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The Number 6

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The Number 6 The number 6 Gabriel Coutinho 2013 Abstract The number 6 has a unique and distinguished property. It is the only natural number n for which there is a construction of n isomor- phic objects on a set with n elements, invariant under all permutations of the set, but not in natural bijection with the set. I will talk about this fact and a closely related result. This is based on Chapter 6 of Cameron and Van Lint [1]. 1 A distinguished property of 6 Theorem 1. The number 6 is the only natural number n that satisfies the following property: If E is an n-set, then there is a construction of exactly n isomor- phic objects on E whose union is invariant under the permuta- tions of E, but none correspond in natural way to an element of E. Some clarification is needed here. Neither the meaning of object nor that of natural is clear. I will make it even worse, and provide a bad definition for one, and an incomplete for the other. An object on a set E is a construction that can be made using the elements of E and set-theoretical operations, such as taking the union. An object on E corresponds naturally to a point of E if the automorphism group of the object is the stabilizer of that point in Sn. Despite such abstract formulation, we will see that this theorem has a very concrete interpretation. In particular, the \shape" of such objects will arise naturally throughout the proof. The idea of the proof is to analyse the automorphism group the objects defined by 1 the automorphisms of the ground set which do not change the object. The skeleton and the muscle of the proof is in Cameron and Van Lint [1]. Here I will add all the fat it deserves. Proof. Let O be one of such objects. Let H be the automorphism group of O, in the sense that H is the subgroup of Sn that fixes O. By hypothesis, there are n − 1 objects isomorphic to O defined on E, and H is the automorphism group of each of them. Because Sn fixes the union of such objects, either an element of Sn fixes O or maps it to some of the other objects, therefore H has n left classes, and hence its index in Sn is n. Fact 1: H is transitive. Proof: If not, consider a natural k satisfying 0 < k < n such that H fixes a k-subset F of E. Let K be the full group of automorphisms n of E that fixes F , K ≤ Sn. Note that H ≤ K. Now K has k left classes on Sn, as Sn acts transitively on k-subsets. Hence its n index is k , and jHj jKj ) [Sn : K] [Sn : H]: Thus n n: k This is only possible for k = 1 or k = n − 1, which is essentially the same case, but is anyway contrary to the assumption that none of the objects is in natural correspondence to an element of E. Fact 2: H is primitive - meaning that it is transitive and it preserves no non-trivial partition of E. Proof: Being transitive, the only possible way H could preserve a par- tition would be if such partition had parts of same size. Now suppose that k is the number of parts of a non-trivial parti- tion P preserved by H, all of the parts of P of same size and 1 < k ≤ n=2. Let K be the full automorphism group of P , hence H ≤ K ≤ Sn. n! Note that there are: such partitions, therefore this is n k k! k ! the index of K in Sn (as it is the number of left cosets). 2 Again, we have that the index of K must divide the index of H: n! n n k k! k ! This is certainly possible for k = 1 and k = n, cases in which the partition is trivial. Now observe that n! ≥ k!(n=k)−1: n k k! k ! For k ≥ 3 and n ≥ 8, we have that p k!(n=k)−1 > k kn ≥ n; hence a contradiction. For k = 3 and n = 6, n! = 15 > 6: n k k! k ! For k = 2 and n ≥ 8, we have that n! > 2(n=2)−1 ≥ n; n k k! k ! again a contradiction. Finally, for n = 6 we have n! = 10 > 6; n k k! k ! and for n = 4, we have n! = 3 n k k! k ! that does not divide 4. We therefore observe that there is no valid choice for k but for 1 or n, concluding that H is primitive. Fact 3: H contains no 3-cycle. Proof: Suppose to the contrary that it does. Consider the relation in E defined by a ∼ b whenever a = b or there is a 3-cycle σ 2 H acting as σ(a) = b: We claim that this is an equivalence relation. It is clearly reflex- ive and symmetric. To see transitivity, note that if a ∼ b and b ∼ c, we have either of the following: 3 { (abx) 2 H and (bcy) 2 H, and therefore (bcy)◦(xba)◦(ycb)◦ (abx) = (abc), hence a ∼ c. { (abx) 2 H and (bcx) 2 H, and therefore (abx) ◦ (bcx) = (cab), hence a ∼ c. Also, if a ∼ b ∼ c, then as we saw above, (abc) 2 H. Now suppose ∼ partitioned H into more than one equivalence class. As we saw above, H cannot fix such classes, so consider a permutation τ that maps some elements of a class C1 to some class C2, and some other elements to a class Ci, where i can be anything but 2. Suppose, for example, a; b; c 2 C1 and τ(b) 2 C2 and τ(a) 2 Ci. Now observe that τ ◦ (abc) ◦ τ −1 = (τ(a)τ(b)τ(c)) 2 H; a contradiction to the fact that τ(a) and τ(b) lie on different classes. Therefore ∼ must be a universal relation, in the sense that any two elements are related. Hence H contains all 3-cycles and must be the alternating subgroup of Sn. Its index in Sn for any n is 2, and hence n = 2. Thus H is trivial, and therefore fixes a point, a contradiction to the fact that H is transitive. Fact 4: H contains no transposition. Proof: Observe that we could follow the same argument as above, being even easier to show that an analogous relation is of equivalence. If, as above, such equivalence relation is not universal, we would find a contradiction to the fact that H is primitive. If it were universal, we would end up finding that H is Sn itself, as all the transpositions generate Sn, again a contradiction. n Now let us analise where the transpositions of Sn are. We have 2 transpositions distributed among the n−1 left (or right) cosets of H. If two transpositions moving a common point lay on the same coset, we would have (ab)(bc)−1 = (abc) 2 H, a contradiction. Therefore each n coset must contain 2 pairwise disjoint transpositions, what defines a factor of E. Fact 5: If (a b) and (c d) are the transpositions in the same left coset, then for any element h 2 H (h(a) h(b)) and (h(c) h(d)) are in the same left coset. 4 Proof: As (cd) 2 (ab)H, let (cd) = (ab)g, g 2 H. We have h(a b)h−1 = (h(a) h(b)): Thus (h(c) h(d)) = h(c d)h−1 = h(a b)gh−1 = (h(a) h(b))hgh−1 2 (h(a) h(b))H: Therefore an element of H either preserves one of the collections of n=2 transpositions or maps one to another. Thus H preserves a factorization of E. There are (n − 1) factors on a factorization of E. We pick one and note that an element of H either fixes it or maps it to some other. So the stabilizer of a factor in H, say F , has index at most n − 1 (using the Orbit-Stabilizer Lemma and Lagrange's theorem). Its index on Sn is therefore at most n(n − 1). However, there are (n − 1)!! = (n − 1)(n − 3):::3:1 factors in E, what can be easily seen by elementary counting argu- ments. They are all permuted transitively in Sn, so this is the index 0 of the stabilizer of the same factor in Sn, say F . Now 0 0 F ≤ F ≤ Sn ) [Sn : F ] ≤ [Sn : F ]; therefore (n − 1)(n − 3):::3:1 ≤ n(n − 1): This implies that n ≤ 6. Remember that n is even. If n = 2 or n = 4, there is just one factorization. Hence these cases are excluded. So n = 6 is the only possible case. We construct 6 isomorphic factorization on a set of size 6. It is easier at this point to think of our 6-set as the complete graph K6. Theorem 2. There are 6 factorizations on a set of size 6, any two of them isomorphic. Furthermore, two disjoint factors are contained in a unique factorization. 5 Proof. Let us build a factorization. The union of two disjoint fac- tors of a complete graph is a graph where every vertex has degree 2. Hence it is a union of disjoint cycles. As n = 6, it must be a 6-cycle. The remaining 9 edges are either between vertices at distance 3 or at distance 2 in such cycle.
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