hcju/proofwikiqa huggingface dataset:
Let $z = \arccot p$. Then: {{begin-eqn}} {{eqn | l = p       | r = \frac {\cos z} {\sin z}       | c = {{Defof|Real Arccotangent}} }} {{eqn | ll= \leadsto       | l = p       | r = i \dfrac {\map \exp {i z} + \map \exp {-i z} } {\map \exp {i z} - \map \exp {-i z} }       | c = Cotangent Exponential Formulation }} {{eqn | r = \frac {i \paren {\map \exp {2 i z} + 1} } {\map \exp {2 i z} - 1}       | c =  }} {{eqn | ll= \leadsto       | l = p \paren {\map \exp {2 i z} - 1}       | r = i \paren {\map \exp {2 i z} + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {p - i} \map \exp {2 i z}       | r = p + i       | c =  }} {{eqn | ll= \leadsto       | l = \map \exp {2 i z}       | r = \frac {p + i} {p - i}       | c =  }} {{end-eqn}} So we have: {{begin-eqn}} {{eqn | l = \map \exp {2 m i \arccot p}       | r = \paren {\map \exp {2 i \arccot p} }^m       | c =  }} {{eqn | r = \paren {\frac {p + i} {p - i} }^m       | c =  }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | l = \paren {\frac {p + i} {p - i} } \paren {\frac {p i + 1} {p i - 1} }       | r = \frac {p^2 i + p i^2 + p + i} {p^2 i - p i^2 - p + i}       | c =  }} {{eqn | r = \frac {p^2 i + i} {p^2 i + i}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} and the result follows. {{qed}}
Let $T$ be the equilateral triangle whose vertices are $z_1$, $z_2$ and $z_3$. We have that $z_2 - z_1$ and $z_3 - z_1$ are two sides of $T$ which meet at $z_1$. From the geometry of $T$ it follows that $z_2 - z_1$ is at an angle of $\pi/3$ to $z_3 - z_1$. Similarly, $z_1 - z_3$ and $z_2 - z_3$ are two sides of $T$ which meet at $z_3$. From the geometry of $T$ it follows that $z_1 - z_3$ is at an angle of $\pi / 3$ to $z_2 - z_3$. From Complex Multiplication as Geometrical Transformation/Corollary: :$(1): \quad z_2 - z_1 = e^{i \pi / 3} \left({z_3 - z_1}\right)$ :$(2): \quad z_1 - z_3 = e^{i \pi / 3} \left({z_2 - z_3}\right)$ Then: {{begin-eqn}} {{eqn | l = \dfrac {z_2 - z_1} {z_1 - z_3}       | r = \dfrac {z_3 - z_1} {z_2 - z_3}       | c = $(1)$ divided by $(2)$ }} {{eqn | ll= \leadsto       | l = \paren {z_2 - z_1} \paren {z_2 - z_3}       | r = \paren {z_3 - z_1} \paren {z_1 - z_3}       | c =  }} {{eqn | ll= \leadsto       | l = {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1       | r = - {z_1}^2 - {z_3}^2 + 2 z_3 z_1       | c =  }} {{eqn | ll= \leadsto       | l = {z_1}^2 + {z_2}^2 + {z_3}^2       | r = z_1 z_2 + z_2 z_3 + z_3 z_1       | c =  }} {{end-eqn}}
Let: :${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$ Then: {{begin-eqn}} {{eqn | l = {z_1}^2 + {z_2}^2 + {z_3}^2       | r = z_1 z_2 + z_2 z_3 + z_3 z_1       | c =  }} {{eqn | ll= \leadsto       | l = {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1       | r = - {z_1}^2 - {z_3}^2 + 2 z_3 z_1       | c =  }} {{eqn | ll= \leadsto       | l = \paren {z_2 - z_1} \paren {z_2 - z_3}       | r = \paren {z_3 - z_1} \paren {z_1 - z_3}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {z_2 - z_1} {z_1 - z_3}       | r = \dfrac {z_3 - z_1} {z_2 - z_3}       | c =  }} {{end-eqn}} Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_3$ and $z_2 - z_1$. Similarly: {{begin-eqn}} {{eqn | l = {z_1}^2 + {z_2}^2 + {z_3}^2       | r = z_1 z_2 + z_2 z_3 + z_3 z_1       | c =  }} {{eqn | ll= \leadsto       | l = - {z_2}^2 - {z_1}^2 + 2 z_1 z_2       | r = {z_3}^2 - z_1 z_3 - z_2 z_3 + z_1 z_2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {z_2 - z_1} \paren {z_1 - z_2}       | r = \paren {z_3 - z_1} \paren {z_3 - z_2}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {z_2 - z_1} {z_3 - z_2}       | r = \dfrac {z_3 - z_1} {z_1 - z_2}       | c =  }} {{end-eqn}} Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_2$ and $z_3 - z_2$. Thus all three angles: :$\angle z_2 z_1 z_3$ :$\angle z_1 z_3 z_2$ :$\angle z_3 z_2 z_1$ are equal. By definition, therefore, $\triangle z_1 z_2 z_3$ is equilateral.
{{begin-eqn}} {{eqn | l = AC^2       | r = AB^2 + BC^2 - 2 (AB) (BC) \cos \angle B       | c = Cosine Rule }} {{eqn | l = BD^2       | r = BC^2 + CD^2 - 2 (CD) (BC) \cos \angle C       | c = Cosine Rule }} {{eqn | r = DA^2 + CD^2 - 2 (AB) (CD) \cos \angle C       | c = as $AB = CD$ and $BC = DA$ }} {{eqn | ll= \leadsto       | l = AC^2 + BD^2       | r = AB^2 + BC^2 + DA^2 + CD^2 - 2 (AB) (BC) \paren {\cos \angle B + \cos \angle C}       | c =  }} {{end-eqn}} But we have that $\angle C$ and $\angle B$ are supplementary: :$\angle C = \angle B = 180 \degrees$ Thus from Cosine of Supplementary Angle: :$\cos \angle B + \cos \angle C = 0$ The result follows. {{Qed}}
From the given parametrization it follows that: {{begin-eqn}} {{eqn | l = E       | r = \mathbf r_\phi \cdot \mathbf r_\phi }} {{eqn | r = \tuple {-a \sin \phi, a \cos \phi, 0} \cdot \tuple {-a \sin \phi, a \cos \phi, 0} }} {{eqn | r = a^2 }} {{eqn | l = F       | r = \mathbf r_\phi \cdot \mathbf r_z }} {{eqn | r = \tuple {-a \sin \phi, a \cos \phi, 0} \cdot \tuple {0, 0, 1} }} {{eqn | r = 0 }} {{eqn | l = G       | r = \mathbf r_z \cdot \mathbf r_z }} {{eqn | r = \tuple {0, 0, 1} \cdot \tuple {0, 0, 1} }} {{eqn | r = 1 }} {{end-eqn}} where $E, F, G$ are the functions of the first fundamental form. Furthermore, all derivatives of $E, F, G$ {{WRT|Differentiation}} $\phi$ and $z$ vanish. Then geodesic equations read: :$\dfrac \d {\d t} \dfrac {a^2 \phi'} {\sqrt {a^2 \phi'^2 + z'^2} } = 0$ :$\dfrac \d {\d t} \dfrac {z'} {\sqrt {a^2 \phi'^2 + z'^2} } = 0$ Integrate these differential equations once: :$\dfrac {a^2 \phi'} {\sqrt {a^2 \phi'^2 + z'^2} } = b_1$ :$\dfrac {z'} {\sqrt {a^2 \phi'^2 + z'^2} } = b_2$ where $b_1, b_2$ are real arbitrary constants. Divide the first equation by the second one: :$\dfrac {a^2 \phi'} {z'} = \dfrac {b_1} {b_2}$ To solve this in terms of $z$ as a function of $\phi$, define: :$C_1 = \dfrac {a^2 b_2} {b_1}$  and use the chain rule: :$\dfrac {\d z} {\d \phi} = C_1$ Integration {{WRT|Integration}} $\phi$ yields the desired result. In other words, geodesics are helical lines. {{qed}}
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$. Then: {{begin-eqn}} {{eqn | l = \cmod {z_1 + z_2}       | r = \cmod {z_1 - z_2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_1 + x_2}^2 + \paren {y_1 + y_2}^2       | r = \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = {x_1}^2 + 2 x_1 x_2 + {x_2}^2 + {y_1}^2 + 2 y_1 y_2 + {y_2}^2       | r = {x_1}^2 - 2 x_1 x_2 + {x_1}^2 + {y_1}^2 - 2 y_1 y_2 + {y_1}^2       | c = Square of Sum, Square of Difference }} {{eqn | ll= \leadsto       | l = 4 x_1 x_2 + 4 y_1 y_2       | r = 0       | c = simplifying }} {{eqn | ll= \leadsto       | l = x_1 x_2 + y_1 y_2       | r = 0       | c = simplifying }} {{end-eqn}} Now we have: {{begin-eqn}} {{eqn | l = \dfrac {z_1} {z_2}       | r = \frac {x_1 + i y_1} {x_2 + i y_2}       | c =  }} {{eqn | r = \frac {\paren {x_1 + i y_1} \paren {x_2 - i y_2} } { {x_2}^2 + {y_2}^2}       | c = {{Defof|Complex Division}} }} {{eqn | r = \frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} + \frac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}       | c = {{Defof|Complex Multiplication}} }} {{end-eqn}} But we have: :$x_1 x_2 + y_1 y_2 = 0$ Thus: :$\dfrac {z_1} {z_2} = \dfrac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}$ which is wholly imaginary. {{qed}}
From Factors of Difference of Two Even Powers: :$x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \displaystyle \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$ Substituting $z + a$ for $x$, $z - a$ for $y$, and $m$ for $n$ we get: {{begin-eqn}} {{eqn | o =        | r = \paren {z + a}^{2 m} - \paren {z - a}^{2 m}       | c =  }} {{eqn | r = \paren {\paren {z + a} - \paren {z - a} } \paren {\paren {z + a} + \paren {z - a} } \prod_{k \mathop = 1}^{m - 1} \paren {\paren {z + a}^2 - 2 \paren {z + a} \paren {z - a} \cos \frac {k \pi} n + \paren {z - a}^2}       | c =  }} {{eqn | r = \paren {2 a} \paren {2 z} \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + 2 a z + a^2 - 2 \paren {z^2 - a^2} \cos \frac {k \pi} m + \paren {z^2 - 2 a z + a^2} }       | c =  }} {{eqn | r = 4 a z \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 + 2 a^2 - 2 \paren {z^2 - a^2} \cos \frac {k \pi} m}       | c =  }} {{eqn | r = 4 a z \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 \paren {1 - \cos \frac {k \pi} m} + 2 a^2 \paren {1 + \cos \frac {k \pi} m} }       | c =  }} {{eqn | r = 4 a z \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 \paren {2 \sin^2 \frac {k \pi} {2 m} } + 2 a^2 \paren {2 \cos^2 \frac {k \pi} {2 m} } }       | c = Double Angle Formula for Cosine: Corollary 1 and Corollary 2 }} {{eqn | r = 4 a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{m - 1} 4 \sin^2 \frac {k \pi} {2 m} }       | c = {{Defof|Cotangent|subdef = Real Function}} }} {{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} {2 m} } \paren {\prod_{k \mathop = 1}^{m - 1} \map \sin {\pi - \frac {k \pi} {2 m} } }       | c = Sine of Supplementary Angle }} {{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} {2 m} } \paren {\prod_{k \mathop = 1}^{m - 1} \sin \frac {\paren {2 m - k} \pi} {2 m} } \sin \frac {m \pi} {2 m}       | c = Sine of Right Angle }} {{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^m \sin \frac {k \pi} {2 m} } \paren {\prod_{k \mathop = m + 1}^{2 m - 1} \sin \frac {k \pi} {2 m} }       | c = Translation of Index Variable of Product }} {{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{2 m - 1} \sin \frac {k \pi} {2 m} } }} {{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\frac {2 m} {2^{2 m - 1} } }       | c = Product of Sines of Fractions of Pi }} {{eqn | r = 4 m a z \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } }} {{end-eqn}} {{qed}}
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$. By definition of complex multiplication: :$z_1 z_2 = x_1 x_2 - y_1 y_2 + i \paren {x_1 y_2 + x_2 y_1}$ Then: {{begin-eqn}} {{eqn | l = \map \Re {z_1 z_2}       | r = x_1 x_2 - y_1 y_2       | c = {{Defof|Real Part}} }} {{eqn | r = \map \Re {z_1} \, \map \Re {z_2} - \map \Im {z_1} \, \map \Im {z_2}       | c = {{Defof|Real Part}} }} {{end-eqn}} {{qed}}
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$. By definition of complex multiplication: :$z_1 z_2 = x_1 x_2 - y_1 y_2 + i \paren {x_1 y_2 + x_2 y_1}$ Then {{begin-eqn}} {{eqn | l = \map \Im {z_1 z_2}       | r = x_1 y_2 + x_2 y_1       | c = {{Defof|Imaginary Part}} }} {{eqn | r = \map \Re {z_1} \, \map \Im {z_2} + \map \Im {z_1} \, \map \Re {z_2}       | c = {{Defof|Imaginary Part}} }} {{end-eqn}} {{qed}}
Let $M$ be the straight line passing through $z_1$ and $z_2$. Let $Q$ be the midpoint of $M$. We have that: :$Q = \tuple {\dfrac {x_1 + x_2} 2, \dfrac {y_1 + y_2} 2}$ The slope of $M$ is $\dfrac {y_2 - y_1} {x_2 - x_1}$. As $L$ is perpendicular to the $M$, its slope is $\dfrac {x_1 - x_2} {y_2 - y_1}$. Thus by Equation of Straight Line in Plane: Point-Slope Form, the equation for $L$ is: :$y - \dfrac {y_1 + y_2} 2 = \dfrac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$ {{qed}} {{improve|I can find no actual page on the web anywhere which gives this result explicitly, so I don't know what the "standard form" may be for this line. Hence I have not tried to simplify it, as any such "simplification" only seems to make it more complicated and less intuitive.}} Category:Straight Lines 2ecrrfe5hfj681ty310ptb9l4v0yfac
Let $H = \gen x$ be the subgroup of $G$ generated by $x$. We have that the index of $H$ in $G$ is $2$. Then by Subgroup of Index 2 is Normal, $H$ is normal in $G$. Let $y = x^{2^{n - 2} }$. We have that $y \in H$, by definition of the construction of $H$. Then the order of $y$ in $G$ is $2$. As $H$ is normal, we have: :$\forall g \in G: g y g^{-1} \in H$ By Order of Conjugate Element equals Order of Element, $g y g^{-1}$ is also of order $2$. We can write $g y g^{-1} = x^m$ for some $m \in \Z$. By Order of Power of Group Element, we have: :$2 = \dfrac {2^{n - 1} } {\gcd \set {m, 2^{n - 1} } }$ This leads to: :$\gcd \set {m, 2^{n - 1} } = 2^{n - 2}$ We have $0 < m < 2^{n - 1}$ and $2^{n - 2} \divides m$. This forces $m = 2^{n - 2}$ and $g y g^{-1} = y$. It follows that $g y = y g$. Hence $y = x^{2^{n - 2} }$ is an element of the center of $G$. {{qed}}
From  We have that $L$ and $M$ are abelian. From Intersection of Subgroups is Subgroup, we have that $L \cap M$ is a subgroup of both $L$ and $M$. From Subgroup of Abelian Group is Normal, we also have that $L \cap M$ is a normal subgroup of both $L$ and $M$. Thus from Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup: :$L \subseteq \map N {L \cap M}$ and: :$M \subseteq \map N {L \cap M}$ Hence: :$L \cup M \subseteq \map N {L \cap M}$ But from Normalizer is Subgroup, $\map N {L \cap M}$ is a subgroup of $G$. So it follows that: :$H \subseteq \map N {L \cap M}$ Hence the result by Subgroup is Normal Subgroup of Normalizer. {{qed}}
By Corollary 1 of Characterisation of Non-Archimedean Division Ring Norms then: :$\sup \set {\norm{n \cdot 1_R}: n \in \N_{> 0}} = 1$ By Norm Axiom $(\text N 1)$ (Positive Definiteness) then: :$\norm {0 \cdot 1_R} = 0 \le 1$ Let $n < 0$ then: {{begin-eqn}} {{eqn | l = \norm {n \cdot 1_R}       | r = \norm {-\underbrace {\paren {1_R + 1_R + \dots + 1_R} }_{\text {$-n$ times} } } }} {{eqn | r = \norm {\underbrace {1_R + 1_R + \dots + 1_R}_{\text {$-n$ times} } }       | c = Norm of Ring Negative }} {{eqn | r = \norm {\paren {-n} \cdot 1_R}       | c =  }} {{eqn | o = \le       | r = 1       | c = Characterisation of Non-Archimedean Division Ring Norms }} {{end-eqn}} The result follows. {{qed}}
We have: {{begin-eqn}} {{eqn | l = \paren {x h}^2 h^{-1} \paren {x^{-1} }^2       | r = x h x h h^{-1} x^{-1} x^{-1}       | c = {{GroupAxiom|1}} }} {{eqn | r = x h x x^{-1} x^{-1}       | c = {{GroupAxiom|3}} }} {{eqn | r = x h x^{-1}       | c = {{GroupAxiom|3}} }} {{end-eqn}} Because $\paren {x h}^2$ and $\paren {x^{-1} }^2$ are in the form $x^2$ for $x \in G$, they are both elements of $H$. Thus: :$x h x^{-1} \in H$ and so $H$ is normal in $G$ by definition. {{qed}}
Let $e$ be the identity element of $G$. From Identity is Only Group Element of Order 1, $\order e = 1$ and so $H \ne \O$. Let $a \in H$. Then by Order of Group Element equals Order of Inverse: :$\order a = \order {a^{-1} }$ and so $a \in H$. Let $a, b \in H$. From Product of Orders of Abelian Group Elements Divides LCM of Order of Product: :$\order {a b} \divides \lcm \set {\order a, \order b}$ where: :$\order a$ denotes the order of $a$ :$\divides$ denotes divisibility :$\lcm$ denotes the lowest common multiple. Thus $a b$ is also of finite order. Thus by definition: :$a b \in H$. By the Two-Step Subgroup Test it follows that $H$ is a subgroup of $G$. By Subgroup of Abelian Group is Normal, $H$ is normal in $G$. {{qed}}
As $G$ is a group, it is by definition a monoid. Hence Product of Commuting Elements with Inverses applies: :$x y x^{-1} y^{-1} = e = x^{-1} y^{-1} x y$ {{qed}}
{{begin-eqn}} {{eqn | l = \sqbrk {x N, y N}       | r = \paren {x N}^{-1} \paren {y N}^{-1} \paren {x N} \paren {y N}       | c = {{Defof|Commutator of Group Elements}} }} {{eqn | r = \paren {x^{-1} N} \paren {y^{-1} N} \paren {x N} \paren {y N}       | c = Quotient Group is Group: inverse of $x N$ is $x^{-1} N$ }} {{eqn | r = \paren {x^{-1} y^{-1} N} \paren {x y N}       | c = {{Defof|Coset Product}} }} {{eqn | r = x^{-1} y^{-1} x y N       | c = {{Defof|Coset Product}} }} {{eqn | r = \sqbrk {x, y} N       | c = {{Defof|Commutator of Group Elements}} }} {{end-eqn}} {{qed}}
Let $x, y \in G$. Then: {{begin-eqn}} {{eqn | n = 1       | l = \paren {x N} \paren {y N}       | r = \paren {y N} \paren {x N}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \sqbrk {x N, y N}       | r = N       | c = {{Defof|Commutator of Group Elements}} }} {{eqn | ll= \leadstoandfrom       | l = \sqbrk {x, y} N       | r = N       | c = Commutator of Quotient Group Elements }} {{end-eqn}} Let $G / N$ be abelian. Then by definition: :$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$ and it follows from $(1)$ that: :$\forall x, y \in G: \sqbrk {x, y} \in N$ Conversely, let: :$\forall x, y \in G: \sqbrk {x, y} \in N$ Again it follows from $(1)$ that: :$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$ That is, that $G / N$ is abelian. {{qed}}
From Intersection of Normal Subgroups is Normal, we have that $N \cap K$ is normal in $G$. We are given that $G / N$ and $G / K$ are abelian. Hence: {{begin-eqn}} {{eqn | ll= \forall x, y \in G:       | l = \sqbrk {x, y}       | o = \in       | r = N       | c = Quotient Group is Abelian iff All Commutators in Divisor }} {{eqn | lo= \land       | l = \sqbrk {x, y}       | o = \in       | r = K       | c =  }} {{eqn | lll=\leadsto       | ll= \forall x, y \in G:       | l = \sqbrk {x, y}       | o = \in       | r = N \cap K       | c = {{Defof|Set Intersection}} }} {{eqn | lll=\leadsto       | l = G / \paren {N \cap K}       | o =        | r = \text {is abelian}       | c = Quotient Group is Abelian iff All Commutators in Divisor }} {{end-eqn}} {{qed}}
Proof by Counterexample: Let $G = \set {e, a, b, c}$ be the Klein $4$-group whose identity element is $e$. Let $N = \set {e, a}$ and $K = \set {e, b}$. By Subgroups of Klein Four-Group, both $N$ and $K$ are subgroups of $G$. By Prime Group is Cyclic, both $N$ and $K$ are cyclic. By Subgroup of Abelian Group is Normal, both $N$ and $K$ are normal in $G$. Then we have that: :$N \cap K = e$ and so by Trivial Quotient Group is Quotient Group: :$G / \paren {N \cap K} \cong G$ But $G$ is the Klein $4$-group, which is not cyclic. {{qed}}
The case where $n = 0$ is clear: :$\order {\set e} = 1$ and $e^2 = e$. {{AimForCont}} $\order G = m \times 2^k$ for some odd integer $m$. Then $m$ itself has an odd prime $p$ as a integer (which may of course equal $m$ if $m$ is itself prime). Then by Cauchy's Lemma (Group Theory) there exists $g \in G$ such that $\order g = p$. Hence it is not the case that $g^2 = e$. Hence $\order G$ has no prime factor which is odd. The result follows. {{qed}}
By Subgroup of Index 2 is Normal, $H$ is normal in $G$. Hence the quotient group $G / H$ exists. Then we have: {{begin-eqn}} {{eqn | lo= \forall x \in G:       | l = \paren {x^2} H       | r = \paren {x H}^2       | c =  }} {{eqn | r = H       | c = as $G / H$ is of order $2$ }} {{end-eqn}} {{qed}}
Proof by Counterexample: Consider $S_3$, the symmetric group on $3$ letters. From Subgroups of Symmetric Group on 3 Letters, the subgroups of $S_3$ are: subsets of $S_3$ which form subgroups of $S_3$ are: {{begin-eqn}} {{eqn | o =        | r = S_3 }} {{eqn | o =        | r = \set e }} {{eqn | o =        | r = \set {e, \tuple {123}, \tuple {132} } }} {{eqn | o =        | r = \set {e, \tuple {12} } }} {{eqn | o =        | r = \set {e, \tuple {13} } }} {{eqn | o =        | r = \set {e, \tuple {23} } }} {{end-eqn}} One such subgroup of $G$ whose index is $3$ is $\set {e, \tuple {12} }$ But $\set {e, \tuple {12} }$ does not contain $\tuple {123}$ or $\tuple {132}$, both of which are of order $3$. {{qed}}
From Center of Group is Normal Subgroup, $\map Z G$ is a normal subgroup of $G$. By Lagrange's Theorem, the order of $\map Z G$ is either $1$, $p$, $q$ or $p q$. Because $G$ is not abelian, $G \ne \map Z G$. Hence $\order {\map Z G} \ne p q$. From Quotient of Group by Center Cyclic implies Abelian: :$G / \map Z G$ cannot be a cyclic group which is non-trivial. Then we have: :$\map C x \subset G$ where $\map C x$ is the centralizer of $x$. Hence by Prime Group is Cyclic, $G / \map Z G$ cannot be cyclic of order $p$ or $q$. {{qed}}
Let $x$ be an element of $G$ of order $p$. From Center of Non-Abelian Group of Order pq is Trivial: :$p \notin \map Z G$ where $\map Z G$ denotes the center of $G$. As $x \notin \map Z G$: :$\map C x \subsetneq G$ where $\map C x$ is the centralizer of $x$. From Order of Element divides Order of Centralizer: :$p \divides \order {\map C x}$ Then from Lagrange's Theorem, it follows directly that: :$\order {\map C x} = p$ By Number of Conjugates is Number of Cosets of Centralizer: :$\card {\conjclass x} = \index G {\map {C_G} x}$ where $\conjclass x$ denotes the conjugacy class of $x$. It follows that: :$\card {\conjclass x} = q$ and so the element of $G$ of order $p$ come in sets whose cardinality is a divisor of $q$. {{qed}}
First we note that: :Center of Group is Normal Subgroup and from Intersection of Normal Subgroups is Normal: :$N \cap \map Z G$ is normal in $G$. Suppose $G$ is abelian. From Group equals Center iff Abelian: :$\map Z G = G$ Then: :$N \cap \map Z G = N$ which is non-trivial. From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$. So, suppose $G$ is non-abelian. From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial. From Union of Conjugacy Classes is Normal: :$N = \displaystyle \bigcup_{x \mathop \in N} \conjclass x$ Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $N$ is partitioned. From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element. From Number of Conjugates is Number of Cosets of Centralizer: :$\order {\conjclass {x_j} } \divides \order G$ for all $j \in \set {1, 2, \ldots, m}$. {{finish|Needs more thought.}}
From Intersection of Normal Subgroup with Center in p-Group: :$\order {N \cap \map Z G} > 1$ From Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $N$. It follows from Lagrange's Theorem that: :$\order {N \cap \map Z G} = p$ and so: :$N \cap \map Z G = N$ But from Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $\map Z G$ That is: :$N$ is a subgroup of $\map Z G$ and the result follows. {{qed}}
From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is not the trivial subgroup. From Quotient of Group by Center Cyclic implies Abelian, $G / \map G Z$ cannot be cyclic and non-trivial. Thus $\order {G / \map G Z}$ cannot be $p$ and so must be $p^2$. Thus $\order {\map G Z} = p$. Let $N$ be a normal subgroup of $G$ of order $p$. Then from Normal Subgroup of p-Group of Order p is Subset of Center: :$N \subseteq \map G Z$ Thus there is no normal subgroup of order $p$ different from $\map G Z$. Hence the result. {{qed}}
From Subgroups of Quaternion Group: {{:Subgroups of Quaternion Group}} From Quaternion Group is Hamiltonian, all these subgroups are normal. For two subgroups to be complementary, they need to have an intersection which is trivial. However, apart from $\set e$ itself, all these subgroups contain $a^2$. Hence none of these subgroups has a complement. {{qed}}
By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $p$ or $2p$. From Identity is Only Group Element of Order 1, $2 p - 1$ elements of $G$ have orders greater than $1$. From Cyclic Group is Abelian, $G$ is not the cyclic group $2 p$. If $g \in G$ was of order $2 p$ then $g$ would generate the cyclic group $C_{2 p}$. Thus $G$ has no element of order $2 p$. So all elements of $G$ except the identity have orders $2$ or $p$. {{AimForCont}} $G$ has no element of order $p$. Then all elements are of order $2$. Then $G$ is a Boolean group. Then by Boolean Group is Abelian, $G$ is abelian. This contradicts the assertion that $G$ is non-abelian. Hence the result by Proof by Contradiction. {{qed}}