By R. Goebel, E. Walker
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Example text
Now, if we conjugate each of t0 , s0 and t0 s0 by d and take all images under the group L, we obtain three 22 The Mathieu group M24 more sets of generators for M24 , making six copies of M24 in total. As is shown later, these are the only ways in which a group isomorphic to L2 7 acting transitively on 24 letters can be extended to a copy of M24 . They are cycled by the element of order 6: zd = 1 2 15 9 13 20 3 6 4 0 5 11 7 8 21 16 14 19 10 18 12 17 22 where z is an element of order 3 commuting with L.
Since every face is joined to just one face of each tern, the sum ui i∈T 30 The Mathieu group M24 for T a tern, is the zero vector. Secondly, note that the sum of the seven uj for j joined to a fixed face, i say, is just ui , for every face other than these eight faces is joined to none or two of them. In particular, we have u = u0 + u18 + u3 + u20 + u8 + u14 + u15 ∈ B As above, we now project onto the top row of the tern array. That is to say, we define X →X∩ +u Certainly, u = 1 1 1 1 1 1 1 1 ∈ Ker ; we need to show that Im consists of the 7-dimensional space of all even subspaces of the top row.
In a Steiner system S(3,4,8), every subset of three points is contained in a unique special tetrad, so there are four dodecads of which can be added to a given octad of to give a further octad. Finally, if e is a dodecad in , then we obtain an octad by adding a dodecad of whose special tetrad of terns is contained in the set of six terns in which e lies. But, since the complement of a special tetrad is itself a special tetrad, the number of special tetrads contained in a fixed set of size 6 is equal to the number of special tetrads containing a given pair of points, namely three.