nyuuzyou/wikis huggingface dataset:
Existence of Subgroup of Dipper Semigroup

== Theorem ==
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ denote the dipper semigroup.
Consider the subset $H \subseteq N_{< \paren {m \mathop + n} }$ defined as:
:$H = \set {k \in \N: m \le k < m + n} = \set {m, m + 1, \ldots, m + n - 1}$
Then the substructure $\struct {H, +_{m, n} }$ is a subgroup of $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
== Proof ==
Recall the definition of the dipper semigroup:
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $+_{m, n}$ be the dipper operation on $\N_{< \paren {m \mathop + n} }$:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Taking the group axioms in turn:
=== ===
Let $a, b \in H$.
Then:
:$2 m \le a + b \le 2 \paren {m + n - 1}$
Thus in all cases $a + b > m$.
Thus by definition of $+_{m, n}$:
:$a +_{m, n} b = a + b - k n$
where $k$ is the largest such that $m + k n < a + b$
Hence:
That is:
:$a +_{m, n} b \in H$
and so $\struct {H, +_{m, n} }$ is closed.
=== ===
From Restriction of Associative Operation is Associative, associativity is inherited by $\struct {H, +_{m, n} }$ from $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$.
=== ===
We have:
:$m \le a < m + n$
Hence
:$a +_{m, n} n = a$
From Dipper Semigroup is Commutative Semigroup we have that:
:$n +_{m, n} a = a +_{m, n} n = n$
Thus $n$ is the identity element of $\struct {H, +_{m, n} }$.
=== ===
We have that $n$ is the identity element of $\struct {H, +_{m, n} }$.
Hence we need to find $x \in H$ such that:
:$a +_{m, n} x = n$
That is:
:$a + x - k n = n$
So:
:$x = \paren {k + 1} n - a$
such that $m \le x < m + n$
where:
:$m - n + a \le k n < m + a$
Thus every element $x$ of $\struct {H, +_{m, n} }$ has an inverse $x^{-1}$ where:
:$\paren {k + 1} n - a$
such that:
:$m - n + a \le k n < m + a$
All the group axioms are thus seen to be fulfilled, and so $\struct {H, +_{m, n} }$ is a group.
== Examples ==
Category:Dipper Semigroups
Category:Examples of Subgroups
Category:Existence of Subgroup of Dipper Semigroup
Existence of Subgroup of Dipper Semigroup/Examples/(m, n) = (3, 4)

== Examples of Use of Existence of Subgroup of Dipper Semigroup ==
Consider the dipper semigroup $\struct {N_{<7}, +_{3, 4} }$.
Let $H = \set {x \in \N: 3 \le x < 7} = \set {3, 4, 5, 6}$.
Then:
:$\struct {H, +_{3, 4} }$ is a subgroup of $\struct {N_{<7}, +_{3, 4} }$
where:
:the identity of $\struct {H, +_{3, 4} }$ is $4$
:the inverse $a^{-1}$ of $a \in H$ is given by:
::$a^{-1} = 4 \paren {k - 1} - a$
:such that:
::$a - 1 \le 4 k < a + 3$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.4$
Category:Existence of Subgroup of Dipper Semigroup
Existence of Subgroup whose Index is Prime Power

== Theorem ==
Let $G$ be a finite group.
Let $H$ be a normal subgroup of $G$ which has a finite index in $G$.
Let:
:$p^k \divides \index G H$
where:
:$p$ is a prime number
:$k \in \Z_{>0}$ is a (strictly) positive integer
:$\divides$ denotes divisibility.
Then $G$ contains a subgroup $K$ such that:
:$\index K H = p^k$
== Proof ==
The order $\order {G / H}$ of the quotient group $G / H$ is $\index G H$.
Hence $p^k$ divides $\order {G / H}$.
By Group has Subgroups of All Prime Power Factors, $G / H$ has a subgroup of order $p^k$.
By Correspondence Theorem, this subgroup is in the form $K / H$ where $H \le K \le G$.
Hence the result.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \delta$
Category:P-Groups
Category:Sylow p-Subgroups
Existence of Support Functional

== Theorem ==
Let $\mathbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\mathbb F$. 
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$. 
Let $x \in X$. 
Then there exists $f \in X^\ast$ such that:
:$(1): \quad$ $\norm f_{X^\ast} = 1$
:$(2): \quad$ $\map f x = \norm x_X$.
That is: 
:there exists a support functional at $x$. 
== Proof ==
Let: 
:$U = \span {\set x}$ 
Then $U$ consists precisely of the $u \in X$ of the form: 
:$u = \alpha x$
for $\alpha \in \mathbb F$.
From Linear Span is Linear Subspace, we have: 
:$U$ is a linear subspace of $X$.
Let $\struct {U^\ast, \norm \cdot_{U^\ast} }$ be the normed dual space of $U$.
Define $f_0 : U \to \R$ by: 
:$\map {f_0} {\alpha x} = \alpha \norm x_X$
for each $\alpha \in \mathbb F$.
In particular, we have: 
:$\map {f_0} x = \norm x$ 
We show that this is a linear functional.
Let $u, v \in U$ and $k, l \in \mathbb F$. 
Then there exists $\alpha, \beta \in \mathbb F$ such that: 
:$u = \alpha x$
and:
:$v = \beta x$
We then have:
so $f_0$ is a linear functional.
Now we show that $f_0 \in X^\ast$ and: 
:$\norm {f_0}_{U^\ast} = 1$
Let $u \in U$ and write: 
:$u = \alpha x$
for $\alpha \in \mathbb F$. 
We then have:
so we have that $f_0$ is bounded.
That is, $f \in U^\ast$. 
We then have: 
:$\ds \sup_{\norm u_X = 1} \size {\map {f_0} u} = 1$ 
That is, from the definition of dual norm, we have: 
:$\norm {f_0}_{U^\ast} = 1$
We apply: 
:Hahn-Banach Theorem: Real Vector Space: Corollary 2 if $\mathbb F = \R$
:Hahn-Banach Theorem: Complex Vector Space: Corollary if $\mathbb F = \C$
to find that there exists $f \in X^\ast$ such that:
:$f$ extends $f_0$ to $X$
and:
:$\ds \norm f_{X^\ast} = \norm {f_0}_{U^\ast} = 1$
Since $f$ extends $f_0$, we have: 
:$\map f x = \map {f_0} x = \norm x$
So $f$ is the required linear functional.
== Sources ==
* $20.1$: Existence of a Support Functional
Category:Normed Dual Spaces
Existence of T4 Space which is not T3 1/2

== Theorem ==
There exists at least one example of a topological space which is a $T_4$ space, but is not also a $T_{3 \frac 1 2}$ space.
== Proof ==
Let $T$ be a Hjalmar Ekdal space.
From Hjalmar Ekdal Space is $T_4$, $T$ is a $T_4$ space.
From Hjalmar Ekdal Space is not $T_{3 \frac 1 2}$ Space, $T$ is not a $T_{3 \frac 1 2}$ space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:Separation Axioms
Category:T4 Spaces
Category:T3 1/2 Spaces
Existence of Topological Space satisfying Countable Chain Condition which is not Separable

== Theorem ==
There exists at least one example of a topological space which satisfies the countable chain condition which is not also a separable space.
== Proof ==
Let $T$ be a countable complement topology on an uncountable set.
From Countable Complement Space Satisfies Countable Chain Condition, $T$ satisfies the countable chain condition.
From Countable Complement Space is not Separable, $T$ is not a separable space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:Countable Chain Condition
Category:Separable Spaces
Existence of Topological Space which satisfies no Separation Axioms

== Theorem ==
There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied.
== Proof ==
Proof by Counterexample:
Let $T$ be the topological space consisting of the double pointed topology on the countable complement topology on an uncountable set.
From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T$ satisfies none of the Tychonoff separation axioms.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:Separation Axioms
Existence of Topological Space which satisfies no Separation Axioms but T0

== Theorem ==
There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_0$ (Kolmogorov) axiom.
== Proof ==
Let $T$ be the overlapping interval space.
From Overlapping Interval Space fulfils no Separation Axioms but $T_0$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_0$ (Kolmogorov) axiom.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:Separation Axioms
Category:T0 Spaces
Existence of Topological Space which satisfies no Separation Axioms but T0 and T1

== Theorem ==
There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_0$ (Kolmogorov) axiom and $T_1$ (Fréchet) axiom.
== Proof ==
Proof by Counterexample:
Let $T$ be the finite complement topology on a countable space.
From Finite Complement Space is $T_1$, $T$ is a $T_1$ (Fréchet) space.
Hence from $T_1$ Space is $T_0$ Space, also a $T_0$ (Kolmogorov) space.
The rest of the result follows from:
:Finite Complement Space is not $T_2$
and:
:Finite Complement Space is not $T_3$, $T_4$ or $T_5$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:Separation Axioms
Category:T0 Spaces
Category:T1 Spaces
Existence of Topological Space which satisfies no Separation Axioms but T3

== Theorem ==
There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_3$ axiom.
== Proof ==
Proof by Counterexample:
Let $T$ be the topological space consisting of the double pointed topology on the Tychonoff corkscrew topology.
From Double Pointed Tychonoff Corkscrew fulfils no Separation Axioms but $T_3$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_3$ axiom.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:Separation Axioms
Category:T3 Spaces
Existence of Topological Space which satisfies no Separation Axioms but T4

== Theorem ==
There exists at least one example of a topological space for which none of the Tychonoff separation axioms are satisfied except for the $T_4$ axiom.
== Proof ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $D = \struct {\set {0, 1}, \vartheta}$ be the indiscrete topology on two points.
Let $T \times D$ be the double pointed topology on $T$.
Let $\paren {T \times D}^*_{\bar p}$ be the open extension topology on $S \times \set {0, 1} \cup \set p$ where $p \notin S \times \set {0, 1}$.
From Open Extension of Double Pointed Countable Complement Topology is $T_4$ Space, we have that $\paren {T \times D}^*_{\bar p}$ satisfies none of the Tychonoff separation axioms except for the $T_4$ axiom.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:Separation Axioms
Category:T4 Spaces
Existence of Transformations whose Commutator equals Identity

== Theorem ==
Let $\map {C^\infty} \R$ be the space of smooth real functions.
Let $\circ$ denote the composition of mappings.
Let $A : \map {C^\infty} \R \to \map {C^\infty} \R$ be the mapping such that:
:$\forall \phi \in \map {C^\infty} \R : \forall x \in \R : \map {\paren{ A \circ \phi } } x := \map {\dfrac {d \phi} {d x} } x$
Let $B : \map {C^\infty} \R \to \map {C^\infty} \R$ be the mapping such that
:$\forall \phi \in \map {C^\infty} \R : \forall x \in \R : \map {\paren{ B \circ \phi } } x := x \map \phi x$
Let $I : \map {C^\infty} \R \to \map {C^\infty} \R$ be the identity such that:
:$\forall \phi \in \map {C^\infty} \R : I \circ \phi = \phi$
Then:
:$\forall \Psi \in \map {C^\infty} \R : \paren {A \circ B - B \circ A} \circ \Psi = I \circ \Psi$
or with slight abuse of notation:
:$A \circ B - B \circ A = I$
== Proof ==
Let $\Psi \in \map {C^\infty} \R$.
Then:
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Smooth Real Functions
Category:Commutativity
Existence of Translation between Each Pair of Points in Euclidean Space

== Theorem ==
Let $\R^n$ denote the real Euclidean space of $n$ dimensions.
Let $\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$ and $\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$ be points in $\R^n$.
There exists an isometry $f: \R^n \to \R^n$ such that $\map f {\mathbf a} = b$.
== Proof ==
Let $\mathbf t = \mathbf a - \mathbf b$.
Then the translation $\tau_\mathbf t$ is such an isometry.
We have that:
:$\map {\tau_\mathbf t} {\mathbf a} = \mathbf a - \paren {\mathbf a - \mathbf b} = \mathbf b$
The result follows from Translation Mapping is Isometry.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Exercise $1$
Category:Translation Mappings
Category:Euclidean Metric
Existence of Tychonoff Space which is not Normal

== Theorem ==
There exists at least one example of a topological space which is a Tychonoff space, but is not also a normal space.
== Proof ==
Let $T$ be a Niemytzki's tangent disc space.
From Niemytzki's Tangent Disc Space is Tychonoff, $T$ is a Tychonoff space.
From Niemytzki's Tangent Disc Space is not Normal, $T$ is not a normal space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:Separation Axioms
Category:Normal Spaces
Category:Tychonoff Spaces
Existence of Uncomputable Mappings/Proof/Overview

== Theorem ==
== Overview ==
 every mapping is computable.
Hence for each mapping there exists a computer program or algorithm which computes it.
This computer program is a finite string of symbols from some alphabet.
Hence the set of all computer programs $\CC$ is countably infinite.
Consider now the set $S$ of mappings which maps the set of integers $\Z$ to the Boolean domain $\set {0, 1}$.
 $S$ is countably infinite.
Let $S$ be placed in one-to-one correspondence with $\Z$.
Let $f_i$ be the mapping in $S$ which corresponds to the $i$th integer.
Now consider the mapping $f: \Z \to \set {0, 1}$ defined as follows:
:$\forall n \in \Z: \map f n = \begin {cases} 0 & : \map {f_n} n = 1 \\ 1 & : \text {otherwise} \end {cases}$
Then $\map f n$ cannot correspond to any integer.
This contradicts our supposition that $S$ is countably infinite.
Hence it cannot be the case that $\CC$ is countably infinite.
So $\CC$ is uncountable.
But this contradicts our deduction that $\CC$ is countably infinite.
Hence by Proof by Contradiction it cannot be the case that every mapping is computable.
== Sources ==
* : Chapter $1$: Preliminaries: $1.6$ Synopsis of the Book
Category:Existence of Uncomputable Mappings
Existence of Unique C* Norm on Unitization of Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\struct {A_+, \norm {\, \cdot \,}_+}$ be the unitization of $A$ as a normed algebra.
Define:
:$\tuple {a, \lambda}^\dagger = \tuple {a^\ast, \overline \lambda}$
for each $\tuple {a, \lambda} \in A_+$. 
Then there exists a unique algebra norm $\norm {\, \cdot \,}_\ast$ such that $\struct {A_+, \dagger, \norm {\, \cdot \,}_\ast}$ is a $\text C^\ast$-algebra and:
:$\norm {\tuple {a, 0} }_\ast = \norm a$
for each $a \in A$. 
== Proof ==
From Unitization of *-Algebra is *-Algebra, we have that $\tuple {A_+, \dagger}$ is a $\ast$-algebra.
Let $A \times \C$ be the direct product of $A$ and $\C$.
Note that while $A_+$ and $A \times \C$ share the same underlying set, their multiplication differs.
For clarity, we will specify $\circ_{A_+}$ and $\circ_{A \times \C}$, where:
:$\tuple {a, \lambda} \circ_{A_+} \tuple {b, \mu} = \tuple {a b + \lambda b + \mu a, \lambda \mu}$
and:
:$\tuple {a, \lambda} \circ_{A \times \C} \tuple {b, \mu} = \tuple {a b, \lambda \mu}$
for each $\tuple {a, \lambda}, \tuple {b, \mu} \in A \times \C$. 
Let $\norm {\, \cdot \,}_{A \times \C}$ be the direct product norm on $A \times \C$. 
From Direct Product of C*-Algebras is C*-Algebra and Complex Numbers form Unital C*-Algebra, $\tuple {A \times \C, \dagger, \norm {\, \cdot \,}_{A \times \C} }$ is a $\text C^\ast$-algebra. 
Further, $\tuple { {\mathbf 1}_A, 1}$ is the identity element for $A \times \C$.
Define $\phi : A_+ \to A \times \C$ by:
:$\map \phi {a, \lambda} = \tuple {a + \lambda {\mathbf 1}_A, \lambda}$
for each $\tuple {a, \lambda} \in A_+$. 
We show that $\phi$ is an $\ast$-algebra isomorphism.
First, $\phi$ is surjective, since we have:
:$\map \phi {a - \lambda {\mathbf 1}_A, \lambda} = \tuple {a, \lambda}$
for each $\tuple {a, \lambda} \in A_+$.
Let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$ and $t \in \C$.
Then we have:
Hence $\phi$ is linear.
We also have:
So $\phi$ is an algebra homomorphism.
We have:
So $\phi$ is a $\ast$-algebra homomorphism.
Hence $\phi$ is a $\ast$-algebra isomorphism.
Define $\norm {\, \cdot \,}_\ast : A_+ \to \hointr 0 \infty$ by:
:$\norm {\tuple {a, \lambda} }_\ast = \norm {\map \phi {a, \lambda} }_{A_+}$
for each $\tuple {a, \lambda} \in A_+$. 
From Direct Product of C*-Algebras is C*-Algebra, $\struct {A_+, \dagger, \norm {\, \cdot \,}_\ast}$ is a $\text C^\ast$-algebra. 
Further from Norm on C*-Algebra is Unique, $\norm {\, \cdot \,}_\ast$ is unique.
Finally, we have:
for each $a \in A$, hence we are done.
Category:C*-Algebras
Existence of Unique Difference between Cuts

== Theorem ==
Let $\alpha$ and $\beta$ be cuts.
Then there exists exactly one cut $\gamma$ such that:
:$\alpha + \gamma = \beta$
== Proof ==
From Ordering on Cuts is Compatible with Addition of Cuts:
:$\gamma_1 \ne \gamma_2 \implies \alpha + \gamma_1 \ne \alpha + \gamma_2$
That demonstrates uniqueness.
Let:
:$\gamma = \beta + \paren {-\alpha}$
where $-\alpha$ is the negative of $\alpha$.
Then by Identity Element for Addition of Cuts:
The result follows.
== Also see ==
* Definition:Subtraction of Cuts: $\gamma$ is defined as $\beta - \alpha$, the difference between $\beta$ and $\alpha$
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.19$. Theorem
Category:Cuts
Category:Subtraction
Existence of Unique Inverse Element for Addition of Cuts

== Theorem ==
Let $\alpha$ be a cut.
Let $0^*$ be the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
Then there exists a unique cut $\beta$ such that:
:$\alpha + \beta = 0^*$
where $+$ denotes the operation of addition of cuts.
== Proof ==
=== Proof of Uniqueness ===
Suppose $\alpha + \beta_1 = \alpha + \beta_2 = 0^*$.
We have:
So if such a $\beta$ exists, it is indeed unique.
=== Proof of Existence ===
Let $\beta$ be the set of all rational numbers $p$ such that $-p$ is an upper number of $\alpha$, but not the smallest upper number, were it to exist.
We verify that the conditions are satisfied for $\beta$ to be a cut.
It follows directly from the definition that $\beta \ne \O$ and $\beta \ne \Q$.
Thus condition $(1)$ is satisfied.
Let $p \in \beta$.
Let $q \in \Q$ be a rational number such that $q < p$.
Then:
:$-p \notin \alpha$
and:
:$-q > -p$
Thus $-q$ is an upper number of $\alpha$, but not the smallest.
Thus $q \in \beta$.
Thus condition $(2)$ is satisfied.
Let $p \in \beta$.
Then by definition $-p$ is an upper number of $\alpha$, but not the smallest.
Thus there exists a rational number $q$ such that $-q < -p$ and such that $-q \notin \alpha$.
Let:
:$r = \dfrac {p + q} 2$
By Mediant is Between:
:$-q < -r < -p$
so $-r$ is also an upper number of $\alpha$, but not the smallest.
So we have found a rational number $r$ such that $r > p$ where $r \in \beta$.
Thus condition $(3)$ is satisfied.
We have demonstrated that $\beta$ is a cut.
We now verify that $\alpha + \beta = 0^*$.
Suppose $p \in \alpha + \beta$.
Then:
:$p = q + r$
for some $q \in \alpha, r \in \beta$.
Hence:
:$-r \notin \alpha$
:$-r < q$
:$q + r < 0$
and:
:$p \in 0^*$
Suppose $p \in 0^*$.
Then:
:$p < 0$
By Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational, there exist $q \in \alpha, r \notin \alpha$ such that:
:$r$ is not the smallest upper number of $\alpha$
$r - q = -p$
Because $-r \in \beta$:
Thus by definition of set equality:
:$\alpha + \beta = 0^*$
and the proof is complete.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.16$. Theorem
Category:Cuts
Category:Addition
Existence of Unique Inverse Element for Multiplication of Cuts

== Theorem ==
Let $0^*$ be the rational cut associated with the (rational) number $0$:
:$0^* = \set {r \in \Q: r < 0}$
Let $\alpha$ be a cut such that $\alpha \ne 0^*$.
Then for every cut $\beta$, there exists a unique cut $\gamma$ such that:
:$\alpha \gamma = \beta$
where $\alpha \gamma$ denotes the operation of product of $\alpha$ and $\gamma$.
In this context, $\gamma$ can be expressed as $\beta / \alpha$.
== Proof ==
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.27$. Theorem
Category:Cuts
Category:Multiplication
Existence of Unique Subgroup Generated by Subset

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\O \subset S \subseteq G$.
Let $\struct {H, \circ}$ be the subgroup generated by $S$.
Then $H = \gen S$ exists and is unique.
Also, $\struct {H, \circ}$ is the intersection of all of the subgroups of $G$ which contain the set $S$:
:$\ds \gen S = \bigcap_i {H_i}: S \subseteq H_i \le G$
=== Singleton Generator ===
== Proof ==
=== Existence ===
First, we prove that such a subgroup exists.
Let $\mathbb S$ be the set of all subgroups of $G$ which contain $S$.
$\mathbb S \ne \O$ because $G$ is itself a subgroup of $G$, and thus $G \in \mathbb S$.
Let $H$ be the intersection of all the elements of $\mathbb S$.
By Intersection of Subgroups is Subgroup, $H$ is the largest element of $\mathbb S$ contained in each element of $\mathbb S$.
Thus $H$ is a subgroup of $G$.
Since $\forall x \in \mathbb S: S \subseteq x$, we see that $S \subseteq H$, so $H \in \mathbb S$.
=== Smallest ===
Now to show that $H$ is the smallest such subgroup.
If any $K \le G: S \subseteq K$, then $K \in \mathbb S$ and therefore $H \subseteq K$.
So $H$ is the smallest subgroup of $G$ containing $S$.
=== Uniqueness ===
Now we show that $H$ is unique.
Suppose $\exists H_1, H_2 \in \mathbb S$ such that $H_1$ and $H_2$ were two such smallest subgroups containing $S$.
Then, by the definition of "smallest", each would be equal in size.
If one is not a subset of the other, then their intersection (by definition containing $S$) would be a smaller subgroup and hence neither $H_1$ nor $H_2$ would be the smallest.
Hence one must be a subset of the other.
By definition of set equality, that means they must be the same set.
So the smallest subgroup, whose existence we have proved above, is unique.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.5$
* : $\S 1.9$
* : Chapter $2$: Subgroups and Cosets: $\S 35 \epsilon$
* : Chapter $4$: Subgroups: Definition $4.7$
Category:Group Theory
Category:Generated Subgroups
Existence of Unique Subgroup Generated by Subset/Singleton Generator

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $a \in G$.
Then $H = \gen a = \set {a^n: n \in \Z}$ is the unique smallest subgroup of $G$ such that $a \in H$.
That is:
: $K \le G: a \in K \implies H \subseteq K$
== Proof ==
From Powers of Element form Subgroup, $H = \set {a^n: n \in \Z}$ is a subgroup of $G$.
Let $K \le G: a \in K$.
Then $\forall n \in \Z: a^n \in K$.
Thus, $H \subseteq K$.
== Sources ==
* : $\S 37.5$ Some important general examples of subgroups
Category:Generated Subgroups
Existence of Unique Subsemigroup Generated by Subset

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\O \subset X \subseteq S$.
Let $\struct {T, \circ}$ be the subsemigroup generated by $X$.
Then $T = \gen X$ exists and is unique.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.7$
Category:Subsemigroups
Category:Intersection of Subsemigroups
Category:Existence of Unique Subsemigroup Generated by Subset