avewright/proofwiki-math huggingface dataset:
Theorem: Identity of Affine Group of One Dimension

Let $\map {\mathrm {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$.
Then $\map {\mathrm {Af}_1} \R$ has $f_{1, 0}$ as an identity element.

Proof:
Let $f_{a b} \in \map {\mathrm {Af}_1} \R$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f_{a b} \circ f_{1, 0} } } x
      | r = a \paren {1 x + 0} + b
      | c = 
}}
{{eqn | r = a x + b
      | c = 
}}
{{eqn | r = f_{a b}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\paren {f_{1, 0} \circ f_{a b} } } x
      | r = 1 \paren {a x + b} + 0
      | c = 
}}
{{eqn | r = a x + b
      | c = 
}}
{{eqn | r = f_{a b}
      | c = 
}}
{{end-eqn}}
Thus $f_{1, 0}$ is the identity element of $\map {\mathrm {Af}_1} \R$.
{{qed}}
Category:Affine Groups
tshbkac0nysbwi84oa6q266ws8cg59d
Theorem: Inverse in Affine Group of One Dimension

Let $\map {\operatorname {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$.
Let $f_{a b} \in \map {\operatorname {Af}_1} \R$.
Let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
Then $f_{c d} \in \map {\operatorname {Af}_1} \R$ is the inverse of $f_{a b}$.

Proof:
{{begin-eqn}}
{{eqn | l = y
      | r = a x + b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y - b
      | r = a x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {y - b} a
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac y a + \frac {- b} a
      | r = x
      | c = 
}}
{{end-eqn}}
As $a \in \R_{\ne 0}$ by definition of $\map {\operatorname {Af}_1} \R$ it follows that $\dfrac 1 a \in \R_{\ne 0}$ and $\dfrac {-b} a \in \R$.
So let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f_{a b} \circ f_{c d} } x
      | r = a \paren {\frac 1 a x + \frac {-b} a} + b
      | c = 
}}
{{eqn | r = \frac a a x + \frac {-a b} a + b
      | c = 
}}
{{eqn | r = 1 x + 0
      | c = 
}}
{{eqn | r = \map {f_{1, 0} } x
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \map {f_{c d} \circ f_{a b} } x
      | r = \frac 1 a \paren {a x + b} + \frac {-b} a
      | c = 
}}
{{eqn | r = \frac a a x + \frac b a + \frac {-b} a
      | c = 
}}
{{eqn | r = 1 x + 0
      | c = 
}}
{{eqn | r = \map {f_{1, 0} } x
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Commutation with Group Elements implies Commuation with Product with Inverse

Let $G$ be a group.
Let $a, b, c \in G$ such that $a$ commutes with both $b$ and $c$.
Then $a$ commutes with $b c^{-1}$.

Proof:
{{begin-eqn}}
{{eqn | l = a b c^{-1}
      | r = b a c^{-1}
      | c = as $a$ commutes with $b$
}}
{{eqn | r = b c^{-1} a
      | c = Commutation with Inverse in Monoid
}}
{{end-eqn}}
{{qed}}
Theorem: Möbius Strip has Euler Characteristic Zero

Let $M$ be a Möbius Strip.
Then:
:$\map \chi M = 0$
where $\map \chi M$ denotes the Euler characteristic of the graph $M$.

Proof:
Let the number of vertices, edges and faces of $M$ be $V$, $E$ and $F$ respectively.
From Möbius Strip has no Vertices:
:$V = 0$
From Möbius Strip has 1 Edge:
:$E = 1$
From Möbius Strip has 1 Face:
:$F = 1$
By definition of the Euler characteristic:
{{begin-eqn}}
{{eqn | l = \map \chi M
      | r = V - E + F
      | c = {{Defof|Euler Characteristic}}
}}
{{eqn | r = 0 - 1 + 1
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Centralizer of Subset is Intersection of Centralizers of Elements

Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Let $\map {C_G} S$ be the centralizer of $S$ in $G$.
Then:
:$\ds \map {C_G} S = \bigcap_{x \mathop \in S} \map {C_G} x$
where $\map {C_G} z$ is the centralizer of $x$ in $G$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {C_G} S
      | r = \set {g \in G: \forall x \in S: g \circ x = x \circ g}
      | c = {{Defof|Centralizer of Group Subset}}
}}
{{eqn | r = \set {g \in G: \forall x \in S: g \in \map {C_G} x}
      | c = {{Defof|Centralizer of Group Element}}
}}
{{eqn | r = \bigcap_{x \mathop \in S} \map {C_G} x
      | c = {{Defof|Intersection of Family}}
}}
{{end-eqn}}
{{qed}}
Theorem: Permutation Representation defines Group Action

Let $G$ be a group whose identity is $e$.
Let $X$ be a set.
Let $\map \Gamma X$ be the symmetric group of $X$.
Let $\rho: G \to \map \Gamma X$ be a permutation representation, that is, a homomorphism.
The mapping $\phi: G \times X \to X$ associated to $\rho$, defined by:
:$\map \phi {g, x} = \map {\map \rho g} x$
is a group action.

Proof:
Let $g, h \in G$ and $x \in X$.
We verify that $g * \paren {h * x} = \paren {g h} * x$:
{{begin-eqn}}
{{eqn | l = g * \paren {h * x}
      | r = \map {\map \rho g} {\map {\map \rho h} x}
      | c = Definition of $\phi$
}}
{{eqn | r = \map {\paren {\map \rho g \circ \map \rho h} } x
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map {\map \rho {g h} } x
      | c = $\rho$ is a homomorphism
}}
{{eqn | r = \paren {g h} * x
      | c = Definition of $\phi$
}}
{{end-eqn}}
We verify that $e * x = x$.
Let $I_X$ denote the identity mapping on $X$.
{{begin-eqn}}
{{eqn | l = e*x
      | r = \map {\map \rho e} x
      | c = Definition of $\phi$
}}
{{eqn | r = \map {I_X} x
      | c = Group Homomorphism Preserves Identity, Set of all Self-Maps under Composition forms Monoid
}}
{{eqn | r = x
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
{{qed}}
Theorem: Group Action of Symmetric Group

Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.
Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.
The mapping $*: S_n \times \N_n \to \N_n$ defined as:
:$\forall \pi \in S_n, \forall n \in \N_n: \pi * n = \map \pi n$
is a group action.

Proof:
The group action axioms are investigated in turn.
Let $\pi, \rho \in S_n$ and $n \in \N_n$.
Thus:
{{begin-eqn}}
{{eqn | l = \pi * \paren {\rho * n}
      | r = \pi * \map \rho n
      | c = Definition of $*$
}}
{{eqn | r = \map \pi {\map \rho n}
      | c = Definition of $*$
}}
{{eqn | r = \map {\paren {\pi \circ \rho} } n
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \paren {\pi \circ \rho} * n
      | c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
Then:
{{begin-eqn}}
{{eqn | l = I_{\N_n} * n
      | r = \map {I_{\N_n} } n
      | c = where $I_{\N_n}$ is the identity mapping on ${\N_n}$
}}
{{eqn | r = n
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
{{qed}}
Theorem: Group Action of Symmetric Group Acts Transitively

Let $S$ be a set.
Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$.
Let $*: \map \Gamma S \times S \to S$ be the group action defined as:
:$\forall \pi \in \map \Gamma S, \forall s \in S: \pi * s = \map \pi s$
Then $*$ is a transitive group action.
In other words, $\struct {\map \Gamma S, \circ}$ acts transitively on $S$ by $*$.

Proof:
By Group Action of Symmetric Group, $*: \map \Gamma S \times S \to S$ is indeed a group action
Let $s, t \in S$.
As $\map \Gamma S$ is the symmetric group on $S$, there exists a permutation $\pi \in \map \Gamma S$ such that:
:$\map \pi s = t$
This holds for any $s, t \in S$.
Thus:
:$\forall t \in S: t \in \Orb s$
and so $S$ consists of a single orbit.
Hence the result by definition of transitive group action.
{{qed}}
Theorem: Trivial Group Action is Group Action

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S$ be a non-empty set.
{{explain|Why does $S$ have to be non-empty?}}
Let $*: G \times S \to S$ be the trivial group action:
:$\forall \tuple {g, s} \in G \times S: g * s = s$
Then $*$ is indeed a group action.

Proof:
The group action axioms are investigated in turn.
Let $g_1, g_2 \in G$ and $s \in S$.
Thus:
{{begin-eqn}}
{{eqn | l = g_1 * \paren {g_2 * s}
      | r = g_1 * s
      | c = {{Defof|Trivial Group Action}}
}}
{{eqn | r = s
      | c = {{Defof|Trivial Group Action}}
}}
{{eqn | r = \paren {g_1 \circ g_2} * s
      | c = {{Defof|Trivial Group Action}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
Then:
{{begin-eqn}}
{{eqn | l = e * s
      | r = s
      | c = {{Defof|Trivial Group Action}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
{{qed}}
Category:Group Actions
Category:Trivial Group Action
jzechsqt61z65dkg1shdmvknibvtxd5
Theorem: Orbit of Trivial Group Action is Singleton

Let $\left({G, \circ}\right)$ be a group whose identity is $e$.
Let $S$ be a set.
Let $*: G \times S \to S$ be the trivial group action:
:$\forall \left({g, s}\right) \in G \times S: g * s = s$
Let $s \in S$.
Then the orbit of $s$ under $*$ is $\left\{{s}\right\}$.

Proof:
By definition:
:$\operatorname{Orb} \left({s}\right) = \left\{{t \in S: \exists g \in G: g * s = t}\right\}$
By definition of the trivial group action:
:$\forall g \in G: g * s = s$
Hence the result.
{{qed}}
Theorem: Right Regular Representation by Inverse is Group Action

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the operation:
:$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$
where $\rho_g$ is the right regular representation of $G$ with respect to $g$.
Then $*$ is a group action.

Proof:
The group action axioms are investigated in turn.
Let $g, h, a \in G$.
Thus:
{{begin-eqn}}
{{eqn | l = g * \paren {h * a}
      | r = g * \map {\rho_{h^{-1} } } a
      | c = Definition of $*$
}}
{{eqn | r = g * \paren {a \circ h^{-1} }
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = \map {\rho_{g^{-1} } } {a \circ h^{-1} }
      | c = Definition of $*$
}}
{{eqn | r = \paren {a \circ h^{-1} } \circ g^{-1}
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = a \circ \paren {h^{-1} \circ g^{-1} }
      | c = {{Group-axiom|1}}
}}
{{eqn | r = a \circ \paren {g \circ h}^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = \map {\rho_{\paren {g \circ h}^{-1} } } a
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = \paren {g \circ h} * a
      | c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
Then:
{{begin-eqn}}
{{eqn | l = e * s
      | r = \map {\rho_{e^{-1} } } s
      | c = Definition of $*$
}}
{{eqn | r = s \circ e^{-1}
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = s \circ e
      | c = Inverse of Identity Element is Itself
}}
{{eqn | r = s
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
{{qed}}
Theorem: Left Regular Representation is Group Action

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the operation:
:$\forall g, h \in G: g * h = \map {\lambda_g} h$
where $\lambda_g$ is the left regular representation of $G$ with respect to $g$.
Then $*$ is a group action.

Proof:
The group action axioms are investigated in turn.
Let $g, h, a \in G$.
Thus:
{{begin-eqn}}
{{eqn | l = g * \paren {h * a}
      | r = g * \map {\lambda_h} a
      | c = Definition of $*$
}}
{{eqn | r = g * \paren {h \circ a}
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = \map {\lambda_g} {h \circ a}
      | c = Definition of $*$
}}
{{eqn | r = g \circ \paren {h \circ a}
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = \paren {g \circ h} \circ a
      | c = {{Group-axiom|1}}
}}
{{eqn | r = \map {\lambda_{\paren {g \circ h} } } a
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = \paren {g \circ h} * a
      | c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
Then:
{{begin-eqn}}
{{eqn | l = e * s
      | r = \map {\lambda_e} s
      | c = Definition of $*$
}}
{{eqn | r = e \circ s
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = s
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
{{qed}}
Theorem: Right Regular Representation by Inverse is Transitive Group Action

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the group action:
:$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$
where $\rho_g$ is the right regular representation of $G$ with respect to $g$.
Then $*$ is a transitive group action.

Proof:
Let $g, h \in G$.
Then:
{{begin-eqn}}
{{eqn | q = \exists a \in G
      | l = h
      | r = a \circ g^{-1}
      | c = Group has Latin Square Property
}}
{{eqn | ll= \leadsto
      | l = h
      | r = \map {\rho_{g^{-1} } } a
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | ll= \leadsto
      | l = h
      | r = g * a
      | c = Definition of $*$
}}
{{end-eqn}}
$h$ is arbitrary, therefore the above holds for all $h \in G$.
By definition of orbit:
:$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$
That is:
:$\Orb h = G$
Hence the result by definition of transitive group action.
{{qed}}
Theorem: Left Regular Representation is Transitive Group Action

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the group action:
:$\forall g, h \in G: g * h = \map {\lambda_g} h$
where $\lambda_g$ is the left regular representation of $G$ with respect to $g$.
Then $*$ is a transitive group action.

Proof:
Let $g, h \in G$.
Then:
{{begin-eqn}}
{{eqn | q = \exists a \in G
      | l = h
      | r = g \circ a
      | c = Group has Latin Square Property
}}
{{eqn | ll= \leadsto
      | l = h
      | r = \map {\lambda_g} a
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | ll= \leadsto
      | l = h
      | r = g * a
      | c = Definition of $*$
}}
{{end-eqn}}
$h$ is arbitrary, therefore the above holds for all $h \in G$.
By definition of orbit:
:$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$
That is:
:$\Orb h = G$
Hence the result by definition of transitive group action.
{{qed}}
Theorem: Conjugacy Action is not Transitive

Let $\struct {G, \circ}$ be a non-trivial group whose identity is $e$.
Let $*: G \times G \to G$ be the conjugacy group action:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Then $*$ is ''not'' a transitive group action.

Proof:
Proof by Counterexample:
For $G$ to be a transitive group action, the orbit of any element of $G$ needs to be the whole of $G$.
Take $h = e$.
Then:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = g * e
      | r = g \circ e \circ g^{-1}
      | c = {{Defof|Conjugacy Action}}
}}
{{eqn | r = g \circ g^{-1}
      | c = {{Group-axiom|2}}
}}
{{eqn | r = e
      | c = {{Group-axiom|3}}
}}
{{end-eqn}}
Thus by definition of orbit:
:$\Orb e = \set e$
Only when $G$ is the trivial group, that is, when:
:$G = \set e$
then:
:$\Orb e = G$.
Hence the result by definition of transitive group action.
{{qed}}
Theorem: Conjugacy Action on Abelian Group is Trivial

Let $\struct {G, \circ}$ be an abelian group whose identity is $e$.
Let $*: G \times G \to G$ be the conjugacy group action:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Then $*$ is a trivial group action.

Proof:
For $G$ to be a trivial group action, the orbit of any element of $G$ is a singleton containing only that element.
Take $h \in G$.
Then:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = g * h
      | r = g \circ h \circ g^{-1}
      | c = 
}}
{{eqn | r = h \circ g \circ g^{-1}
      | c = {{Defof|Abelian Group}}: $g$ commutes with $h$
}}
{{eqn | r = h \circ e
      | c = {{Group-axiom|3}}
}}
{{eqn | r = h
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
Thus by definition of orbit:
:$\Orb h = \set h$
Hence the result by definition of trivial group action.
{{qed}}
Theorem: Group Action on Subgroup by Left Regular Representation

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the operation defined as:
:$\forall h \in H: \forall g \in G: h * g = \map {\lambda_h} g$
where $\map {\lambda_h} g$ is the left regular representation of $g$ by $h$.
Then $*$ is a group action.

Proof:
The group action axioms are investigated in turn.
Let $h_1, h_2 \in H$ and $g \in G$.
Thus:
{{begin-eqn}}
{{eqn | l = h_1 * \paren {h_2 * g}
      | r = h_1 * \paren {\map {\lambda_{h_2} } g}
      | c = Definition of $*$
}}
{{eqn | r = h_1 * \paren {h_2 \circ g}
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = \map {\lambda_{h_1} } {h_2 \circ g}
      | c = Definition of $*$
}}
{{eqn | r = h_1 \circ \paren {h_2 \circ g}
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = \paren {h_1 \circ h_2} \circ g
      | c = {{Group-axiom|1}}
}}
{{eqn | r = \map {\lambda_{h_1 \circ h_2} } g
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = \paren {h_1 \circ h_2} * g
      | c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
{{qed|lemma}}
Then:
{{begin-eqn}}
{{eqn | l = e * g
      | r = \map {\lambda_e} g
      | c = Definition of $*$
}}
{{eqn | r = e \circ g
      | c = {{Defof|Left Regular Representation}}
}}
{{eqn | r = g
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
{{qed|lemma}}
The group action axioms are thus seen to be fulfilled, and so $*$ is a group action.
{{qed}}
Theorem: Group Action on Subgroup by Right Regular Representation

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the operation defined as:
:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$.
Then $*$ is a group action.

Proof:
The group action axioms are investigated in turn.
Let $h_1, h_2 \in H$ and $g \in G$.
Thus:
{{begin-eqn}}
{{eqn | l = h_1 * \paren {h_2 * g}
      | r = h_1 * \paren {\map {\rho_{h_2^{-1} } } g}
      | c = Definition of $*$
}}
{{eqn | r = h_1 * \paren {g \circ h_2^{-1} }
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = \map {\rho_{h_1^{-1} } } {h_2^{-1} \circ g}
      | c = Definition of $*$
}}
{{eqn | r = \paren {g \circ h_2^{-1} } \circ h_1^{-1}
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = g \circ \paren {h_2^{-1} \circ h_1^{-1} }
      | c = {{Group-axiom|1}}
}}
{{eqn | r = g \circ \paren {h_1 \circ h_2}^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = \map {\rho_{\paren {h_1 \circ h_2}^{-1} } } g
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = \paren {h_1 \circ h_2} * g
      | c = Definition of $*$
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|1}} holds.
Then:
{{begin-eqn}}
{{eqn | l = e * g
      | r = \map {\rho_e} g
      | c = Definition of $*$
}}
{{eqn | r = g \circ e^{-1}
      | c = {{Defof|Right Regular Representation}}
}}
{{eqn | r = g \circ e
      | c = Inverse of Identity Element is Itself
}}
{{eqn | r = g
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
demonstrating that {{GroupActionAxiom|2}} holds.
The group action axioms are thus seen to be fulfilled, and so $*$ is a group action.
{{qed}}
Theorem: Group Action on Subgroup by Right Regular Representation is not Transitive

Let $G$ be a group.
Let $H$ be a proper subgroup of $G$.
Let $*: H \times G \to G$ be the group action defined as:
:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$.
Then $*$ is not transitive.

Proof:
From Group Action on Subgroup by Right Regular Representation it is established that $*$ is a group action.
From Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset:
:$\forall x \in G: \Orb x = H x$
where $H x$ is the right coset of $H$ by $x$.
From Right Coset Space forms Partition it is apparent that $H x \ne G$ unless $H = G$.
Thus $\Orb x \ne G$ and the result follows by definition of transitive group action.
{{qed}}
Theorem: Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset

Let $G$ be a group.
Let $H$ be a proper subgroup of $G$.
Let $*: H \times G \to G$ be the group action defined as:
:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$.
Let $x \in G$.
Then the orbit of $x$ under $*$ is given by:
:$\forall x \in G: \Orb x = H x$
where $H x$ is the right coset of $H$ by $x$.

Proof:
We are given $G$ is a group and $H$ is a proper subgroup of $G$, so:
:$H < G$
We are given $*: H \times G \to G$ is the group action defined as:
:$\forall \tuple {h, g} \in H \times G: h * g = \map {\rho_{h^{-1} } } g$
where $\map {\rho_{h^{-1} } } g$ is the right regular representation of $g$ by $h^{-1}$.
We are also given $x \in G$. 
So there exists $x \in G$ and $h \in H$ with $H < G$ such that:
:$h * x = \map {\rho_{h^{-1} } } x \in G$
{{ProofWanted}}
Therefore the orbit of $x$ is given by:
:$\Orb x = \set {y \in G: \exists h \in H: y = h * x}$
Hence the result.
{{qed}}
Theorem: Left Coset by Identity

Then:
: $e H = H$
where $e H$ is the left coset of $H$ by $e$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = e H
      | r = \set {y \in G: \exists h \in H: y = e h}
      | c = {{Defof|Left Coset}} of $H$ by $e$
}}
{{eqn | r = \set {y \in G: \exists h \in H: y = h}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \set {y \in G: y \in H}
      | c = 
}}
{{eqn | r = H
      | c = 
}}
{{end-eqn}}
So $e H = H$.
{{qed}}
Theorem: Right Coset by Identity

Then:
: $H = H e$
where $H e$ is the right coset of $H$ by $e$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = H e
      | r = \set {x \in G: \exists h \in H: x = h e}
      | c = {{Defof|Right Coset}} of $H$ by $e$
}}
{{eqn | r = \set {x \in G: \exists h \in H: x = h}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \set {x \in G: x \in H}
      | c = 
}}
{{eqn | r = H
      | c = 
}}
{{end-eqn}}
So $H = H e$.
{{qed}}
Theorem: Inversion Mapping is Permutation

Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$.
Then $\iota$ is a permutation on $G$.

Proof:
The inversion mapping on $G$ is the mapping $\iota: G \to G$ defined by:
:$\forall g \in G: \map \iota g = g^{-1}$
where $g^{-1}$ is the inverse or $g$.
By Inversion Mapping is Involution, $\iota$ is an involution:
:$\forall g \in G: \map \iota {\map \iota g} = g$
The result follows from Involution is Permutation.
{{qed}}
Theorem: Inversion Mapping is Permutation

Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$.
Then $\iota$ is a permutation on $G$.

Proof:
=== Proof of Surjection ===
Let $a \in G$.
By definition of $\iota$:
:$\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$
By Inverse of Inverse:
:$\left({a^{-1}}\right)^{-1} = a$
Hence $a$ has a preimage.
Since $a$ was arbitrary, $\iota$ is a surjection.
=== Proof of Injection ===
Suppose for some $a, b \in G$ that:
:$\iota \left({a}\right) = \iota \left({b}\right)$
Then by the definition of $\iota$:
:$a^{-1} = b^{-1}$
It follows from Inverse in Group is Unique that:
:$a = b$
Hence $\iota$ is an injection.
{{qed|lemma}}
Hence by definition $\iota$ is a bijection.
A bijection from a set to itself is by definition a permutation.
{{qed}}
Theorem: Inversion Mapping is Mapping

Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$.
Then $\iota$ is indeed a mapping.

Proof:
To show that $\iota$ is a mapping, it is sufficient to show that:
:$\map \iota a \ne \map \iota b \implies a \ne b$:
{{begin-eqn}}
{{eqn | l = \map \iota a
      | o = \ne
      | r = \map \iota b
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = a^{-1}
      | o = \ne
      | r = b^{-1}
      | c = Definition of $\iota$
}}
{{eqn | ll= \leadsto
      | l = a \circ a^{-1} \circ b
      | o = \ne
      | r = a \circ b^{-1} \circ b
      | c = Cancellation Laws and {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
      | l = e \circ b
      | o = \ne
      | r = a \circ e
      | c = {{Group-axiom|3}}
}}
{{eqn | ll= \leadsto
      | l = b
      | o = \ne
      | r = a
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
{{qed}}
Category:Inversion Mappings
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