Since Pr 1,s, k s = Pr 1,s,k s + 1, we obtain by induction that k s r 1s 1 1 = rs r s. Hence k rs r + s/ >rs 1/, a contradiction. Therefore by induction the result holds for all r 4. Proposition.8 Let s 4 and s k s 1. Then P4,s,k= P4,s,k+ 1 if and only if k = s. Proof Since r = 4 is fixed, for this proof we will abbreviate Ar, s, k by As, k and Br,s,k by Bs,k. We first show that for all s 4, P4,s,s = P4,s,s 1. We need to evaluate Bs,k more precisely. Now Bs, k ={a,a,b,c: s a b c 0, a + b + c = k}. Now 0 b + c a implies a k 4a. Hence k 4 a k. Thus Bs,k = k a= k 4 P,a,k a. By Lemma.5, thevalueofp,a,k a depends on whether 0 k a a or a k a a. Nowa k a = 4a k 0. Also k a a whenever a k. Therefore by Lemma.5 It follows that Bs,s 1 = Bs,k = s 1 a= s 1 4 s 1 k a= k 4 a k a +1 + a s 1 a +1 + s 1 = a s a= s = 1 s s 1 s +1 +s s 1 s 1 a= s 1 +1 k a= k +1 s 1 a= s 1 +1 s a + s 1 s 1 1 s 1s + 1 s 1 k a s 1 a s s 1 s 1 +1

9 J Algebr Comb 009 9: 15 7 = s 1 s s s s s +s s + s s 1 s + 1 s Bs,s 1 = 1 s 1 s s + s s s + 1 s s + }{{}}{{} X B X B We now work out As, s in a similar fashion. Firstly note that As, s = P,s,s = P,s,s, and P,s,s = #{a,b,c : a b c 0,a+ b + c = s }. This implies that s a s. Thus As, s = s a= s P,a, s a. From Lemma.5, and noting that s a a when a s,we make the following calculation. As, s = = = s a= s s 1 a= s s 1 a s a +1 + a s a + a= s a s + + s a= s s a= s +1 s a= s s a s a +1 s a { { } } 1 # i,...,s s : i odd Now the number of odd numbers in the range {,...,x} is { } s+ odd numbers in,...,s s 1 is = As, s = s s + s s s s s s s 1 6 x 1 = + s s 1 s s s 1 s 1, so the number of. Therefore s 1 = s s + s 1 + s s s s 1 + s s s s 1 s 1 s 1

10 4 J Algebr Comb 009 9: 15 7 As, s = s s + 1 s s 1s + }{{} + 1 s 4 X A s 5 s 1 s 1 }{{} X A Now P4,s,s = P4,s,s 1 if and only if Bs,s 1 = As, s, which is if and only if X B X A = X A X B. Wehave X B X A = s s s + 1 s s + s s + 1 s s 1s +. A simple calculation shows that regardless of whether s is odd or even, X B X A = 1 4 s 9s + 6. X A X B = 1 s 1 14 s s s s 5.. s 1 s 1 Calculating for each possible value of s modulo shows that in each case, X A X B = 1 4 s 9s + 6 = X B X A. Therefore, for all s 4, P4,s,s = P4,s,s 1. We now show that P4,s,s 1 <P4,s,s for all s 4. Since P4,s, s = P4,s,s 1 for all s 4, by substituting s + 1fors in Lemma.4 we have As + 1, s = Bs + 1, s + 1. Now As, s 1 = P,s,s 1 = P,s 1,s 1 as no part of a partition of s 1 can exceed s 1. Similarly As + 1, s = P,s + 1,s 1 = P,s 1,s 1. Hence As, s 1 = As + 1, s. Now we consider Bs + 1, s + 1 compared to Bs,s. Setting k + 1 = s and k + 1 = s + 1in1, respectively, gives: Bs,s = = s a= s a s a s a= s a s s a= s +1 s a + 1 s a= s +1 s a + 1 ;

11 J Algebr Comb 009 9: Bs + 1, s + 1 = If s = s+1 If s < s+1 = s+1 a= s+1 4 s+1 a= s+1 a s a a s + s a= s+1 +1 s a= s+1 +1 s a + 1., then Bs,s Bs+1, s +1 s+1/ a= s/ then s s =, s+1 = s+1 and s Bs,s Bs + 1, s + 1 s+1 s + a= s 1 s s s s 1 4s+ + s s s a s 1 s 1 > 0. = 1 6 s + 8s 4 + 1s = 1 6 s 1>0. Thus in any case Bs,s > Bs + 1, s + 1. Therefore by equations and, Bs,s As, s 1>Bs+ 1, s + 1 As + 1, s = 0. Hence by Lemma.4 P4,s,s 1<P4,s,s. Proposition.9 Let s 4 and 1 k s 1. Then P4,s,k= P4,s,k + 1 if and only if k = s or s = k = 4. Proof In the case s = k = 4 it can be easily computed that P4, 4, 4 = P4, 4, 5 = 5. Otherwise, by Theorem.7, ifp4,s,k= P4,s,k+ 1, then 4s 1 1 k and, since k is an integer, s k. We may now apply Proposition.8 to get the result. Theorem.1 now follows immediately from Lemma.5, Proposition.6 and Proposition.9. 4 Primitive groups with equality Primitive groups which are not k + 1-homogeneous but achieve equality in the Livingstone-Wagner Theorem for some k<n 1/ are fairly rare. Searching the database of primitive groups in GAP [6] for degrees up to 8 produced the list given below. The lack of any primitive examples in the more special situation in [4] for degrees greater than 4 suggests very tentatively that this may be the complete list.

12 6 J Algebr Comb 009 9: 15 7 It might be possible to prove such a result using the O Nan-Scott Theorem and the Classification of Finite Simple Groups, but such a proof would certainly be very labourious and we do not attempt this here. The difficulty in finding a more enlightening approach to this problem is that one would like to exploit the high degree of transitivity in the examples below. In particular it would be useful to have a result relating the number of orbits of a group to that of its point-stabilizers, but a simple relationship in general does not appear to exist. We therefore leave this problem open. Remark 4.1 The known primitive but not k + 1-homogeneous groups G such that σ k G = σ k+1 G, forsomek<n 1/, are: a AGLm,, form 4,n= m,k= 4, b ASL, or AGL,, forn = 9,k=, c Sym5, Sym6, P GL, 9 or PƔL, 9, forn = 10,k= 4, d M 11,PSL, 11, P GL, 11, forn = 1,k= 4, e PSL,,forn = 1,k= 4, f PGL, 1,forn = 14,k= 4, g 4 : Alt6, 4 : Sym6, 4 : Alt7,forn = 16,k= 6, h PGL, 17,forn = 18,k= 6or8, i M or AutM,forn =,k= 8, j M,forn =,k= 8, 9, k M 4,forn = 4,k= 6, 8, 9or10. Observe that many of these groups are subgroups of M 4. Regarding case a, we prove the following. Proposition 4. Let G = AGLm,, for m 4, acting naturally on an m- dimensional vector space V over GF. Then σ 4 G = σ 5 G =. Proof Observe that the stabiliser in G of any three points of V fixes the fourth point in the unique affine plane containing these three points and is transitive on the remaining points of V. It follows that σ 4 G = and also G has a single orbit on the set of 5- subsets which contain affine planes. Let be any set of five distinct points in V which does not contain any affine plane. Then is not contained in an affine -dimensional subspace of V. Furthermore the stabiliser in G of an affine -dimensional subspace W is transitive on pairs α,, where α is a point not in W and is any set of four points in W which is not an affine plane. Therefore G has a single orbit on 5-subsets which do not contain any affine plane. Thus σ 5 G =. References 1. Cameron, P.J.: Transitivity of permutation groups on unordered sets. Math. Z. 148, Cameron, P.J.: Orbits of permutation groups on unordered sets. J. London Math. Soc. 17, Cameron, P.J.: Orbits of permutation groups on unordered sets. II. J. London Math. Soc.,

13 J Algebr Comb 009 9: Cameron, P.J., Neumann, P.M., Saxl, J.: An interchange property in finite permutation groups. Bull. London Math. Soc. 11, Cameron, P.J., Thomas, S.: Groups acting on unordered sets. Proc. London Math. Soc. 59, GAP Groups, Algorithms and Programming, Version Livingstone, D., Wagner, A.: Transitivity of finite permutation groups on unordered sets. Math. Z. 90, Robinson, G. de B.: Note on a theorem of Livingstone and Wagner. Math. Z. 10, Schur, I.: Vorlesungen Über Invariantentheorie. Bearbeitet und herausgegeben von Helmut Grunsky. Die Grundlehren der mathematischen Wissenschaften, Band 14. Springer, Berlin 1968