A construction for the outer automorphism of S 6

Transcription

1 A construction for the outer automorphism of S 6 Padraig Ó Catháin joint work with Neil Gillespie and Cheryl Praeger University of Queensland 5 August 2013

2 Automorphisms of finite groups G a finite group. ρ : G Aut(V ) a representation. Let σ Aut(G). When can σ be realised as an element of Aut(V )? I.e. when does there exist S Aut(V ) such that, for all g G: ρ(x σ ) = S 1 ρ(x)s (Obvious) sufficient condition: σ is inner. (Obvious) necessary condition: χ(x σ ) = χ(x) for all x G.

3 Motivation: M 12 M 12 has two conjugacy classes of subgroups M 11. Coset action on either class is 5-transitive on 12 points. The outer automorphism of M 12 swaps classes of M 11 s, and hence the two actions. These actions cannot be (linearly) equivalent: the traces are different, e.g. P = P σ = ( ) ( )

4 Motivation: M 12 Theorem (M. Hall, 1962) Let H be a Hadamard matrix of order 12. Modulo the centre of Aut(H), the automorphisms of H are the Mathieu group M 12 of order = Here M 12 is represented as a quintuply transitive group of monomial permutations on the columns or rows of H. The row and column representations of H are isomorphic, but the correspondence given by P = HQH 1 determines an outer automorphism of M 12 of order 2. (Conway and Elkies also note that (P, Q) (Q, P) exhibits the outer automorphism of M 12.)

5 Let H be Hadamard of order 12. Aut(H) = {(P, Q) PHQ = H, where P, Q are ± 1-monomial} Consider the representations α : 2.M 12 P and β : 2.M 12 Q: α(x) = Hβ(x)H 1. By Hall: β(x) = α(x σ ) for some outer automorphism σ of M 12. So for every x 2.M 12, we have α(x σ ) = H 1 α(x)h, a linear representation of the outer automorphism of 2.M 12.

6 Construction for the outer automorphism of M 12 While the outer automorphism of M 12 (acting on 12 points) cannot be realised linearly, it almost can. Given an element of M 12, 1 lift to 2.M 12 2 conjugate by H 3 project back onto M 12. Sample automorphism: ( π (P) = ( ) π 1 (P σ ) = = π 1 (P) H )

7 The outer automorphism of S 6 Possibly the most famous outer automorphism of any group: (1, 2) (1, 2)(3, 6)(4, 5) (1, 2, 3) (1, 5, 6)(2, 3, 4) (1, 2, 3, 4, 5, 6) (1, 5)(2, 3, 6) etc. 1 Described by Sylvester: duads, synthemes, totals, etc. 2 Exhibited by the actions of Aut(K 6 ) on points and a set of one-factors 3 Exhibited by action of S 6 on certain 2-colourings of K 5 4 Via the isomorphism S 6 = Sp4 (2) 5 As a subgroup of M 12, etc. Can we find a matrix H which intertwines (lifts of) these representations of S 6?

8 A fragment from Moorhouse Moorhouse: classification of the complex Hadamard matrices with doubly transitive automorphism groups. A sporadic example: H = ω ω ω ω 1 ω 1 ω ω ω 1 ω ω 1 ω ω 1 ω ω ω 1 ω 1 ω ω ω ω 1 which has automorphism group 3.A 6. Note: HH = 6I 6.

9 Aside: what we were actually doing... Attempting to construct codes with interesting automorphism groups from the rows of Hadamard matrices. The rows of the matrix (considered over F 4 ) generate the hexacode, used by Conway and Sloane to describe the Golay code. Considered over F 3, we obtained a new nonlinear uniformly packed code, and a new frequency permutation array, etc. These ideas should extend to larger families of Hadamard matrices...

10 H intertwines two representations of 3.A 6. Restricting to A 6 we get inequivalent representations of A 6. So every automorphism of H is of the form ρ(x)hρ(x σ ). Now, ρ extends to a representation of 3.S 6. If we take y to be an odd permutation, then ρ(y)hρ(y σ ) = H. But H is not Hermitian... In any case, H could never intertwine representations of 3.S 6 : involutions of shape have non-zero trace over C, but are mapped to elements of shape 2 3. (Cf. elements of order 3.)

11 The split-quaternions, Ξ Discovered by Hamilton Cockle in dimensional over R: generated by {1, i, τ, iτ}, where i 2 = 1 τ 2 = 1 τiτ = i. Isomorphic to M 2 (R): ( ) 0 1 i 1 0 τ ( ) Not a division algebra: (i + τi) 2 = 0 (Cockle s "impossibles"). Contains C = 1, i as a subalgebra.

12 Take S = τi 6, then, for any odd permutation y S 6, we have: Sρ(y)Hρ(y σ )S = SH S = H So odd permutations can be made into Ξ-linear automorphisms of H. Note that τ and τi have trace 0: so all odd permutations lift to elements of equal trace. Even permutations are defined over C, odd permutations involve τ. We have 3.S 6 acting on H with the row and column actions differing by an outer automorphism of S 6. To compute the image of σ S 6 under an outer automorphism: lift σ to 3.S 6, if odd, multiply by S, conjugate by H, restrict to S 6.

13 Conclusion A construction of the outer automorphism for S 6 which does not depend on finding two non-conjugate subgroups of index 6. Could be realised as a C-linear 12-dimensional representation. Question: Which automorphisms of which representations of which groups can be realised linearly in this sense? Question: Can we find other interesting combinatorial objects hiding in the rings of intertwiners of representations?

14 Thank you!