The Group of Symmetries of the Square

The Group of Symmetries of the Square

Recall from The Group of Symmetries of the Equilateral Triangle page that if we have an equilateral triangle whose vertices are labelled $1$, $2$, and $3$ and if $G = \{ \rho_0, \rho_1, \rho_2, \mu_1, \mu_2, \mu_3 \}$ is the group of symmetries where $\rho_0, \rho_1, \rho_2 : \{1, 2, 3 \} \to \{1, 2, 3 \}$ are functions respectively defining a counterclockwise rotation about the center of the equilateral triangle by $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$ and $\mu_1\ \mu_2, \mu_3 : \{ 1, 2, 3 \} \to \{ 1, 2, 3 \}$ are functions defining axial flips along lines that intersect the triangle's vertices and perpendicularly bisect the adjacent edge, and if $\circ : G \to G$ defines for each $f_1, f_2 \in G$ the composition of $f_1$ and $f_2$, i.e., $(f_1 \circ f_2)(x) = f_1(f_2(x))$ then $(G, \circ)$ defines a group.

We will now see that the group of symmetries of the square also form a group with respect to the operation of composition $\circ$. Consider a square and label the vertices $1$, $2$, $3$, and $4$: One type of symmetry we can define are once again, rotational symmetries of $0^{\circ}$, $90^{\circ}$, $180^{\circ}$ and $270^{\circ}$ which produce: Another type of symmetry we can define are axial flips along perpendicular bisectors of the square to which there are two perpendicular bisectors that we can flip across: Upon applying these perpendicular axial flips on the original square, we get: The last type of symmetry we can define are axial flips along diagonal bisectors of the square to which there are, once again, two diagonal bisectors that we can flip across: Upon applying these diagonal axial flips on the original square, we get: Let $\rho_0, \rho_1, \rho_2, \rho_3 : \{1, 2, 3, 4 \} \to \{1, 2, 3, 4 \}$ be functions that represent the rotational symmetries, let $\mu_1, \mu_2 : \{1, 2, 3, 4 \} \to \{1, 2, 3, 4 \}$ be functions that represent the perpendicular bisector axial symmetries, and let $\delta_1, \delta_2 : \{1, 2, 3, 4 \} \to \{1, 2, 3, 4 \}$ be functions that represent the diagonal bisector axial symmetries. Each of these functions can be described as a $2 \times 4$ matrix where the first row lists the original arrangement of the vertices on the square and the second row lists the change in the positions of the vertices after the particular symmetries are applied. We have that:

(1)
\begin{align} \quad \rho_0 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & 2 & 3 & 4 \end{pmatrix} \quad , \quad \rho_1 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 1 & 2 & 3 \end{pmatrix} \quad , \quad \rho_2 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 3 & 4 & 1 & 2 \end{pmatrix} \quad , \quad \rho_3 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix} \end{align}
(2)
\begin{align} \quad \mu_1 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1 \end{pmatrix} \quad , \quad \mu_2 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3 \end{pmatrix} \end{align}
(3)
\begin{align} \quad \delta_1 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 1 & 4 & 3 & 2 \end{pmatrix} \quad , \quad \delta_2 = \begin{pmatrix} 1 & 2 & 3 & 4\\ 3 & 2 &1 & 4 \end{pmatrix} \end{align}

Let $G = \{ \rho_0, \rho_1, \rho_2, \rho_3, \mu_1, \mu_2, \delta_1, \delta_2 \}$ and let $\circ : G \to G$ be defined for all $f_1, f_2 \in G$ as function composition, that is, $(f_1 \circ f_2)(x) = f_1(f_2(x))$. Then $(G, \circ)$ is yet again, a group. The following table illustrates all possible compositions of elements in $G$: The table above shows us that $G$ is closed under $\circ$, that $\rho_0$ is the identity element with respect to $\circ$, and that each element in $G$ has an inverse (since every row and every column has a composition that equals $\rho_0$. We've already proven that $\circ$ is associative, so indeed $(G, \circ)$ is a group. Furthermore, from the table above we see that $(G, \circ)$ is not abelian since:

(4)
\begin{align} \quad \mu_1 \circ \rho_3 = \delta_2 \neq \delta_1 \rho_3 \circ \mu_1 \end{align}