Construction of the outer automorphism of S6 via a complex Hadamard matrix

We give a new construction of the outer automorphism of the symmetric group on six points. Our construction features a complex Hadamard matrix of order six containing third roots of unity and the algebra of split quaternions over the real numbers.


Introduction
Sylvester showed that the fifteen two-subsets of a six element set can be formed into 5 parallel classes in six different ways and that the action of S 6 on these synthematic totals is essentially different from its natural action on six points, [13]. To our knowledge this was the first construction for the outer automorphism of S 6 .
Miller attributes the result that for n = 6, S n has no outer automorphisms to Hölder, and Sylvester's construction of the outer automorphism of S 6 to Burnside, [11]. He also gives a by-hand construction of the outer automorphism. The papers of Janusz and Rotman, and of Ward provide easily readable accounts which are similar to Sylvester's, [10,14]. Cameron and van Lint devoted an entire chapter (their sixth!) to the outer automorphism of S 6 , [2]. They build on Sylvester's construction to construct the 5-(12, 6, 1) Witt design, the projective plane of order 4, and the Hoffman-Singleton graph.
Via consideration of the cube in R 3 , Fournelle gives a heuristic for the existence of an outer automorphism of S 6 , and constructs it with the aid of a computer, [7]. Howard, Millson, Snowden and Vakil give two constructions of the outer automorphism of S 6 , and use this to describe the invariant theory of six points in certain projective spaces, [9].
In this note we give a construction which we believe has not previously been described, using a complex Hadamard matrix of order 6 and a representation of the triple cover of A 6 over the complex numbers. This note is inspired by a construction of Marshall Hall Jr [8] for the outer automorphism of M 12 via a real Hadamard matrix of order 12, and by Moorhouse's classification of the complex Hadamard matrices with doubly transitive automorphism groups, [12]. It was in the latter paper that we first became aware of the complex Hadamard matrix of order 6 discussed in this article, where it is described as corresponding to the distance transitive triple cover of the complete bipartite graph K 6,6 .

Hadamard matrices
Let ω be a primitive complex third root of unity. Then the matrix H 6 is complex Hadamard.
This means that H 6 satisfies the identity H 6 H † 6 = 6I 6 , where for an invertible complex matrix A, A † is the complex conjugate transpose of A. Equivalently, H 6 reaches equality in Hadamard's determinant bound. We refer the reader to [6] for a comprehensive discussion of Hadamard matrices and their generalisations.
An automorphism of a complex Hadamard matrix is a pair of monomial matrices (P, Q) such that P −1 HQ = H . The set of all automorphisms of H forms a group under composition. In this note we will work with the subgroup of automorphisms (P, Q) where all non-zero entries are third roots of unity, we denote this group Aut(H). Consider now the projection maps ρ 1 (P, Q) → P and ρ 2 (P, Q) → Q. Since 1 √ 6 H 6 is unitary, and for any automorphism (P, Q) of H the identity HQH −1 = P holds, it follows that ρ 1 and ρ 2 are conjugate representations of Aut(H). Note further that ρ i is a faithful representation, since Q = I forces P = I . Thus Aut(H) is isomorphic to a finite subgroup of monomial matrices of GL n (C). Furthermore, if Aut(H) contains a subgroup isomorphic to G, then the projections ρ 1 and ρ 2 onto the first and second components of Aut(H) give two conjugate representations of G by monomial matrices.
Every monomial matrix has a unique factorisation P = DK where D is diagonal and K is a permutation matrix. The projection π : P → K is a homomorphism for any group of monomial matrices. In general, the representation Aut(H) ρ 1 π is not linearly equivalent to the representation Aut(H) ρ 2 π . As mentioned above, this phenomenon was first observed by Hall, who showed that the automorphism group of a Hadamard matrix of order 12 is isomorphic to 2.M 12 , and that ρ 1 π and ρ 2 π realise the two inequivalent actions of M 12 on 12 points, [8]. This interpretation of the outer automorphism of M 12 was also used by Elkies, Conway and Martin in their analysis of the Mathieu groupoid M 13 , [4].
Throughout this note we use the following shorthand for monomial matrices: we list the elements of the diagonal matrix D , and give the cycle notation for K as a permutation of the columns of the identity matrix (i.e. a right action).
Consider the following pairs of monomial matrices.
We define * to be the entry-wise complex conjugation map, and consider the group X = τ 1 , τ 2 , * .
shows that all the relations in this presentation hold for these elements s, t, and hence Y = τ 1 , τ ′ 2 is isomorphic to a quotient of S 6 . On the other hand, Y ρ 1 π is easily seen to be isomorphic to S 6 , so we conclude that Y ∼ = S 6 . Now let N be the subgroup of X consisting of all elements for which each component is a diagonal matrix. Since τ ρ i 1 and τ ρ i 2 have determinants in {±1}, every element of the projection X ρ i 0 has this property. However all the elements of N ρ i have third roots of unity along the diagonal, and so must have determinant 1. As a result, X ρ i 0 is isomorphic to a subgroup of M ⋊ S 6 where M ∼ = 3 5 is the group of unimodular diagonal matrices with entries from ω , and S 6 acts as Y ρ i π . The only non-trivial S 6 -submodule of M is the constant module of order 3.
Define n i+1 := [τ 2 , * ] τ i 1 for each i ≥ 1. (We shift subscripts because the action of τ 1 on [τ 2 , * ] gives elements of N which have the non-initial rows of H 6 as the diagonal of the first component.) Since [τ 2 , * ] ∈ X 0 , we have n i ∈ X 0 for 2 ≤ i ≤ 6. Observe that So neither of the projections N ρ 1 , N ρ 2 are onto the constant module, and the kernel of N ρ 1 is neither trivial nor the constant module. It follows that N ∼ = M × M . Finally, we observe that monomial matrices normalise diagonal matrices, and that X 0 acts as a group of monomial matrices in each component. It follows that N ⊳ X 0 , and that Y is a complement of N in X 0 . Since * acts on N by inversion, N ⊳ X .
The group X has a natural action on 6 × 6 matrices over C where (P, Q) ∈ X 0 acts as H (P,Q) = P −1 HQ, and * acts by complex conjugation. We compute the stabiliser of H 6 under this action. We denote this group Aut * (H 6 ) to emphasise that this is a group of semilinear transformations in its action on the normal subgroup N . We require the subgroups X 0 , Y and N defined in Proposition 1 in the proof of the following. Proposition 2. The group Aut * (H 6 ) is isomorphic to the nonsplit extension 3.S 6 , and Aut * (H 6 ) contains a C-linear subgroup isomorphic to 3.A 6 .
First, we show that the intersection Aut * (H 6 ) ∩ N has order 3. To prove this, suppose that (D, E) ∈ N , and that D −1 H 6 E = H 6 , or equivalently DH 6 = H 6 E . Since the first column of H 6 is constant, D must be a scalar matrix. So D commutes with H 6 , and we have DH 6 = H 6 D = H 6 E . Hence D = E , so (D, E) = (ω i I, ω i I) for some i. Since these elements do leave H 6 invariant, the claim is proved.
We next claim that there is no element (D, E) of N such that DH * 6 = H 6 E ; suppose to the contrary that such a (D, E) exists. Precisely the same argument as before shows that D must be scalar. This implies that H * 6 = H 6 ED −1 , but this equation has no solution in diagonal matrices: since the first row of H * 6 is equal to the first row of H 6 , we would require ED −1 = I 6 , from which we derive H 6 = H * 6 , a contradiction. Consider the subgroup K := τ 1 , τ 2 * , N of X . Since X = K, * and * ∈ K , we have |X : K| = 2 and X = K ∪ (K * ). It follows, moreover, from the previous arguments that no element of K sends H 6 to H * 6 , and hence no element of the right coset K * can fix H 6 . Therefore, Aut * (H 6 ) ⊆ K , and from the first paragraph of the proof we also have Aut * (H 6 )N = K . The quotient Aut * (H 6 )/(Aut * (H 6 ) ∩ N ) is isomorphic to K/N , an index 2 subgroup of X/N ∼ = S 6 .2. In particular K/N contains A 6 as a normal subgroup of index 2. Since the element N τ 2 * does not lie in A 6 and does not centralise A 6 it follows that K/N ∼ = S 6 .
We have shown that Aut * (H 6 ) has a normal subgroup of order 3 with quotient isomorphic to S 6 . The elements (τ 2 * ) τ i 1 for 0 ≤ i ≤ 4 project onto a set of Coxeter generators for S 6 . With these generators, it is straightforward to construct a Sylow 3-subgroup of Aut * (H 6 ). One such subgroup is generated by A computation shows that [x, y] = ([ω, ω, ω, ω, ω, ω], [ω, ω, ω, ω, ω, ω]) . This shows that the commutator subgroup contains the normal subgroup of order 3, hence the extension is non-split. Elements of Aut * (H) which map onto odd permutations act on [x, y] by inversion. So the centraliser of this normal subgroup is of index 2 in Aut * (H): this is necessarily a non-split central extension 3.A 6 .
A perfect group S has a largest non-split central extensionŜ which is unique up to isomorphism. The center ofŜ is the Schur multiplier of S , and every non-split central extension of S is a quotient ofŜ . The number of generators of the Schur multiplier is bounded by g − r where g is the number of generators in a presentation of S and r is the number of relations. We refer the reader to Wiegold's survey on the Schur multiplier for proofs of all these results [15]. Since A 6 is shown in [3] to have the presentation it follows that the Schur multiplier of A 6 is cyclic. Hence the non-split extension 3.A 6 is unique up to isomorphism. Now, since Aut * (H) splits over 3.A 6 , we have that 3.A 6 < Aut * (H) < Aut(3.A 6 ). Suppose that ξ ∈ Aut(3.A 6 ) such that the image of ξ in Aut(A 6 ) is the trivial automorphism. Let σ ∈ 3.A 6 be an element of order 15, projecting onto a 5-cycle in A 6 . Then σ 5 generates the central subgroup of order 3. Each coset of σ 5 contains a unique element of order 5, which is fixed by hypothesis. So either σ is fixed element-wise, or ξ = * . Moreover, any two subgroups of order 15 intersect in σ 5 , so the action of ξ is identical on all 5-cycles. Since the 5-cycles generate A 6 , the action of ξ is completely determined.
So each choice of actions on 3 and on A 6 determines at most one isomorphism class of groups. It follows that Aut * (H) is uniquely described as the group of shape 3.S 6 with trivial center.
The projection of ρ 1 (Aut * (H) ∩ X 0 ) is clearly a faithful linear representation of 3.A 6 over the complex numbers, completing the proof.
The kernel of X in this action is the subgroup of N of order 3 5 consisting of pairs with trivial first component. The restriction to Aut * (H 6 ) is faithful, however. One could construct a faithful action of X by taking the permutation action induced by its action on the rows of H 6 together with the induced action on columns.
Remark 3. The matrix H 6 and the group 3.A 6 can be realised over any field k for which k × has a subgroup of order 3. In the case that k is the finite field of order 4, the rows of H 6 span the Hexacode, introduced by Conway as part of a construction for the group M 12 . It is discussed in detail in Section 11.2 of [5]. In particular, this code is the extended quadratic residue code with parameters (6,3,4). Uniqueness can easily be verified by hand: observe that the punctured code is the Hamming (5, 3, 3) code, which is unique, and that any pair of one-bit extensions which increase the minimum distance are isomorphic. The 6-dimensional C-representation of 3 · A 6 has been previously described in the literature, normally via its action on a set of vectors in C 6 derived from the hexacode. In particular, Wilson gives the action of 3 · A 6 on certain vectors of weight 4 in Section 2.7.4 of [16].
3 The outer automorphism of S 6 Finally we construct the outer automorphism of S 6 over the split-quaternions. Recall that the split-quaternions are a 4-dimensional R-algebra with basis [1, i, β, βi] where [1, i] generates the usual algebra of complex numbers and β 2 = 1, i β = −i. We denote the split quaternions by B. They admit an R-linear representation generated by Observe that Aut * (H 6 ) admits a B-linear representation if and only if * does, and that the latter is realised by (βI 6 , βI 6 ).
Since H 6 is invertible over C, it is invertible over B. Now, rearranging the matrix equation H τ 2 * 6 = H 6 , and using the same notation as before for monomial matrices, we obtain that Note that (βω) 2 = (βω) 2 = 1 so that the matrix on the right hand side of the above equation is an involution. As was the case over the complex numbers, H 6 intertwines the projections ρ 1 and ρ 2 . We observe that for any g ∈ Aut * (H), we have that g ρ 1 = H 6 g ρ 2 H −1 6 . But, as illustrated above, τ ρ 1 π 2 is a 2-cycle, while the projection τ ρ 2 π 2 is a product of 3 disjoint 2-cycles. We conclude that the representations ρ 1 π and ρ 2 π of S 6 cannot be conjugate. Thus whereas the permutation representations of S 6 on 6 points are not equivalent, and the monomial representations of 3.A 6 are not equivalent, we have constructed two explicit B-linear representations of 3.S 6 which are equivalent under conjugation by H 6 . Moreover, although the representation is not defined over C, the intertwiner H 6 is. Theorem 4. There exists an irreducible 6-dimensional monomial representation of 3.S 6 over the split-quaternions. Two conjugate representations of 3.S 6 intertwined by the complex Hadamard matrix H 6 give an explicit construction for the outer automorphism of S 6 .