hcju/proofwikiqa huggingface dataset:
By lemma 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 $x = \displaystyle \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$. Let $y = \displaystyle \lim_{n \mathop \to \infty} y_n$ for some sequence $\sequence {y_n} \subseteq R_1$. By Sum Rule for Sequences in Normed Division Ring then: :$x + y = \displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n}$ By Product Rule for Sequences in Normed Division Ring then: :$x y = \displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n}$ 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:Completion of Normed Division Ring 9x58937p8qd7tqgcynrn16eb87pn045
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 === Lemma 1 === {{:Normed Division Ring Completions are Isometric and Isomorphic/Lemma 1}} === Lemma 2 === {{:Normed Division Ring Completions are Isometric and Isomorphic/Lemma 2}} === Lemma 3 === {{:Normed Division Ring Completions are Isometric and Isomorphic/Lemma 3}} === Lemma 4 === {{:Normed Division Ring Completions are Isometric and Isomorphic/Lemma 4}} === Lemma 5 === {{:Normed Division Ring Completions are Isometric and Isomorphic/Lemma 5}}{{qed}} Category:Normed Division Rings Category:Complete Metric Spaces Category:Completion of Normed Division Ring pmio9r7oj1x0usm2h8v3hcam5m5lnvb
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:Definitions/Metric Spaces Category:Definitions/Isometries mzp3gc1t8oxve33zhyembmh0h7gwq7z
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
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
By Existence of Laurent Series, there exists a Laurent series:  :$\displaystyle \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$  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:  :$\displaystyle \paren {z - a}^N \map f z = \sum_{n \mathop = -N}^\infty c_n \paren {z - a}^{n + N}$  Which can be rewritten:  :$\displaystyle \paren {z - a}^N \map f z = \sum_{n \mathop = 0}^\infty c_{n - N} \paren {z - a}^n$ Note that this is a Taylor series with centre $a$.  By the definition of a residue:  :$\displaystyle \Res f a = c_{-1}$ 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:  :$\displaystyle c_{-1} = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac { \d^{N - 1} } { \d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$ Hence the result.  {{qed}} Category:Complex Analysis hsgyngvd2qagmuo7wyg4mdhcjcyobuj
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}}
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}}
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}}
Let $\phi: \powerset {\N_n} \to \N_{2^n}$ be defined as: :$\forall A \in \powerset {\N_n}: \map \phi A = \begin{cases} \displaystyle \sum_{k \mathop \in A} 2^{k - 1} & : A \ne \O \\ 2^k & : A = \O \end{cases}$ Apart from $\O$, every $A \in \powerset {\N_n}$ consists of a set of integers between $1$ and $n$. The expression $\displaystyle \sum_{k \mathop \in A} 2^{k - 1}$ 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}}
Let $G = \struct {\set e, \circ}$ be an algebraic structure. === G0: Closure === For $G$ to be a group, it must be closed. So it must be the case that: :$\forall e \in G: e \circ e = e$ {{qed|lemma}} === G1: Associativity === $\circ$ is associative: :$e \circ \paren {e \circ e} = e = \paren {e \circ e} \circ e$ trivially. {{qed|lemma}} === G2: Identity Element === $e$ is the identity: :$\forall e \in G: e \circ e = e$ {{qed|lemma}} === G3: Inverse Elements === Every element of $G$ (all one of them) has an inverse: This follows from the fact that the Identity is Self-Inverse, and the only element of $G$ is indeed the identity: :$e \circ e = e \implies e^{-1} = e$ {{qed}}
Let us refer to this group as $D_1$. Taking the group axioms in turn: === G0: Closure === From the Cayley table it is seen directly that $D_1$ is closed. {{qed|lemma}} === G1: Associativity === Composition of Mappings is Associative. {{qed|lemma}} === G2: Identity === The identity is $e$. {{qed|lemma}} === G3: Inverses === Each element is seen to be self-inverse: :$r^{-1} = r$ {{qed|lemma}} No more need be done. $D_1$ is seen to be a group. {{qed}}
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 symmetry of $P$. Hence the result. {{qed}}
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}}
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}}
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}}
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}}
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}}
By Summation Formula: Lemma, there exists a constant $A$ such that:  :$\displaystyle \cmod {\cot \paren {\pi z} } < A$  for all $z$ on $C_N$.  Let $X_N$ be the poles of $f$ contained in $C_N$.  From Poles of Cotangent Function, $\cot \paren {\pi z}$ 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 \cot \paren {\pi z} \map f z \rd z 	| r = 2 \pi i \sum_{z_0 \mathop \in X_N \cap A_N} \Res {\pi \cot \paren {\pi z} \map f z} {z_0} 	| c = Residue Theorem  }} {{eqn	| r = 2 \pi i \paren {\sum_{n \mathop \in A_N \setminus X_N} \Res {\pi \cot \paren {\pi z} \map f z} n + \sum_{z_0 \in X_N} \Res {\pi \cot \paren {\pi z} \map f z} {z_0} } }} {{end-eqn}} We then have:  {{begin-eqn}} {{eqn	| l = \Res {\pi \cot \paren {\pi z} \map f z} n 	| r = \lim_{z \mathop \to n} \paren {\paren {z - n} \pi \cot \paren {\pi z} \map f z} 	| c = Residue at Simple Pole }} {{eqn	| r = \map f n \lim_{z_0 \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}} We also have:  {{begin-eqn}} {{eqn	| l = \cmod {\oint_{C_N} \pi \cot \paren {\pi z} \map f z \rd z} 	| o = \le 	| r = \frac {\pi A M \paren {8 N + 4} } {N^k} 	| c = Estimation Lemma }} {{eqn	| r = \frac {8 \pi A M} {N^{k - 1} } + \frac {4 \pi A M} {N^k} }} {{eqn	| o = \to 	| r = 0 	| c = as $N \to \infty$, as $k - 1 > 0$ }} {{end-eqn}} So, taking $N \to \infty$:  :$\displaystyle 0 = 2 \pi i \paren {\sum_{n \mathop \in \Z \setminus X} \map f n + \sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0} }$ which gives: :$\displaystyle \sum_{n \mathop = -\infty}^\infty \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}$ {{qed}}
Let $z = x + iy$ for real $x, y$.  === Case 1: $y > \frac 1 2$ ===  We have:  {{begin-eqn}} {{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert 	| r = \left\vert{\frac {e^{i \pi z} + e^{-i \pi z} } {e^{i \pi z} - e^{-i \pi z} } }\right\vert  	| c = Cotangent Exponential Formulation }} {{eqn	| r = \left\vert{\frac {e^{i \pi x - \pi y} + e^{-i \pi x + \pi y} } {e^{i \pi x - \pi y} - e^{-i \pi x + \pi y} } }\right\vert 	| c = as $z = x + iy$ }} {{eqn	| o = \le 	| r = \frac {\left\vert{e^{i \pi x - \pi y} }\right\vert + \left\vert{e^{-\pi i x + \pi y} }\right\vert} {\left\vert{e^{-i \pi x + \pi y} }\right\vert - \left\vert{e^{i \pi x - \pi y} }\right\vert}  	| c = Triangle Inequality }} {{eqn	| r = \frac {e^{-\pi y} + e^{\pi y} } {e^{\pi y} - e^{-\pi y} } }} {{eqn	| r = \frac {e^{-2 \pi y} + 1} {1 - e^{-2 \pi y} } 	| c = multiplying through $\dfrac {e^{-\pi y} } {e^{-\pi y} }$ }} {{eqn	| o = \le 	| r = \frac {1 + e^{-\pi} } {1 - e^{-\pi} } }} {{eqn	| r = A_1 }} {{end-eqn}} === Case 2: $y < -\frac 1 2$ ===  Similarly:  {{begin-eqn}} {{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert 	| r = \frac {\left\vert{e^{i \pi x - \pi y} }\right\vert + \left\vert{e^{-\pi i x + \pi y} }\right\vert} {\left\vert{e^{-i \pi x + \pi y} }\right\vert - \left\vert{e^{i \pi x - \pi y} }\right\vert}  }} {{eqn	| r = \frac {e^{-\pi y} + e^{\pi y} } {e^{-\pi y} - e^{\pi y} }  }} {{eqn	| r = \frac {1 + e^{2 \pi y} } {1 - e^{2 \pi y} } 	| c = multiplying through $\dfrac {e^{\pi y} } {e^{\pi y} }$ }} {{eqn	| o = \le 	| r = \frac {1 + e^{-\pi} } {1 - e^{-\pi} } }} {{eqn	| r = A_1 }} {{end-eqn}} === Case 3: $-\frac 1 2 \le y \le \frac 1 2$ ===  First consider $z = N + \frac 1 2 + iy$.  Then:  {{begin-eqn}} {{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert 	| r = \left\vert{\cot \pi \left({N + \frac 1 2 + iy}\right)}\right\vert }} {{eqn	| r = \left\vert{\cot \left({\frac \pi 2 + i \pi y}\right)}\right\vert 	| c = Cotangent Function is Periodic on Reals }} {{eqn	| r = \left\vert{\tanh \pi y}\right\vert 	| c = Cotangent of Complement equals Tangent, Hyperbolic Tangent in terms of Tangent }} {{eqn	| r = \left\vert{\tanh \frac \pi 2}\right\vert }} {{eqn	| r = A_2 }} {{end-eqn}} Similarly in the case of $z = -N - \frac 1 2 + iy$, we have:  {{begin-eqn}} {{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert 	| r = \left\vert{\cot \pi \left({-N - \frac 1 2 + iy}\right)}\right\vert }} {{eqn	| r = \left\vert{\cot \left({-\frac 1 2 + iy}\right)}\right\vert 	| c = Cotangent Function is Periodic on Reals }} {{eqn	| r = \left\vert{\tanh \pi y}\right\vert 	| c = Cotangent of Complement equals Tangent, Hyperbolic Tangent in terms of Tangent }} {{eqn	| r = \left\vert{\tanh \frac \pi 2}\right\vert }} {{eqn	| r = A_2 }} {{end-eqn}} Picking $A = \max \left({A_1, A_2}\right)$ gives the desired bound.  {{qed}}
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. === Basis for the Induction === We have: {{begin-eqn}} {{eqn | l = \beta \alpha \beta        | r = \alpha^{-1}       | c = Group Presentation of Dihedral Group }} {{eqn | ll= \leadsto       | l = \beta \alpha \beta^2       | r = \alpha^{-1} \beta       | c =  }} {{eqn | ll= \leadsto       | l = \beta \alpha       | r = \alpha^{-1} \beta       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\beta \alpha^r = \alpha^{n - r} \beta$ from which it is to be shown that: :$\beta \alpha^{r + 1} = \alpha^{n - r - 1} \beta$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \beta \alpha^{r + 1}       | r = \alpha^{n - r} \beta \alpha       | c =  }} {{eqn | r = \alpha^{n - r} \alpha^{-1} \beta       | c = Basis for the Induction }} {{eqn | r = \alpha^{n - r - 1} \beta       | c =  }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall k \in \Z_{\ge 0}: \beta \alpha^k = \alpha^{n - k} \beta$ {{qed}} Category:Dihedral Groups 3x2tn8gwziutax0mwls4l9m97plr2oq
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 Group equals Center iff Abelian $\map Z {D_n} = D_n$ 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}}
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}}
Consider the map $\phi: \struct {m \Z, +} \times \struct {n \Z, +} \mapsto \struct {\Z, +} \times \struct {\Z, +}$ defined by: :$\forall c, d \in \Z: \map \phi {m c, n d} = \tuple {c, d}$ which is a group isomorphism. {{explain|Prove the above statement}} Hence, $\struct {m \Z, +} \times \struct {n \Z, +}$ is a free abelian group of rank $2$. Therefore, any subgroup generated by a singleton, for example, $\set {\tuple {x, 0}: x \in \Z}$ is a subgroup not in the form $\struct {m \Z, +} \times \struct {n \Z, +}$. {{refactor|Implement "singly generated" as a definition page in its own right and then link to it, hence keep this page clean}} {{qed}}
We claim that $\sim$ is left congruence modulo $S_{n - 1}$, the symmetric group on $n - 1$ letters $\set {1, \dots, n - 1}$. Notice that every element of $S_{n - 1}$ fixes $n$. For all $\pi, \tau \in S_n$ such that $\pi \sim \tau$: {{begin-eqn}} {{eqn | l = \map {\paren {\pi^{-1} \circ \tau} } n       | r = \map {\pi^{-1} } {\map \tau n}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\pi^{-1} } {\map \pi n}       | c = Definition of $\sim$ }} {{eqn | r = \map {\paren {\pi^{-1} \circ \pi} } n       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = n       | c = {{Defof|Inverse Element}} }} {{end-eqn}} so $\pi^{-1} \circ \tau$ fixes $n$ as well. This shows that $\pi^{-1} \circ \tau \in S_{n - 1}$. By definition of Left Congruence Modulo Subgroup: :$\pi \equiv^l \tau \pmod {S_{n - 1} }$ Now we show the converse. Suppose $\pi \equiv^l \tau \pmod {S_{n - 1} }$. Then $\pi^{-1} \circ \tau \in S_{n - 1}$. Hence $\map {\paren {\pi^{-1} \circ \tau} } n = n$. Then: {{begin-eqn}} {{eqn | l = \map \pi n       | r = \map \pi {\map {\paren {\pi^{-1} \circ \tau} } n} }} {{eqn | r = \map {\paren {\pi \circ \pi^{-1} \circ \tau} } n       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map \tau n       | c = {{Defof|Inverse Element}} }} {{eqn | ll = \leadsto       | l = \pi       | o = \sim       | r = \tau }} {{end-eqn}} Therefore $\sim$ and left congruence modulo $S_{n - 1}$ are equivalent. The fact that $\sim$ is an equivalence relation follows from Left Congruence Modulo Subgroup is Equivalence Relation. {{qed}}