Symmetry Group is Group

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 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 symmetries on $P$.

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

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

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.3$
• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \eta$
• 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 34$. Examples of groups: $(5)$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $1$: Definitions and Examples: Example $1.9$