Permutation Group/Examples/Example on Real Numbers

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Example of Permutation Group

Let $\R$ denote the set of real numbers.

Let $\R_{\ge 0}$ denote the set of positive real numbers.


Let $S = \R_{\ge 0} \times \R$ denote the Cartesian product of $\R_{\ge 0}$ and $\R$.


Let $\tuple {a, b} \in S$.

Let $f_{a, b}: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map {f_{a, b} } x := a x + b$

Let $\GG$ be the set defined as:

$\GG = \set {f_{a, b}: \tuple {a, b} \in S}$


Let $\struct {S, \oplus}$ be the group where $\oplus$ is defined as:

$\forall \tuple {a, b}, \tuple {c, d} \in S: \tuple {a, b} \oplus \tuple {c, d} := \tuple {a c, a d + b}$


Then $\struct {\GG, \circ}$ is a permutation group on $\R$ which is isomorphic to $\struct {S, \oplus}$.


Proof

From Group Example: $\tuple {a c, a d + b}$ on Positive Reals by Reals, $\struct {S, \oplus}$ has been shown to be a group.


From Bijection Example: $a x + b$ on Real Numbers, we have that $f_{a, b}$ is a bijection from $\R$ to $\R$.

Hence by definition $f_{a, b}$ is a permutation on $\R$.

From Group Example: Linear Functions, $\struct {\GG, \circ}$ is a group.

Hence as $\struct {\GG, \circ}$ is a group of permutations on $\R$, it is by definition a permutation group on $\R$.


Let $\phi: \struct {\GG, \circ} \to \struct {S, \oplus}$ be the identity mapping on $S$:

$\forall \tuple {a, b} \in S: \map \phi {a, b} = \tuple {a, b}$

From Identity Mapping is Bijection we have that $\phi$ is a bijection.


Consider $f_{a, b}, f_{c, d} \in \GG$.

We have:

\(\ds \forall x \in \R: \, \) \(\ds \map \phi {\map {f_{a, b} \circ f_{c, d} } x}\) \(=\) \(\ds \map {f_{a, b} \circ f_{c, d} } x\) Definition of Identity Mapping
\(\ds \) \(=\) \(\ds \map {f_{a, b} } {\map {f_{c, d} } x}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map {f_{a, b} } {c x + d}\)
\(\ds \) \(=\) \(\ds \map {f_{a, b} } {a \paren {c x + d} + b}\)
\(\ds \) \(=\) \(\ds \map {f_{a, b} } {a c x + a d + b}\)
\(\ds \) \(=\) \(\ds \tuple {a, b} \oplus \tuple {c, d}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \map \phi {a, b} \oplus \map \phi {c, d}\) Definition of Identity Mapping

This demonstrates that $\phi$ is a (group) homomorphism.

So $\phi$ is a bijection which is a homomorphism.

Hence by definition $\phi$ is an isomorphism.


The result follows.

$\blacksquare$


Sources

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