nyuuzyou/wikis huggingface dataset:
Symmetric Difference with Self is Empty Set

== Theorem ==
The symmetric difference of a set with itself is the empty set:
:$S \symdif S = \O$
== Proof ==
This follows directly from Symmetric Difference of Equal Sets:
:$S \symdif T = \O \iff S = T$
substituting $S$ for $T$.
== Sources ==
* : $\S 5$: Complements and Powers
* : Chapter $1$. Sets: Exercise $7 \ \text{(ii)}$
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $6$
* : Chapter $1$: Sets and Logic: Exercise $9 \ \text{(i)}$
* : $\text {(i)}$
Category:Symmetric Difference
Category:Empty Set
Symmetric Difference with Union does not form Ring

== Theorem ==
Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cup$ denote the set union operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cup}$ does not form a ring.
== Proof ==
For $\struct {S, \symdif, \cup}$ to be a ring, it is a necessary condition that $\cup$ be distributive over $*$.
Also, the identity element for set union and symmetric difference must be different.
However:
:$(1): \quad$ the identity for union and symmetric difference is $\O$ for both operations
:$(2): \quad$ set union is not distributive over symmetric difference:
From Symmetric Difference of Unions:
:$\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
The result follows.
== Also see ==
* Symmetric Difference with Intersection forms Ring
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Some 'non-examples': $\text {(d)}$
Category:Set Union
Category:Symmetric Difference
Category:Power Set
Category:Examples of Rings
Symmetric Difference with Universal Set

== Theorem ==
:$\mathbb U \symdif S = \map \complement S$
where:
:$\mathbb U$ denotes the universal set
:$\symdif$ denotes symmetric difference.
== Proof ==
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $9 \ \text{(i)}$
* : $\text {(i)}$
Category:Symmetric Difference
Category:Universal Set
Category:Set Complement
Symmetric Function Theorem

== Theorem ==
Let $f$ be a polynomial in $n$ variables.
Let $f$ be of degree $r$ in each of its $n$ variables.
Then $f$ is equal to a polynomial of total degree $r$ with integer coefficients in the elementary symmetric functions:
:$\ds \sum x_i, \sum x_i x_j, \dotsb, \prod x_j$
and the coefficients of $f$.
== Proof ==
== Sources ==
* : Preface to first edition: Prerequisites: $\text {(ii)}$
Category:Polynomial Theory
Category:Named Theorems
Symmetric Group has Non-Normal Subgroup

== Theorem ==
Let $S_n$ be the (full) symmetric group on $n$ elements, where $n \ge 3$.
Then $S_n$ contains at least one subgroup which is not normal.
== Proof ==
Let $S_n$ act on the set $S$.
Let $e$ be the identity of $S_n$, by definition the identity mapping $I_S$ on $S$.
As $S$ has at least three elements, three can be arbitrary selected and called $a$, $b$ and $c$.
Let $\rho$ be a transposition of $S_n$, transposing elements $a$ and $b$.
$\rho$ can be described in cycle notation as $\paren {a \ b}$.
From Transposition is Self-Inverse it follows that $\set {e, \rho}$ is a subgroup of $S_n$.
Let $\pi$ be the permutation on $S$ described in cycle notation as $\paren {a \ b \ c}$.
By inspection it is found that $\pi^{-1} = \paren {a \ c \ b}$.
Then we have:
So, by definition, $\set {e, \rho}$ is not a normal subgroup.
== Sources ==
* : Chapter $\text{II}$: Groups: Morphisms
Category:Normal Subgroups
Category:Symmetric Groups
Symmetric Group is Generated by Transposition and n-Cycle

== Theorem ==
Let $n \in \Z: n > 1$.
Let $S_n$ denote the symmetric group on $n$ letters.
Then the set of cyclic permutations:
:$\set {\begin {pmatrix} 1 & 2 \end{pmatrix}, \begin {pmatrix} 1 & 2 & \cdots & n \end{pmatrix} }$
is a generator for $S_n$.
== Proof ==
Denote:
:$s = \begin {pmatrix} 1 & 2 \end{pmatrix}$
:$r = \begin {pmatrix} 1 & 2 & \cdots & n \end{pmatrix}$
By Cycle Decomposition of Conjugate,:
:$r s r^{-1} = r \begin {pmatrix} 1 & 2 \end{pmatrix} r^{-1} = \begin {pmatrix} \map r 1 & \map r 2 \end{pmatrix} = \begin {pmatrix} 2 & 3 \end{pmatrix}$.
By repeatedly using Cycle Decomposition of Conjugate:
:$r^2 s r^{-2} = \begin {pmatrix} 3 & 4 \end{pmatrix}$
:$r^3 s r^{-3} = \begin {pmatrix} 4 & 5 \end{pmatrix}$
:$\cdots$
:$r^{n - 2} s r^{-\paren {n - 2} } = \begin {pmatrix} n - 1 & n \end{pmatrix}$
The result then follows from Transpositions of Adjacent Elements generate Symmetric Group.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 80 \delta$
Category:Symmetric Groups
Category:Examples of Generators of Groups
Symmetric Group is Group

== Theorem ==
Let $S$ be a set.
Let $\map \Gamma S$ denote the set of all permutations on $S$.
Then $\struct {\map \Gamma S, \circ}$, the symmetric group on $S$, forms a group.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Symmetric Group is Subgroup of Monoid of Self-Maps
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 26 \alpha$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : $\S 1.1$ Example $6$
Category:Symmetric Group is Group
Category:Symmetric Groups
Symmetric Group is Group/Proof 1

== Theorem ==
== Proof ==
Taking the group axioms in turn:
=== ===
By Composite of Permutations is Permutation, $S$ is itself a permutation on $S$.
Thus $\struct {\map \Gamma S, \circ}$ is closed.
=== ===
From Set of all Self-Maps under Composition forms Monoid, we have that $\struct {\map \Gamma S, \circ}$ is associative.
=== ===
From Set of all Self-Maps under Composition forms Monoid, we have that $\struct {\map \Gamma S, \circ}$ has an identity, that is, the identity mapping.
=== ===
By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
Thus all the group axioms have been fulfilled, and the result follows.
== Sources ==
* : $\S 4.5$. Examples of groups: Example $79$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.5$
* : $\S 1.6$: Theorem $5$
* : $\S 1$: Some examples of groups: Example $1.13$
* : Chapter $2$: Examples of Group Structure: $\S 30$
* : Chapter $2$: The Symmetric Groups: $\S 76$
* : $\S 34$. Examples of groups: $(4)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(iii)}$
* : Chapter $2$: Maps and relations on sets: Corollary $2.18$
Category:Symmetric Group is Group
Symmetric Group is Subgroup of Monoid of Self-Maps

== Theorem ==
Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself
Let $\struct {\Gamma \paren S, \circ}$ denote the symmetric group on $S$.
Let $\struct {S^S, \circ}$ be the monoid of self-maps under composition of mappings.
Then $\struct {\Gamma \paren S, \circ}$ is a subgroup of $\struct {S^S, \circ}$.
== Proof ==
By Symmetric Group is Group, $\struct {\Gamma \paren S, \circ}$ is a group.
Let $\phi \in \Gamma \paren S$ be a permutation on $S$.
As a permutation is a self-map, it follows that $\phi \in S^S$.
Thus by definition $\Gamma \paren S$ is a subset of $S^S$.
So by definition, $\Gamma \paren S$, is a subgroup of $\struct {S^S, \circ}$.
Category:Symmetric Groups
Category:Subgroups
Category:Monoids
Symmetric Group is not Abelian

== Theorem ==
Let $S_n$ be the symmetric group of order $n$ where $n \ge 3$.
Then $S_n$ is not abelian.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4.5$. Examples of groups: Example $79$
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.3$
* : Chapter $\text{II}$: Groups: The Group Property
* : Chapter $2$: Examples of Group Structure: $\S 30 \beta$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP
* 
Category:Symmetric Groups
Category:Symmetric Group is not Abelian
Symmetric Group is not Abelian/Proof 1

== Theorem ==
== Proof ==
Let $\alpha \in S_n$ such that $\alpha$ is not the identity mapping.
From Center of Symmetric Group is Trivial, $\alpha$ is not in the center $\map Z {S_n}$ of $S_n$.
Thus $S_n \ne \map Z {S_n}$.
The result follows by definition of abelian group.
Category:Symmetric Group is not Abelian
Symmetric Group is not Abelian/Proof 2

== Theorem ==
== Proof ==
Let $a, b, c \in S$.
Let $\alpha$ be the transposition on $S$ which exchanges $a$ and $b$.
Let $\beta$ be the transposition on $S$ which exchanges $b$ and $c$.
Then:
:$\alpha \circ \beta$ maps $\tuple {a, b, c}$ to $\tuple {c, a, b}$
while:
:$\beta \circ \alpha$ maps $\tuple {a, b, c}$ to $\tuple {b, c, a}$
Thus $\alpha, \beta \in S_n$ such that $\alpha$ does not commute with $\beta$.
Hence the result by definition of abelian group.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(iii)}$
Category:Symmetric Group is not Abelian
Symmetric Group on 3 Letters

== Group Example ==
Let $S_3$ denote the set of permutations on $3$ letters.
The symmetric group on $3$ letters is the algebraic structure:
:$\struct {S_3, \circ}$
where $\circ$ denotes composition of mappings.
It is usually denoted, when the context is clear, without the operator: $S_3$.
=== Cycle Notation ===
It can be expressed in the form of permutations given in cycle notation as follows:
=== Cayley Table ===
=== Group Presentation ===
Its group presentation is:
== Order of Elements ==
== Subgroups ==
== Normal Subgroups ==
== Generators ==
== Centralizers ==
== Normalizers of Subgroups ==
== Center ==
== Conjugacy Classes ==
== Also see ==
* Symmetric Group is Group, which demonstrates that this is a (finite) group.
== Sources ==
* 
* 
Category:Symmetric Group on 3 Letters
Category:Groups of Order 6
Symmetric Group on 3 Letters/Cayley Table

== Cayley Table of Symmetric Group on $3$ Letters ==
The Cayley table of the symmetric group on $3$ letters can be presented in cycle notation as:
:$\begin{array}{c|cccccc}\circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$
It can also often be seen presented as:
:$\begin{array}{c|cccccc} & e & p & q & r & s & t \\ \hline e & e & p & q & r & s & t \\ p & p & q & e & s & t & r \\ q & q & e & p & t & r & s \\ r & r & t & s & e & q & p \\ s & s & r & t & p & e & q \\ t & t & s & r & q & p & e \\ \end{array}$
== Sources ==
* : Tables: $1$. Symmetric Group $\map S 3$
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.6$
* : Chapter $\text{II}$: Groups: Exercise $\text{A}$
* : Chapter $2$: Maps and relations on sets: Example $2.19$
Category:Examples of Cayley Tables
Category:Symmetric Group on 3 Letters
Symmetric Group on 3 Letters/Center

== Center of the Symmetric Group on 3 Letters ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
The center of $S_3$ is given by:
:$\map Z {S_3} = \set e$
== Proof ==
From Center of Dihedral Group:
:$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$
Hence the result.
== Sources ==
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $5$
Category:Centers of Groups
Category:Symmetric Group on 3 Letters
Symmetric Group on 3 Letters/Centralizers

== Centralizers of the Symmetric Group on 3 Letters ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
The centralizers of each element of $S_3$ are given by:
== Proof ==
By definition, the centralizer of $a$ (in $S_3$) is defined as:
:$\map {C_{S_3} } a = \set {x \in S_3: x \circ a = a \circ x}$
The centralizer are then apparent by inspection of the Cayley table.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $16$
Category:Symmetric Group on 3 Letters
Category:Examples of Centralizers
Symmetric Group on 3 Letters/Conjugacy Classes

== Conjugacy Classes of the Symmetric Group on 3 Letters ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
The conjugacy classes of $S_3$ are:
== Proof ==
We have that Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3.
This is the group presentation of $D_3$:
First we have that Conjugacy Class of Element of Center is Singleton.
From Center of Symmetric Group is Trivial, only $e$ is in the center.
Thus $\set e$ forms one of the conjugacy classes.
We then have:
and so $a^2$ is in the same conjugacy class as $a$.
From Centralizer of Group Element is Subgroup, the centralizer $\map {C_{S_3} } a$ of $a$ is a subgroup of $S_3$.
Thus $\map {C_{S_3} } a$ also contains $a^2$.
Thus the order of $\map {C_{S_3} } a$ is either $3$ or $6$, from Lagrange's Theorem.
But we have that $b a \ne a b$ and so $b \notin \map {C_{S_3} } a$.
Thus $\map {C_{S_3} } a$ contains $3$ elements.
By the Orbit-Stabilizer Theorem $a$ has $2$ conjugates.
Hence the conjugacy class of $a$ is $\set {a, a^2}$.
By a similar argument, $\map {C_{S_3} } b = \gen b$ and so contains $2$ elements.
Thus $b$ has $3$ conjugates.
As the conjugacy classes form a partition, $\card {\map {C_{S_3} } b} = 3$.
Hence the result.
== Sources ==
* : $\S 6.5$. Orbits: Example $119$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.17$
Category:Symmetric Group on 3 Letters
Category:Examples of Conjugacy Classes
Symmetric Group on 3 Letters/Generators

== Generators of the Symmetric Group on 3 Letters ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
Let:
Then:
where $\gen G$ denotes the group generated by a subset $G$ of $S_3$.
== Proof ==
For $G_1$:
For $G_2$:
== Sources ==
* : $\S 5.3$. Subgroup generated by a subset: Example $97$
* : Chapter $5$: Subgroups: Exercise $8$
* : Chapter $4$: Subgroups: Example $4.9$
Category:Symmetric Group on 3 Letters
Category:Examples of Generators of Groups
Symmetric Group on 3 Letters/Group Presentation

== Group Presentation of Symmetric Group on $3$ Letters ==
The group presentation of the symmetric group on $3$ letters is given by:
:$S_3 := \gen {a, b: a^3 = b^2 = \paren {a b}^2 = e}$
Hence:
:$\begin{array}{c|cccccc} & e & a & a^2 & b & a b & a^2 b \\ \hline e & e & a & a^2 & b & a b & a^2 b \\ a & a & a^2 & e & a b & a^2 b & b \\ a^2 & a^2 & e & a & a^2 b & b & a b \\ b & b & a^2 b & a b & e & a^2 & a \\ a b & a b & b & a^2 b & a & e & a^2 \\ a^2 b & a^2 b & a b & b & a^2 & a & e \\ \end{array}$
== Proof ==
Let $G = \gen {a, b: a^3 = b^2 = \paren {a b}^2 = e}$.
It is to be demonstrated that $S_3$ is isomorphic to $G$.
Consider the Cayley table for $S_3$:
We demonstrate that $S_3$ is isomorphic to $G$ as follows:
Let $\phi: G \to S_3$ be a mapping that sends:
:$\phi: a \mapsto p$
:$\phi: b \mapsto r$
We have:
:$p^3 = e$
:$r^2 = e$
:$\paren {p r}^2 = s^2 = e$
demonstrating that $S_3$ has the same group presentation as $G$.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
* : $1$: Subgroups: Problem $1.1$
* : Chapter $5$: Cosets and Lagrange's Theorem: Example $5.2$
Category:Symmetric Group on 3 Letters
Symmetric Group on 3 Letters/Normal Subgroups

== Normal Subgroups of the Symmetric Group on 3 Letters ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
Consider the subgroups of $S_3$:
Of those, the normal subgroups in $S_3$ are:
:$S_3, \set e, \set {e, \tuple {123}, \tuple {132} }$
=== Cayley table of Quotient Group ===
== Proof ==
$S_3$ itself is normal in $S_3$ by Group is Normal in Itself.
$\set e$ is normal in $S_3$ by Trivial Subgroup is Normal.
$\set {e, \tuple {12} }$:
Hence $\set {e, \tuple {12} }$ is not normal in $S_3$.
$\set {e, \tuple {23} }$:
Hence $\set {e, \tuple {23} }$ is not normal in $S_3$.
$\set {e, \tuple {13} }$:
Hence $\set {e, \tuple {13} }$ is not normal in $S_3$.
$\set {e, \tuple {123}, \tuple {132} }$:
We have that $\set {e, \tuple {123}, \tuple {132} }$ is the set of even permutations of $S_3$.
Any permutation of the form $\alpha \pi \alpha^{-1}$, for $\pi$ even, is also even.
Thus:
:$\forall \alpha \in S_3: \alpha \pi \alpha^{-1} \in \set {e, \tuple {123}, \tuple {132} }$
Hence $\set {e, \tuple {123}, \tuple {132} }$ is normal in $S_3$.
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $122$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.5 \ \text{(a)}$
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $2$
Category:Symmetric Group on 3 Letters
Category:Examples of Normal Subgroups
Symmetric Group on 3 Letters/Normal Subgroups/Cayley table of Quotient Group

== Normal Subgroups of the Symmetric Group on 3 Letters ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
Let $H$ denote the normal subgroup $\set {e, \tuple {123}, \tuple {132} }$.
Let $K$ denote the coset of $H$ in $S_3$.
The Cayley table of the quotient of $S_3$ by $H$ is given as:
:$\begin {array} {c|cc} S_3 / H & H & K \\ \hline H & H & K \\ K & K & H \end {array}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.5 \ \text{(b)}$
Category:Symmetric Group on 3 Letters
Category:Examples of Normal Subgroups
Symmetric Group on 3 Letters/Normalizers

== Normalizers of the Subgroups of the Symmetric Group on 3 Letters ==
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
The normalizers of each subgroup of $S_3$ are given by:
== Proof ==
The subgroups of $S_3$ are as follows:
By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$\map {N_{S_3} } {\set e}$ is the largest subgroup $N$ of $S_3$ in which $H$ is normal in $N$.
From Trivial Subgroup is Normal:
:$\map {N_{S_3} } {\set e} = S_3$
From Group is Normal in Itself:
:$\map {N_{S_3} } {S_3} = S_3$
The index of $\set {e, \tuple {123}, \tuple {132} }$ is $2$.
Hence from Subgroup of Index 2 is Normal, $\set {e, \tuple {123}, \tuple {132} }$ is normal in $S_3$.
It follows that:
:$\map {N_{S_3} } {\set {e, \tuple {123}, \tuple {132} } } = S_3$
From Normal Subgroups of Symmetric Group on 3 Letters, none of $\set {e, \tuple {12} }$, $\set {e, \tuple {13} }$ and $\set {e, \tuple {23} }$ is normal in $S_3$.
There are no larger subgroup of $S_3$ containing any of them, so they are their own normalizers.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.11$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $6$
Category:Symmetric Group on 3 Letters
Category:Examples of Normalizers
Symmetric Group on 3 Letters/Order of Elements

== Orders of Elements of the Symmetric Group on $3$ Letters ==
Let $S_3$ denote the Symmetric Group on $3$ Letters, whose Cayley table is given as:
The orders of the various elements of $S_3$ are:
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Exercise $4$
Category:Symmetric Group on 3 Letters
Symmetric Group on 3 Letters/Subgroups

== Subgroups of the Symmetric Group on $3$ Letters ==
Let $S_3$ denote the Symmetric Group on $3$ Letters, whose Cayley table is given as:
The subsets of $S_3$ which form subgroups of $S_3$ are:
== Examples ==
== Sources ==
* : $\S 5.2$. Subgroups: Example $93$
* : $1$: Subgroups: Problem $1.1$
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $2$
Category:Symmetric Group on 3 Letters
Symmetric Group on 3 Letters/Subgroups/Examples/Non-Subgroup

== Example of Subset of Symmetric Group on 3 Letters which is not a Subgroup ==
Let $S_3$ denote the Symmetric Group on $3$ Letters, whose Cayley table is given as:
Consider the subset $H$ of $S_3$:
:$H = \set {e, \tuple {12}, \tuple {13}, \tuple {23} }$
Then $H$ is not a subgroup of $S_3$.
== Proof ==
We have:
:$\tuple {12} \circ \tuple {13} = \tuple {123}$
But $\tuple {123} \notin H$.
Thus $H$ is not closed under $\circ$.
Hence the result by definition of subgroup.
== Sources ==
* : Chapter $4$: Subgroups: Exercise $1 \ \text{(b)}$
Category:Symmetric Group on 3 Letters
Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3

== Theorem ==
Let $S_3$ denote the Symmetric Group on 3 Letters.
Let $D_3$ denote the dihedral group $D_3$.
Then $S_3$ is isomorphic to $D_3$.
== Proof ==
Consider $S_3$ as presented by its Cayley table:
Consider $D_3$ as presented by its group presentation:
and its Cayley table:
Let $\phi: S_3 \to D_3$ be specified as:
Then by inspection, we see:
and the result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.6$
* : Chapter $8$: The Homomorphism Theorem: Example $8.3$
Category:Symmetric Group on 3 Letters
Category:Dihedral Group D3
Category:Examples of Group Isomorphisms
Symmetric Group on 4 Letters

== Group Example ==
Let $S_4$ denote the set of permutations on $4$ letters.
The symmetric group on $4$ letters is the algebraic structure:
:$\struct {S_4, \circ}$
where $\circ$ denotes composition of mappings.
It is usually denoted, when the context is clear, without the operator: $S_4$.
=== Cycle Notation ===
It can be expressed in the form of permutations given in cycle notation as follows:
=== Cayley Table ===
The Cayley table of $S_4$ can be written:
== Subgroups ==
== Normalizers ==
== Conjugacy Classes ==
== Also see ==
* Symmetric Group is Group, which demonstrates that this is a (finite) group.
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $1$
* : Chapter $9$: Permutations
Category:Groups of Order 24
Category:Symmetric Group on 4 Letters
Symmetric Group on 4 Letters/Conjugacy Classes

== Conjugacy Classes of the Symmetric Group on 4 Letters ==
Let $S_4$ denote the Symmetric Group on 4 Letters, whose Cayley table is given as:
The conjugacy classes of $S_4$ are:
== Proof ==
From Conjugacy Classes of Symmetric Group, the conjugacy classes of $S_4$ are determined by the cycle types of the elements of $S_4$.
Hence the result.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $3$
Category:Symmetric Group on 4 Letters
Category:Examples of Conjugacy Classes
Symmetric Group on 4 Letters/Cycle Notation

== Cycle Notation for Symmetric Group on $4$ Letters ==
The alternating group on $4$ letters can be given in cycle notation, and its elements assigned arbitrary labels from $1$ to $23$ as follows:
== Sources ==
* : Chapter $9$: Permutations: Exercise $2$
Category:Symmetric Group on 4 Letters
Category:Examples of Cycle Notation
Symmetric Group on 4 Letters/Normalizers

== Normalizers of the Symmetric Group on 4 Letters ==
Let $S_4$ denote the Symmetric Group on 4 Letters, whose Cayley table is given as:
Let $\alpha$ denote the permutation in $S_4$ given in cycle notation as $\tuple {1234}$.
The normalizer of $S = \set {\alpha, \alpha^{-1} }$ in $S_4$ is given by:
:$\map {N_{S_4} } S = \set {e, \alpha, \alpha^2, \alpha^3, \beta, \alpha \beta, \alpha^2 \beta, \alpha^3 \beta}$
where $\beta$ denotes the permutation in $S_4$ given in cycle notation as $\tuple {24}$.
== Proof ==
== Sources ==
* : Chapter $5$: Subgroups: Exercise $17$
Category:Symmetric Group on 4 Letters
Category:Examples of Normalizers