# Symmetry Group of Square is Group

## Theorem

### Definition

Recall the definition of the symmetry group of the square:

Let $\mathcal S = ABCD$ be a square. The various symmetry mappings of $\mathcal S$ are:

The identity mapping $e$
The rotations $r, r^2, r^3$ of $90^\circ, 180^\circ, 270^\circ$ counterclockwise respectively about the center of $\mathcal S$.
The reflections $t_x$ and $t_y$ are reflections about the $x$ and $y$ axis respectively.
The reflection $t_{AC}$ is a reflection about the diagonal through vertices $A$ and $C$.
The reflection $t_{BD}$ is a reflection about the diagonal through vertices $B$ and $D$.

This group is known as the symmetry group of the square.

## Proof

Let us refer to this group as $D_4$.

Taking the group axioms in turn:

### G0: Closure

From the Cayley table it is seen directly that $D_4$ is closed.

$\Box$

### G1: Associativity

$\Box$

### G2: Identity

The identity is $e$ as defined.

$\Box$

### G3: Inverses

Each element can be seen to have an inverse:

$r^{-1} = r^3$ and so $\left({r^3}\right)^{-1} = r$
$r^2$, $t_{AC}$, $t_{BD}$, $t_x$ and $t_y$ are all self-inverse.

$\Box$

Thus $D_4$ is seen to be a group.

Note that from the Cayley table it can be observed directly that:

$r \circ t_x = t_{BD}$
$t_x \circ r = t_{AC}$

thus illustrating by counterexample that $D_4$ is not abelian.

$\blacksquare$