# Symmetry Group of Equilateral Triangle is Group

## Theorem

### Definition

Recall the definition of the symmetry group of the equilateral triangle:

Let $\triangle ABC$ be an equilateral triangle.

We define in cycle notation the following symmetry mappings on $\triangle ABC$:

 $\displaystyle e$ $:$ $\displaystyle \tuple A \tuple B \tuple C$ Identity mapping $\displaystyle p$ $:$ $\displaystyle \tuple {ABC}$ Rotation of $120 \degrees$ anticlockwise about center $\displaystyle q$ $:$ $\displaystyle \tuple {ACB}$ Rotation of $120 \degrees$ clockwise about center $\displaystyle r$ $:$ $\displaystyle \tuple {BC}$ Reflection in line $r$ $\displaystyle s$ $:$ $\displaystyle \tuple {AC}$ Reflection in line $s$ $\displaystyle t$ $:$ $\displaystyle \tuple {AB}$ Reflection in line $t$

Note that $r, s, t$ can equally well be considered as a rotation of $180 \degrees$ (in $3$ dimensions) about the axes $r, s, t$.

Then these six operations form a group.

This group is known as the symmetry group of the equilateral triangle.

## Proof

Let us refer to this group as $D_3$.

Taking the group axioms in turn:

### G0: Closure

From the Cayley table it is seen directly that $D_3$ is closed.

$\Box$

### G1: Associativity

$\Box$

### G2: Identity

The identity is $e = (A) (B) (C)$.

$\Box$

### G3: Inverses

Each element can be seen to have an inverse:

$p^{-1} = q$ and so $q^{-1} = p$
$r$, $s$ and $t$ are all self-inverse.

$\Box$

No more need be done. $D_3$ is seen to be a group.

$\blacksquare$