nyuuzyou/wikis huggingface dataset:
Interior of Simple Closed Contour is Well-Defined

== Theorem ==
Let $C$ be a simple closed contour in the complex plane.
Let $f : \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $\phi : \R^2 \to \C$ be defined by:
:$\map \phi {x, y} = x + i y$
Let $\Img C = \map \phi {\Img f}$, where $\Img C$ denotes the image of $C$, and $\Img f$ denotes the image of $f$.
Then the interior of $C$:
:$\Int C = \map \phi {\Int f}$
is well-defined.
Here, $\Int f$ denotes the interior of $f$.
== Proof ==
Let $C$ be defined as a concatenation of a (finite) sequence of directed smooth curves $\sequence {C_1, \ldots, C_n}$.
We show existence of a Jordan curve $f : \closedint 0 1 \to \R^2$ that fulfills the criteria $\map \phi {\Img f} = \Img C$.
Reparameterization of Directed Smooth Curve with Given Domain shows that $C_k$ can be reparameterized by a smooth path:
:$\gamma_k: \closedint {\dfrac {k - 1} n} {\dfrac k n} \to \C$ for all $k \in \set {1, \ldots, n}$
Define $\gamma: \closedint 0 1 \to \C$ by:
:$\map \gamma t = \map {\gamma_k} t$ for all $t \in \closedint {\dfrac {k - 1} n} {\dfrac k n}$
Reparameterization of Directed Smooth Curve Preserves Image shows that $\Img \gamma = \Img C$.
Pasting Lemma for Pair of Continuous Mappings on Closed Sets shows that $\gamma$ is continuous.
By definition of simple contour, it follows that:
:$\map \gamma {t_1} \ne \map \gamma {t_2}$ for all $t_1 \in \hointr 0 1, t_2 \in \hointr 0 1$ with $t_1 \ne t_2$
By definition of closed contour, it follows that:
:$\map \gamma 0 = \map \gamma 1$
Complex Plane is Homeomorphic to Real Plane shows that $\phi$ is a homeomorphism.
Composite of Continuous Mappings between Metric Spaces is Continuous shows that $\phi^{-1} \circ \gamma : \closedint 0 1 \to \R^2$ is continuous.
As $\phi^{-1}$ is a homeomorphism, it follows by the definition of Jordan curve that $\phi^{-1} \circ \gamma$ is a Jordan curve.
Set $f = \phi^{-1} \circ \gamma$.
We now have $\Img C = \map \phi {\Img f}$.
We show uniqueness of the definition of $\Int C$.
Let $g: \closedint 0 1 \to \R^2$ be a Jordan curve such that $\Img g = \Img f$.
Jordan Curve Theorem shows that the definition of the interior of a Jordan curve only depends on the image of the Jordan curve.
Then:
:$\Int g = \Int f$
Category:Complex Contour Integrals
Interior of Singleton in Real Number Line is Empty

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $a \in \R$ be a real number.
Then:
:$\set a^\circ = \O$
where $\set a^\circ$ denotes the interior of $\set a$ in $\R$.
== Proof ==
Category:Examples of Set Interiors
Category:Singletons
Category:Real Number Line with Euclidean Topology
Interior of Subset

== Theorem ==
Let $\left({S, \tau}\right)$ be a topological space.
Let $X$ and $Y$ be subsets of $S$, and suppose that $X \subseteq Y$.
Then:
:$X^\circ \subseteq Y^\circ$
where $X^\circ$ denotes the interior of $X$.
== Proof ==
By definition of interior, $X^\circ$ is open in $\tau$, and:
:$Y^\circ \subseteq Y$
Hence, by Subset Relation is Transitive:
:$X^\circ \subseteq Y$
is an open set contained in $Y$.
The result follows by Set Interior is Largest Open Set.
Category:Subsets
Category:Set Interiors
Interior of Subset in Subspace

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T_H = \struct {H, \tau_H}$ be a subspace of $T$ where $H \subseteq S$.
Let $A \subseteq H$ be an arbitrary subset of $H$.
Then:
:$\map {\mathrm {Int}_H} A \supseteq H \cap \map {\mathrm {Int} } A$
where:
:$\map {\mathrm {Int}_H} A$ denotes the interior of $A$ in $T_H$
:$\map {\mathrm {Int} } A$ denotes the interior of $A$ in $T$.
== Proof ==
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces
Category:Set Interiors
Category:Topological Subspaces
Interior of Subset of Double Pointed Topological Space

== Theorem ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Let $X \subseteq S \times D$ be a subset of $S \times D$.
Define $A \subseteq S$ by:
:$A := \set {s \in S: \paren {\forall d \in D: \tuple {s, d} \in X} }$
Then the interior of $X$ in $\tau$ is:
:$X^\circ = A^\circ \times D$
== Proof ==
By Open Sets of Double Pointed Topology, $X^\circ$ must be of the form:
:$X^\circ = U \times D$
with $U$ open in $\tau_S$.
By Set Interior is Largest Open Set, we have for any open set $U' \times D$ of $\tau$ that:
:$U' \times D \subseteq X \iff U' \times D \subseteq X^\circ = U \times D$
By Cartesian Product of Subsets, we have for open sets $V, V'$:
:$V \subseteq V' \iff V \times D \subseteq V' \times D$
since $D$ is non-empty.
Combining these two equivalences, we have:
:$U' \times D \subseteq X \iff U' \subseteq U$
The condition that $U' \times D \subseteq X$ can be expressed by:
:$\forall s \in U': \forall d \in D: \tuple {s, d} \in X$
By definition of the set $A$, this is equivalent to:
:$U' \subseteq A$
Combining this with the above, it follows that:
:$U' \subseteq A \iff U' \subseteq U$
Recall that $U$ is open in $\tau_S$.
By Set Interior is Largest Open Set, we conclude:
:$U = A^\circ$
and so:
:$X^\circ = A^\circ \times D$
as desired.
== Also see ==
* Closure of Subset of Double Pointed Topological Space
Category:Double Pointed Topologies
Category:Examples of Set Interiors
Interior of Subset of Indiscrete Space

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subsetneqq S$ (that is, let $H$ be a proper subset of $S$).
Then:
:$H^\circ = H^{\circ -} = H^{\circ - \circ} = \O$
where:
:$H^\circ$ denotes the interior of $H$
:$H^-$ denotes the closure of $H$.
== Proof ==
As $H \subsetneqq S$, it follows that $H \ne S$.
So the only open subset of $H$ is $\O$.
So by definition:
:$H^\circ = \O$
From Empty Set is Closed in Topological Space, $\O$ is closed in $T$.
From Closed Set Equals its Closure:
:$\O^- = \O$
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Interior of Open Set:
:$\O^\circ = \O$
The result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $6$
Category:Indiscrete Topology
Category:Examples of Set Interiors
Interior of Translation of Set in Topological Vector Space is Translation of Interior

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
Let $E \subseteq X$. 
Let $x \in X$.
Then:
:$\paren {E + x}^\circ = E^\circ + x$
== Proof ==
Let $y \in \paren {E + x}^\circ$.
Then there exists an open neighborhood $U$ of $y$ such that $U \subseteq E + x$.
Then from Translation of Open Set in Topological Vector Space is Open, $U - x$ is an open neighborhood of $y - x$ such that $U - x \subseteq E$.
Hence $y - x \in E^\circ$.
So $y \in E^\circ + x$.
We therefore obtain $\paren {E + x}^\circ \subseteq E^\circ + x$.
Now let $y \in E^\circ + x$.
Then $y - x \in E^\circ$.
So there exists an open neighborhood $U$ of $y - x$ such that $U \subseteq E$.
Then from Translation of Open Set in Topological Vector Space is Open, $U + x$ is an open neighborhood of $y$ such that $U + x \subseteq E + x$. 
Hence $y \in \paren {E + x}^\circ$.
We therefore obtain $E^\circ + x \subseteq \paren {E + x}^\circ$.
So $\paren {E + x}^\circ = E^\circ + x$.
Category:Translation of Subsets of Vector Spaces
Category:Set Interiors
Category:Topological Vector Spaces
Interior of Union is not necessarily Union of Interiors

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H_1$ and $H_2$ be subsets of $S$.
Let ${H_1}^\circ$ and ${H_2}^\circ$ denote the interiors of $H_1$ and $H_2$ respectively.
Then it is not necessarily the case that:
:$\left({H_1 \cup H_2}\right)^\circ = {H_1}^\circ \cup {H_2}^\circ$
== Proof ==
From Union of Interiors is Subset of Interior of Union:
:$\paren {H_1 \cup H_2}^\circ \supseteq {H_1}^\circ \cup {H_2}^\circ$
It remains to be shown that it is not necessarily the case that:
:$\paren {H_1 \cup H_2}^\circ = {H_1}^\circ \cup {H_2}^\circ$
Proof by Counterexample:
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $H_1 = \closedint 0 {\dfrac 1 2}$ and $H_2 = \closedint {\dfrac 1 2} 1$.
Then:
and:
Hence the result.
== Also see ==
* Closure of Intersection may not equal Intersection of Closures
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $6$
Category:Set Interiors
Category:Set Union
Interior of Union of Adjacent Open Intervals

== Theorem ==
Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the two adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A = A^\circ$
where $A^\circ$ is the interior of $A$.
== Proof ==
From Open Sets in Real Number Line, $A$ is open in $\R$.
The result follows from Interior of Open Set.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(a)}$
Category:Examples of Set Interiors
Category:Union of Adjacent Open Intervals
Intermediate Value Theorem

== Theorem ==
Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be continuous on $I$.
Then $f$ is a Darboux function.
That is:
Let $a, b \in I$.
Let $k \in \R$ lie between $\map f a$ and $\map f b$.
That is, either:
:$\map f a < k < \map f b$
or:
:$\map f b < k < \map f a$
Then $\exists c \in \openint a b$ such that $\map f c = k$.
== Corollary ==
== Proof ==
This theorem is a restatement of Image of Interval by Continuous Function is Interval.
From Image of Interval by Continuous Function is Interval, the image of $\openint a b$ under $f$ is also a real interval (but not necessarily open).
Thus if $k$ lies between $\map f a$ and $\map f b$, it must be the case that:
:$k \in \Img {\openint a b}$
The result follows.
== Also known as ==
This result is also known as Bolzano's theorem, for .
Some sources attribute it to , and call it the Weierstrass Intermediate Value Theorem.
== Also see ==
* Intermediate Value Theorem (Topology), of which this is a corollary
== Historical Note ==
== Sources ==
* : $1$: Review of some real analysis: $\S 1.4$: Continuity: Theorem $1.4.5$
* : $\S 9.10$
* : Chapter $\text {A}.33$: Weierstrass ()
* 
* 
* 
Category:Named Theorems
Category:Analysis
Intermediate Value Theorem/Corollary

== Theorem ==
Let $I$ be a real interval.
Let $a, b \in I$ such that $\openint a b$ is an open interval.
Let $f: I \to \R$ be a real function which is continuous on $\openint a b$.
Let $0 \in \R$ lie between $\map f a$ and $\map f b$.
That is, either:
:$\map f a < 0 < \map f b$
or:
:$\map f b < 0 < \map f a$
Then $f$ has a root in $\openint a b$.
== Proof ==
Follows directly from the Intermediate Value Theorem and from the definition of root.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Path-Connectedness
Category:Analysis
Intermediate Value Theorem/Historical Note

== Historical Note on Intermediate Value Theorem ==
This result rigorously proves the intuitive truth that:
: if a continuous real function defined on an interval is sometimes positive and sometimes negative, then it must have the value $0$ at some point.
 was the first to provide this proof in $1817$, but because of incomplete understanding of the nature of the real numbers it was not completely satisfactory.
Hence many sources refer to this as Bolzano's Theorem.
The first completely successful proof was provided by , hence its soubriquet the Weierstrass Intermediate Value Theorem.
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes
Intermediate Value Theorem (Topology)

== Theorem ==
Let $X$ be a connected topological space.
Let $\struct {Y, \preceq, \tau}$ be a totally ordered set equipped with the order topology.
Let $f: X \to Y$ be a continuous mapping.
Let $a$ and $b$ are two points of $a, b \in X$ such that:
:$\map f a \prec \map f b$
Let:
:$r \in Y: \map f a \prec r \prec \map f b$
Then there exists a point $c$ of $X$ such that:
:$\map f c = r$
== Proof ==
Let $a, b \in X$, and let $r \in Y$ lie between $\map f a$ and $\map f b$.
Define the sets:
:$A = f \sqbrk X \cap r^\prec$ and $B = f \sqbrk X \cap r^\succ$
where $r^\prec$ and $r^\succ$ denote the strict lower closure and strict upper closure respectively of $r$ in $Y$.
$A$ and $B$ are disjoint by construction.
$A$ and $B$ are also non-empty since one contains $\map f a$ and the other contains $\map f b$.
$A$ and $B$ are also both open by definition as the intersection of open sets.
Suppose there is no point $c$ such that $\map f c = r$.
Then:
:$f \sqbrk X = A \cup B$
so $A$ and $B$ constitute a separation of $X$.
But this contradicts the fact that Continuous Image of Connected Space is Connected.
Hence by Proof by Contradiction:
:$\exists c \in X: \map f c = r$
which is what was to be proved.
== Also see ==
* Intermediate Value Theorem of calculus, which follows as a corollary from this by considering $\R$ under the order topology and noting that Subset of Real Numbers is Interval iff Connected.
== Sources ==
* : $\S 24$: Theorem $24.3$
Category:Connected Spaces
Category:Continuous Mappings
Category:Order Topology
Intermediate Value Theorem for Derivatives

== Theorem ==
Let $I$ be an open interval.
Let $f : I \to \R$ be everywhere differentiable.
Then $f'$ satisfies the Intermediate Value Property.
== Proof ==
Since $\forall \set {a, b \in I: a < b}: \openint a b \subseteq I$, the result follows from Image of Interval by Derivative.
== Source ==
* 
Category:Intermediate Value Theorem
Category:Real Analysis
Internal Angle of Equilateral Triangle

== Theorem ==
The internal angles of an equilateral triangle measure $60^\circ$ or $\dfrac \pi 3$ radians.
== Proof ==
By definition, an equilateral triangle is a regular polygon with $3$ sides.
From Internal Angles of Regular Polygon, the size $A$ of each internal angle of a regular $n$-gon is given by:
:$A = \dfrac {\paren {n - 2} 180^\circ} n$
Thus:
:$A = \dfrac {180^\circ} n = 60^\circ$
From Value of Degree in Radians:
:$1^\circ = \dfrac {\pi} {180^\circ} \mathrm {rad}$
and so:
:$A = 60^\circ \times \dfrac {\pi} {180^\circ} = \dfrac \pi 3 \mathrm {rad}$
== Sources ==
* : $60$
* : $60$
Category:Equilateral Triangles
Internal Angles of Regular Polygon

== Theorem ==
The size $A$ of each internal angle of a regular $n$-gon is given by:
:$A = \dfrac {\paren {n - 2} 180 \degrees} n$
=== Corollary ===
== Proof ==
From Sum of Internal Angles of Polygon, we have that the sum $S$ of all internal angles of a $n$-gon is:
:$S = \paren {n - 2} 180 \degrees$
From the definition of a regular polygon, all the internal angles of a regular polygon are equal.
Therefore, the size $A$ of each internal angle of a regular polygon with $n$ sides is:
:$A = \dfrac {\paren {n - 2} 180 \degrees} n$
== Also presented as ==
This formula can also be seen presented as:
:$A = 180 \degrees - \dfrac {360 \degrees} n$
Category:Regular Polygons
Internal Angles of Square

== Theorem ==
The internal angles of a square are right angles.
== Proof ==
By definition, a square is a regular quadrilateral.
From Internal Angles of Regular Polygon, the internal angles of a square measure $\dfrac {180 \degrees \paren {4 - 2} } 4 = 90 \degrees$.
The result follows from Measurement of Right Angle.
Category:Squares
Category:Internal Angles
Internal Direct Product/Examples/Non-Example 1

== Example of External Direct Product which is not Internal Direct Product ==
Let $m$ and $n$ be integers such that $m, n > 1$.
Let $S$ be a set with $n$ elements.
Let $A$ and $B$ be subsets of $S$ which have $m$ and $n$ elements respectively.
Let $\struct {S, \gets}$ be the algebraic structure formed from $S$ with the left operation.
Then:
:$\struct {S, \gets}$ is isomorphic with the external direct product of $\struct {A, \gets_A}$ and $\struct {B, \gets_B}$
but:
:$\struct {S, \gets}$ is not the internal direct product of $\struct {A, \gets_A}$ and $\struct {B, \gets_B}$.
== Proof ==
From Cardinality of Cartesian Product of Finite Sets, $S$ has the same number of elements as the Cartesian product of $A$ and $B$.
That is:
:$\card {\struct {S, \gets} } = \card {\struct {A, \gets_A} \times \struct {B, \gets_B} }$
Hence by definition of cardinality, there exists a bijection between $S$ and $A \times B$.
Indeed, from Cardinality of Set of Bijections, there are $m n!$ such bijections.
First we demonstrate that $\struct {S, \gets}$ is isomorphic with the external direct product of $\struct {A, \gets_A}$ and $\struct {B, \gets_B}$.
Let $\phi: A \times B \to S$ be an arbitrary bijection.
We have:
demonstrating isomorphism.
Let $\phi: A \times B \to S$ be the mapping defined as:
:$\map \phi {a, b} = a \gets b$
Let $\tuple {a, b}$ and $\tuple {c, b}$ be arbitrary elements of $A \times B$ such that $a \ne c$.
As the cardinality of $A$ is greater than $1$, it is apparent that this is possible.
Thus:
:$\tuple {a, b} \ne \tuple {c, b}$
But we have:
demonstrating that $\phi$ is not an injection.
Thus $\phi$ is not a bijection.
Hence by definition $\phi$ is not an isomorphism.
It follows that there can be no isomorphism from $\struct {A, \gets_A} \times \struct {B, \gets_B}$ to $\struct {S, \gets}$.
That is, $\struct {S, \gets}$ is not the internal direct product of $\struct {A, \gets_A}$ and $\struct {B, \gets_B}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.8$
Category:Internal Direct Products
Internal Direct Product Generated by Subgroups

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $\sequence {H_n}$ be a sequence of subgroups of $G$.
Then:
:the subgroup generated by $\ds \bigcup_{k \mathop = 1}^n H_k$ is the internal group direct product of $\sequence {H_n}$
:
:$\sequence {H_n}$ is an independent sequence of subgroups such that every element of $H_i$ commutes with every element of $H_j$ whenever $1 \le i < j \le n$.
== Proof ==
In the following, the notation $\closedint m n$ is to be understood to mean a (closed) integer interval:
:$\closedint m n := \set {x \in \Z: m \le x \le n}$
for $m, n \in \Z$.
For each $k \in \closedint 1 n$, let $\ds L_k = \prod_{j \mathop = 1}^k H_j$ be the cartesian product of the subgroups $H_1, H_2, \ldots, H_k$ of $G$.
Let $\ds C_k: L_k \to G: \map {C_k} {x_1, x_2, \ldots, x_k} = \prod_{j \mathop = 1}^k x_j$.
=== Necessary Condition ===
Suppose that the subgroup generated by $\ds \bigcup_{k \mathop = 1}^n H_k$ is the internal group direct product of $\sequence {H_n}$.
We have that Internal Group Direct Product is Injective.
Hence by definition $C_k$ is a monomorphism.
It follows from Kernel is Trivial iff Monomorphism that the kernel of $C_n$ is $\set {\tuple {e, e, \ldots, e} }$.
Therefore $\sequence {H_n}$ is an independent sequence.
Let $x \in H_i$ and $y \in H_j$ where $1 \le i < j \le n$.
For each $k \in \closedint 1 n$, let $x_k$ and $y_k$ be defined as:
:$x_k = \begin{cases}
e & : k \ne i \\
x & : k = i
\end{cases}
\qquad
y_k = \begin{cases}
e & : k \ne j \\
y & : k = j
\end{cases}$
Then:
=== Sufficient Condition ===
Suppose that $\sequence {H_n}$ is an independent sequence of subgroups such that every element of $H_i$ commutes with every element of $H_j$ whenever $1 \le i < j \le n$.
Let $S$ be the set of all $k \in \closedint 1 n$ such that $C_k: L_k \to G$ is a (group) homomorphism.
Clearly $1 \in S$.
Now let $k \in S$ such that $k < n$.
Let $\tuple {x_1, \ldots, x_k, x_{k + 1} }, \tuple {y_1, \ldots, y_k, y_{k + 1} } \in L_k$.
By the General Associativity Theorem and Element Commutes with Product of Commuting Elements:
Thus $C_{k + 1}: L_{k + 1} \to G$ is a homomorphism, so $k + 1 \in S$.
So by induction:
:$S = \closedint 1 n$
In particular, $C_n: L_n \to G$ is a homomorphism.
By Morphism Property Preserves Closure and Homomorphism Preserves Subsemigroups, the codomain $\ds \prod_{k \mathop = 1}^n H_k$ of $C_n$ is therefore a subgroup of $G$ containing $\ds \bigcup_{k \mathop = 1}^n H_k$.
However, any subgroup containing $\ds \bigcup_{k \mathop = 1}^n H_k$ must clearly contain $\ds \prod_{k \mathop = 1}^n H_k$.
Therefore, $\ds \prod_{k \mathop = 1}^n H_k$ is the subgroup of $G$ generated by $\ds \bigcup_{k \mathop = 1}^n H_k$.
As $\sequence {H_n}$ is an independent sequence of subgroups, the kernel of $C_n$ is $\set {\tuple {e, \ldots, e} }$.
Hence by the Quotient Theorem for Group Epimorphisms, $C_n$ from $L_n$ to the subgroup of $G$ generated by $\ds \bigcup_{k \mathop = 1}^n H_k$ is an isomorphism.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.13$
Category:Internal Group Direct Products
Internal Direct Product Theorem/Examples/Additive Group of Integers Modulo 6

== Example of Use of Internal Direct Product Theorem ==
Consider the additive group of integers modulo $6$ $\struct {\Z_6, \times_6}$, illustrated by Cayley Table:
Let $H := \set {0, 2, 4}$.
Let $K := \set {0, 3}$.
We have that:
:$H +_6 K = \struct {\Z_6, +_6}$
and:
:$H \cap K = \set 0$
Hence $H$ and $K$ are subgroups of $\struct {\Z_6, +_6}$ which fulfil the conditions of the Internal Direct Product Theorem.
Thus $\struct {\Z_6, +_6}$ is the internal group direct product of $H$ and $K$.
Because:
:$H$ is isomorphic to $\struct {\Z_3, +_3}$
:$K$ is isomorphic to $\struct {\Z_2, +_2}$
it follows by Isomorphism of External Direct Products that:
:$\struct {\Z_6, +_6}$ is isomorphic to $\struct {\Z_3, +_3} \times \struct {\Z_2, +_2}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Example $13.3$
Category:Internal Direct Product Theorem
Internal Direct Product Theorem/Examples/Multiplicative Monoid of Integers Modulo 6

== Example of Use of Internal Direct Product Theorem ==
Consider the multiplicative monoid of integers modulo $6$ $\struct {\Z_6, +_6}$, illustrated by Cayley Table:
Let $H := \set {0, 2, 4}$.
Let $K := \set {0, 3}$.
We have that:
:$H \times_6 K = \set 0$
so $\struct {\Z_6, \times_6}$ is not the internal group direct product of $H$ and $K$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Example $13.3$
Category:Internal Direct Product Theorem
Internal Direct Product Theorem/Examples/Symmetry Group of Rectangle

== Example of Use of Internal Direct Product Theorem ==
Consider the symmetry group of the rectangle $D_2$:
Let $H := \set {e, r}$.
Let $K := \set {e, h}$.
Then $H$ and $K$ are subgroups of $D_2$ which fulfil the conditions of the Internal Direct Product Theorem, as:
:$r \circ h = v = h \circ r$
Thus $D_2$ is the internal group direct product of $H$ and $K$.
Both $H$ and $K$ are isomorphic to $\struct {\Z_2, +_2}$, the additive group of integers modulo $2$.
Hence by Isomorphism of External Direct Products:
:$D_2$ is isomorphic to $\struct {\Z_2, +_2} \times \struct {\Z_2, +_2}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Example $13.2$
Category:Internal Direct Product Theorem
Internal Direct Product Theorem/General Result

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.
Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ :
:$(1): \quad G = H_1 H_2 \cdots H_n$
:$(2): \quad \sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups
:$(3): \quad \forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$
where $H_k \lhd G$ denotes that $H_k$ is a normal subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
Category:Internal Direct Product Theorem