nyuuzyou/wikis huggingface dataset:
Symmetric Group on 4 Letters/Subgroups/Examples/Disjoint Transpositions

== Example of Subgroup of Symmetric Group on 4 Letters ==
Let $H$ be the subset of the Symmetric Group on $4$ Letters $S_4$ which consists of the $3$ products of disjoint transpositions of $S_4$, and the identity:
:$V := \set {e, \tuple {1 2} \tuple {3 4}, \tuple {1 3} \tuple {2 4}, \tuple {1 4} \tuple {2 3} }$
Then $V$ forms a subgroup of $S_4$.
The Cayley table of $V$ can be presented as:
:$\begin{array}{c|cccc}
\circ & e & \tuple {1 2} \tuple {3 4} & \tuple {1 3} \tuple {2 4} & \tuple {1 4} \tuple {2 3} \\
\hline
 e & e & \tuple {1 2} \tuple {3 4} & \tuple {1 3} \tuple {2 4} & \tuple {1 4} \tuple {2 3} \\
\tuple {1 2} \tuple {3 4} & \tuple {1 2} \tuple {3 4} & e & \tuple {1 4} \tuple {2 3} & \tuple {1 3} \tuple {2 4} \\
\tuple {1 3} \tuple {2 4} & \tuple {1 3} \tuple {2 4} & \tuple {1 4} \tuple {2 3} & e & \tuple {1 2} \tuple {3 4} \\
\tuple {1 4} \tuple {2 3} & \tuple {1 4} \tuple {2 3} & \tuple {1 3} \tuple {2 4} & \tuple {1 2} \tuple {3 4} & e \\
\end{array}$
This is the Klein $4$-group.
$V$ is normal in $S_4$.
The quotient group $S_4 / V$ is the Symmetric Group on $3$ Letters $S_3$.
The (left) cosets of $V$ are:
and its resulting Cayley table is:
:$\begin{array}{c|ccc|ccc}
\circ & E & A & B & C & D & F \\
\hline
E & E & A & B & C & D & F \\
A & A & B & E & F & C & D \\
B & B & E & A & D & F & C \\
\hline
C & C & D & F & E & A & B \\
D & D & F & C & B & E & A \\
F & F & C & D & A & B & E \\
\end{array}$
== Proof ==
We have that $V$ contains all the elements of $S_4$ of the same cycle type.
From Cycle Decomposition of Conjugate, the conjugate of a permutation is another permutation of the same cycle type.
Hence the conjugate of an element of $V$ is an element of $V$.
That is, $V$ is normal in $S_4$.
The quotient group $S_4 / V$ is of order $\dfrac {24} 4 = 6$, so must be either $C_6$ or $S_3$.
From the Cayley table above it is apparent that it is $S_3$.
== Sources ==
* : Chapter $9$: Permutations
* : Chapter $9$: Permutations: Exercise $2$
Category:Symmetric Group on 4 Letters
Symmetric Group on 4 Letters/Subgroups/Examples/Even Permutations

== Example of Subgroup of Symmetric Group on 4 Letters ==
The subset of the Symmetric Group on $4$ Letters $S_4$ which consists of all the even permutations of $S_4$ forms a subgroup of $S_4$.
From Alternating Group is Set of Even Permutations, this is by definition the alternating group on $4$ letters $A_4$
Its Cayley table can be presented as follows:
As $A_4$ has index $2$, it is normal in $S_4$ from Subgroup of Index 2 is Normal.
Hence the quotient group $S_4 / A_4$ is cyclic of order $2$.
== Sources ==
* : Chapter $9$: Permutations
Category:Symmetric Group on 4 Letters
Symmetric Group on Greater than 4 Letters is Not Solvable

== Theorem ==
Let $n \in \N$ such that $n > 4$.
Let $S_n$ denote the symmetric group on $n$ letters.
Then $S_n$ is not a solvable group.
== Proof ==
As stated, let $n > 4$ in the below.
Recall the definition of solvable group:
:A finite group $G$ is a solvable group it has a composition series in which each factor is a cyclic group.
:A composition series for $G$ is a normal series for $G$ which has no proper refinement.
:A normal series for $G$ is a sequence of (normal) subgroups of $G$:
::$\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$
:where $G_{i - 1} \lhd G_i$ denotes that $G_{i - 1}$ is a proper normal subgroup of $G_i$.
Consider the alternating group on $n$ letters $A_n$.
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is a proper normal subgroup of $S_n$.
But from Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group, the only normal subgroup of $S_n$ is $A_n$
From Alternating Group is Simple except on 4 Letters, $A_n$ is simple.
That is, $A_n$ has no proper normal subgroup.
It follows that the only composition series for $S_n$ is:
:$\set e \lhd A_n \lhd S_n$
From Alternating Group on More than 3 Letters is not Abelian, $A_n$ is not an abelian group.
From Cyclic Group is Abelian, $A_n$ is not cyclic.
Thus we have demonstrated that the only composition series for $S_n$ contains a factor which is not cyclic.
Hence the result.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 84$
Category:Symmetric Groups
Symmetric Group on n Letters is Isomorphic to Symmetric Group

== Theorem ==
The symmetric group on $n$ letters $\struct {S_n, \circ}$ is isomorphic to the symmetric group on the $n$ elements of any set $T$ whose cardinality is $n$.
That is:
:$\forall T \subseteq \mathbb U, \card T = n: \struct {S_n, \circ} \cong \struct {\Gamma \paren T, \circ}$
where:
:$\map \Gamma T$ denotes the set of permutations on $T$
:$\struct {\map \Gamma T, \circ}$ denotes an algebraic structure with $1$ operation
:$\circ$ denotes composition of mappings.
== Proof ==
The fact that $\struct {S_n, \circ}$ is a group is a direct implementation of the result Symmetric Group is Group.
By definition of cardinality, as $\card T = n$ we can find a bijection between $T$ and $\N_n$.
From Number of Permutations, it is immediate that $\order {\paren {\Gamma \paren T, \circ} } = n! = \order {\struct {S_n, \circ} }$.
Again, we can find a bijection $\phi$ between $\struct {\Gamma \paren T, \circ}$ and $\struct {S_n, \circ}$.
The result follows directly from the Transplanting Theorem.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 77$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\S 1.1$ Example $6$
Category:Symmetric Groups
Symmetric Groups of Same Order are Isomorphic

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $T_1$ and $T_2$ be sets whose cardinality $\card {T_1}$ and $\card {T_2}$ are both $n$.
Let $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ be the symmetric group on $S$ and $T$ respectively.
Then $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ are isomorphic.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.6$
Category:Symmetric Groups of Same Order are Isomorphic
Category:Symmetric Groups
Category:Group Isomorphisms
Symmetric Groups of Same Order are Isomorphic/Proof 1

== Theorem ==
== Proof ==
Consider the symmetric group on $n$ letters $S_n$.
From Symmetric Group on n Letters is Isomorphic to Symmetric Group we have that:
:$\struct {\map \Gamma {T_1}, \circ}$ is isomorphic to $S_n$
:$\struct {\map \Gamma {T_2}, \circ}$ is isomorphic to $S_n$
and hence from Isomorphism is Equivalence Relation:
:$\struct {\map \Gamma {T_1}, \circ}$ is isomorphic to $\struct {\map \Gamma {T_2}, \circ}$.
Category:Symmetric Groups of Same Order are Isomorphic
Symmetric Groups of Same Order are Isomorphic/Proof 2

== Theorem ==
== Proof ==
Let us define a bijection:
:$\alpha: T_1 \to T_2$
Let $\theta: \struct {\map \Gamma {T_1}, \circ} \to \struct {\map \Gamma {T_2}, \circ}$ be defined as:
:$\forall f \in \struct {\map \Gamma {T_1}, \circ}: \map \theta f = \alpha \circ f \circ \alpha^{-1}$
Let $f, g \in \map \Gamma {T_1}$.
We have:
That is, $\theta$ is a group homomorphism.
Let $f, g \in \map \Gamma {T_1}$ such that $\map \theta f = \map \theta g$.
Then:
Thus it is seen that $\theta$ is an injection.
Let $g \in \map \Gamma {T_2}$ be arbitrary.
Let $\alpha^{-1} \circ g \circ \alpha = f$
Then:
and so $g$ is the image of $f$ under $\theta$.
As $g$ is arbitrary, it follows that $\theta$ is a surjection.
As $\theta$ is an injection and a surjection, it follows that $\theta$ is a bijection by definition.
So $\theta$ is a bijective group homomorphism, and so a group isomorphism by definition.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $1$
Category:Symmetric Groups of Same Order are Isomorphic
Symmetric Integral of Even Function over One Plus Even Function to Power of Odd Function

== Theorem ==
<onlyinclude>
:$\ds \int_{-a}^a \frac {\map e x} {1 + \map t x^{\map o x} } \rd x = \int_0^a \map e x \rd x$
where:
:$a$ is a positive real number
:$\map e x$ and $\map t x$ are arbitrary even functions
:$\map o x$ is an arbitrary odd function.
<onlyinclude>
== Proof ==
Let the integrand be denoted by:
:$\map f x = \dfrac {\map e x} {1 + \map t x^{\map o x} }$
Using Real Function is Expressible as Sum of Even Function and Odd Function, let us express $\map f x$ as:
:$(1): \quad \map f x = \map E x + \map O x$
where:
:$\map E x$ is an even function
:$\map O x$ is an odd function
defined as:
Then:
Hence the result.
Category:Odd Functions
Category:Even Functions
Category:Definite Integrals
Symmetric Preordering is Equivalence Relation

== Theorem ==
Let $\RR \subseteq S \times S$ be a preordering on a set $S$.
Let $\RR$ also be symmetric.
Then $\RR$ is an equivalence relation on $S$.
== Proof ==
By definition, a preordering on $S$ is a relation on $S$ which is:
:$(1): \quad$ reflexive
and:
:$(2): \quad$ transitive.
Thus $\RR$ is a relation on $S$ which is reflexive, transitive and symmetric.
Thus by definition $\RR$ is an equivalence relation on $S$.
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 7$: Exercise $7$
* : Chapter $1$: Sets and mappings: Further exercises: $5$
Category:Preorder Theory
Category:Equivalence Relations
Category:Symmetric Relations
Symmetric Relation/Examples/Cousin

== Example of Symmetric Relation ==
Let $P$ be the set of people.
Let $\sim$ be the relation on $P$ defined as:
:$\forall \tuple {x, y} \in P \times P: x \sim y \iff \text { $x$ is a cousin of $y$}$
Then $\sim$ is a symmetric relation.
== Sources ==
* 
* 
Category:Examples of Symmetric Relations
Symmetric Relation/Examples/Distance Less than 1

== Example of Symmetric Relation ==
Let $\sim$ be the relation on the set of real numbers $\R$ defined as:
:$x \sim y \iff \size {x - y} < 1$
Then $\sim$ is symmetric and reflexive, but not transitive.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations
Category:Examples of Symmetric Relations
Symmetric Relation/Examples/Is a Brother of

== Example of Symmetric Relation ==
Let $P$ be the set of male people.
Let $\sim$ be the relation on $P$ defined as:
:$\forall \tuple {x, y} \in P \times P: x \sim y \iff \text { $x$ is a brother of $y$}$
Then $\sim$ is a symmetric relation.
This does not hold if $P$ is the set of all people.
Because if $a$ is male and $b$ are brother and sister, then:
:$a \sim b$
but:
:$b \not \sim a$
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
Category:Examples of Symmetric Relations
Symmetric Relation/Examples/Is of Opposite Gender

== Example of Symmetric Relation ==
Let $P$ be the set of people.
Let $\sim$ be the relation on $P$ defined as:
:$\forall \tuple {x, y} \in P \times P: x \sim y \iff \text { $x$ of the opposite gender to $y$}$
(This assumes that gender is binary and well-defined.)
Then $\sim$ is a symmetric relation.
However, $\sim$ is antireflexive and antitransitive.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $9$
Category:Examples of Symmetric Relations
Symmetric Transitive and Serial Relation is Reflexive

== Theorem ==
Let $\RR$ be a relation which is:
:symmetric
:transitive
:serial.
Then $\RR$ is reflexive.
Thus such a relation is an equivalence.
== Proof ==
Let $S$ be a set on which $\RR$ is a relation which is symmetric, transitive and serial.
As $\RR$ is symmetric:
:$x \mathrel \RR y \implies y \mathrel \RR x$
As $\RR$ is transitive:
:$x \mathrel \RR y \land y \mathrel \RR x \implies x \mathrel \RR x$
As $\RR$ is serial:
:$\forall x \in S: \exists y \in S: x \mathrel \RR y$
Let $x \in S$.
Then
Thus:
:$\forall x: x \mathrel \RR x$
and by definition $\RR$ is reflexive.
It follows by definition that such a relation is an equivalence relation.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $2 \ \text{(c)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Exercise $10.5$
Category:Reflexive Relations
Category:Symmetric Relations
Category:Transitive Relations
Category:Serial Relations
Category:Equivalence Relations
Symmetric and Antisymmetric Relation is Transitive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$ which is both symmetric and antisymmetric.
Then $\RR$ is transitive.
== Proof ==
Let $\tuple {x, y}, \tuple {y, z} \in \RR$.
By Relation is Symmetric and Antisymmetric iff Coreflexive:
:$x = y, y = z$
and so trivially:
:$\tuple {x, z} = \tuple {x, x} \in \RR$
Thus $\RR$ is transitive.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $5$
Category:Symmetric Relations
Category:Antisymmetric Relations
Category:Transitive Relations
Symmetric and Transitive Relation is not necessarily Reflexive

== Theorem ==
Let $S$ be a set.
Let $\alpha \subseteq S \times S$ be a relation on $S$.
Let $\alpha$ be both symmetric and transitive.
Then it is not necessarily the case that $\alpha$ is also reflexive.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Examples ==
== Sources ==
* : Chapter $1$: Preliminaries: Exercises: $1.9$
Category:Reflexive Relations
Category:Symmetric Relations
Category:Transitive Relations
Category:Symmetric and Transitive Relation is not necessarily Reflexive
Symmetric and Transitive Relation is not necessarily Reflexive/Examples/Subset of Cartesian Plane

== Examples of Use of Symmetric and Transitive Relation is not necessarily Reflexive ==
The subset of the Cartesian plane defined as:
:$\RR := \set {\tuple {x, y} \in \R^2: -1 \le x \le 1, -1 \le y \le 1}$
determines a relation on $\R^2$ which is symmetric and transitive but not reflexive.
== Proof ==
=== Non-Reflexive Relation ===
We note that, for example, $\tuple {2, 2} \notin \RR$.
Hence $\RR$ is non-reflexive.
=== Symmetric Relation ===
thus demonstrating that $\RR$ is symmetric.
=== Transitive Relation ===
thus demonstrating that $\RR$ is transitive.
The relation $\RR$ is illustrated below:
:
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations
Category:Symmetric and Transitive Relation is not necessarily Reflexive
Symmetric and Transitive Relation is not necessarily Reflexive/Examples/Subset of Cartesian Plane/Mistake

== Source Work ==
:
:Chapter $\text{I}$: Sets and Functions
::Relations
== Mistake ==
:The shaded regions represent relations
::
Symmetric and transitive, but not reflexive
== Correction ==
The annotations to the lower corner points of the shaded region are incorrect.
This is what they should look like:
::
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations
Category:Mistakes/Topology: An Introduction with Application to Topological Groups
Symmetric and Transitive Relation is not necessarily Reflexive/Proof 1

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b, c}$.
Let:
: $\alpha = \set {\tuple {a, a}, \tuple {a, b}, \tuple {b, a}, \tuple {b, b} }$
By inspection it is seen that $\alpha$ is both symmetric and transitive.
However, we have:
:$\neg c \mathrel \alpha c$
Hence $\alpha$ is both symmetric and transitive but not reflexive.
Category: Symmetric and Transitive Relation is not necessarily Reflexive
Symmetric and Transitive Relation is not necessarily Reflexive/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $S = \Z$ be the set of integers.
Let $\alpha$ be the relation on $S$ defined as:
:$\forall x, y \in S: x \mathrel \alpha y \iff x = y = 0$
Thus $\alpha$ is symmetric.
Now let $x \mathrel \alpha y$ and $y \mathrel \alpha z$
Then:
:$x = y = 0, y = z = 0$
and so
:$x \mathrel \alpha z$
Now let $x = \Z$ such that $x \ne 0$.
Then it is not the case that:
:$x \mathrel \alpha x$
and so $\alpha$ is not reflexive.
Hence $\alpha$ is both symmetric and transitive but not reflexive.
== Sources ==
* : Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $3$
Category:Symmetric and Transitive Relation is not necessarily Reflexive
Symmetric and Transitive Relation is not necessarily Reflexive/Proof 3

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $S = \set {1, 2}$ be a set.
Let $\RR$ be the relation on $S$ defined as:
:$\forall x, y \in S: x \mathrel \RR y \iff x = y = 2$
Thus $\alpha$ is symmetric.
Now let $x \mathrel \RR y$ and $y \mathrel \RR z$
Then:
:$x = y = 2, y = z = 2$
and so:
:$x \mathrel \RR z$
Now let $x = 1$.
Then it is not the case that:
:$x \mathrel \RR x$
and so $\RR$ is not reflexive.
Hence $\RR$ is both symmetric and transitive but not reflexive.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $5$
Category:Symmetric and Transitive Relation is not necessarily Reflexive
Symmetric and Transitive therefore Reflexive

== Fallacy ==
Let $\RR \subseteq S \times S$ be a relation which is symmetric and transitive.
Then $\RR$ is also always reflexive.
Consider $x, y \in S$.
Suppose $x \mathrel \RR y$.
Then as $\RR$ is symmetric, it follows that $y \mathrel \RR x$.
As $\RR$ is transitive, it follows that $x \mathrel \RR x$.
Therefore $x \mathrel \RR x$ and so $\RR$ is reflexive.
== Resolution ==
For $\RR$ to be reflexive, it is necessary for $x \mathrel \RR x$ for all $x \in S$.
Unless it is the case that $\forall x \in S: \exists y \in S: x \mathrel \RR y$, it is not necessarily the case that also $y \mathrel \RR x$, and so the reasoning does not follow.
Take the set $S = \set {0, 1}$ and the relation:
:$\RR \subseteq S \times S: x \mathrel \RR y \iff x = y = 1$
It is seen that:
:$\RR$ is symmetric
:$\RR$ is transitive
but:
:$\RR$ is not reflexive, as $\neg \paren {0 \mathrel \RR 0}$.
== Also see ==
* Definition:Equivalence Relation, which is the usual motivator of this frequently-met fallacy.
* Symmetric Transitive and Serial Relation is Reflexive which shows that the condition under which a symmetric and transitive relation is guaranteed to be reflexive.
== Sources ==
* : Exercise $1.2: 6$
* : $\S 0.3$: Example $5$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{F}$
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $6$
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations: Exercise $4$
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.3$
Category:Fallacies and Mistakes
Symmetry (Geometry)/Examples/Rotations of Square through 90 Degrees

== Example of Symmetry in context of Geometry ==
Let $S$ be a square embedded in the plane centered at the origin $O$.
A rotation of the plane through an angle of $90 \degrees$ either clockwise or anticlockwise is a symmetry of $S$.
== Sources ==
* : $\S 34$. Examples of groups: $(5)$
Category:Examples of Symmetries (Geometry)
Symmetry Group is Group

== Theorem ==
Let $P$ be a geometric figure.
Let $S_P$ be the set of all symmetries of $P$.
Let $\circ$ denote composition of mappings.
The symmetry group $\struct {S_P, \circ}$ is indeed a group.
== Proof ==
By definition, a symmetry is a bijection, and hence a permutation.
From Symmetric Group is Group, the set of all permutations on $P$ form the symmetric group $\struct {\map \Gamma P, \circ}$ on $P$.
Thus $S_P$ is a subset of $\struct {\map \Gamma P, \circ}$.
Let $A$ and $B$ be symmetries on $P$.
From Composition of Symmetries is Symmetry, $A \circ B$ is also a symmetry on $P$.
Also, by the definition of symmetry, $A^{-1}$ is also a symmetry on $P$.
Thus we have:
:$A, B \in S_P \implies A \circ B \in S_P$
:$A \in S_P \implies A^{-1} \in \S_P$
The result follows from the Two-Step Subgroup Test.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.3$
* : Chapter $2$: The Definition of Group Structure: $\S 26 \eta$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\S 34$. Examples of groups: $(5)$
* : Chapter $1$: Definitions and Examples: Example $1.9$
* 
* 
Category:Symmetry Groups
Symmetry Group of Equilateral Triangle/Cayley Table

== Cayley Table of Symmetry Group of Equilateral Triangle ==
The Cayley table of the symmetry group of the equilateral triangle can be written:
:$\begin{array}{c|ccc|ccc}
\circ & e & p & q & r & s & t \\
\hline
e & e & p & q & r & s & t \\
p & p & q & e & s & t & r \\
q & q & e & p & t & r & s \\
\hline
r & r & t & s & e & q & p \\
s & s & r & t & p & e & q \\
t & t & s & r & q & p & e \\
\end{array}$
where:
:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.11$
* : $\S 1.4$: Exercise $2.1$
* : $\S 34$. Examples of groups: $(5)$
* : Chapter $1$: Definitions and Examples: Example $1.9$
Category:Examples of Cayley Tables
Category:Symmetry Group of Equilateral Triangle
Symmetry Group of Equilateral Triangle is Group

== Theorem ==
The symmetry group of the equilateral triangle is a group.
=== Definition ===
Recall the definition of the symmetry group of the equilateral triangle:
== Proof ==
Let us refer to this group as $D_3$.
Taking the group axioms in turn:
=== ===
From the Cayley table it is seen directly that $D_3$ is closed.
=== ===
Composition of Mappings is Associative.
=== ===
The identity is $e = (A) (B) (C)$.
=== ===
Each element can be seen to have an inverse:
:$p^{-1} = q$ and so $q^{-1} = p$
:$r$, $s$ and $t$ are all self-inverse.
Thus $D_3$ is seen to be a group.
== Sources ==
* : $\S 1.3$: Example $9$
* : Chapter $2$: The Definition of Group Structure: $\S 26 \eta$
* : Chapter $1$: Definitions and Examples: Example $1.9$
Category:Symmetry Group of Equilateral Triangle
Symmetry Group of Equilateral Triangle is Solvable

== Theorem ==
The symmetry group $D_3$ of the equilateral triangle is a solvable group.
== Proof ==
From Symmetry Group of Equilateral Triangle is Symmetric Group, $D_3$ is the symmetric group on $3$ letters $S_3$.
The result follows from Examples of Composition Series: Symmetric Group $S_3$.
== Sources ==
* 
Category:Symmetry Group of Equilateral Triangle
Category:Examples of Solvable Groups
Symmetry Group of Equilateral Triangle is Symmetric Group

== Theorem ==
Let $D_3$ denote the symmetry group of the equilateral triangle.
Let $S_3$ denote the symmetric group on $3$ letters.
Then $D_3$ is isomorphic to $S_3$.
== Proof ==
Follows from Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3.
== Sources ==
* : Chapter $2$: Examples of Group Structure: $\S 30 \gamma$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(iv)}$
* : Chapter $2$: Maps and relations on sets: Example $2.19$
Category:Symmetry Group of Equilateral Triangle
Category:Symmetric Group on 3 Letters
Symmetry Group of Isosceles Triangle/Cayley Table

== Cayley Table of Symmetry Group of Isosceles Triangle ==
The Cayley table of the symmetry group of the isosceles triangle can be written:
:$\begin{array}{c|cccccc}
 & e & d \\
\hline
e & e & d \\
d & d & e \\
\end{array}$
where:
:$e$ is the identity mapping
:$d$ is the reflection $d$ in the line through $A$ and the midpoint of $BC$.
:
Category:Examples of Cayley Tables
Category:Symmetry Group of Isosceles Triangle
Symmetry Group of Line Segment/Cayley Table

== Cayley Table of Symmetry Group of Line Segment ==
The Cayley table of the symmetry group of the line segment can be written:
:$\begin{array}{c|cccccc} & e & r \\ \hline e & e & r \\ r & r & e \\ \end{array}$
where:
:$e$ is the identity mapping
:$r$ is the plane rotation of $180^\circ$ about the midpoint of $AB$.
:
Category:Examples of Cayley Tables
Category:Symmetry Group of Line Segment
Symmetry Group of Line Segment is Group

== Theorem ==
The symmetry group of the line segment is a group.
=== Definition ===
Recall the definition of the symmetry group of the line segment:
== Proof ==
Let us refer to this group as $D_1$.
Taking the group axioms in turn:
=== ===
From the Cayley table it is seen directly that $D_1$ is closed.
=== ===
Composition of Mappings is Associative.
=== ===
The identity is $e$.
=== ===
Each element is seen to be self-inverse:
:$r^{-1} = r$
No more need be done. $D_1$ is seen to be a group.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 26 \eta$
Category:Symmetry Group of Line Segment