📂Abstract Algebra

Symmetry groups in Abstract Algebra

Definition ¹

A permutation is a bijection $\phi : A \to A$ for a set $A$. $S_{A}$ is the set of all permutations of $A$, which forms a group $\left< S_{A} , \circ \right>$ with respect to function composition $\circ$, and is called the symmetric group.

Explanation

The fact that symmetric groups indeed satisfy the conditions of a group can be easily ascertained, given that a permutation is defined as a bijection. The main interest lies in when $A$ is a finite set, namely when it is $|A| = n$, which is commonly denoted as $S_{A} = S_{n}$.

Symmetry of a Triangle

The concept of permutation is not essentially different from what is learned in high school. Consider $A = \left\{ 1,2,3 \right\}$. Since the number of elements is $3$, the order of the symmetric group $S_{3}$ is $3! = 6$. Since it is not a large number, let’s list them all. The representation is similar to a matrix, where the element in row $1$ is mapped to the element in row $2$. $$\rho_{0} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{bmatrix} \qquad \rho_{1} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \qquad \rho_{2} = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{bmatrix} \\ \mu_{1} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{bmatrix} \qquad \mu_{2} = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{bmatrix} \qquad \mu_{3} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix}$$

Represented graphically, it appears as follows:

• $\rho_{k}$ is rotating the triangle in a counterclockwise direction by $\displaystyle {{2 k \pi} \over {n}}$ with respect to $|A| = n$. Here, $\rho_{0}$ represents a rotation that has no effect on the shape, serving as the identity element in the symmetric group. The reason for using $\rho$ is that it stems from Rotation.
• $\mu_{k}$ is swapping the two points while fixing $k$, or it could be described as flipping over a symmetry line that bisects $1$, just like a mirror image. The reason for using $\mu$ is that it originates from Mirror Image.

Not Commutative

Concerning $n \ge 3$, $S_{n}$ is not a commutative group.

One interesting property of a symmetric group is that it is not a commutative group.

The table above shows all the operations in $S_{3}$, for example, $\rho_{1} \circ \mu_{1} = \mu_{3}$ while $\mu_{1} \circ \rho_{1} = \mu_{2}$. Thus, $$\rho_{1} \circ \mu_{1} \ne \mu_{1} \circ \rho_{1}$$ hence, $S_{3}$ is not a commutative group. A true proof is sufficient by showing through mathematical induction that all $S_{n}$ have such exceptions.