# Symmetry Group of Square is Group

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## Theorem

The symmetry group of the square is a non-abelian group.

### Definition

Recall the definition of the symmetry group of the square:

Let $\SS = ABCD$ be a square.

The various symmetry mappings of $\SS$ are:

- the identity mapping $e$
- the rotations $r, r^2, r^3$ of $90^\circ, 180^\circ, 270^\circ$ around the center of $\SS$ anticlockwise respectively
- the reflections $t_x$ and $t_y$ are reflections in the $x$ and $y$ axis respectively
- the reflection $t_{AC}$ in the diagonal through vertices $A$ and $C$
- the reflection $t_{BD}$ in the diagonal through vertices $B$ and $D$.

This center 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:

### Group Axiom $\text G 0$: Closure

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

$\Box$

### Group Axiom $\text G 1$: Associativity

Composition of Mappings is Associative.

$\Box$

### Group Axiom $\text G 2$: Existence of Identity Element

The identity is $e$ as defined.

$\Box$

### Group Axiom $\text G 3$: Existence of Inverse Element

Each element can be seen to have an inverse:

- $r^{-1} = r^3$ and so $\paren {r^3}^{-1} = r$
- $r^2$, $t_{AC}$, $t_{BD}$, $t_x$ 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.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.3$