hcju/proofwikiqa huggingface dataset:
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$: :$\sqbrk {x, y} := x^{-1} y^{-1} x y$ We have that: {{begin-eqn}} {{eqn | n = 1       | l = y x \sqbrk {x, y}       | r = y x x^{-1} y^{-1} x y       | c = {{Defof|Commutator of Group Elements}} }} {{eqn | r = y y^{-1} x y       | c = {{GroupAxiom|3}} }} {{eqn | r = x y       | c = {{GroupAxiom|3}} }} {{end-eqn}} Let $h_1 \in H_1$, $h_2 \in H_2$. We have: {{begin-eqn}} {{eqn | l = \sqbrk {h_1, h_2}       | r = {h_1}^{-1} {h_2}^{-1} h_1 h_2       | c =  }} {{eqn | r = {h_1}^{-1} \paren { {h_2}^{-1} h_1 h_2}       | c = {{GroupAxiom|1}} }} {{eqn | o = \in       | r = {h_1}^{-1} H_1       | c = {{Defof|Normal Subgroup}} }} {{eqn | ll= \leadsto       | l = \sqbrk {h_1, h_2}       | o = \in       | r = H_1       | c = as ${h_1}^{-1} H_1 = H_1$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \sqbrk {h_1, h_2}       | r = {h_1}^{-1} {h_2}^{-1} h_1 h_2       | c =  }} {{eqn | r = \paren { {h_1}^{-1} {h_2}^{-1} h_1} h_2       | c = {{GroupAxiom|1}} }} {{eqn | o = \in       | r = H_2 h_2       | c = {{Defof|Normal Subgroup}} }} {{eqn | ll= \leadsto       | l = \sqbrk {h_1, h_2}       | o = \in       | r = H_2       | c = as $H_2 h_2 = H_2$ }} {{end-eqn}} Thus: :$\sqbrk {h_1, h_2} \in H_1 \cap H_2$ But as $H_1 \cap H_2 = \set e$, it follows that: :$\sqbrk {h_1, h_2} = e$ It follows from Commutator is Identity iff Elements Commute that: :$h_1 h_2 = h_2 h_1$ and the result follows. {{qed}}
First note that by Group Homomorphism Preserves Inverses: :$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1} = \map \phi x^{-1}$ and so there is no amphiboly in the notation used. It is established that $\tuple {x, \map \theta x^{-1} } \in G \times H$: :$x \in G$ :$\map \theta x \in H$ and so $\map \theta x^{-1} \in H$. Let: :$e_G$ be the identity element of $G$ :$e_H$ be the identity element of $H$ We have that $e_G \in Z$, and so: {{begin-eqn}} {{eqn | l = \tuple {e_G, \map \theta {e_G}^{-1} }       | r = \tuple {e_G, \map \theta {e_G} }       | c = as $e_G^{-1} = e_G$ }} {{eqn | r = \tuple {e_G, e_H}       | c = Group Homomorphism Preserves Identity }} {{end-eqn}} Thus $\tuple {e_G, e_H} \in X$ and so $X \ne \O$. Then we note that: {{begin-eqn}} {{eqn | l = \tuple {x, \map \theta x^{-1} } \tuple {x^{-1}, \map \theta x}       | r = \tuple {x x^{-1}, \map \theta x^{-1} \map \theta x}       | c = {{Defof|Group Direct Product}} }} {{eqn | r = \tuple {e_G, e_H}       | c = as $\theta$ is a group isomorphism }} {{eqn | r = \tuple {x^{-1} x, \map \theta x \map \theta x^{-1} }       | c = as $\theta$ is a group isomorphism }} {{eqn | r = \tuple {x^{-1}, \map \theta x} \tuple {x, \map \theta x^{-1} }       | c = {{Defof|Group Direct Product}} }} {{end-eqn}} and so $\tuple {x^{-1}, \map \theta x}$ is the inverse of $\tuple {x, \map \theta x^{-1} }$ in $H$. Let $x, y \in G$. Then $x^{-1} \in G$ from {{GroupAxiom|0}}. Hence if: :$\tuple {x, \map \theta x^{-1} } \in X$ it follows that: :$\tuple {x^{-1}, \map \theta x} \in X$ Let $x, y^{-1} \in G$. Then: :$\tuple {x, \map \theta x^{-1} } \in X$ and: :$\tuple {y^{-1}, \map \theta y} \in X$ Thus: {{begin-eqn}} {{eqn | l = \tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y}       | r = \tuple {x y^{-1}, \map \theta x^{-1} \map \theta y}       | c = {{Defof|Group Direct Product}} }} {{eqn | r = \tuple {x y^{-1}, \map \theta {x^{-1} y} }       | c = as $\theta$ is a group isomorphism }} {{eqn | r = \tuple {x y^{-1}, \map \theta {\paren {y^{-1} x}^{-1} } }       | c = Inverse of Group Product }} {{eqn | r = \tuple {x y^{-1}, \map \theta {\paren {x y^{-1} }^{-1} } }       | c = $x$ and $y$ are in the center of $G$, so commute }} {{eqn | r = \tuple {x y^{-1}, \map \theta {x y^{-1} }^{-1} }       | c =  }} {{end-eqn}} So: :$\tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y} \in X$ and it follows from the One-Step Subgroup Test that $X$ is a subgroup of $G \times H$. {{qed|lemma}} We have that $Z$ is a central subgroup of $G$. We also have that the image of $\theta$ is $W$. From Direct Product of Central Subgroups, $Z \times W$ is a central subgroup of $G \times H$. It follows that $X$ is a central subgroup of $G \times H$. {{qed}}
This result is proved by an application of the Two-Step Subgroup Test: === Condition $(1)$ === From the definition of pullback: :$\tuple {e_G, e_H} \in G \times^\theta H$ {{iff}}: :$\map \theta {e_G \circ N} = e_H * K$ By Coset by Identity, $e_G \circ N, e_H * K$ are the identities of $G / N$ and $H / K$ From Group Homomorphism Preserves Identity: :$\map \theta {e_G \circ N} = e_H * K$ So $\tuple {e_G, e_H} \in G \times^\theta H$ Thus $G \times^\theta H$ is non-empty. {{qed|lemma}} === Condition $(2)$ === Let $\tuple {g, h}$ and  $\tuple {g', h'}$ be elements of $G \times^\theta H$. It follows by definition of $\theta$ that: :$\map \theta {g \circ N} = h * K$ and: :$\map \theta {g' \circ N} = h' * K$ By the morphism property: :$\map \theta {g \circ N} * \map \theta {g' \circ N} = \map \theta {g \circ N \circ g' \circ N} = \map \theta {\paren {g \circ g'} \circ N}$ Hence: {{begin-eqn}} {{eqn | l = \paren {h * h'} * K       | r = h * K * h' * K }} {{eqn | l =        | r = \map \theta {g \circ N} * \map \theta {g' \circ N} }} {{eqn | l =        | r = \map \theta {\paren {g \circ g'} \circ N} }} {{end-eqn}} Thus: :$\tuple {g \circ g', h * h'} \in G \times^\theta H$ Hence $G \times^\theta H$ is closed under the operation. {{qed|lemma}} === Condition $(3)$ === Let $\tuple {g, h} \in G \times^\theta H$. Then: {{begin-eqn}} {{eqn | l = \map \theta {g \circ N}       | r = h * K       | c =  }} {{eqn | ll= \leadsto       | l = \map \theta {g \circ N}^{-1}       | r = h^{-1} * K       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map \theta {g \circ N}^{-1}       | r = \map \theta {g^{-1} \circ N}       | c = Group Homomorphism Preserves Inverses }} {{eqn | ll= \leadsto       | l = \map \theta {g^{-1} \circ N}       | r = h^{-1} * K       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {g^{-1}, h^{-1} }       | o = \in       | r = G \times^\theta H       | c =  }} {{end-eqn}} Thus $G \times^\theta H$ is closed under inverses. {{qed|lemma}} Therefore by the Two-Step Subgroup Test: :$G \times^\theta H \le G \times H$ {{qed}}
Let $-1 < x, y, z < 1$. We check the group axioms in turn: === $\text G 1$: Associativity === {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ z       | r = \frac {\frac {x + y} {1 + x y} + z} {1 + \frac {x + y} {1 + xy} z}       | c =  }} {{eqn | r = \frac {x + y + z + x y z} {1 + x y + x z + y z}       | c =  }} {{eqn | l = x \circ \paren {y \circ z}       | r = \frac {x + \frac {y + z} {1 + y z} } {1 + x \frac {y + z} {1 + y z} }       | c =  }} {{eqn | r = \frac {x + y + z + x y z} {1 + x y + x z + y z}       | c =  }} {{eqn | r = \paren {x \circ y} \circ z       | c =  }} {{end-eqn}} Thus $\circ$ has been shown to be associative. {{qed|lemma}} === $\text G 2$: Identity === {{begin-eqn}} {{eqn | l = x       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \frac {x + y} {1 + x y}       | r = \frac {0 + y} {1 + 0 y}       | c =  }} {{eqn | r = \frac y 1       | c =  }} {{eqn | r = y       | c =  }} {{end-eqn}} Similarly, putting $y = 0$ we find $x \circ y = x$. So $0$ is the identity. {{qed|lemma}} === $\text G 3$: Inverses === {{begin-eqn}} {{eqn | l = x \circ -x       | r = \frac {x + \paren {-x} } {1 + x \paren {-x} }       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Similarly, putting $x = -y$ gives us $\paren {-y} \circ y = 0$. So each $x$ has an inverse $-x$. {{qed|lemma}} === $\text G 0$: Closure === First note that: :$-1 < x, y < 1 \implies x y > -1 \implies 1 + x y > 0$ Next: {{begin-eqn}} {{eqn | l = -1       | o = <       | r = x, y < 1       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \paren {1 - x} \paren {1 - y}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = 1 + x y - \paren {x + y}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \frac {1 + x y - \paren {x + y} } {1 + x y}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \frac {1 + x y} {1 + x y} - \frac {x + y} {1 + x y}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = 1 - \frac {x + y} {1 + x y}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {x + y} {1 + x y}       | o = <       | r = 1       | c =  }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | l = -1       | o = <       | r = x, y < 1       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \paren {1 + x} \paren {1 + y}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = 1 + x y + \paren {x + y}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \frac {1 + x y + \paren {x + y} } {1 + x y}       | c =  }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \frac {1 + x y} {1 + x y} + \frac {x + y} {1 + x y}       | c =  }} {{eqn | ll=\leadsto       | l = 0       | o = <       | r = 1 + \frac {x + y} {1 + x y}       | c =  }} {{eqn | ll= \leadsto       | l = -1       | o = <       | r = \frac {x + y} {1 + x y}       | c =  }} {{end-eqn}} Thus: :$-1 < x, y < 1 \implies -1 < x \circ y < 1$ and we see that in this range, $\circ$ is closed. {{qed|lemma}} Thus the given set and operation form a group. {{Qed}}
Define $f: \openint 0 1 \to \R$ by: :$\map f x := \map \ln {\dfrac {1 - x} x}$ Let us show that $f$ is a bijection by constructing an inverse mapping $g: \R \to \openint 0 1$: :$\map g z := \dfrac 1 {1 + \exp z}$ === Lemma 1 === {{:Group/Examples/inv x = 1 - x/Lemma 1}} === Lemma 2 === {{:Group/Examples/inv x = 1 - x/Lemma 2}} {{finish|Might also be necessary to demonstrate the domain of $f$ is the image set of $g$ and vice versa.}} Thus $f$ is a bijection. Let $\struct {\R, +}$ be the additive group on $\R$. Now define $\circ := +_f$ to be the operation induced on $\openint 0 1$ by $f$ and $+$: :$x \circ y := \map {f^{-1} } {\map f x + \map f y}$ Let us determine the behaviour of $\circ$ more explicitly: {{begin-eqn}} {{eqn | l = x \circ y       | r = \map g {\map f x + \map f y} }} {{eqn | r = \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} + \map \log {\frac {1 - y} y} } }       | c =  }} {{eqn | r = \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} } \map \exp {\map \log {\frac {1 - y} y} } }       | c = Exponential of Sum }} {{eqn | r = \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac {1 - y} y} }       | c = Exponential of Natural Logarithm }} {{end-eqn}} {{MissingLinks}} We see that $\circ$ is commutative. Let $x = \dfrac 1 2$, $\dfrac {1 - x} x = 1$ so that: {{begin-eqn}} {{eqn | l = \dfrac 1 2 \circ y       | r = \dfrac 1 {1 + \paren {\frac {1 - y} y} } }} {{eqn | r = \frac 1 {\frac y y + \frac {1 - y} y} }} {{eqn | r = \frac 1 {\frac 1 y} }} {{eqn | r = y }} {{end-eqn}} so that $\dfrac 1 2$ is the identity element for $\circ$. Furthermore, putting $y = 1 - x$, the following obtains: {{begin-eqn}} {{eqn | l = \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac x {1 - x} } }       | r = \frac 1 {1 + 1} }} {{eqn | r = \frac 1 2 }} {{end-eqn}} establishing $1 - x$ to be the inverse of $x$, as desired. That $\circ$ in fact determines a group on $\openint 0 1$ follows from Pullback of Group is Group. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === Complex Addition is Closed. {{qed|lemma}} === $\text G 1$: Associativity === Complex Addition is Associative. {{qed|lemma}} === $\text G 2$: Identity === From Complex Addition Identity is Zero, we have that the identity element of $\struct {\C, +}$ is the complex number $0 + 0 i$: :$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$ and similarly for $\paren {0 + 0 i} + \paren {x + i y}$. {{qed|lemma}} === $\text G 3$: Inverses === From Inverse for Complex Addition, the inverse of $x + i y \in \struct {\C, +}$ is $-x - i y$. {{qed|lemma}} === $\text C$: Commutativity === 	 Complex Addition is Commutative. {{qed|lemma}} === Infinite === Complex Numbers are Uncountable. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === Non-Zero Complex Numbers Closed under Multiplication. {{qed|lemma}} === $\text G 1$: Associativity === Complex Multiplication is Associative. {{qed|lemma}} === $\text G 2$: Identity === From Complex Multiplication Identity is One, the identity element of $\struct {\C_{\ne 0}, \times}$ is the complex number $1 + 0 i$. {{qed|lemma}} === $\text G 3$: Inverses === From Inverse for Complex Multiplication, the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is: :$\dfrac 1 z = \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$ where $\overline z$ is the complex conjugate of $z$. {{qed|lemma}} === $\text C$: Commutativity === 	 Complex Multiplication is Commutative. {{qed|lemma}} === Infinite === Complex Numbers are Uncountable. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === Real Addition is Closed. {{qed|lemma}} === $\text G 1$: Associativity === Real Addition is Associative. {{qed|lemma}} === $\text G 2$: Identity === From Real Addition Identity is Zero, we have that the identity element of $\struct {\R, +}$ is the real number $0$. {{qed|lemma}} === $\text G 3$: Inverses === From Inverses for Real Addition, we have that the inverse of $x \in \struct {\R, +}$ is $-x$. {{qed|lemma}} === $\text C$: Commutativity === 	 Real Addition is Commutative. {{qed|lemma}} === Infinite === Real Numbers are Uncountably Infinite. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === From Non-Zero Real Numbers Closed under Multiplication: Proof 2, $\R_{\ne 0}$ is closed under multiplication. Note that proof 2 needs to be used specifically here, as proof 1 rests on this result. {{qed|lemma}} === $\text G 1$: Associativity === Real Multiplication is Associative. {{qed|lemma}} === $\text G 2$: Identity === {{:Real Multiplication Identity is One}} {{qed|lemma}} === $\text G 3$: Inverses === {{:Inverses for Real Multiplication}} {{qed|lemma}} === $\text C$: Commutativity === 	 Real Multiplication is Commutative. {{qed|lemma}} === Infinite === Real Numbers are Uncountably Infinite. {{qed}}
We have Real Numbers under Multiplication form Monoid. From Inverses for Real Multiplication, the non-zero numbers are exactly the invertible elements of real multiplication. Thus from Invertible Elements of Monoid form Subgroup of Cancellable Elements, the non-zero real numbers under multiplication form a group. From: :Real Multiplication is Commutative :Subset Product within Commutative Structure is Commutative it follows that this group is also Abelian. {{qed}}
From Non-Zero Real Numbers under Multiplication form Group, $\struct {\R_{\ne 0}, \times}$ forms a group. {{qed|lemma}} From Real Multiplication is Commutative it follows that $\struct {\R_{\ne 0}, \times}$ is abelian. {{qed|lemma}} From Real Numbers are Uncountably Infinite it follows that $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group. {{qed}}
The rational numbers are, by definition, the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. Hence by definition, $\struct {\Q, +, \times}$ is a field. The fact that $\struct {\Q, +}$ forms an abelian group follows directly from the definition of a field. From Rational Numbers are Countably Infinite, we have that $\struct {\Q, +}$ is countably infinite. {{Qed}}
From the definition of rational numbers, the structure $\struct {\Q, + \times}$ is constructed as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. Hence from Multiplicative Group of Field is Abelian Group, $\struct {\Q_{\ne 0}, \times}$ is an abelian group. From Rational Numbers are Countably Infinite, we have that $\struct {\Q_{\ne 0}, \times}$ is a countably infinite group. {{qed}}
First we note that Integers under Multiplication form Monoid. {{qed|lemma}} Then we have:  === Commutativity === Integer Multiplication is Commutative. {{qed|lemma}} === Infinite === Integers are Countably Infinite. {{Qed}}
From the definition of real numbers, $\Q$ is a subset of $\R$. As $\struct {\R, +}$ is a group, and $\struct {\Q, +}$ is a group, it follows from the definition of subgroup that $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$. As $\struct {\R, +}$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\struct {\Q, +}$ is normal in $\struct {\R, +}$. {{Qed}}
Recall that Integers form Integral Domain. The set $\Q$ of rational numbers is defined as the field of quotients of the integers. The fact that the integers are a subgroup of the rationals follows from the work done in proving the Existence of Field of Quotients from an integral domain. {{qed}}
From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$. Then from Complex Numbers under Addition form Abelian Group, $\struct {\C, +}$ is abelian. The result follows from Subgroup of Abelian Group is Normal. {{Qed}} Category:Additive Group of Real Numbers Category:Additive Group of Complex Numbers Category:Examples of Normal Subgroups 1u96lp0tx8g2ml9xdr96nzf90lau0jp
Let $x, y \in \C_{\ne 0}$ such that $x = x_1 + 0 i, y = y_1 + 0 i$. As $x$ and $y$ are wholly real, we have that $x, y \in \R_{\ne 0}$. Then $x + y = x_1 y_1 - 0 \cdot 0 + \left({0 \cdot y_1 + x_1 \cdot 0}\right) i = x_1 y_1 + 0i$ which is also wholly real. Also, the inverse of $x$ is $\dfrac 1 x = \dfrac 1 {x_1} + 0 i$ which is also wholly real. Thus by the Two-Step Subgroup Test, $\struct {\R_{\ne 0}, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$. From Non-Zero Complex Numbers under Multiplication form Abelian Group, $\struct {\C_{\ne 0}, \times}$ is abelian. The result follows from Subgroup of Abelian Group is Normal. {{Qed}}
From the definition of real numbers, it follows that $\Q_{\ne 0}$ is a subset of $\R_{\ne 0}$. As $\struct {\R_{\ne 0}, \times}$ is a group, and $\struct {\Q_{\ne 0}, \times}$ is a group, it follows from the definition of subgroup that $\struct {\Q_{\ne 0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$. As $\struct {\R_{\ne 0}, \times}$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\struct {\Q_{\ne 0}, \times}$ is normal in $\struct {\R_{\ne 0}, \times}$. {{qed}}
We note that $K \ne \varnothing$ as the identity element $1 + 0 i \in K$. Since all $z \in K$ have modulus $1$, they have, for some $\theta \in \hointr 0 {2 \pi}$, the polar form: :$z = \map \exp {i \theta} = \cos \theta + i \sin \theta$ Conversely, if a complex number has such a polar form, it has modulus $1$. Observe the following property of the complex exponential function: :$\forall a, b \in \C: \map \exp {a + b} = \map \exp a \map \exp b$ We must show that if $x, y \in K$ then $x \cdot y^{-1} \in K$. Let $x, y \in K$ be arbitrary. Choose suitable $s, t \in \hointr 0 {2 \pi}$ such that: :$x = \map \exp {i s}$ :$y = \map \exp {i t}$ We compute: {{begin-eqn}} {{eqn | l = \map \exp {i t} \map \exp {-i t}       | r = \map \exp {i \paren {t - t} } }} {{eqn | r = \map \exp 0 }} {{eqn | r = 1 }} {{end-eqn}} So $y^{-1} = \map \exp {-i t}$. We note that this lies in $K$. Furthermore, we have: {{begin-eqn}} {{eqn | l = x y       | r = \map \exp {i s} \map \exp {-2 \pi i t} }} {{eqn | r = \map \exp {i \paren {s - t} } }} {{end-eqn}} We conclude that $x y \in K$. By the Two-Step Subgroup Test, $K$ is a subgroup of $\C$ under complex multiplication. {{handwaving|actually, instead of $\hointr 0 {2 \pi}$ we should use $\R / 2 \pi \Z$ for formal correctness, but I haven't thought of a nice way to do so. Also the property of $\exp$ is only established for the reals so far.}} That the operation $\times$ on $K$ is commutative follows from Complex Multiplication is Commutative and Restriction of Commutative Operation is Commutative. That is, $\times$ is commutative on $K$ because it is already commutative on $\C$. Finally we have that the Circle Group is Uncountably Infinite. {{qed}}
First we note that $K \subseteq \C$. So to show that $K$ is a group it is sufficient to show that $K$ is a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$. From Complex Multiplication Identity is One, the identity element $1 + 0 i$ is in $K$. Thus $K \ne \O$. We now show that $z, w \in K \implies z w \in K$: {{begin-eqn}} {{eqn | l = z, w       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = \cmod z       | r = 1 = \cmod w       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {z w}       | r = \cmod z \cmod w       | c =  }} {{eqn | ll= \leadsto       | l = z w       | o = \in       | r = K       | c =  }} {{end-eqn}} Next we see that $z \in K \implies z^{-1} \in K$: :$\cmod z = 1 \implies \cmod {\dfrac 1 z} = 1$ Thus by the Two-Step Subgroup Test: :$K \le \C_{\ne 0}$ Thus $K$ is a group. From Complex Multiplication is Commutative it also follows from Subgroup of Abelian Group is Abelian that $K$ is an abelian group. Finally we have that the Circle Group is Uncountably Infinite. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === {{begin-eqn}} {{eqn | l = z, w       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = \cmod z       | r = 1 = \cmod w       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {z w}       | r = \cmod z \cmod w       | c =  }} {{eqn | ll= \leadsto       | l = z w       | o = \in       | r = K       | c =  }} {{end-eqn}} So $\struct {\mathbb S, \cdot}$ is closed. {{qed|lemma}} === $\text G 1$: Associativity === Complex Multiplication is Associative. {{qed|lemma}} === $\text G 2$: Identity === From Complex Multiplication Identity is One we have that the identity element of $K$ is $1 + 0 i$. {{qed|lemma}} === $\text G 3$: Inverses === We have that: :$\cmod z = 1 \implies \dfrac 1 {\cmod z} = \cmod {\dfrac 1 z} = 1$ But: :$z \times \dfrac 1 z = 1 + 0 i$ So the inverse of $z$ is $\dfrac 1 z$. {{qed|lemma}} === $\text C$: Commutative === We have that Complex Multiplication is Commutative. We also have from Restriction of Commutative Operation is Commutative that $\times$ is likewise commutative on $K$. {{qed|lemma}} === Uncountably Infinite === Circle Group is Uncountably Infinite. {{Qed|lemma}} All the criteria are satisfied, and the result follows. {{qed}}
Consider the complex modulus function $\left\vert{\cdot}\right\vert: \C \to \R, z \mapsto \left\vert{z}\right\vert$. By Complex Modulus is Norm, we have that $\left\vert{z}\right\vert \ge 0$ for all $z \in \C$, and: :$\left\vert{z}\right\vert = 0 \iff z = 0$ From Group of Units of Field and Complex Numbers form Field, we have $\C^\times = \C \setminus \left\{{0}\right\}$. By above observation, the modulus has a restriction to $\C^\times$: :$\left\vert{\cdot}\right\vert: \C^\times \to \R_{>0}, z \mapsto \left\vert{z}\right\vert$ From $\left\vert{1}\right\vert = 1$ and Modulus of Product, it follows that it is in fact a group homomorphism: :$\phi: \left({C^\times, \cdot}\right) \to \left({\R_{>0}, \cdot}\right), z \mapsto \left\vert{z}\right\vert$ Now $K$ is by definition the kernel of $\phi$. Hence, by Kernel of Group Homomorphism is Subgroup, $\left({K, \times}\right)$ is a subgroup of $\left({\C^\times, \cdot}\right)$. By Subgroup of Abelian Group is Abelian and Complex Multiplication is Commutative, it is also abelian. Finally, note that $\left({K, \times}\right)$ is uncountable from Circle Group is Uncountably Infinite. {{qed}}
By Epimorphism from Integers to Cyclic Group and integer multiplication: :$\forall n \in \Z: n = \map {+^n} 1 \in \gen 1$ Thus: :$\struct {\Z, +} = \gen 1$ and thus, by the definition of a cyclic group, is cyclic. {{qed}}
From Integers under Addition form Infinite Cyclic Group, $\struct {\Z, +}$ is an infinite cyclic group generated by $1$. From Generators of Infinite Cyclic Group, there is only one other generator of such a group, and that is the inverse of that generator. The result follows. {{qed}}
Let $\gen g = G$. Then from Set of Words Generates Group: :$\map W {\set {g, g^{-1} } } = G$ But: :$\gen {g^{-1} } = \map W {\set {g, g^{-1} } }$ and the result follows. {{qed}}