nyuuzyou/wikis huggingface dataset:
Order of Differential Equation/Examples/Second Order

== Examples of Orders of Differential Equations ==
The following ordinary differential equations are of the $2$nd order:
:$\dfrac {\d^2 y} {\d x^2} = \dfrac 1 {1 - x^2}$
:$\map {f'} x = \map {f' '} x$
:$y' ' + \paren {3 y'}^3 + 2 x = 7$
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation
Category:Examples of Differential Equations
Order of Differential Equation/Examples/Third Order

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

== Theorem ==
The dihedral group $D_n$ is of order $2 n$.
== Proof ==
By definition, $D_n$ is the symmetry group of the regular polygon of $n$ sides.
: 
Let $P$ be a regular $n$-gon.
By inspection, it is seen that:
:$(1): \quad$ there are $n$ symmetries of the vertices of $P$ by rotation
:$(2): \quad$ there are a further $n$ symmetries of the vertices of $P$ by rotation after reflected in any of the axes of bilateral symmetry of $P$.
Hence the result.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 26 \theta$
Category:Dihedral Groups
Order of Divisor Count Function

== Theorem ==
For all $x \ge 1$:
:$\ds \sum_{n \mathop \le x} \map {\sigma_0} n = x \log x + \paren {2 \gamma - 1} x + \map \OO {\sqrt x}$
where:
:$\gamma$ is the Euler-Mascheroni constant
:$\map {\sigma_0} n$ is the divisor count function.
== Proof ==
Category:Number Theory
Order of Element Divides Order of Finite Group

== Theorem ==
In a finite group, the order of a group element divides the order of its group:
:$\forall x \in G: \order x \divides \order G$
== Proof ==
Let $G$ be a group.
Let $x \in G$.
By definition, the order of $x$ is the order of the subgroup generated by $x$.
Therefore, by Lagrange's Theorem, $\order x$ is a divisor of $\order G$.
== Also see ==
* Element to Power of Group Order is Identity
== Sources ==
* : $\S 6.4$. Lagrange's theorem: Example $117$
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.7$
* : $\S 1.9$: Theorem $12$ Corollary
* : Chapter $\text{II}$: Groups: Problem $\text{GG}$
* : Chapter $2$: Subgroups and Cosets: $\S 41$
* : $\S 44.1$ Some consequences of Lagrange's Theorem
* : $\S 3.4$: Cyclic groups
* : Chapter $5$: Cosets and Lagrange's Theorem: Corollary $5.12$
Category:Finite Groups
Category:Order of Group Elements
Order of Element divides Order of Centralizer

== Theorem ==
Let $G$ be a finite group.
Let $x \in G$ be an element of $G$.
Let $\map {C_G} x$ denote the centralizer of $x$.
Then:
:$\order x \divides \order {\map {C_G} x}$
where:
:$\order x$ denotes the order of $x$ in $G$
:$\divides$ denotes divisibility
:$\order {\map {C_G} x}$ denotes the order of $\map {C_G} x$.
== Proof ==
From Order of Cyclic Group equals Order of Generator:
:$\order x = \order {\gen x}$
where $\gen x$ denotes the subgroup of $G$ generated by $x$.
By definition, $\gen x$ is a cyclic group.
By Cyclic Group is Abelian, all elements of $\gen x$ commute with $x$.
Thus by definition of centralizer:
:$\gen x \subseteq \map {C_G} x$
From Centralizer of Group Element is Subgroup, $\map {C_G} x$ is a subgroup of $G$.
So $\gen x$ is a subgroup of $\map {C_G} x$.
The result follows from Lagrange's theorem.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $14$
Category:Centralizers
Category:Order of Group Elements
Order of Element in Group equals its Order in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H \le G$, where $\le$ denotes the property of being a subgroup.
Let $x \in H$.
Then the order of $x$ in $H$ equals the order of $x$ in $G$.
== Proof ==
Let $\gen x$ be the subgroup of $G$ generated by $x$.
By definition, $\gen x \le G$.
All the elements of $\gen x$ are powers of $x$.
As $x \in H$ it follows by that all the powers of $x$ are elements of $H$.
That is:
:$\gen x \le H$
By Order of Cyclic Group equals Order of Generator:
:$\order x = \order {\gen x}$
But $\gen x$ is the same cyclic group whether it is considered as embedded in $H$ or $G$.
Hence the result.
== Sources ==
* : $0$: Some Conventions and some Basic Facts
Category:Order of Group Elements
Category:Subgroups
Order of Element in Quotient Group

== Theorem ==
Let $G$ be a group, and let $H$ be a normal subgroup of $G$.
Let $G / H$ be the quotient group of $G$ by $H$.
The order of $a H \in G / H$ divides the order of $a \in G$.
== Proof ==
Let $G$ be a group with normal subgroup $H$.
Let $G / H$ be the quotient of $G$ by $H$.
From Quotient Group Epimorphism is Epimorphism, $G / H$ is a homomorphic image of $G$.
Let $q_H: G \to G / H$ given by $\map f a = a H$ be that quotient epimorphism.
Let $a \in G$ such that $a^n = e$ for some integer $n$.
Then, by the morphism property of $q_H$:
Hence $\order H$ divides $n$.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \delta$
Category:Quotient Groups
Order of Elements in Quaternion Group

== Theorem ==
Let $Q = \Dic 2$ be the quaternion group, whose group presentation is given by:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
Then $\Dic 2$ has:
:$1$ element of order $2$
and:
:$6$ elements of order $4$.
== Proof ==
From Identity is Only Group Element of Order 1, the identity element $e$ , and only $e$, is of order $1$.
From here, we inspect the Cayley table:
It is immediately seen that:
:$\paren {a^2}^2 = e$
and so by definition $a^2$ is of order $2$.
As can be seen from inspection of the main diagonal, for all other $x \in Q$, we have:
:$x^2 = a^2 \ne e$
and so:
:$x^4 = \paren {a^2}^2 = e$
There are $6$ such elements.
Hence the result by definition of order of group element.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $14$
Category:Quaternion Group
Order of External Direct Product

== Theorem ==
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.
Then the order of $\struct {S \times T, \circ}$ is $\card S \times \card T$.
== Proof ==
By definition the order of $\struct {S \times T, \circ}$ is $\card S \times \card T$ is the cardinality of the underlying set $S \times T$.
The result follows directly from Cardinality of Cartesian Product of Finite Sets.
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
Category:External Direct Products
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a finite abelian group whose identity is $e$.
Let $G$ have at least $p$ elements of order $p$.
Then:
: $p^2 \divides \order G$
where:
:$\divides$ denotes divisibility
:$\order G$ denotes the order of $G$.
== Proof ==
Let $x \in G$ be of order $p$.
Consider $\gen x$, the subgroup generated by $x$.
By Group of Prime Order p has p-1 Elements of Order p, the elements of $\gen x$ are all of order $p$ except $e$.
Thus, by hypothesis, there must exist another $y \in G$ of order $p$.
Consider the subset of $G$:
:$S := \set {x^i y^j: 0 \le i, j < p}$
By the Finite Subgroup Test, $S$ is a subgroup of $G$ which has $p^2$ elements.
By Lagrange's theorem:
:$\order S \divides \order G$
But $\order S = p^2$ and so $p^2 \divides \order G$.
== Examples ==
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $20$
Category:Order of Group Elements
Category:Examples of Order of Group Elements
Category:Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Category:Abelian Groups
Category:Finite Groups
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2/Examples

== Examples of Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$ ==
=== Order $3$ ===
Category:Examples of Order of Group Elements
Category:Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2/Examples/Order 3

== Example of Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$ ==
Let $G$ be a finite abelian group whose identity is $e$.
Let $G$ have more than $2$ elements of order $3$.
Then:
: $9 \divides \order G$
where:
:$\divides$ denotes divisibility
:$\order G$ denotes the order of $G$.
== Proof 1 ==
== Proof 2 ==
Category:Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2/Examples/Order 3/Proof 1

== Example of Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$ ==
== Proof ==
By hypothesis there are elements $x, y$ of order $3$ in $G$ such that $x, y, x^2$ are all different.
Consider the subset of $G$:
:$S := \set {x^i y^j: 0 \le i, j \le 2}$
By the Finite Subgroup Test, $S$ is a subgroup of $G$ which has $9$ elements.
By Lagrange's theorem:
:$\order S \divides \order G$
But $\order S = 9$ and so $9 \divides \order G$.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $20$
Category:Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2/Examples/Order 3/Proof 2

== Example of Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$ ==
== Proof ==
An example of Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$, setting $p = 3$.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $20$
Category:Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
Order of Finite p-Group is Power of p

== Theorem ==
Let $G$ be a finite group.
Let $p$ be a prime number.
Let all elements of $G$ have order a power of $p$.
Then $G$ is a $p$-group.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
Category:Order of Finite p-Group is Power of p
Category:P-Groups
Category:Finite Groups
Order of Finite p-Group is Power of p/Proof 1

== Theorem ==
== Proof ==
:
:$\order G = k p^n: p \nmid k$
where $\order G$ denotes the order of $G$.
By Divisors of Power of Prime:
:$k \nmid p^n$
From the First Sylow Theorem:
:$\exists H \le G: \order H = k$
where $H \le G$ denotes that $H$ is a subgroup of $G$.
Thus:
:$\exists h \in H: \order h \divides k \implies \order h \nmid p$
where $\divides$ denotes divisibility.
Thus:
:$\exists h \in G: \order h \ne p^n: n \in \Z$
Thus by Proof by Contradiction, $\order G$ must be a power of $p$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Definition $11.8$
Category:Order of Finite p-Group is Power of p
Order of Finite p-Group is Power of p/Proof 2

== Theorem ==
== Proof ==
Let every element of $G$ be a $p$-element.
Let $q$ be a prime number which is a divisor of the order $\order G$ of $G$.
By Cauchy's Lemma (Group Theory), there exists an element of $G$ whose order is a divisor of $q$.
But as the order of all elements of $G$ divide $p^n$ it follows that $q = p$.
Thus $G$ is a group whose order is $p^n$ for some $n \in \Z_{>0}$.
Category:Order of Finite p-Group is Power of p
Order of Floor Function

== Theorem ==
Let $\floor x$ denote the floor function of $x$. 
Then:
:$\floor x = x + \map \OO 1$
where $\OO$ is big-O notation.
== Proof ==
From Floor is between Number and One Less: 
:$\floor x \le x < \floor x + 1$
so:
:$0 \le x - \floor x < 1$
By the definition of the absolute value function, we have: 
:$\size {\floor x - x} < 1$
so by the definition of Big-O notation, we have:
:$\floor x - x = \map \OO 1$
We can conclude that:
:$\floor x = x + \map \OO 1$
Category:Analytic Number Theory
Category:Floor Function
Order of General Linear Group over Galois Field

== Theorem ==
Let $\GF$ be a Galois field with $p$ elements.
Then the order of the general linear group $\GL {n, \GF}$ is:
:$\ds \prod_{j \mathop = 1}^n \paren {p^n - p^{j - 1} }$
== Proof ==
Let $\GF$ be a Galois field with $p$ elements: $\card \GF = p$.
Let $A = \sqbrk {a_{i j} }_{n, n}$ be a matrix such that $\size A \ne 0$ and $a_{i j} \in \GF$.
How many such matrices can be constructed?
In order to avoid a zero determinant, the top row of the matrix, $\sequence {a_{1 j} }_{j \mathop = 1, \dotsc, n}$ must have at least one non-zero element.
Therefore there are $p^n - 1$ possibilities for the top row:
:the $p^n$ possible sequences of $n$ values from $\GF$, minus the one sequence $\paren {0, 0, \dotsc, 0}$.
The only restriction on the second row is that it not be a multiple of the first.
Therefore, there are the $p^n$ possible sequences again, minus the $p$ sequences which are multiples of the first row.
Thus, continuing in this fashion, the $j^{th}$ row can be any of the $p^n$ possible sequences, minus the $p^{\paren {j - 1} }$ sequences which are linear combinations of previous rows.
The number of possible matrices satisfying the conditions of $A$, then, is:
:$\ds \prod_{j \mathop = 1}^n \paren {p^n - p^{j - 1} }$
Category:Combinatorics
Category:General Linear Group
Category:Galois Fields
Order of Graph/Examples/Arbitrary Order 4

== Example of Order of Graph ==
Let $G = \struct {V, E}$ be the graph defined as:
:$V = \set {v_1, v_2, v_3, v_4}$
:$E = \set {\set {\tuple {v_1, v_2}, \tuple {v_2, v_1} }, \set {\tuple {v_1, v_3}, \tuple {v_3, v_1} }, \set {\tuple {v_2, v_3}, \tuple {v_3, v_2} }, \set {\tuple {v_3, v_4}, \tuple {v_4, v_3} } }$
Then the order of $G$ is the cardinality of $V$:
:$\card V = 4$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs
Category:Examples of Order of Graph
Order of Group Element/Examples

== Examples of Order of Group Element ==
=== Order of $2$ in $\struct {\R_{\ne 0}, \times}$ ===
=== Order of $i$ in $\struct {\C_{\ne 0}, \times}$ ===
=== Order of $\begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix}$ in General Linear Group ===
=== Rotation Through the $n$th Part of a Full Angle ===
=== Possible Orders of $x$ when $x^2 = x^{12}$ ===
=== Non-Cyclic Group of Order $55$ has Order $5$ Element and Order $11$ Element ===
Category:Examples of Order of Group Elements
Order of Group Element/Examples/Element of Multiplicative Group of Real Numbers

== Examples of Order of Group Element ==
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
The order of $2$ in $\struct {\R_{\ne 0}, \times}$ is infinite.
== Proof ==
From Real Multiplication Identity is One, the identity of $\struct {\R_{\ne 0}, \times}$ is $1$.
There exists no $n \in \Z_{\ge 0}$ such that $2^n = 1$.
Hence the result by definition of infinite order element.
== Sources ==
* : $\S 38$. Period of an element: Illustrations: $\text{(i)}$
Category:Examples of Order of Group Elements
Category:Multiplicative Group of Real Numbers
Order of Group Element/Examples/Imaginary Unit in Multiplicative Group of Complex Numbers

== Examples of Order of Group Element ==
Consider the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.
The order of $i$ in $\struct {\C_{\ne 0}, \times}$ is $4$.
== Proof ==
We have:
Hence the result by definition of order of group element.
== Sources ==
* : $\S 38$. Period of an element: Illustrations: $\text{(ii)}$
Category:Examples of Order of Group Elements
Order of Group Element/Examples/Matrix (1 1, 0 1) in General Linear Group

== Examples of Order of Group Element ==
Consider the general linear group $\GL 2$.
Let $\mathbf A := \begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix} \in \GL 2$
The order of $\mathbf A$ in $\GL 2$ is infinite.
== Proof ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then we see that:
:$\mathbf A^n = \begin{bmatrix} 1 & n \cr 0 & 1 \end{bmatrix}$
and so:
:$\forall n \in \Z_{>0}: \mathbf A^n \ne \begin{bmatrix} 1 & 0 \cr 0 & 1 \end{bmatrix}$
Hence the result.
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Definition $3.9$: Remark $1$
Category:Examples of Order of Group Elements
Category:General Linear Group
Order of Group Element/Examples/Possible Orders of x when x^2 = x^12

== Examples of Order of Group Element ==
Let $G$ be a group.
Let $x \in G \setminus \set e$ be such that $x^2 = x^{12}$.
Then the possible orders of $x$ are $2$, $5$ and $10$.
== Proof ==
Hence:
:$\order x \in \set {2, 5, 10}$
as required.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $11$
Category:Examples of Order of Group Elements
Order of Group Element/Examples/Rotation Through nth Part of Full Angle

== Example of Order of Group Element ==
Let $G$ denote the group of isometries in the plane under composition of mappings.
Let $r$ be the rotation of the plane about a given point $O$ through an angle $\dfrac {2 \pi} n$, for some $n \in \Z_{> 0}$.
Then $r$ is the generator of a subgroup $\gen r$ of $G$ which is of order $n$.
== Proof ==
== Sources ==
* : $\S 38$. Period of an element
Category:Examples of Order of Group Elements
Order of Group Element equals Order of Coprime Power

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let $\order g$ denote the order of $g$ in $G$.
Then:
:$\forall m \in \Z: \order {g^m} = \order g \iff m \perp \order g$
where:
:$g^m$ denotes the $m$th power of $g$ in $G$
:$\perp$ denotes coprimality.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Order of Group Elements
Category:Order of Group Element equals Order of Coprime Power
Order of Group Element equals Order of Coprime Power/Corollary

== Corollary to Order of Group Element equals Order of Coprime Power ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let $\order g$ denote the order of $g$ in $G$.
Let $H \le G$ be a subgroup of $G$.
Let $\order g = n$.
Let $g^m \in H$.
Let $m$ and $n$ be coprime.
Then $g \in H$.
== Proof ==
From Order of Group Element equals Order of Coprime Power:
:$m \perp n \iff \order {g^m} = \order g$
As $\order g = n$ and $n \perp m$, from Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order:
:$\gen g = \gen {g^m}$
Thus as $g^m \in H$ it follows that $\gen {g^m} \le H$
Hence $\gen g \le H$.
The result follows.
== Sources ==
* : $0$: Some Conventions and some Basic Facts: Exercise $12$
Category:Order of Group Element equals Order of Coprime Power
Category:Order of Group Elements
Order of Group Element equals Order of Coprime Power/Proof 1

== Theorem ==
== Proof ==
Let $\order g = n$.
Then from Order of Power of Group Element:
:$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n}}$
where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$.
Thus:
:$\order {g^m} = \order g \iff \gcd \set {m, n} = 1$
The result follows by definition of coprime integers.
Category:Order of Group Element equals Order of Coprime Power
Order of Group Element equals Order of Coprime Power/Proof 2

== Theorem ==
== Proof ==
Let $\order g = n$.
=== Necessary Condition ===
Let $m \perp n$.
Then by Bézout's Identity:
:$\exists x, y \in \Z: x m + y n = 1$
Let $h = g^m$.
Then:
:$h^x = g^{m x} = g^{1 - y n} = g g^{- y n} = g e = g$
and so $g$ is also a power of $h$.
Hence from Order of Group Element not less than Order of Power:
:$\order g \le \order h \le \order g$
and it follows that:
:$\order g = \order h$
=== Sufficient Condition ===
We prove the contrapositive.
Suppose $m \not \perp n$.
Then:
:$\exists d \in \Z_{>1}: \exists a, b \in \Z: m = a d \land n = b d$
Let $h = g^m$.
Then:
:$h^b = \paren {g^m}^{n/d} = g^{a n} = \paren {g^n}^a = e^a = e$
Hence:
:$\order h \le b < n$
and it follows that:
:$\order g \ne \order h$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: $\text{(iv)}$
Category:Order of Group Element equals Order of Coprime Power
Order of Group Element equals Order of Inverse

== Theorem ==
Let $G$ be a group whose identity is $e$.
Then:
: $\forall x \in G: \order x = \order {x^{-1} }$
where $\order x$ denotes the order of $x$.
== Proof ==
By Powers of Group Elements: Negative Index:
: $\paren {x^k}^{-1} = x^{-k} = \paren {x^{-1} }^k$
Hence:
Similarly:
A similar argument shows that if $x$ is of infinite order, then so must $x^{-1}$ be.
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: $\text{(ii)}$
* : $\S 1.9$: Exercise $5.6$
* : Chapter $6$: An Introduction to Groups: Exercise $10$
Category:Order of Group Elements
Category:Inverse Elements
Order of Group Element in Group Direct Product

== Theorem ==
Let $G$ and $H$ be finite groups.
Let $g \in G: \order g = m, h \in H: \order h = n$.
Then $\order {\tuple {g, h} }$ in $G \times H$ is $\lcm \set {m, n}$.
== Proof ==
Let $G$ and $H$ be a groups whose identities are $e_G$ and $e_H$.
Let $l = \lcm \set {m, n}$.
From the definition of lowest common multiple:
:$\exists x, y \in \Z: l = m x = n y$
From the definition of order of an element:
:$g^m = e_G, h^n = e_H$
Thus:
Now suppose $\exists k \in \Z: \tuple {g, h}^k = \tuple {e_G, e_H}$.
It follows that:
:$\tuple {g, h}^k = \tuple {g^k, h^k} = \tuple {e_G, e_H}$
Thus:
:$g^k = e_G, h^k = e_H$
It follows from $g^k = e_G$ and $\order g = m$, by Absolute Value of Integer is not less than Divisors, that:
:$m \divides k$
Similarly it follows that:
:$n \divides k$
So $k$ is a positive common multiple of both $m$ and $n$.
Since $l$ is the lowest common multiple:
:$l \le k$
Therefore $l$ is the smallest such that $\tuple {g, h}^l = \tuple {e_G, e_H}$.
The result follows by definition of the order of an element.
Category:Group Direct Products
Order of Group Element not less than Order of Power

== Theorem ==
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let $\left\lvert{g}\right\rvert$ denote the order of $g$ in $G$.
Then:
:$\forall m \in \Z: \left\lvert{g^m}\right\rvert \le \left\lvert{g}\right\rvert$
where $g^m$ denotes the $m$th power of $g$ in $G$.
== Proof 1 ==
== Proof 2 ==
Category:Order of Group Elements
Category:Order of Group Element not less than Order of Power
Order of Group Element not less than Order of Power/Proof 1

== Theorem ==
== Proof ==
Let $\left\lvert{g}\right\rvert = n$.
Then from Order of Power of Group Element:
:$\forall m \in \Z: \left\lvert{g^m}\right\rvert = \dfrac n {\gcd \left\{ {m, n}\right\} }$
where $\gcd \left\{ {m, n}\right\}$ denotes the greatest common divisor of $m$ and $n$.
The result follows from Greatest Common Divisor is at least 1.
Category:Order of Group Element not less than Order of Power
Order of Group Element not less than Order of Power/Proof 2

== Theorem ==
== Proof ==
Let $g^n = e$.
Let $h = g^m$.
Then:
:$h^n = g^{m n} = \paren {g^n}^m = e^m = e$
Hence by definition of order of group element:
:$\order h \le n$
Hence the result.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: $\text{(iv)}$
Category:Order of Group Element not less than Order of Power
Order of Group of Units of Integers Modulo n

== Theorem ==
Let $n \in \Z_{\ge 0}$ be an integer.
Let $\struct {\Z / n \Z, +, \cdot}$ be the ring of integers modulo $n$.
Let $U = \struct {\paren {\Z / n \Z}^\times, \cdot}$ denote the group of units of this ring.
Then:
:$\order U = \map \phi n$
where $\phi$ denotes the Euler $\phi$-function.
== Proof ==
By Reduced Residue System under Multiplication forms Abelian Group, $U$ is equal to the set of integers modulo $n$ which are coprime to $n$.
It follows by Cardinality of Reduced Residue System:
:$\order U = \map \phi n$
== Sources ==
* : $\S 2.3$: Congruences
Category:Ring of Integers Modulo m
Category:Euler Phi Function
Order of Homomorphic Image of Group Element

== Theorem ==
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.
Let $\phi: G \to H$ be a homomorphism.
Let $g \in G$ be of finite order.
Then:
:$\forall g \in G: \order {\map \phi g} \divides \order g$
where $\divides$ denotes divisibility.
== Proof ==
Let $\phi: G \to H$ be a homomorphism.
Let $\order g = n, \order {\map \phi g} = m$.
It follows from Element to Power of Multiple of Order is Identity that $m \divides n$.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $4 \ \text{(ii)}$
* : Chapter $8$: The Homomorphism Theorem: Exercise $4$
Category:Group Homomorphisms