Set of Isometries in Euclidean Space under Composition forms Group

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {\R^n, d}$ be a real Euclidean space of $n$ dimensions.

Let $S$ be the set of all mappings $f: \R^n \to \R^n$ which preserve distance:

That is:

$\map d {\map f a, \map f b} = \map d {a, b}$

Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.


Then $\struct {S, \circ}$ is a group.


Proof

From Euclidean Metric on Real Vector Space is Metric, $\R^n$ is a metric space.

Hence it is seen that a complex function $f: \C \to \C$ which preserves distance is in fact an isometry on $\C$.


Taking the group axioms in turn:


Group Axiom $\text G 0$: Closure

Let $f$ and $g$ be isometries on $\C$.

We have:

\(\ds \forall a, b \in \C: \, \) \(\ds \map d {\map {\paren {g \circ f} } a, \map {\paren {g \circ f} } b}\) \(=\) \(\ds \map d {\map g {\map f a}, \map g {\map f b} }\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map d {\map g a, \map g b}\) $f$ is an Isometry
\(\ds \) \(=\) \(\ds \map d {a, b}\) $g$ is an Isometry

Thus $g \circ f$ is an isometry, and so $\struct {S, \circ}$ is closed.

$\Box$


Group Axiom $\text G 1$: Associativity

By Composition of Mappings is Associative, $\circ$ is an associative operation.

$\Box$


Group Axiom $\text G 2$: Existence of Identity Element

The identity mapping is the identity element of $\struct {S, \circ}$.

$\Box$


Group Axiom $\text G 3$: Existence of Inverse Element

By definition, an isometry is a bijection.

For $f \in S$, we have that $f^{-1}$ is also an isometry.

Thus every element of $f$ has an inverse $f^{-1}$.

$\Box$


All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Set_of_Isometries_in_Euclidean_Space_under_Composition_forms_Group&oldid=527959"