nyuuzyou/wikis huggingface dataset:
Elements in Same Coset iff Product with Inverse in Subgroup

== Theorem ==
Let $G$ be a group and let $H$ be a subgroup of $G$.
Let $x, y \in G$.
==== Elements in Same Left Coset iff Product with Inverse in Subgroup ====
==== Elements in Same Right Coset iff Product with Inverse in Subgroup ====
Category:Cosets
Elements in Same Left Coset iff Product with Inverse in Subgroup

== Theorem ==
Let $G$ be a group
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Then:
: $x, y$ are in the same left coset of $H$ $x^{-1} y \in H$.
== Proof ==
Let: $x H$ denote the left coset of $H$ by $x$
First we note that, from Congruence Class Modulo Subgroup is Coset, we have that the left cosets of $G$ form a partition of $G$.
=== Sufficient Condition ===
Suppose $x, y$ are in the same left coset of $H$.
It follows from Congruence Class Modulo Subgroup is Coset that:
: $x \in y H$ and $y \in x H$ $x H = y H$
From Cosets are Equal iff Product with Inverse in Subgroup, we have that:
: $x H = y H \iff x^{-1} y \in H$
So if $x, y$ are in the same left coset of $H$ then $x^{-1} y \in H$.
=== Necessary Condition ===
Suppose that $x^{-1} y \in H$.
From Left Cosets are Equal iff Product with Inverse in Subgroup, we have that:
: $x H = y H \iff x^{-1} y \in H$
Again, it follows from Congruence Class Modulo Subgroup is Coset that:
: $x \in y H$ and $y \in x H$ iff $x H = y H$
and so $x, y$ are in the same left coset of $H$.
== Also see ==
* Elements in Same Right Coset iff Product with Inverse in Subgroup
== Sources ==
* : $\S 42.6 \ \text {(4L)}$ Another approach to cosets
Category:Cosets
Elements in Same Right Coset iff Product with Inverse in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Then:
: $x, y$ are in the same right coset of $H$ $x y^{-1} \in H$
== Proof ==
Let $H x$ denote the right coset of $H$ by $x$.
First we note that, from Congruence Class Modulo Subgroup is Coset, we have that the right cosets of $G$ form a partition of $G$.
=== Sufficient Condition ===
Suppose $x, y$ are in the same left coset of $H$.
It follows from Congruence Class Modulo Subgroup is Coset that:
: if $x \in H y$ and $y \in H x$ iff $H x = H y$
From Cosets are Equal iff Product with Inverse in Subgroup, we have that:
: $H x = H y \iff x y^{-1} \in H$
So if $x, y$ are in the same right coset of $H$ then $x y^{-1} \in H$.
=== Necessary Condition ===
Now suppose that $x y^{-1} \in H$.
From Right Cosets are Equal iff Product with Inverse in Subgroup, we have that:
: $H x = H y \iff x y^{-1} \in H$
It follows from Congruence Class Modulo Subgroup is Coset that:
: $x \in H y$ and $y \in H x$ iff $H x = H y$
and so:
: $x, y$ are in the same left coset of $H$
Hence $x, y$ are in the same right coset of $H$ if $x y^{-1} \in H$.
== Also see ==
* Elements in Same Left Coset iff Product with Inverse in Subgroup
== Sources ==
* : $\S 42.6 \ \text {(4R)}$ Another approach to cosets
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $7 \ \text{(ii)}$
Category:Cosets
Elements of 5th Cyclotomic Ring with Field Norm 1

== Theorem ==
Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
The only elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose field norm equals $1$ are the units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$.
== Proof ==
From Units of 5th Cyclotomic Ring, $1$ and $-1$ are the only units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z \in \Z \sqbrk {i \sqrt 5}$ such that $\map N z = 1$.
Let $z = x + i y$.
Then:
The result follows.
== Sources ==
* : Chapter $9$: Rings: Exercise $19 \ \text {(i)}$
Category:Cyclotomic Rings
Elements of Abelian Group whose Order Divides n is Subgroup

== Theorem ==
Let $G$ be an abelian group whose identity element is $e$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer .
Let $G_n$ be the subset of $G$ defined as:
:$G_n = \set {x \in G: \order x \divides n}$
where:
:$\order x$ denotes the order of $x$
:$\divides$ denotes divisibility.
Then $G_n$ is a subgroup of $G$.
== Proof ==
From Identity is Only Group Element of Order 1:
:$\order e = 1$
and so from One Divides all Integers:
:$\order e \divides n$
Thus $G_n \ne \O$.
Then:
Let $a, b \in G_n$ such that $\order a = r, \order b = s$.
Then from Order of Product of Abelian Group Elements Divides LCM of Orders of Elements:
:$\order {a b} = \divides \lcm \set {a, b}$
But $r \divides n$ and $s \divides n$ by definition of $G_n$.
Therefore, by definition of lowest common multiple:
:$\order {a b} \divides n$
Thus we have:
:$G_n \ne \O$
:$x \in G_n \implies x^{-1} \in G_n$
:$a, b \in G_n \implies a b \in G_n$
and the result follows by the Two-Step Subgroup Test.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 41 \epsilon$
Category:Order of Group Elements
Category:Subgroups
Category:Abelian Groups
Elements of Cross-Relation Equivalence Class

== Theorem ==
Let $\eqclass {\tuple {x, y} } \boxtimes$ be the $\boxtimes$-equivalence class of $\tuple {x, y}$, where $\tuple {x, y} \in S_1 \times S_2$.
Then:
$\forall x, y \in S_1, a, b \in S_2:$
:$(1): \quad \eqclass {\tuple {x \circ a, a} } \boxtimes = \eqclass {\tuple {y \circ b, b} } \boxtimes \iff x = y$
:$(2): \quad \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$
== Proof ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Cross-Relations
Elements of Finite Support form Submagma of Direct Product

== Theorem ==
Let $\struct {S_i, \circ_i}_{i \mathop \in I}$ be a family of magmas with identity.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be their direct product.
Let $T$ be the subset of elements of $S$ whose support is finite:
:$T = \set {s \in S: \map \supp s \text{ is finite} }$
Then $T$ is a submagma of $S$.
== Proof ==
From Finite Subsets form Ideal, the set of finite subsets of $I$ form an ideal of $I$.
From Elements with Support in Ideal form Submagma of Direct Product, $T$ is a submagma of $S$.
Category:Direct Products
Elements of Geometric Sequence between Coprime Numbers

== Theorem ==
Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n + 1$.
Let $a_0$ be coprime to $a_n$.
Then there exist geometric sequences of integers $Q_1$ and $Q_2$ of length $n + 1$ such that:
:the initial term of both $Q_1$ and $Q_2$ is $1$
:the final term of $Q_1$ is $a_0$
:the final term of $Q_2$ is $a_n$.
== Proof ==
Let the common ratio of $P$ be $r$.
By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $P$ is given by:
:$a_j = q^j p^{n - j}$
such that:
:$a_0 = p^n$
:$a_n = q^n$
For $j \in \set {0, 1, \ldots, n}$, let $r_j = q^j$.
Let the finite sequence $Q_2 = \sequence {t_j}_{0 \mathop \le j \mathop \le n}$ be defined as:
:$\forall j \in \set {0, 1, \ldots, n}: t_j = p^{n - j}$
Then $Q_1$ is a geometric sequence of integers of length $n$ such that:
:$t_0 = 1$
:$t_n = p^n$
Let the finite sequence $Q_1 = \sequence {s_j}_{0 \mathop \le j \mathop \le n}$ be defined as:
:$\forall j \in \set {0, 1, \ldots, n}: s_j = q^j$
Then $Q_1$ is a geometric sequence of integers of length $n$ such that:
:$s_0 = 1$
:$s_n = q^n$
Hence the result.
== Sources ==
* : Book $\text{VIII}$. Propositions
Category:Geometric Sequences of Integers
Elements of Geometric Sequence from One Divisible by Prime

== Theorem ==
Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $p$ be a prime number such that:
:$p \divides a_n$
where $\divides$ denotes divisibility.
Then $p \divides a_1$.
== Proof ==
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as:
:$a_j = q^j$
for some $q \in \Z$.
Thus by hypothesis:
:$p \divides q^n$
From Euclid's Lemma for Prime Divisors: General Result:
:$p \divides q$
Hence the result.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences
Elements of Geometric Sequence from One where First Element is Power of Number

== Theorem ==
Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $m \in \Z_{> 0}$.
Let $a_1$ be the $m$th power of an integer.
Then all the terms of $G_n$ are $m$th powers of integers.
== Proof ==
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as:
:$a_j = q^j$
for some $q \in \Z$.
Let $a_2 = k^m$.
By definition of geometric sequence:
:$\forall j \in \set {1, 2, \ldots, n}: a_j = r a_{j - 1}$
where $r$ is the common ratio.
This holds specifically for $j = 1$:
:$k^m = r \cdot 1$
Thus:
Hence the result.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences
Elements of Geometric Sequence from One where First Element is not Power of Number

== Theorem ==
Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $k \in \Z_{> 1}$.
Let $a_1$ not be a power of $k$.
Then $a_m$ is not a power of $k$ except for:
:$\forall m, k \in \set {1, 2, \ldots, n}: k \divides m$
where $\divides$ denotes divisibility.
== Proof ==
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as:
:$a_j = q^j$
for some $q \in \Z$.
Let $k \nmid m$.
Then by the Division Theorem there exists a unique $q \in \Z$ such that:
:$m = k q + b$
for some $b$ such that $0 < b < k$.
Thus:
:$a_m = a^{k q} a^b$
which is not a power of $k$.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences
Elements of Geometric Sequence from One which Divide Later Elements

== Theorem ==
Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $m \in \Z_{> 0}$.
Then:
:$\forall r \in \set {0, 1, \ldots, m}: a_k \divides a_m$
where $\divides$ denotes divisibility.
=== Porism ===
== Proof ==
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as:
:$a_j = q^j$
for some $q \in \Z$.
Hence the result from Divisors of Power of Prime.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences
Elements of Geometric Sequence from One which are Powers of Number

== Theorem ==
Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Then:
:$\forall m, k \in \set {1, \ldots, n}: k \divides m \implies a_m$ is a power of $k$
where $\divides$ denotes divisibility.
== Proof ==
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as:
:$a_j = q^j$
for some $q \in \Z$.
Let $k, m \in \set {1, 2, \ldots, n}$ such that $k \divides m$.
By definition of divisibility:
:$\exists r \in \Z: m = r k$
Then:
That is, $a_m$ is a power of $k$.
== Sources ==
* : Book $\text{IX}$. Propositions
Category:Geometric Sequences of Integers
Elements of Group with Equal Images under Homomorphisms form Subgroup

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $f: G \to H$ and $g: G \to H$ be group homomorphisms.
Then the set:
:$S = \set {x \in G: \map f x = \map g x}$
is a subgroup of $G$.
== Proof ==
Let the identities of $\struct {G, \circ}$ and $\struct {H, *}$ be $e_G$ and $e_H$ respectively.
By Homomorphism to Group Preserves Identity:
:$\map f {e_G} = \map g {e_G} = e_H$
Thus $e_G \in S$, and so $S \ne \O$.
Similarly, from Homomorphism to Group Preserves Inverses, $x \in S \implies x^{-1} \in S$.
Let $x, y \in S$. Then:
Thus $x \circ y \in S$.
So, by the Two-Step Subgroup Test:
:$S \le G$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.8$
Category:Group Homomorphisms
Elements of Inverse of Hilbert Matrix are Integers

== Theorem ==
Let $H_n$ be the Hilbert matrix of order $n$:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Consider its inverse $H_n^{-1}$.
All the elements of $H_n^{-1}$ are integers.
== Proof ==
From Inverse of Hilbert Matrix, $H_n^{-1} = \begin {bmatrix} b \end{bmatrix}_n$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j}! \paren {i + j - 1} } \end{bmatrix}$
Thus:
All of the factors of the above expression are integers, from Binomial Coefficient is Integer.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $45$
Category:Hilbert Matrices
Elements of Minimally Inductive Set are Well-Ordered

== Theorem ==
Let $\omega$ be the minimally inductive set.
Let $a \in \omega$.
Then $a$ is well-ordered by $\subseteq$.
== Proof ==
== Sources ==
* : $\S 17$: Well Ordering
Category:Total Orderings
Category:Well-Orderings
Category:Minimally Inductive Set
Elements of Module with Equal Images under Linear Transformations form Submodule

== Theorem ==
Let $G$ and $H$ be $R$-modules.
Let $\phi$ and $\psi$ be linear transformations from $G$ into $H$.
Then the set $S = \set {x \in G: \map \phi x = \map \psi x}$ is a submodule of $G$.
== Proof ==
Let $x, y \in S$.
Let $\lambda \in R$.
Then:
Hence $x + y, \lambda \circ x \in S$.
By Submodule Test, $S$ is a submodule of $G$.
== Also see ==
* Elements of Group with Equal Images under Homomorphisms form Subgroup
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.3$
Category:Linear Transformations
Elements of Ordered Pair do not Commute

== Theorem ==
Let $\set {a, b}$ be a doubleton, so that $a$ and $b$ are distinct objects.
Let $\tuple {a, b}$ denote the ordered pair such that the first coordinate is $a$ and the second coordinate is $b$.
Then:
:$\tuple {a, b} \ne \tuple {b, a}$
== Proof ==
By the Kuratowski formalization of $\tuple {a, b}$:
:$\tuple {a, b} = \set {\set a, \set {a, b} }$
and by Equality of Ordered Pairs:
:$\tuple {a, b} = \tuple {b, a} \iff a = b$
But $a \ne b$ and so:
:$\tuple {a, b} \ne \tuple {b, a}$
== Sources ==
* : $\S 3$. Ordered pairs; cartesian product sets: Theorem $3.1$: Corollary
Category:Ordered Pairs
Elements of Primitive Pythagorean Triple are Pairwise Coprime

== Theorem ==
Let $\tuple {x, y, z}$ be a primitive Pythagorean triple.
Then:
:$x \perp y$
:$y \perp z$
:$x \perp z$
That is, all elements of $\tuple {x, y, z}$ are pairwise coprime.
== Proof ==
We have that $x \perp y$ by definition.
Suppose there is a prime divisor $p$ of both $x$ and $z$.
That is:
:$\exists p \in \mathbb P: p \divides x, p \divides z$
Then from Prime Divides Power:
:$p \divides x^2, p \divides z^2$
Then from Common Divisor Divides Integer Combination:
:$p \divides \paren {z^2 - x^2} = y^2$
So from Prime Divides Power again:
:$p \divides y$
and:
:$x \not \perp y$
This contradicts our assertion that $\tuple {x, y, z}$ is a primitive Pythagorean triple.
Hence $x \perp z$.
The same argument shows that $y \perp z$.
Category:Pythagorean Triples
Category:Coprime Integers
Elements of Primitive Pythagorean Triples Modulo 4

== Theorem ==
Let $x \in \Z: x > 2$.
Then $x$ is an element of some primitive Pythagorean triple $x \not \equiv 2 \pmod 4$.
=== Corollary ===
== Proof ==
Let $m = k + 1, n = k$ where $k \in \Z: k \ge 1$.
From Consecutive Integers are Coprime, $m \perp n$.
Then we have:
:$m, n \in \Z$ are positive integers
:$m \perp n$, that is, $m$ and $n$ are coprime
:$m$ and $n$ are of opposite parity
:$m > n$.
From Solutions of Pythagorean Equation, these conditions are necessary and sufficient for $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ to be a primitive Pythagorean triple.
Substituting for $m$ and $n$, we get:
:$\tuple {2 k^2 + 2 k, 2 k + 1, 2 k^2 + 2 k + 1}$ is a primitive Pythagorean triple.
So we see that for all $k \ge 1$, $2 k + 1$ is an element of a primitive Pythagorean triple.
So every odd integer from $3$ upwards is an element of some primitive Pythagorean triple.
That is, any integer $x$ such that $x \equiv 1$ or $x \equiv 3 \pmod 4$.
Now, consider $m = 2 k, n = 1$ where $k \in \Z: k \ge 1$.
These also fit the criteria for $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ to be a primitive Pythagorean triple.
Substituting for $m$ and $n$, we get:
:$\tuple {4 k, 4 k^2 - 1, 4 k^2 + 1}$ is a primitive Pythagorean triple.
This means that for all $k \ge 1$, $4 k$ is an element of a primitive Pythagorean triple.
So every multiple of $4$ is an element of some primitive Pythagorean triple.
That is, any integer $x$ such that $x \equiv 0 \pmod 4$.
Now, consider the general primitive Pythagorean triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$.
As $m$ and $n$ are of opposite parity, they can be expressed as $2 r$ and $2 s + 1$.
So:
:$2 m n = 2 \paren {2 r} \paren {2 s + 1} = 4 r \paren {2 s + 1}$
So $2 m n$ is divisible by $4$.
As the only even elements of a primitive Pythagorean triple are of this form $2 m n$ from Parity of Smaller Elements of Primitive Pythagorean Triple, there can be no such elements $x$ of the form $x = 2 \pmod 4$.
Hence the result.
Category:Pythagorean Triples
Elements of Primitive Pythagorean Triples Modulo 4/Corollary

== Corollary to Elements of Primitive Pythagorean Triples Modulo 4 ==
In every Pythagorean triple, at least one element is a multiple of $4$.
== Proof ==
From Solutions of Pythagorean Equation, any Pythagorean triple is such that all elements are a multiple of the elements in some primitive Pythagorean triple.
As the latter must have an even element which is a multiple of $4$, such an element when multiplied by some integer will still be a multiple of $4$.
Category:Pythagorean Triples
Elements of Semigroup with Equal Images under Homomorphisms form Subsemigroup

== Theorem ==
Let $\struct {A, \circ}$ and $\struct {B, *}$ be semigroups.
Let $f: A \to B$ and $g: A \to B$ be semigroup homomorphisms.
Then the set:
:$S = \set {x \in A: \map f x = \map g x}$
is a subsemigroup of $A$.
== Proof ==
Let $x, y \in A$. Then:
Thus $x \circ y \in A$.
So, by the Subsemigroup Closure Test:
:$S$ is a subsemigroup of $A$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.13$
Category:Semigroup Homomorphisms
Elements of Subgroup of Dipper Semigroup are not Invertible in Dipper

== Theorem ==
Let $m, n \in \Z$ be integers such that $m, n > 0$.
Let $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ denote the dipper semigroup.
Let $\struct {H, +_{m, n} }$ be the subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ where $H = \set {k \in \N: m \le k < m + n}$
Then the elements of $\struct {H, +_{m, n} }$ are not invertible in $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
== Proof ==
From Identity of Subgroup of Dipper Semigroup is not Identity of Dipper, the identity of $\struct {H, +_{m, n} }$ is not an identity of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
Hence (indirectly) from Identity of Submagma containing Identity of Magma is Same Identity, $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ can have no identity.
Hence there can be no elements of $\struct {H, +_{m, n} }$ that are invertible, by definition.
== Examples ==
Category:Dipper Semigroups
Category:Examples of Inverse Elements
Category:Elements of Subgroup of Dipper Semigroup are not Invertible in Dipper
Elements of Subgroup of Dipper Semigroup are not Invertible in Dipper/Examples/(m, n) = (3, 4)

== Examples of Use of Elements of Subgroup of Dipper Semigroup are not Invertible in Dipper ==
Consider the dipper semigroup $\struct {N_{<7}, +_{3, 4} }$.
Let $H = \set {x \in \N: 3 \le x < 7} = \set {3, 4, 5, 6}$.
From Existence of Subgroup of Dipper Semigroup Example: $\struct {H, +_{3, 4} }$
:$\struct {H, +_{3, 4} }$ is a subgroup of $\struct {N_{<7}, +_{3, 4} }$
whose identity is $4$.
But from Elements of Subgroup of Dipper Semigroup are not Invertible in Dipper, $\struct {H, +_{3, 4} }$ has no invertible elements in $\struct {N_{<7}, +_{3, 4} }$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.4$
Category:Elements of Subgroup of Dipper Semigroup are not Invertible in Dipper
Elements of Submodule form Subgroup

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {G, +_G}$ be an abelian group.
Let $\struct {G, +_G, \circ_G}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ_H}_R$ be a submodule of $\struct {G, +_G, \circ_G}_R$.
Then $\struct {H, +_H}$ is a subgroup of $\struct {G, +_G}$.
== Proof ==
By definition of submodule, $\struct {H, +_H}$ is an abelian group.
The result follows by definition of subgroup.
Category:Submodules
Elements of Unital C*-Subalgebra of C*-Algebra is Positive in Subalgebra iff Positive in whole Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra with identity element ${\mathbf 1}_A$.
Let $B \subseteq A$ be a unital $\text C^\ast$-subalgebra of $A$. 
Let $x \in B$.
Then $x$ is positive in $B$ it is positive in $A$.
== Proof ==
Since $\ast$ on $B$ is obtained by restriction, we have that $x$ is Hermitian in $B$ it is Hermitian in $A$.
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra, we have:
:$\map {\sigma_B} x = \map {\sigma_A} x$
Hence $\map {\sigma_B} x \subseteq \hointr 0 \infty$ $\map {\sigma_A} x \subseteq \hointr 0 \infty$.
Hence $x$ is positive in $B$ it is positive in $A$.
Category:Positive Elements of C*-Algebras
Elements with Support in Ideal form Submagma of Direct Product

== Theorem ==
Let $\family {S_i, \circ_i}_{i \mathop \in I}$ be a family of magmas with identity.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be their direct product.
Let $J \subset I$ be an ideal of $I$.
Let $T = \set {s \in S: \map \supp s \in J}$ where $\supp$ denotes support.
Then $T$ is a submagma of $S$.
== Proof ==
== Also see ==
* Elements of Finite Support form Submagma of Direct Product
Category:Abstract Algebra
Eliminant/Examples/Arbitrary Example 1

== Example of Eliminant ==
Consider the system of simultaneous linear equations:
The eliminant of the above is:
:$\begin {vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end {vmatrix}$
== Sources ==
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Category:Examples of Eliminants
Elimination/Examples/Arbitrary Example 1

== Example of Elimination ==
Consider the system of simultaneous linear equations:
This can be solved by multiplying $(2)$ by $3$ and then subtracting $(1)$ from the resulting equation:
Hence $y$ has been eliminated, and we have:
:$x = 4$
from which we can substitute for $x$ in either $(1)$ or $(2)$ and obtain $y = 1$.
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Category:Examples of Elimination
Elimination/Examples/Arbitrary Example 2

== Example of Elimination ==
Consider the system of simultaneous linear equations:
This can be solved by putting $(1)$ in the form:
:$x = 7 - 3 y$
from which we can substitute for $x$ in either $(2)$ to obtain:
:$2 \paren {7 - 3 y} + y = 9$
from which it follows that $y = 1$ and $x = 4$.
== Sources ==
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Category:Examples of Elimination
Elimination of Constants by Partial Differentiation

== Theorem ==
Let $x_1, x_2, \dotsc, x_m$ be independent variables.
Let $c_1, c_2, \dotsc, c_n$ be arbitrary constants.
Let this equation:
:$(1): \quad \map f {x_1, x_2, \dotsc, x_m, z, c_1, c_2, \dotsc, c_n} = 0$
define a dependent variable $z$ via the implicit function $f$.
Then it may be possible to eliminate the constants by successive partial differentiation of $(1)$.
== Proof ==
We differentiate $(1)$ partially each of $x_j$ for $1 \le j \le m$:
:$(2): \quad \dfrac {\partial f} {\partial x_j} + \dfrac {\partial f} {\partial z} \cdot \dfrac {\partial z} {\partial x_j} = 0$
Suppose $m \ge n$.
Then there exist sufficient equations of the form of $2$ for the constants $c_1, c_2, \dotsc, c_n$ to be eliminated.
Otherwise suppose $m > n$.
Then we differentiate $(2)$ partially for each $j$ each of $x_j$ for $1 \le j \le m$, to obtain:
and continue the operation until enough equations have been obtained so that the remaining constants can be eliminated.
== Sources ==
* : Chapter $\text I$: Introductory: $\S 1.21$ Formation of Partial Differential Equations through the Elimination of Arbitrary Constants
Category:Partial Differentiation