avewright/proofwiki-math huggingface dataset:
Theorem: Power of Coset Product is Coset of Power

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a \in G$.
Then:
:$\forall n \in \Z: \paren {a \circ N}^n = \paren {a^n} \circ N$

Proof:
From Quotient Group is Group, the operation:
:$\forall a, b \in G: \paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is the group operation in the quotient group $\struct {G / N, \circ}$.
The result follows directly by definition of power of group element.
{{qed}}
Theorem: Condition for Power of Element of Quotient Group to be Identity

Let $G$ be a group whose identity is $e$.
Let $N$ be a normal subgroup of $G$.
Let $a \in G$.
Then:
:$\paren {a N}^n$ is the identity of the quotient group $G / N$
{{iff}}:
:$a^n \in N$

Proof:
Let $\paren {a N}^n$ be the identity of $G / N$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {a N}^n
      | r = N
      | c = Quotient Group is Group: {{Group-axiom|2}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {a^n} N
      | r = N
      | c = Power of Coset Product is Coset of Power
}}
{{eqn | ll= \leadstoandfrom
      | l = a^n
      | o = \in
      | r = N
      | c = Coset Equals Subgroup iff Element in Subgroup
}}
{{end-eqn}}
{{qed}}
Theorem: Additive Group of Integers is Normal Subgroup of Reals

Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$.

Proof:
From Additive Group of Integers is Subgroup of Reals, $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$.
As the additive group of real numbers is abelian, from Subgroup of Abelian Group is Normal it follows that $\struct {\Z, +} \lhd \struct {\R, +}$.
{{qed}}
Theorem: Condition for Element of Quotient Group of Additive Group of Reals by Integers to be of Finite Order

Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\R / \Z$ denote the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$.
Let $x + \Z$ denote the coset of $\Z$ by $x \in \R$.
Then $x + \Z$ is of finite order {{iff}} $x$ is rational.

Proof:
From Additive Group of Integers is Normal Subgroup of Reals, we have that $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$.
Hence $\R / \Z$ is indeed a quotient group.
By definition of rational number, what is to be proved is:
:$x + \Z$ is of finite order {{iff}}:
:$x = \dfrac m n$
for some $m \in \Z, n \in \Z_{> 0}$.
Let $x + \Z$ be of finite order in $\R / \Z$.
Then:
{{begin-eqn}}
{{eqn | q = \exists n \in \Z_{\ge 0}
      | l = \paren {x + \Z}^n
      | r = \Z
      | c = {{Defof|Quotient Group}}: {{Group-axiom|2}}
}}
{{eqn | ll= \leadstoandfrom
      | l = n x
      | o = \in
      | r = \Z
      | c = Condition for Power of Element of Quotient Group to be Identity
}}
{{eqn | ll= \leadstoandfrom
      | l = n x
      | r = m
      | c = for some $m \in \Z$
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | r = \dfrac m n
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Additive Group of Integers is Normal Subgroup of Rationals

Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Then $\struct {\Z, +}$ is a normal subgroup of $\struct {\Q, +}$.

Proof:
From Additive Group of Integers is Subgroup of Rationals, $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.
From Rational Numbers under Addition form Infinite Abelian Group, $\struct {\Q, +}$ is an abelian group.
From Subgroup of Abelian Group is Normal it follows that $\struct {\Z, +}$ is a normal subgroup of $\struct {\Q, +}$.
{{qed}}
Category:Additive Group of Integers
Category:Additive Group of Rational Numbers
Category:Examples of Normal Subgroups
o0mha3730xf7oqu9spbe60zdzw5rvx2
Theorem: Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism

Let $\struct {G, \circ}$ be a group.
Let $g \in G$.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\phi_g: \struct {\Z, +} \to \struct {G, \circ}$ be the mapping defined as:
:$\forall k \in \Z: \map {\phi_g} k = g^k$
Then $\phi_g$ is a (group) homomorphism.

Proof:
Let $k, l \in \Z$.
{{begin-eqn}}
{{eqn | l = \map {\phi_g} {k + l}
      | r = a^{k + l}
      | c = 
}}
{{eqn | r = a^k a^l
      | c = 
}}
{{eqn | r = \map {\phi_g} k \circ \map {\phi_g} l
      | c = 
}}
{{end-eqn}}
thus proving that $\phi_g$ is a homomorphism as required.
{{qed}}
Theorem: Inner Automorphisms form Subgroup of Symmetric Group

Let $G$ be a group.
Let $\struct {\map \Gamma G, \circ}$ be the symmetric group on $G$.
Let $\Inn G$ denote the inner automorphism group of $G$.
Then:
:$\Inn G \le \struct {\map \Gamma G, \circ}$
where $\le$ denotes the relation of being a subgroup.

Proof:
An inner automorphism is a permutation on $G$ by definition.
From Inner Automorphisms form Subgroup of Automorphism Group:
:$\Inn G \le \Aut G$
where $\Aut G$ denotes the set of automorphisms of $G$.
From Automorphism Group is Subgroup of Symmetric Group:
:$\Aut G \le \struct {\map \Gamma G, \circ}$
Thus $\Inn G \le \struct {\map \Gamma G, \circ}$ as required.
{{qed}}
Theorem: Group Isomorphism Preserves Order of Group

Let $G$ and $H$ be groups.
Let $\phi: G \to H$ be a (group) isomorphism.
Then:
:$\order G = \order H$
where $\order {\, \cdot \,}$ denotes the order of a group.

Proof:
By definition, an isomorphism is a bijection.
By definition, the order of a group is the cardinality of its underlying set.
The result follows by definition of set equivalence.
{{qed}}
Theorem: Additive Group of Reals is Subgroup of Complex

Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.

Proof:
Let $x, y \in \C$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
As $x$ and $y$ are wholly real, we have that $x, y \in \R$.
Then $x + y = \paren {x_1 + y_1} + \paren {0 + 0} i$ which is also wholly real.
Also, the inverse of $x$ is $-x = -x_1 + 0 i$ which is also wholly real.
We have that $\R$ is non-empty.
Thus by the Two-Step Subgroup Test, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
{{Qed}}
Theorem: C6 is not Isomorphic to S3

Let $C_6$ denote the cyclic group of order $6$.
Let $S_3$ denote the symmetric group on $3$ letters.
Then $C_6$ and $S_3$ are not isomorphic.

Proof:
Note that both $C_6$ and $S_3$ are of order $6$.
From Cyclic Group is Abelian, $C_6$ is abelian.
From Symmetric Group is not Abelian, $S_6$ is not abelian.
From Isomorphism of Abelian Groups, if two groups are isomorphic, they are either both abelian or both not abelian.
Hence $C_6$ and $S_3$ are not isomorphic.
{{qed}}
Theorem: Order of Alternating Group

Let $n \in \Z$ be an integer such that $n > 1$.
Let $A_n$ be the alternating group on $n$ letters.
Then:
:$\order {A_n} = \dfrac {n!} 2$
where $\order {A_n}$ denotes the order of $A_n$.

Proof:
Let $S_n$ denote the symmetric group on $n$ letters.
From Alternating Group is Normal Subgroup of Symmetric Group:
:$\index {S_n} {A_n} = 2$
where $\index {S_n} {A_n}$ denotes the index of $A_n$ in $S_n$.
From Order of Symmetric Group:
:$\order {S_n} = n!$
The result follows from Lagrange's Theorem.
{{qed}}
Theorem: Mapping to Power is Endomorphism iff Abelian

Let $\struct {G, \circ}$ be a group.
Let $n \in \Z$ be an integer.
Let $\phi: G \to G$ be defined as:
:$\forall g \in G: \map \phi g = g^n$
Then $\struct {G, \circ}$ is abelian {{iff}} $\phi$ is a (group) endomorphism.

Proof:
=== Necessary Condition ===
Let $\struct {G, \circ}$ be an abelian group.
Let $a, b \in G$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {a \circ b}
      | r = \paren {a \circ b}^n
      | c = Definition of $\phi$
}}
{{eqn | r = a^n \circ b^n
      | c = Power of Product of Commutative Elements in Group
}}
{{eqn | r = \map \phi a \circ \map \phi b
      | c = Definition of $\phi$
}}
{{end-eqn}}
As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$.
Thus $\phi$ is a group homomorphism from $G$ to $G$.
So by definition, $\phi$ is a group endomorphism.
{{qed|lemma}}
Theorem: Additive Groups of Integers and Integer Multiples are Isomorphic

Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $\struct {n \Z, +}$ denote the additive group of integer multiples.
Let $\struct {\Z, +}$ denote the additive group of integers.
Then $\struct {n \Z, +}$ is isomorphic to $\struct {\Z, +}$.

Proof:
We have that:
:Infinite Cyclic Group is Isomorphic to Integers.
:Integer Multiples under Addition form Infinite Cyclic Group.
:Infinite Cyclic Group is Unique up to Isomorphism
Hence the result.
{{qed}}
Theorem: Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers

Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.
Then $\struct {\R, +}$ is not isomorphic to $\struct {\R_{\ne 0}, \times}$.

Proof:
Consider the element $-1 \in \struct {\R_{\ne 0}, \times}$.
We have that:
:$-1 \times -1 = 1$
From Real Multiplication Identity is One it follows that $-1$ is of order $2$ in $\struct {\R_{\ne 0}, \times}$.
From Group Isomorphism Preserves Order of Group Element, it is sufficient to demonstrate that there exists no element of order $2$ in $\struct {\R, +}$.
From Real Addition Identity is Zero, that means finding $x$ such that $x + x = 0$.
But:
:$x + x = 0 \implies x = 0$
and as $0$ is the identity of $\struct {\R, +}$, there is indeed no element of order $2$ in $\struct {\R, +}$.
The result follows from Group Isomorphism Preserves Order of Group Element.
{{qed}}
Theorem: Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers

Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.
Then $\struct {\R, +}$ is not isomorphic to $\struct {\R_{\ne 0}, \times}$.

Proof:
There are two element of $\struct {\R_{\ne 0}, \times}$ which are self-inverse, that is, $-1$ and $1$.
However, there is only one element of $\struct {\R, +}$ which is self-inverse, that is, $0$.
{{AimForCont}} there exists an isomorphism $f: \struct {\R_{\ne 0}, \times} \to \struct {\R, +}$.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | r = \map f 1
      | c = 
}}
{{eqn | r = \map f {-1 \times -1}
      | c = 
}}
{{eqn | r = \map f {-1} + \map f {-1}
      | c = 
}}
{{eqn | r = 2 \map f {-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map f {-1}
      | r = 0
      | c = 
}}
{{eqn | r = \map f 1
      | c = 
}}
{{end-eqn}}
So we have demonstrated that $f$ is not an injection.
Hence $f$ is not a bijection and so not an isomorphism.
It follows from Proof by Contradiction that there can be no such isomorphism.
Hence the result.
{{qed}}
Theorem: Normal Subgroup is Kernel of Group Homomorphism

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Then there exists a group homomorphism of which $N$ is the kernel.

Proof:
Let $G / N$ be the quotient group of $G$ by $N$.
Let $q_N: G \to G / N$ be the quotient epimorphism from $G$ to $G / N$:
:$\forall x \in G: \map {q_N} x = x N$
Then from Quotient Group Epimorphism is Epimorphism, $N$ is the kernel of $q_n$
Thus $q_N$ is that group homomorphism of which $N$ is the kernel.
{{qed}}
Theorem: Homomorphic Image of Cyclic Group is Cyclic Group

Let $G$ be a cyclic group with generator $g$.
Let $H$ be a group.
Let $\phi: G \to H$ be a (group) homomorphism.
Let $\Img G$ denote the homomorphic image of $G$ under $\phi$.
Then $\Img G$ is a cyclic group with generator $\map \phi g$.
That is:
:$\phi \sqbrk {\gen g} = \gen {\map \phi g}$

Proof:
Let $y \in \Img G$.
Then $\exists x \in G: y = \map \phi x$.
As $G$ be a cyclic group with generator $g$, $x = g^n$ for some $n \in \Z$.
Thus by Homomorphism of Power of Group Element:
:$y = \paren {\map \phi g}^n$
and so is a power of $\map \phi g$.
As $y$ is arbitrary, it follows that all elements of $\Img G$ are powers of $\map \phi g$.
Hence the result, by definition of cyclic group.
{{qed}}
Theorem: Power of Group Element in Kernel of Homomorphism iff Power of Image is Identity

Let $G$ be a group whose identity is $e_G$.
Let $H$ be a group whose identity is $e_H$.
Let $\phi: G \to H$ be a (group) homomorphism.
Let $x^n \in \map \ker \phi$ for some integer $n$.
Then:
:$\paren {\map \phi x}^n = e_H$

Proof:
{{begin-eqn}}
{{eqn | l = x^n
      | o = \in
      | r = \map \ker \phi
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi {x^n}
      | r = e_H
      | c = {{Defof|Kernel of Group Homomorphism}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {\map \phi x}^n
      | r = e_H
      | c = Homomorphism of Power of Group Element
}}
{{end-eqn}}
{{qed}}
Theorem: Kernel of Homomorphism on Cyclic Group

Let $G = \gen g$ be a cyclic group with generator $g$.
Let $H$ be a group.
Let $\phi: G \to H$ be a (group) homomorphism.
Let $\map \ker \phi$ denote the kernel of $\phi$.
Let $\Img G$ denote the homomorphic image of $G$ under $\phi$.
Then:
:$\map \ker \phi = \gen {g^m}$
where:
:$m = 0$ if $\Img \phi$ is an infinite cyclic group
:$m = \order {\Img \phi}$ if $\Img \phi$ is a finite cyclic group.

Proof:
From Kernel of Group Homomorphism is Subgroup and Subgroup of Cyclic Group is Cyclic:
:$\exists m \in \N: \map \ker \phi = \gen {g^m}$
From Homomorphic Image of Cyclic Group is Cyclic Group:
:$\Img \phi$ is a cyclic group generated by $\map \phi g$.
Theorem: Mapping from Group Element to Inner Automorphism is Homomorphism

Let $G$ be a group.
Let $\kappa: G \to \Aut G$ be the mapping from $G$ to the automorphism group of $G$ defined as:
:$\forall x \in G: \map \kappa x := \kappa_x$
where $\kappa_x$ is the inner automorphism on $x$:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
Then $\kappa$ is a homomorphism.

Proof:
Let $x, y \in G$.
By definition of automorphism group, we have that:
:$\map \kappa x \map \kappa y = \kappa_x \circ \kappa_y$
where $\circ$ denotes composition of mappings.
Then $\forall g \in G$:
{{begin-eqn}}
{{eqn | l = \map {\kappa_x \circ \kappa_y} g
      | r = \map {\kappa_x} {\map {\kappa_y} g}
      | c = for all $g \in G$
}}
{{eqn | r = \map {\kappa_x} {y g y^{-1} }
      | c = Definition of $\kappa_y$
}}
{{eqn | r = x \paren {y g y^{-1} } x^{-1}
      | c = Definition of $\kappa_x$
}}
{{eqn | r = \paren {x y} g \paren {y^{-1} x^{-1} }
      | c = {{Group-axiom|1}}
}}
{{eqn | r = \paren {x y} g \paren {x y}^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = \map {\kappa_{x y} } g
      | c = Definition of $\kappa_{x y}$
}}
{{end-eqn}}
And so:
:$\forall g \in G: \map {\kappa_{x y} } g = \map {\kappa_x \circ \kappa_y} g$
Thus by definition of $\kappa$:
:$\map \kappa x \map \kappa y = \map \kappa{x y}$
demonstrating that $\kappa$ is a homomorphism.
{{qed}}
Theorem: Image of Mapping from Group Element to Inner Automorphism is Inner Automorphism Group

Let $G$ be a group.
Let $\kappa: G \to \Aut G$ be the mapping from $G$ to the automorphism group of $G$ defined as:
:$\forall x \in G: \map \kappa x := \kappa_x$
where $\kappa_x$ is the inner automorphism on $x$:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
Then $\Img \kappa$ is the inner automorphism group of $G$.

Proof:
Let $\Inn G$ denote the inner automorphism group of $G$.
For all $x \in G$, $\map \kappa x = \kappa_x \in \Inn G$.
Hence $\Img \kappa \subseteq \Inn G$. 
Let $\phi \in \Inn G$. Then:
:$\exists y \in G: \forall g \in G: \map \phi g = y g y^{-1}$
Then $\map \kappa y = \phi$.
Hence $\Inn G \subseteq \Img \kappa$. 
Therefore $\Img \kappa = \Inn G$.
{{qed}}
Theorem: Order of Monomorphic Image of Group Element

Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.
Let $\phi: G \to H$ be a monomorphism.
Let $g \in G$ be of finite order.
Then:
:$\forall g \in G: \order {\map \phi g} = \order g$

Proof:
By definition of monomorphism, $\phi$ is a homomorphism which is also an injection.
From Order of Homomorphic Image of Group Element:
:$\forall g \in G: \order {\map \phi g} \divides \order g$
{{begin-eqn}}
{{eqn | l = \map \phi {g^m}
      | r = \paren {\map \phi g}^m
      | c = Homomorphism of Power of Group Element
}}
{{eqn | r = e_H
      | c = 
}}
{{eqn | r = \map \phi {e_G}
      | c = Homomorphism to Group Preserves Identity
}}
{{end-eqn}}
So $g^m = e$, as $\phi$ is injective.
From the definition of order of group element, that means $n \le m$ since $n$ is the ''smallest'' such power.
Thus $m = n$ and the result holds.
{{Qed}}
Theorem: Image under Epimorphism of Center is Subset of Center

Let $G$ and $H$ be groups.
Let $\theta: G \to H$ be an epimorphism.
Let $\map Z G$ denote the center of $G$.
Then:
:$\theta \sqbrk {\map Z G} \subseteq \map Z H$

Proof:
Let $y \in \theta \sqbrk {\map Z G}$.
Let $t \in H$.
We have that:
:$y = \map \theta z$
for some $z \in \map Z G$
As $\theta$ is an epimorphism, it is by definition surjective.
Then:
:$t = \map \theta s$
for some $s \in G$.
Hence:
{{begin-eqn}}
{{eqn | l = y t
      | r = \map \theta z \map \theta s
      | c = 
}}
{{eqn | r = \map \theta {z s}
      | c = {{Defof|Group Epimorphism}}
}}
{{eqn | r = \map \theta {s z}
      | c = {{Defof|Center of Group}}
}}
{{eqn | r = \map \theta s \map \theta z
      | c = {{Defof|Group Epimorphism}}
}}
{{eqn | r = t y
      | c = {{Defof|Group Epimorphism}}
}}
{{end-eqn}}
As $t$ is arbitrary, it follows that $y \in \map Z H$.
The result follows.
{{qed}}
Theorem: Metric Subspace Induces Subspace Topology

Let $M = \struct {A,d}$ be a metric space.
Let $H \subseteq A$.
Let $\tau$ be the topology induced by the metric $d$.
Let $\tau_H$ be the subspace topology induced by $\tau$ on $H$.
Let $d_H$ be the subspace metric induced by $d$ on $H$.
Let $\tau_{d_H}$ be the topology induced by the metric $d_H$.
Then:
:$\tau_{d_H} = \tau_H$

Proof:
Let $\BB$ be the set of open $\epsilon$-balls in $M$.
Let $\BB_H$ be the set of open $\epsilon$-balls in $\struct {H, d_H}$.
Let $U \in \tau_{d_H}$.
By the definition of the topology induced by the metric $d_H$:
:$\exists \AA_H \subseteq \BB_H: U = \bigcup \AA_H$
Let $\AA = \set {B': B' \in \BB, B' \cap H \in \AA_H}$.
Let $V = \bigcup \set {B' : B' \in \AA}$.
By the definition of the topology induced by the metric $d$:
:$V \in \tau$
By the definition of the subspace metric:
:$\forall B \in \BB_H: \exists B' \in \BB: B = B' \cap H$
Hence:
{{begin-eqn}}
{{eqn | l = U
      | r = \bigcup \set {B' \cap H : B' \in \AA}
}}
{{eqn | r = \paren {\bigcup \set {B' : B' \in \AA} } \cap H
      | c = Intersection Distributes over Union
}}
{{eqn|r= V \cap H
}}
{{end-eqn}}
By the definition of the subspace topology induced by $\tau$ on $H$:
:$U \in \tau_H$
Hence:
:$\tau_{d_H} \subseteq \tau_H$.
{{qed|lemma}}
Let $U \in \tau_H$.
By the definition of the subspace topology induced by $\tau$ on $H$:
:$\exists V \in \tau : U = V \cap H$
By the definition of the topology induced by the metric $d$:
:$\exists \AA \subseteq \BB: V = \bigcup \set {B' : B' \in \AA}$
Hence:
{{begin-eqn}}
{{eqn | l = U
      | r = \paren {\bigcup \set {B' : B' \in \AA} } \cap H
}}
{{eqn | r = \bigcup \set {B' \cap H : B' \in \AA}
      | c = Intersection Distributes over Union
}}
{{end-eqn}}
By the definition of the subspace metric:
:$\forall B' \in \BB: B' \cap H \in \BB_H$
By the Definition of the topology induced by the metric $d_H$:
:$U \in \tau_{d_H}$
Hence:
:$\tau_H \subseteq \tau_{d_H}$
By Equivalence of Definitions of Set Equality:
:$\tau_H = \tau_{d_H}$
{{qed}}
Theorem: Index of Intersection of Subgroups/Corollary

Let $H$ be a subgroup of $G$.
Let $K$ be a subgroup of finite index of $G$.
Then:
:$\index H {H \cap K} \le \index G K$

Proof:
Note that $H \cap K$ is a subgroup of $H$.
From Index of Intersection of Subgroups, we have:
:$\index G {H \cap K} \le \index G H \index G K$
Setting $G = H$, we have:
:$\index H {H \cap K} \le \index H H \index H K$
{{finish|This does not get us where we want}}
{{CircularStructure|This is Index in Subgroup, which is used to prove the main theorem}}
Theorem: Quotient of Cauchy Sequences is Metric Completion/Lemma 1

:$\quad \CC \,\big / \NN = \tilde \CC$

Proof:
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $\CC$.
Then:
{{begin-eqn}}
{{eqn | l = \sequence {x_n} + \NN = \sequence {y_n} + \NN
      | o = \leadstoandfrom
      | r = \sequence {x_n} - \sequence {y_n} \in \NN
      | c = Cosets are Equal iff Product with Inverse in Subgroup
}}
{{eqn | r = \lim_{n \mathop \to \infty} x_n - y_n = 0_R
      | o = \leadstoandfrom
      | c = Definition of $\NN$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n - y_n} = 0
      | o = \leadstoandfrom
      | c = {{Defof|Convergent Sequence in Normed Division Ring}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map d {x_n - y_n} = 0
      | o = \leadstoandfrom
      | c = {{Defof|Metric Induced by Norm on Division Ring}}
}}
{{eqn | r = \sequence {x_n} \sim \sequence {y_n}
      | o = \leadstoandfrom
      | c = {{Defof|Equivalence Relation}} $\sim$
}}
{{end-eqn}}
Hence:
:$\sequence {x_n}$ and $\sequence {y_n}$ belong to the same equivalence class in $\CC \,\big / \NN$ {{iff}} $\sequence {x_n}$ and $\sequence {y_n}$ belong to the same equivalence class in $\tilde \CC$.
The result follows.
{{qed}}
Theorem: Quotient of Cauchy Sequences is Metric Completion/Lemma 2

:$\quad d' = \tilde d$

Proof:
By Lemma 1 of Quotient of Cauchy Sequences is Metric Completion we have that:
:$\CC \,\big / \NN = \tilde {\CC}$
Let $\eqclass {x_n} {}$ and $\eqclass {x_n} {}$ be equivalence classes in $\CC \,\big / \NN = \tilde {\CC}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d'} {\eqclass {x_n}{}, \eqclass {x_n}{} }
      | r = \norm {\eqclass {x_n}{} - \eqclass {x_n}{} }
      | c = Definition of $d'$
}}
{{eqn | r = \norm {\eqclass {x_n - y_n}{} }
      | c = {{Defof|Quotient Ring}} Addition
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n - y_n}
      | c = {{Defof|Norm on Division Ring}}: $\norm {\,\cdot\,}$ on $\CC \,\big / \NN$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map d {x_n, y_n}
      | c = {{Defof|Metric Induced by Norm on Division Ring}}: by the norm $\norm {\,\cdot\,}$ on $R$
}}
{{eqn | r = \map {\tilde d} {\eqclass {x_n}{}, \eqclass {x_n}{} }
      | c = Definition of $\tilde d$
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Group of Order 15 is Cyclic Group

Let $G$ be a group whose order is $15$.
Then $G$ is cyclic.

Proof:
We have that $15 = 3 \times 5$.
Thus:
:$15$ is square-free
:$5 \equiv 2 \pmod 3$
:$3 \equiv 3 \pmod 5$
The conditions are fulfilled for Condition for Nu Function to be 1.
Thus $\map \nu {15} = 1$ and so all groups of order $15$ are cyclic.
{{Qed}}
Theorem: Group of Order 15 is Cyclic Group

Let $G$ be a group whose order is $15$.
Then $G$ is cyclic.

Proof:
From Number of Sylow p-Subgroups in Group of Order 15:
:the number of Sylow $3$-subgroups is in the set $\set {1, 4, 7, \ldots}$
:the number of Sylow $5$-subgroups is in the set $\set {1, 6, 11, \ldots}$.
From the Fifth Sylow Theorem
:the number of Sylow $3$-subgroups is a divisor of $15$
:the number of Sylow $5$-subgroups is a divisor of $15$.
Combining the above:
:there is a unique Sylow $3$-subgroup in $G$, which we will denote $P$
:there is a unique Sylow $5$-subgroup in $G$, which we will denote $Q$.
Let $x \in G$ be of order $3$.
Then by Non-Trivial Group has Non-Trivial Cyclic Subgroup, $\gen x$ has $3$ elements.
Thus $\gen x$ is a Sylow $3$-subgroup and so:
:$\gen x = P$
and so $x$ is one of the $2$ non-identity elements of $P$.
Similarly, let $y \in G$ be of order $5$.
Then by a similar argument:
:$\gen y = Q$
and so $y$ is one of the $4$ non-identity elements of $Q$.
So $G$ has:
:$1$ element of order $1$ (that is, the identity)
:$2$ elements of order $3$
:$4$ elements of order $5$
which leaves $8$ elements whose order is still to be determined.
From Order of Element Divides Order of Finite Group, they are all of order $1$, $3$, $5$ or $15$.
As the elements of order $1$, $3$ and $5$ have been accounted for, they must all be of order $15$.
So $G$ has $8$ distinct elements of order $15$.
Hence $G$ must be cyclic.
{{Qed}}