avewright/proofwiki-math huggingface dataset:
Theorem: Subgroup Generated by Infinite Order Element is Infinite

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:
{{AimForCont}} $\gen a$ 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.
{{qed}}
Theorem: Element of Cyclic Group is not necessarily Generator

Let $\gen g = G$ be a cyclic group.
Let $a \in G$
Then it is not necessarily the case that $a$ is also a generator of $G$.

Proof:
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
Consider the subgroup $\gen 2$ of $\struct {\R_{\ne 0}, \times}$ generated by $2$.
By definition, $\gen 2$ is a cyclic group.
Consider the element $4 \in \struct {\R_{\ne 0}, \times}$.
We have that $4 = 2^2$.
Thus $4 \in \gen 2$.
There exists no $n \in \Z$ such that $4^n = 2$.
But $2 \in \gen 2$.
Thus $\gen 2$ is not generated by $4$.
That is, $4$ is not a generator of $\gen 2$.
Hence the result, by Proof by Counterexample.
{{qed}}
Theorem: Graph of Nonlinear Additive Function is Dense in the Plane

Let $f: \R \to \R$ be an additive function which is not linear.
Then the graph of $f$ is dense in the real number plane.

Proof:
From Additive Function is Linear for Rational Factors:
:$\map f q = q \map f 1$
for all $q \in \Q$.
{{WLOG}}, let:
:$\map f q = q$
for all $q \in \Q$.
Since $f$ is not linear, let $\alpha \in \R \setminus \Q$ be such that:
:$\map f \alpha = \alpha + \delta$
for some $\delta \ne 0$.
Consider an arbitrary nonempty circle in the plane.
Let:
:its centre be $\tuple {x, y}$ where $x \ne y$ and $x, y \in \Q$
:its radius be $r > 0$.
We will show how to find a point of the graph of $f$ inside this circle.
As $x \ne y$ and $r$ can be arbitrarily small, this will prove the theorem.
Since $\delta \ne 0$, let $\beta = \dfrac {y - x} \delta$.
Since $x \ne y$:
:$\beta \ne 0$
As Rationals are Everywhere Dense in Topological Space of Reals,
there exists a rational number $b \ne 0$ such that:
:$\size {\beta - b} < \dfrac r {2 \size \delta}$
As Rationals are Everywhere Dense in Topological Space of Reals, there also exists a rational number $a$ such that:
:$\size {\alpha - a} < \dfrac r {2 \size b}$
Now put:
:$X = x + b \paren {\alpha - a}$
:$Y = \map f X$
Then:
:$\size {X - x} = \size {b \paren {\alpha - a} } < \frac r 2$
so $X$ is in the circle.
Then:
{{begin-eqn}}
{{eqn | l = Y 
      | r = \map f {x + b \paren {\alpha - a} } 
      | c = Definition of $Y$ and $X$ 
}}
{{eqn | r = \map f x + \map f {b \alpha} - \map f {b a}
      | c = {{Defof|Cauchy Functional Equation}}
}}
{{eqn | r = x + b \map f \alpha - b \map f a
      | c = Additive Function is Linear for Rational Factors
}}
{{eqn | r = y - \delta \beta + b \map f \alpha - b \map f a
      | c = Definition of $y$
}}
{{eqn | r = y - \delta \beta + b \paren {\alpha + \delta} - b a
      | c = Additive Function is Linear for Rational Factors
}}
{{eqn | r = y + b \paren {\alpha - a} - \delta \paren {\beta - b}
}}
{{end-eqn}}
Therefore 
:$\size {Y - y} = \size {b \paren {\alpha - a} - \delta \paren {\beta - b} } \le \size {b \paren {\alpha - a} } + \size {\delta \paren {\beta - b} } \le r$
so $Y$ is in the circle as well.
Hence the point $\tuple {X, Y}$ is inside the circle.
{{qed}}
Theorem: Order of Elements in Quaternion Group

Let $Q = \Dic 2$ be the quaternion group, whose group presentation is given by:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
Then $\Dic 2$ has:
:$1$ element of order $2$
and:
:$6$ elements of order $4$.

Proof:
From Identity is Only Group Element of Order 1, the identity element $e$ , and only $e$, is of order $1$.
From here, we inspect the Cayley table:
{{:Quaternion Group/Cayley Table}}
It is immediately seen that:
:$\paren {a^2}^2 = e$
and so by definition $a^2$ is of order $2$.
As can be seen from inspection of the main diagonal, for all other $x \in Q$, we have:
:$x^2 = a^2 \ne e$
and so:
:$x^4 = \paren {a^2}^2 = e$
There are $6$ such elements.
Hence the result by definition of order of group element.
{{qed}}
Theorem: Multiplicative Group of Reduced Residues Modulo 7 is Cyclic

Let $\struct {\Z'_7, \times_7}$ denote the multiplicative group of reduced residues modulo $7$.
Then $\struct {\Z'_7, \times_7}$ is cyclic.

Proof:
From Reduced Residue System under Multiplication forms Abelian Group it is noted that $\struct {\Z'_7, \times_7}$ is a group.
It remains to be shown that $\struct {\Z'_7, \times_7}$ is cyclic.
It will be demonstrated that:
:$\gen {\eqclass 3 7} = \struct {\Z'_7, \times_7}$
That is, that $\eqclass 3 7$ is a generator of $\struct {\Z'_7, \times_7}$.
We note that $\eqclass 1 7$ is the identity element of $\struct {\Z'_7, \times_7}$.
Thus successive powers of $\eqclass 3 7$ are taken, until $n \in \Z$ is found such that $\eqclass 3 7^n = \eqclass 1 7$:
{{begin-eqn}}
{{eqn | l = \eqclass 3 7^2
      | r = \eqclass {3 \times 3} 7
      | c = 
}}
{{eqn | r = \eqclass 9 7
      | c = 
}}
{{eqn | r = \eqclass 2 7
      | c = 
}}
{{eqn | l = \eqclass 3 7^3
      | r = \eqclass 3 7^2 \times \eqclass 3 7
      | c = 
}}
{{eqn | r = \eqclass {2 \times 3} 7
      | c = 
}}
{{eqn | r = \eqclass 6 7
      | c = 
}}
{{eqn | l = \eqclass 3 7^4
      | r = \eqclass 3 7^3 \times \eqclass 3 7
      | c = 
}}
{{eqn | r = \eqclass {6 \times 3} 7
      | c = 
}}
{{eqn | r = \eqclass {18} 7
      | c = 
}}
{{eqn | r = \eqclass 4 7
      | c = 
}}
{{eqn | l = \eqclass 3 7^5
      | r = \eqclass 3 7^4 \times \eqclass 3 7
      | c = 
}}
{{eqn | r = \eqclass {4 \times 3} 7
      | c = 
}}
{{eqn | r = \eqclass {12} 7
      | c = 
}}
{{eqn | r = \eqclass 5 7
      | c = 
}}
{{eqn | l = \eqclass 3 7^6
      | r = \eqclass 3 7^5 \times \eqclass 3 7
      | c = 
}}
{{eqn | r = \eqclass {5 \times 3} 7
      | c = 
}}
{{eqn | r = \eqclass {15} 7
      | c = 
}}
{{eqn | r = \eqclass 1 7
      | c = 
}}
{{end-eqn}}
All elements of $\struct {\Z'_7, \times_7}$ are seen to be in $\gen {\eqclass 3 7}$.
Hence the result by definition of cyclic group.
{{qed}}
Theorem: Reduced Residues Modulo 5 under Multiplication form Cyclic Group

Let $\struct {\Z'_5, \times_5}$ denote the multiplicative group of reduced residues modulo $5$.
Then $\struct {\Z'_5, \times_5}$ is cyclic.

Proof:
From Reduced Residue System under Multiplication forms Abelian Group it is noted that $\struct {\Z'_5, \times_5}$ is a group.
It remains to be shown that $\struct {\Z'_5, \times_5}$ is cyclic.
It will be demonstrated that:
:$\gen {\eqclass 2 5} = \struct {\Z'_5, \times_5}$
That is, that $\eqclass 2 5$ is a generator of $\struct {\Z'_5, \times_5}$.
We note that $\eqclass 1 5$ is the identity element of $\struct {\Z'_5, \times_5}$.
Thus successive powers of $\eqclass 2 5$ are taken, until $n \in \Z$ is found such that $\eqclass 2 5^n = \eqclass 1 5$:
{{begin-eqn}}
{{eqn | l = \eqclass 2 5^2
      | r = \eqclass {2 \times 2} 5
      | c = 
}}
{{eqn | r = \eqclass 4 5
      | c = 
}}
{{eqn | l = \eqclass 2 5^3
      | r = \eqclass 2 5^2 \times \eqclass 2 5
      | c = 
}}
{{eqn | r = \eqclass {2 \times 4} 5
      | c = 
}}
{{eqn | r = \eqclass 8 5
      | c = 
}}
{{eqn | r = \eqclass 3 5
      | c = 
}}
{{eqn | l = \eqclass 2 5^4
      | r = \eqclass 2 5^3 \times \eqclass 2 5
      | c = 
}}
{{eqn | r = \eqclass {3 \times 2} 5
      | c = 
}}
{{eqn | r = \eqclass 6 5
      | c = 
}}
{{eqn | r = \eqclass 1 5
      | c = 
}}
{{end-eqn}}
All elements of $\struct {\Z'_5, \times_5}$ are seen to be in $\gen {\eqclass 2 5}$.
Hence the result by definition of cyclic group.
{{qed}}
Category:Multiplicative Groups of Reduced Residues
Category:Multiplicative Group of Reduced Residues Modulo 5
Category:Examples of Cyclic Groups
Category:Cyclic Group of Order 4
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Theorem: Number of Generators of Cyclic Group whose Order is Power of 2

Let $G$ be a finite cyclic group.
Let the order of $G$ be $2^k$ for some $k \in \Z_{>0}$.
Then $G$ has $2^{n - 1}$ distinct generators.

Proof:
From Finite Cyclic Group has Euler Phi Generators, $G$ has $\map \phi {2^n}$ generators.
The result follows from {{Corollary|Euler Phi Function of Prime Power}}.
{{qed}}
Theorem: Equivalent Statements for Congruence Modulo Subgroup/Left

Let $x \equiv^l y \pmod H$ denote that $x$ is left congruent modulo $H$ to $y$.
Then the following statements are equivalent:
{{begin-eqn}}
{{eqn | n = 1
      | l = x
      | o = \equiv^l
      | r = y \pmod H
}}
{{eqn | n = 2
      | l = x^{-1} y
      | o = \in
      | r = H
}}
{{eqn | n = 3
      | q = \exists h \in H
      | l = x^{-1} y
      | r = h
}}
{{eqn | n = 4
      | q = \exists h \in H
      | l = y
      | r = x h
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \equiv^l
      | r = y \pmod H
}}
{{eqn | ll= \leadstoandfrom
      | l = x^{-1} y
      | o = \in
      | r = H
      | c = {{Defof|Left Congruence Modulo Subgroup}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists h \in H
      | l = x^{-1} y
      | r = h
      | c = {{Defof|Element}} of $H$
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists h \in H
      | l = y
      | r = x h
      | c = Division Laws for Groups
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalent Statements for Congruence Modulo Subgroup/Right

Let $x \equiv^r y \pmod H$ denote that $x$ is right congruent modulo $H$ to $y$.
Then the following statements are equivalent:
{{begin-eqn}}
{{eqn | n = 1
      | l = x
      | o = \equiv^r
      | r = y \pmod H
}}
{{eqn | n = 2
      | l = x y^{-1}
      | o = \in
      | r = H
}}
{{eqn | n = 3
      | q = \exists h \in H
      | l = x y^{-1}
      | r = h
}}
{{eqn | n = 4
      | q = \exists h \in H
      | l = x
      | r = h y
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \equiv^r
      | r = y \pmod H
}}
{{eqn | ll= \leadstoandfrom
      | l = x y^{-1}
      | o = \in
      | r = H
      | c = {{Defof|Right Congruence Modulo Subgroup}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists h \in H
      | l = x y^{-1}
      | r = h
      | c = {{Defof|Element}} of $H$
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists h \in H
      | l = x
      | r = h y
      | c = Division Laws for Groups
}}
{{end-eqn}}
{{qed}}
Theorem: Element of Group is in its own Coset/Left

Let:
: $x H$ be the left coset of $x$ modulo $H$.
Then:
: $x \in x H$

Proof:
Let $e$ be the identity of $G$.
Then:
{{begin-eqn}}
{{eqn | l = e
      | o = \in
      | r = H
      | c = Identity of Subgroup
}}
{{eqn | l = x
      | r = x e
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | q = \exists h \in H
      | l = x
      | r = x h
      | c = Existential Generalisation
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = x H
      | c = {{Defof|Left Coset}}
}}
{{end-eqn}}
{{Qed}}
Theorem: Element of Group is in its own Coset/Right

Let:
: $H x$ be the right coset of $x$ modulo $H$.
Then:
: $x \in H x$

Proof:
Let $e$ be the identity of $G$.
Then:
{{begin-eqn}}
{{eqn | l = e
      | o = \in
      | r = H
      | c = Identity of Subgroup
}}
{{eqn | l = x
      | r = e x
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | q = \exists h \in H
      | l = x
      | r = h x
      | c = Existential Generalisation
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = H x
      | c = {{Defof|Right Coset}}
}}
{{end-eqn}}
{{Qed}}
Theorem: Element of Group is in Unique Coset of Subgroup/Left

There exists a exactly one left coset of $H$ containing $x$, that is: $x H$

Proof:
Follows directly from:
* Left Congruence Modulo Subgroup is Equivalence Relation
* Element in its own Equivalence Class.
{{qed}}
Theorem: Element of Group is in Unique Coset of Subgroup/Right

There exists a exactly one right coset of $H$ containing $x$, that is: $H x$

Proof:
Follows directly from:
* Right  Congruence Modulo Subgroup is Equivalence Relation
* Element in its own Equivalence Class.
{{qed}}
Theorem: Equivalence of Definitions of Infinite Order Element

{{TFAE|def = Infinite Order Element}}
Let $G$ be a group whose identity is $e_G$.
Let $x \in G$ be an element of $G$.

Proof:
=== $(1)$ implies $(2)$ ===
Let $x$ be an infinite order element of $G$ by definition 1.
Then by definition there exists no $k \in \Z_{>0}$ such that $x^k = e_G$.
{{AimForCont}} not all $x^r$ are distinct for all $r \in \Z_{>0}$.
Then $x^m = x^n$ for some $m, n \in \Z$ where $m \ne n$.
{{WLOG}}, let $m > n$.
Then:
: $x^{m - n} = e^G$
As $m > n$ we have that $m - n \in \Z_{>0}$.
Let $k = m - n$.
and so:
:$\exists k \in \Z_{>0}$ such that $x^k = e_G$.
But this contradicts the statement that no such $k$ exists.
Hence no such distinct $m$ and $n$ exist.
Thus $x$ is an infinite order element of $G$ by definition 2.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Finite Order Element

{{TFAE|def = Finite Order Element}}
Let $G$ be a group whose identity is $e_G$.
Let $x \in G$ be an element of $G$.

Proof:
=== $(1)$ implies $(2)$ ===
Let $x$ be a finite order element of $G$ by definition 1.
Then by definition there exists $k \in \Z_{>0}$ such that $x^k = e_G$.
Consider some $m, n \in \Z_{>0}$ such that $m = n + k$.
{{begin-eqn}}
{{eqn | l = x^m
      | r = x^{n + k}
      | c = {{hypothesis}}
}}
{{eqn | r = x^n x^k
      | c = Powers of Group Elements/Sum of Indices
}}
{{eqn | r = x^n e_G
      | c = {{hypothesis}}
}}
{{eqn | r = x^n
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
Thus $x$ is a finite order element of $G$ by definition 2.
{{qed|lemma}}
Theorem: Order of Element in Group equals its Order in Subgroup

Let $G$ be a group.
Let $H \le G$, where $\le$ denotes the property of being a subgroup.
Let $x \in H$.
Then the order of $x$ in $H$ equals the order of $x$ in $G$.

Proof:
Let $\gen x$ be the subgroup of $G$ generated by $x$.
By definition, $\gen x \le G$.
All the elements of $\gen x$ are powers of $x$.
As $x \in H$ it follows by {{Group-axiom|0}} that all the powers of $x$ are elements of $H$.
That is:
:$\gen x \le H$
By Order of Cyclic Group equals Order of Generator:
:$\order x = \order {\gen x}$
But $\gen x$ is the same cyclic group whether it is considered as embedded in $H$ or $G$.
Hence the result.
{{qed}}
Theorem: Order of Product of Commuting Group Elements of Coprime Order is Product of Orders

Let $G$ be a group.
Let $g_1, g_2 \in G$ be commuting elements such that:
{{begin-eqn}}
{{eqn | l = \order {g_1}
      | r = n_1
}}
{{eqn | l = \order {g_2}
      | r = n_2
}}
{{end-eqn}}
where $\order {g_1}$ denotes the order of $g_1$ in $G$.
Let $n_1$ and $n_2$ be coprime.
Then:
:$\order {g_1 g_2} = n_1 n_2$

Proof:
Let $g_1 g_2 = g_2 g_1$.
We have:
:$\paren {g_1 g_2}^{n_1 n_2} = e$
Thus:
:$\order {g_1 g_2} \le n_1 n_2$
Suppose $\order {g_1 g_2}^r = e$.
Then:
{{begin-eqn}}
{{eqn | l = {g_1}^r
      | r = {g_2}^{-r}
      | c = 
}}
{{eqn | o = \in
      | r = \gen {g_1} \cap \gen {g_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = {g_1}^r = {g_2}^r
      | r = e
      | c = 
}}
{{end-eqn}}
Thus $r$ is divisible by both $m$ and $n$.
The result follows.
{{qed}}
Theorem: Right Cosets are Equal iff Left Cosets by Inverse are Equal

Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $g_1, g_2 \in G$.
Then:
:$H g_1 = H g_2 \iff {g_1}^{-1} H = {g_2}^{-1} H$
where:
:${g_1}^{-1}$ and ${g_2}^{-1}$ denote the inverses of $g_1$ and $g_2$ in $G$
:$H g_1$ and $H g_2$ denote the right cosets of $H$ by $g_1$ and $g_2$ respectively
:${g_1}^{-1} H$ and ${g_2}^{-1} H$ denote the left cosets of $H$ by ${g_1}^{-1}$ and ${g_2}^{-1}$ respectively.

Proof:
{{begin-eqn}}
{{eqn | l = H g_1
      | r = H g_2
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = g_1 {g_2}^{-1}
      | o = \in
      | r = H
      | c = Right Cosets are Equal iff Product with Inverse in Subgroup
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren { {g_1}^{-1} }^{-1} {g_2}^{-1}
      | o = \in
      | r = H
      | c = Inverse of Group Inverse
}}
{{eqn | ll= \leadstoandfrom
      | l = {g_1}^{-1} H
      | r = {g_2}^{-1} H
      | c = Left Cosets are Equal iff Product with Inverse in Subgroup
}}
{{end-eqn}}
{{qed}}
Theorem: Subgroup of Subgroup with Prime Index

Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let $K$ be a subgroup of $H$.
Let:
:$\index G K = p$
where:
:$p$ denotes a prime number
:$\index G K$ denotes the index of $K$ in $G$.
Then either:
:$H = K$
or:
:$H = G$

Proof:
From the Tower Law for Subgroups:
:$\index G K = \index G H \index H K$
As $\index G K = p$ is prime, either $\index G H = p$ or $\index H K = p$.
Thus either $\index G H = 1$ or $\index H K = 1$.
The result follows from Index is One iff Subgroup equals Group.
{{qed}}
Theorem: Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Reverse Implication

:$\vdash \paren {\paren {p \lor q} \land \paren {p \lor r} } \implies \paren {p \lor \paren {q \land r} }$

Proof:
{{finish|Use formulation 1 + Modus Ponens}}
Theorem: Completeness Theorem for Hilbert Proof System Instance 2 and Boolean Interpretations

Instance 2 of the Hilbert proof systems is a complete proof system for boolean interpretations.
That is, for every WFF $\mathbf A$:
:$\models_{\mathrm{BI}} \mathbf A$ implies $\vdash_{\mathscr H_2} \mathbf A$

Proof:
{{ProofWanted|This is going to be fun}}
Theorem: Symmetry Group of Rectangle is Klein Four-Group

The symmetry group of the rectangle is the Klein $4$-group.

Proof:
Comparing the Cayley tables of the symmetry group of the rectangle with the Klein $4$-group the isomorphism can be seen: 
{{:Symmetry Group of Rectangle/Cayley Table}}
{{:Klein Four-Group/Cayley Table}}
Thus the required isomorphism $\phi$ can be set up as:
{{begin-eqn}}
{{eqn | l = \map \phi e
      | r = e
}}
{{eqn | l = \map \phi r
      | r = a
}}
{{eqn | l = \map \phi h
      | r = b
}}
{{eqn | l = \map \phi v
      | r = c
}}
{{end-eqn}}
{{qed}}
Theorem: Klein Four-Group as Subgroup of S4

Let $G$ be the following subset of the symmetric group on $4$ letters $S_4$, expressed in two-row notation:
{{begin-eqn}}
{{eqn | l = e
      | r = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{bmatrix}
      | c = 
}}
{{eqn | l = a
      | r = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{bmatrix}
      | c = 
}}
{{eqn | l = b
      | r = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{bmatrix}
      | c = 
}}
{{eqn | l = c
      | r = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{bmatrix}
      | c = 
}}
{{end-eqn}}
Then $G$ is an example of the Klein $4$-group.

Proof:
By inspection, the Cayley table is constructed:
:<nowiki>$\begin{array}{c|cccc}
  & e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c & e & a \\
c & c & b & a & e \\
\end{array}$</nowiki>
Again by inspection this can be seen to be the same as the Cayley table for the Klein $4$-group.
{{qed}}
Theorem: Quotient of Cauchy Sequences is Metric Completion

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $d$ be the metric induced by $\struct {R, \norm {\, \cdot \,} }$.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences in $R$.
Let $\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.
Let $\norm {\, \cdot \,}: \CC \,\big / \NN \to \R_{\ge 0}$ be the norm on the quotient ring $\CC \,\big / \NN$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN} = \lim_{n \mathop \to \infty} \norm{x_n}$
Let $d'$ be the metric induced by $\struct {\CC \,\big / \NN, \norm {\, \cdot \,} }$
Let $\phi: R \to \CC \,\big / \NN$ be the mapping from $R$ to the quotient ring $\CC \,\big / \NN$ defined by:
:$\forall a \in R: \map \phi a = \sequence {a, a, a, \ldots} + \NN$
where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {a, a, a, \ldots}$.
Then:
:$\struct {\CC \,\big / \NN, d'}$ is the metric completion of $\struct {R,d}$
and:
:$\map \phi R$ is a dense subset of $\CC \,\big / \NN$

Proof:
By the definition of the metric induced by a norm then:
:a sequence $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, d}$.
So $\CC$ is the set of Cauchy sequences in $\struct {R, d}$.
Let $\sim$ be the equivalence relation on $\CC$ defined by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
Let $\tilde {\CC} = \CC  / \sim$ denote the set of equivalence classes under $\sim$.
For $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the equivalence class containing $\sequence {x_n}$.
Theorem: Completion of Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,}_R }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences.
Let $Q = \CC  / \NN$ where $\CC  / \NN$ denotes a quotient ring.
Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_Q = \lim_{n \mathop \to \infty} \norm{x_n}_R$
Then:
:$\struct {Q, \norm {\, \cdot \,}_Q}$ is a normed division ring completion of $\struct {R, \norm {\, \cdot \,}_R }$

Proof:
From Quotient Ring of Cauchy Sequences is Normed Division Ring:
:$\struct {Q, \norm {\, \cdot \,}_Q}$ is a normed division ring.
Let $d_R$ be the metric induced by $\struct {R, \norm {\, \cdot \,}_R }$.
Let $d_Q$ be the metric induced by $\struct {Q, \norm {\, \cdot \,}_Q}$.
From Quotient of Cauchy Sequences is Metric Completion, $\struct {Q, d_Q}$ is the metric completion of $\struct {R, d_R}$.
Let $\phi: R \to Q$ be the mapping from $R$ to the quotient ring $Q$ defined by:
:$\forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$
where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $Q$ that contains the constant sequence $\sequence {a, a, a, \ldots} $.
From Quotient of Cauchy Sequences is Metric Completion, $\map \phi R$ is a dense subset of $\struct {Q, d_Q }$.
From Embedding Division Ring into Quotient Ring of Cauchy Sequences, $\phi$ is a distance-preserving ring monomorphism.
By the definition of a normed division ring completion, $\struct {Q, \norm {\, \cdot \,}_Q }$ is a normed division ring completion of $\struct {R, \norm {\, \cdot \,}_R }$.
{{qed}}
Theorem: Group of Order 27 has Subgroup of Order 3

Let $G$ be a group whose identity element is $e$.
Let $G$ be of order $27$.
Then $G$ has at least one subgroup of order $3$.

Proof:
Let $x \in G \setminus \set e$.
From Identity is Only Group Element of Order 1:
:$\order x > 1$
where $\order x$ denotes the order of $x$.
From Lagrange's Theorem, $\order x$ is $3$, $9$ or $27$.
Thus one of the following applies:
{{begin-eqn}}
{{eqn | l = \order x
      | r = 3
}}
{{eqn | l = \order {x^3}
      | r = 3
}}
{{eqn | l = \order {x^9}
      | r = 3
}}
{{end-eqn}}
Hence one of the following holds:
{{begin-eqn}}
{{eqn | l = \order {\gen x}
      | r = 3
}}
{{eqn | l = \order {\gen {x^3} }
      | r = 3
}}
{{eqn | l = \order {\gen {x^9} }
      | r = 3
}}
{{end-eqn}}
where $\gen x$ denotes the subgroup generated by $x$.
Thus one of these: $\gen x$, $\gen {x^3}$ or $\gen {x^9}$ is a subgroup of $G$ of order $3$.
{{qed}}
Theorem: Group does not Necessarily have Subgroup of Order of Divisor of its Order

Let $G$ be a finite group whose order is $n$.
Let $d$ be a divisor of $n$.
Then it is not necessarily the case that $G$ has a subgroup of order $d$.

Proof:
Proof by Counterexample:
Consider $S_5$, the symmetric group on $5$ letters.
By Order of Symmetric Group, $\order {S_5} = 5! = 120$.
We have that $120 = 8 \times 15$ and so $15$ is a divisor of $120$.
However, Symmetric Group on 5 Letters has no Subgroup of Order 15.
{{qed}}
Theorem: Group does not Necessarily have Subgroup of Order of Divisor of its Order

Let $G$ be a finite group whose order is $n$.
Let $d$ be a divisor of $n$.
Then it is not necessarily the case that $G$ has a subgroup of order $d$.

Proof:
Proof by Counterexample:
Consider the symmetric group $S_4$.
Then the order of the alternating group $A_4$ is $12$.
We list the subgroups of $A_4$:
{{:Alternating Group on 4 Letters/Subgroups}}
Now $6$ divides $12$.
But there is no subgroup of $A_4$ of order $6$.
{{qed}}