C2 - symmetry groups

symmetry group of degree

n

let $A = {1, 2, \dots, n}$ , then $(sym (A), \circ) =: S_{n}$ is the symmetry group of degree $n$

α = [\begin{matrix} 1 & 2 & \dots & n \\ α (1) & α (2) & \dots & α (n) \end{matrix}]

the order, $| S_{n} | = n (n - 1) \dots 2 \cdot 1 = n!$
this is because there are $n$ choices of $α (1)$ , $n - 1$ of $a_{2}$ and so on as $α$ is one-to-one and $α (1) \neq α (2) \neq \dots$
the identity is:

ϵ = [\begin{matrix} 1 & 2 & \dots & n \\ 1 & 2 & \dots & n \end{matrix}]

\begin{matrix} ϵ = [\begin{matrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{matrix}] \\ α = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix}] \\ α^{2} = [\begin{matrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{matrix}] \\ β = [\begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{matrix}] \\ α β = [\begin{matrix} 2 & 1 & 3 \\ 1 & 3 & 2 \end{matrix}] \\ α^{2} β = [\begin{matrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{matrix}] \end{matrix}

note that $S_{3}$ is non-abelian as $β α = α^{2} β \neq α β$
in general, $S_{n}$ is non-abelian
the symmetric groups are rich in subgroups, $S_{4}$ has 30, $S_{5}$ has 100
symmetries of a square, $D_{4} :$ each motion an be associated with the permutation of the locations of each of the corners, and a $90 °$ CCW rotation and a horizontal reflection can generate the entire group
by labeling each corner with a number from 1 to 4, $D_{4}$ can be seen as a subgroup of $S_{4}$