avewright/proofwiki-math huggingface dataset:
Theorem: Center of Group is Subgroup

Let $G$ be a group.
The center $\map Z G$ of $G$ is a subgroup of $G$.

Proof:
For brevity, suppress the symbol for the group operation (which may be $\circ$, or $+$).
Apply the Two-Step Subgroup Test:
=== Condition $(1)$ ===
By the definition of identity, $e g = g e = g$ for all $g \in G$.
So, $e \in \map Z G$, meaning $\map Z G$ is non-empty.
{{qed|lemma}}
=== Condition $(2)$ ===
Suppose $a, b \in \map Z G$.
Using the associative property and the definition of center, we have:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = \paren {a b} g 
      | r = a \paren {b g}
      | c = {{Defof|Associative Operation}}
}}
{{eqn | r = a \paren {g b}
      | c = {{Defof|Center of Group}}
}}
{{eqn | r = \paren {a g} b
      | c = {{Defof|Associative Operation}}
}}
{{eqn | r = \paren {g a} b
      | c = {{Defof|Center of Group}}
}}
{{eqn | r = g \paren {a b}
      | c = {{Defof|Associative Operation}}
}}
{{end-eqn}}
Thus, $a b \in \map Z G$.
{{qed|lemma}}
=== Condition $(3)$ ===
Suppose $c \in \map Z G$. Then:
{{begin-eqn}}
{{eqn | l = c
      | o = \in
      | r = \map Z G
}}
{{eqn | ll= \leadsto
      | q = \forall g \in G
      | l = c g
      | r = g c
      | c = {{Defof|Center of Group}}
}}
{{eqn | ll= \leadsto
      | l = c^{-1} \paren {c g} c^{-1}
      | r = c^{-1} \paren {g c} c^{-1}
      | c = {{Group-axiom|0}}
}}
{{eqn | ll= \leadsto
      | l = \paren {c^{-1} c} g c^{-1}
      | r = c^{-1} g \paren {c c^{-1} }
      | c = {{Defof|Associative Operation}}
}}
{{eqn | ll= \leadsto
      | l = e g c^{-1}
      | r = c^{-1} g e
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = g c^{-1}
      | r = c^{-1} g
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | l = c^{-1}
      | o = \in
      | r = \map Z G
      | c = {{Defof|Center of Group}}
}}
{{end-eqn}}
{{qed|lemma}}
Therefore:
:$\map Z G \le G$
{{qed}}
Theorem: Center of Group is Subgroup

Let $G$ be a group.
The center $\map Z G$ of $G$ is a subgroup of $G$.

Proof:
We have the result Center is Intersection of Centralizers.
That is, $\map Z G$ is the intersection of all the centralizers of $G$.
All of these are subgroups of $G$ by Centralizer of Group Element is Subgroup.
Thus from Intersection of Subgroups is Subgroup, $\map Z G$ is also a subgroup of $G$.
{{qed}}
Theorem: Center of Group is Abelian Subgroup

Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Then $\map Z G$ is an abelian subgroup of $G$.

Proof:
By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.
The definition of the center $\map Z G$ grants that all elements of $\map Z G)$ commute with all elements of $G$.
In particular, all elements of $\map Z G$ commute with all elements of $\map Z G$ as $\map Z G \subseteq G$.
Therefore $\map Z G$ is abelian.
{{qed}}
Theorem: Dihedral Group/Group Presentation

The dihedral group $D_n$ has the group presentation:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
That is, the dihedral group $D_n$ is generated by two elements $\alpha$ and $\beta$ such that:
:$(1): \quad \alpha^n = e$
:$(2): \quad \beta^2 = e$
:$(3): \quad \beta \alpha = \alpha^{n - 1} \beta$

Proof:
By definition, the '''dihedral group''' $D_n$ of order $2 n$ is the group of symmetries of the regular $n$-gon.
So, let $P$ denote a regular polygon with $n$ sides.
Let $\alpha$ be a rotation of $P$ by $\dfrac {2 \pi} n$.
Let $\beta$ be a reflection $P$ whose axis of reflection is the $y$ axis.
It takes $n$ rotations by $\dfrac {2 \pi} n$ to return $P$ to its original position.
Thus $\alpha$ is of order $n$, and so:
:$\alpha^n = e$
Two reflections of $P$ in the same axis of reflection also returns $P$ to its original position.
Thus $\beta$ is of order $2$, and so:
:$\beta^2 = e$
It remains to be shown that $\beta \alpha \beta = \alpha^{-1}$.
Let the vertices of $P$ be labelled in order anticlockwise from the $y$-axis: $A_1, A_2, \ldots, A_n$.
Let $P$ be arranged so that $A_1$ is on the $y$-axis.
From here, $\beta$ on $P$ exchanges the places of $A_2$ and $A_n$.
From here, $\alpha$ sends $A_2$ to the position of $A_1$, $A_1$ to the original position of $A_2$, and $A_3$ to the original position of $A_n$.
From here, $\beta$ sends exchanges the positions of $A_3$ and $A_1$.
So $\beta \alpha \beta$ on $P$ results in $A_1$ moving to the original position of $A_n$, $A_2$ moving to the original position of $A_1$ and $A_3$ moving to the original position of $A_2$.
That is, $\beta \alpha \beta$ is a rotation of $P$ by $-\dfrac {2 \pi} n$, or $\alpha^{-1}$.
{{qed}}
Theorem: Intersection with Subgroup Product of Superset

Let $X, Y, Z$ be subgroups of a group $\struct {G, \circ}$.
Let $Y \subseteq X$.
Then:
:$X \cap \paren {Y \circ Z} = Y \circ \paren {X \cap Z}$
where $Y \circ Z$ denotes subset product.

Proof:
By definition of set equality, it suffices to prove:
:$X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$
and:
:$Y \circ \paren {X \cap Z} \subseteq X \cap \paren {Y \circ Z}$
Theorem: Properties of Semi-Inner Product

Let $V$ be a vector space over $\Bbb F \in \set {\R, \C}$.
Let $\innerprod \cdot \cdot$ be a semi-inner product on $V$.
Denote, for $x \in V$, $\norm x := \innerprod x x^{1 / 2}$.
Then, $\forall x, y \in V, a \in \Bbb F$:
:$(1): \quad \norm {x + y} \le \norm x + \norm y$
:$(2): \quad \norm {a x} = \size a \norm x$

Proof:
=== Proof of $(1)$ ===
For $x, y \in V$, compute:
{{begin-eqn}}
{{eqn | l = \norm {x + y}^2
      | r = \innerprod {x + y} {x + y}
      | c = Definition of $\norm \cdot$
}}
{{eqn | r = \innerprod x x + \innerprod x y + \innerprod y x + \innerprod y y
      | c = Linearity of $\innerprod \cdot \cdot$
}}
{{eqn | o = \le
      | r = \innerprod x x + \sqrt {\innerprod x x \innerprod y y} + \sqrt {\innerprod y y \innerprod x x} + \innerprod y y
      | c = Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
}}
{{eqn | r = \norm x^2 + 2 \norm x \norm y + \norm y^2
      | c = Definition of $\norm \cdot$
}}
{{eqn | r = \paren {\norm x + \norm y}^2
}}
{{end-eqn}}
Taking square roots on either side gives the result.
{{qed|lemma}}
Theorem: Inner Product Norm is Norm

Let $V$ be an inner product space over a subfield $\Bbb F$ of $\C$.
Let $\norm {\, \cdot \,}$ denote the inner product norm on $V$.
Then $\norm {\, \cdot \,}$ is a norm on $V$.

Proof:
Let us verify the norm axioms in turn.
Theorem: Metric Induced by Norm is Metric

Let $V$ be a normed vector space.
Let $\norm{\,\cdot\,}$ denote its norm.
Let $d$ be the metric induced by $\norm {\,\cdot\,}$.
Then $d$ is a metric.

Proof:
=== Proof of {{Metric-space-axiom|1}} and {{Metric-space-axiom|4}} ===
Let $x, y \in V$. 
Then:
:$\map d {x, y} = \norm {x - y} \ge 0$
and furthermore:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | l = \norm {x - y}
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | l = x - y
      | r = \mathbf 0_V
      | c = {{NormAxiomVector|1}}
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | r = y
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Product of Sums of Four Squares/Corollary

Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, c_1, c_2, \ldots, c_n, d_1, d_2, \ldots, d_n$ be integers.
Then:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
That is, the product of any number of sums of four squares is also a sum of four squares.

Proof:
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
$\map P 1$ is true, as this just says:
:$\exists w, x, y, z \in \Z: a^2 + b^2 + c^2 + d^2 = w^2 + x^2 + y^2 + z^2$
which is trivially true.
Theorem: Subgroup Subset of Subgroup Product

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$H \subseteq H \circ K \supseteq K$
where $H \circ K$ denotes the subset product of $H$ and $K$.

Proof:
By definition of subset product:
:$H \circ K = \set {h \circ k: h \in H, k \in K}$
So:
{{begin-eqn}}
{{eqn | l=x
      | o=\in
      | r=H
      | c=
}}
{{eqn | ll=\leadsto
      | l=x
      | r=x \circ e
      | c={{Defof|Identity Element}}
}}
{{eqn | o=\in
      | r=H \circ K
      | c=Identity of Subgroup
}}
{{eqn | ll=\leadsto
      | l=H
      | o=\subseteq
      | r=H \circ K
      | c={{Defof|Subset}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l=y
      | o=\in
      | r=K
      | c=
}}
{{eqn | ll=\leadsto
      | l=y
      | r=e \circ y
      | c={{Defof|Identity Element}}
}}
{{eqn | o=\in
      | r=H \circ K
      | c=Identity of Subgroup
}}
{{eqn | ll=\leadsto
      | l=K
      | o=\subseteq
      | r=H \circ K
      | c={{Defof|Subset}}
}}
{{end-eqn}}
{{qed}}
Category:Subset Products
Category:Subgroups
ppygaf0xnusccs9g6u30wulbqww4zo0
Theorem: Product Space is Path-connected iff Factor Spaces are Path-connected

Let $\SS = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces for $i$ in some indexing set $I$ such that $\forall i \in I: S_i \ne \O$.
Let $\ds T = \struct {S, \tau} = \prod_{i \mathop \in I} \struct {S_i, \tau_i}$ be the product space of $\SS$.
Then $T$ is a path-connected space {{iff}} each of $\struct {S_i, \tau_i}$ is a path-connected space.

Proof:
=== Necessary Condition ===
Suppose $S_i$ is path-connected for each $i \in I$. 
Let $x \ne y \in S$ be arbitrary.
Since each $S_i$ is path-connected, we have a continuous mapping:
:$f_i: \closedint 0 1 \to S_i$
such that $\map {f_i} 0 = x_i$ and $\map {f_i} 1  = y_i$ for all $i \in I$.
We then define:
: $f : \closedint 0 1 \to S$
by setting:
:$\forall t \in \closedint 0 1: \map f t_i = \map {f_i} t$
We have that $f_i = \pr_i \circ f$ are all continuous.
So, by Continuous Mapping to Product Space, it follows that $f$ is continuous.
Also:
: $\map f 0_i = \map {f_i} 0 = x_i$
and:
: $\map f 1_i = \map {f_i} 1 = y_i$
Therefore $\map f 0 = x$ and $\map f 1 = y$.
So there exists a continuous mapping:
: $f: \closedint 0 1 \to S$ 
such that $\map f 0 = x$ and $\map f 1 = y$.
Since $x$ and $y$ were arbitrary, we have shown that $S$ is path-connected.
{{qed|lemma}}
Theorem: Normal Subgroup of Subset Product of Subgroups

Let $G$ be a group whose identity is $e$.
Let:
: $H$ be a subgroup of $G$
: $N$ be a normal subgroup of $G$.
Then:
: $N \lhd N H$
where:
: $\lhd$ denotes normal subgroup
: $N H$ denotes subset product.

Proof:
From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N$ is a subgroup of $G$.
By definition of subset product all elements of $H N$ can be written in the form:
:$h n \in H N$
where $h \in H, n \in N$.
Let $h n \in H N$.
Let $n_1 \in N$.
From Inverse of Group Product:
:$\paren {h n} n_1 \paren {h n}^{-1} = h n n_1 n^{-1} h^{-1}$
We have that:
:$n n_1 n \in N$
:$h, h^{-1} \in G$.
Then, since $N$ is a normal subgroup of $G$:
:$\paren {h n} n_1 \paren {n^{-1} h^{-1} } \in N$
Thus:
: $N \lhd N H$
{{qed}}
Theorem: Subset Products of Normal Subgroup with Normal Subgroup of Subgroup

Let $G$ be a group.
Let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad K$ be a normal subgroup of $H$
:$(3): \quad N$ be a normal subgroup of $G$
Then:
:$N K \lhd N H$
where:
: $N K$ and $N H$ denote subset product
: $\lhd$ denotes the relation of being a normal subgroup.

Proof:
Consider arbitrary $x_n \in N, x_h \in H$.
Thus:
:$x_n x_h \in N H$
We aim to show that:
:$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$
thus demonstrating $N K \lhd N H$ by the Normal Subgroup Test.
We have:
{{begin-eqn}}
{{eqn | l = x_n x_h N K \paren {x_n x_h}^{-1}
      | r = x_n x_h N K {x_h}^{-1} {x_n}^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = x_n N x_h K {x_h}^{-1} {x_n}^{-1}
      | c = as $N \lhd G$ and $H \le G$
}}
{{eqn | r = x_n N K {x_n}^{-1}
      | c = as $K \lhd H$
}}
{{eqn | r = N K {x_n}^{-1}
}}
{{eqn | r = K N {x_n}^{-1}
      | c = as $N \lhd G$ and $K \le G$
}}
{{eqn | r = K N
}}
{{eqn | r = N K
      | c = as $N \lhd G$ and $K \le G$
}}
{{end-eqn}}
{{qed}}
Theorem: Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group

Let $G$ be a group with the following properties:
:$(1): \quad G$ is non-abelian.
:$(2): \quad G$ is of order $8$.
:$(3): \quad G$ has precisely one element of order $2$.
Then $G$ is isomorphic to the quaternion group $Q$.

Proof:
By Groups of Order 8, the only group satisfying all three properties is $Q = \Dic 2$.
{{qed}}
Theorem: Quaternion Group not Dihedral Group

Let $Q$ be the quaternion group.
Then $Q$ is not isomorphic to the dihedral group $D_4$.

Proof:
From Group Presentation of Dihedral Group, $D_4$ is generated by two elements of orders $4$ and $2$ respectively.
Let these generators be $\alpha$ and $\beta$ where:
:$\alpha^4 = e$
:$\beta^2 = e$
Hence $\alpha^2$ and $\beta$ are different elements of $D_4$ which both have order $2$.
But the quaternion group:
:$Q = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$
has only one element of order $2$, which is $-\mathbf 1$.
The rest have order one or four.
Thus $Q$ and $D_4$ cannot be isomorphic because there are not the same number of elements of order $2$ in both groups.
{{qed}}
Theorem: Subset Product of Normal Subgroups with Trivial Intersection

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H, K$ be normal subgroups of $G$.
Let $H \cap K = e$.
Then $H K$ is isomorphic to $H \times K$ where:
: $H K$ denotes the subset product of $H$ and $K$
: $H \times K$ denotes the direct product of $H$ and $K$.

Proof:
Let $G' = H K$.
From Subset Product of Normal Subgroups is Normal, $G'$ is a normal subgroup of $G$.
That is $G'$ is itself a group.
So by the Internal Direct Product Theorem, $G'$ is the internal group direct product of $H$ and $K$.
The result follows by definition of the internal group direct product.
{{qed}}
Theorem: Quotient Group of Direct Products

Let $G$ and $G'$ be groups.
Let:
{{begin-eqn}}
{{eqn | l = H
      | o = \lhd
      | r = G
}}
{{eqn | l = H'
      | o = \lhd
      | r = G'
}}
{{end-eqn}}
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
:$\paren {G \times G'} / \paren {H \times H'}$ is isomorphic to $\paren {G / H} \times \paren {G' / H'}$
where:
:$G \times G'$ denotes the group direct product of $H$ and $H'$
:$G / H$ denotes the quotient group of $G$ by $H$.

Proof:
Let $\phi_1: G \to G / H$ and $\phi_2: G' \to G' / H'$ be the quotient epimorphisms with $H$ and $H'$ as their kernels, respectively.
Now define a homomorphism $\phi: G \times G' \to \paren {G / H} \times \paren {G' / H'}$ by:
:$\phi = \phi_1 \times \phi_2$
so:
:$\map \phi {x, x'} = \tuple {\map {\phi_1} x, \map {\phi_2} {x'} }$
The kernel of $\phi$ is clearly $H \times H'$, and $\phi$ is surjective.
So by the First Isomorphism Theorem for Groups:
:$\paren {G / H} \times \paren {G' / H'} \cong \paren {G \times G'} / \paren {H \times H'}$
{{qed}}
Theorem: Completion Theorem (Inner Product Space)

Let $\GF \in \set {\R, \C}$.
Let $\struct {V, \innerprod \cdot \cdot_V}$ be an inner product space over a subfield $\GF$.
Let $\norm {\, \cdot \,}_V$ be the inner product norm induced by $\innerprod \cdot \cdot_V$.
Let $\struct {H, \norm {\, \cdot \,}_H}$ be a completion of $V$ with respect to $\norm {\, \cdot \,}$ given by Completion Theorem (Normed Vector Space).
Then $\innerprod \cdot \cdot_V$ can be extended to an inner product $\innerprod \cdot \cdot_H$ on $H$.
By definition, $H$ will be a Hilbert space.
Therefore, the theorem can alternatively be stated as:
:any inner product space may be completed to a Hilbert space.

Proof:
We first show that $\norm {\, \cdot \,}_H$ is induced by an inner product by using Norm satisfying Parallelogram Law induced by Inner Product.
From Completion Theorem (Normed Vector Space), there exists a linear isometry $\phi : V \to H$ such that $\phi \sqbrk V$ is dense in $H$. 
From Parallelogram Law (Inner Product Space), we have:
:$\norm {u + v}_V^2 + \norm {u - v}_V^2 = 2 \paren {\norm u_V^2 + \norm v_V^2}$
for each $u, v \in V$. 
Since $\phi$ is a linear isometry, we have:
:$\norm {u + v}_V = \norm {\map \phi {u + v} }_H = \norm {\map \phi u + \map \phi v}_H$
:$\norm {u - v}_V = \norm {\map \phi {u - v} }_H = \norm {\map \phi u - \map \phi v}_H$
:$\norm u_V = \norm {\map \phi u}_H$
:$\norm v_V = \norm {\map \phi v}_H$
So we have:
:$\norm {\map \phi u + \map \phi v}_H^2 + \norm {\map \phi u - \map \phi v}_H^2 = 2 \paren {\norm {\map \phi u}_H^2 + \norm {\map \phi v}_H^2}$
for each $u, v \in V$.
In other words, we have:
:$\norm {x + y}_H^2 + \norm {x - y}_H^2 = 2 \paren {\norm x^2_H + \norm y^2_H}$ 
for each $x, y \in \phi \sqbrk V$. 
Now let $x, y \in H$.
Since $\phi \sqbrk V$ is dense in $H$, there exists sequences $\sequence {x_n}_{n \mathop \in \N}$, $\sequence {y_n}_{n \mathop \in \N}$ in $\phi \sqbrk V$ such that:
:$x_n \to x$
and:
:$y_n \to x$ 
in $\struct {H, \norm {\, \cdot \,}_H}$, by Point in Closure of Subset of Metric Space iff Limit of Sequence.
For each $n \in \N$, we have:
:$\norm {x_n + y_n}_H^2 + \norm {x_n - y_n}_H^2 = 2 \paren {\norm {x_n}^2_H + \norm {y_n}^2_H}$
By using Norm is Continuous, we have:
:$\norm {x + y}_H^2 + \norm {x - y}^2_H = 2 \paren {\norm x^2_H + \norm y^2_H}$
by taking $n \to \infty$.
Theorem: Identity Mapping is Group Endomorphism

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a group endomorphism.

Proof:
From Identity Mapping is Group Automorphism, $G$ is a group automorphism.
The result follows from Group Automorphism is Endomorphism.
{{qed}}
Theorem: Pythagoras's Theorem (Inner Product Space)

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $\norm \cdot$ be the inner product norm on $\struct {V, \innerprod \cdot \cdot}$. 
Let $f_1, \ldots, f_n \in V$ form an orthogonal set.
Then:
:$\ds \norm {\sum_{i \mathop = 1}^n f_i}^2 = \sum_{i \mathop = 1}^n \norm {f_i}^2$

Proof:
{{begin-eqn}}
{{eqn | l = \norm {\sum_{i \mathop = 1}^n f_i}^2
      | r = \innerprod {\sum_{i \mathop = 1}^n f_i} {\sum_{j \mathop = 1}^n f_j}
      | c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \innerprod {f_i} {f_j}
      | c = since the Inner Product is linear in the first argument
}}
{{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \begin {cases} \innerprod {f_i} {f_i} & i = j \\ 0 & i \ne j \end {cases}
      | c = the $f_i$ are pairwise orthogonal
}}
{{eqn | r = \sum_{i \mathop = 1}^n \norm {f_i}^2
      | c = {{Defof|Inner Product Norm}}
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared plus a squared/Arctangent Form

:$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$

Proof:
Let:
:$a \tan \theta = x$
for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Shape of Tangent Function, this substitution is valid for all real $x$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = a \tan \theta 
      | c = from above
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \sec^2 \theta
      | c = Derivative of Tangent Function
}}
{{eqn | ll= \leadsto
      | l = \int \frac 1 {x^2 + a^2} \rd x
      | r = \int \frac {a \ \sec^2 \theta} {a^2 \tan^2 \theta + a^2} \rd \theta
      | c = Integration by Substitution
}}
{{eqn | r = \frac a {a^2} \int \frac {\sec^2 \theta} {\tan^2 \theta + 1} rd \theta
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 a \int \frac {\sec^2 \theta} {\sec^2 \theta} \rd \theta
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \frac 1 a \int \rd \theta
}}
{{eqn | r = \frac 1 a \theta + C
      | c = Integral of Constant
}}
{{end-eqn}}
As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
:$\tan \theta = \dfrac x a \iff \theta = \arctan \dfrac x a$
Thus:
:$\ds \int \frac 1 {x^2 + a^2} \rd x = \frac 1 a \arctan \frac x a + C$
{{qed|lemma}}
When $a = 0$, both $\dfrac x a$ and $\dfrac 1 a$ are undefined.
However, consider the limit of the above primitive as $a \to 0$:
{{begin-eqn}}
{{eqn | l = \lim_{a \mathop \to 0} \frac 1 a \arctan {\frac x a}
      | r = \lim_{a \mathop \to 0} \frac {\arctan {\frac x a} } a
}}
{{eqn | ll= \leadsto
      | r = \lim_{a \mathop \to 0} \frac {-x a^{-2} } {1 + \frac {x^2} {a^2} }
      | c = L'Hôpital's Rule, Derivative of Arctangent Function
}}
{{eqn | r = \lim_{a \mathop \to 0} \frac {a^{-2} } {a^{-2} } \frac {-x} {x^2 + a^2}
}}
{{eqn | r = -\frac 1 x
}}
{{end-eqn}}
This corresponds with the result:
:$\ds \int \frac 1 {x^2} \rd x = \frac {-1} x + C$
which follows from Primitive of Power.
{{qed}}
Theorem: Primitive of Reciprocal of x squared plus a squared/Arctangent Form

:$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$

Proof:
We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$.
Thus from Primitive of $\dfrac 1 {a x^2 + b x + c}$ for $b^2 - 4 a c > 0$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
setting $a := 1, b := 0, c := a^2$:
{{begin-eqn}}
{{eqn | l = \int \frac 1 {x^2 + a^2} \rd x
      | r = \dfrac 2 {\sqrt {4 a^2 - 0} } \map \arctan {\dfrac {2 x + 0} {\sqrt {4 a^2} } } + C
      | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$
}}
{{eqn | r = \frac 1 a \arctan {\frac x a} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared plus a squared/Arctangent Form

:$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 + a^2}
      | r = \frac 1 a \int \frac {\d t} {t^2 + 1}
      | c = Substitution of $x \to a t$
}}
{{eqn | r = \frac 1 a \int \frac {\d t} {\paren {1 + i t} \paren {1 - i t} }
      | c = factoring
}}
{{eqn | r = \frac 1 {2 a} \paren {\int \frac {\d t} {1 + i t} + \int \frac {\d t} {1 - i t} }
      | c = {{Defof|Partial Fractions Expansion}}
}}
{{eqn | r = \frac 1 {2 a} \paren {i \map \ln {1 - i t} - i \map \ln {1 + i t} } + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac i {2 a} \map \ln {\frac {1 - i t} {1 + i t} } + C
      | c = Sum of Logarithms
}}
{{eqn | r = \frac 1 a \arctan \frac x a + C
      | c = Arctangent Logarithmic Formulation and substituting back $t \to \dfrac x a$
}}
{{end-eqn}}
{{qed}}
Theorem: Endomorphism from Integers to Multiples

Let $\struct {\Z, +}$ be the additive group of integers.
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be a mapping.
Then $\phi$ is a group endomorphism {{iff}}:
:$\exists k \in \Z: \forall n \in \Z: \map \phi n = k n$

Proof:
=== Necessary Condition ===
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be an endomorphism.
Let $k = \map \phi 1$.
We have that $n = \overbrace {1 + \cdots + 1}^n$ for any positive integer $n$.
Thus:
{{begin-eqn}}
{{eqn | l = \map \phi n
      | r = \map \phi {\overbrace {1 + \cdots + 1}^n}
}}
{{eqn | r = \overbrace {\map \phi 1 + \cdots + \map \phi 1}^n
      | c = 
}}
{{eqn | r = \overbrace {k + \cdots + k}^n
      | c = 
}}
{{eqn | r = k n
      | c = 
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | l = \map \phi 1
      | r = \map \phi {1 + 0}
      | c = 
}}
{{eqn | r = \map \phi 1 + \map \phi 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi 0
      | r = 0
      | c = 
}}
{{eqn | r = k \cdot 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi {-1}
      | r = -k
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \N_{>0}
      | l = \map \phi {-n}
      | r = -k n
      | c = similar with above
}}
{{end-eqn}}
Thus:
:$\forall n \in \Z: \map \phi n = k n$
{{qed|lemma}}