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 group is known as the symmetry group of the square, and can be denoted $D_4$.
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 $7.3$