nyuuzyou/wikis huggingface dataset:
Almost-Everywhere Equality Relation for Measurable Sets is Equivalence Relation/Proof

== Theorem ==
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
By Symmetric Difference with Self is Empty Set, we have:
We have:
:$A \symdif A = \O$
By Measure of Empty Set is Zero:
:$\map \mu \O = 0$
So:
:$\map \mu {A \symdif A} = 0$
Therefore:
:$A \sim_\mu A$
Hence $\sim_\mu$ is a reflexive relation.
=== Symmetry ===
Suppose:
:$A \sim_\mu B$
Then:
:$\map \mu {A \symdif B} = 0$
We have:
:$\map \mu {B \symdif A} = 0$
Therefore:
:$B \sim_\mu A$
Hence $\sim_\mu$ is a symmetric relation.
=== Transitivity ===
Suppose:
:$A \sim_\mu B$
and:
:$B \sim_\mu C$
Then:
:$\map \mu {A \symdif B} = 0$
And:
:$\map \mu {B \symdif C} = 0$
Therefore:
:$\map \mu {A \symdif C} = 0$
Hence $\sim_\mu$ is a transitive relation.
All the criteria are therefore seen to hold for $\sim_\mu$ to be an equivalence relation.
== Sources ==
* Chapter $2$: Isomorphism, Conjugacy, and Spectral Isomorphism
Category:Definitions/Almost-Everywhere Equality Relation
Almost-Everywhere Equality Relation for Real-Valued Functions is Equivalence Relation

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
<onlyinclude>
Let $\map {\mathcal M} {X, \Sigma, \R}$ be the real-valued $\Sigma$-measurable functions on $X$. 
Let $\sim_\mu$ be the $\mu$-almost-everywhere equality relation on $\map {\mathcal M} {X, \Sigma, \R}$ by: 
:$f \sim_\mu g$ $\map f x = \map g x$ for $\mu$-almost all $x \in X$.
That is: 
:$\map \mu {\set {x \in X : \map f x \ne \map g x} } = 0$
Then $\sim_\mu$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
From Equality is Reflexive, we have:
:$\map f x = \map f x$ for $\mu$-almost all $x \in X$.
Therefore:
:$f \sim_\mu f$
Hence $\sim_\mu$ is a reflexive relation.
=== Symmetry ===
Suppose:
:$f \sim_\mu g$
Then:
:$\map f x = \map g x$ for $\mu$-almost all $x \in X$.
Then by Equality is Symmetric:
:$\map g x = \map f x$ for $\mu$-almost all $x \in X$.
Therefore:
:$g \sim_\mu f$
Hence $\sim_\mu$ is a symmetric relation.
=== Transitivity ===
Suppose:
:$f \sim_\mu g$
And:
:$g \sim_\mu h$
Then:
:$\map f x = \map g x$ for $\mu$-almost all $x \in X$.
And:
:$\map g x = \map h x$ for $\mu$-almost all $x \in X$.
Then by Equality is Transitive:
:$\map f x = \map h x$ for $\mu$-almost all $x \in X$.
Hence $\sim_\mu$ is a transitive relation.
All the criteria are therefore seen to hold for $\sim_\mu$ to be an equivalence relation.
Category:Definitions/Almost-Everywhere Equality Relation
Almost All Horizontal Sections of Integrable Function are Integrable

== Theorem ==
Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.
Let $f: X \times Y \to \overline \R_{\ge 0}$ be a $\mu \times \nu$-integrable function.
Then: 
:$f^y$ is $\mu$-integrable for $\nu$-almost all $y \in Y$
where $f^y$ is the $y$-horizontal section of $f$.
== Proof ==
From Horizontal Section of Measurable Function is Measurable, we have: 
:$f^y$ is $\Sigma_X$-measurable for each $y \in Y$.
From Function Measurable iff Positive and Negative Parts Measurable, we have: 
:$\paren {f^y}^+$ is $\Sigma_X$-measurable for each $y \in Y$
and:
:$\paren {f^y}^-$ is $\Sigma_X$-measurable for each $y \in Y$.
Define a function $g : Y \to \overline \R$ by: 
:$\ds \map g y = \int \paren {f^y}^+ \rd \mu$
for each $y \in Y$. 
Define $h : Y\to \overline \R$ by: 
:$\ds \map h y = \int \paren {f^y}^- \rd \mu$ 
for each $y \in Y$. 
From Integral of Horizontal Section of Measurable Function gives Measurable Function, we have: 
:$g$ and $h$ are $\Sigma_Y$-measuarble.
From Positive Part of Horizontal Section of Function is Horizontal Section of Positive Part, we have: 
:$\paren {f^y}^+ = \paren {f^+}^y$
From Negative Part of Horizontal Section of Function is Horizontal Section of Negative Part, we have:
:$\paren {f^y}^- = \paren {f^-}^y$
We then have: 
Similarly:
Since $f$ is $\mu \times \nu$-integrable, we have: 
:$\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} < \infty$
and:
:$\ds \int_{X \times Y} f^- \map \rd {\mu \times \nu} < \infty$
So:
:$g$ and $h$ are $\mu$-integrable.
From Integrable Function is A.E. Real-Valued, we therefore have: 
:$\ds \int \paren {f^y}^+ \rd \mu < \infty$
and:
:$\ds \int \paren {f^y}^- \rd \mu < \infty$
for $\nu$-almost all $y \in Y$. 
That is, there exists a $\nu$-null set $N_1 \subseteq Y$ such that whenever: 
:$\ds \int \paren {f^y}^+ \rd \mu = \infty$
we have $y \in N_1$. 
Similarly, there exists a $\nu$-null set $N_2 \subseteq Y$ such that whenever: 
:$\ds \int \paren {f^y}^- \rd \mu = \infty$
we have $y \in N_2$.
If $f^y$ is $\mu$-integrable, we have either: 
:$\ds \int \paren {f^y}^+ \rd \mu < \infty$
or:
:$\ds \int \paren {f^y}^- \rd \mu < \infty$
that is: 
:$y \in N_1$ or $y \in N_2$.
That is, if $f^y$ is not $\mu$-integrable, we have: 
:$y \in N_1 \cup N_2$
From Null Sets Closed under Countable Union, we have: 
:$N_1 \cup N_2$ is $\nu$-null.
So:
:$f^y$ is $\mu$-integrable for $\nu$-almost all $y \in Y$.
== Sources ==
* : $5.2$: Fubini's Theorem: Proposition $5.2.2 \text{ (a)}$
Category:Horizontal Section of Functions
Almost All Real Numbers are Transcendental

== Theorem ==
Almost all real numbers are transcendental.
== Proof ==
By definition, a transcendental number (in this context) is a real number which is not an algebraic number.
Recall that the Real Numbers are Uncountable.
Also recall that the Algebraic Numbers are Countable.
Thus the subset of all real numbers which are not transcendental is countable.
The result follows by the definition of almost all.
== Sources ==
* : $0 \cdotp 11000 10000 00000 00000 00010 00000 00000 00000 0 \ldots$
* : $0 \cdotp 11000 \, 10000 \, 00000 \, 00000 \, 00010 \, 00000 \, 00000 \, 00000 \, 0 \ldots$
Category:Transcendental Numbers
Almost All Vertical Sections of Integrable Function are Integrable

== Theorem ==
Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be $\sigma$-finite measure spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y, \mu \times \nu}$ be the product measure space of $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$.
Let $f: X \times Y \to \overline \R_{\ge 0}$ be a $\mu \times \nu$-integrable function.
Then: 
:$f_x$ is $\nu$-integrable for $\mu$-almost all $x \in X$
where $x$ is the $x$-vertical section of function.
== Proof ==
From Vertical Section of Measurable Function is Measurable, we have: 
:$f_x$ is $\Sigma_Y$-measurable for each $x \in X$.
From Function Measurable iff Positive and Negative Parts Measurable, we have: 
:$\paren {f_x}^+$ is $\Sigma_Y$-measurable for each $x \in X$
and:
:$\paren {f_x}^-$ is $\Sigma_Y$-measurable for each $x \in X$.
Define a function $g : X \to \overline \R$ by: 
:$\ds \map g x = \int \paren {f_x}^+ \rd \nu$
for each $x \in X$. 
Define $h : X \to \overline \R$ by: 
:$\ds \map h x = \int \paren {f_x}^- \rd \nu$ 
for each $x \in X$.
From Integral of Vertical Section of Measurable Function gives Measurable Function, we have: 
:$g$ and $h$ are $\Sigma_X$-measuarble.
From Positive Part of Vertical Section of Function is Vertical Section of Positive Part, we have: 
:$\paren {f_x}^+ = \paren {f^+}_x$
From Negative Part of Vertical Section of Function is Vertical Section of Negative Part, we have:
:$\paren {f_x}^- = \paren {f^-}_x$
We then have: 
Similarly:
Since $f$ is $\mu \times \nu$-integrable, we have: 
:$\ds \int_{X \times Y} f^+ \map \rd {\mu \times \nu} < \infty$
and:
:$\ds \int_{X \times Y} f^- \map \rd {\mu \times \nu} < \infty$
So:
:$g$ and $h$ are $\mu$-integrable.
From Integrable Function is A.E. Real-Valued, we therefore have: 
:$\ds \int \paren {f_x}^+ \rd \nu < \infty$
and:
:$\ds \int \paren {f_x}^- \rd \nu < \infty$
for $\mu$-almost all $x \in X$. 
That is, there exists a $\mu$-null set $N_1 \subseteq X$ such that whenever: 
:$\ds \int \paren {f_x}^+ \rd \nu = \infty$
we have $x \in N_1$. 
Similarly, there exists a $\mu$-null set $N_2 \subseteq X$ such that whenever: 
:$\ds \int \paren {f_x}^- \rd \nu = \infty$
we have $x \in N_2$.
If $f_x$ is not $\nu$-integrable, we have either: 
:$\ds \int \paren {f_x}^+ \rd \nu = \infty$
or:
:$\ds \int \paren {f_x}^- \rd \nu = \infty$
that is:
:$x \in N_1$ or $x \in N_2$
That is, if $f_x$ is not $\nu$-integrable, we have: 
:$x \in N_1 \cup N_2$
From Null Sets Closed under Countable Union, we have: 
:$N_1 \cup N_2$ is $\mu$-null.
So:
:$f_x$ is $\nu$-integrable for $\mu$-almost all $x \in X$.
== Sources ==
* : $5.2$: Fubini's Theorem: Proposition $5.2.2 \text{ (a)}$
Category:Vertical Section of Functions
Almost Convergent Sequence/Examples/Sequence of alternating zeros and ones converges almost to one half

== Theorem ==
Let $\sequence {x_n}_{n \in \N}$ be the sequence defined by:
:$x_n = \begin{cases} 0 & : n \equiv 0 \pmod 2 \\ 1 & : n \equiv 1 \pmod 2 \end{cases}$
where $\bmod$ denotes the congruence modulo.
Then $\sequence {x_n}_{n \in \N}$ almost converges to $1/2$.
== Proof ==
Let $\phi$ be a Banach limit.
Let $S$ be the left shift operator on $\map {\ell^\infty} \R$.
Let $\mathbf 1 := \sequence {1, 1, 1, \ldots}$.
Then:
Thus by definition of Banach limit:
That is:
:$\ds \map \phi {\sequence {x_n} } = \frac 1 2$
As $\phi$ is an arbitrary Banach limit, the claim follows.
Category:Definitions/Banach Limits
Almost Perfect Numbers/Examples/16/Proof 1

== Example of Almost Perfect Number ==
== Proof ==
The divisors of $16$, not including $16$ itself are:
:$1, 2, 4, 8$
Thus we have:
:$1 + 2 + 4 + 8 = 15 = 16 - 1$
Hence the result by definition of almost perfect.
== Sources ==
* : $16$
* : $16$
Category:Almost Perfect Numbers/Examples/16
Alphabetic Substitution is Semantically Equivalent

== Theorem ==
Let $\map \phi x$ be a WFF of predicate logic.
Let $z$ be free for $x$ in $\phi$.
Let $z$ not occur freely in $\phi$.
Let $\map \phi z$ be the result of the alphabetic substitution of $z$ for $x$.
=== Universal Quantifier ===
=== Existential Quantifier ===
Category:Model Theory for Predicate Logic
Alphabetic Substitution is Semantically Equivalent/Existential Quantifier

== Theorem ==
Then:
:$\exists x: \map \phi x$ and $\exists z: \map \phi z$ are semantically equivalent
== Proof ==
== Sources ==
* : $\text{II}.8$ Further Semantic Notions: Exercise $\text{II}.8.25$
Category:Model Theory for Predicate Logic
Alphabetic Substitution is Semantically Equivalent/Universal Quantifier

== Theorem ==
Then:
:$\forall x: \map \phi x$ and $\forall z: \map \phi z$ are semantically equivalent
== Proof ==
== Sources ==
* : $\text{II}.8$ Further Semantic Notions: Exercise $\text{II}.8.25$
Category:Model Theory for Predicate Logic
Alternant (Linear Algebra)/Examples/Arbitrary Example 1

== Example of Alternant in the context of Linear Algebra ==
An example of an alternant:
:$\begin {vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end {vmatrix}$
Here we have:
:$r_1$, $r_2$ and $r_3$ are identified with $x$, $y$ and $z$
:$f_1$ is identified with the constant mapping: $\map {f_1} {x_i} = 1$
:$f_2$ is identified with the identity mapping: $\map {f_2} {x_i} = x_i$
:$f_3$ is identified with the square function: $\map {f_3} {x_i} = {x_i}^2$
== Sources ==
* 
* 
Category:Alternants
Alternate Ratios of Equal Fractions

== Theorem ==
== Proof ==
Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$, and another $D$ be the same aliquot part of another, $EF$, that $A$ is of $BC$.
We need to show that whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$.
:
We have that whatever aliquot part $A$ is of $BC$, the same aliquot part $D$ is of $EF$.
Therefore, as many numbers as there are in $BC$ equal to $A$, so many also are there in $EF$ equal to $D$.
Let $BC$ be divided into the numbers equal to $A$, namely $BG, GC$.
Let $EF$ be divided into the numbers equal to $D$, namely $EH, HF$.
Thus the multitude of $BG, GC$ equals the multitude of $EH, HF$.
We have that $BG = GC$ and $EH = HF$.
Therefore whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part or the same aliquant part is $GC$ of $HF$ also.
So, from and , we have that whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part also, or the same aliquant part, is the sum $BC$ of the sum $EF$.
But $BG = A$ and $EH = D$.
So whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$ also.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Ratios
Alternate Ratios of Multiples

== Theorem ==
== Proof ==
Let the unit $A$ measure any (natural) number $BC$.
Let another (natural) number $D$ measure any other (natural) number $EF$ the same number of times.
We need to show that $A$ measures $D$ the same number of times that $BC$ measures $EF$.
:
We have that $A$ measures $BC$ the same number of times that $D$ measures $EF$.
So as many units as there are in $BC$ there are numbers equal to $D$ in $EF$.
Let $BC$ be divided into the units in it: $BG, GH, HC$.
Let $EF$ be divided into the numbers in it equal to $D$: $EK, KL, LF$.
So the multitude of $BG, GH, HC$ will equal the multitude of $EK, KL, LF$.
We have that $BG = GH = HC$ and $EK = KL = LF$.
So $BG : EK = GH : KL = HC : LF$.
So from , $BG : EK = BG + GH + HC : EK + KL + LF$.
That is, $BG : EK = BC : EF$.
But $BG = A$ and $EK = D$.
So $A : D = BC : EF$.
So $A$ measures $D$ the same number of times that $BC$ measures $EF$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Ratios
Alternating Bilinear Form is Reflexive

== Theorem ==
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $b$ be a bilinear form on $V$.
Let $b$ be alternating.
Then $b$ is reflexive.
== Proof ==
Let $\tuple {v, w} \in V \times V$ with $\map b {v, w} = 0$.
We have:
Because $\tuple {v, w}$ was arbitrary, $b$ is reflexive.
== Also see ==
* Bilinear Form is Reflexive iff Symmetric or Alternating
Category:Bilinear Forms (Linear Algebra)
Alternating Even-Odd Digit Palindromic Prime

== Theorem ==
Let the notation $\paren {abc}_n$ be interpreted to mean $n$ consecutive repetitions of a string of digits $abc$ concatenated in the decimal representation of an integer.
The integer:
:$\paren {10987654321234567890}_{42} 1$
has the following properties:
: it is a palindromic prime with $841$ digits
: its digits are alternately odd and even.
== Proof ==
This number has $20 \times 42 + 1 = 841$ digits.
The remaining properties of this number is obvious by inspection.
== Sources ==
* 
* : $\left({10987654321234567890}\right)_{42} 1$
Category:Prime Numbers
Alternating Group has no Subgroup of Order 6

== Theorem ==
Let $A_4$ denote the alternating group on $4$ letters.
$A_4$ has no subgroup of order $6$.
== Proof ==
We list the subgroups of $A_4$:
So there is no subgroup of $A_4$ of order $6$.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 83 \beta$
Category:Alternating Group on 4 Letters
Alternating Group is Normal Subgroup of Symmetric Group

== Theorem ==
Let $n \ge 2$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ be the alternating group on $n$ letters.
Then $A_n$ is a normal subgroup of $S_n$ whose index is $2$.
== Proof ==
We have that $\map \sgn {S_n}$ is onto $C_2$.
Thus from the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$.
The result follows from Subgroup of Index 2 is Normal.
== Examples ==
When $n = 2$, we have that:
:$S_2 \cong C_2$
and:
:$A_n = \set {e_{S_2} }$
== Sources ==
* : $\S 7.4$. Kernel and image: Example $142$
* : Chapter $2$: The Symmetric Groups: $\S 81$
* : Chapter $9$: Permutations: Definition $9.19$: Remark
* 
* : Chapter $13$: $32 \, \text{a}$
* 
Category:Symmetric Groups
Category:Alternating Groups
Alternating Group is Set of Even Permutations

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ be the alternating group on $n$ letters.
Then $A_n$ consists of the set of even permutations of $S_n$.
== Proof ==
Let $\sgn$ denote the sign of permutation on n Letters.
We have that $\map \sgn {S_n}$ is onto $C_2$.
Thus from the First Isomorphism Theorem, $A_n$ consists of the set of even permutations of $S_n$.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 81$
* : Chapter $9$: Permutations: Definition $9.19$: Remark
Category:Symmetric Groups
Category:Alternating Groups
Alternating Group is Simple except on 4 Letters/Lemma 1

== Theorem ==
Let $n$ be an integer such that $n \ge 5$.
Let $A_n$ denote the alternating group on $n$ letters.
Let $\alpha \in A_n$ be a permutation on $\N_n$ such that $\map \alpha 1 = 2$.
Let $\beta$ be the $3$-cycle $\tuple {3, 4, 5}$.
Then the permutation $\beta^{-1} \alpha^{-1} \beta \alpha$ fixes $1$.
== Proof ==
We let $\beta^{-1} \alpha^{-1} \beta \alpha$ act on $1$.
By construction:
:$\map \alpha 1 = 2$
Thus:
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 83$
Category:Alternating Group is Simple except on 4 Letters
Alternating Group is Simple except on 4 Letters/Lemma 2

== Theorem ==
Let $n$ be an integer such that $n \ge 5$.
Let $A_n$ denote the alternating group on $n$ letters.
Let $\alpha \in A_n$ be the permutation on $\N_n$ in the form:
:$\alpha = \tuple {1, 2} \tuple {3, 4}$
Let $\beta$ be the $3$-cycle $\tuple {3, 4, 5}$.
Then:
:$\beta^{-1} \alpha^{-1} \beta \alpha = \beta$
== Proof ==
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 83$
Category:Alternating Group is Simple except on 4 Letters
Alternating Group on 3 Letters

== Group Example ==
Let $S_3$ denote the symmetric group on $3$ letters.
The alternating group on $3$ letters $A_3$ is the kernel of the mapping $\sgn: S_3 \to C_2$.
$A_3$ consists of the $3$ elements:
:$A_3 = \set {e, \tuple {123}, \tuple {132} }$
where the tuples denote cycle notation.
Thus $A_3$ is an instance of the cyclic group of order $3$.
== Sources ==
* : Chapter $9$: Permutations: Definition $9.19$: Remark
Category:Groups of Order 3
Category:Examples of Alternating Groups
Alternating Group on 4 Letters

== Group Example ==
Let $S_4$ denote the symmetric group on $4$ letters.
The alternating group on $4$ letters $A_4$ is the kernel of the mapping $\sgn: S_4 \to C_2$.
=== Cycle Notation ===
It can be expressed in the form of permutations given in cycle notation as follows:
=== Cayley Table ===
The Cayley table of $A_4$ can be written:
== Order of Elements ==
== Subgroups ==
== Normality of Subgroups ==
== Conjugacy Classes ==
Category:Groups of Order 12
Category:Alternating Group on 4 Letters
Category:Examples of Alternating Groups
Alternating Group on 4 Letters/Cayley Table

== Cayley Table of Alternating Group on $4$ Letters ==
The Cayley table of the alternating group on $4$ letters can be written:
:$\begin{array}{c|cccc|cccc|cccc}
\circ & e & t & u & v & a & b & c & d & p & q & r & s \\
\hline
e & e & t & u & v & a & b & c & d & p & q & r & s \\
t & t & e & v & u & b & a & d & c & q & p & s & r \\
u & u & v & e & t & c & d & a & b & r & s & p & q \\
v & v & u & t & e & d & c & b & a & s & r & q & p \\
\hline
a & a & c & d & b & p & r & s & q & e & u & v & t \\
b & b & d & c & a & q & s & r & p & t & v & u & e \\
c & c & a & b & d & r & p & q & s & u & e & t & v \\
d & d & b & a & c & s & q & p & r & v & t & e & u \\
\hline
p & p & s & q & r & e & v & t & u & a & d & b & c \\
q & q & r & p & s & t & u & e & v & b & c & a & d \\
r & r & q & s & p & u & t & v & e & c & b & d & a \\
s & s & p & r & q & v & e & u & t & d & a & c & b \\
\end{array}$
where the expression for $A_4$ in cycle notation is given as:
== Sources ==
* : Tables: $3$. Alternating group $\map A 4$
Category:Examples of Cayley Tables
Category:Alternating Group on 4 Letters
Alternating Group on 4 Letters/Conjugacy Classes

== Conjugacy Classes of the Alternating Group on 4 Letters ==
Let $A_4$ denote the Alternating Group on 4 Letters, whose Cayley table is given as:
The conjugacy classes of $A_4$ are:
== Proof ==
The elements of the Alternating Group on 4 Letters can be expressed in cycle notation thus:
== Sources ==
* : Chapter $6$: Cosets: Exercise $7$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $3$
Category:Alternating Group on 4 Letters
Category:Examples of Conjugacy Classes
Alternating Group on 4 Letters/Normality of Subgroups

== Normality of Subgroups of the Alternating Group on $4$ Letters ==
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
The normality status of the non-trivial proper subgroups of $A_4$ is as follows:
Order $2$ subgroups:
Order $3$ subgroups:
Order $4$ subgroup:
== Proof ==
Testing one of the left cosets of $T = \set {e, t}$ against its corresponding right coset:
The left coset does not equal the right coset and so $T$ is not normal in $A_4$.
Testing one of the left cosets of $U = \set {e, u}$ against its corresponding right coset:
The left coset does not equal the right coset and so $U$ is not normal in $A_4$.
Testing one of the left cosets of $V = \set {e, v}$ against its corresponding right coset:
The left coset does not equal the right coset and so $V$ is not normal in $A_4$.
Testing one of the left cosets of $P = \set {e, a, p}$ against its corresponding right coset:
The left coset does not equal the right coset and so $P$ is not normal in $A_4$.
Testing one of the left cosets of $Q = \set {e, c, q}$ against its corresponding right coset:
The left coset does not equal the right coset and so $Q$ is not normal in $A_4$.
Testing one of the left cosets of $R = \set {e, d, r}$ against its corresponding right coset:
The left coset does not equal the right coset and so $R$ is not normal in $A_4$.
Testing one of the left cosets of $S = \set {e, b, s}$ against its corresponding right coset:
The left coset does not equal the right coset and so $S$ is not normal in $A_4$.
The cosets of $K = \set {e, t, u, v}$ are as follows:
The left cosets equal the right cosets and so $K$ is normal in $A_4$.
== Sources ==
* : Chapter $6$: Cosets: Exercise $11$
* : Chapter $2$: The Symmetric Groups: $\S 83 \alpha$
Category:Alternating Group on 4 Letters
Category:Examples of Normal Subgroups
Alternating Group on 4 Letters/Order of Elements

== Order of Elements of Alternating Group on $4$ Letters ==
The order of each element of the Alternating Group on $4$ Letters is given below:
where the expression for $A_4$ in cycle notation is given as:
The above can be determined by inspecting its Cayley table:
== Sources ==
* : Chapter $5$: Subgroups: Exercise $13$
* : Chapter $9$: Permutations: Exercise $3$
Category:Alternating Group on 4 Letters
Alternating Group on 4 Letters/Subgroups

== Subgroups of the Alternating Group on $4$ Letters ==
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
The subsets of $A_4$ which form subgroups of $A_4$ are as follows:
Trivial:
Order $2$ subgroups:
Order $3$ subgroups:
Order $4$ subgroup:
== Examples of Subgroups ==
== Sources ==
* : Chapter $9$: Permutations: Exercise $3$
Category:Alternating Group on 4 Letters
Alternating Group on 4 Letters/Subgroups/Examples/Order 3

== Subgroups of the Alternating Group on $4$ Letters ==
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
Let $P$ denote the subset of $A_4$:
:$P := \set {e, a, p}$
Then $P$ is a subgroup of $A_4$.
Its left cosets are:
Its right cosets are:
== Proof ==
We have that:
Thus $\set {e, a, p}$ forms a cyclic group generated by $a$.
Thus $\set {e, a, p}$ is a group which is a subset of $A_4$.
Hence by definition $\set {e, a, p}$ is a subgroup of $A_4$.
Then:
and:
== Sources ==
* : Chapter $6$: Cosets: Exercise $1$
Category:Alternating Group on 4 Letters
Alternating Group on More than 3 Letters is not Abelian

== Theorem ==
Let $n$ be an integer such that $n > 3$.
Then the $n$th alternating group $A_n$ is not abelian.
== Proof ==
Let $\tuple {1, 2, 3}$ and $\tuple {2, 3, 4}$ be elements of $A_n$.
Then we have:
So:
:$\tuple {1, 2, 3} \tuple {2, 3, 4} \ne \tuple {2, 3, 4} \tuple {1, 2, 3}$
and so $A_n$ is not abelian.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 84$
Category:Alternating Groups
Category:Cyclic Groups