hcju/proofwikiqa huggingface dataset:
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Product of Indices: :$\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$ where in this context: :the group operation is $+$ :the $n$th power of $g$ is denoted $n g$ {{qed}}
By definition of abelian group, $x$ and $y$ commute. That is: :$x + y = y + x$ The result follows from Power of Product of Commutative Elements in Group. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition: :$s^n = s$ $P \left({1}\right)$ is the case: :$s^1 = s$ which holds by definition. Thus $P \left({1}\right)$ is seen to hold. === Basis for the Induction === $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = s^2       | r = s^1 \circ s       | c = Definition of $s^n$ }} {{eqn | r = s \circ s       | c = Definition of $s^1$ }} {{eqn | r = s       | c = {{Defof|Idempotent Element}} }} {{end-eqn}} Thus $P \left({2}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$s^k = s$ from which it is to be shown that: :$s^{k + 1} = s$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = s^{k + 1}       | r = s^k \circ s       | c =  }} {{eqn | r = s \circ s       | c = Induction Hypothesis }} {{eqn | r = s       | c = Basis for the Induction }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{> 0}: s^n = s$ </onlyinclude> Category:Idempotence d7rj1kcs86m65c7frk9092f545vlpib
By definition, a group is also a semigroup. The result follows as a special case of Powers of Semigroup Element Commute {{Qed}}
From Non-Zero Real Numbers under Multiplication form Abelian Group we have that $\struct {\R_{\ne 0}, \times}$ is a group. We know that $\R_{>0} \ne \O$, as (for example) $1 \in \R_{>0}$. Now, verify that the conditions for Two-Step Subgroup Test are satisfied: === Closure under $\times$ === Let $a, b \in \R_{>0}$. We take on board the fact that the Real Numbers form Ordered Integral Domain. Then: :$a b \in \R_{\ne 0}$ From Positive Elements of Ordered Ring : :$a \times b > 0$ so $a b \in \R_{>0}$. {{qed|lemma}} === Closure under Inverse === Let $a \in \R_{>0}$.  Then $a^{-1} = \dfrac 1 a \in \R_{>0}$. {{qed|lemma}} Hence, by the Two-Step Subgroup Test, $\struct {\R_{>0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$. {{qed}}
From Non-Zero Rational Numbers under Multiplication form Abelian Group we have that $\struct {\Q_{\ne 0}, \times}$ is a group. We know that $\Q_{> 0} \ne \O$, as (for example) $1 \in \Q_{> 0}$. Let $a, b \in \Q_{> 0}$. Then: :$a b \in \Q_{\ne 0}$ and $ab > 0$ Hence: :$a b \in \Q_{> 0}$ Let $a \in \Q_{> 0}$. Then: :$a^{-1} = \dfrac 1 a \in \Q_{> 0}$ So, by the Two-Step Subgroup Test, $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$. {{qed}}
Let $x \in \bigcup A, y \in S$ such that :$x \preceq y$ By definition of union: :$\exists Y \in A: x \in Y$ By assumption: :$Y$ is an upper set. By definition of upper set: :$y \in Y$ Thus by definition of union: :$y \in \bigcup A$ Hence :$\bigcup A$ is an upper set. {{qed}}
Let $\SS$ be a square lamina. Let $O$ be the center of $\SS$. Recall the definition of the symmetry group of the square $D_4$: {{:Definition:Symmetry Group of Square}} We have that: :Reflection $t_x$ can be achieved by a rotation of $\SS$ of $\pi$ radians about $x$. :Reflection $t_y$ can be achieved by a rotation of $\SS$ of $\pi$ radians about $y$. Thus $D_4$ forms a subgroup of $\GG$. From Symmetry Group of Square is Group we have that $D_4$ is not abelian. From Subgroup of Abelian Group is Abelian it follows that $\GG$ is also not abelian. {{qed}}
Let $x, y \in \bigcup A$. By definition of union: :$\exists X \in A: x \in X$ and :$\exists Y \in A: y \in Y$ By assumption: :$\exists Z \in A: X \cup Y \subseteq Z$ By definition of union: :$x, y \in X \cup Y$ By definition of subset: :$x, y \in Z$ By assumption: :$Z$ is filtered. By definition of filtered: :$\exists z \in Z: z \preceq x \land z \preceq y$ Thus by definition of union: :$z \in \bigcup A$ Hence :$\bigcup A$ is filtered. {{qed}}
A direct corollary of Special Orthogonal Group is Subgroup of Orthogonal Group. {{qed}}
By definition, a unit matrix is a diagonal matrix. From Inverse of Diagonal Matrix, the inverse of a diagonal matrix: :$\mathbf D = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$ is the diagonal matrix: : $\mathbf D^{-1} = \begin{bmatrix} \dfrac 1 {a_{11}} & 0 & \cdots & 0 \\ 0 & \dfrac 1 {a_{22}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\ \end{bmatrix}$ When $\mathbf D$ is the unit matrix $\mathbf I_n$, all elements $a_{kk}$ are equal to $1$: :$\mathbf I_n = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix}$ Hence: :$\mathbf I_n^{-1} = \begin{bmatrix} \dfrac 1 1 & 0 & \cdots & 0 \\ 0 & \dfrac 1 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \dfrac 1 1 \\ \end{bmatrix}$ Hence the result. {{qed}} Category:Unit Matrices Category:Inverse Matrices rgio764lguz6z2g6wb4yn16z1brin39
Let $y \in x^\gg$. By definition of way above closure: :$x \ll y$ where $\ll$ denotes the way below relation. By Way Below implies Preceding: :$x \preceq y$ Thus by definition of upper closure of element: :$y \in x^\succeq$ {{qed}}
By definition of diagonal matrix: :$\forall j, k: j \ne k \implies a_{jk} = 0 = a_{kj}$ So by definition of transpose of $D$: :$D = D^\intercal$ where $D^\intercal$ denotes the transpose. Hence the result, by definition of symmetric matrix. {{qed}} Category:Symmetric Matrices Category:Diagonal Matrices dv7n9kzvzryg6tjumxlrly9jp6semgo
By Unit Matrix is its own Inverse the inverse $I_n^{-1}$ of $I_n$ is $I_n$. By definition a unit matrix is a diagonal matrix. Hence by Diagonal Matrix is Symmetric: :$I_n = I_n^\intercal$ where $I_n^\intercal$ is the transpose of $I_n$. Thus: :$I_n^{-1} = I_n^\intercal$ and the result follows by definition of orthogonal. {{qed}} Category:Unit Matrices Category:Orthogonal Matrices thfuv0wq4j5hsyxq6htlcafz1z4aotx
By Unit Matrix is Orthogonal, $\mathbf I_n$ is an orthogonal matrix. By Determinant of Unit Matrix, the determinant of $\mathbf I_n$ is equal to $1$. The result follows by definition of proper orthogonal. {{qed}} Category:Unit Matrices Category:Orthogonal Matrices jukceu1rs4hk2cse2pnlfcn8ztarw3v
By definition of orthogonal matrix: :$\mathbf A^\intercal = \mathbf A^{-1}$ where $\mathbf A^\intercal$ is the transpose of $\mathbf A$. By Inverse of Inverse of Matrix: :$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$ By Transpose of Transpose of Matrix: :$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$ Thus we have: :$\paren {\mathbf A^{-1} }^{-1} = \mathbf A = \paren {\mathbf A^\intercal}^\intercal = \paren {\mathbf A^{-1} }^\intercal$ and so by definition $\mathbf A^{-1}$ is orthogonal. {{qed}} Category:Orthogonal Matrices Category:Inverse Matrices 8qgfo01t3z6dht0dvrmkf741k1527ht
By definition of inverse matrix: :$\mathbf A^{-1} \mathbf A = \mathbf I$ where $\mathbf I$ is the unit matrix. Thus the inverse of $\mathbf A^{-1}$ is $\mathbf A$. Hence the result. {{qed}} Category:Inverse Matrices c92664gz3mew2c1geggrxuvcjmj5c4x
By definition of proper orthogonal matrix: :$\mathbf A$ is orthogonal :the determinant of $\mathbf A$ is equal to $1$. By Inverse of Orthogonal Matrix is Orthogonal, $\mathbf A^{-1}$ is orthogonal. By Determinant of Inverse Matrix: :$\det \mathbf A^{-1} = \dfrac 1 {\det \mathbf A} = 1$ Hence, by definition, $\mathbf A^{-1}$ is proper orthogonal. {{qed}} Category:Orthogonal Matrices Category:Inverse Matrices sxpg2zjfvvqirrj1e87ohv80ib8a4k4
We have that Unit Matrix is Proper Orthogonal, so $\map {\operatorname {SO} } {n, k}$ is not empty. Let $\mathbf A, \mathbf B \in \map {\operatorname {SO} } {n, k}$. Then, by definition, $\mathbf A$ and $\mathbf B$ are proper orthogonal. Then by Inverse of Proper Orthogonal Matrix is Proper Orthogonal: :$\mathbf B^{-1}$ is a proper orthogonal matrix. By Product of Proper Orthogonal Matrices is Proper Orthogonal Matrix: :$\mathbf A \mathbf B^{-1}$ is a proper orthogonal matrix. Thus by definition of special orthogonal group: :$\mathbf A \mathbf B^{-1} \in \map {\operatorname {SO} } {n, k}$ Hence the result by One-Step Subgroup Test. {{qed}} Category:Orthogonal Groups c84yl527lbmgzjz3uiuiri8mcwzpl8t
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_G} {m, n}$. Then: {{begin-eqn}} {{eqn | l = \mathbf A \circ \paren {-\mathbf A}       | r = \sqbrk a_{m n} \circ \paren {-\sqbrk a_{m n} }       | c = Definition of $\mathbf A$ }} {{eqn | r = \sqbrk a_{m n} \circ \sqbrk {a^{-1} }_{m n}       | c = {{Defof|Negative Matrix|subdef = General Group}} }} {{eqn | r = \sqbrk {a \cdot \paren {a^{-1} } }_{m n}       | c = {{Defof|Hadamard Product}} }} {{eqn | r = \sqbrk e_{m n}       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = \mathbf A \circ \paren {-\mathbf A}       | r = \mathbf e       | c = {{Defof|Zero Matrix over General Monoid}} }} {{end-eqn}} The result follows from Zero Matrix is Identity for Hadamard Product. {{qed}}
Consider the semigroup $\struct {\N, +}$. From Natural Numbers under Addition form Commutative Semigroup, $\struct {\N, +}$ forms a semigroup. From Natural Numbers are Infinite, the underlying set of $\struct {\N, +}$ is infinite. From Natural Number Addition is Cancellable, all elements of $\struct {\N, +}$ are cancellable. But from Natural Numbers under Addition do not form Group, $\struct {\N, +}$ is not a group.
By Left Cancellable iff Left Regular Representation Injective, $\lambda_a$ is an injection. By hypothesis, $S$ is finite. From Injection from Finite Set to Itself is Surjection, $\lambda_a$ is a surjection. Thus $\lambda_a$ is injective and surjective, and therefore a bijection. {{qed}} Category:Semigroups Category:Regular Representations Category:Cancellability lkm98i64qmrup50xs7e5y34akkjnvud
It suffices to prove that :$\left\{ {x}\right\}$ is connected Let $y, z \in \left\{ {x}\right\}$ By definition of singleton: :$y = x$ and $z = x$ By definition of reflexivity; :$y \preceq z$ Thus :$y \preceq z$ or $z \preceq y$ {{qed}}
By definition of inverse row form, the rows of $\CC$ are headed by the inverse elements of the elements which head the corresponding columns. The entries in the main diagonal of $\CC$ have the same column number as row number. Let $\sqbrk c_{k k}$ denote the entry of $\CC$ corresponding to the element where the $k$th row intersects the $k$th column. Let $a$ be the element which heads column $k$. Then, by definition, $a^{-1}$ is the element which heads row $k$. Thus the element which occupies entry $\struct c_{k k}$ is $a^{-1} \circ a = e$. Hence the result. {{qed}}
Proof by Counterexample: Let $\struct {S, \circ}$ be the algebraic structure defined by the following Cayley table: :$\begin{array}{c|ccccc} \circ & e & a & b & c & d \\ \hline e & e & a & b & c & d  \\ a & a & e & d & b & c  \\ b & b & c & e & d & a  \\ c & c & d & a & e & b  \\ d & d & b & c & a & e  \\ \end{array}$ By inspection it can be seen that the Cayley table for $\struct {S, \circ}$ is a Latin square. However, we have that: :$\paren {a \circ b} \circ c = d \circ c = a$ :$a \circ \paren {b \circ c} = a \circ d = c$ Thus $\circ$ is not an associative operation. So, by definition, $\struct {S, \circ}$ is not a group. {{qed}}
Let $x \equiv y \pmod m$. Then by definition of congruence: :$\exists k \in Z: x - y = k m$ Hence: :$c x - c y = c k m$ and so by definition of congruence: :$c x \equiv c y \pmod m$ {{qed}} Category:Modulo Addition 5ro93s15864yankv4wl8fdte2utm0ix
By Scaling preserves Modulo Addition: :$c_1 x_1 \equiv c_1 y_1 \pmod m$ :$c_2 x_2 \equiv c_2 y_2 \pmod m$ and so by Modulo Addition is Well-Defined: :$c_1 x_1 + c_2 x_2 \equiv c_1 y_1 + c_2 y_2 \pmod m$ {{qed}}
From the definition of a prime number, the only (strictly) positive integer less than or equal to a prime $p$ which is ''not'' prime to $p$ is $p$ itself. Thus it follows directly that: :$\map \phi p = p - 1$ {{qed}}
Let $x, y \in C$. By definition of connected relation: :$x \preceq y$ or $y \preceq x$ {{WLOG}}, suppose that :$x \preceq y$ Define $z = y$. Thus by definition of reflexivity :$x \preceq z$ and $y \preceq z$ Hence $C$ is directed. {{qed}}
From Prime not Divisor implies Coprime, each of $1, 2, \ldots, p - 1$ is coprime to $p$. The result follows by definition of reduced residue system modulo $p$. {{qed}}
Let $\eqclass r m, \eqclass s m \in \Z'_m$. Then by definition of reduced residue system: :$r, s \perp m$ By Bézout's Lemma: :$\exists u_1, v_1 \in \Z: u_1 r + v_1 m = 1$ :$\exists u_2, v_2 \in \Z: u_2 s + v_2 m = 1$ Then: {{begin-eqn}} {{eqn | l = \paren {u_1 r + v_1 m} \paren {u_2 s + v_2 m}       | r = u_1 u_2 r s + v_1 u_2 s m + u_1 v_2 r m + v_1 v_2 m^2       | c =  }} {{eqn | r = \paren {u_1 u_2} r s + \paren {v_1 u_2 s + u_1 v_2 r + v_1 v_2 m} m       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} So, again by Bézout's Lemma, $r s$ is coprime to $m$. So the product of two elements of $\struct {\Z'_m, \times_m}$ is again in $\struct {\Z'_m, \times_m}$. That is, $\struct {\Z'_m, \times_m}$ is closed. {{qed}}
Let $a, b, c \in \Z'_m$ such that $a \times_m c = b \times_m c$ Let $p, q, r$ be integers such that: :$p \in a$ :$q \in b$ :$r \in c$ By definition of residue class, this means: {{begin-eqn}} {{eqn | l = p r       | o = \equiv       | r = q r       | rr= \pmod m       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \equiv       | r = q       | rr= \pmod m       | c = Cancellability of Congruences: Corollary 1 }} {{end-eqn}} Thus as $p \in a$ and $q \in b$ it follows that $a = b$. Hence the result. {{qed}}
=== Sufficient Condition === Let $X$ be order generating. Let $x \in S$. By definition of order generating: :$x = \map \inf {x^\succeq \cap X}$ Thus :$\exists Y \subseteq X: x = \inf Y$ {{qed|lemma}} === Necessary Condition === Let :$\forall x \in S: \exists Y \subseteq X: x = \inf Y$ Let $x \in S$. By assumption: :$\exists Y \subseteq X: x = \inf Y$ Define $Z := x^\succeq \cap X$. We will prove that :$\forall b \in S: b$ is lower bound for $Z \implies b \preceq x$ Let $b \in S$ such that :$b$ is lower bound for $Z$. Let $c \in Y$. By definition of infimum: :$x$ is lower bound for $Y$. By definition of lower bound: :$x \preceq c$ By definition of upper closure of element: :$c \in x^\succeq$ By definition of subset: :$c \in X$ By definition of intersection: :$c \in Z$ Thus by definition of lower bound: :$b \preceq c$ By definition of lower bound: :$b$ is lower bound for $Y$ Thus by definition of infimum: :$b \preceq x$ {{qed|lemma}} We will prove that :$x$ is lower bound for $Z$ Let $a \in Z$ By definition of intersection: :$a \in x^\succeq$ Thus by definition of upper closure of element: :$x \preceq a$ {{qed|lemma}} Thus by definitions of complete lattice and infimum: :$x^\succeq \cap X$ admits an infimum and $x = \map \inf {x^\succeq \cap X}$ {{qed}}
From Weierstrass Product Theorem, the function: :$\displaystyle \map h z = z^m \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$ defines an entire function that has the same zeroes as $f$ counting multiplicity.  Thus $f / h$ is both an entire function and non-vanishing. As $f / h$ is both holomorphic and nowhere zero there exists a holomorphic function $g$ such that: :$e^g = f / h$ Therefore: :$f = e^g h$ as desired. {{qed}}
Let $S$ be an algebraic structure of finite order which fulfils the group axioms. It is to be shown that $S$ also fulfils all the finite group axioms. We have that $\text {FG} 0$ and $\text {FG} 1$ are the same as $\text G 0$ and $\text G 1$ and so a priori $\text {FG} 0$ and $\text {FG} 1$ are fulfilled. By hypothesis $S$ is of finite order. That is, there exists some (strictly) positive integer $n$ such that $s$ has exactgly $n$ elements. That is, $\text {FG} 2$ is fulfilled. As $S$ fulfils the group axioms, it is by definition a group. Therefore the Cancellation Laws hold: :$b a = c a \implies b = c$ :$a b = a c \implies b = c$ That is, $\text {FG} 3$ is fulfilled. Thus, $S$ fulfils all the finite group axioms. {{qed|lemma}} Let $S$ be an algebraic structure which fulfils the finite group axioms. It is to be shown that $S$ also fulfils all the group axioms. We have that $\text G 0$ and $\text G 1$ are the same as $\text {FG} 0$ and $\text {FG} 1$ and so a priori $\text G 0$ and $\text G 1$ are fulfilled. By definition, an algebraic structure which fulfils group axioms $\text G 0$ and $\text G 1$ is a semigroup. From Cancellable Finite Semigroup is Group, it then follows that $S$ is a group. That is, $S$ fulfils all the group axioms $\text G 0$, $\text G 1$, $\text G 2$ and $\text G 3$. Hence the result. {{qed}} Category:Finite Groups j7vmvepggdd73ql70gasxnpgizbyf6k
=== Sufficient Condition === Let $X$ be order generating. Let $Y \subseteq S$ such that :$Y \supseteq X \land \left({\forall Z \subseteq Y: \inf Z \in Y}\right)$ We will prove that :$S \subseteq Y$ Let $s \in S$. By Order Generating iff Every Element is Infimum: :$\exists Z \subseteq X: s = \inf Z$ By Subset Relation is Transitive: :$Z \subseteq Y$ Thus by assumption: :$s \in Y$ {{qed|lemma}} By definition of set equality: :$Y = S$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall Y \subseteq S: Y \supseteq X \land \left({\forall Z \subseteq Y: \inf Z \in Y}\right) \implies Y = S$ Define $Y := \left\{ {\inf Z: Z \subseteq X}\right\}$ By definition of subset: :$Y \subseteq S$ We will prove that :$X \subseteq Y$ Let $x \in X$. By definitions of singleton and subset: :$\left\{ {x}\right\} \subseteq X$ By Infimum of Singleton: :$\inf \left\{ {x}\right\} = x$ Thus by definition of $Y$: :$x \in Y$ {{qed|lemma}} We will prove that :$\forall x \in S: \exists Z \subseteq X: x = \inf Z$ Let $x \in S$. We will prove that :$\forall Z \subseteq Y: \inf Z \in Y$ Let $Z \subseteq Y$. Define $T := \bigcup \left\{ {A \subseteq X: \inf A \in Z}\right\}$ By Union of Subsets is Subset: :$T \subseteq X \subseteq S$ Define $N := \left\{ {\inf A: A \subseteq X \land \inf A \in Z}\right\}$ We will prove that :$Z \subseteq N$ Let $y \in Z$. By definition of subset: :$y \in Y$ By definition of $Y$: :$\exists Z \subseteq X: y = \inf Z$ Thus by definition of $N$: :$y \in N$ {{qed|lemma}} By definition of subset: :$N \subseteq Z$ By definition of set equality: :$Z = N$ By Infimum of Infima: :$\inf Z = \inf T$ Thus :$\inf Z \in Y$ {{qed|lemma}} By assumption: :$Y = S$ Then :$x \in Y$ Thus by definition of $Y$: :$\exists Z \subseteq X: x = \inf Z$ {{qed|lemma}} Hence $X$ is order generating by Order Generating iff Every Element is Infimum. {{qed}}
We have by definition of composition of $\RR_1$ and $\RR_2$: :$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$ Let $\left({x, z}\right) \in \RR_2 \circ \RR_1$. By definition of image: :$z \in \Img {\RR_2 \circ \RR_1}$ By definition of composition of $\RR_1$ and $\RR_2$: :$z \in \Img {\RR_2}$ But also: :$z \in \Img {\Img {\RR_1} }$ We also have that $\Img {\RR_2} = \Img {\Preimg {\RR_2} }$ That is: $z \in \Img {\Img {\RR_1} } \cap \Img {\Preimg {\RR_2} }$ {{finish|Not convinced this is going to work.}} Category:Composite Relations b5ge9pk84j3e46bqba3xyq39i1lm3y0
By: :Locally Uniformly Absolutely Convergent Product is Locally Uniformly Convergent :Infinite Product of Analytic Functions is Analytic :Zeroes of Infinite Product of Analytic Functions it suffices to show that the product $\displaystyle \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$ converges locally uniformly absolutely. By Bounds for Weierstrass Elementary Factors and Weierstrass M-Test, this is the case. {{stub}} {{qed}}
Establish the mapping $f: U_C \to R_4$ as follows: {{begin-eqn}} {{eqn | l = \map f 1       | r = r_0 }} {{eqn | l = \map f i       | r = r_1 }} {{eqn | l = \map f {-1}       | r = r_2 }} {{eqn | l = \map f {-i}       | r = r_3 }} {{end-eqn}} From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $f$ is an isomorphism: === Cayley Table of Gaussian Integer Units === The Cayley table for $\struct {U_\C, \times}$ is as follows: {{:Group of Gaussian Integer Units/Cayley Table}}</onlyinclude> === Group of Rotation Matrices Order $4$ === The Cayley table for $\struct {R_4, \times}$ is as follows: {{:Group of Rotation Matrices Order 4/Cayley Table}}
Establish the mapping $f: U_C \to \Z'_5$ as follows: {{begin-eqn}} {{eqn | l = \map f 1       | r = \eqclass 1 5 }} {{eqn | l = \map f i       | r = \eqclass 2 5 }} {{eqn | l = \map f {-1}       | r = \eqclass 4 5 }} {{eqn | l = \map f {-i}       | r = \eqclass 3 5 }} {{end-eqn}} From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $f$ is an isomorphism: === Cayley Table of Gaussian Integer Units === The Cayley table for $\struct {U_\C, \times}$ is as follows: {{:Group of Gaussian Integer Units/Cayley Table}}</onlyinclude> === Multiplicative Group of Reduced Residues Modulo $5$ === The Cayley table for $\struct {\Z'_5, \times_5}$ is as follows: {{:Multiplicative Group of Reduced Residues Modulo 5/Cayley Table}}
Establish the mapping $\phi: K_4 \to S$ as follows: {{begin-eqn}} {{eqn | l = \map \phi e       | r = z }} {{eqn | l = \map \phi a       | r = -z }} {{eqn | l = \map \phi b       | r = \dfrac 1 z }} {{eqn | l = \map \phi c       | r = -\dfrac 1 z }} {{end-eqn}} From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism: === Cayley Table of Klein $4$-Group === The Cayley table for $K_4$ is as follows: {{:Klein Four-Group/Cayley Table}}</onlyinclude> === Group Generated by $1 / z$ and $-z$ === The Cayley table for $S$ is as follows: {{:Group Generated by Reciprocal of z and Minus z/Cayley Table}}
Establish the mapping $\phi: K_4 \to R_4$ as follows: {{begin-eqn}} {{eqn | l = \map \phi e       | r = r_0 }} {{eqn | l = \map \phi a       | r = r_1 }} {{eqn | l = \map \phi b       | r = r_2 }} {{eqn | l = \map \phi c       | r = r_3 }} {{end-eqn}} From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism: === Cayley Table of Klein $4$-Group === The Cayley table for $K_4$ is as follows: {{:Klein Four-Group/Cayley Table}}</onlyinclude> === Group of Reflection Matrices Order $4$ === The Cayley table for $S$ is as follows: {{:Group of Reflection Matrices Order 4/Cayley Table}}
Establish the mapping $\phi: K_4 \to R_4$ as follows: {{begin-eqn}} {{eqn | l = \map \phi e       | r = \eqclass 1 8 }} {{eqn | l = \map \phi a       | r = \eqclass 3 8 }} {{eqn | l = \map \phi b       | r = \eqclass 5 8 }} {{eqn | l = \map \phi c       | r = \eqclass 7 8 }} {{end-eqn}} From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $\phi$ is an isomorphism: === Cayley Table of Klein $4$-Group === The Cayley table for $K_4$ is as follows: {{:Klein Four-Group/Cayley Table}}</onlyinclude> === Multiplicative Group of Reduced Residues Modulo $8$ === The Cayley table for $S$ is as follows: {{:Multiplicative Group of Reduced Residues Modulo 8/Cayley Table}}
Recall the Cayley table for $K_4$: {{:Klein Four-Group/Cayley Table}} From Finite Cyclic Group is Isomorphic to Integers under Modulo Addition, $C_4$ can be exemplified using the additive group of integers modulo $4$. Recall the Cayley table for $\struct {\Z_4, +_4}$: {{:Modulo Addition/Cayley Table/Modulo 4}} Note that all elements of $K_4$ are self-inverse. However, for example: :$\eqclass 1 4 +_4 \eqclass 1 4 = \eqclass 2 4$ and so $\eqclass 1 4$ is not self-inverse. Thus there can be no bijection between $K_4$ and $\struct {\Z_4, +_4}$ such that: :$\forall a, b \in K_4: \map \phi a +_4 \map \phi b = \map \phi {a b}$ Hence the result. {{qed}}