Set of Isometries in Euclidean Space under Composition forms Group

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:

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

We have:

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

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

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

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 $f^{-1}$.

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