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

|H| |K| ⇒ [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 √ 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 ∈ 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) ∈ H and (bcy) ∈ H, and therefore (bcy)◦(xba)◦(ycb)◦ (abx) = (abc), hence a ∼ c. – (abx) ∈ H and (bcx) ∈ H, and therefore (abx) ◦ (bcx) = (cab), hence a ∼ c. Also, if a ∼ b ∼ c, then as we saw above, (abc) ∈ 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 ∈ C1 and τ(b) ∈ C2 and τ(a) ∈ Ci. Now observe that

τ ◦ (abc) ◦ τ −1 = (τ(a)τ(b)τ(c)) ∈ 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) ∈ 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 ∈ H

(h(a) h(b)) and (h(c) h(d))

are in the same left coset.

4 Proof: As (cd) ∈ (ab)H, let (cd) = (ab)g, g ∈ 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 ∈ (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. There are 3 of the first type and 6 of the second. It is easy to see that 3 edges of the second type cannot form a factor, hence any factor now must have 1 edge of the first type and 2 of the second. It is also easy to see that fixed an edge of the first type, the other two are determined. Therefore all factorizations have the same structure, hence they are isomorphic. Each factorization contains 10 pairs of disjoint factors, and each pair of factors determines a factorization. There are 15 factors in the graph, and for each we can build 8 disjoint factors. So there ate 60 pairs of disjoint factors in total. Easy double-counting implies that there are 6 factorizations.

2 The outer automorphisms of Sn

In this section, we will prove that S6 is the only symmetric group that admits an outer automorphism. First, we will understand what this means. Then we will see how this is related to the past section. Given a group G and an element a ∈ G, we call conjugation by a the mapping:

ϕa : G → G g 7→ aga−1.

Observe that each ϕa is a group automorphism. Any automor- phism of this form will be called an inner automorphism of G. More- over, ϕa ◦ ϕb = ϕab, so the set of all such automorphisms form a subgroup of the automorphisms group of G, denoted by Inn(G) Any automorphism of a group which is not an inner automorphism is called an outer automorphism. A very important fact about Sn is that an inner automorphism preserves the cycle type of a permutation. The following fact, whose proof is trivial, is the key to prove this statement:

Proposition 1. Let a ∈ Sn and consider the transposition (i j) ∈ Sn. Then a(i j)a−1 = (a(i) a(j)).

6 Corollary 2.1. An inner automorphism preserves cycle type. More- over, if two permutations have the same cycle type, there is an inner automorphism mapping one to the other. Using the corollary above, we have:

Corollary 2.2. Let θ be an automorphism of Sn. Then permutations of same cycle type are mapped by θ to permutations of same cycle type. In other words, θ naturally induces a bijection between the cycle types.

Proof. Let a and b be permutations of same cycle type. Let ϕg be the inner automorphism that maps a to b. Then:

θ(b) = θ(ϕg(a)) = ϕθ(g)θ(a) Hence θ(b) and θ(a) have the same cycle type.

2.1 Number 6 again It is easier to prove the theorem below than to prove Theorem 1. This is compensated by the fact that it is easier to construct the factorizations of K6 than an outer automorphism of S6. Here I will prove the theorem below using Theorem 1, probably to understand better the connection between these results.

Theorem 3. Except for n = 6, no symmetric group Sn admits an outer automorphism. Proof. Let θ be an outer automorphism. Observe that it maps the transpositions containing a fixed element i ∈ [n] to permutations of same cycle type. The collection of such permutations is an object Oi, and the permutation θ(i j) maps a permutation of Oi to one of −1 Oj. That is because θ(i j) θ(i k)θ(i j) = θ(k j). Moreover, any permutation g ∈ Sn can be written as g = θ(h), and h is a product of transpositions, hence the collection of the objects Oi is invariant under Sn. The automorphism θ maps transpositions to permutations of order 2 which are not transpositions, that is, disjoint union of transpositions. If θ(i j) is a permutation that fixes i and j, then (i j) is a permutation that fixes θ(i j) but not i. Suppose θ(i j) does not fix i. Because it is a union of at least two transposition, let k be such that

θ(i j)[i] = i1 and θ(i j)[k] = i2,

7 with i 6= i1 and k 6= i2. Then (i1 i2) maps θ(i j) to θ(k j) but fixes i. At any case, the stabilizer of i in Sn is not equal to the automorphism group of Oi. So the objects Oi with i ∈ [n] are isomorphic, their union is in- variant under Sn, but their automorphism group is not the stabilizer of any point. That all together implies that n = 6 by Theorem 1.

Theorem 4. S6 admits an outer automorphism. Observe that this outer automorphism must map each transposi- tion to a permutation that resembles a 1-factor, hence the transpo- sitions containing a particular element will define a 1-factorization. This implies that for any two outer automorphisms θ and σ, there will be an inner automorphism ϕ such that θ = σϕ. In particular, |Aut(S6)/Inn(S6)| = 2. There are many constructions of an outer automorphism for S6. Inspired by what we did before, we could just select a 1-factorization of [6] and associate its 1-factors to the transpositions containing 1. As such transpositions generate Sn, we would need only to prove that this is a well-defined bijection. There is an easier and nicer way though.

Proof. In general, the group Sn can be embedded in Sn+1 by con- sidering the stabilizer of an element. The great thing here is that there is a copy of S5 in S6 which acts transitively on [6]: it does not fix an element at all. To construct such embedding, consider the 24 permutations of S5 that are cycles of length 5. The order of such cy- cles is 5, therefore these 24 permutations can be partitioned into the non-identity elements of six distinct subgroups of S5. Voil`a: S5 acts transitively by conjugation on these 6 subgroups. Each element of S5 can be hence identified with a permutation of S6 once such six subgroups are numbered. Moreover, the kernel of this homomorphism is a normal subgroup of S5, hence the only non-trivial possible choice would be A5. But clearly (123) is not on the kernel, hence the kernel is trivial and this homomorphism is injective. Let H be the image of S5 in S6 which is a transitive subgroup. Observe that H has index 6, and call L the set of its right cosets. As |L| = 6, the full group of symmetries of L is S6. However S6 acts on L in other way: by right multiplication, permuting its elements. Therefore the correspondence φ between an element of S6 and the permutation that it induces on the right cosets of H is a endomor- phism of S6. It is injective because any permutation fixing H must

8 be contained in H, so the kernel of this endomorphism is a subgroup of H. But H has index 6, and S6 has no normal subgroup of index greater or equal than 6. Thus φ is an automorphism of S6. Finally, observe that φ maps every element of H to a permutation that fixes the coset He = H, so H is mapped to the subgroup stabilizer of a point. If φ were an inner automorphism, say ϕa, it would have to map a transitive group to a transitive group. This is because given elements i and j of [6], there is a permutation of h ∈ H that maps a−1(i) to a−1(j) because H is transitive. Hence aha−1 maps i to j, and ϕa(H) is transitive. Therefore φ is not an inner automorphism, and we are done.

References

[1] Peter J Cameron and Jacobus Hendricus Van Lint. Designs, graphs, codes, and their links. Cambridge University Press, 1992.

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