Symmetry Group of Equilateral Triangle is Group

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The symmetry group of the equilateral triangle is a group.


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.


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.


G1: Associativity

Composition of Mappings is Associative.


G2: Identity

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


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.


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