Symmetry Group of Equilateral Triangle
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Group Example
Let $\triangle ABC$ be an equilateral triangle.
We define in cycle notation the following transformations on $\triangle ABC$:
| \(\displaystyle \) | \(\displaystyle e\) | \(:\) | \(\displaystyle (A) (B) (C)\) | \(\displaystyle \) | Identity transformation | ||
| \(\displaystyle \) | \(\displaystyle p\) | \(:\) | \(\displaystyle (ABC)\) | \(\displaystyle \) | Rotation of $120^\circ$ anticlockwise about center | ||
| \(\displaystyle \) | \(\displaystyle q\) | \(:\) | \(\displaystyle (ACB)\) | \(\displaystyle \) | Rotation of $120^\circ$ clockwise about center | ||
| \(\displaystyle \) | \(\displaystyle r\) | \(:\) | \(\displaystyle (BC)\) | \(\displaystyle \) | Reflection in line $r$ | ||
| \(\displaystyle \) | \(\displaystyle s\) | \(:\) | \(\displaystyle AC)\) | \(\displaystyle \) | Reflection in line $s$ | ||
| \(\displaystyle \) | \(\displaystyle t\) | \(:\) | \(\displaystyle (AB)\) | \(\displaystyle \) | Reflection in line $t$ |
Note that $r, s, t$ can equally well be considered as a rotation of $180^\circ$ (in three 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.
Its Cayley table can be written:
- $\begin{array}{c|cccccc} & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$
... or directly in cycle notation:
- $\begin{array}{c|cccccc} & e & (ABC) & (ACB) & (BC) & (AC) & (AB) \\ \hline e & e & (ABC) & (ACB) & (BC) & (AC) & (AB) \\ (ABC) & (ABC) & (ACB) & e & (AC) & (AB) & (BC) \\ (ACB) & (ACB) & e & (ABC) & (AB) & (BC) & (AC) \\ (BC) & (BC) & (AB) & (AC) & e & (ACB) & (ABC) \\ (AC) & (AC) & (BC) & (AB) & (ABC) & e & (ACB) \\ (AB) & (AB) & (AC) & (BC) & (ACB) & (ABC) & e \\ \end{array}$
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.
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 inverses.
$\blacksquare$
No more need be done. $D_3$ is seen to be a group.
Sources
- Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.3$: Example $9$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 30 \gamma$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 34 \ (5)$
- John F. Humphreys: A Course in Group Theory (1996): $\S 1$: Example $1.9$
