avewright/proofwiki-math huggingface dataset:
Theorem: Torus as Surface of Revolution

Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
{{explain|Surface of revolution of what?}}
Let $C$ be a circle defined by $x^2 + \paren {y - 2}^2 = 1$ in the open upper half-plane.
Let the smooth local parametrization of $C$ be:
:$\map \gamma t = \tuple {\sin t, 2 + \cos t}$
Then the induced metric on $S_C$ is:
:$g = d t^2 + \paren {2 + \cos t}^2 \, d \theta^2$

Proof:
We have that:
:$\map {\gamma'} t = \tuple {\cos t, - \sin t}$
Furthermore:
:$\paren {\cos t}^2 + \paren {- \sin t}^2 = 1$
Hence, $\map \gamma t$ is a unit-speed curve.
By {{Corollary|Induced Metric on Surface of Revolution}}:
:$g = d t^2 + \paren {2 + \cos t}^2 \, d \theta^2$
{{qed}}
Theorem: Canonical P-adic Expansion of Rational is Eventually Periodic/Lemma 1

:$\forall n \in \N: \exists r_n \in \Z : \dfrac a b - \paren{p^{n + 1} \dfrac {r_n} b} \in \set{0, 1, \ldots, p^{n + 1} - 1}$

Proof:
Let $n \in \N$.
From Integer Coprime to all Factors is Coprime to Whole:
:$b, p^{n + 1}$ are coprime
From Integer Combination of Coprime Integers:
:$\exists c_n, d_n \in \Z : c_n b + d_n p^{n + 1} = 1$
Multiplying both sides by $a$:
:$a = a c_n b + a d_n p^{n + 1}$
Let $A_n$ be the least positive residue of $a c_n \pmod {p^{n + 1} }$.
By definition of least positive residue:
:$0 \le A_n \le p^{n + 1} - 1$
:$p^{n + 1} \divides \paren{a c_n - A_n}$
By definition of divisor: 
:$\exists x_n \in \Z : x_n p^{n + 1} = a c_n - A_n$
Re-arranging terms:
:$a c_n = x_n p^{n + 1} + A_n$
We have:
{{begin-eqn}}
{{eqn | l = a
      | r = \paren {x_n p^{n + 1} + A_n} b + a d_n p^{n + 1}
}}
{{eqn | r = A_n b + \paren {b x_n + a d_n} p^{n + 1}
      | c = Rearranging terms
}}
{{end-eqn}}
Let $r_n = b x_n + a d_n$.
Then:
{{begin-eqn}}
{{eqn | l = a
      | r = A_n b + r_n p^{n + 1}
}}
{{eqn | ll= \leadsto
      | l = \dfrac a b
      | r = A_n + p^{n + 1} \dfrac {r_n} b
      | c = dividing both sides by $b$
}}
{{eqn | ll= \leadsto
      | l = A_n
      | r = \dfrac a b - \paren{p^{n + 1} \dfrac {r_n} b}
      | c = rearranging terms
}}
{{end-eqn}}
By definition of least positive residue:
:$A_n \in \set{0, 1, \ldots, p^{n + 1} - 1}$
The result follows.
{{qed}}
Category:Canonical P-adic Expansion of Rational is Eventually Periodic
h0f24g8ao8otxl9fajtzhjhbsdp6uo8
Theorem: Product Inverse Operation Properties/Lemma 1

:$\forall x, y, z \in G: \paren {x \circ z} \oplus \paren {y \circ z} = x \oplus y$

Proof:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
      | l = \paren {x \circ z} \oplus \paren {y \circ z} 
      | r = \paren {x \oplus \paren {e \oplus z} } \oplus \paren {y \oplus \paren {e \oplus z} } 
      | c = Definition of $\circ$
}}
{{eqn | r = x \oplus y
      | c = $\text {PI} 4$: Cancellation Property
}}
{{end-eqn}}
{{qed}}
Theorem: Product Inverse Operation Properties/Lemma 2

:$\forall x, y, z \in G: \paren {x \oplus z} \circ \paren {z \oplus y} = x \oplus y$

Proof:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
      | l = \paren {x \oplus z} \circ \paren {z \oplus y} 
      | r = \paren {x \oplus z} \oplus \paren {e \oplus \paren {z \oplus y} }
      | c = Definition of $\circ$
}}
{{eqn | r = \paren {x \oplus z} \oplus \paren {y \oplus z}
      | c = $\text {PI} 3$: Product Inverse with Right Identity
}}
{{eqn | r = x \oplus y
      | c = $\text {PI} 4$: Cancellation Property
}}
{{end-eqn}}
{{qed}}
Theorem: Product Inverse Operation Properties/Lemma 3

:$\forall x, y \in G: \quad x \oplus y = e \implies x = y$

Proof:
Let $x \oplus y = e$.
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x, y \in G
      | l = \paren {x \oplus y} \oplus \paren {e \oplus y}
      | r = e \oplus \paren {e \oplus y}
      | c = from $x \oplus y = e$
}}
{{eqn | r = y \oplus e
      | c = $\text {PI} 3$: Product Inverse with Right Identity
}}
{{eqn | r = y
      | c = $\text {PI} 2$: Right Identity
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | q = \forall x, y \in G
      | l = \paren {x \oplus y} \oplus \paren {e \oplus y}
      | r = x \oplus e
      | c = $\text {PI} 4$: Cancellation Property
}}
{{eqn | r = x
      | c = $\text {PI} 2$: Right Identity
}}
{{end-eqn}}
That is:
:$x = y$
Hence the result.
{{qed}}
Theorem: Product Inverse Operation Properties/Lemma 4

:$\forall x, y, z \in G: x \oplus z = y \oplus z \implies x = y$

Proof:
Let $x \oplus z = y \oplus z$.
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
      | l = \paren {x \oplus z} \oplus \paren {y \oplus z}
      | r = \paren {x \oplus z} \oplus \paren {x \oplus z}
      | c = from $x \oplus z = y \oplus z$
}}
{{eqn | r = x \oplus x
      | c = $\text {PI} 4$: Cancellation Property
}}
{{eqn | r = e
      | c = $\text {PI} 1$: Self-Inverse Property
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in G
      | l = \paren {x \oplus z} \oplus \paren {y \oplus z}
      | r = x \oplus y
      | c = $\text {PI} 4$: Cancellation Property
}}
{{end-eqn}}
Thus we have:
:$x \oplus y = e$
as both are equal to $\paren {x \oplus z} \oplus \paren {y \oplus z}$.
Then from Lemma $3$:
:$x = y$
Hence the result.
{{qed}}
Theorem: Product Inverse Operation Properties/Lemma 5

:$\forall x, y \in G: \paren {x \circ y} \oplus y = x$

Proof:
{{begin-eqn}}
{{eqn | q = \forall x, y \in G
      | l = \paren {x \circ y} \oplus y
      | r = \paren {x \oplus \paren {e \oplus y} } \oplus y
      | c = Definition of $\circ$
}}
{{eqn | r = \paren {x \oplus \paren {e \oplus y} } \oplus \paren {y \oplus e}
      | c = $\text {PI} 2$: Right Identity
}}
{{eqn | r = \paren {x \oplus \paren {e \oplus y} } \oplus \paren {e \oplus \paren {e \oplus y} }
      | c = $\text {PI} 3$: Product Inverse with Right Identity
}}
{{eqn | r = x \oplus e
      | c = $\text {PI} 4$: Cancellation Property
}}
{{eqn | r = x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Operation Induced by Permutation on Magma is Closed

Let $\struct {S, \circ}$ be a magma.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\circ_\sigma$ be the operation on $S$ induced by $\sigma$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\circ_\sigma$ is closed on $S$

Proof:
{{Proofread|Check the validity of the proof for the case when S is the empty set}}
Suppose $S$ is the empty set.
Let $\sigma: S \to S$ be a permutation on $S$.
Since $S$ is empty, by definition, $\sigma$ is the empty map. 
Since $\circ_\sigma$ is the operation on $S$ induced by $\sigma$, it follows that $\circ_\sigma$  is the empty map.
It is vacuously true that $\circ_\sigma$ is closed on $S$, as required.
Suppose $S$ is non-empty.
Let $a, b \in S$.
By definition of magma, $\circ$ is closed on $S$.
Thus:
:$a \circ b \in S$
Hence:
:$a \circ b \in \Dom \sigma$
As $\sigma$ is a permutation on $S$, it follows directly that:
:$\map \sigma {a \circ b} \in S$
The operation on $S$ induced by $\sigma$ is well-defined.
In other words, $\circ_\sigma$ is well-defined.
Hence:
:$a \circ_\sigma b \in S$
{{qed}}
Category:Operations Induced by Permutations
Category:Magmas
ataui7gwii3u9ihhvf6fb9tit7a2ibx
Theorem: Structure Induced by Permutation on Semigroup is not necessarily Semigroup

Let $\struct {S, \circ}$ be a semigroup.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is not necessarily itself a semigroup.

Proof:
From Operation Induced by Permutation on Magma is Closed we have that $\struct {S, \circ_\sigma}$ is a closed structure.
Hence {{Semigroup-axiom|0}} holds.
However, we have that Operation Induced by Permutation on Semigroup is not necessarily Associative.
Hence {{Semigroup-axiom|1}} does not necessarily hold for $\struct {S, \circ_\sigma}$ 
Hence $\struct {S, \circ_\sigma}$ is not necessarily a semigroup.
{{qed}}
Theorem: Operation Induced by Permutation on Semigroup is not necessarily Associative

Let $\struct {S, \circ}$ be a semigroup.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\circ_\sigma$ be the operation on $S$ induced by $\sigma$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\circ_\sigma$ is not necessarily associative on $S$.

Proof:
;Proof by Counterexample
Let $S = \set {a, b, c}$.
Let $\circ$ denote the right operation on $S$:
:$\forall x, y \in S: x \to y = y$
From Structure under Right Operation is Semigroup, $\struct {S, \circ}$ is a semigroup.
Hence we have:
:$a \circ \paren {b \circ c} = c = \paren {a \circ b} \circ c$
Let $\sigma$ denote the permutation on $S$ defined as:
{{begin-eqn}}
{{eqn | l = \map \sigma a
      | r = b
}}
{{eqn | l = \map \sigma b
      | r = c
}}
{{eqn | l = \map \sigma c
      | r = a
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = a \circ_\sigma \paren {b \circ_\sigma c}
      | r = a \circ_\sigma \map \sigma {b \circ c}
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{eqn | r = a \circ_\sigma \map \sigma c
      | c = {{Defof|Right Operation}}
}}
{{eqn | r = a \circ_\sigma a
      | c = Definition of $\sigma$
}}
{{eqn | r = \map \sigma {a \circ a}
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{eqn | r = \map \sigma a
      | c = {{Defof|Right Operation}}
}}
{{eqn | r = b
      | c = Definition of $\sigma$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \paren {a \circ_\sigma b} \circ_\sigma c
      | r = \map \sigma {a \circ b} \circ_\sigma c
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{eqn | r = \map \sigma b \circ_\sigma c
      | c = {{Defof|Right Operation}}
}}
{{eqn | r = c \circ_\sigma c
      | c = Definition of $\sigma$
}}
{{eqn | r = \map \sigma {c \circ c}
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{eqn | r = \map \sigma c
      | c = {{Defof|Right Operation}}
}}
{{eqn | r = a
      | c = Definition of $\sigma$
}}
{{end-eqn}}
So $a \circ_\sigma \paren {b \circ_\sigma c} \ne \paren {a \circ_\sigma b} \circ_\sigma c$ and the result follows.
{{qed}}
Category:Operations Induced by Permutations
Category:Semigroups
Category:Associativity
qikyz9t5j46vvgqgzl8h1qcpyd95k0j
Theorem: Structure under Right Operation is Semigroup

Let $\struct {S, \to}$ be an algebraic structure in which the operation $\to$ is the right operation.
Then $\struct {S, \to}$ is a semigroup.

Proof:
We need to verify the semigroup axioms:
{{:Axiom:Semigroup Axioms}}
By the nature of the right operation, $\struct {S, \to}$ is closed:
:$\forall x, y \in S: x \to y = y \in S$
whatever $S$ may be.
Hence {{Semigroup-axiom|0}} holds.
From Right Operation is Associative, $\to$ is associative.
Hence {{Semigroup-axiom|1}} holds.
So $\struct {S, \to}$ is a semigroup.
{{qed}}
Theorem: Structure under Left Operation is Semigroup

Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
Then $\struct {S, \gets}$ is a semigroup.

Proof:
We need to verify the semigroup axioms:
{{:Axiom:Semigroup Axioms}}
By the nature of the right operation, $\struct {S, \to}$ is closed:
:$\forall x, y \in S: x \gets y = y \in S$
whatever $S$ may be.
Hence {{Semigroup-axiom|0}} holds.
From Right Operation is Associative, $\gets$ is associative.
Hence {{Semigroup-axiom|1}} holds.
So $\struct {S, \gets}$ is a semigroup.
{{qed}}
Theorem: Right Identity in Semigroup may not be Unique

Let $\struct {S, \circ}$ be a semigroup.
Let $e_R$ be a right identity of $\struct {S, \circ}$.
Then it is not necessarily the case that $e_R$ is unique.

Proof:
;Proof by Counterexample
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
From Structure under Left Operation is Semigroup, $\struct {S, \gets}$ is a semigroup.
From Element under Left Operation is Right Identity, every element of $\struct {S, \gets}$ is a right identity.
The result follows.
{{qed}}
Category:Identity Elements
Category:Semigroups
epw1apctnwhwp7ima8hh50oj58ojz0l
Theorem: Left Operation has no Left Identities

Let $S$ be a set with more than $1$ element.
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
Then $\struct {S, \gets}$ has no left identities.

Proof:
From Element under Left Operation is Right Identity, every element of $\struct {S, \gets}$ is a right identity.
Because there are at least $2$ elements in $\struct {S, \gets}$, it follows that $\struct {S, \gets}$ has more than one right identity.
From More than one Right Identity then no Left Identity, it follows that $\struct {S, \gets}$ has no left identity.
{{qed}}
Theorem: Right Operation has no Right Identities

Let $S$ be a set with more than $1$ element.
Let $\struct {S, \to}$ be an algebraic structure in which the operation $\to$ is the right operation.
Then $\struct {S, \to}$ has no right identities.

Proof:
From Element under Right Operation is Left Identity, every element of $\struct {S, \to}$ is a left identity.
Because there are at least $2$ elements in $\struct {S, \to}$, it follows that $\struct {S, \to}$ has more than one left identity.
From More than one Left Identity then no Right Identity, it follows that $\struct {S, \to}$ has no right identity.
{{qed}}
Theorem: Left Identity in Semigroup may not be Unique

Let $\struct {S, \circ}$ be a semigroup.
Let $e_L$ be a left identity of $\struct {S, \circ}$.
Then it is not necessarily the case that $e_L$ is unique.

Proof:
;Proof by Counterexample
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\to$ is the right operation.
From Structure under Right Operation is Semigroup, $\struct {S, \to}$ is a semigroup.
From Element under Right Operation is Left Identity, every element of $\struct {S, \to}$ is a left identity.
The result follows.
{{qed}}
Theorem: Structure Induced by Permutation on Quasigroup is Quasigroup

Let $\struct {S, \circ}$ be a quasigroup.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is also a quasigroup.

Proof:
By definition of quasigroup:
:$\forall a, b \in S: \exists ! x \in S: x \circ a = b$
:$\forall a, b \in S: \exists ! y \in S: a \circ y = b$
Let $a, b \in S$.
As $\sigma$ is a permutation, it is by definition both surjective and injective.
We have that:
:$\exists ! x: x \circ a = b$
Thus:
{{begin-eqn}}
{{eqn | q = \exists ! x \in S
      | l = \map \sigma {x \circ a}
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \circ_\sigma a
      | r = b
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{end-eqn}}
Similarly, we have that:
:$\exists ! x: a \circ x = b$
Thus:
{{begin-eqn}}
{{eqn | q = \exists ! x \in S
      | l = \map \sigma {a \circ x}
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \circ_\sigma x
      | r = b
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Structure Induced by Permutation on Algebra Loop is not necessarily Algebra Loop

Let $\struct {S, \circ}$ be an algebra loop.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is not necessarily also an algebra loop.

Proof:
Consider the Cayley table of the algebra loop on $S = \set {e, a, b}$:
:$\begin{array}{r|rrr}
\circ & e & a & b \\
\hline
e & e & a & b
\\
a & a & b & e
\\
b & b & e & a
\\
\end{array}$
Consider the permutation on $S$:
Let $\sigma$ denote the permutation on $S$ defined as:
{{begin-eqn}}
{{eqn | l = \map \sigma e
      | r = a
}}
{{eqn | l = \map \sigma a
      | r = e
}}
{{eqn | l = \map \sigma b
      | r = b
}}
{{end-eqn}}
Then the Cayley table of the structure induced by $\sigma$ on $\circ$ is seen to be:
:$\begin{array}{r|rrr}
\circ & e & a & b \\
\hline
e & a & e & b
\\
a & e & b & a
\\
b & b & a & e
\\
\end{array}$
It is apparent by inspection that this is the Cayley table of a quasigroup.
However, there is no identity element.
Hence by definition $\struct {S, \circ_\sigma}$ is not an algebra loop.
{{qed}}
Theorem: Structure Induced by Permutation on Commutative Quasigroup is Commutative Quasigroup

Let $\struct {S, \circ}$ be a quasigroup such that $\circ$ is a commutative operation.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is also a quasigroup such that $\circ_\sigma$ is a commutative operation.

Proof:
From definition of a quasigroup, $\struct {S, \circ}$ is closed under $\circ$.
From Structure Induced by Permutation on Quasigroup is Quasigroup, $\struct {S, \circ_\sigma}$ is a quasigroup.
Again, from definition of a quasigroup, $\struct {S, \circ_\sigma}$ is closed under $\circ_\sigma$.
Hence:
{{begin-eqn}}
{{eqn | q = \forall a, b \in S
      | l = a \circ_\sigma b
      | r = \map \sigma {a \circ b}
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{eqn | r = \map \sigma {b \circ a}
      | c = {{Defof|Commutative Operation}}
}}
{{eqn | r = b \circ_\sigma a
      | c = {{Defof|Operation Induced by Permutation}}
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Isomorphism between Structures Induced by Permutations

Let $S$ be a set.
Let $\oplus$ and $\otimes$ be closed operations on $S$ such that both $\oplus$ and $\otimes$ have the same identity.
Let $\sigma$ and $\tau$ be permutations on $S$.
Let $\oplus_\sigma$ and $\otimes_\tau$ be the operations on $S$ induced on $\oplus$ by $\sigma$ and on $\otimes$ by $\tau$ respectively:
:$\forall x, y \in S: x \oplus_\sigma y := \map \sigma {x \oplus y}$
:$\forall x, y \in S: x \otimes_\tau y := \map \tau {x \otimes y}$
Let $f: S \to S$ be a mapping.
Then:
:$f$ is an isomorphism from $\struct {S, \oplus_\sigma}$ to $\struct {S, \otimes_\tau}$
{{iff}}:
:$f$ is an isomorphism from $\struct {S, \oplus}$ to $\struct {S, \otimes}$ satisfying the condition:
:::$f \circ \sigma = \tau \circ f$
:where $\circ$ denotes composition of mappings.

Proof:
Recall that:
:an isomorphism is a bijection which is a homomorphism
:a permutation is a bijection from a set to itself.
Hence on both sides of the double implication:
:$f$ is a permutation on $S$
:both $f \circ \sigma$ and $\tau \circ f$ are permutations on $S$.
So bijectivity of all relevant mappings can be taken for granted throughout the following.
Theorem: Count of Commutative Quasigroups on Set given Count of Commutative Algebra Loops

Let $S$ be a finite set of cardinality $n$.
Let $e \in S$.
Let there be $m$ commutative operations $\oplus$ on $S$ such that $\struct {S, \oplus}$ is an algebra loop whose identity is $e$.
Then there are $n! m$ commutative binary operations $\otimes$ on $S$ such that $\struct {S, \otimes}$ is a quasigroup.

Proof:
Consider a commutative algebra loop $\struct {S, \oplus}$ whose identity is $e$.
Consider the row of the Cayley table of $S$ which is headed by $e$.
There are $n!$ permutations of the elements of this row.
Each of these corresponds to a different commutative operation on $S$, as the corresponding column headed by $e$ is constrained to be the same as the row.
The result follows by the Product Rule for Counting.
{{qed}}
Theorem: Regular Representations wrt Element are Permutations then Element is Invertible

Let $\struct {S, \circ}$ be a semigroup.
Let $\lambda_a: S \to S$ and $\rho_a: S \to S$ be the left regular representation and right regular representation with respect to $a$ respectively:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \map {\lambda_a} x
      | r = a \circ x
}}
{{eqn | q = \forall x \in S
      | l = \map {\rho_a} x
      | r = x \circ a
}}
{{end-eqn}}
Let both $\lambda_a$ and $\rho_a$ be permutations on $S$.
Then there exists an identity element for $\circ$ and $a$ is invertible.

Proof:
We have that $\rho_a$ is a permutation on $S$.
Hence:
{{begin-eqn}}
{{eqn | q = \exists g \in S
      | l = a
      | r = \map {\rho_a} g
      | c = 
}}
{{eqn | r = g \circ a
      | c = {{Defof|Right Regular Representation}}
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | q = \forall b \in S
      | l = \paren {b \circ g} \circ a
      | r = b \circ \paren {g \circ a}
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | r = b \circ a
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = b \circ g
      | r = b
      | c = Right Cancellable iff Right Regular Representation Injective
}}
{{end-eqn}}
which demonstrates that $g$ is a right identity for $\circ$.
In the same way, we have that $\lambda_a$ is also a permutation on $S$.
Hence:
{{begin-eqn}}
{{eqn | q = \exists g \in S
      | l = a
      | r = \map {\lambda_a} g
      | c = 
}}
{{eqn | r = a \circ g
      | c = {{Defof|Left Regular Representation}}
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | q = \forall b \in S
      | l = \paren {a \circ g} \circ b
      | r = a \circ \paren {g \circ b}
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | r = a \circ b
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = g \circ b
      | r = b
      | c = Left Cancellable iff Left Regular Representation Injective
}}
{{end-eqn}}
which demonstrates that $g$ is a left identity for $\circ$.
So, by definition, $g$ is an identity element for $\circ$.
Again, we have that $\rho_a$ is a permutation on $S$, and so:
{{begin-eqn}}
{{eqn | q = \exists h \in S
      | l = g
      | r = \map {\rho_a} h
      | c = 
}}
{{eqn | r = h \circ a
      | c = {{Defof|Right Regular Representation}}
}}
{{end-eqn}}
and that $\lambda_a$ is also a permutation on $S$, and so:
{{begin-eqn}}
{{eqn | q = \exists h \in S
      | l = g
      | r = \map {\lambda_a} h
      | c = 
}}
{{eqn | r = a \circ h
      | c = {{Defof|Right Regular Representation}}
}}
{{end-eqn}}
So $h$ is both a left inverse and a right inverse for $a$.
Hence by definition $h$ is an inverse for $a$.
Hence $a$ is invertible by definition.
{{qed}}
Theorem: Regular Representations in Semigroup are Permutations then Structure is Group

Let $\struct {S, \circ}$ be a semigroup.
For $a \in S$, let $\lambda_a: S \to S$ and $\rho_a: S \to S$ denote the left regular representation and right regular representation with respect to $a$ respectively:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \map {\lambda_a} x
      | r = a \circ x
}}
{{eqn | q = \forall x \in S
      | l = \map {\rho_a} x
      | r = x \circ a
}}
{{end-eqn}}
For all $a$ in $S$, let $\lambda_a$ be a permutation on $S$.
Let there exist $b$ in $S$ such that $\rho_b$ is a permutation on $S$.
Then $\struct {S, \circ}$ is a group.

Proof:
We have that $\lambda_a$ be a permutation on $S$ for all $a \in S$.
In particular this applies to $b$.
So:
:$\lambda_b$ is a permutation on $S$
:$\rho_b$ is a permutation on $S$
and so from Regular Representations wrt Element are Permutations then Element is Invertible:
:$\struct {S, \circ}$ has an identity element
:$b$ is invertible in $\struct {S, \circ}$.
It remains to be shown that $a$ is invertible in $S$ for all $a \in S$.
Let the identity element of $\struct {S, \circ}$ be $e$.
We have that $\lambda_a$ be a permutation on $S$ for all $a \in S$.
Hence:
{{begin-eqn}}
{{eqn | q = \forall a \in S: \exists a' \in S
      | l = e
      | r = \map {\lambda_a} {a'}
      | c = 
}}
{{eqn | r = a \circ a'
      | c = 
}}
{{end-eqn}}
Thus for all $a$ in $S$, $a$ has a left inverse $a'$.
Now as $e$ is an identity element of $S$, it is by definition a left identity.
Hence from Left Inverse for All is Right Inverse, $a'$ is also a right inverse for $a$.
That is, every $a \in S$ has an element which is both a left inverse and a right inverse.
That is, every $a \in S$ has an inverse.
So we have that $\struct {S, \circ}$ is a semigroup such that:
:$\struct {S, \circ}$ has an identity element
:every element of $S$ has an inverse.
Hence, by definition, $\struct {S, \circ}$ is a group.
{{qed}}