Consider the ordered sequence $(1,2,3,4,5,6,7,8)$ organized as in the figure above. We are interested in potential generators of the permutation group $S_8$. The generators in question are the following (which also explains the diagram):
$F = (1 2 3 4), B = (5 6 7 8), L = (1 4 8 5), R = ( 2 6 7 3), U = (1 5 6 2), D = (3 7 8 4)$.
These generators shift the numbers on their corresponding boundaries in a clock-wise direction. These generators $4$-cycles and thus of order 4 and $F^{-1} = F^3$ and so forth for the other generators. For convenience we list the inverses:
$F^{-1} = (1 4 3 2), B^{-1} = (5 8 7 6), L^{-1} = (1 5 8 4), R^{-1} = ( 2 3 7 6), U^{-1} = (1 2 6 5), D^{-1} = (3 4 8 7)$.
We have the following elementary facts about permutations and transpositions: $(a b c) = (a c) (a b)$ and $(c d) = (a c b) (a b c d)$ (for $a,b,c,d$ distinct). We use the notation $[X,Y] = X^{-1}Y^{-1}XY$. Note that $[X,Y]^{-1} = [Y,X]$ and that if $[X,Y]$ can be written as a composition of disjoint transpositions then $[X,Y]^2 = I$ and $[X,Y] = [Y,X]$.
Let us calculate:
$[D^{-1},F^{-1}] = (14)(37) \quad [D,R^{-1}]= (37)(42)$
Hence $T = [D^{-1},F^{-1}][D,R^{-1}] = (14)(37)(37)(42) = (4 2 1)$.
Then $TF = DFD^{-1}F^{-1}D^{-1}RDR^{-1}F = (4 2 1)(1 2 3 4) = (2 3)$.
It is now easy to see that $F,B,L,R,U,D$ generate $S_{8}$.
Let $\Sigma = \{id,\sigma,\sigma^2\}$ be the cyclic permutation group on $(1,2,3)$ (of order 3).
Consider 8 copies of $\Sigma$ denoted by $\Sigma_1,...,\Sigma_8$ and let $W = \Pi^8_{i =1} \Sigma_i$.
Suppose we had a map $S^i$ acting on $W$ which for $i = (1,2,3)$ multiplies components $1$,$2$ and $3$ of $W$ by $\sigma$. And likewise we have $S^i$ for $i = (2,3,4), (3,4,1),(4,1,2)$ and also $i = (1,2,5)$, etc.
Then $Y = (S^{(2,3,4)})^2 S^{(1,2,3)}$ is equivalent to applying $\sigma$ to the first component and $\sigma^2$ to the fourth component. For instance for $w = (id,id,id,id,id,id,id,id) \in W$ we have $Y(w) = (\sigma,id,id,\sigma^2,id,id,id,id)$.
To be continued...
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