nyuuzyou/wikis huggingface dataset:
Order of Monomorphic 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 monomorphism.
Let $g \in G$ be of finite order.
Then:
:$\forall g \in G: \order {\map \phi g} = \order g$
== Proof ==
By definition of monomorphism, $\phi$ is a homomorphism which is also an injection.
From Order of Homomorphic Image of Group Element:
:$\forall g \in G: \order {\map \phi g} \divides \order g$
So $g^m = e$, as $\phi$ is injective.
From the definition of order of group element, that means $n \le m$ since $n$ is the smallest such power.
Thus $m = n$ and the result holds.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $4 \ \text{(ii)}$
* : Chapter $8$: The Homomorphism Theorem: Exercise $4$
Category:Group Monomorphisms
Order of Multiplicative Group of Reduced Residues

== Theorem ==
Let $\struct {\Z'_m, \times_m}$ denote the multiplicative group of reduced residues modulo $m$.
The order of $\struct {\Z'_m, \times_m}$ is $\map \phi m$, where $\phi$ denotes the Euler $\phi$ function.
== Proof ==
By definition, the order of a group is the cardinality of its underlying set.
By definition, the underlying set of $\struct {\Z'_m, \times_m}$ is the reduced residue system $\Z'_m$:
:$\Z'_m = \set {\eqclass {a_1} m, \eqclass {a_2} m, \ldots, \eqclass {a_{\map \phi m} } m}$
where:
:$\forall k: a_k \perp m$
From Cardinality of Reduced Residue System, $\Z'_m$ has $\map \phi m$ elements.
Hence the result.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 38$
Category:Multiplicative Groups of Reduced Residues
Order of Möbius Function

== Theorem ==
Let $\mu$ denote the Möbius function .
Then:
:$\ds \sum_{n \mathop \le N} \map \mu n = \map o N$
where $o$ denotes little-o notation.
== Proof ==
Let $\map \Re z$ be the real part of a complex variable $z$.
By Dirichlet Series is Analytic, the Riemann zeta function is analytic.
By Trivial Zeroes of Riemann Zeta Function are Even Negative Integers, the Riemann zeta function has no zeroes in $\map \Re z > 1$.
Thus the reciprocal of the Riemann zeta function is analytic in $\map \Re z > 1$.
By Reciprocal of Riemann Zeta Function, this means that $\ds \sum_{n \mathop = 1}^\infty \map \mu n n^{-z}$ converges to an analytic function in $\map \Re z > 1$.
By taking $a_n = \map \mu n$ in Ingham's Theorem on Convergent Dirichlet Series:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z}$ converges for $\map \Re z \ge 1$.
Taking $z = 1$, we are given a convergent sum:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map \mu n} n$
Clearly:
:$\ds \sum_{n \mathop = 1}^N \frac {\map \mu n} n \ge \sum_{n \mathop = 1}^N \frac {\map \mu n} N$
but:
:$\ds \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \frac {\map \mu n} n = \lim_{z \mathop \to 1} \frac 1 {\map \zeta z}$
Since Harmonic Series is Divergent, $\dfrac 1 {\map \zeta z}$ goes to $0$ by Reciprocal of Null Sequence.
Hence also:
:$\ds \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \frac {\map \mu n} N = 0$
Category:Möbius Function
Category:Analytic Number Theory
Order of Natural Logarithm Function

== Theorem ==
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Then: 
:$\ln x = \map \OO {x^\epsilon}$ as $x \to \infty$
and:
:$\ln x = \map \OO {x^{-\epsilon} }$ as $x \to 0^+$
where $\OO$ is big-O notation.
== Proof ==
We first show that: 
:$\ln x = \map \OO {x^\epsilon}$ as $x \to \infty$
We show that for $x \ge 1$, we have: 
:$\ds 0 \le \ln x \le \frac 1 \epsilon x^\epsilon$
We first show that for $t \ge 0$, we have: 
:$\ds t \le \frac 1 \epsilon e^{\epsilon t}$
The claim then follows taking $t = \ln x$.
Define a real function $f : \openint 0 \infty \to \R$ by: 
:$\ds \map f t = \frac 1 \epsilon e^{\epsilon t} - t$
for $t \ge 0$. 
Then $f$ is differentiable with derivative:
:$\map {f'} t = e^{\epsilon t} - 1$
by Derivative of Exponential Function.
So:
:$\map {f'} t \ge 0$ for all $t \ge 0$. 
So, from Real Function with Positive Derivative is Increasing:
:$f$ is increasing for $t \ge 0$. 
That is, for $t \ge 0$:
:$\ds \map f t \ge \map f 0 = \frac 1 \epsilon > 0$ 
So:
:$\ds \frac 1 \epsilon e^{\epsilon t} - t > 0$
for $t \ge 0$. 
So:
:$\ds \frac 1 \epsilon e^{\epsilon t} \ge t \ge 0$
for $t \ge 0$ as required.
Hence, for $x \ge 1$ we have: 
:$\ds 0 \le \ln x \le \frac 1 \epsilon x^\epsilon$
From the definition big-O notation, we have: 
:$\ln x = \map \OO {x^\epsilon}$ as $x \to \infty$
We now show that: 
:$\ln x = \map \OO {x^{-\epsilon} }$ as $x \to 0^+$
In particular, we can show that: 
:$\ds \size {\ln x} \le \frac 1 \epsilon x^{-\epsilon}$
for $0 < x \le 1$. 
Note that since: 
:$\ds \ln x \le \frac 1 \epsilon x^\epsilon$
for $x \ge 1$ we have: 
:$\ds \map \ln {\frac 1 x} \le \frac 1 \epsilon x^{-\epsilon}$
for $0 < x \le 1$. 
So, by Logarithm of Power:
:$\ds -\ln x \le \frac 1 \epsilon x^{-\varepsilon}$
For $0 < x \le 1$, we have: 
:$\ln x \le 0$
so:
:$\size {\ln x} = -\ln x$
from the definition of the absolute value. 
So we obtain: 
:$\ds \size {\ln x} \le \frac 1 \epsilon x^{-\epsilon}$
for $0 < x \le 1$ as required.
From the definition big-O notation, we therefore have:
:$\ln x = \map \OO {x^{-\epsilon} }$ as $x \to 0^+$
Category:Natural Logarithms
Order of Power of Group Element

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $G$ such that:
:$\order g = n$
where $\order g$ denotes the order of $g$.
Then:
:$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n} }$
where $\gcd \set {m, n}$ denotes the greatest common divisor of $m$ and $n$.
== Proof ==
Let $\gcd \set {m, n} = d$.
From Integers Divided by GCD are Coprime: there exists $m', n' \in \Z$ such that $m = d m'$, $n = d n'$.
Then:
By Element to Power of Multiple of Order is Identity:
:$\order {g^m} \divides n'$.
 $\order {g^m} = n' ' < n'$.
By Bézout's Identity:
:$\exists x, y \in \Z: m x + n y = d$
But $d n' ' < d n' = n$, contradicting the fact that $n$ is the order of $g$.
Therefore:
:$\order {g^m} = n'$
Recalling the definition of $n'$:
:$\order {g^m} = \dfrac n {\gcd \set {m, n} }$
as required.
== Examples ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 41 \alpha$
* : $0$: Some Conventions and some Basic Facts: Exercise $5 \ \text{(i)}$
Category:Order of Group Elements
Order of Power of Group Element/Examples/Powers of Element of Order 20

== Example of Order of Power of Group Element ==
Let $G$ be a group.
Let $x \in G$ be such that:
:$\order x = 20$
where $\order x$ denotes the order of $x$ in $G$.
Then:
== Proof ==
From Order of Power of Group Element:
:$\order {x^m} = \dfrac {\order x} {\gcd \set {m, \order x} }$
Here we have that $\order x = 20$.
Thus:
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $13$
Category:Examples of Order of Power of Group Element
Order of Product of Abelian Group Elements Divides LCM of Orders of Elements

== Theorem ==
Let $G$ be an abelian group.
Let $a, b \in G$.
Then:
:$\order {a b} \divides \lcm \set {\order a, \order b}$
where:
:$\order a$ denotes the order of $a$
:$\divides$ denotes divisibility
:$\lcm$ denotes the lowest common multiple.
== Proof ==
Let $\order a = m, \order b = n$.
Let $c = \lcm \set {m, n}$.
Then:
So:
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 41 \beta$
Category:Order of Group Elements
Category:Abelian Groups
Order of Product of Commuting Group Elements of Coprime Order is Product of Orders

== Theorem ==
Let $G$ be a group.
Let $g_1, g_2 \in G$ be commuting elements such that:
where $\order {g_1}$ denotes the order of $g_1$ in $G$.
Let $n_1$ and $n_2$ be coprime.
Then:
:$\order {g_1 g_2} = n_1 n_2$
== Proof ==
Let $g_1 g_2 = g_2 g_1$.
We have:
:$\paren {g_1 g_2}^{n_1 n_2} = e$
Thus:
:$\order {g_1 g_2} \le n_1 n_2$
Suppose $\order {g_1 g_2}^r = e$.
Then:
Thus $r$ is divisible by both $m$ and $n$.
The result follows.
== Also see ==
* Unique Composition of Group Element whose Order is Product of Coprime Integers, a converse of this
== Sources ==
* : $0$: Some Conventions and some Basic Facts: Exercise $6$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $17$
Category:Order of Group Elements
Order of Product of Disjoint Permutations

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi$ be a product of disjoint permutations of orders $k_1, k_2, \ldots, k_r$.
Then:
:$\order \pi = \lcm \set {k_1, k_2, \ldots, k_r}$
where:
:$\order \pi$ denotes the order of $\pi$ in $S_n$
:$\lcm$ denotes lowest common multiple.
== Proof ==
Suppose $\pi$ is a cycle.
Then from Order of Cycle is Length of Cycle, $\order \pi$ is its length.
As the LCM of $n \in \Z$ is $n$ itself, the result follows.
Let $\pi = \rho_1 \rho_2 \cdots \rho_r$ where:
: each $\rho_s$ is of order $k_s$
: $\rho_1$ to $\rho_r$ are mutually disjoint permutations.
Let $t = \lcm \set {k_1, k_2, \ldots, k_r}$.
From Disjoint Permutations Commute:
:$\pi^t = \rho_1^t \rho_2^t \cdots \rho_r^t$
We have that:
:$\forall s: 1 \le s \le r: \exists m \in \Z: t = m s$
Hence:
:$\forall s: \rho_s^t = e$
Thus $\pi^t = e$ and certainly $\order \pi \divides t$.
Let $\pi^u = e$ for some $u \in \Z_{>0}$.
We have that $\rho_1, \rho_2, \cdots, \rho_r$ are mutually disjoint permutations.
Thus $\rho_1^u, \rho_2^u, \cdots, \rho_r^u$ are also mutually disjoint permutations.
Then we have that each $\rho_s^u = e$.
Hence $k_s \divides u$.
Thus, if $\pi^u = e$, then:
:$t \divides u$
Thus, by choosing $u = \order \pi$:
:$\order \pi = t = \lcm \set {k_1, k_2, \ldots, k_r}$
== Examples ==
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 80$: Corollary
* : Chapter $9$: Permutations: Proposition $9.8$
Category:Order of Product of Disjoint Permutations
Category:Disjoint Permutations
Order of Product of Disjoint Permutations/Examples/Non-Disjoint Permutations in S9

== Example of Order of Product of Non-Disjoint Permutations ==
Consider the permutation given in cycle notation as
:$\rho = \begin{pmatrix} 1 & 2 & 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 6 & 7 \end{pmatrix} \begin{pmatrix} 3 & 9 \end{pmatrix}$
Its order is given by:
:$\order \rho = 7$
and not $\lcm \set {4, 3, 2} = 12$.
== Proof ==
$\rho$ is the product of $3$ cyclic but not disjoint permutations.
Thus we cannot use Order of Product of Disjoint Permutations to determine its order.
By compositing cycles, we obtain:
:$\rho = \begin{pmatrix} 1 & 2 & 6 & 7 & 3 & 9 & 4 \end{pmatrix}$
which is of order $7$.
== Sources ==
* : Chapter $9$: Permutations: Example $9.9$
Category:Order of Product of Disjoint Permutations
Order of Product of Disjoint Permutations/Examples/Permutations in S9

== Example of Order of Product of Disjoint Permutations ==
Consider the permutation given in cycle notation as
:$\rho = \begin{pmatrix} 1 & 2 & 3 & 4 \end{pmatrix} \begin{pmatrix} 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} 8 & 9 \end{pmatrix}$
Its order is given by:
:$\order \rho = 12$
== Proof ==
$\rho$ is the product of $3$ disjoint permutations, of orders $4$, $3$ and $2$.
From Order of Product of Disjoint Permutations:
:$\order \rho = \lcm \set {4, 3, 2} = 12$
== Sources ==
* : Chapter $9$: Permutations: Example $9.9$
Category:Order of Product of Disjoint Permutations
Order of Product of Entire Function with Polynomial

== Theorem ==
Let $f: \C \to \C$ be an entire function of order $\omega$.
Let $P: \C \to \C$ be a nonzero polynomial.
Then $f \cdot P$ has order $\omega$.
== Proof ==
If $f=0$, then $f \cdot P = 0$.
Thus the claim is trivial.
Therefore suppose that $f$ is not identically zero.
By Order of Product of Entire Functions and Polynomial has Order Zero, $f\cdot P$ has order at most $\omega$.
By Limit at Infinity of Polynomial, there exist $r, \delta > 0$ such that $\size {\map P z} \ge \delta$ for $\size z \ge r$.
 $\ds \map \log {\max_{\size z \mathop \le R} \size {\map f z \map P z} } = \map \OO {R^\beta}$ for some $\beta < \omega$.
By the Maximum Modulus Principle:
:$\ds \max_{\size z \mathop \le R} \size {\map f z} \le \dfrac 1 \delta \max_{\size z \mathop \le R} \size {\map f z \map P z}$
for $R \ge r$.
Thus:
:$\ds \map \log {\max_{\size z \mathop \le R} \size {\map f z} } = \map \OO {R^\beta}$
By , this means:
:$\omega \le \beta$
This is a contradiction.
Thus $f \cdot P$ has order $\omega$.
== Also see ==
* Order of Product of Entire Functions
* Polynomial has Order Zero
Category:Entire Functions
Order of Product of Entire Functions

== Theorem ==
Let $f, g: \C \to \C$ be entire functions of order $\alpha$ and $\beta$.
Then $f g$ has order at most $\map \max {\alpha, \beta}$.
== Proof ==
If $\map \max {\alpha, \beta} = +\infty$, the claim is trivial.
Thus we may assume that $\alpha < +\infty$ and $\beta < +\infty$.
Let $\epsilon > 0$ be arbitrary.
By , we have:
:$\map f z = \map \OO {\map \exp {\cmod z^{\alpha + \epsilon} } }$
and:
:$\map g z = \map \OO {\map \exp {\cmod z^{\beta + \epsilon} } }$
By , there are $c_1,c_2,r_1,r_2 \in \R$ such that:
:$\cmod z \ge r_1 \implies \cmod {\map f z} \le c_1 \map \exp {\cmod z^{\alpha + \epsilon} }$
and:
:$\cmod z \ge r_2 \implies \cmod {\map g z} \le c_2 \map \exp {\cmod z^{\beta + \epsilon} }$
Let $r_0 := \max \set {r_1, r_2, 2^{1 / \epsilon } }$.
Then, if $\cmod z \ge r_0$, we have:
Therefore the order of $fg$ is:
:$\le \map \max {\alpha, \beta} + 2 \epsilon$
Letting $\epsilon \to 0$, the claim follows.
== Also see ==
* Order of Sum of Entire Functions
* Order of Product of Entire Function with Polynomial
Category:Entire Functions
Order of Quotient Group

== Theorem ==
Let $G$ be a finite group.
Let $N$ be a normal subgroup of $G$.
Let $G / N$ be the quotient group of $G$ by $N$.
Then:
:$\dfrac {\order G} {\order N} = \order {G / N}$
where $\order G$ denotes the order of $G$.
== Proof ==
From Lagrange's Theorem:
:$\dfrac {\order G} {\order N} = \index G N$
where $\index G N$ is the index of $N$ in $G$.
By definition of index:
:$\index G N = \order {G / N}$
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.7$
Category:Quotient Groups
Category:Order of Groups
Order of Real Numbers is Dual of Order of their Negatives

== Theorem ==
:$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(d)}$
Category:Real Numbers
Category:Inequalities
Category:Order of Real Numbers is Dual of Order of their Negatives
Order of Reciprocal of Entire Function

== Theorem ==
Let $f: \C \to \C$ be an entire function of order $\rho$.
Let $f$ have no zeroes.
Then $1/f$ has order $\rho$.
== Proof ==
By Zerofree Analytic Function on Simply Connected Set has Logarithm, there exists an entire function $g$ with $f = \exp g$.
== Also see ==
* Zerofree Entire Function of Finite Order is Exponential of Polynomial
Category:Entire Functions
Order of Second Chebyshev Function

== Theorem ==
Let $x \ge 2$ be a real number.
Then:
:$\ds \map \psi x = \sum_{p \mathop \le x} \ln p + \map \OO {\sqrt x \paren {\ln x}^2}$
where:
:$\OO$ is big-O notation
:$\psi$ is the Second Chebyshev Function
:the sum runs over the primes less than or equal to $x$.
== Proof ==
From the definition of the Second Chebyshev Function, we have: 
:$\ds \map \psi x = \sum_{k \mathop = 1}^\infty \paren {\sum_{p^k \mathop \le x} \ln p}$
where the inner sum runs over the primes $p$ with $p^k \le x$.
That is, the primes $p$ with $p \le x^{1/k}$, so we can write: 
:$\ds \sum_{p^k \mathop \le x} \ln p = \sum_{p \mathop \le x^{1/k} } \ln p$
Note that there are no primes with $p < 2$, so if: 
:$x^{1/k} < 2$
we have: 
:$\ds \sum_{p^k \mathop \le x} \ln p = 0$
That is, if: 
:$\ds k > \frac {\ln x} {\ln 2}$
So, we have: 
From Logarithm is Strictly Increasing, for $2 \le p \le x^{1/k}$ we have: 
:$\ds 0 < \ln p \le \map \ln {x^{1/k} }$
So:
Then, for $k \ge 2$, we have: 
:$\ds \frac 1 k x^{1/k} \ln x \le \frac 1 2 \sqrt x \ln x$
so:
From the definition of big-O notation, we have: 
:$\ds \sum_{2 \mathop \le k \mathop \le \frac {\ln x} {\ln 2} } \paren {\sum_{p \mathop \le x^{1/k} } \ln p} = \map \OO {\sqrt x \paren {\ln x}^2}$
so:
:$\ds \map \psi x = \sum_{p \mathop \le x} \ln p + \map \OO {\sqrt x \paren {\ln x}^2}$
Category:Second Chebyshev Function
Category:Analytic Number Theory
Order of Shifted Entire Function

== Theorem ==
Let $f: \C \to \C$ be an entire function of order $\alpha$.
Let $a \in \C$.
Then $\map f {z + a}$ has order $\alpha$.
== Proof ==
We shall verify .
Let $\map g z := \map f {z + a}$ for $z \in \C$.
Then for all $R > \cmod a$:
:$\ds \max_{\cmod z \mathop \le R \mathop - \cmod a} \cmod {\map f z} \le \max_{\cmod z \mathop \le R} \cmod {\map g z} \le \max_{\cmod z \mathop \le R \mathop + \cmod a} \cmod {\map f z}$
Thus:
:$\paren 1 : \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R \mathop - \cmod a} \cmod {\map f z} }{\ln R} \le \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R} \cmod {\map g z} }{\ln R} \le \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R \mathop + \cmod a} \cmod {\map f z} }{\ln R}$
On the other hand, for each $c \in \R$:
:$\ds \paren 2 : \lim_{R \mathop \to +\infty} \dfrac {\map \ln {R + c} } {\ln R} = 1$
Thus:
Similarly, we have:
:$\ds \paren 4 : \limsup_{R \mathop \to +\infty} \dfrac {\ds \ln \ln \max_{\cmod z \mathop \le R + \cmod a} \cmod {\map f z} }{\ln R} = \alpha$
In view of $\paren 3$ and $\paren 4$, the claim follows from $\paren 1$ by Sandwich Rule.
== Also see ==
* Order of Product of Entire Functions
Category:Entire Functions
Order of Simple Group divides Factorial of Index of Subgroup

== Theorem ==
Let $G$ be a simple group, and let $H$ be a subgroup of $G$ with index $n$. Then $|G|$ divides $n!$.
== Proof ==
Let $G$ act on the cosets of $H$, $G/H$ by the left regular action.
By Group Action defines Permutation Representation, this defines a homomorphism $\tilde \phi: G \to \map \Gamma X$.
By Kernel is Normal Subgroup of Domain, $\ker \tilde \phi$ is a normal subgroup of $G$.
Since $G$ is simple, $\ker \tilde \phi = \{e\}$ or $G$.
If $\ker \tilde \phi = G$, then the action is trivial, but the action consists of an orbit of size $n$, so it is not trivial.
Hence $\ker \tilde \phi = \{e\}$, so $\tilde \phi$ is injective. So $G$ is isomorphic to $\tilde \phi (G)$ which is a subgroup of $\Gamma (X)$.
By Lagrange's Theorem (Group Theory), $|G|$ divides $n!$.
Order of Squares in Ordered Field

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered field whose zero is $0_R$ and whose unity is $1_R$.
Suppose that $\forall a \in R: 0 < a \implies 0 < a^{-1}$.
Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$.
Then $x \le y \iff x \circ x \le y \circ y$.
That is, the square function is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself.
When $R$ is one of the standard fields of numbers $\Q$ and $\R$, then this translates into:
:If $x, y$ are positive then $x \le y \iff x^2 \le y^2$.
== Proof ==
From Order of Squares in Ordered Ring, we have:
:$x \le y \implies x \circ x \le y \circ y$
To prove the reverse implication, suppose that $x \circ x \le y \circ y$.
Thus:
As $0_R \le x, y$ we have $0_R \le x + y$.
Hence by the premise we have $0_R \le \paren {x + y}^{-1}$.
So as $0_R \le \paren {y + \paren {-x} } \circ \paren {y + x}$ we can multiply both sides by $\paren {x + y}^{-1}$ and get $0_R \le \paren {y + \paren {-x} }$.
Adding $-x$ to both sides gives us $x \le y$.
== Also see ==
* Order of Squares in Ordered Ring
* Order of Squares in Totally Ordered Ring without Proper Zero Divisors
Category:Ordered Fields
Order of Squares in Ordered Ring

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered ring whose zero is $0_R$ and whose unity is $1_R$.
Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$.
Then:
:$x \le y \implies x \circ x \le y \circ y$
When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, then this translates into:
:If $x, y$ are positive then $x \le y \implies x^2 \le y^2$.
== Proof ==
Assume $x \le y$.
As $\le$ is compatible with the ring structure of $\struct {R, +, \circ, \le}$, we have:
:$x \ge 0 \implies x \circ x \le x \circ y$
:$y \ge 0 \implies x \circ y \le y \circ y$
and thus as $\le$ is transitive, it follows that $x \circ x \le y \circ y$.
== Also see ==
* Order of Squares in Ordered Field
* Order of Squares in Totally Ordered Ring without Proper Zero Divisors
* Power Function is Strictly Increasing on Positive Elements
Category:Ordered Rings
Order of Squares in Totally Ordered Ring without Proper Zero Divisors

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring without proper zero divisors whose zero is $0_R$.
Let $x, y \in R$ be positive, that is, $0_R \le x, y$.
Then $x \le y \iff x \circ x \le y \circ y$.
That is, the square mapping is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself.
When $R$ is one of the standard sets of numbers $\Z, \Q, \R$, then this translates into:
:If $x, y$ are positive, then $x \le y \iff x^2 \le y^2$.
== Proof ==
From Order of Squares in Ordered Ring, we have:
:$x \le y \implies x \circ x \le y \circ y$
To prove the opposite implication, we use a Proof by Contradiction.
 $x \circ x \le y \circ y$ but $x \not\le y$.
Since $\le$ is a total ordering, this means $y < x$.
Since $0_R \le y$, Extended Transitivity shows that $0_R < x$, so in particular $x ≠ 0_R$.
As $\le$ is compatible with the ring structure $\struct {R, +, \circ}$ and $x$ and $y$ are both positive, we have:
:$y \circ y \le y \circ x \le x \circ x$
Since we assume $x \circ x \le y \circ y$, Ordering Cycle implies Equality shows that:
:$y \circ y = x \circ x = y \circ x$
Therefore:
Since $x \ne y$, we have that $x + \paren {-y} \ne 0_R$; otherwise, by , we would have $x = \paren {x + \paren {-y} } + y = y$.
We have already shown that $x \ne 0_R$.
But then $x$ and $x + \paren {-y}$ are proper zero divisors of $R$, contradicting the hypothesis that $R$ has no proper zero divisors.
== Also see ==
* Order of Squares in Ordered Ring
* Order of Squares in Ordered Field
Category:Totally Ordered Rings
Order of Strictly Positive Real Numbers is Dual of Order of their Reciprocals

== Theorem ==
:$\forall x, y \in \R: x > y > 0 \implies \dfrac 1 x < \dfrac 1 y$
== Proof ==
From Reciprocal of Strictly Positive Real Number is Strictly Positive:
:$(1): \quad x > 0 \implies \dfrac 1 x > 0$
:$(2): \quad y > 0 \implies \dfrac 1 y > 0$
Then:
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(j)}$
Category:Real Numbers
Order of Subgroup Product

== Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes subset product
:$\order H$ denotes the order of $H$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 38 \alpha$
* : $\S 3.3$: Group actions and coset decompositions: Exercise $9$
Category:Subgroups
Category:Subset Products
Category:Order of Subgroup Product
Order of Subgroup Product/Corollary

== Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be finite subgroups of $G$.
Then:
:$\size {H \vee K} \ge \dfrac {\order H \order K} {\order {H \cap K} }$
or
:$\dfrac {\size {H \vee K} } {\order H} \ge \dfrac {\order K} {\order {H \cap K} }$
where $H \vee K$ denotes join and $\order H$ denotes the order of $H$.
== Proof ==
From Order of Subgroup Product:
: $(1): \quad \size {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
From Subset Product is Subset of Generator, we have that:
: $H K \subseteq H \vee K$
where $H K$ is the subset product of $H$ and $K$.
Thus:
: $(2): \quad \size {H \vee K} \ge \size {H K}$
The result follows by substituting for $\size {H K}$ from $(1)$ into $(2)$.
== Sources ==
* : $\S 1.9$: Theorem $21$: Corollary
Category:Order of Subgroup Product
Order of Subgroup Product/Lemma

== Lemma for Order of Subgroup Product ==
Let $h_1, h_2 \in H$.
Then:
:$h_1 K = h_2 K$
:
:$h_1$ and $h_2$ are in the same left coset of $H \cap K$.
== Proof ==
Let $h_1, h_2 \in H$.
Then:
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $10$
Category:Order of Subgroup Product