nyuuzyou/wikis huggingface dataset:
Subalgebra Generated by Inclusion and Conjugate is Everywhere Dense in Space of Continuous Functions on Compact Subset of Complex Numbers

== Theorem ==
Let $K \subseteq \C$ be compact. 
Let $\map \CC K$ be the space of continuous functions on $K$.
Let $\iota : K \to \C$ be the inclusion mapping.
Let $\overline \iota : K \to \C$ be the complex conjugate.
Let $\AA$ be the subalgebra generated by $1$, $\iota$ and $\overline \iota$.
That is, by Explicit Form for Generated Subalgebra:
:$\ds \AA = \set {\sum_{j \mathop = 1}^n \alpha_j z^{n_j} \overline z^{m_j} : n \in \N, \, \alpha_j \in \C, \, n_j, m_j \ge 0}$
Then $\AA$ is everywhere dense in $\map \CC K$. 
== Proof ==
From Subalgebra Generated by Self-Adjoint Set is Self-Adjoint, $\AA$ is a $\ast$-subalgebra of $\map \CC K$.
We use the Complex-Valued Stone-Weierstrass Theorem: Compact Space.
We clearly have $1, \iota \in \AA$ by the definition of the generated algebra.
We have $\map \iota z = \map \iota w$ for $z, w \in K$ with $z \ne w$. 
Hence $\AA$ separates points. 
Hence from the Complex-Valued Stone-Weierstrass Theorem: Compact Space, $\AA$ is dense in $\map \CC K$.
Category:Space of Continuous Functions on Compact Hausdorff Space
Subalgebra Generated by Self-Adjoint Set is Self-Adjoint

== Theorem ==
Let $\tuple {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $S \subseteq A$ be self-adjoint.
Let $K \sqbrk S$ be the subalgebra generated by $S$.
Then $\C \sqbrk S$ is a $\ast$-subalgebra of $A$.
== Proof ==
From Explicit Form for Generated Subalgebra, we have:
:$\C \sqbrk S = \map \span \AA$
where:
:$\AA = \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
From Star of Product of Elements in *-Algebra we have:
:$\paren {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} }^\ast = x_n^{k_n} \ldots x_2^{k_2} x_1^{k_1} \in S$
Hence $\AA$ is self-adjoint.
Then $\C \sqbrk S$ is self-adjoint from Linear Span of Self-Adjoint Subset of *-Algebra is Self-Adjoint.
Hence $\C \sqbrk S$ is a $\ast$-subalgebra of $A$.
Category:*-Algebras
Subalgebra of Algebraic Field Extension is Field

== Theorem ==
Let $E / F$ be an algebraic field extension.
Let $A \subseteq E$ be a unital subalgebra over $F$.
Then $A$ is a field.
== Proof ==
By Integral Ring Extension is Integral over Intermediate Ring, $E$ is integral over $A$.
Let $a \in A$ be nonzero.
Because $E$ is a field, $a$ is a unit of $E$.
By Ring Element is Unit iff Unit in Integral Extension, $a$ is a unit of $A$.
Thus $A$ is a field.
== Also see ==
=== Weaker statement ===
* Subalgebra of Finite Field Extension is Field
Category:Integral Ring Extensions
Subalgebra of Finite Field Extension is Field

== Theorem ==
Let $E / F$ be an algebraic field extension.
Let $A \subseteq E$ be a unital subalgebra over $F$.
Then $A$ is a field.
== Proof 1 ==
By Subring of Integral Domain is Integral Domain, $A$ is an integral domain.
By Subspace of Finite Dimensional Vector Space is Finite Dimensional, $A$ is finite dimensional over $F$.
By Finite-Dimensional Integral Domain over Field is Field, $A$ is a field.
== Proof 2 ==
By Finite Field Extension is Algebraic, $E/F$ is algebraic.
The result follows from Subalgebra of Algebraic Field Extension is Field
Category:Integral Ring Extensions
Subband iff Idempotent under Induced Operation

== Theorem ==
Let $\struct {S, \circ}$ be a band.
Let $\struct {\powerset S, \circ_\PP}$ be the algebraic structure consisting of the power set of $S$ and the operation induced on $\powerset S$ by $\circ$.
Let $X \in \powerset S$.
Then $X$ is idempotent $\struct {X, \circ}$ is a subband of $\struct {S, \circ}$.
== Proof ==
=== Subbandhood implies Idempotency ===
==== Proving $\paren {X \circ_PP X} \subseteq X$ ====
Let $c \in X \circ_\PP X$.
By definition of subset product for some $a, b \in X$ we have:
:$a \circ b = c$
Suppose $c \notin X$.
Then:
:$a \circ b \notin X$
which contradicts that $\struct {X, \circ}$ is a subband.
==== Proving $X \subseteq \paren {X \circ_\PP X}$ ====
Let $a \in X$.
By definition of subset product:
:$X \circ_\PP X = \set {a \circ b: a, b \in X}$
As $\circ$ is idempotent:
:$a \circ a = a$.
Thus:
:$a \in \set {X \circ_\PP X}$
Hence by definition of subset:
:$X \subseteq \paren {X \circ_\PP X}$
=== Idempotency implies Subbandhood ===
Let $X \in \powerset S$.
Suppose $X$ is idempotent:
That is suppose:
:$X \circ_\PP X = X$
Let $a, b \in X$.
By the definition of subset product:
:$X \circ_\PP X = \set {a \circ b: a, b \in X}$
Then $a \circ b \in X$.
Hence $\struct {X, \circ}$ is a magma.
By Restriction of Associative Operation is Associative it is a semigroup.
Finally by Restriction of Idempotent Operation is Idempotent it is a band.
Category:Idempotent Semigroups
Subband of Induced Operation is Set of Subbands

== Theorem ==
Let $\struct {S, \circ}$ be a band.
Let $\struct {\powerset S, \circ_\PP}$ be the algebraic structure consisting of:
: the power set $\powerset S$ of $S$
and
: the operation $\circ_\PP$ induced on $\powerset S$ by $\circ$.
Let $T \subseteq \powerset S$.
Let $\struct {T, \circ_\PP}$ be a subband of $\struct {\powerset S, \circ_\PP}$.
Then every element of $T$ is a subband of $\struct {S, \circ}$.
== Proof ==
=== Case 1: $T$ is the Empty Set ===
By:
: Empty Set is Submagma of Magma
: Restriction of Associative Operation is Associative
: Restriction of Idempotent Operation is Idempotent
it follows that $\struct {\O, \circ_\PP}$ is a subband of $\struct {T, \circ_\PP}$.
Let $X \in \O$.
Then by the definition of the empty set it follows that $\struct {X, \circ}$ is a subband of $\struct {S, \circ}$ vacuously.
=== Case 2: $T$ is Non-Empty ===
Let $X \in T$.
Then by definition of a subband $X$ is idempotent under $\circ_\PP$.
That is:
:$X \circ_\PP X = X$
By Subband iff Idempotent under Induced Operation we have that $\struct {X, \circ}$ is a subband of $\struct {S, \circ}$.
Hence the result.
Category:Idempotent Semigroups
Subclass of Set is Set

== Theorem ==
Let $A$ be a set.
Let $\map \phi x$ be a condition in which $x$ is taken to be a set.
Then there exists a set that consists of all of the elements of $A$ that satisfies this condition.
In ZF, this result is known as the Axiom of Specification.
== Proof ==
By the axiom of class comprehension, let $B$ be the class defined as:
 that $B$ is not a set.
Then $B$ must be a proper class.
It is easily seen that $B \subseteq A$.
So by the Axiom of Powers:
:$B \in \powerset A$
where $\powerset A$ is denotes the power set of $A$.
But by Proper Class is not Element of Class, this is a contradiction.
Therefore by Proof by Contradiction it follows that $B$ is a set.
== Sources ==
* : Chapter $1$: Separation Schema
Category:Gödel-Bernays Class Theory
Subclass of Subclass is Subclass

== Theorem ==
Let $A$, $B$ and $C$ be classes.
Let $A$ be a subclass of $B$.
Let $B$ be a subclass of $C$.
Then $A$ is a subclass of $C$.
== Proof ==
Let $x \in A$ be arbitrary.
It follows by definition of subclass that $x \in B$.
It further follows by definition of subclass that $x \in C$.
So we have that $x \in A$ implies that $x \in C$.
As $x$ is arbitrary, the result follows.
Category:Subclasses
Subclass of Well-Ordered Class is Well-Ordered

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $A$ be a subclass of $V$ which is well-ordered under $\RR$.
Let $B$ be a subclass of $A$.
Then $B$ is also well-ordered under $\RR$.
== Proof ==
First suppose $B$ is the empty class.
From Empty Class is Subclass of All Classes, $B$ is a subclass of $A$.
Then by Empty Class is Well-Ordered, $\O$ is well-ordered under $\RR$.
Otherwise, let $X$ be an arbitrary non-empty class subclass of $B$.
By Subclass of Subclass is Subclass, $X$ is a subclass of $A$.
Hence by definition of well-ordered class, $X$ has a smallest element under $\RR$.
As $X$ is arbitrary, it follows that every non-empty class subclass of $B$ has a smallest element under $\RR$.
Hence the result by definition of well-ordered class.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.1$
Category:Well-Orderings
Category:Subclasses
Subcover is Refinement of Cover

== Theorem ==
Let $S$ be a set.
Let $\CC$ be a cover for $S$.
Let $\VV$ be a subcover of $\CC$.
Then $\VV$ is a refinement of $\CC$.
=== Corollary ===
== Proof ==
From definition of subcover:
:$\VV \subseteq \CC$
That is, every element of $\VV$ is an element of $\CC$.
From definition of subset, every element of $\VV$ is the subset of some element of $\CC$.
This is precisely the definition of refinement.
Category:Subcovers
Category:Covers
Subcover is Refinement of Cover/Corollary

== Corollary to Subcover is Refinement of Cover ==
Let $T = \struct {X, \tau}$ be a topological space.
Let $\UU$ be an open cover for $S$.
Let $\VV$ be a subcover of $\UU$.
Then $\VV$ is an open refinement of $\UU$.
== Proof ==
By definition of open cover and subcover:
:$\VV \subseteq \UU \subseteq \tau$
As all the elements of $\UU$ are open, all the elements of $\VV$ are likewise open.
Hence the result from definition of open refinement.
Category:Subcovers
Category:Covers
Subdomain Test

== Theorem ==
Let $S$ be a subset of an integral domain $\struct {R, +, \circ}$.
Then $\struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subdomain of $\struct {R, +, \circ}$ these conditions hold:
:$(1): \quad \struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subring of $\struct {R, +, \circ}$
:$(2): \quad$ The unity of $R$ is also in $S$, that is $1_R = 1_S$.
== Proof ==
By Idempotent Elements of Ring with No Proper Zero Divisors, it follows that the unity of a subdomain is the unity of the integral domain it's a subdomain of.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.4$
Category:Integral Domains
Category:Subrings
Subextensions of Separable Field Extension are Separable

== Theorem ==
Let $E / K / F$ be a tower of fields.
Let $E / F$ be separable.
Then $E / K$ and $K / F$ are separable.
== Proof ==
=== Upper extension ===
We prove that $E / K$ is separable.
Let $\alpha \in E$.
Let $f$ be its minimal polynomial over $F$.
Let $g$ be its minimal polynomial over $K$.
Then , $f$ is separable.
On the other hand:
:$f \in K \sqbrk x$
and:
:$\map f \alpha = 0$
Hence by definition $g$ divides $f$.
Consider the roots of $f$ and $g$ in the algebraic closure of $K$.
Since $f$ is separable, it has no repeated roots.
Since $g$ divides $f$, the roots of $g$ are included in the roots of $f$,
we see that $g$ has no repeated roots in the algebraic closure of $K$,
so $g$ is separable.
=== Lower extension ===
It follows immediately by definition that $K / F$ is a separable extension.
== Also see ==
* Transitivity of Separable Field Extensions, the converse
== Source ==
* : Chapter V, $\S4$ Separable Extensions, the paragraph before Theorem $4.3$
Category:Separable Field Extensions
Subfactorial/Examples/5/Mistake

== Source Work ==
:
:The Dictionary
::$44$
== Mistake ==
:Subfactorial $5 = 5! \paren {1 - 1 / {1!} + 1 / {2!} - 1 / {3!} - 1 / {4!} - 1 / {5!} } = 44$
== Correction ==
There is an incorrect sign.
It should read:
:Subfactorial $5 = 5! \paren {1 - 1 / {1!} + 1 / {2!} - 1 / {3!} + 1 / {4!} - 1 / {5!} } = 44$
This typo does not exist in the first edition of .
== Sources ==
* : $44$
Category:Mistakes/Curious and Interesting Numbers
Subfield Test/Four-Step

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $K$ be a subset of $F$.
$\struct {K, +, \circ}$ is a subfield of $\struct {F, +, \times}$ these all hold:
:$(1): \quad K^* \ne \O$
:$(2): \quad \forall x, y \in K: x + \paren {-y} \in K$
:$(3): \quad \forall x, y \in K: x \times y \in K$
:$(4): \quad x \in K^* \implies x^{-1} \in K^*$
where $K^*$ denotes $K \setminus \set {0_F}$.
== Proof ==
=== Necessary Condition ===
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \circ}$.
Then the conditions $(1)$ to $(4)$ all hold by virtue of the field axioms.
=== Sufficient Condition ===
Suppose the conditions $(1)$ to $(4)$ hold.
From the Division Subring Test, it follows that $\struct {K, +, \times}$ is a division ring.
As $\struct {F, +, \times}$ is a field, then $\times$ is commutative on all of $F$.
From Restriction of Commutative Operation is Commutative, $\times$ is commutative also on $K$.
Thus $\struct {K, +, \times}$ is a field.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : $\S 56.3$ Subrings and subfields
Category:Subfield Test
Subfield Test/Three-Step

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $K$ be a subset of $F$.
$\struct {K, +, \times}$ is a subfield of $\struct {F, +, \times}$ these all hold:
:$(1): \quad K^* \ne \O$
:$(2): \quad \forall x, y \in K: x - y \in K$
:$(3): \quad \forall x \in K: \forall y \in K^*: \dfrac x y \in K$
where $K^*$ denotes $K \setminus \set {0_F}$.
== Proof ==
=== Necessary Condition ===
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
Then the conditions $(1)$ to $(3)$ all hold by virtue of the field axioms.
=== Sufficient Condition ===
Suppose the conditions $(1)$ to $(3)$ hold.
As $K^* \ne \O$, it follows that $K \ne \O$.
From the One-Step Subgroup Test, it follows that $\struct {K, +}$ is a subgroup of $\struct {F, +}$.
Also from the One-Step Subgroup Test, it follows that $\struct {K^*, \times}$ is a subgroup of $\struct {F^*, \times}$.
As $\struct {F, +, \times}$ is a field, then $\times$ is commutative on all of $F$.
From Restriction of Commutative Operation is Commutative, $\times$ is commutative also on $K$.
As $\struct {F, +, \times}$ is a field, then $\times$ is distributive over $+$ on all of $F$.
From Restriction of Operation Distributivity $\times$ is distributive over $+$ also on $K$.
Thus $\struct {K, +, \times}$ is a field.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88 \alpha$
* : Chapter $4$: Fields: $\S 16$. Subfields: Theorem $26$
Category:Subfield Test
Subfield of Subfield is Subfield

== Theorem ==
Let $R$ be a ring with unity.
Let $K_1, K_2$ be fields, such that:
: $K_1$ is a subfield of $R$
: $K_2$ is a subfield of $K_1$
Then $K_2$ is a subfield of $R$.
== Proof ==
Let $K_1$ be a subfield of $R$ and $K_2$ be a subfield of $K_1$.
Then by definition:
:$K_1 \subseteq R$
:$K_2 \subseteq K_1$
From Subset Relation is Transitive it follows that $K_2 \subseteq R$
So by definition $K_2$ is a subfield of $R$.
Category:Subfields
Subgroup/Examples/Matrices (1 a, 0 1) in General Linear Group

== Example of Subgroup ==
Let $\GL 2$ denote the general linear group of order $2$.
Let $H$ be the set of square matrices of the form $\begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix}$ for $a \in \R$.
Then $\struct {H, \times}$ is a subgroup of $\GL 2$, where $\times$ is used to denote (conventional) matrix multiplication.
== Proof ==
We have that the unit matrix $\begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix}$ is in $H$, so $H \ne \O$.
Then we have:
:$\forall a, b \in \R: \begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \cr 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a + b \cr 0 & 1 \end{bmatrix} \in H$
Then:
:$\forall a \in \R: \begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -a \cr 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 + b \cr 0 & 1 \end{bmatrix} \in H$
Thus $\begin{bmatrix} 1 & -a \cr 0 & 1 \end{bmatrix}$ is the inverse of $\begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix}$ and so $\begin{bmatrix} 1 & a \cr 0 & 1 \end{bmatrix}^{-1} \in H$.
Hence the result from the Two-Step Subgroup Test.
== Sources ==
* : Chapter $4$: Subgroups: Exercise $1 \ \text{(c)}$
Category:Examples of Subgroups
Category:General Linear Group
Subgroup/Examples/Natural Numbers in Multiplicative Group of Real Numbers

== Example of Closed Subset which is not a Subgroup ==
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
Consider the algebraic structure $\struct {\N_{> 0}, \times}$ formed by the non-zero natural numbers under multiplication.
Then $\struct {\N_{> 0}, \times}$ is not a subgroup of $\struct {\R_{\ne 0}, \times}$.
== Proof ==
From Non-Zero Natural Numbers under Multiplication form Commutative Monoid, $\struct {\N_{> 0}, \times}$ is a monoid whose identity is $1$.
But there exists no $x \in \N_{> 0}$ such that, for example, $x \times 2 = 1$.
So $\N_{> 0}$ is not a subgroup of $\struct {\R_{\ne 0}, \times}$.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $8$
Category:Examples of Subgroups
Subgroup Action is Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\struct {H, \circ}$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the subgroup action defined for all $h \in H, g \in G$ as:
:$\forall h \in H, g \in G: h * g := h \circ g$
Then $*$ is a group action.
== Proof ==
Let $g \in G$.
First we note that since $G$ is closed, and $h \circ g \in G$, it follows that $h * g \in G$.
Next we note:
:$e * g = e \circ g = g$
and so is satisfied.
Now let $h_1, h_2 \in G$.
We have:
and so is satisfied.
Hence the result.
== Sources ==
* : $\S 5.5$. Groups acting on sets: Example $106$
* : Chapter $2$: The Sylow Theorems: $\S 54 \alpha$
Category:Subgroup Action
Subgroup Containing all Squares of Group Elements is Normal

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$ with the property that:
:$\forall x \in G: x^2 \in H$
Then $H$ is normal in $G$.
=== Corollary ===
== Proof ==
We have:
Because $\paren {x h}^2$ and $\paren {x^{-1} }^2$ are in the form $x^2$ for $x \in G$, they are both elements of $H$.
Thus:
:$x h x^{-1} \in H$
and so $H$ is normal in $G$ by definition.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $13$
Category:Normal Subgroups
Subgroup Containing all Squares of Group Elements is Normal/Corollary

== Corollary to Subgroup Containing all Squares of Group Elements is Normal ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$ with the property that:
:$\forall x \in G: x^2 \in H$
The quotient group $G / H$ is abelian.
== Proof ==
From Subgroup Containing all Squares of Group Elements is Normal, $H$ is normal in $G$.
For every $x \in G$, we have:
Thus by definition $G / H$ is boolean.
The result follows by Boolean Group is Abelian.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $13$
Category:Normal Subgroups
Subgroup Generated by Commuting Elements is Abelian

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$ such that:
:$\forall x, y \in S: x \circ y = y \circ x$
Then the subgroup generated by $S$ is abelian.
== Proof ==
Let $H = \gen S$ denote the subgroup generated by $S$.
Let $a, b \in H$.
Then:
:$a = s_1$
:$b = s_2$
for some words $s_1, s_2$ of the set of words $\map W S$ of $S$.
Then:
Hence the result, by definition of abelian group.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $9$
Category:Abelian Groups
Subgroup Generated by Infinite Order Element is Infinite

== Theorem ==
Let $G$ be a group.
Let $a \in G$ be of infinite order.
Let $\gen a$ be the subgroup generated by $a$.
Then $\gen a$ is of infinite order.
== Proof ==
 $\gen a$ is of finite order.
We have that $a \in \gen a$ by definition.
From Element of Finite Group is of Finite Order it follows that $a$ is of finite order.
From this contradiction it follows that $\gen a$ must be of infinite order after all.
== Sources ==
* : $\S 38.2$ Period of an element
Category:Infinite Groups
Category:Order of Group Elements
Category:Generated Subgroups
Subgroup Generated by One Element is Cyclic

== Theorem ==
Let $G$ be a group.
Let $a \in G$.
Then $\gen a$, the subgroup generated by $a$, is cyclic:
== Proof ==
By Subgroup Generated by One Element is Set of Powers:
:$\gen a = \set {a^n : n \in \Z}$
The result follows by definition of cyclic group.
== Also see ==
* Definition:Cyclic Group
== Sources ==
* : $\S 39$. Cyclic groups 
Category:Generated Subgroups
Subgroup Generated by One Element is Set of Powers

== Theorem ==
Let $G$ be a group.
Let $a \in G$.
Then the subgroup generated by $a$ is the set of powers:
:$\gen a = \set {a^n : n \in \Z}$
== Proof ==
By definition, the subgroup generated by $a$ is the intersection of all subgroups containing $a$.
By Powers of Element form Subgroup, the set $H = \set {a^n : n \in \Z}$ is a subgroup.
Thus $\gen a \subseteq H$.
By Power of Element in Subgroup, $H \subseteq \gen a$.
By definition of set equality, $\gen a = H$.
== Also see ==
* Definition:Cyclic Group
Category:Generated Subgroups
Subgroup Generated by Subgroup

== Theorem ==
Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Then:
:$H = \gen H$
where $\gen H$ denotes the subgroup generated by $H$.
== Proof ==
By definition of generated subgroup, $\gen H$ is the smallest subgroup of $H$ containing $H$.
Hence the result.
== Sources ==
* : $\S 5.3$. Subgroup generated by a subset: Example $97$
Category:Generated Subgroups
Subgroup Product is Internal Group Direct Product iff Surjective

== Theorem ==
Let $G$ be a group.
Let $\sequence {H_n}$ be a sequence of subgroups of $G$.
Let $\ds \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a mapping defined by:
:$\ds \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
Then $\phi$ is surjective :
:$\ds G = \prod_{k \mathop = 1}^n H_k$
That is, $G$ is the internal group direct product of $H_1, H_2, \ldots, H_n$.
== Proof ==
=== Necessary Condition ===
Let $\phi$ be a surjection.
Then $\Img \phi$ consists of all the products $\ds \prod_{k \mathop = 1}^n h_k$ such that:
:$\forall k \in \closedint 1 n: h_k \in H_k$
Thus, as $\phi$ is surjective, every element of $G$ must be representable in this form.
Using the notation:
:$\ds \prod_{k \mathop = 1}^n H_k = \set {\prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k}$
the result follows immediately.
Thus every element of $G$ is the product of an element of each of $H_k$.
=== Sufficient Condition ===
Let $\ds G = \prod_{k \mathop = 1}^n H_k$.
If this is the case, then $g$ can be written as:
:$\ds g = \prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k$
Thus:
:$g = \map \phi {h_1, h_2, \ldots, h_n}$
and $\phi$ is surjective.
Category:Internal Group Direct Products
Subgroup Subset of Subgroup Product

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$H \subseteq H \circ K \supseteq K$
where $H \circ K$ denotes the subset product of $H$ and $K$.
== Proof ==
By definition of subset product:
:$H \circ K = \set {h \circ k: h \in H, k \in K}$
So:
and:
Category:Subset Products
Category:Subgroups
Subgroup equals Conjugate iff Normal

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) :
:$\forall g \in G: g \circ N \circ g^{-1} = N$
:$\forall g \in G: g^{-1} \circ N \circ g = N$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Equivalence of Definitions of Normal Subgroup
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $124$
* : $\S 1.10$: Theorem $26$: Corollary
* : Chapter $\text{II}$: Groups: Morphisms
Category:Conjugacy
Category:Normal Subgroups
Category:Subgroup equals Conjugate iff Normal
Subgroup equals Conjugate iff Normal/Proof 1

== Theorem ==
== Proof ==
By definition, a subgroup is normal in $G$ :
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} = N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g = N}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessary Condition ===
Suppose that $N$ is normal in $G$.
Then:
Similarly:
=== Sufficient Condition ===
Let $N$ be a subgroup of $G$ such that:
: $\forall g \in G: g \circ N \circ g^{-1} = N$
and so from $(1)$ above:
: $\forall g \in G: g^{-1} \circ N \circ g = N$
Then:
== Sources ==
* : $\S 6.6$. Normal subgroups
* : $\S 11$: Theorem $11.2: \ 2^\circ, 3^\circ$
* : $\text{II}$: Morphisms
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.4 \ \text{(c)}$
Category:Subgroup equals Conjugate iff Normal