nyuuzyou/wikis huggingface dataset:
Period of Revolution of Conical Pendulum

== Theorem ==
Let $P$ be a conical pendulum.
Let $h$ be the height of the pivot of $P$ above the circle in which $P$ is moving.
Then the period of revolution of $P$ around the circle is given by:
:$T = 2 \pi \sqrt {\dfrac h g}$
where $g$ is the acceleration on $P$ caused by the gravitational field in which $P$ is suspended.
== Proof ==
== Sources ==
* 
* 
Category:Conical Pendulums
Period of Rotation of Plane of Oscillation of Foucault's Pendulum

== Theorem ==
Let $P$ be a Foucault pendulum.
The plane of oscillation of $P$ rotates through a full circle in time $T$, where:
:$T = \dfrac D {\sin \lambda}$
where:
:$\lambda$ denotes the latitude on Earth at which $P$ is located
:$D$ denotes the length of a sidereal day.
== Proof ==
From Angular Speed of Rotation of Plane of Oscillation of Foucault's Pendulum, the angular speed of rotation $\alpha$ of the plane of oscillation of $P$ is given by:
:$\alpha = \omega \sin \lambda$
where:
:$\lambda$ denotes the latitude on Earth at which $P$ is located
:$\omega$ denotes the angular speed of rotation of Earth.
We have:
:$D = \dfrac {2 \pi} \omega$
Hence:
== Sources ==
* 
* 
Category:Foucault's Pendulum
Period of Rotation of Plane of Oscillation of Foucault's Pendulum/Mistake

== Source Work ==
:
:Foucault's pendulum
== Mistake ==
:To an observer on earth the plane of oscillation [of Foucault's pendulum] makes one rotation every $24$ hours (approximately).
== Correction ==
An oversimplification.
At Earth's poles, the period of rotation of the plane of oscillation of Foucault's pendulum is $1$ sidereal day, which is $23$ hours, $56$ minutes and $4.0916$ seconds.
Elsewhere on Earth, the period $T$ of rotation of the plane of oscillation depends on the latitude at which the pendulum is located:
:$T = \dfrac D {\sin \lambda}$
where:
:$\lambda$ is that latitude
:$D$ denotes the length of a sidereal day.
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Periodic Element is Multiple of Antiperiod

== Theorem ==
Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.
Let $L$ be a periodic element of $f$.
Then $A \divides L$.
== Proof ==
Consider $A + L$:
Hence $A + L$ is an anti-periodic element of $f$.
Combining Antiperiodic Element is Multiple of Antiperiod, Divides is Reflexive, and Common Divisor Divides Difference yields: 
:$A \divides \paren {A + L} \land A \divides A \implies A \divides L$
Hence the result.
Category:Antiperiodic Functions
Periodic Element is Multiple of Period

== Theorem ==
Let $f: \R \to \R$ be a real periodic function with period $P$.
Let $L$ be a periodic element of $f$.
Then $P \divides L$.
== Proof ==
 that $P \nmid L$.
Then by the Division Theorem we have $L = q P + r$ where $q \in \Z$ and $0 < r < P$.
And so:
But then $r$ is a periodic element of $f$ that is less than $P$.
Therefore $P$ cannot be the period of $f$.
The result follows from Proof by Contradiction.
Category:Periodic Functions
Periodic Function as Mapping from Unit Circle

== Theorem ==
Let $\SS$ denote the unit circle whose center is at the origin of the Cartesian plane $\R^2$.
Let $p: \R \to \SS$ be the mapping defined as:
:$\forall x \in \R: \map p x = \tuple {\cos x, \sin x}$
Let $f: \R \to \R$ be a periodic real function whose period is $2 \pi$.
Then there exists a well-defined real-valued function $f': \SS \to \R$ such that:
:$f = f' \circ p$
where $f' \circ p$ denotes the composition of $f'$ with $p$.
== Proof ==
Let $f': \SS \to \R$ be defined as:
:$\forall \tuple {x, y} \in \SS: \map {f'} {x, y} = \map f x$
Consider the inverse $p^{-1}: \SS \to \R$ of $p$:
:$\forall \tuple {x', y'} \in \SS: p^{-1} \sqbrk {x', y'} = \set {x \in \R: \cos x = x', \sin x = y'}$
Let $\RR$ be the equivalence relation on $\R$ induced by $p$:
:$\forall \tuple {x, y} \in \R \times \R: \tuple {x, y} \in \RR \iff \map p x = \map p y$
That is:
Let $f'$ be defined as:
:$f' = f \circ p^{-1}$
Then by the Quotient Theorem for Sets:
:$f'$ is well-defined $f$ is periodic with period $2 \pi$.
It follows from Conditions for Commutative Diagram on Quotient Mappings between Mappings that $f$ and $f'$ are related by the commutative diagram:
::$\begin{xy} \xymatrix@L+2mu@+1em{
 \R \ar[r]^*{p}
 \ar@{-->}[rd]_*{f := f' \circ p}
&
 \SS \ar[d]^*{f'}
\\ &
 \R
}\end{xy}$
== Motivation ==
== Also see ==
* Mapping from Unit Circle defines Periodic Function
* Periodic Function is Continuous iff Mapping from Unit Circle is Continuous
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces
Category:Periodic Functions
Periodic Function is Continuous iff Mapping from Unit Circle is Continuous

== Theorem ==
Let $f: \R \to \R$ be a periodic real function whose period is $2 \pi$.
Let $\SS$ denote the unit circle whose center is at the origin of the Cartesian plane $\R^2$.
Let $p: \R \to \SS$ be the mapping defined as:
:$\forall x \in \R: \map p x = \tuple {\cos x, \sin x}$
Let $f': \SS \to \R$ be the well-defined real-valued function such that:
:$f = f' \circ p$
where $f' \circ p$ denotes the composition of $f'$ with $p$.
Then $f$ is continuous $f'$ is continuous.
== Proof ==
The existence and well-definedness of $f'$ are demonstrated in Periodic Function as Mapping from Unit Circle.
=== Necessary Condition ===
Let $f'$ be continuous.
We have that Real Sine Function is Continuous and Cosine Function is Continuous.
Hence from Continuity of Composite with Inclusion: Inclusion on Mapping and Continuous Mapping to Product Space it follows that $p$ is continuous.
Hence $f = f \circ p$ is continuous.
=== Sufficient Condition ===
:
Let $f$ be continuous.
Let $U \subseteq \SS$.
Let $p^{-1} \sqbrk U$ be open in $\R$.
Let $b \in U$.
Then:
:$\exists a \in p^{-1} \sqbrk U: b = \tuple {\cos a, \sin a}$
Because $p^{-1} \sqbrk U$ be open:
:$\exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq p^{-1} \sqbrk U$
Let us assume that $\delta < \pi$.
Thus:
:$b \in p \sqbrk {\map {B_\delta} a} \subseteq U$
But we have that:
:$p \sqbrk {\map {B_\delta} a} = S \cap \map {B_\delta} b$
as can be seen from the diagram on the right
It follows by calculation that:
:$\epsilon = 2 \sin \dfrac \delta 2$
Thus $p \sqbrk {\map {B_\delta} a}$ is an open ball in $\SS$ which has a center at $b$.
Hence it follows that $U$ is open in $\SS$.
We have asserted that $f$ is continuous.
Let $W$ be open in $\RR$.
Then $p^{-1} \sqbrk {f'^{-1} \sqbrk W} = f^{-1} \sqbrk W$ is open in $\RR$.
Hence $f'^{-1} \sqbrk W$ is open in $\SS$.
Thus it follows that $f'$ is continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces
Category:Periodic Functions
Periodic Function plus Constant

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $k \in \R$ be constant.
Then $f$ is periodic with period $L$ $f + k$ is periodic with period $L$.
== Proof ==
=== Sufficient Condition ===
Let $f$ be periodic with period $L$.
Then:
Thus $f + k$ has been shown to be periodic with period $L$.
=== Necessary Condition ===
Let $f + k$ be periodic with period $L$.
Then:
Thus $f$ has been shown to be periodic with period $L$.
Category:Periodic Functions
Periodic Point/Examples/Complex Cube Function

== Examples of Periodic Points ==
Let $T: \C \to \C$ be the complex function defined such that:
:$\forall z \in \C: \map T z = z^3$
Consider the element $i \in \C$.
We have that:
where $i$ is the imaginary unit:
:$i^2 = -1$
Hence $i$ is a periodic point of period $2$ whose periodic orbit is $\set {i, -i}$.
== Sources ==
* 
* 
Category:Examples of Periodic Points
Periodic Real Function/Examples/Sawtooth Function

== Example of Periodic Real Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x - \floor x$
where $\floor x$ denotes the floor function.
:
$f$ is periodic with period $1$.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(i)}$ Periodic Functions
Category:Examples of Periodic Real Functions
Periodicity of Complex Exponential Function

== Theorem ==
For all $k \in \Z$:
:$e^{i \paren {\theta + 2 k \pi} } = e^{i \theta}$
== Proof ==
== Sources ==
* : $\S 7$: Periodicity of Exponential Functions: $7.23$
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $25$
Category:Exponential Function
Periodicity of Hyperbolic Cosecant

== Theorem ==
Let $k \in \Z$.
Then:
:$\map \csch {x + 2 k \pi i} = \csch x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.89$: Periodicity of Hyperbolic Functions
Category:Hyperbolic Cosecant Function
Periodicity of Hyperbolic Cosine

== Theorem ==
Let $k \in \Z$.
Then:
:$\map \cosh {x + 2 k \pi i} = \cosh x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.87$: Periodicity of Hyperbolic Functions
Category:Hyperbolic Cosine Function
Periodicity of Hyperbolic Cotangent

== Theorem ==
Let $k \in \Z$.
Then:
:$\map \coth {x + 2 k \pi i} = \coth x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.91$: Periodicity of Hyperbolic Functions
Category:Hyperbolic Tangent Function
Periodicity of Hyperbolic Secant

== Theorem ==
Let $k \in \Z$.
Then:
:$\map \sech {x + 2 k \pi i} = \sech x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.90$: Periodicity of Hyperbolic Functions
Category:Hyperbolic Secant Function
Periodicity of Hyperbolic Tangent

== Theorem ==
Let $k \in \Z$.
Then:
:$\map \tanh {x + 2 k \pi i} = \tanh x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.88$: Periodicity of Hyperbolic Functions
Category:Hyperbolic Tangent Function
Permanent/Examples/3 x 3 Matrix

== Example of Permanent ==
The square matrix:
:$\begin {pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end {pmatrix}$
has a permanent of $450$.
== Proof ==
Let $P$ denote the permanent of $\begin {pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end {pmatrix}$.
We have:
== Sources ==
* 
Category:Examples of Permanents
Permanent/Examples/Matrix whose Entries are Product of Row and Column Indices

== Example of Permanent ==
The square matrix of the form:
:$\begin{pmatrix}
1 \times 1 & 1 \times 2 & \cdots & 1 \times m \\
2 \times 1 & 2 \times 2 & \cdots & 2 \times m \\
\vdots & \vdots & \ddots & \vdots \\
m \times 1 & m \times 2 & \cdots & m \times m
\end{pmatrix}$
has a permanent of $\paren {n!}^3$.
== Proof ==
There are $n!$ terms in a permanent.
By its structure, each one of these has one element from each row multitplied by one element from each column.
Thus each term of the permanent consists of:
:$\paren {1 \times 2 \times \cdots \times n} \times \paren {1 \times 2 \times \cdots \times n}$
in some order, that is:
:$\paren {n!}^2$
As has been stated, there are $n!$ of these.
Thus the permanent of this matrix is $n! \paren {n!}^2$.
Hence the result.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $15$
Category:Examples of Permanents
Permutable Prime with more than 3 Digits is Probably Repunit

== Conjecture ==
Let $p$ be a permutable prime with more than $3$ digits.
Then $p$ is very probably repunit.
Hence the next permutable prime after $991$ is the repunit prime $R_{19}$:
:$1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$
== Historical Note ==
== Sources ==
* 
* 
* 
* 
* : $113$
Category:Permutable Primes
Category:Repunit Primes
Category:Unsolved Problems
Permutable Prime with more than 3 Digits is Probably Repunit/Historical Note

== Historical Note on Permutable Prime with more than 3 Digits is Probably Repunit ==
It is trivially true, as shown in Digits of Permutable Prime, that a permutable prime with more than $1$ digit can contain only the digits from the set $\set {1, 3, 7, 9}$.
 and demonstrated in a $1974$ paper that a permutable prime cannot contain all of $1$, $3$, $7$ and $9$.
Subsequently, showed that a non-repunit permutable prime is of the form $\sqbrk {aaa \cdots aab}$ and have more than $9$ billion digits.
In $1995$, determined further restrictions on the structure of permutable primes.
However, it still has not been proven that a permutable prime with more than $3$ digits is always a repunit prime.
== Sources ==
* 
* 
* 
* 
Category:Historical Notes
Permutation/Examples/Addition of Constant on Integers

== Examples of Permutations ==
Let $\Z$ denote the set of integers.
Let $a \in \Z$.
Let $f: \Z \to \Z$ denote the mapping defined as:
:$\forall x \in \Z: \map f x = x + a$
Then $f$ is a permutation on $\Z$.
== Proof ==
demonstrating that $f$ is injective.
Then:
As $y - a \in \Z$ it follows that $f$ is a surjection.
So $f$ is both an injection and a surjection.
By definition, then, $f$ is a bijection.
As the domain and codomain of $f$ is the same, $f$ is by definition a permutation.
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations: Example $55$
Category:Examples of Permutations
Permutation/Examples/Arbitrary Example 1

== Example of Permutation ==
Let $A = \set {a_1, a_2, a_3, a_4}$.
Let $f: A \to A$ be the mapping defined as:
:$f = \set {\tuple {a_1, a_3}, \tuple {a_2, a_4}, \tuple {a_3, a_1}, \tuple {a_4, a_2} }$
Then $f$ is a permutation on $A$.
== Sources ==
* : Appendix $\text{A}.4$: Functions
Category:Examples of Permutations
Permutation/Examples/Arbitrary Example 2

== Example of Permutation ==
Let $A = \set {a_1, a_2, a_3}$.
Using two-row notation, a permutation on $A$ can be denoted:
:$f = \begin {pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end {pmatrix}$
This can also be denoted:
:$f = \set {\tuple {a_1, a_3}, \tuple {a_2, a_1}, \tuple {a_3, a_2} }$
== Sources ==
* 
* 
Category:Examples of Permutations
Permutation/Ordered Selection/Examples/2 from 3

== Example of Permutations ==
There are $6$ permutations of $3$ objects taken $2$ at a time:
:$a \, b \quad a \, c \quad b \, a \quad b \, c \quad c \, a \quad c \, b$
== Sources ==
* : $\text {3-1}$ Permutations and Combinations: Example $\text {3-4}$
Category:Examples of Permutations
Permutation/Ordered Selection/Examples/2 from 4

== Example of Permutations ==
There are $12$ permutations of $4$ objects taken $2$ at a time:
:$a \, b \quad a \, c \quad a \, d \quad b \, c \quad b \, d \quad c \, d$
and then the same reversed:
:$b \, a \quad c \, a \quad d \, a \quad c \, b \quad d \, b \quad d \, c$
== Sources ==
* 
* 
Category:Examples of Permutations
Permutation/Ordered Selection/Examples/3 from 3

== Example of Permutations ==
There are $6$ permutations of $3$ objects taken $3$ at a time:
:$a \, b \, c \quad a \, c \, b \quad b \, a \, c \quad b \, c \, a \quad c \, a \, b \quad c \, b \, a$
In two-row notation on the permutations on $3$ letters:
:$\dbinom {1 \ 2 \ 3} { 1 \ 2 \ 3}, \dbinom {1 \ 2 \ 3} { 1 \ 3 \ 2}, \dbinom {1 \ 2 \ 3} { 2 \ 1 \ 3}, \dbinom {1 \ 2 \ 3} { 2 \ 3 \ 1}, \dbinom {1 \ 2 \ 3} { 3 \ 1 \ 2}, \dbinom {1 \ 2 \ 3} { 3 \ 2 \ 1}$
== Sources ==
* : $\text {3-1}$ Permutations and Combinations: Example $\text {3-4}$
* : $\S 1.2.5$: Permutations and Factorials: $(1)$
Category:Examples of Permutations
Permutation/Ordered Selection/Examples/4 from 4

== Example of Permutations ==
There are $24$ permutations of $4$ objects taken $4$ at a time:
In two-row notation on the permutations on $4$ letters:
:$\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_1 & a_2 & a_3 & a_4 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_1 & a_2 & a_4 & a_3 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_1 & a_3 & a_2 & a_4 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_1 & a_3 & a_4 & a_2 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_1 & a_4 & a_2 & a_4 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_1 & a_4 & a_3 & a_2 \end{pmatrix},$
:$\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_2 & a_1 & a_3 & a_4 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_2 & a_1 & a_4 & a_3 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_2 & a_3 & a_1 & a_4 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_2 & a_3 & a_4 & a_1 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_2 & a_4 & a_1 & a_3 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_2 & a_4 & a_3 & a_1 \end{pmatrix},$
:$\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_3 & a_1 & a_2 & a_4 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_3 & a_1 & a_4 & a_2 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_3 & a_2 & a_1 & a_4 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_3 & a_2 & a_4 & a_1 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_3 & a_4 & a_1 & a_2 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_3 & a_4 & a_2 & a_1 \end{pmatrix},$
:$\begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_4 & a_1 & a_2 & a_3 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_4 & a_1 & a_3 & a_2 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_4 & a_2 & a_1 & a_3 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_4 & a_2 & a_3 & a_1 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_4 & a_3 & a_1 & a_2 \end{pmatrix}, \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ a_4 & a_3 & a_2 & a_1 \end{pmatrix}$
== Sources ==
* : Chapter $\text I$ Permutations and Combinations: The $r$, $s$ Principle: Example $2$.
* : Appendix $\text{A}.4$: Functions: Problem Set $\text{A}.4$: $24$
Category:Examples of Permutations
Permutation Group/Examples/Example on Real Numbers

== Example of Permutation Group ==
Let $\R$ denote the set of real numbers.
Let $\R_{\ge 0}$ denote the set of positive real numbers.
Let $S = \R_{\ge 0} \times \R$ denote the Cartesian product of $\R_{\ge 0}$ and $\R$.
Let $\tuple {a, b} \in S$.
Let $f_{a, b}: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map {f_{a, b} } x := a x + b$
Let $\GG$ be the set defined as:
:$\GG = \set {f_{a, b}: \tuple {a, b} \in S}$
Let $\struct {S, \oplus}$ be the group where $\oplus$ is defined as:
:$\forall \tuple {a, b}, \tuple {c, d} \in S: \tuple {a, b} \oplus \tuple {c, d} := \tuple {a c, a d + b}$
Then $\struct {\GG, \circ}$ is a permutation group on $\R$ which is isomorphic to $\struct {S, \oplus}$.
== Proof ==
From Group Example: $\tuple {a c, a d + b}$ on Positive Reals by Reals, $\struct {S, \oplus}$ has been shown to be a group.
From Bijection Example: $a x + b$ on Real Numbers, we have that $f_{a, b}$ is a bijection from $\R$ to $\R$.
Hence by definition $f_{a, b}$ is a permutation on $\R$.
From Group Example: Linear Functions, $\struct {\GG, \circ}$ is a group.
Hence as $\struct {\GG, \circ}$ is a group of permutations on $\R$, it is by definition a permutation group on $\R$.
Let $\phi: \struct {\GG, \circ} \to \struct {S, \oplus}$ be the identity mapping on $S$:
:$\forall \tuple {a, b} \in S: \map \phi {a, b} = \tuple {a, b}$
From Identity Mapping is Bijection we have that $\phi$ is a bijection.
Consider $f_{a, b}, f_{c, d} \in \GG$.
We have:
This demonstrates that $\phi$ is a (group) homomorphism.
So $\phi$ is a bijection which is a homomorphism.
Hence by definition $\phi$ is an isomorphism.
The result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.7 \ \text {(b)}$
Category:Examples of Permutation Groups
Permutation Group is Subgroup of Symmetric Group

== Theorem ==
Let $S$ be a set.
Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$, where $\circ$ denotes the composition operation.
Let $\struct {H, \circ}$ be a set of permutations of $S$ which forms a group under $\circ$.
Then $\struct {H, \circ}$ is a subgroup of $\struct {\map \Gamma S, \circ}$.
== Proof ==
Follows directly from the definition of subgroup:
$H$ is a subset of $\map \Gamma S$, and $\struct {H, \circ}$ is a group.
Hence the result.
== Sources ==
* : $\S 36$: Subgroups: Simple illustrations: $(3)$
Category:Symmetric Groups
Category:Subgroups
Permutation Induces Equivalence Relation

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$.
Let $\RR_\pi$ be the relation defined by:
:$i \mathrel {\RR_\pi} j \iff \exists k \in \Z: \map {\pi^k} i = j$
Then $\RR_\pi$ is an equivalence relation.
=== Corollary ===
== Proof ==
Let $\pi \in S_n$.
From Element of Finite Group is of Finite Order, every element of a finite group has finite order.
Thus $\pi$ has finite order.
So:
:$\exists r \in \Z: \pi^r = e$
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
From above:
:$\exists r \in \Z: \pi^r = e$
Therefore:
:$\exists k \in \Z: \map {\pi^k} i = i$
So $\RR_\pi$ is reflexive.
=== Symmetric ===
Let $\map {\pi^k} i = j$.
Because $\pi^r = e$, we have:
:$\map {\pi^r} j = j$ 
(from above).
Thus:
:$\map {\pi^{r - k} } j = i$
So $\RR_\pi$ is symmetric.
=== Transitive ===
Let $\map {\pi^{s_1} } i = j$ and $\map {\pi^{s_2} } j = k$.
Then:
:$\map {\pi^{s_1 + s_2} } i = k$
So $\RR_\pi$ is transitive.
All criteria are met, and so $\RR_\pi$ is an equivalence relation.
Category:Symmetric Groups
Category:Examples of Equivalence Relations
Permutation Induces Equivalence Relation/Corollary

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$.
Let $\RR_\pi$ be the relation defined as:
:$i \mathrel {\RR_\pi} j \iff \exists k \in \Z: \map {\pi^k} i = j$
Then:
:$i \mathrel {\RR_\pi} j$ $i$ and $j$ are in the same cycle of $\pi$.
== Proof ==
We have that Permutation Induces Equivalence Relation.
The equivalence classes of that equivalence relation are the cycles of $\pi$.
Hence the result by definition of equivalence class.
Category:Symmetric Groups
Category:Examples of Equivalence Relations
Permutation Matrix/Examples/3 x 3

== Example of Permutation Matrix ==
An example of a $3 \times 3$ permutation matrix:
:$\mathbf A = \begin {pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 
\end {pmatrix}$
This corresponds to the permutation $\begin {pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end {pmatrix}$ (expressed in two-row notation).
== Proof ==
We have:
:$\mathbf A = \begin {pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end {pmatrix} \begin {pmatrix} 1 \\ 2 \\ 3 \end {pmatrix} = \begin {pmatrix} 2 \\ 3 \\ 1 \end {pmatrix}$ 
== Sources ==
* 
* 
Category:Examples of Permutation Matrices
Permutation Matrix/Examples/Full Rook Matrix

== Example of Permutation Matrix ==
An $8 \times 8$ permutation matrix is known as a full rook matrix.
For example:
:$\mathbf A = \begin{bmatrix}
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$
That is, it is a rook matrix in which each row and column has a $1$ in it.
== Also see ==
* Definition:Rook Matrix
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.4$: Multiplication and inverses: Ponderable $2.2.2$
Category:Examples of Permutation Matrices
Permutation Representation defines Group Action

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $X$ be a set.
Let $\map \Gamma X$ be the symmetric group of $X$.
Let $\rho: G \to \map \Gamma X$ be a permutation representation, that is, a homomorphism.
The mapping $\phi: G \times X \to X$ associated to $\rho$, defined by:
:$\map \phi {g, x} = \map {\map \rho g} x$
is a group action.
== Proof ==
Let $g, h \in G$ and $x \in X$.
We verify that $g * \paren {h * x} = \paren {g h} * x$:
We verify that $e * x = x$.
Let $I_X$ denote the identity mapping on $X$.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions
Category:Group Actions