By American Mathematical Society

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**Example text**

1 The Golay code group M. as defined above is preserved by the Proof The code is certainly preserved by the group L and, since the seven generators in are conjugate to one another under the action of L, it will suffice to show that is preserved by t0 . From the preceding remarks, we need only show that the image under t0 of each member of a basis of intersects each member of that basis evenly. 4, and the three dodecads corresponding to the special tetrads of terns 1 2 4 , 2 3 5 and 3 4 6 . For now only, we shall denote the dodecad corresponding to the four terns i j k l by d(i, j, k, l).

Proof Let be the permutation, let U = a1 a2 a8 be the octad which is fixed pointwise by , and let b be the fixed point not in U . Let x = b be any other point not in U . We shall show that x must be fixed by and so is the identity. Consider firstly the octad containing a1 a2 a3 b x . It must contain a further point of U as all octads intersect evenly, so, without loss of generality, let this be a4 . It can contain no further members of U , and so V = a1 a2 a3 a4 b x y z is an octad, where y and z are not in U .

2 The group M = t0 t1 shapes 38 12 112 1 23 and 18 28 . t6 , it possesses elements of cycle- Proof It is readily checked that the element ti tj for i = j has cycle-shape 64 , and that ti tj tk , for i j k distinct, has cycle-shape 24 44 if i j k is a line, and 12 112 if not. Moreover, the element aˆ t0 has cycle-shape 1 3 5 15. 1 states that L ⊂ M and, since L acts transitively on , so does M. The point-stabilizer in M contains elements of cycle-shapes 1 112 and 3 5 15 and, since any orbit of this point-stabilizer must have length a sum of cycle lengths in each case, we see that M must act doubly transitively.