nyuuzyou/wikis huggingface dataset:
Strong Twin Prime Conjecture/Historical Note

== Historical Note on Strong Twin Prime Conjecture ==
The Strong Twin Prime Conjecture was proposed by and in $1923$, as a special case of the First Hardy-Littlewood Conjecture.
 and present this as:
:Un argument probabiliste montre que, s'il existe une infinité de nombres premiers jumeaux, alors de nombre de ceux qui sont situés dans l'intervalle $\sqbrk {x, x + a}$ est de l'ordre de $C \cdot \dfrac a {\paren {\Log x}^2}$ avec $C = 1,32 \ldots$
In English:
:A probabilistic argument shows that, if there exists an infinite number of twin primes, then the number of those which are situated in the interval $\closedint x {x + a}$ is of the order of $C \cdot \dfrac a {\paren {\Log x}^2}$ where $C = 1,32 \ldots$
They also appear to attribute it to .
== Sources ==
* 
* : $1,32032 36316 \ldots$
* 
Category:Historical Notes
Strongly Locally Compact Space is Weakly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a strongly locally compact space.
Then $T$ is weakly locally compact.
== Proof==
Let $T = \struct {S, \tau}$ be strongly locally compact.
Let $x \in S$.
By definition, there exists an open set $U_x$ of $T$ such that:
:$x \in U_x$
:${U_x}^-$ (the closure of $U_x$) is compact.
From Set is Subset of its Topological Closure, $U_x \subseteq {U_x}^-$ and so $x \in {U_x}^-$.
Thus $x$ is contained in a compact neighborhood.
As this holds for all $x$, $T$ is a weakly locally compact.
== Also see ==
* Locally Compact Space is Weakly Locally Compact
* Sequence of Implications of Local Compactness Properties
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Weakly Locally Compact Spaces
Category:Strongly Locally Compact Spaces
Structure Induced by Abelian Group Operation is Abelian Group

== Theorem ==
Let $\struct {G, \circ}$ be an abelian group whose identity is $e$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.
Then $\struct {G^S, \oplus}$ is an abelian group.
== Proof ==
From Structure Induced by Group Operation is Group, $\struct {G^S, \oplus}$ is a group.
From Structure Induced by Commutative Operation is Commutative, so is the pointwise operation it induces on $G^S$.
Hence the result.
Category:Abelian Groups
Category:Pointwise Operations
Structure Induced by Absorbing Operations is Absorbing

== Theorem ==
Let $\struct {T, \circ, *}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus, \otimes}$ be the structure on $T^S$ induced by $\circ$ and $*$.
Let $\circ$ and $*$ satisfy the absorption law:
:$\forall a, b \in S: a \circ \paren {a * b} = a$
Then the pointwise operations $\oplus$ and $\otimes$ on $T^S$ also satisfy the absorption law:
:$\forall f, g \in T^S: f \oplus \paren {f \otimes g} = f$
== Proof ==
Let $f, g \in T^S$.
Then:
From Equality of Mappings:
:$f \oplus \paren {f \otimes g} = f$
Since $f, g$ were arbitrary:
:$\forall f, g \in T^S : f \oplus \paren {f \otimes g} = f$
Hence $\oplus$ absorbs $\otimes$ by definition.
Category:Absorption Laws
Category:Pointwise Operations
Structure Induced by Associative Operation is Associative

== Theorem ==
Let $\struct {T, \circ}$ be an algebraic structure.
Let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be associative.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also associative.
== Proof ==
Let $f, g, h \in T^S$.
Let $\struct {T, \circ}$ be an associative algebraic structure.
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.6: \ 1^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.4$
Category:Associativity
Category:Pointwise Operations
Structure Induced by Commutative Operation is Commutative

== Theorem ==
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be a commutative operation.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also commutative.
== Proof ==
Let $\struct {T, \circ}$ be a commutative algebraic structure.
Let $f, g \in T^S$.
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.6: \ 1^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.4$
Category:Pointwise Operations
Category:Commutativity
Structure Induced by Commutative Ring Operations is Commutative Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the structure on $R^S$ induced by $+'$ and $\circ'$.
Then $\struct {R^S, +', \circ'}$ is a commutative ring.
== Proof ==
By Structure Induced by Ring Operations is Ring then $\struct {R^S, +', \circ'}$ is a ring.
From Structure Induced by Commutative Operation is Commutative, so is the pointwise operation $\circ$ induces on $R^S$.
The result follows by definition of commutative ring.
Category:Rings of Mappings
Structure Induced by Distributive Operation is Distributive

== Theorem ==
Let $\struct {T, +, \times}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus, \otimes}$ be the structure on $T^S$ induced by $+$ and $\times$.
Let $\times$ be distributive over $+$:
:$\forall a, b, c \in S$: 
::$a \times \paren {b + c} = \paren{a \times b} + \paren{a \times c}$
:and:
::$\paren {b + c} \times a= \paren{b \times a} + \paren{c \times a}$
Then the pointwise operation $\otimes$ is distributive over the pointwise operation $\oplus$ on $T^S$:
:$\forall f, g, h \in T^S$:
::$f \otimes \paren {g \oplus h} = \paren{f \otimes g} \oplus \paren{f \otimes h}$
:and:
::$\paren {g \oplus h} \otimes f = \paren{g \otimes f} \oplus \paren{h \otimes f}$
== Proof ==
By definition of distributive operation:
:$\times$ is left distributive over $+$
and
:$\times$ is right distributive over $+$
From Structure Induced by Left Distributive Operation is Left Distributive:
:$\otimes$ is left distributive over $\oplus$
From Structure Induced by Right Distributive Operation is Right Distributive:
:$\otimes$ is right distributive over $\oplus$
It follws that $\otimes$ is distributive over $\oplus$ by definition.
Category:Pointwise Operations
Category:Distributive Operations
Structure Induced by Group Operation is Group

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.
Then $\struct {G^S, \oplus}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
As $\struct {G, \circ}$ is a group, it is closed by .
From Closure of Pointwise Operation on Algebraic Structure it follows that $\struct {G^S, \oplus}$ is likewise closed.
=== ===
As $\struct {G, \circ}$ is a group, $\circ$ is associative.
So from Structure Induced by Associative Operation is Associative, $\struct {G^S, \oplus}$ is also associative.
=== ===
We have from Induced Structure Identity that the constant mapping $f_e: S \to T$ defined as:
:$\forall x \in S: \map {f_e} x = e$
is the identity for $\struct {G^S, \oplus}$.
=== ===
Let $f \in G^S$.
Let $f^* \in G^S$ be defined as follows:
:$\forall f \in G^S: \forall x \in S: \map {f^*} x = \paren {\map f x}^{-1}$
From Pointwise Inverse in Induced Structure, $f^*$ is the inverse of $f$ for the pointwise operation $\oplus$ induced on $G^S$ by $\circ$.
All the group axioms are thus seen to be fulfilled, and so $\struct {G^S, \oplus}$ is a group.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.6: \ 3^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.4$
Category:Group Theory
Category:Pointwise Operations
Structure Induced by Idempotent Operation is Idempotent

== Theorem ==
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be an idempotent operation.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also idempotent.
== Proof ==
Let $f \in T^S$.
Then:
From Equality of Mappings:
:$f \oplus f = f$
Since $f$ was arbitrary:
:$\forall f \in T^S : f \oplus f = f$
Hence $\oplus$ is idempotent by definition.
Category:Pointwise Operations
Category:Idempotence
Structure Induced by Lattice Operations is Lattice

== Theorem ==
Let $\struct {L, \vee, \wedge, \preceq}$ be a lattice.
Let $S$ be a set.
Let $\struct {L^S, \veebar, \barwedge}$ be the structure on $T^S$ induced by $\vee$ and $\wedge$.
Let $\precsim$ be the ordering on $S$ defined by:
:$\forall f, g \in L^S: f \precsim g$ $f \veebar g = g$
as on Semilattice Induces Ordering.
Then $\struct {L^S, \veebar, \barwedge, \precsim}$ is a lattice.
== Proof ==
=== $\struct {L^S, \veebar}$ and $\struct {L^S, \barwedge}$ are Semilattices ===
By definition of lattice:
:$\struct {L, \vee}$ and $\struct {L, \wedge}$ are semilattices.
From Structure Induced by Semilattice Operation is Semilattice:
:$\struct {L^S, \veebar}$ and $\struct {L^S, \veebar}$ are semilattices.
=== Absorption Laws ===
By definition of lattice:
:$\vee$ absorbs $\wedge$ 
and:
:$\wedge$ absorbs $\vee$
From Structure Induced by Absorbing Operations is Absorbing
:$\veebar$ absorbs $\barwedge$
and:
:$\barwedge$ absorbs $\veebar$
It follows that $\struct {L^S, \veebar, \barwedge, \precsim}$ is a lattice by definition.
Category:Lattices (Order Theory)
Category:Pointwise Operations
Structure Induced by Left Distributive Operation is Left Distributive

== Theorem ==
Let $\struct {T, +, \times}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus, \otimes}$ be the structure on $T^S$ induced by $+$ and $\times$.
Let $\times$ be left distributive over $+$:
:$\forall a, b, c \in S: a \times \paren {b + c} = \paren{a \times b} + \paren{a \times c}$
Then the pointwise operation $\otimes$ is left distributive over the pointwise operation $\oplus$ on $T^S$:
:$\forall f, g, h \in T^S: f \otimes \paren {g \oplus h} = \paren{f \otimes g} \oplus \paren{f \otimes h}$
== Proof ==
Let $f, g, h \in T^S$.
Then:
From Equality of Mappings:
:$f \otimes \paren {g \oplus h} = \paren{f \otimes g} \oplus \paren{f \otimes h}$
Since $f, g, h$ were arbitrary:
:$\forall f, g, h \in T^S: f \otimes \paren {g \oplus h} = \paren{f \otimes g} \oplus \paren{f \otimes h}$
Hence $\otimes$ is left distributive over $\oplus$ by definition.
Category:Pointwise Operations
Category:Distributive Operations
Structure Induced by Permutation on Algebra Loop is not necessarily Algebra Loop

== Theorem ==
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:
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.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(a)}$
Category:Operations Induced by Permutations
Category:Algebra Loops
Structure Induced by Permutation on Commutative Quasigroup is Commutative Quasigroup

== Theorem ==
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:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(a)}$
Category:Operations Induced by Permutations
Category:Quasigroups
Category:Commutativity
Structure Induced by Permutation on Quasigroup is Quasigroup

== Theorem ==
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:
Similarly, we have that:
:$\exists ! x: a \circ x = b$
Thus:
The result follows.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(a)}$
Category:Operations Induced by Permutations
Category:Quasigroups
Structure Induced by Permutation on Semigroup is not necessarily Semigroup

== Theorem ==
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 holds.
However, we have that Operation Induced by Permutation on Semigroup is not necessarily Associative.
Hence does not necessarily hold for $\struct {S, \circ_\sigma}$ 
Hence $\struct {S, \circ_\sigma}$ is not necessarily a semigroup.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.9 \ \text {(a)}$
Category:Operations Induced by Permutations
Category:Semigroups
Structure Induced by Right Distributive Operation is Right Distributive

== Theorem ==
Let $\struct {T, +, \times}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus, \otimes}$ be the structure on $T^S$ induced by $+$ and $\times$.
Let $\times$ be right distributive over $+$:
:$\forall a, b, c \in S: \paren {b + c} \times a= \paren{b \times a} + \paren{c \times a}$
Then the pointwise operation $\otimes$ is left distributive over the pointwise operation $\oplus$ on $T^S$:
:$\forall f, g, h \in T^S: \paren {g \oplus h} \otimes f = \paren{g \otimes f} \oplus \paren{h \otimes f}$
== Proof ==
Let $f, g, h \in T^S$.
Then:
From Equality of Mappings:
:$\paren {g \oplus h} \otimes f = \paren{g \otimes f} \oplus \paren{h \otimes f}$
Since $f, g, h$ were arbitrary:
:$\forall f, g, h \in T^S: \paren {g \oplus h} \otimes f = \paren{g \otimes f} \oplus \paren{h \otimes f}$
Hence $\otimes$ is right distributive over $\oplus$ by definition.
Category:Pointwise Operations
Category:Distributive Operations
Structure Induced by Ring Operations is Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a set.
Then $\struct {R^S, +', \circ'}$ is a ring, where $+'$ and $\circ'$ are the pointwise operations induced on $R^S$ by $+$ and $\circ$.
== Proof ==
As $R$ is a ring, both $+$ and $\circ$ are closed on $R$ by definition.
From Closure of Pointwise Operation on Algebraic Structure, it follows that both $+'$ and $\circ'$ are closed on $R^S$:
:$\forall f, g \in R^S: f +' g \in R^S$
:$\forall f, g \in R^S: f \circ' g \in R^S$
By Structure Induced by Abelian Group Operation is Abelian Group, $\struct {R^S, +'}$ is an abelian group.
By Structure Induced by Associative Operation is Associative, $\struct {R^S, \circ'}$ is a semigroup.
From Pointwise Operation on Distributive Structure is Distributive, $\circ'$ is distributive over $+'$.
The result follows by definition of ring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.10$
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $8$
Category:Rings of Mappings
Category:Examples of Rings
Structure Induced by Ring with Unity Operations is Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the structure on $R^S$ induced by $+'$ and $\circ'$.
Then $\struct {R^S, +', \circ'}$ is a ring with unity whose unity is $f_{1_R}: S \to R$, defined by:
:$\forall s \in S: \map {f_{1_R} } s = 1_R$
== Proof ==
By Structure Induced by Ring Operations is Ring then $\struct {R^S, +', \circ'}$ is a ring.
We have from Induced Structure Identity that the constant mapping $f_{1_R}: S \to R$ defined as:
:$\forall x \in S: \map {f_{1_R} } x = 1_R$
is the identity for $\struct {R^S, \circ'}$.
The result follows by definition of ring with unity and unity of ring.
Category:Rings of Mappings
Category:Rings with Unity
Structure Induced by Semigroup Operation is Semigroup

== Theorem ==
Let $\struct {T, \circ}$ be a semigroup.
Let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Then $\struct {T^S, \oplus}$ is a semigroup.
== Proof ==
Taking the semigroup axioms in turn:
=== ===
As $\struct {T, \circ}$ is a semigroup, it is closed by .
From Closure of Pointwise Operation on Algebraic Structure it follows that $\struct {T^S, \oplus}$ is likewise closed.
=== ===
As $\struct {T, \circ}$ is a semigroup, $\circ$ is associative.
So from Structure Induced by Associative Operation is Associative, $\struct {T^S, \oplus}$ is also associative.
All the semigroup axioms are thus seen to be fulfilled, and so $\struct {T^S, \oplus}$ is a semigroup.
Category:Semigroups
Category:Pointwise Operations
Structure Induced by Semilattice Operation is Semilattice

== Theorem ==
Let $\struct {T, \circ}$ be a semilattice.
Let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Then $\struct {T^S, \oplus}$ is a semilattice.
== Proof ==
Taking the semilattice axioms in turn:
=== ===
As $\struct {T, \circ}$ is a semilattice, it is closed by .
From Closure of Pointwise Operation on Algebraic Structure it follows that $\struct {T^S, \oplus}$ is likewise closed.
=== ===
As $\struct {T, \circ}$ is a semilattice, $\circ$ is associative.
So from Structure Induced by Associative Operation is Associative, $\struct {T^S, \oplus}$ is also associative.
=== ===
As $\struct {T, \circ}$ is a semilattice, $\circ$ is commutative.
So from Structure Induced by Commutative Operation is Commutative, $\struct {T^S, \oplus}$ is also commutative.
=== ===
As $\struct {T, \circ}$ is a semilattice, $\circ$ is idempotent.
So from Structure Induced by Idempotent Operation is Idempotent, $\struct {T^S, \oplus}$ is also idempotent.
All the semilattice axioms are thus seen to be fulfilled, and so $\struct {T^S, \oplus}$ is a semilattice.
Category:Semilattices
Category:Pointwise Operations
Structure Induced on Set of All Mappings to Ordered Semigroup is Ordered Semigroup

== Theorem ==
Let $S$ be a set.
Let $\struct {T, \odot, \preccurlyeq}$ be an ordered semigroup.
Let $\struct {T^S, \otimes, \preccurlyeq}$ denote the algebraic structure on $T^S$ induced by $\odot$.
Then $\struct {T^S, \otimes, \preccurlyeq}$ is an ordered semigroup.
== Proof ==
From Structure Induced by Semigroup Operation is Semigroup, $\struct {T^S, \otimes}$ is a semigroup
From Ordered Set of All Mappings is Ordered Set, $\struct {T^S, \preccurlyeq}$ is an ordered set.
It remains to be demonstrated that $\preccurlyeq$ is compatible with $\odot$.
Let $f, g \in T^S$ such that $f \preccurlyeq g$.
That is:
:$\forall x \in S: \map f x \preccurlyeq \map g x$
We are given that $\struct {T, \odot, \preccurlyeq}$ be an ordered semigroup.
Hence $\preccurlyeq$ is compatible with $\odot$.
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.7$
Category:Examples of Ordered Semigroups
Category:Pointwise Operations
Structure Induced on Set of Self-Maps on Entropic Structure is Entropic

== Theorem ==
Let $\struct {S, \odot}$ be a magma.
Let $\struct {S, \odot}$ be an entropic structure.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the algebraic structure on $S^S$ induced by $\odot$.
Then $\struct {S^S, \oplus}$ is an entropic structure.
== Proof ==
Recall the definition of algebraic structure on $S^S$ induced by $\odot$:
Let $f: S \to S$ and $g: S \to S$ be self-maps on $S$, and thus elements of $S^S$.
The pointwise operation on $S^S$ induced by $\odot$ is defined as:
:$\forall x \in S: \map {\paren {f \oplus g} } x = \map f x \odot \map g x$
Let $f, g, p, q \in S^S$ be arbitrary.
Let $x \in S$ be arbitrary.
Then:
Hence the result by definition of entropic structure.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(g)}$
Category:Entropic Structures
Category:Pointwise Operations
Category:Magmas
Structure is Group iff Semigroup and Quasigroup

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Then:
:$\struct {S, \circ}$ is a group
:$\struct {S, \circ}$ is both a semigroup and a quasigroup.
== Proof ==
=== Sufficient Condition ===
Let $\struct {S, \circ}$ be a group.
Then a fortiori $\struct {S, \circ}$ is a semigroup.
From Regular Representations in Group are Permutations:
:for all $a \in S$, the left regular representation and the rightt regular representation are permutations of $S$.
Hence by definition $\struct {S, \circ}$ is a quasigroup.
=== Necessary Condition ===
Let $\struct {S, \circ}$ be both a semigroup and a quasigroup.
By definition of quasigroup:
:$\forall a \in S$, the left and right regular representations $\lambda_a$ and $\rho_a$ are permutations on $S$.
It follows from Regular Representations in Semigroup are Permutations then Structure is Group that $\struct {S, \circ}$ is a group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.14$
Category:Quasigroups
Category:Semigroups
Category:Group Theory
Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent

== Theorem ==
Let $S$ be a set whose cardinality is at least $3$.
Let $\struct {S, \circ}$ be an algebraic structure on $S$ such that every permutation on $S$ is an automorphism on $\struct {S, \circ}$.
Then $\circ$ is an idempotent operation.
== Proof ==
 $\circ$ is not idempotent.
Then there exists $x \in S$ such that:
:$\exists y \in S: x \circ x = y$
where $x \ne y$.
Because there are at least $3$ distinct elements of $S$ :
:$\exists z \in S: z \ne x, z \ne y$
Let $f: S \to S$ be a permutation on $S$ such that:
:$\map f x = x$
:$\map f y = z$
We have:
This contradicts our assertion that $x$ and $z$ are distinct.
From Proof by Contradiction it follows that our assumption that $\circ$ is not idempotent must have been false.
Hence $\circ$ is idempotent.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.11 \ \text {(c)}$
Category:Idempotence