Let an 8-set $X$ represent the vertices of a cube: {$1,2,3,4$} is the label assigned to the four vertices at the "top" and {$5,6,7,8$} is the label assigned to the four vertices at the "bottom"
a rotation of the cube wrt vertices can be defined as the permutation: $$ \sigma = (1,2,3,4)(5,6,7,8). $$ Like a cube kebab spinning! The permutation group $G$ generated by sigma has order 4: $$ \sigma^2 = (1,3)(2,4)(5,7)(6,8); \sigma^3 = (1,4,3,2)(5,8,7,6); \sigma^4 = (1)(2)(3)(4)(5)(6)(7)(8). $$ Is this group $G$ isomorphic to $C_4$? I can understand the action of $G$ on the 8-set $X$ as the natural action of a permutation group. But how would one define $C_4$ acting on $X$? Does $C_4$ act pairwise on $S =$ { {$1,5$},{$2,6$},{$3,7$},{$4,8$} }?
Thanks.
$G=\langle \sigma \rangle$ where $\sigma$ has order 4. Thus $G\cong C_4$.
I suppose you could say that the action of $c^i\in C_4$ on $G$ is $c^i\cdot \sigma^j = \sigma^i\sigma^j$.