Symmetry Group is Group

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Theorem

Let $P$ be a geometric figure.

Let $S_P$ be the set of all symmetries of $P$.

Let $\circ$ denote composition of mappings.


The symmetry group $\struct {S_P, \circ}$ is indeed a group.


Proof

By definition, a symmetry mapping is a bijection, and hence a permutation.

From Symmetric Group is Group, the set of all permutations on $P$ form the symmetric group $\struct {\map \Gamma P, \circ}$ on $P$.

Thus $S_P$ is a subset of $\struct {\map \Gamma P, \circ}$.


Let $A$ and $B$ be symmetry mappings on $P$.

From Composition of Symmetries is Symmetry, $A \circ B$ is also a symmetry mapping on $P$.

Also, by the definition of symmetry, $A^{-1}$ is also a symmetry mapping on $P$.

Thus we have:

$A, B \in S_P \implies A \circ B \in S_P$
$A \in S_P \implies A^{-1} \in \S_P$

The result follows from the Two-Step Subgroup Test.

$\blacksquare$


Sources

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