Set of Isometries in Complex Plane under Composition forms Group

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Theorem

Let $S$ be the set of all complex functions $f: \C \to \C$ which preserve distance when embedded in the complex plane.

That is:

$\size {\map f a - \map f b} = \size {a - b}$

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 Complex Plane is Metric Space, $\C$ can be treated as 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:

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

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