avewright/proofwiki-math huggingface dataset:
Theorem: Normed Division Ring Completions are Isometric and Isomorphic/Lemma 5

:$\psi$ is a ring isomorphism.

Proof:
By {{Lemma|Normed Division Ring Completions are Isometric and Isomorphic|4}}, $\psi$ is an isometry.
By the definition of an isometry, $\psi$ is a bijection.
By the definition of a ring isomorphism, all that remains is to show that $\psi$ is a ring homomorphism.
That is:
:$(1): \quad \forall x, y \in S_1: \map \psi {x + y} = \map \psi x + \map \psi y$
:$(2): \quad \forall x, y \in S_1: \map \psi {x y} = \map \psi x \map \psi y$
Let $x, y \in S_1$.
Let $x = \ds \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$.
Let $y = \ds \lim_{n \mathop \to \infty} y_n$ for some sequence $\sequence {y_n} \subseteq R_1$.
By Sum Rule for Sequences in Normed Division Ring then:
:$x + y = \ds \lim_{n \mathop \to \infty} \paren {x_n + y_n}$
By Product Rule for Sequences in Normed Division Ring then:
:$x y = \ds \lim_{n \mathop \to \infty} \paren {x_n y_n}$
Then:
{{begin-eqn}}
{{eqn | l = \map \psi {x + y}
      | r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n + y_n}
      | c = Definition of $\psi$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\map {\psi'} {x_n} + \map {\psi'} {y_n} }
      | c = $\psi'$ is a ring isomorphism
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n} + \lim_{n \mathop \to \infty} \map {\psi'} {y_n}
      | c = Sum Rule for Sequences in Normed Division Ring
}}
{{eqn | r = \map \psi x + \map \psi y
      | c = Definition of $\psi$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map \psi {x y}
      | r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n y_n}
      | c = Definition of $\psi$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \paren {\map {\psi'} {x_n} \map {\psi'} {y_n} }
      | c = $\psi'$ is a ring isomorphism
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n} \lim_{n \mathop \to \infty} \map {\psi'} {y_n}
      | c = Product Rule for Sequences in Normed Division Ring
}}
{{eqn | r = \map \psi x \map \psi y
      | c = Definition of $\psi$
}}
{{end-eqn}}
{{qed}}
Category:Normed Division Ring Completions are Isometric and Isomorphic
nmsddwdutqe1dwahr5lwqz1xv6q6vfx
Theorem: Normed Division Ring Completions are Isometric and Isomorphic

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be normed division ring completions of $\struct {R, \norm {\, \cdot \,} }$
Then there exists an isometric isomorphism:
:$\psi: \struct {S_1, \norm {\, \cdot \,}_1 } \to \struct {S_2, \norm {\, \cdot \,}_2 }$

Proof:
By the definition of a normed division ring completion then:
:there exists a distance-preserving ring monomorphisms $\phi_1: R \to S_1$
:$R_1 = \map {\phi_1} R$ is a dense subring of $S_1$
:$S_1$ is a complete metric space
:there exists a distance-preserving ring monomorphisms $\phi_2: R \to S_2$
:$R_2 = \map {\phi_2} R$ is a dense subring of $S_2$
:$S_2$ is a complete metric space
Theorem: Distance-Preserving Image Isometric to Domain for Metric Spaces

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a distance-preserving mapping.
Then:
:$\phi: M_1 \to \Img \phi$
is an isometry.

Proof:
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi$ be a distance-preserving mapping from $M_1$ to $M_2$.
Let $A = \Img \phi$ be the image of $\phi$.
By Subspace of Metric Space is Metric Space, $\struct {A, d_2}$ is a metric space.
As $\phi$ is a distance-preserving mapping, by Distance-Preserving Mapping is Injection of Metric Spaces it is injective.
From Restriction of Injection is Injection, $\phi: M_1 \to \Img \phi$ is an injection.
From Restriction of Mapping to Image is Surjection, $\phi: M_1 \to \Img \phi$ is a surjection.
Thus $\phi \to \Img \phi$ is by definition a bijection.
Thus $\phi: M_1 \to \Img \phi$ is a bijective distance-preserving mapping.
Hence, by definition, $\phi: M_1 \to \Img \phi$ is an isometry.
{{qed}}
Category:Isometries (Metric Spaces)
Category:Metric Spaces
8wd3c8cwbakcddt14oj8f24wxkaro7m
Theorem: Distance-Preserving Mapping is Injection of Metric Spaces

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a distance-preserving mapping.
Then $\phi$ is an injection.

Proof:
Let $a, b \in A_1$ and suppose that $\map \phi a = \map \phi b$.
Then by the definition of a metric space:
:$\map {d_2} {\map \phi a, \map \phi b} = 0$
By the definition of a distance-preserving mapping then:
:$\map {d_1} {a, b} = 0$
Thus by the definition of a metric space:
:$a = b$
Hence $\phi$ is injective.
{{qed}}
Category:Metric Spaces
psyxnk22y9f84t2gcazrzrfubf3j17p
Theorem: Odd Power Function is Surjective

Let $n \in \Z_{\ge 0}$ be an odd positive integer.
Let $f_n: \R \to \R$ be the real function defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is a surjection.

Proof:
From Existence of Positive Root of Positive Real Number we have that:
:$\forall x \in \R_{\ge 0}: \exists y \in \R: y^n = x$
From Power of Ring Negative:
:$\paren {-x}^n = -\paren {x^n}$
and so:
:$\forall x \in \R_{\le 0}: \exists y \in \R: y^n = x$
Thus:
:$\forall x \in \R: \exists y \in \R: y^n = x$
and so $f_n$ is a surjection.
Hence the result.
{{qed}}
Category:Powers
Category:Surjections
Category:Real Functions
0ejl6jmukqy21eb92502g3vn2r5quxu
Theorem: Residue at Multiple Pole

Let $f: \C \to \C$ be a function meromorphic on some region, $D$, containing $a$. 
Let $f$ have a single pole in $D$, of order $N$, at $a$.
Then the residue of $f$ at $a$ is given by:
:$\ds \Res f a = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac {\d^{N - 1} } {\d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$

Proof:
By Existence of Laurent Series, there exists a Laurent series: 
:$\ds \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$ 
convergent on $D \setminus \set a$. 
As $f$ has a pole of order $N$ at $a$, we have $c_n = 0$ for $n < -N$. 
So: 
:$\ds \paren {z - a}^N \map f z = \sum_{n \mathop = -N}^\infty c_n \paren {z - a}^{n + N}$ 
Which can be rewritten: 
:$\ds \paren {z - a}^N \map f z = \sum_{n \mathop = 0}^\infty c_{n - N} \paren {z - a}^n$
Note that this is a Taylor series with centre $a$. 
By the definition of a residue: 
:$\ds \Res f a = c_{-1}$
This corresponds to the $\paren {N - 1}$th in the Taylor series of $\paren {z - a}^N \map f z$ about $a$. 
We therefore have by Taylor Series of Holomorphic Function: 
:$\ds c_{-1} = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac {\d^{N - 1} } {\d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$
Hence the result. 
{{qed}}
Category:Complex Analysis
b12b17gzllmsn9ilh04lkdzwfbknhk7
Theorem: Bijection from Cartesian Product of Initial Segments to Initial Segment

Let $\N_k$ be used to denote the set of the first $k$ non-zero natural numbers:
:$\N_k := \set {1, 2, \ldots, k}$
Then a bijection can be established between $\N_k \times \N_l$ and $\N_{k l}$, where $\N_k \times \N_l$ denotes the Cartesian product of $\N_k$ and $\N_l$.

Proof:
Let $\phi: \N_k \times \N_l \to \N_{k l}$ be defined as:
:$\forall \tuple {m, n} \in \N_k \times \N_l: \map \phi {m, n} = \paren {m - 1} \times l + n$
First it is confirmed that the codomain of $\phi$ is indeed $\N_{k l}$.
{{finish|fiddly and tedious, can't think of an elegant way to prove it}}
Theorem: Bijection between S x T and T x S

Let $S$ and $T$ be sets.
Let $S \times T$ be the Cartesian product of $S$ and $T$.
Then there exists a bijection from $S \times T$ to $T \times S$.

Proof:
Let $\phi: S \times T \to T \times S$ be the mapping defined as:
:$\forall \tuple {s, t} \in S \times T: \map \phi {s, t} = \tuple {t, s}$
Then $\phi$ is the bijection required, as follows:
The domain of $\phi$ is $S \times T$.
Let $\tuple {t, s} \in T \times S$.
Then there exists $\tuple {s, t} \in S \times T$ such that $\map \phi {s, t} = \tuple {t, s}$.
Thus $\phi$ is a surjection.
Let $\map \phi {s_1, t_1} = \map \phi {s_2, t_2}$ for some $\tuple {s_1, t_1}$ and $\tuple {s_2, t_2}$ in $S \times T$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {s_1, t_1}
      | r = \map \phi {s_2, t_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {t_1, s_1}
      | r = \tuple {t_2, s_2}
      | c = Definition of $\phi$
}}
{{eqn | ll= \leadsto
      | l = \tuple {s_1, t_1}
      | r = \tuple {s_2, t_2}
      | c = {{Defof|Ordered Pair}}
}}
{{end-eqn}}
and so $\phi$ is an injection.
Hence the result by definition of bijection.
{{qed}}
Theorem: Bijection between R x (S x T) and (R x S) x T

Let $R$, $S$ and $T$ be sets.
Let $S \times T$ be the Cartesian product of $S$ and $T$.
Then there exists a bijection from $R \times \paren {S \times T}$ to $\paren {R \times S} \times T$.
Hence:
:$\card {R \times \paren {S \times T} } = \card {\paren {R \times S} \times T}$

Proof:
Let $\phi: R \times \paren {S \times T} \to \paren {R \times S} \times T$ be the mapping defined as:
:$\forall \tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}: \map \phi {s, t} = \tuple {\tuple {r, s}, t}$
Then $\phi$ is the bijection required, as follows:
The domain of $\phi$ is $R \times \paren {S \times T}$.
Let $\tuple {\tuple {r, s}, t} \in \paren {R \times S} \times T$.
Then there exists $\tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}$ such that $\map \phi {r, \tuple {s, t} } = \tuple {\tuple {r, s}, t}$.
Thus $\phi$ is a surjection.
Let $\map \phi {r_1, \tuple {s_1, t_1} } = \map \phi {r_2, \tuple {s_2, t_2} }$ for some $\tuple {r_1, \tuple {s_1, t_1} }$ and $\tuple {r_2, \tuple {s_2, t_2} }$ in $R \times \paren {S \times T}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {r_1, \tuple {s_1, t_1} }
      | r = \map \phi {r_2, \tuple {s_2, t_2} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {\tuple {r_1, s_1}, t_1}
      | r = \tuple {\tuple {r_2, s_2}, t_2}
      | c = Definition of $\phi$
}}
{{eqn | ll= \leadsto
      | l = \tuple {r_1, \tuple {s_1, t_1} }
      | r = \tuple {r_2, \tuple {s_2, t_2} }
      | c = {{Defof|Ordered Pair}}
}}
{{end-eqn}}
and so $\phi$ is an injection.
Hence the result by definition of bijection.
{{qed}}
Theorem: Bijection between Power Set of nth Initial Section and Initial Section of nth Power of 2

Let $\N_n$ be used to denote the first $n$ non-zero natural numbers:
:$\N_n = \set {1, 2, \ldots, n}$
Then there exists a bijection between the power set of $\N_n$ and $\N_{2^n}$.

Proof:
Let $\phi: \powerset {\N_n} \to \N_{2^n}$ be defined as:
:$\forall A \in \powerset {\N_n}: \map \phi A = \begin{cases} \ds \sum_{k \mathop \in A} 2^{k - 1} & : A \ne \O \\ 2^k & : A = \O \end{cases}$
Apart from $\O$, every $A \in \powerset {\N_n}$ consists of a set of integers between $1$ and $n$.
The expression $\ds \sum_{k \mathop \in A} 2^{k - 1}$ is the summation of a set of powers of $2$ between $2^0$ and $2^{n - 1}$.
Hence $A$ is seen to be left-total and many-to-one and so is a mapping.
By the Basis Representation Theorem, every integer between $1$ and $2^n - 1$ can be expressed uniquely as the sum of powers of $2$ between $2^0$ and $2^{n - 1}$.
The final integer $2^n$ is mapped to from $\O$.
Thus it is seen that $\phi$ is both an injection and a surjection.
Hence the result.
{{qed}}
Theorem: Trivial Group is Group

The trivial group is a group.

Proof:
Let $G = \struct {\set e, \circ}$ be an algebraic structure.
Theorem: Symmetry Group of Line Segment is Group

The symmetry group of the line segment is a group.

Proof:
Let us refer to this group as $D_1$.
Taking the group axioms in turn:
Theorem: Order of Dihedral Group

The dihedral group $D_n$ is of order $2 n$.

Proof:
By definition, $D_n$ is the symmetry group of the regular polygon of $n$ sides.
:500px 500px
Let $P$ be a regular $n$-gon.
By inspection, it is seen that:
:$(1): \quad$ there are $n$ symmetries of the vertices of $P$ by rotation
:$(2): \quad$ there are a further $n$ symmetries of the vertices of $P$ by rotation after reflected in any of the axes of bilateral symmetry of $P$.
Hence the result.
{{qed}}
Theorem: Klein Four-Group is Group

The Klein $4$-group $K_4$ is a group.

Proof:
From Klein Four-Group as Subgroup of $S_4$ it is demonstrated that $K_4$ is a subgroup of the $4$th symmetric group.
Hence the result.
{{qed}}
Theorem: Dihedral Group is Non-Abelian

Let $n \in \N$ be a natural number such that $n > 2$.
Let $D_n$ denote the dihedral group of order $2 n$.
Then $D_n$ is not abelian.

Proof:
From Group of Order less than 6 is Abelian we have that $D_1$ and $D_2$ are abelian, which is why the condition on $n$.
From Group Presentation of Dihedral Group we have:
:$\beta \alpha = \alpha^{n - 1} \beta$
for some $\alpha, \beta \in D_n$ such that $\alpha \ne \beta$.
We also have:
:$\alpha^n = e$
But if $D_n$ were abelian, that would mean:
:$\beta \alpha = \beta \alpha^{n - 1}$
and so by the Cancellation Laws:
:$\alpha = \alpha^{n - 1}$
and so:
:$\alpha^2 = e$
But by definition of the dihedral group, this is not the case.
{{qed}}
Theorem: Matrix Entrywise Addition is Commutative

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is commutative on $\map \MM {m, n}$.
That is:
:$\mathbf A + \mathbf B = \mathbf B + \mathbf A$
for all $\mathbf A$ and $\mathbf B$ in $\map \MM {m, n}$.

Proof:
From:
:Integers form Ring
:Rational Numbers form Ring
:Real Numbers form Ring
:Complex Numbers form Ring
the standard number systems $\Z$, $\Q$, $\R$ and $\C$ are rings.
Hence we can apply Matrix Entrywise Addition over Ring is Commutative.
{{qed|lemma}}
The above cannot be applied to the natural numbers $\N$, as they do not form a ring.
However, from Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\struct {\N, +}$ is a commutative monoid.
By definition, matrix entrywise addition is the '''Hadamard product''' with respect to addition of numbers.
The result follows from Commutativity of Hadamard Product.
{{qed}}
Theorem: Matrix Entrywise Addition is Commutative

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
The operation $+$ is commutative on $\map \MM {m, n}$.
That is:
:$\mathbf A + \mathbf B = \mathbf B + \mathbf A$
for all $\mathbf A$ and $\mathbf B$ in $\map \MM {m, n}$.

Proof:
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be matrices whose order is $m \times n$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A + \mathbf B
      | r = \sqbrk a_{m n} + \sqbrk b_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf B$
}}
{{eqn | r = \sqbrk {a + b}_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \sqbrk {b + a}_{m n}
      | c = Commutative Law of Addition
}}
{{eqn | r = \sqbrk b_{m n} + \sqbrk a_{m n}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{eqn | r = \mathbf B + \mathbf A
      | c = Definition of $\mathbf A$ and $\mathbf B$
}}
{{end-eqn}}
{{qed}}
Theorem: Special Linear Group is not Abelian

Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.
Then $\SL {n, K}$ is not an abelian group.

Proof:
From Special Linear Group is Subgroup of General Linear Group we have that $\SL {n, K}$ is a group.
From Matrix Multiplication is not Commutative it follows that $\SL {n, K}$ is not abelian.
{{qed}}
Theorem: Summation Formula (Complex Analysis)

:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$

Proof:
By Summation Formula: Lemma, there exists a constant $A$ such that: 
:$\cmod {\map \cot {\pi z} } < A$ 
for all $z$ on $C_N$.
Since $f$ has only finitely many poles, we can take $N$ large enough so that no poles of $f$ lie on $C_N$. 
Let $X_N$ be the set of poles of $f$ contained in the region bounded by $C_N$. 
From Poles of Cotangent Function, $\map \cot {\pi z}$ has poles at $z \in \Z$. 
Let $A_N = \set {n \in \Z : -N \le n \le N}$
We then have: 
{{begin-eqn}}
{{eqn | l = \oint_{C_N} \pi \map \cot {\pi z} \map f z \rd z
      | r = 2 \pi i \sum_{z_0 \mathop \in X_N \mathop \cap A_N} \Res {\pi \map \cot {\pi z} \map f z} {z_0}
      | c = Cauchy's Residue Theorem 
}}
{{eqn | r = 2 \pi i \paren {\sum_{n \mathop \in A_N \mathop \setminus X_N} \Res {\pi \map \cot {\pi z} \map f z} n + \sum_{z_0 \mathop \in X_N} \Res {\pi \map \cot {\pi z} \map f z} {z_0} }
}}
{{end-eqn}}
We then have, for each integer $n$: 
{{begin-eqn}}
{{eqn | l = \Res {\pi \map \cot {\pi z} \map f z} n
      | r = \lim_{z \mathop \to n} \paren {\paren {z - n} \pi \map \cot {\pi z} \map f z}
      | c = Residue at Simple Pole
}}
{{eqn | r = \map f n \lim_{z \mathop \to n} \paren {\frac {z - n} z + 2 \sum_{k \mathop = 1}^\infty \frac {z \paren {z - n} } {z^2 - k^2} }
      | c = Mittag-Leffler Expansion for Cotangent Function
}}
{{eqn | r = \map f n \cdot 2 \lim_{z \mathop \to n} \paren {\frac z {z + n} }
}}
{{eqn | r = \map f n \frac {2 n} {2 n}
}}
{{eqn | r = \map f n
}}
{{end-eqn}}
Note that {{hypothesis}}:
:$\ds \int_{C_N} \paren {\pi \cot \pi z} \map f z \rd z \to 0$
So, taking $N \to \infty$: 
:$\ds 0 = 2 \pi i \paren {\sum_{n \mathop \in \Z \mathop \setminus X} \map f n + \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0} }$
which gives:
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
{{qed}}
Theorem: Summation Formula (Complex Analysis)/Lemma

Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.
Then there exists a constant real number $A$ independent of $N$ such that: 
:$\cmod {\map \cot {\pi z} } < A$ 
for all $z \in C_N$.

Proof:
Let $z = x + iy$ for real $x, y$.
Theorem: Product of Generating Elements of Dihedral Group

Let $D_n$ be the dihedral group of order $2 n$.
Let $D_n$ be defined by its group presentation:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Then for all $k \in \Z_{\ge 0}$:
:$\beta \alpha^k = \alpha^{n - k} \beta$

Proof:
The proof proceeds by induction.
For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition:
:$\beta \alpha^k = \alpha^{n - k} \beta$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \beta \alpha^0
      | r = \beta e
      | c = 
}}
{{eqn | r = e \beta
      | c = 
}}
{{eqn | r = \alpha^n \beta
      | c = 
}}
{{eqn | r = \alpha^{n - 0} \beta
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
Theorem: Center of Dihedral Group

Let $n \in \N$ be a natural number such that $n \ge 3$.
Let $D_n$ be the dihedral group of order $2 n$, given by:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Let $\map Z {D_n}$ denote the center of $D_n$.
Then:
:$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$

Proof:
By definition, the center of $D_n$ is:
:$\map Z {D_n} = \set {g \in D_n: g x = x g, \forall x \in D_n}$
For $n \le 2$ we have that $\order {D_n} \le 4$ and so by Group of Order less than 6 is Abelian $D_n$ is abelian for $n < 3$.
Hence by definition of abelian group:
:$\map Z {D_n} = D_n$
for $n < 3$.
So, let $n \ge 3$.
By Group Presentation of Dihedral Group:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
From Product of Generating Elements of Dihedral Group:
:$\beta \alpha^k = \alpha^{n - k} \beta$
for all $k \in \Z_{\ge 0}$.
We have that $D_n$ is generated by $\alpha$ and $\beta$.
Thus:
:$x \in \map Z {D_n} \iff x \alpha = \alpha x \land x \beta = \beta x$
Let $x \in \map Z {D_n}$.
We have that $x$ can be expressed in the form:
:$x = \alpha^i \beta^j$
As $x \in \map Z {D_n}$, we have:
{{begin-eqn}}
{{eqn | l = x \alpha
      | r = \alpha x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha^i \beta^j \alpha
      | r = \alpha^{i + 1} \beta^j
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \beta^j \alpha
      | r = \alpha \beta^j
      | c = 
}}
{{end-eqn}}
But for $j = 1$ this means:
{{begin-eqn}}
{{eqn | l = \alpha \beta
      | r = \beta \alpha
      | c = 
}}
{{eqn | r = \alpha^{-1} \beta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha
      | r = \alpha^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha^2
      | r = e
      | c = 
}}
{{end-eqn}}
But the order of $\alpha$ is $n$ and $n > 2$, and hence:
:$\alpha^2 \ne e$
So if $x \in \map Z {D_n}$ it follows that $x$ has to be in the form:
:$x = \alpha^i$
for some $i \in \Z_{\ge 0}$.
Again, as $x \in \map Z {D_n}$, we have:
{{begin-eqn}}
{{eqn | l = x \beta
      | r = \beta x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha^i \beta
      | r = \beta \alpha^i
      | c = 
}}
{{eqn | r = \alpha^{n - i} \beta
      | c = Product of Generating Elements of Dihedral Group
}}
{{eqn | ll= \leadsto
      | l = \alpha^i
      | r = \alpha^{n - i}
      | c = 
}}
{{eqn | r = \alpha^n \alpha^{-i}
      | c = 
}}
{{eqn | r = \alpha^{-i}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha^{2 i}
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n
      | o = \divides
      | r = 2 i
      | c = {{Defof|Order of Group Element}}
}}
{{end-eqn}}
So either $i = 0$ or $n = 2 i$, as $0 \le i \le n$.
If $i = 0$ then $x = \alpha^0 = e$.
If $2 i = n$ then $n$ is even and so:
:$x = \alpha^{n / 2}$
Hence the result.
{{qed}}
Theorem: Intersection of Additive Groups of Integer Multiples

Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {m \Z, +}$ and $\struct {n \Z, +}$ be the corresponding additive groups of integer multiples.
Then:
:$\struct {m \Z, +} \cap \struct {n \Z, +} = \struct {\lcm \set {m, n} \Z, +}$

Proof:
By definition:
:$m \Z = \set {x \in \Z: m \divides x}$
Thus:
{{begin-eqn}}
{{eqn | l = m \Z \cap n \Z
      | r = \set {x \in \Z: n \divides x} \cap \set {x \in \Z: m \divides x}
      | c = 
}}
{{eqn | r = \set {x \in \Z: n \divides x \land m \divides x}
      | c = 
}}
{{eqn | r = \set {x \in \Z: \lcm \set {m, n} \divides x}
      | c = 
}}
{{eqn | r = \lcm \set {m, n} \Z
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Subgroups of Additive Group of Integers Modulo m

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
The subgroups of $\struct {\Z_m, +_m}$ are the additive groups of integers modulo $k$ where:
:$k \divides m$

Proof:
From Integers Modulo m under Addition form Cyclic Group, $\struct {\Z_m, +_m}$ is cyclic.
Let $H$ be a subgroup of $\struct {\Z_m, +_m}$
From Subgroup of Cyclic Group is Cyclic, $H$ is of the form $\struct {\Z_k, +_k}$ for some $k \in \Z$.
From Lagrange's Theorem, it follows that $k \divides m$.
Hence the result.
{{qed}}
Theorem: Subgroup of Additive Group of Integers Generated by Two Integers

Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\gen {m, n}$ be the subgroup of $\struct {\Z, +}$ generated by $m$ and $n$.
Then:
:$\gen {m, n} = \struct {\gcd \set {m, n} \Z, +}$
That is, the additive groups of integer multiples of $\gcd \set {m, n}$, where $\gcd \set {m, n}$ is the greatest common divisor of $m$ and $n$.

Proof:
By definition:
:$\gen {m, n} = \set {x \in \Z: \gcd \set {m, n} \divides x}$
{{Handwaving|Sorry, I would make the effort, but it's tedious.}}
Hence the result.
{{qed}}