Generators of the group $2^6 \rtimes 3 \cdot S_6$ in the Miracle Octad Generator (MOG) I am studying the large Mathieu groups and more specifically I have arrived at the Golay code and the Miracle Octad Generator. My question comes from Robert Wilson's book "The Finite Simple Groups". The following paragraph is from page 185:[Golay code][1][1]: http://i.stack.imgur.com/zapXD.jpg
I am trying to describe the generators, which are pictured in this paragraph, as elements. The left generator seems to only show vertical changes in the MOG, which makes me think it's an element of $2^6$ (=$C_2 \times C_2 \times C_2 \times C_2 \times C_2 \times C_2$). I think it's the following element: $((0, 1, 0, 1, \omega, \bar{\omega}), (e, e))$. Can anyone confirm wheter this is correct or false?
The other two generators seem to be a lot harder, since they have vertical as well as horizontal components. I can't seem to figure out how I can find these elements. For the last generator (which seems the simplest of the two), I think it would be something like this: $((1, 1, 1, 1, ?, ?), (?, ?))$. Can anyone help me fill in the question marks for this element and the second generator?
 A: So first, the normal subgroup $2^6$ will correspond specifically to automorphisms of each column that correspond to addition by an element of the hexacode, so you are correct in your description of the first element.
The second subgroup is the group of semilinear automorphisms and will preserve the $0$ row; both of the latter two automorphisms are of this form, and should be written $((0,0,0,0,0,0),x)$ for $x \in 3 \cdot S_6$.
The second element can be seen as a five-cycle, with the projection into $S_6$ being the five-cycle $(12345)$.  Because the five-cycle is even, this is an element of the Valentiner group and will be a linear morphism of the hexacode, and we can see which it must be to see what each coordinate goes to; in fact, it will be the map
$$(u,v,w,x,y,z) \mapsto (\omega v,w,\overline{\omega}x,\omega y,\overline{\omega}u,z). $$
The third element will be the transposition $(56)$ in $S_6$, which is odd, and the semilinear map should be strictly conjugate-linear: note that on each column it is in fact the transposition $(\omega \overline{\omega})$ which is simply $x \mapsto \overline{x}$, so the map on the vector space $\mathbf{F}_4^6$ with the hexacode as a subgroup is
$$(u,v,w,x,y,z) \mapsto (\overline{u},\overline{v},\overline{w},\overline{x},\overline{z},\overline{y}). $$
