The "obvious" symmetry group $C_3 \times S_4$ related to the hexacode I am studying the large Mathieu groups and more specifically the hexacode from Robert Wilson's book "The Finite Simple Groups". The following paragraph is from page 184:The hexacode
My question is related to what is underlined: How should I interpret this? What does the group "$C_3 \times S_4$ generated by scalar multiplications and those three coordinate permutations" look like?
I figured that the "$C_3$-part" of the group is there because of the (cyclic) multiplicative group {$1, \omega, \overline{\omega}$} of the field $\mathbb{F}_4$, but how come $<(1,2)(3,4); (1,3,5)(2,4,6); (1,3)(2,4)> \cong S_4$? I tried constructing $<(1,2)(3,4); (1,3,5)(2,4,6); (1,3)(2,4)>$ element by element and became only 16 different elements to be in this group. Can anyone help me to get some more insight in how this group looks like and perhaps how this is important in relation to the hexacode and the large Mathieu groups?

Write $a:=(12)(34)$, $b:=(135)(246)$, and $c:=(13)(24)$. Let $G=\langle a,b,c\rangle$. Note that $a^c=a$ while $(a^b)^c=a^{b^2}$. It follows that $N:=\{1,a,a^b,a^{b^2}\}$ is a normal subgroup of $G$ of order $4$. (Isomorphic to a Klein group.) Moreover, $b^c=b^{-1}$ so $H:=\langle b,c\rangle$ is isomorphic to $S_3$.  Moreover, $H\cap N=1$, so $G=N\rtimes H$ and $G$ has order $24$.
