nyuuzyou/wikis huggingface dataset:
Order Topology on Natural Numbers is Discrete Topology

== Theorem ==
Let $\le$ be the standard ordering on the natural numbers $\N$.
Then the order topology $\tau$ on $\N$ is the discrete topology.
== Proof ==
By Topology is Discrete iff All Singletons are Open, it suffices to show that for all $n \in \N$, the singleton $\set n$ is an open of $\tau$.
Now observe that $\map {\mathop{\downarrow} } 1 = \set 0$, since for all $n \in \N$, $n < 1 \implies n = 0$.
It follows that $\set 0$ is an open set of $\tau$.
Suppose now that $n \in \N$ and $n \ne 0$.
Then it is known that for all $m \in \N$, $n - 1 < m < n + 1$ implies $m = n$.
Thus, $\map {\mathop{\uparrow} } {n - 1} \cap \map {\mathop{\downarrow} } {n + 1} = \set n$.
It follows that $\set n$ is an open set of $\tau$.
Hence the result, from Proof by Cases.
Category:Natural Numbers
Category:Order Topology
Category:Discrete Topology
Order Type Addition is Associative

== Theorem ==
Let $\alpha$, $\beta$ and $\gamma$ be order types of ordered sets.
Then:
:$\paren {\alpha + \beta} + \gamma = \alpha + \paren {\beta + \gamma}$
where $+$ denotes order type addition.
== Proof ==
Let $\struct {S_1, \preccurlyeq_1}$, $\struct {S_2, \preccurlyeq_2}$ and $\struct {S_3, \preccurlyeq_3}$ be ordered structures such that:
where $\ot$ denotes order type.
Thus by definition of order type we are required to show that:
:$\paren {\struct {S_1, \preccurlyeq_1} \oplus \struct {S_2, \preccurlyeq_2} } \oplus \struct {S_3, \preccurlyeq_3} \cong \struct {S_1, \preccurlyeq_1} \oplus \paren {\struct {S_2, \preccurlyeq_2} \oplus \struct {S_3, \preccurlyeq_3} }$
where:
:$\oplus$ denotes order sum
:$\cong$ denotes order isomorphism.
Let:
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $34 \ \text {(a)}$
Category:Order Sums
Category:Examples of Associative Operations‎
Order Type Addition is Well-Defined Operation

== Theorem ==
The addition operation on order types is well-defined.
== Proof ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {T_1, \preccurlyeq_{1'} }$ and $\struct {T_2, \preccurlyeq_{2'} }$ be ordered sets such that:
:$\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {T_1, \preccurlyeq_{1'} }$
:$\struct {S_2, \preccurlyeq_2}$ is isomorphic to $\struct {T_2, \preccurlyeq_{2'} }$
Let $\alpha := \map \ot {S_1, \preccurlyeq_1}$ and $\beta := \map \ot {S_2, \preccurlyeq_2}$ denote the order types of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ respectively.
Let $\alpha' := \map \ot {T_1, \preccurlyeq_{1'} }$ and $\beta' := \map \ot {T_2, \preccurlyeq_{2'} }$ denote the order types of $\struct {T_1, \preccurlyeq_{1'} }$ and $\struct {T_2, \preccurlyeq_{2'} }$ respectively.
It is required to show that:
:$\alpha + \beta$
is the same as:
:$\alpha' + \beta'$
We have that:
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Order Types
Category:Addition
Category:Order Sums
Order Type Addition is not Commutative

== Theorem ==
The operation of order type addition is not commutative.
== Proof ==
Consider the order type $\omega^* + \omega$, where:
:$\omega$ denotes the order type of the natural numbers $\N$
:$\omega^*$ denotes the dual of $\omega$
:$+$ denotes addition of order types.
From Order Type of Integers under Usual Ordering, this is the order type of $\struct {\Z, \le}$, the set of integers under the usual ordering.
Now consider order type $\omega + \omega^*$.
From Ordered Set with Order Type of Natural Numbers plus Dual has Minimum Element, this is the order type of an ordered structure $\struct {S, \preccurlyeq}$ which has a smallest element.
But $\struct {\Z, \le}$ has no such smallest element.
It follows that $\struct {\Z, \le}$ is not isomorphic to $\struct {S, \preccurlyeq}$.
Hence $\omega + \omega^*$ and $\omega^* + \omega$ are different order types.
The result follows.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $28 \ \text {(c)}$
Category:Order Types
Category:Examples of Commutative Operations
Order Type Multiplication Distributes over Addition

== Theorem ==
Let $\alpha$, $\beta$ and $\gamma$ be order types of ordered sets.
Then:
:$\alpha \cdot \paren {\beta + \gamma} = \paren {\alpha \cdot \beta} + \paren {\alpha \cdot \gamma}$
where:
:$+$ denotes order type addition
:$\cdot$ denotes order type multiplication.
== Proof ==
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $34 \ \text {(c)}$
Category:Order Products
Category:Order Sums
Category:Examples of Distributive Operations
Order Type Multiplication is Associative

== Theorem ==
Let $\alpha$, $\beta$ and $\gamma$ be order types of ordered sets.
Then:
:$\paren {\alpha \cdot \beta} \cdot \gamma = \alpha \cdot \paren {\beta \cdot \gamma}$
where $\cdot$ denotes order type multiplication.
== Proof ==
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $34 \ \text {(b)}$
Category:Order Products
Category:Examples of Associative Operations‎
Order Type Multiplication is Well-Defined Operation

== Theorem ==
The multiplication operation on order types is well-defined.
== Proof ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {T_1, \preccurlyeq_{1'} }$ and $\struct {T_2, \preccurlyeq_{2'} }$ be ordered sets such that:
:$\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {T_1, \preccurlyeq_{1'} }$
:$\struct {S_2, \preccurlyeq_2}$ is isomorphic to $\struct {T_2, \preccurlyeq_{2'} }$
Let $\alpha := \map \ot {S_1, \preccurlyeq_1}$ and $\beta := \map \ot {S_2, \preccurlyeq_2}$ denote the order types of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ respectively.
Let $\alpha' := \map \ot {T_1, \preccurlyeq_{1'} }$ and $\beta' := \map \ot {T_2, \preccurlyeq_{2'} }$ denote the order types of $\struct {T_1, \preccurlyeq_{1'} }$ and $\struct {T_2, \preccurlyeq_{2'} }$ respectively.
It is required to show that:
:$\alpha \cdot \beta$
is the same as:
:$\alpha' \cdot \beta'$
We have that:
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $33$
Category:Order Types
Category:Multiplication
Category:Order Products
Order Type of Integers under Usual Ordering

== Theorem ==
Consider the ordered structure $\struct {\Z, \le}$ that is the set of integers under the usual ordering.
Then:
:$\map \ot {\Z, \le} = \omega^* + \omega$
where:
:$\ot$ denotes order type
:$\omega$ denotes the order type of the natural numbers $\N$
:$\omega^*$ denotes the dual of $\omega$
:$+$ denotes addition of order types.
== Proof ==
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $28 \ \text {(a)}$
Category:Examples of Order Types
Order Types of Duals of Isomorphic Sets are Equal

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let:
:$\map \ot {S_1, \preccurlyeq_1} = \map \ot {S_2, \preccurlyeq_2}$
where $\ot$ denotes the order type operator.
Let $\struct {S_1, \succcurlyeq_1}$ and $\struct {S_2, \succcurlyeq_2}$ denote the dual ordered sets of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Then:
:$\map \ot {S_1, \succcurlyeq_1} = \map \ot {S_2, \succcurlyeq_2}$
== Proof ==
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $27 \ \text {(d)}$
Category:Order Types
Category:Dual Orderings
Order and Size of Graph do not determine Degrees of Vertices

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph.
Let both the order $\card V$ and size $\card E$ of $G$ be given.
Then it is not always possible to determine the degrees of each of the vertices of $G$.
== Proof ==
The following $2$ simple graphs both have order $4$ and size $4$:
:
But:
:the graph on the left has vertices with degrees $1, 2, 2, 3$
:the graph on the right has vertices with degrees $2, 2, 2, 2$.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $6$
Category:Degrees of Vertices
Order is Maximum of Exponent of Convergence and Degree

== Theorem ==
Let $f: \C \to \C$ be an entire function.
Let $\omega$ be its order.
Let $\tau$ be its exponent of convergence.
Let $h$ be the degree of the polynomial in its canonical factorization.
Then:
:$\omega = \map \max {\tau, h}$
== Proof ==
By Exponent of Convergence is Less Than Order:
:$\tau \le \omega$
By Hadamard Factorization Theorem:
:$h \le \omega$
Thus:
:$\map \max {\tau, h} \le \omega$
By Order is Less Than Maximum of Exponent of Convergence and Degree:
:$\omega \le \map \max {\tau, h}$
== Sources ==
* : Chapter $\text {III}$: Further Theory of $\map \zeta s$. Applications: $\S 7$: Integral Functions: Theorem $\text F$
Category:Entire Functions
Order is Preserved on Positive Reals by Squaring

== Theorem ==
Let $x, y \in \R: x > 0, y >0$.
Then:
:$x < y \iff x^2 < y^2$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Real Analysis
Category:Inequalities
Category:Order is Preserved on Positive Reals by Squaring
Order is Preserved on Positive Reals by Squaring/Proof 1

== Theorem ==
== Proof ==
=== Necessary Condition ===
Assume $x < y$.
Then:
So:
:$x < y \implies x^2 < y^2$
=== Sufficient Condition ===
Assume $x^2 < y^2$.
Then:
So:
:$x^2 < y^2 \implies x < y$
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.6$: Example
Category:Order is Preserved on Positive Reals by Squaring
Order is Preserved on Positive Reals by Squaring/Proof 3

== Theorem ==
== Proof ==
From Real Numbers form Totally Ordered Field, the real numbers form a totally ordered field.
By definition, a totally ordered field is a totally ordered ring without proper zero divisors.
The result follows from Order of Squares in Totally Ordered Ring without Proper Zero Divisors.
Category:Order is Preserved on Positive Reals by Squaring
Order is Preserved on Positive Reals by Squaring/Proof 4

== Theorem ==
== Proof ==
=== Necessary Condition ===
Let $x < y$.
Then:
So:
:$x < y \implies x^2 < y^2$
=== Sufficient Condition ===
Let $x^2 < y^2$.
 $x \ge y$.
Then:
But this contradicts our assertion that $x^2 < y^2$.
Hence by Proof by Contradiction it follows that:
:$x < y$
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.6$: Example
Category:Order is Preserved on Positive Reals by Squaring
Order of 5 in Units of Ring of Integers Modulo 2^n

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer and $n \ge 2$.
Let $\struct {\Z / 2^n \Z, +, \times}$ be the ring of integers modulo $2^n$.
Let $U = \struct {\paren {\Z / 2^n \Z}^\times, \times}$ denote the group of units of $\struct {\Z / 2^n \Z, +, \times}$.
The order of $5$ in $U$ is $2^{n - 2}$.
== Proof 1 ==
Proof by induction:
=== Induction Hypothesis ===
For all $k \ge 3$, we can write
where $a_k$ is odd.
=== Basis for the Induction ===
For $n = 3$, it means $5 ^2 = 1 + 3 \cdot 8$
=== Induction Step ===
This is our induction step:
Suppose for some $k \ge 3$, (1) is satisfied.
Consider $n = k + 1$.
This proves the inductive step.
Hence the result by induction.
Now we prove the order of $5$ in $U$ is $2^{n - 2}$.
By (1)
:$5^{2^{n - 2}} \equiv 1 \pmod {2^n}$
so the order of $5$ in $U$ divides $2^{n - 2}$.
 it is less than $2^{n - 2}$, then it divides $2^{n - 3}$, so:
:$5^{2^{n - 3}} = 1 + m \cdot 2^n$ 
for some integer $m$. Then:
:$5^{2^{n - 2}} = \paren{1 + m \cdot 2^n}^2 = 1 + \paren{2 m + m^2 2^n} 2^n$
where the coefficient $2 m + m^2 2^n$ is even, contradicting lemma.
== Proof 2 ==
=== Lemma ===
For all $k \ge 3$,
:$5^{2^{k - 3}} \equiv 1 + 2^{k - 1} \pmod{ 2^k}$
=== Proof of lemma === 
Write $\map {\nu_2} n$ for the highest exponent of a power of $2$ dividing $n$.
For all $i \ge 2$,
:$i \le 2^{i - 1}$
so
:$\map {\nu_2} i \le i - 1$
so
From definition of binomial coefficient:
:$\ds \binom {2^{k - 3} } i = \frac {2^{k - 3} \binom {2^{k - 3} - 1} {i - 1} } i$
we have
so
By the lemma,
:$5^{2^{n - 3}} \not \equiv 1 \pmod{ 2^n}$
so the order of $5$ in $U$ does not divide $2^{n - 3}$.
Squaring both sides of the lemma:
:$5^{2^{n - 2}} \equiv \paren{1 + 2^{n - 1}}^2 \equiv 1 \pmod{ 2^n}$
so the order of $5$ in $U$ divides $2^{n - 2}$.
Hence the order of $5$ in $U$ is $2^{n - 2}$.
== Generalization ==
Let $U$ be the same as above.
Let $u$ be any odd number of the form $8k + 3$ or $8k + 5$.
The order of $u$ in $U$ is $2^{n - 2}$.
== Sources ==
* , arts. $90$
* 
Category:Ring of Integers Modulo m
Category:Proofs by Induction
Order of Additive Group of Integers Modulo m

== Theorem ==
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
The order of $\struct {\Z_m, +_m}$ is $m$.
== Proof ==
By definition, the order of a group is the cardinality of its underlying set.
By definition, the underlying set of $\struct {\Z_m, +_m}$ is the set of residue classes $\Z_m$:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
From Cardinality of Set of Residue Classes, $\Z_m$ has $m$ elements.
Hence the result.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 38$
Category:Additive Groups of Integers Modulo m
Order of Alternating Group

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 1$.
Let $A_n$ be the alternating group on $n$ letters.
Then:
:$\order {A_n} = \dfrac {n!} 2$
where $\order {A_n}$ denotes the order of $A_n$.
== Proof ==
Let $S_n$ denote the symmetric group on $n$ letters.
From Alternating Group is Normal Subgroup of Symmetric Group:
:$\index {S_n} {A_n} = 2$
where $\index {S_n} {A_n}$ denotes the index of $A_n$ in $S_n$.
From Order of Symmetric Group:
:$\order {S_n} = n!$
The result follows from Lagrange's Theorem.
== Sources ==
* : $\S 7.4$. Kernel and image: Example $142$
* : Chapter $2$: The Symmetric Groups: $\S 81$
* 
* 
Category:Alternating Groups
Order of Automorphism Group

== Theorem ==
Let $G$ be a finite group whose order is greater than $2$.
Let $\Aut G$ be the automorphism group of $G$.
Then the order of $\Aut G$ is greater than $1$.
== Proof ==
There are $3$ cases, and for each case we find an automorphism that is not the identity automorphism.
=== Case $1$: $G$ is non-Abelian ===
There exists some elements $g, h \in G$ such that $g h \ne h g$.
Then the inner automorphism:
:$x \mapsto g x g^{-1}$
gives an automorphism that is not the identity automorphism, since:
:$g h g^{-1} \ne h$
Hence $\order {\Aut G} > 1$.
=== Case $2$: $G$ has an element with order greater than $2$ ===
Suppose $g \in G$ has order $k$, where $k > 2$.
Then $g \ne g^{-1}$.
Thus the inverse automorphism:
:$x \mapsto x^{-1}$
is not the identity automorphism.
Hence $\order {\Aut G} > 1$.
=== Case $3$: $G$ is Abelian and has no element with order greater than $2$ ===
By Fundamental Theorem of Finite Abelian Groups, write:
:$G = C_{a_1} \times \dots \times C_{a_k}$
where $C_{a_i}$ are cyclic groups of prime power order.
Since all elements of $G$ has order not more than $2$, $a_i = 2$ for all $i$, that is:
:$G = C_2^k$
As $\order G = 2^k > 2$:
:$k > 1$
Writing each element of $G$ as $\tuple {g_1, \dots, g_k}$ where $g_i = 0$ or $1$, the swapping automorphism:
:$\tuple {g_1, g_2, \dots, g_k} \mapsto \tuple {g_2, g_1, \dots, g_k}$
is an automorphism that is not the identity automorphism. 
Hence $\order {\Aut G} > 1$.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \gamma$
Category:Automorphism Groups
Order of Automorphism Group of Cyclic Group

== Theorem ==
Let $C_n$ denote the cyclic group of order $n$.
Let $\Aut {C_n}$ denote the automorphism group of $C_n$.
Then:
:$\order {\Aut {C_n} } = \map \phi n$
where:
:$\order {\, \cdot \,}$ denotes the order of a group
:$\map \phi n$ denotes the Euler $\phi$ function.
== Proof ==
Let $g$ be a generator of $C_n$.
Let $\varphi$ be an automorphism on $C_n$.
By Homomorphic Image of Cyclic Group is Cyclic Group, $\map \varphi g$ is a generator of $C_n$.
By Homomorphism of Generated Group, $\varphi$ is uniquely determined by $\map \varphi g$.
By Finite Cyclic Group has Euler Phi Generators, there are $\map \phi n$ possible values for $\map \varphi g$.
Therefore there are $\map \phi n$ automorphisms on $C_n$:
:$\order {\Aut {C_n} } = \map \phi n$
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \delta$
Category:Automorphism Groups
Category:Cyclic Groups
Order of Automorphism Group of Dihedral Group

== Theorem ==
Let $D_n$ denote the dihedral group of order $n$.
Let $\Aut {D_n}$ denote the automorphism group of $D_n$.
Then:
:$\order {\Aut {D_n} } = n \cdot \map \phi n$
where:
:$\order {\, \cdot \,}$ denotes the order of a group
:$\map \phi n$ is the Euler $\phi$ function.
== Proof ==
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \delta$
Category:Automorphism Groups
Category:Dihedral Groups
Order of Automorphism Group of Prime Group

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a group of order $p$.
Let $\Aut G$ denote the automorphism group of $G$.
Then:
:$\order {\Aut G} = p - 1$
where $\order {\, \cdot \,}$ denotes the order of a group.
== Proof ==
From Prime Group is Cyclic we have that $G$ is a cyclic group.
From Order of Automorphism Group of Cyclic Group:
:$\order {\Aut G} = \map \phi p$
where $\map \phi n$ denotes the Euler $\phi$ function.
The result follows from Euler Phi Function of Prime.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $27 \ \text {(i)}$
Category:Prime Groups
Category:Automorphism Groups
Order of Boolean Group is Power of 2

== Theorem ==
Let $G$ be a Boolean group.
Let $\order G$ denote the order of $G$.
Then:
:$\order G = 2^n$
where $n \in \Z_{\ge 0}$ is a positive integer.
== Proof ==
The case where $n = 0$ is clear:
:$\order {\set e} = 1$
and $e^2 = e$.
 $\order G = m \times 2^k$ for some odd integer $m$.
Then $m$ itself has an odd prime $p$ as a integer (which may of course equal $m$ if $m$ is itself prime).
Then by Cauchy's Lemma (Group Theory) there exists $g \in G$ such that $\order g = p$.
Hence it is not the case that $g^2 = e$.
Hence $\order G$ has no prime factor which is odd.
The result follows.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $17$
Category:Boolean Groups
Order of Conjugate Element equals Order of Element

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then
:$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$
where $\order a$ denotes the order of $a$ in $G$.
=== Corollary ===
== Proof ==
Let $\order a = k$.
Then $a^k = e$, and:
:$\forall n \in \N_{>0}: n < k \implies a^n \ne e$
by definition of the order of $a$ in $G$
We have:
Thus $\order {x \circ a \circ x^{-1} } \le \order a$.
Now suppose $a^n = y, y \ne e$.
Then:
:$x \circ a^n \circ x^{-1} = x \circ y \circ x^{-1}$
If $x \circ y = e$, then:
:$x \circ a^n \circ x^{-1} = x^{-1}$
If $y \circ x^{-1} = e$, then:
:$x \circ a^n \circ x^{-1} = x$
So:
:$a^n \ne e \implies x \circ a^n \circ x^{-1} = \paren {x \circ a \circ x^{-1} }^n \ne e$
Thus:
:$\order {x \circ a \circ x^{-1} } \ge \order a$
and the result follows.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: $\text{(iii)}$
* : Chapter $2$: Subgroups and Cosets: $\S 41 \gamma$
* : Chapter $6$: An Introduction to Groups: Exercise $12$
* : Chapter $3$: Elementary consequences of the definitions: Exercise $7$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.18$
Category:Order of Group Elements
Category:Conjugacy
Order of Conjugate Element equals Order of Element/Corollary

== Corollary to Order of Conjugate Element equals Order of Element ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then:
:$\forall a, x \in \struct {G, \circ}: \order {x \circ a} = \order {a \circ x}$
where $\order a$ denotes the order of $a$ in $G$.
== Proof ==
From Order of Conjugate Element equals Order of Element, putting $a \circ x$ for $a$:
:$\order {x \circ \paren {a \circ x} \circ x^{-1} } = \order {a \circ x}$
from which the result follows by $x \circ x^{-1} = e$.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: Chapter $1$: The Group Concept: Examples: $(5)$
* : Chapter $2$: Subgroups and Cosets: $\S 41 \delta$
* : $0$: Some Conventions and some Basic Facts: Exercise $4$
* : Chapter $6$: An Introduction to Groups: Exercise $12 \ \text {(i)}$
* : $\S 3.4$: Cyclic groups: Exercise $6$
Category:Order of Group Elements
Category:Conjugacy
Order of Conjugate of Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$ such that $H$ is of finite order.
Then $\order {H^a} = \order H$.
== Proof ==
From the definition of Conjugate of Group Subet we have $H^a = a H a^{-1}$.
From Set Equivalence of Regular Representations:
:$\order {a H a^{-1} } = \order {a H} = \order H$
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $5$
* : $\S 3.3$: Group actions and coset decompositions: Exercise $1$
Category:Conjugacy
Order of Convergence Implies Convergence

== Theorem ==
Let $\sequence {x_n}$ be an real sequence that converges to $\alpha$ with order $p$, where $p \ge 1$.
Then, $\sequence {x_n}$ converges to $\alpha$.
== Proof ==
By definition of order of convergence, there exist a sequence $\sequence {\epsilon_n}$ and $c > 0$ such that:
:$\size {x_n - \alpha} \le \epsilon_n$
:$\ds \lim_{n \to \infty} \frac {\epsilon_{n + 1}} {\epsilon_n^p} = c$
:$p = 1 \implies c < 1$
=== Case $p = 1$ ===
Suppose $p = 1$.
Then, we have:
:$\ds \lim_{n \to \infty} \frac {\epsilon_{n + 1}} {\epsilon_n} = c < 1$
By definition of limit, there exists some $N \in \N$ such that:
:$\forall n > N: \size {\dfrac {\epsilon_{n + 1}} {\epsilon_n} - c} < \dfrac {1 - c} 2$
In particular, for $n > N$:
:$\dfrac {\epsilon_{n + 1}} {\epsilon_n} < c + \dfrac {1 - c} 2 = \dfrac {c + 1} 2 < 1$
Let $c' = \dfrac {c + 1} 2 < 1$.
Then:
:$\epsilon_{n + 1} < c' \epsilon_n$
It follows that, for every $n \in \N$:
:$\epsilon_{N + 1 + n} \le \paren {c'}^n \epsilon_{N + 1}$
By Sequence of Powers of Number less than One, Multiple Rule for Real Sequences, Squeeze Theorem for Real Sequences:
:$\sequence {\epsilon_n}$ is a null sequence
Therefore, by Corollary to Squeeze Theorem for Real Sequences:
:$x_n \to \alpha$
=== Case $p > 1$ ===
Order of Cycle is Length of Cycle

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$ be a cyclic permutation of length $k$.
Then:
:$\order \pi = k$
where:
:$\order \pi$ denotes the order of $\pi$ in $S_n$.
== Proof ==
Let $\pi = \tuple {a_0, a_1, \ldots, a_{k - 1} }$.
Observe that:
Hence the result, by definition of order of group element.
== Sources ==
* 
Category:Symmetric Groups
Order of Cyclic Group equals Order of Generator

== Theorem ==
Let $G$ be a finite cyclic group which is generated by $a \in G$.
Then:
:$\order a = \order G$
where:
:$\order a$ denotes the order of $a$ in $G$
:$\order G$ denotes the order of $G$.
== Proof ==
Let $\order a = n$.
From List of Elements in Finite Cyclic Group:
:$G = \set {a_0, a_1, \ldots, a_{n - 1} }$
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 9$: Cyclic Groups
* : Chapter $2$: Subgroups and Cosets: $\S 43$
* : $0$: Some Conventions and some Basic Facts
* : $\S 39.2$ Cyclic Groups
Category:Order of Group Elements
Category:Order of Groups
Category:Cyclic Groups
Order of Differential Equation/Examples/First Order

== Examples of Orders of Differential Equations ==
The following ordinary differential equations are of the $1$st order:
:$\dfrac {\d y} {\d x} + y = 0$
:$y' = e^x$
:$x y' = 2 y$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
Category:Examples of Differential Equations
Order of Differential Equation/Examples/Fourth Order

== Examples of Orders of Differential Equations ==
The following ordinary differential equation is of the $4$th order:
:$x y^{\paren 4} + 2 y' ' + \paren {x y'}^5 = x^3$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
Category:Examples of Differential Equations