avewright/proofwiki-math huggingface dataset:
Theorem: P-Sequence Space with P-Norm forms Banach Space

Let $\ell^p$ be a p-sequence space.
Let $\norm {\, \cdot \,}_p$ be a p-norm.
Then $\struct {\ell^p, \norm {\, \cdot \,}_p}$ is a Banach space.

Proof:
A Banach space is a normed vector space, where a Cauchy sequence converges {{WRT}} the supplied norm.
To prove the theorem, we need to show that a Cauchy sequence in $\struct {\ell^p, \norm {\,\cdot\,}_p}$ converges.
We take a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\ell^p, \norm {\,\cdot\,}_p}$.
Then we consider the $k$th component and show, that a real Cauchy sequence $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$ with the limit $x^{\paren k}$ and denote the entire set as $\mathbf x$.
Finally, we show that $\sequence {\mathbf x_n}_{n \in \N}$, composed of components $x_n^{\paren k},$ converges in $\struct {\ell^p, \norm {\,\cdot\,}_p}$ with the limit $\mathbf x$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\ell^p, \norm{\, \cdot \,}_p}$.
Denote the $k$th component of $\mathbf x_n$ by $x_n^{\paren k}$.
Theorem: Size of y-1 lt n and Size of y+1 gt 1 over n

Let $T_n \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:
:$T_n = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}$
Then:
:$T_n = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}$

Proof:
:500pxthumbright
First note that:
{{begin-eqn}}
{{eqn | l = T_n
      | r = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}
      | c = 
}}
{{eqn | r = \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}
      | c = 
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = \set {y: \size {y - 1} < n}
      | r = \openint {1 - n} {1 + n}
      | c = Open Interval Defined by Absolute Value
}}
{{eqn | l = \set {y: \size {y + 1} > \dfrac 1 n}
      | r = \R \setminus \closedint {-1 - \dfrac 1 n} {-1 + \dfrac 1 n}
      | c = Complement of Closed Interval Defined by Absolute Value
}}
{{eqn | ll= \leadsto
      | l = \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}
      | r = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Unions of Families
Category:Intersections of Families
tmrlmfh1k5doftvxlyvw1fpx34g0rab
Theorem: Real Part as Mapping is Surjection

Let $f: \C \to \R$ be the projection from the complex numbers to the real numbers defined as:
:$\forall z \in \C: \map f z = \map \Re z$
where $\map \Re z$ denotes the real part of $z$.
Then $f$ is a surjection.

Proof:
Let $x \in \R$ be a real number.
Let $y \in \R$ be an arbitrary real number.
Let $z \in \C$ be the complex number defined as:
:$z = x + i y$
Then we have:
:$\map \Re z = x$
That is:
:$\exists z \in \C: \map f z = x$
The result follows by definition of surjection.
{{qed}}
Theorem: Imaginary Part as Mapping is Surjection

Let $f: \C \to \R$ be the projection from the complex numbers to the real numbers defined as:
:$\forall z \in \C: \map f z = \map \Im z$
where $\map \Im z$ denotes the imaginary part of $z$.
Then $f$ is a surjection.

Proof:
Let $y \in \R$ be a real number.
Let $x \in \R$ be an arbitrary real number.
Let $z \in \C$ be the complex number defined as:
:$z = x + i y$
Then we have:
:$\map \Im z = y$
That is:
:$\exists z \in \C: \map f z = y$
The result follows by definition of surjection.
{{qed}}
Theorem: Condition for Mapping from Quotient Set to be Well-Defined

Let $S$ and $T$ be sets.
Let $\RR$ be an equivalence relation on $S$.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $S / \RR$ be the quotient set of $S$ induced by $\RR$.
Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Then:
:there exists a mapping $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$
{{iff}}:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
::<nowiki>$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 S \ar[r]^*{f}
   \ar[d]_*{q_\RR}
& T \\
S / \RR \ar@{-->}[ur]_*{\phi} } \end {xy}$</nowiki>

Proof:
From Condition for Composite Mapping on Left, we have:
:$\exists \phi: S / \RR \to T$ such that $\phi$ is a mapping and $\phi \circ q_\RR = f$
{{iff}}:
:$\forall x, y \in S: \map {q_\RR} x = \map {q_\RR} y \implies \map f x = \map f y$
But by definition of the quotient mapping induced by $\RR$:
:$\map {q_\RR} x = \map {q_\RR} y \iff \tuple {x, y} \in \RR$
Hence the result.
{{qed}}
Theorem: Automorphism Maps Generator to Generator

Let $G$ be a cyclic group.
Let $g$ be a generator of $G$.
Let $\phi$ be an automorphism on $G$.
Then $\map \phi g$ is also a generator of $G$.

Proof:
By definition of automorphism, $\phi$ is a homomorphism
It follows that this result is a specific instance of Homomorphic Image of Cyclic Group is Cyclic Group.
{{qed}}
Category:Cyclic Groups
Category:Group Automorphisms
kwnpb54bt14so7j5unset7a5q7styla
Theorem: Condition for Mapping from Quotient Set to be Surjection

Let the mapping $\phi: S / \RR \to T$ defined as:
:$\phi \circ q_\RR = f$
be well-defined.
Then:
:$\phi$ is a surjection
{{iff}}:
:$f$ is a surjection.

Proof:
We are given that:
:$\phi \circ q_\RR = f$
is well-defined.
Note that from Quotient Mapping is Surjection, $q_\RR$ is a surjection.
Theorem: Condition for Mapping from Quotient Set to be Injection

Let the mapping $\phi: S / \RR \to T$ defined as:
:$\phi \circ q_\RR = f$
be well-defined.
Then:
:$\phi$ is an injection
{{iff}}:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff \map f x = \map f y$

Proof:
From Condition for Mapping from Quotient Set to be Well-Defined, $\phi$ is well-defined {{iff}}:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
By definition of injection, $\phi$ is injective {{iff}}:
:$\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$
that is:
:$\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \tuple {x, y} \in \RR$
By definition of $\phi$:
:$\map \phi {\eqclass x \RR} = \map f x$
and so:
:$\map f x = \map f y \implies \tuple {x, y} \in \RR$
Hence the result.
{{qed}}
Theorem: Mapping from Quotient Set when Defined is Unique

Let the mapping $\phi: S / \RR \to T$ defined as:
:$\phi \circ q_\RR = f$
be well-defined.
Then $\phi$ is unique.

Proof:
From Condition for Mapping from Quotient Set to be Well-Defined, $\phi$ is well-defined {{iff}}:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
From Quotient Mapping is Surjection, $q_\RR$ is a surjection.
Suppose $\psi: S / \RR \to T$ is another well-defined mapping defined as:
:$\psi \circ q_\RR = f$
Then we have:
{{begin-eqn}}
{{eqn | l = \phi \circ q_\RR
      | r = \psi \circ q_\RR
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \phi
      | r = \psi
      | c = Surjection iff Right Cancellable
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Limit of Subsequence equals Limit of Sequence/Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\sequence {x_n}$ be a sequence in $X$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\sequence {x_{n_r} }$ is convergent and $\ds \lim_{r \mathop \to \infty} x_{n_r} = l$

Proof:
Let $\epsilon > 0$.
Since $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that:
:$\exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\norm {x_{n_r} - l} < \epsilon$
The result follows.
{{qed}}
Category:Normed Vector Spaces
Category:Convergence
Category:Limits of Sequences
bwn4z6rernjot58za4nm8x6cg0ge799
Theorem: Conditions for Commutative Diagram on Quotient Mappings between Mappings

Let $A$ and $B$ be sets.
Let $\RR_S$ and $\RR_T$ be equivalence relations on $S$ and $T$ respectively.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $S / \RR_S$ and $T / \RR_T$ be the quotient sets of $S$ and $T$ induced by $\RR_S$ and $\RR_T$ respectively.
Let $q_S: S \to S / \RR_S$ and $q_T: T \to T / \RR_T$ be the quotient mappings induced by $\RR_S$ and $\RR_T$ respectively.
Then a mapping $g: S / \RR_S \to T / \RR_T$ exists such that:
:$q_T \circ f = g \circ q_S$
{{iff}}:
:$\forall x, y \in S: x \mathrel {\RR_S} y \implies \map f x \mathrel {\RR_T} \map f y$
::<nowiki>$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 S \ar[r]^*{f}
   \ar[d]_*{q_S}
& T \ar[d]^*{q_T} \\
S / \RR_S \ar@{-->}[r]_*{g} & T / \RR_T } \end {xy}$</nowiki>

Proof:
Consider the commutative diagram:
::<nowiki>$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 S \ar[rr]^*{f}
   \ar[dd]_*{q_S}
   \ar[ddrr]^*{q_T \circ f}
& & T \ar[dd]^*{q_T} \\
& & \\
S / \RR_S \ar@{-->}[rr]_*{g} & & T / \RR_T } \end {xy}$</nowiki>
We consider the mapping $q_T \circ f: S \to T / \RR_T$.
From Condition for Mapping from Quotient Set to be Well-Defined:
:there exists a mapping $g: S / \RR_S \to T / \RR_T$ such that $g \circ q_S = q_T \circ f$
{{iff}}:
:$\forall x, y \in S: \tuple {x, y} \in \RR_S \implies \map {\paren {q_T \circ f} } x = \map {\paren {q_T \circ f} } y$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\paren {q_T \circ f} } x
      | r = \map {\paren {q_T \circ f} } y
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map {q_t} {\map f x}
      | r = \map {q_t} {\map f y}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \map f x
      | o = \RR_T
      | r = \map f y
      | c = {{Defof|Quotient Mapping}}
}}
{{end-eqn}}
{{Qed}}
Theorem: Subset of Finite Dimensional Normed Vector Space is Compact iff Closed and Bounded/Sufficient Condition

Let $\struct {X, \norm {\,\cdot\,}}$ be a finite-dimensional normed vector space.
Let $K \subset X$ be a compact subset.
Then $K$ is closed and bounded.

Proof:
=== Closedness ===
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$.
Suppose, $\sequence {x_n}_{n \mathop \in \N}$ converges to $L \in K$. 
Then there is a subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ convergent to $L' \in K$. 
But $\sequence {x_{n_k}}_{k \mathop \in \N}$ is a subsequence $\sequence {x_n}_{n \mathop \in \N}$.
Hence, $\sequence {x_{n_k}}_{k \mathop \in \N}$ converges to $L$.
By uniqueness of limits, $L = L'$.
By definition, $K$ is closed.
=== Boundedness ===
Suppose, $K$ is not bounded.
Then:
:$\forall n \in \N : \exists x_n \in K: \norm {x_n} > n$
Therefore, no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ is convergent.
Hence, $K$ is not compact.
This is in contradiction with the assumption.
Thus, $K$ is bounded.
Theorem: Geometrical Interpretation of Complex Modulus

Let $z \in \C$ be a complex number expressed in the complex plane.
Then the modulus of $z$ can be interpreted as the distance of $z$ from the origin.

Proof:
Let $z = x + i y$.
By definition of the complex plane, it can be represented by the point $\tuple {x, y}$.
By the Distance Formula, the distance $d$ of $z$ from the origin is:
{{begin-eqn}}
{{eqn | l = d
      | r = \sqrt {\paren {x - 0}^2 + \paren {y - 0}^2}
      | c = 
}}
{{eqn | r = \sqrt {x^2 + y^2}
      | c = 
}}
{{end-eqn}}
which is precisely the modulus of $z$.
{{qed}}
Theorem: Condition for Factoring of Quotient Mapping between Modulo Addition Groups

Let $m, n \in \Z_{>0}$ be strictly positive integers.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\struct {\Z_m, +_m}$ and $\struct {\Z_n, +_n}$ denote the additive groups of integers modulo $m$ and $n$ respectively.
Let $f: \Z \to \Z_n$ be the quotient epimorphism from $\struct {\Z, +}$ to $\struct {\Z_n, +_n}$.
Let $q: \Z \to \Z_m$ be the quotient epimorphism from $\struct {\Z, +}$ to $\struct {\Z_m, +_m}$.
Then:
:there exists a group homomorphism $\psi: \struct {\Z_m, +_m} \to \struct {\Z_n, +_n}$
{{iff}} 
:$m \divides n$
where $\divides$ denotes divisibility.

Proof:
An example of the use of Third Isomorphism Theorem/Groups/Corollary.
{{ProofWanted|details}}
Theorem: Factors of Group Direct Product are not Subgroups

Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups.
Let $\struct {G \times H, \circ}$ be the group direct product of $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Then neither $\struct {G, \circ_1}$ nor $\struct {H, \circ_2}$ is a subgroup of $\struct {G \times H, \circ}$.

Proof:
A subgroup is by definition a subset which is a group.
But neither $G$ nor $H$ are actually subsets of their cartesian product $G \times H$.
Hence the result.
{{qed}}
Category:Group Direct Products
Category:Subgroups
ninzsg4wxsaaotzi11ntmic8a0nz9sn
Theorem: Canonical Injection of Real Number Line into Complex Plane

Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $f: \R \to \C$ be the mapping from the real numbers to the complex numbers defined as:
:$\forall x \in \R: \map f z = x + 0 y$
Then $f: \struct {\R, +} \to \struct {\C, +}$ is a monomorphism.

Proof:
Consider the mapping $g: \C \to \R$ defined as:
:$\forall z \in \C: \map f z = \map \Re z$
where $\map \Re z$ denotes the real part of $z$.
From Real Part as Mapping is Endomorphism for Complex Addition, this is a projection from $\C$ to $\R$.
The result follows from Canonical Injection is Right Inverse of Projection.
{{Qed}}
Theorem: Additive Group of Complex Numbers is Direct Product of Reals with Reals

Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then the direct product $\struct {\R, +} \times \struct {\R, +}$ is isomorphic with $\struct {\C, +}$.

Proof:
Let us define the mapping $\phi: \R^2 \to \C$ as:
:$\forall \tuple {x, y} \in \R^2: \phi: \tuple {x, y} = x + y i$
We will show that $\phi$ is a group isomorphism.
Theorem: Multiplicative Group of Complex Numbers is not Direct Product of Reals with Reals

Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Then the direct product $\struct {\R_{\ne 0}, \times} \times \struct {\R_{\ne 0}, \times}$ is not isomorphic with $\struct {\C_{\ne 0}, \times}$.

Proof:
Let $\tuple {a, b}$ and $\tuple {c, d}$ be pairs of non-zero real numbers:
:$\tuple {a, b} \in \R_{\ne 0} \times \R_{\ne 0}$
:$\tuple {c, d} \in \R_{\ne 0} \times \R_{\ne 0}$
Then by definition of group direct product:
:$\tuple {a, b} \times \tuple {c, d} = \tuple {a \times c, b \times d}$
However, by interpreting $\tuple {a, b}$ and $\tuple {c, d}$ as complex numbers:
:$\tuple {a, b} \times \tuple {c, d} = \tuple {a \times c - b \times d, b \times c + a \times d}$
by definition of complex multiplication.
Hence the result.
{{Qed}}
Theorem: Imaginary Numbers under Addition form Group

Let $\II$ denote the set of complex numbers of the form $0 + i y$
That is, let $\II$ be the set of all wholly imaginary numbers.
Then the algebraic structure $\struct {\II, +}$ is a group.

Proof:
We have that $\II$ is a non-empty subset of the complex numbers $\C$.
Indeed, for example:
:$0 + 0 i \in \II$
Now, let $0 + i x, 0 + i y \in \II$.
Then we have:
{{begin-eqn}}
{{eqn | l = \paren {0 + i x} + \paren {-\paren {0 + i y} }
      | r = \paren {0 + i x} - \paren {0 + i y}
      | c = Inverse for Complex Addition
}}
{{eqn | r = i \paren {x - y}
      | c = {{Defof|Complex Subtraction}}
}}
{{eqn | o = \in
      | r = \II
      | c = 
}}
{{end-eqn}}
Hence the result by the One-Step Subgroup Test.
{{qed}}
Theorem: Imaginary Numbers under Multiplication do not form Group

Let $\II$ denote the set of complex numbers of the form $0 + i y$ for $y \in \R_{\ne 0}$.
That is, let $\II$ be the set of all wholly imaginary non-zero numbers.
Then the algebraic structure $\struct {\II, \times}$ is not a group.

Proof:
Let $0 + i x \in \II$.
We have:
{{begin-eqn}}
{{eqn | l = \paren {0 + i x} \times \paren {0 + i x}
      | r = \paren {0 - x^2} + i \paren {0 \times x + 0 \times x}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = -x^2
      | c = 
}}
{{eqn | o = \notin
      | r = \II
      | c = 
}}
{{end-eqn}}
So $\struct {\II, \times}$ is not closed.
Hence $\struct {\II, \times}$ is not a group.
{{qed}}
Theorem: Set of Isometries in Complex Plane under Composition forms Group

Let $S$ be the set of all bijective complex functions $f: \C \to \C$ which preserve distance when embedded in the complex plane.
That is:
:$\size {\map f a - \map f b} = \size {a - b}$
where $\size z$ denotes the complex modulus of $z \in \C$.
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.

Proof:
From Complex Plane is Metric Space, $\C$ can be treated as a metric space.
We have {{hypothesis}} that $f$ is:
:a distance-preserving mapping.
:a bijection.
Hence $f$ is an isometry on $\C$.
Taking the group axioms in turn:
Theorem: Set of Affine Mappings on Real Line under Composition forms Group

Let $S$ be the set of all real functions $f: \R \to \R$ of the form:
:$\forall x \in \R: \map f x = r x + s$
where $r \in \R_{\ne 0}$ and $s \in \R$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.

Proof:
We note that $S$ is a subset of the set of all real functions on $\R$.
From Set of all Self-Maps under Composition forms Semigroup, we have that $\circ$ is associative.
Consider the real function $I: \R \to \R$ defined as:
:$\forall x \in \R: \map I x = 1 \times x + 0$
We have that:
:$I \in S$
:$I$ is the identity mapping.
So $S$ is not empty.
Then we note:
Let $f, g \in S$ such that:
:$\map f x = r_1 x + s_1$
:$\map g x = r_2 x + s_2$
For all $x \in \R$, we have:
{{begin-eqn}}
{{eqn | l = \map {g \circ f} x
      | r = \map g {\map f x}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map g {r_1 x + s_1}
      | c = Definition of $f$
}}
{{eqn | r = r_2 {r_1 x + s_1} + s_2
      | c = Definition of $f$
}}
{{eqn | r = \paren {r_2 r_1} x + \paren {r_2 s_1 + s_2}
      | c = 
}}
{{eqn | r = r x + s
      | c = where $r = r_2 r_2$ and $s = r_2 s_1 + s_2$
}}
{{end-eqn}}
This demonstrates that $\struct {S, \circ}$ is closed.
Let $\phi: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map \phi x = p x + q$
where $p \in \R_{\ne 0}$ and $q \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | r = p x + q
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \dfrac {y - q} p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \dfrac 1 p y + \paren {-\dfrac q p}
      | c = 
}}
{{end-eqn}}
Thus we define $\psi: \R \to \R$ be the real function defined as:
:$\forall y \in \R: \map \psi y = \dfrac 1 p y + \paren {-\dfrac q p}$
where $\dfrac 1 p \in \R_{\ne 0}$ and $-\dfrac q p \in \R$.
It is seen that $\psi \in S$.
For all $x \in \R$, we have:
{{begin-eqn}}
{{eqn | l = \map {\phi \circ \psi} x
      | r = \map \phi {\map \psi x}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map \phi {\dfrac 1 p x - \dfrac q p}
      | c = Definition of $\psi$
}}
{{eqn | r = p {\dfrac 1 p x -\dfrac q p} + q
      | c = Definition of $\phi$
}}
{{eqn | r = \dfrac p p x - \dfrac {p q} p + q
      | c = 
}}
{{eqn | r = x
      | c = 
}}
{{end-eqn}}
Hence:
:$\phi \circ \psi = I$
and so $\psi$ is the inverse of $\phi$.
We have then that $\struct {S, \circ}$ is closed under inversion.
The result follows by the Two-Step Subgroup Test.
{{qed}}
Theorem: Arbitrary Cyclic Group of Order 4

Let $S = \set {1, 2, 3, 4}$.
Consider the algebraic structure $\struct {S, \circ}$ given by the Cayley table:
:$\begin{array}{r|rrrr}
\circ & 2 & 3 & 4 & 1 \\
\hline
2 & 2 & 3 & 4 & 1 
\\
3 & 3 & 4 & 1 & 2
\\
4 & 4 & 1 & 2 & 3
\\
1 & 1 & 2 & 3 & 4
\\
\end{array}$
Then $\struct {S, \circ}$ is a group.
Specifically, $\struct {S, \circ}$ is the cyclic group of order $4$.

Proof:
Let $S' = \set {0, 1, 2, 3}$.
Let $\phi: S \to S'$ be the bijection:
{{begin-eqn}}
{{eqn | l = \map \phi 2
      | r = 0
}}
{{eqn | l = \map \phi 3
      | r = 1
}}
{{eqn | l = \map \phi 4
      | r = 2
}}
{{eqn | l = \map \phi 1
      | r = 3
}}
{{end-eqn}}
By inspection, the Cayley table presented above is in the same form as the Cayley table for the cyclic group of order $4$:
:$\begin{array}{r|rrrr}
+_4 & 0 & 1 & 2 & 3 \\
\hline
0 & 0 & 1 & 2 & 3 
\\
1 & 1 & 2 & 3 & 0
\\
2 & 2 & 3 & 0 & 1
\\
3 & 3 & 0 & 1 & 2
\\
\end{array}$
{{qed}}
Theorem: Group Product Identity therefore Inverses/Part 1

:$g h = e \implies h = g^{-1}$ and $g = h^{-1}$

Proof:
From the Division Laws for Groups:
:$g h = e \implies g = e h^{-1} = h^{-1}$
Also by the Division Laws for Groups:
:$g h = e \implies h = g^{-1} e = g^{-1}$
{{qed}}
Theorem: Group Product Identity therefore Inverses/Part 1

:$g h = e \implies h = g^{-1}$ and $g = h^{-1}$

Proof:
Let $g h = e$.
Then:
{{begin-eqn}}
{{eqn | l = h
      | r = e h
      | c = {{Group-axiom|2}}
}}
{{eqn | r = \paren {g^{-1} g} h
      | c = {{Group-axiom|3}}
}}
{{eqn | r = g^{-1} \paren {g h}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = g^{-1} e
      | c = {{hypothesis}}
}}
{{eqn | r = g^{-1}
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = g
      | r = g e
      | c = {{Group-axiom|2}}
}}
{{eqn | r = g \paren {h h^{-1} }
      | c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {g h} h^{-1}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = e h^{-1}
      | c = {{hypothesis}}
}}
{{eqn | r = h^{-1}
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
{{qed}}
Theorem: Group Product Identity therefore Inverses/Part 2

:$h g = e \implies h = g^{-1}$ and $g = h^{-1}$

Proof:
From the Division Laws for Groups:
:$h g = e \implies g = e h^{-1} = h^{-1}$
Also by the Division Laws for Groups:
:$h g = e \implies h = g^{-1} e = g^{-1}$
{{qed}}
Theorem: Group Product Identity therefore Inverses/Part 2

:$h g = e \implies h = g^{-1}$ and $g = h^{-1}$

Proof:
Let $h g = e$.
Then:
{{begin-eqn}}
{{eqn | l = g
      | r = e g
      | c = {{Group-axiom|2}}
}}
{{eqn | r = \paren {h^{-1} h} g
      | c = {{Group-axiom|3}}
}}
{{eqn | r = h^{-1} \paren {h g}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = h^{-1} e
      | c = {{hypothesis}}
}}
{{eqn | r = h^{-1}
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = h
      | r = h e
      | c = {{Group-axiom|2}}
}}
{{eqn | r = h \paren {g g^{-1} }
      | c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {h g} g^{-1} 
      | c = {{Group-axiom|1}}
}}
{{eqn | r = e g^{-1}
      | c = {{hypothesis}}
}}
{{eqn | r = g^{-1}
      | c = {{Group-axiom|2}}
}}
{{end-eqn}}
{{qed}}
Theorem: Factors of Sums of Powers of 100,000/General Result

All integers $n$ of the form:
:$n = \ds \sum_{k \mathop = 0}^m 10^{r k}$ for $m \in \Z_{> 0}$
are composite for $r \ge 2$.
The only exceptions are:
:$r = 2^k, m = 1$ for some $k \in \N$
:$r = m + 1 =$ some odd prime
in which cases $n$ may be prime.

Proof:
=== Case $1$: $m + 1$ is Composite ===
Suppose $m + 1$ is composite.
Then:
:$\exists p, q > 1: m + 1 = p q$
By Division Theorem, for each $k$ with $0 \le z \le m$:
:$\exists i, j \in \N: 0 \le i \le q - 1, \, 0 \le j \le p - 1: k = i + q j$
Thus:
{{begin-eqn}}
{{eqn | l = n
      | r = \sum_{k \mathop = 0}^m 10^{r k}
}}
{{eqn | r = \sum_{i \mathop = 0}^{q - 1} \sum_{j \mathop = 0}^{p - 1} 10^{r \paren {i + q j} }
}}
{{eqn | r = \paren {\sum_{i \mathop = 0}^{q - 1} 10^{r i} } \paren {\sum_{j \mathop = 0}^{p - 1} 10^{r q j} }
}}
{{end-eqn}}
Both sums are greater than $1$, so $n$ is composite.
{{qed|lemma}}
Theorem: Odd Integers under Addition do not form Subgroup of Integers

Let $S$ denote the set of odd integers.
Then $\struct {S, +}$ is not a subgroup of the additive group of integers $\struct {\Z, +}$.

Proof:
Consider the odd integers $1$ and $3$.
We have that $1 + 3 = 4$.
But $4$ is not odd.
Thus addition on $\struct {S, +}$ is not closed.
Hence $\struct {S, +}$ is not a group, let alone a subgroup of $\struct {\Z, +}$
{{qed}}