7 - Elements of the Mathieu Group M24

Summary

What is the minimal test to decide whether a permutation π ∈ S₂₄ lies in our preferred copy of M₂₄? The space C is 12 dimensional and so if we choose a basis of 12 codewords of C, apply π to each codeword in the basis and verify that the image is also in C then π ∈ M₂₄. The 12-dimensional subspace C is self-orthogonal with respect to the usual inner product, and so C = C⊥. Thus a vector is in C if, and only if, it is orthogonal to every codeword in a basis of C. Now one codeword in our basis may be chosen to be the all 1s vector that is clearly fixed by any permutation; the other 11 can be chosen to be octads. In this chapter we show that we can do much better than this. In fact we show that we can choose 8 octads that are contained in one, and only one, copy of C, but that any set of 7 octads is contained in no copy of C or in more than one. To this set of 8 octads we add a further 3 to form a basis together with the all 1s codeword. We now have a minimal test for membership of M₂₄: apply π to each of the 8 octads; if the image in each case intersects each of the 11 octads in the basis evenly, then π is in M₂₄, otherwise it is not. When working with M₂₄ we often require an element possessing certain properties. In this chapter we show how to construct elements of shape 1⁸.2⁸, 2¹² and 1⁶.3⁶. We also reproduce a diagram due to Todd and Conway showing the orbits of M₂₄ on the subsets of Ω.