The Miracle Octad Generator (MOG) is a construction of the [24,12,8] extended binary Golay code $\gc$ . It makes use of the hexacode. The notation [24,12,8] indicates that this linear code has length 24, dimension 12, and minimum weight 8.

The construction is originally due to R. T. Curtis ([3], [4]). The description of the MOG below is taken from [1], Chapter 11. A proof that the construction gives a [24,12,8] code is given below. A proof of the uniqueness of the binary Golay code (up to permutation of the coordinates) can be found in [2], Chapter 5.

Construction

The MOG consists of a rectangular array of 4 rows and 6 columns. The rows of the MOG are labelled with elements of $\F_4$ (the field of 4 elements). The labels are $0,1,\w,\W$ from top to bottom, where $\w$ is a cube root of unity. The code $\gc$ consists of subsets $S$ of the 24 squares of the array which satisfy the following conditions:

• the parities of the number of elements in $S$ in each of the 6 columns, and in the top row, are the same
• the sums of the row labels for elements of $S$ in each column form an element of the hexacode.

Notation: For $S$ any subset of the MOG, let us write $\Sigma(S)$ for the element of $\F_4^6$ corresponding to the column sums. Hence, the second condition above says that $\Sigma(S)$ is in the hexacode for any $S \in \gc$ .

When we say that a column is ``even'' or ``odd'', we are referring to the parity of the number of elements of $S$ in the column. When we speak of the ``sum'' of a column, we are referring to the sum of the row labels of $S$ for that column, i.e., the component of $\Sigma(S)$ for that column.

Proof that the construction gives a [24,12,8] code

Let us note a few simple facts about $\gc$ and the MOG.

From these facts, we can show that $\gc$ is a doubly even code (the weight of any element is divisible by 4). Let $S$ be an element of $\gc$ . If the columns of $S$ are even, then it is clear that $|S|$ is divisible by 4. Suppose the columns of $S$ are odd. Let $t_1$ (resp., $t_3$ ) denote the number of columns consisting of 1 (resp., 3) elements, for which the top row is filled. Let $u_1$ (resp., $u_3$ ) denote the number of columns consisting of 1 (resp., 3) elements, for which the top row is empty. The defining conditions of $\gc$ imply that $t_1 + t_3$ is odd. By Fact 2, the number of 0's in $\Sigma(S)$ is equal to $t_1 + u_3$ , which is therefore even by a property of the hexacode. So $t_3 + u_3$ is odd. Since there are 6 columns, we have $t_1 + t_3 + u_1 + u_3 = 6$ . Hence \begin{eqnarray*} |S|& = & 3t_3 + 3u_3 + t_1 + u_1 \\ & = & 3t_3 + 3u_3 + t_1 + (6 - t_1 - t_3 - u_3)\\ & = & 2(t_3 + u_3) + 6\\ & \equiv & 0 \pmod{4}. \end{eqnarray*} Furthermore, it is not difficult to see that an element of weight 4 is impossible, hence the minimum weight of $\gc$ is 8. So the only possible weights of elements of $\gc$ are 0, 8, 12, 16, and 24.

To calculate the dimension of $\gc$ , we need to count the number of elements of each weight. Weights 0 and 24 are obvious, and the counts for weight 8 and weight 16 are the same. So we need only count weights 8 and 12. This is not difficult using the defining conditions above, and knowledge of the hexacode; the final result is that the number of elements of weights 0,8,12,16,and 24 are 1, 759, 2576, 759, and 1, respectively, for a total of 4096 = $2^{12}$ . Hence $\gc$ has dimension 12, so it is a [24,12,8] code.

Further facts about the binary Golay code

Elements of $\gc$ of weight 8 are called octads. Elements of $\gc$ of weight 12 are called dodecads.

The octads of $\gc$ form a (5,8,24) Steiner system. In other words, given any 5 squares in the MOG, there is a unique way to add 3 squares to form an octad. To complete an octad, it is necessary to solve 3-problems or 5-problems for the hexacode. For a detailed procedure, see ([1], 11.6).

The automorphism group of $\gc$ is the largest Mathieu group $M_{24}$ , one of the sporadic simple groups.

The code $\gc$ is used in the construction of the Leech lattice, whose automorphism group is the largest Conway group $Co_0$ (sometimes written $\cdot 0$ ). The quotient of $Co_0$ by its center, called $Co_1$ , is a sporadic simple group. The group $Co_1$ plays an important role in the construction of the monster group, the largest sporadic simple group of all.

Bibliography

1

J. H. Conway and N. J. A. Sloane. Sphere Packings, Lattices, and Groups. Springer-Verlag, 1999.

2

Robert L. Griess, Jr. Twelve Sporadic Groups. Springer-Verlag, 1998.

3

R. T. Curtis. On subgroups of $\cdot 0$ , I: lattice stabilizers. Journal of Algebra, 27 (1973), 549-573.

4

R. T. Curtis. A new combinatorial approach to $M_{24}$ . Proceedings of the Cambridge Philosophical Society 79 (1976), 25-42.